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Documents This page intentionally left blank Statistics for Business Derek L Waller AMSTERDAM • BOSTON • HEIDELBERG • LONDON • NEW YORK • OXFORD PARIS • SAN DIEGO • SAN FRANCISCO • SYDNEY • TOKYO Butterworth-Heinemann is an imprint of Elsevier Butterworth-Heinemann is an imprint of Elsevier Linacre House, Jordan Hill, Oxford OX2 8DP, UK 30 Corporate Drive, Suite 400, Burlington, MA 01803, USA First edition 2008 Copyright © 2008, Derek L Waller Published by Elsevier Inc.All rights reserved The right of Derek L Waller to be identified as the author of this work has been asserted in accordance with the Copyright, Designs and Patents Act 1988 No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means electronic, mechanical, photocopying, recording or otherwise without the prior written permission of the publisher Permissions may be sought directly from Elsevier’s Science & Technology Rights Department in Oxford, UK: phone (�44) (0) 1865 843830; fax (�44) (0) 1865 853333; email: [email protected] .

Alternatively you can submit your request online by visiting the Elsevier web site at /locate/permissions, and selecting Obtaining permission to use Elsevier material Notice No responsibility is assumed by the publisher for any injury and/or damage to persons or property as a matter of products liability, negligence or otherwise, or from any use or operation of any methods, products, instructions or ideas contained in the material herein Library of Congress Cataloging-in-Publication Data A catalog record for this book is available from the Library of Congress British Library Cataloguing in Publication Data A catalogue record for this book is available from the British Library ISBN: 978-0-7506-8660-0 Typeset by Charon Tec Ltd (A Macmillan Company), Chennai, India Printed and bound in Great Britain 08 09 10 10 9 8 7 6 5 4 3 2 1 For information on all Butterworth-Heinemann publications visit our web site at This textbook is dedicated to my family, Christine, Delphine, and Guillaume **Who can help me write corruption powerpoint presentation 86 pages / 23650 words 1 hour Freshman Business**.Alternatively you can submit your request online by visiting the Elsevier web site at /locate/permissions, and selecting Obtaining permission to use Elsevier material Notice No responsibility is assumed by the publisher for any injury and/or damage to persons or property as a matter of products liability, negligence or otherwise, or from any use or operation of any methods, products, instructions or ideas contained in the material herein Library of Congress Cataloging-in-Publication Data A catalog record for this book is available from the Library of Congress British Library Cataloguing in Publication Data A catalogue record for this book is available from the British Library ISBN: 978-0-7506-8660-0 Typeset by Charon Tec Ltd (A Macmillan Company), Chennai, India Printed and bound in Great Britain 08 09 10 10 9 8 7 6 5 4 3 2 1 For information on all Butterworth-Heinemann publications visit our web site at This textbook is dedicated to my family, Christine, Delphine, and Guillaume.

To the many students who have taken a course in business statistics with me … You might find that your name crops up somewhere in this text! This page intentionally left blank Contents About this book ix 1 Presenting and organizing data 1 Numerical Data 3 Categorical Data 15 Chapter Summary 23 Exercise Problems 25 2 Characterizing and defining data 45 Central Tendency of Data 47 Dispersion of Data 53 Quartiles 57 Percentiles 60 Chapter Summary 63 Exercise Problems 65 3 Basic probability and counting rules 79 Basic Probability Rules 81 System Reliability and Probability 93 Counting Rules 99 Chapter Summary 103 Exercise Problems 105 4 Probability analysis for discrete data 119 Distribution for Discrete Random Variables 120 Binomial Distribution 127 Poisson Distribution 130 Chapter Summary 134 Exercise Problems 136 5 Probability analysis in the normal distribution 149 Describing the Normal Distribution 150 Demonstrating That Data Follow a Normal Distribution 161 Using a Normal Distribution to Approximate a Binomial Distribution 169 Chapter Summary 172 Exercise Problems 174 6 Theory and methods of statistical sampling 185 Statistical Relationships in Sampling for the Mean 187 Sampling for the Means from an Infinite Population 196 Sampling for the Means from a Finite Population 199 Sampling Distribution of the Proportion 203 Sampling Methods 206 Chapter Summary 211 Exercise Problems 213 7 Estimating population characteristics 229 Estimating the Mean Value 231 Estimating the Mean Using the Student-t Distribution 237 Estimating and Auditing 243 Estimating the Proportion 245 Margin of Error and Levels of Confidence 248 Chapter Summary 251 Exercise Problems 253 8 Hypothesis testing of a single population 263 Concept of Hypothesis Testing 264 Hypothesis Testing for the Mean Value 265 Hypothesis Testing for Proportions 272 The Probability Value in Testing Hypothesis 274 Risks in Hypothesis Testing 276 Chapter Summary 279 Exercise Problems 281 9 Hypothesis testing for different populations 301 Difference Between the Mean of Two Independent Populations 302 Differences of the Means Between Dependent or Paired Populations 309 Difference Between the Proportions of Two Populations with Large Samples 311 Chi-Square Test for Dependency 313 Chapter Summary 319 Exercise Problems 321 10 Forecasting and estimating from correlated data 333 A Time Series and Correlation 335 Linear Regression in a Time Series Data 339 Linear Regression and Causal Forecasting 345 Forecasting Using Multiple Regression 347 Forecasting Using Non-linear Regression 351 Seasonal Patterns in Forecasting 353 Considerations in Statistical Forecasting 360 Chapter Summary 364 Exercise Problems 366 11 Indexing as a method for data analysis 383 Relative Time-Based Indexes 385 Relative Regional Indexes 391 Weighting the Index Number 392 Chapter Summary 397 Exercise Problems 398 Appendix I: Key Terminology and Formula in Statistics 413 Appendix II: Guide for Using Microsoft Excel 2003 in This Textbook 429 Appendix III: Mathematical Relationships 437 Appendix IV: Answers to End of Chapter Exercises 449 Bibliography 509 Index 511 viii Contents This textbook, Statistics for Business, explains clearly in a readable, step-by-step approach the fundamentals of statistical analysis particularly oriented towards business situations.Much of the information can be covered in an intensive semester course or alternatively, some of the material can be eliminated when a programme is on a quarterly basis.The following paragraphs outline the objectives and approach of this book.The subject of statistics Statistics includes the collecting, organizing, and analysing of data for describing situations and often for the purposes of decision-making ecology.

The subject of statistics Statistics includes the collecting, organizing, and analysing of data for describing situations and often for the purposes of decision-making.

Usually the data collected are quantitative, or numerical, but information can also be categor- ical or qualitative.However, any qualitative data can subsequently be made quantitative by using a numerically scaled questionnaire where sub- jective responses correspond to an established number scale.Statistical analysis is fundamental in the business environment as logical decisions are based on quantitative data.Quite simply, if you cannot express what you know, your current situation, or the future outlook, in the form of numbers, you really do not know much about it.

And, if you do not know much about it, you can- not manage it.

Without numbers, you are just another person with an opinion! This is where statistics plays a role and why it is important to study the subject.For example, by simply dis- playing statistical data in a visual form you can convince your manager or your client.By using probability analysis you can test your company’s strategy and importantly evaluate expected financial risk.Market surveys are useful to eval- uate the probable success of new products or innovative processes.Operations managers in services and manufacturing, use statistical process control for monitoring and controlling performance.

In all companies, historical data are used to develop sales forecasts, budgets, capacity requirements, or personnel needs.In finance, managers analyse company stocks, financial performance, or the economic outlook for investment purposes.For firms like General Electric, Motorola, Caterpillar, Gillette (now a sub- sidiary of Procter & Gamble), or AXA (Insurance), six-sigma quality, which is founded on statistics, is part of the company management culture! Chapter organization There are 11 chapters and each one presents a subject area – organization of information, characteristics of data, probability basics, discrete data, the normal distribution, sampling, estimat- ing, hypothesis testing for single, and multiple populations, forecasting and correlation, and data indexing.Each chapter begins with a box opener illustrating a situation where the particular subject area might be encountered.Following the box opener are the learning objectives, which highlight the principal themes that you will study in the chapter indicating also the subtopics of each theme.

These subtopics underscore the ele- ments that you will cover.Finally, at the end of each chapter is a summary organized according to the principal themes.Thus, the box opener, the learning objectives, the chapter itself, and the About this book chapter summary are logically and conveniently linked that will facilitate navigation and retention of each chapter subject area.Glossary Like many business subjects, statistics contains many definitions, jargon, and equations that are highlighted in bold letters throughout the text.These definitions and equations, over 300, are all compiled in an alphabetic glossary in Appendix I.

Microsoft excel This text is entirely based on Microsoft Excel with its interactive spreadsheets, graphical capabilities, and built-in macro-functions.These functions contain all the mathematical and statistical relationships such as the normal, binomial, Poisson, and Student-t distributions.For this reason, this textbook does not include any of the classic statistical tables such as the standardized normal distribution, Student-t, or chi-square val- ues as all of these are contained in the Microsoft Excel package.As you work through the chap- ters in this book, you will find reference to all the appropriate statistical functions employed.A guide of how to use these Excel functions is contained in Appendix II, in the paragraph “Using the Excel Functions”.

The related Table E-2 then gives a listing and the purpose of all the functions used in this text.The 11 chapters in this book contain numer- ous tables, line graphs, histograms and pie charts.All these have been developed from data using an Excel spreadsheet and this data has then been converted into the desired graph.What I have done with these Excel screen graphs (or screen dumps as they are sometimes dis- paragingly called) is to tidy them up by remov- ing the tool bar, the footers, and the numerical column and the alphabetic line headings to give an uncluttered graph.These Excel graphs in PowerPoint format are available on the Web.

A guide of how to make these Excel graphs is given also in Appendix II in the paragraph, “Generating Excel Graphs”.Associated with this paragraph are several Excel screens giving the stepwise procedure to develop graphs from a particular set of data.I have chosen Excel as the cornerstone of this book, rather than other statistical packages, as in my experience Excel is a major working tool in business.Thus, when you have completed this book you will have gained a double compe- tence – understanding business statistics and versatility in using Excel! Basic mathematics You may feel a little rusty about your basic mathematics that you did in secondary school.In this case, in Appendix III is a section that cov- ers all the arithmetical terms and equations that provide all the basics (and more) for statistical analysis.

Worked examples and end-of-chapter exercises In every chapter there are worked examples to aid comprehension of concepts.Further there are numerous multipart end-of-chapter exer- cises and a case.All of these examples and exer- cises are based on Microsoft Excel.The emphasis of this textbook, as underscored by these chap- ter exercises, is on practical business applica- tions.The answers for the exercises are given in Appendix IV and the databases for these exer- cises and the worked examples are contained on the enclosed CD.

(Note, in the text you may find that if you perform the application examples and test exercises on a calculator you may find slightly different answers than those presented in the textbook.This is because all the examples and exercises have been calculated using Excel, which carries up to 14 figures after the decimal point.) x About this book International The business environment is global.This text- book recognizes this by using box openers, exam- ples, exercises, and cases from various countries where the $US, Euro, and Pound Sterling are employed.

Learning statistics Often students become afraid when they realize that they have to take a course in statistics as part of their college or university curriculum.I often hear remarks like: “I will never pass this course.” “I am no good at maths and so I am sure I will fail the exam.” “I don’t need a course in statistics as I am going to be in marketing.” “What good is statistics to me, I plan to take a job in human resources?” All these remarks are unwarranted and the knowledge of statistics is vital in all areas of business.

The subject is made easier, and more fun, by using Microsoft Excel.To aid comprehension, the textbook begins with fundamental ideas and then moves into more complex areas.The author I have been in industry for over 20 years using statistics and then teaching the subject for the last 21 with considerable success using the subject material, and the approach given in this text.You will find the book shorter than many of the texts on the market but I have only presented those subject areas that in my experience give a solid foundation of statistical analysis for busi- ness, and that can be covered in a reasonable time frame.

This text avoids working through tedious mathematical computations, often found in other statistical texts that I find which confuse students.

You should not have any qualms about studying statistics – it really is not a difficult subject to grasp.If you need any further information, or have questions to ask, please do not hesitate to get in touch through the Elsevier website or at my e-mail address: @ .xiAbout this book This page intentionally left blank 1Presenting and organizing data How not to present data Steve was an undergraduate business student and currently performing a 6-month internship with Telephone Co.Today he was feeling nervous as he was about to present the results of a marketing study that he had performed on the sales of mobile telephones that his firm produced.There were 10 people in the meeting including Roger, Susan, and Helen three of the regional sales directors, Valerie Jones, Steve’s manager, the Head of Marketing, and representatives from production and product development.

Steve showed his first slide as illustrated in Table 1.1 with the comment that “This is the 200 pieces of raw sales data that I have collected”.At first there was silence and then there were several very pointed comments.“What does all that mean?” “I just don’t understand the significance of those figures?” “Sir, would you kindly interpret that data”.After the meeting Valerie took Steve aside and said, “I am sorry Steve but you just have to remember that all of our people are busy and need to be presented information that gives them a clear and concise picture of the situation.

The way that you presented the information is not at all what we expect”.35,378 170,569 104,985 134,859 120,958 107,865 127,895 106,825 130,564 108,654 109,785 184,957 96,598 121,985 63,258 164,295 97,568 165,298 113,985 124,965 108,695 91,864 120,598 47,865 162,985 83,964 103,985 61,298 104,987 184,562 89,597 160,259 55,492 152,698 92,875 56,879 151,895 88,479 165,698 89,486 85,479 64,578 103,985 81,980 137,859 126,987 102,987 116,985 45,189 131,958 73,598 161,895 132,689 120,654 67,895 87,653 58,975 103,958 124,598 168,592 95,896 52,754 114,985 62,598 145,985 99,654 76,589 113,590 80,459 107,865 109,856 101,894 80,157 78,598 86,785 97,562 136,984 89,856 96,215 163,985 83,695 75,894 98,759 133,958 74,895 37,856 90,689 64,189 107,865 123,958 105,987 93,832 58,975 102,986 102,987 144,985 101,498 101,298 103,958 71,589 59,326 121,459 82,198 60,128 86,597 91,786 56,897 112,854 54,128 152,654 99,999 78,562 110,489 86,957 99,486 132,569 134,987 76,589 135,698 118,654 90,598 156,982 87,694 117,895 85,632 104,598 77,654 105,987 78,456 149,562 68,976 50,128 106,598 63,598 123,564 47,895 100,295 60,128 141,298 84,598 100,296 77,498 77,856 134,890 79,432 100,659 95,489 122,958 111,897 129,564 71,458 88,796 110,259 72,598 140,598 125,489 69,584 89,651 70,598 93,876 112,987 123,895 65,847 128,695 66,897 82,459 133,984 98,459 153,298 87,265 72,312 81,456 124,856 101,487 73,569 138,695 74,583 136,958 115,897 142,985 119,654 96,592 66,598 81,490 139,584 82,456 150,298 106,859 68,945 122,654 70,489 94,587 85,975 138,597 97,498 143,985 92,489 146,289 84,592 69,874 As the box opener illustrates, in the business environment, it is vital to show data in a clear and precise manner so that everyone concerned under- stands the ideas and arguments being presented.Management people are busy and often do not have the time to make an in depth analysis of information.

Thus a simple and coherent presen- tation is vital in order to get your message across.Numerical data provide information in a quan- titative form.For example, the house has 250 m2 of living space.My gross salary last year was £70,000 and this year it has increased to £76,000.He ran the Santa Monica marathon in 3 hours and 4 minutes.

The firm’s net income last year was $14,500,400.All these give infor- mation in a numerical form and clearly state a particular condition or situation.When data is collected it might be raw data, which is collected information that has not been organized.The next step after you have raw data is to organize this information and present it in a meaningful form.This section gives useful ways to present numerical data.

Types of numerical data Numerical data are most often either univariate or bivariate.Univariate data are composed of indi- vidual values that represent just one random vari- able, x.Bivariate data involves two vari- ables, x and y, and any data that is subsequently put into graphical form would be bivariate since a value on the x-axis has a corresponding value on the y-axis.

Frequency distribution One way of organizing univariate data, to make it easier to understand, is to put it into a frequency distribution.A frequency distribution is a table, that can be converted into a graph, where the data are arranged into unique groups, categories, or classes according to the frequency, or how often, data values appear in a given class.By grouping data into classes, the data are more manageable than raw data and we can demon- strate clearly patterns in the information.Usually the greater the quantity of data then there should be more classes to clearly show the profile.A guide is to have at least 5 classes but no more than 15 although it really depends on the amount of data Numerical Data 3Chapter 1: Presenting and organizing data After you have studied this chapter you will be able to logically organize and present statistical data in a visual form so that you can convince your audience and objectively get your point across.

You will learn how to develop the following support tools for both numerical and categorical data accordingly as follows.✔ Numerical data • Types of numerical data • Frequency distribution • Absolute frequency his- togram • Relative frequency histogram • Frequency polygon • Ogive • Stem-and-leaf display • Line graph ✔ Categorical data • Questionnaires • Pie chart • Vertical histogram • Parallel histogram • Horizontal bar chart • Parallel bar chart • Pareto diagram • Cross-classification or contingency table • Stacked histogram • Pictograms L e a r n i n g o b j e c t i v e s available and what we are trying to demonstrate.In the frequency distribution, the class range or width should be the same such that there is coherency in data analysis.The class range or class width is given by the following relationship: 1(i) The range is the difference between the highest and the lowest value of any set of data.Let us consider the sales data given in Table 1.

When we develop a frequency distribution we want to be sure that all of the data is contained within the bound- aries that we establish.Thus, to develop a fre- quency distribution for these sales data, a logical maximum value for presenting this data is $185,000 (the nearest value in ’000s above $184,957) and a minimum value is $35,000 (the nearest value in ’000s below $35,378).

By using these upper and lower boundary limits we have included all of the 200 data items.If we want 15 classes then the class range or class width is given as follows using equation 1(i): The tabulated frequency distribution for the sales data using 15 classes is shown in Table 1.The 1st column gives the number of the class range, the 2nd gives the limits of the class range, and the 3rd column gives the amount of data in each range.

The lower limit of the distribution is $35,000 and each class increase by intervals of $10,000 to the upper limit of $185,000.

In selecting a lower value of $35,000 and an upper Class range or class width 15 � �$ , $ ,185 000 35 000 �� $ ,10 000 Class range or class width Desired range of the co � mmplete frequency distribution Number of grroups selected 4 Statistics for Business Table 1.Class range ($) Amount of Percentage Midpoint of data in class of data class range �25,000 to �35,000 0 0.00 value of $185,000 we have included all the sales data values, and so the frequency distribution is called a closed-ended frequency distribution as all data is contained within the limits.2 we have included a line below $35,000 of a class range �25,000 to �35,000 and a line above $185,000 of a class range �185,000 to �195,000.The reason for this will be explained in the later section entitled, “Frequency polygon”.

) In order to develop the frequency distribution using Excel, you first make a single column of the class limits either in the same tab as the dataset or if you prefer in a separate tab.In this case the class limits are $35,000 to $185,000 in increments of $10,000.You then highlight a virgin column, immediately adjacent to the class limits, of exactly the same height and with exactly the corresponding lines as the class limits.1, and the class limits you developed that are demanded by the Excel screen.

When these have been selected, you press the three keys, control-shift-enter Ctrl - ↑ - 8 simultane- ously and this will give a frequency distribution of the amount of the data as shown in the 3rd column of Table 1.Note in the fre- quency distribution the cut-off points for the class limits.The value of $45,000 falls in the class range, �$35,000 and �$45,000, whereas $45,001 is in the class range �$45,000 to �$55,000.The percentage, or proportion of data, as shown in the 4th column of Table 1.

2, is obtained by dividing the amount of data in a par- ticular class by the total amount of data.For example, in the class width �$45,000 to �$55,000, there are six pieces of data and 6/200 is 3.This is a relative frequency distribution meaning that the percentage value is relative to the total amount of data available.Note that once you have created a frequency table or graph you are now making a presenta- tion in bivariate form as all the x values have a corresponding y value.

Note that in this example, when we calcu- lated the class range or class width using the maximum and the minimum values for 15 classes we obtained a whole number of $10,000.Whole numbers such as this make for clear presentations.However, if we wanted 16 classes then the class range would be $9,375 (185,000 – 35,000)/16 which is not as con- venient.In this case we can modify our max- imum and minimum values to say 190,000 and 30,000 which brings us back to a class range of $10,000 (190,000 – 30,000)/16 .Alternatively, we can keep the minimum value at $35,000 and make the maximum value $195,000 which again gives a class range of $10,000 (195,000 – 35,000)/16 .

In either case we still maintain a closed-limit frequency distribution.Absolute frequency histogram Once a frequency distribution table has been developed we can convert this into a histogram, which is a visual presentation of the information, using the graphics capabilities in Excel.An absolute frequency histogram is a vertical bar chart drawn on an x- and y-axes.The horizontal, or x-axis, is a numerical scale of the desired class width where each class is of equal size.

The verti- cal bars, defined by the y-axis, have a length pro- portional to the actual quantity of data, or to the frequency of the amount of data that occurs in a given class range.

1 gives an absolute frequency histogram for the sales data using the 3rd column from Table 1.Here we have 15 vertical bars whose lengths are proportional to the amount of contained data.

The first bar contains data in the range �$35,000 to �$45,000, the second bar has data in the range �$45,000 to �$55,000, the third in the range �$55,000 to �$65,000, etc.Above each bar is indicated the amount of 5Chapter 1: Presenting and organizing data data that is included in each class range.There is no space shown between each bar since the class ranges move from one limit to another though each limit has a definite cut-off point.In present- ing this information to say, the sales department, we can clearly see the pattern of the data and specifically observe that the amount of sales in each class range increases and then decreases beyond $105,000.We can see that the greatest amount of sales of the sample of 200, 30 to be exact, lies in the range �$95,000 to �$ 105,000.

Relative frequency histogram Again using the graphics capabilities in Excel we can develop a relative frequency histogram, which is an alternative to the absolute fre- quency histogram where now the vertical bar, represented by the y-axis, is the percentage or proportion of the total data rather than the absolute amount.The relative frequency histo- gram of the sales data is given in Figure 1.2 where we have used the percent of data from the 4th column of Table 1.The shape of this histogram is identical to the histogram in Figure 1.

We now see that for revenues in the range �$95,000 to �$105,000 the proportion of the total sales data is 15%.Frequency polygon The absolute frequency histogram, or the rela- tive frequency histogram, can be converted into 6 Statistics for Business Figure 1.1 Absolute frequency distribution of sales data.0 2 6 14 18 22 24 30 20 18 14 12 8 6 4 2 0 0 � 35 � 35 to � 45 � 45 to � 55 � 55 to � 65 � 65 to � 75 � 75 to � 85 � 85 to � 95 � 95 to � 10 5 � 10 5 t o � 11 5 � 11 5 t o � 12 5 � 12 5 t o � 13 5 � 13 5 t o � 14 5 � 14 5 t o � 15 5 � 15 5 t o � 16 5 � 16 5 t o � 17 5 � 17 5 t o � 18 5 � 18 5 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 $’000s Am ou nt o f d at a in th is ra ng e 7Chapter 1: Presenting and organizing data Figure 1.

2 Relative frequency distribution of sales data.00 0 � 35 � 35 to � 45 � 45 to � 55 � 55 to � 65 � 65 to � 75 � 75 to � 85 � 85 to � 95 � 95 to � 10 5 � 10 5 t o � 11 5 � 11 5 t o � 12 5 � 12 5 t o � 13 5 � 13 5 t o � 14 5 � 14 5 t o � 15 5 � 15 5 t o � 16 5 � 16 5 t o � 17 5 � 17 5 t o � 18 5 � 18 5 2 1 3 4 5 6 7 8 9 10 11 12 13 14 15 16 $’000s Pe rc en t o f d at a in th is ra ng e a line graph or frequency polygon.The fre- quency polygon is developed by determin- ing the midpoint of the class widths in the respective histogram.

The midpoint of a class range is, For example, the midpoint of the class range, �$95,000 to �$105,000 is, The midpoints of all the class ranges are given in the 5th column of Table 1.Note that we ( , , ) , , 95 000 105 000 2 200 000 2 100 000 � � � ( )maximum value minimum value� 2 have given an entry, �$25,000 to �$35,000 and an entry of �$185,000 to �$195,000 where here the amount of data in these class ranges is zero since in these ranges we are beyond the limits of the closed-ended frequency distribution.In doing this we are able to con- struct a frequency polygon, which cuts the x-axis for a y-value of zero.3 gives the absolute frequency polygon and the relative frequency polygon is shown in Figure 1.These polygons are developed using the graphics cap- abilities in Excel where the x-axis is the mid- point of the class width and the y-axis is the frequency of occurrence.Note that the relative frequency polygon has an identical form as the absolute frequency polygon of Figure 1.

3 but the 8 Statistics for Business Figure 1.

3 Absolute frequency polygon of sales data.0 5 10 15 20 25 30 35 Fr eq ue nc y Average between upper and lower values (midpoint of class) 30 ,00 0 40 ,00 0 50 ,00 0 60 ,00 0 70 ,00 0 80 ,00 0 90 ,00 0 10 0,0 00 11 0,0 00 12 0,0 00 13 0,0 00 14 0,0 00 15 0,0 00 16 0,0 00 17 0,0 00 18 0,0 00 19 0,0 00 Figure 1.4 Relative frequency polygon of sales data.0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Average between upper and lower values (midpoint of range) Fr eq ue nc y (% ) 30 ,00 0 40 ,00 0 50 ,00 0 60 ,00 0 70 ,00 0 80 ,00 0 90 ,00 0 10 0,0 00 11 0,0 00 12 0,0 00 13 0,0 00 14 0,0 00 15 0,0 00 16 0,0 00 17 0,0 00 18 0,0 00 19 0,0 00 y-axis is a percentage, rather than an absolute scale.The difference between presenting the data as a frequency polygon rather than a histo- gram is that you can see the continuous flow of the data.

Ogive An ogive is an adaptation of a frequency distri- bution, where the data values are progressively totalled, or cumulated, such that the resulting table indicates how many, or the proportion of, observations that lie above or below certain limits.There is a less than ogive, which indicates the amount of data below certain limits.This ogive, in graphical form, has a positive slope such that the y values increase from left to right.The other is a greater than ogive that illustrates data above certain values.It has a negative slope, where the y values decrease from left to right.

The frequency distribution data from Table 1.2 has been converted into an ogive format and this is given in Table 1.3, which shows the cumulated data in an absolute form and a rela- tive form.The relative frequency ogives, devel- oped from this data, are given in Figure 1.The usefulness of these graphs is that interpret- ations can be easily made.For example, from the greater than ogive we can see that 80.00% of the sales revenues are at least $75,000.Alternatively, from the less than ogive, we can 9Chapter 1: Presenting and organizing data Table 1.Class Range of class Ogive using absolute data Ogive using relative data No.� class Number Percentage Percentage Percentage limit, n limits (‘000s) but � (n � 1) limit, n � limit age � n age � class � limit but � (n � 1) limit, n 25,000 35,000 �35 0 200 0 0.00% of the sales are no more than $145,000.

The ogives can also be presented as an absolute frequency ogive by indicating on the y-axis the number of data entries which lie above or below given values.This is shown for the sales data in Figure 1.Here we see, for example, that 60 of the 200 data points are sales data that are less than $85,000.The rela- tive frequency ogive is probably more useful than the absolute frequency ogive as propor- tions or percentages are more meaningful and easily understood than absolute values.

In the latter case, we would need to know to what base we are referring.In this case a sample of 200 pieces of data.Stem-and-leaf display Another way of presenting data according to the frequency of occurrence is a stem-and-leaf dis- play.This organizes data showing how values are distributed and cluster around the range of obser- vations in the dataset.The display separates data entries into leading digits, or stems and trailing digits, or leaves.

A stem-and-leaf display shows all individual data entries whereas a frequency distri- bution groups data into class ranges.Let us consider the raw data that is given in Table 1.4, which is the sales receipts, in £’000s for one particular month for 60 branches of a supermarket in the United Kingdom.First the 10 Statistics for Business Figure 1.5 Relative frequency ogives of sales data.

0 35 ,00 0 45 ,00 0 55 ,00 0 65 ,00 0 75 ,00 0 85 ,00 0 95 ,00 0 10 5,0 00 11 5,0 00 12 5,0 00 13 5,0 00 14 5,0 00 15 5,0 00 16 5,0 00 17 5,0 00 18 5,0 00 10 20 30 40 50 60 70 80 90 100 Sales ($) Pe rc en ta ge (% ) Greater than Less than 11Chapter 1: Presenting and organizing data Figure 1.6 Absolute frequency ogives of sales data.0 35 ,00 0 45 ,00 0 55 ,00 0 65 ,00 0 75 ,00 0 85 ,00 0 95 ,00 0 10 5,0 00 11 5,0 00 12 5,0 00 13 5,0 00 14 5,0 00 15 5,0 00 16 5,0 00 17 5,0 00 18 5,0 00 20 40 60 80 100 120 140 160 180 200 Sales ($) Un its o f d at a Greater than Less than Table 1.4 Raw data of sales revenue from a supermarket (£’000s).5 data is sorted from lowest to the highest value using the Excel command SORT from the menu bar Data.

This gives an ordered dataset as shown in Table 1.Here we see that the lowest values are in the seven thousands while the highest are in the sixteen thousands.For the stem and leaf we have selected the thousands as the stem, or those values to the left of the decimal point, and the leaf as the hundreds, or those values to the right of the decimal point.The stem-and-leaf display appears in Figure 1.

The stem that has a value of 11 indicates the data that occurs most 12 Statistics for Business Table 1.5 Ordered data of sales revenue from a supermarket (£’000s).1 Stem items 7 8 1 3 6 8 11 10 7 6 5 3 60 9 10 11 12 13 14 15 16 8 1 5 0 0 0 0 1 1 2 0 2 8 1 2 0 1 2 2 4 0 3 9 2 4 1 1 5 4 5 1 4 6 5 2 2 6 5 6 5 6 6 2 4 7 8 8 6 8 7 5 5 8 9 7 8 5 5 8 8 9 5 6 9 7 7 10 7 9 11 9 Total Leaf Figure 1.7 Stem-and-leaf display for the sales revenue of a supermarket (£’000s).

frequently or in this case, those sales from £11,000 to less than £12,000.The frequency distribution for the same data is shown in Figure 1.The pattern is similar to the stem-and-leaf display but the individual val- ues are not shown.Note that in the frequency distribution, the x-axis has the range greater than the lower thousand value while the stem- and-leaf display includes this value.

For example, in the stem-and-leaf display, 11.0 appears in the stem 11 to less than 12.Alternatively, in the stem that has a value of 16 there are three values (16.

1), whereas in the frequency distribution for the class �16 to �17 there is only one value (16.If you have no add-on stem-and-leaf display in Excel (a separate package) then the following is a way to develop the display using the basic Excel program: ● Arrange all the raw data in a horizontal line.● Sort the data in ascending order by line.) ● Select the stem values and place in a column.

● Transpose the ordered data into their appro- priate stem giving just the leaf value.75 then the stem is 9, and the leaf value is 75.Another approach to develop a stem-and-leaf display is not to sort the data but to keep it in its raw form and then to indicate the leaf values in chronological order for each stem.

This has a disadvantage in that you do not see immedi- ately which values are being repeated.

A stem- and-leaf display is one of the techniques in exploratory data analysis (EDA), which are those methods that give a sense or initial feel about data being studied.A box and whisker plot discussed in Chapter 2 is also another technique in EDA.Line graph A line graph, or usually referred to just as a graph, gives bivariate data on the x- and y-axes.It illustrates the relationship between the variable on the x-axis and the corresponding value on the y-axis.If time represents part of the data this is always shown in the x-axis.

A line graph is not necessarily a straight line but can be curvilinear.Attention has to be paid to the scales on the axes as the appearance of the graph can change and decision-making can be distorted.Consider for example, the sales revenues given in Table 1.6 for the 12-year period from 1992 to 2003.9 gives the graph for this sales data where the y-axis begins at zero and the increase on the axis is in increments of $500,000.Here the slope of the graph, illustrating the increase in sales each year, is moderate.10 now shows the same information except that the y-axis starts at the value of $1,700,000 and the 13Chapter 1: Presenting and organizing data Figure 1.8 Frequency distribution of the sales revenue of a supermarket (£).

0 0 1 4 6 6 9 9 7 10 � 6 t o � 7 � 7 t o � 8 � 8 t o � 9 � 9 t o � 10 � 10 to � 11 � 11 to � 12 � 12 to � 13 � 13 to � 14 � 14 to � 15 � 15 to � 16 � 16 to � 17 2 1 3 4 5 6 7 8 9 10 11 Class limits (£) N um be r o f v al ue s in th is ra ng e 1 7 Period Year Sales ($‘000s) 1 1992 1,775 2 1993 2,000 3 1994 2,105 4 1995 2,213 5 1996 2,389 6 1997 2,415 7 1998 2,480 8 1999 2,500 9 2000 2,665 10 2001 2,810 11 2002 2,940 12 2003 3,070 Table 1.9 Sales data for the last 12 years for “Company A”.0 500 1,000 1,500 2,000 2,500 3,000 3,500 1992 1993 1994 1995 1996 1997 1998 1999 2000 2001 2002 2003 2004 $’0 00 s Year Figure 1.

10 Sales data for the last 12 years for “Company B”.1,700 1,900 2,100 2,300 2,500 2,700 2,900 3,100 1992 1993 1994 1995 1996 1997 1998 1999 2000 2001 2002 2003 2004 $’0 00 s Year 15Chapter 1: Presenting and organizing data Category Very Poor Satisfactory Good Very poor good Score 1 2 3 4 5 Table 1.This gives the impression that the sales growth is very rapid, which is why the two figures are labelled “Company A” and “Company B”.Line graphs are treated further in Chapter 10.Information that includes a qualitative response is categorical data and for this information there may be no quantitative data.

For example, the house is the largest on the street.He ran the Santa Monica marathon in a fast time.Here the categories are large, increased, and fast.The responses, “Yes” or “No”, to a survey are also categorical data.

Alternatively categorical data may be developed from numerical data, which is then organized and given a label, a category, or a name.For example, a firm’s sales revenues, which are quantitative data, may be presented according to geographic region, product type, sales agent, business unit, etc.A presentation of this type can be important to show the strength of the firm.Questionnaires Very often we use questionnaires in order to evalu- ate customers’ perception of service level, stu- dents’ appreciation of a university course, or subscribers’ opinion of a publication.We do this Categorical Data because we want to know if we are “doing it right” and if not what changes should we make.

A ques- tionnaire may take the form as given in Table 1.The first line is the category of the response.This is obviously subjective information.For example with a university course, Student A may have a very different opinion of the same programme as Student B.

We can give the categorical response a score, or a quantitative value for the subjective response, as shown in the second line.Then, if the number of responses is sufficiently large, we can analyse this data in order to obtain a reasonable opinion of say the university course.The analysis of this type of questionnaire is illustrated in Chapter 2, and there is additional information in Chapter 6.Pie chart If we have numerical data, this can be converted into a pie chart according to desired categories.

A pie chart is a circle representing the data and divided into segments like portions of a pie.

Each segment of the pie is proportional to the total amount of data it represents and can be labelled accordingly.The complete pie represents 100% of the data and the usefulness of the pie chart is that we can see clearly the pattern of the data.As an illustration, the sales data of Table 1.1 has now been organized by country and this tabular information is given in Table 1.8 together with the percentage amount of data for each country.

This information, as a pie chart, is shown in Figure 1.We can clearly see now what the data represents and the contribution from each geographical territory.Here for example, the United Kingdom has the greatest contribution to sales revenues, and Austria the least.When you develop a pie chart for data, if you have a cat- egory called “other” be sure that this proportion is small relative to all the other categories in the pie chart; otherwise, your audience will ques- tion what is included in this mysterious “other” slice.

When you develop a pie chart you can only have two columns, or two rows of data.One column, or row, is the category, and the adja- cent column, or row, is the numerical data.Note that in developing a pie chart in Excel you do not have to determine the percentage amount in the table.The graphics capability in Excel does this automatically.Vertical histogram An alternative to a pie chart is to illustrate the data by a vertical histogram where the vertical bars on the y-axis show the percentage of data, and the x-axis the categories.

12 gives an absolute histogram of the above pie chart sales information where the vertical bars show the absolute total sales and the x-axis has now been given a category according to geographic region.13 gives the relative frequency histogram for this same information where the y-axis is now a percentage scale.Note, in these histograms, the bars are separated, as one cate- gory does not directly flow to another, as is the case of a histogram of a complete numerically based frequency distribution.

Parallel histogram A parallel or side-by-side histogram is useful to compare categorical data often of different time periods as illustrated in Figure 1.The figure shows the unemployment rate by country for two different years.From this graph we can com- pare the change from one period to another.1 Horizontal bar chart A horizontal bar chart is a type of histogram where the x- and y-axes are reversed such that the data are presented in a horizontal, rather than a vertical format.

Horizontal bar charts are sometimes referred to as Gantt charts after the American engineer Henry L.Parallel bar chart Again like the histogram, a parallel or side-by-side bar chart can be developed.

16 shows a 16 Statistics for Business 1 Economic and financial indicators, The Economist, 15 February 2003, p.Group Country Sales Percentage revenues ($) 1 Austria 522,065 2.8 Raw sales data according to country ($).side-by-side bar chart for the unemployment data of Figure 1.Pareto diagram Another way of presenting data is to combine a line graph with a categorical histogram as shown in Figure 1.This illustrates the problems, according to categories, that occur in the distribu- tion by truck of a chemical product.The x-axis gives the categories and the left-hand y-axis is the percent frequency of occurrence according to each of these categories with the vertical bars indi- cating their magnitude.The line graph that is shown now uses the right-hand y-axis and the same x-axis.

This is now the cumulative frequency of occurrence of each category.If we assume that the categories shown are exhaustive, meaning that all possible problems are included, then the line graph increases to 100% as shown.Usually the presentation is illustrated so that the bars are in descending order from the most important on the left to the least important on the right so that we have an organized picture of our situation.This type of presentation is known as a Pareto dia- gram, (named after the Italian economist, Vilfredo Pareto (1848–1923), who is also known for coining the 80/20 rule often used in business).The Pareto diagram is a visual chart used often in quality management and operations auditing as it shows those categorical areas that are the most critical and perhaps should be dealt with first.

17Chapter 1: Presenting and organizing data Figure 1.12 Histogram of sales – absolute revenues.0 200,000 400,000 600,000 800,000 1,000,000 1,200,000 1,400,000 1,600,000 1,800,000 2,000,000 2,200,000 2,400,000 2,600,000 2,800,000 3,000,000 3,200,000 3,400,000 3,600,000 3,800,000 4,000,000 4,200,000 4,400,000 4,600,000 4,800,000 Au str ia Be lgiu m Fin lan d Fra nc e Ge rm an y Ita ly Ne the rla nd s Po rtu ga l Sw ed en UK Country R ev en ue s ($) 18 Statistics for Business Figure 1.0 Au str alia Au str ia Be lgiu m Ca na da De nm ark Eu ro zo ne Fra nc e Ge rm an Ita ly Ja pa n Ne the rla nd s Sp ain Sw ed en Sw itze rla nd UK US A Pe rc en ta ge ra te End of 2002 Country End of 2001 Figure 1.0 2 4 6 8 10 12 14 16 18 20 22 24 Au str ia Be lgiu m Fin lan d Fra nc e Ge rm an y Ita ly Ne the rla nd s Po rtu ga l Sw ed en UK Country To ta l r ev en ue s (% ) 19Chapter 1: Presenting and organizing data Figure 1.0 500,000 1,000,000 1,500,000 2,000,000 2,500,000 3,000,000 3,500,000 4,000,000 Austria Belgium Finland France Germany Italy Netherlands Portugal Sweden UK Co un try Sales revenues ($) Figure 1.0 1 2 3 4 5 6 7 8 9 10 11 12 13 Australia Austria Belgium Canada Denmark Euro zone France Germany Italy Japan Netherlands Spain Sweden Switzerland UK USA Unemployment rate (%) End of 2002 End of 2001 Co un try Cross-classification or contingency table A cross-classification or contingency table is a way to present data when there are several variables and we are trying to indicate the relationship between one variable and another.

9 gives a cross-classi- fication table for a sample of 1,550 people in the United States and their professions according to certain states.From this table we can say, for example, that 51 of the teachers are contingent of residing in Vermont.Alternatively, we can say that 24 of the residents of South Dakota are contingent of working for the government.

(Contingent means that values are dependent or conditioned on something else.

) Stacked histogram Once you have developed the cross-classific- ation table you can present this visually by devel- oping a stacked histogram.18 gives a stacked histogram for the cross-classification in Table 1.Portions of the histogram indicate the profes- sion.

19 gives a stacked histogram for the same table but now according to profession.Portions of the histogram now give the state of residence.17 Pareto analysis for the distribution of chemicals.

0 5 10 15 20 25 30 35 40 45 Reasons for poor service Fr eq ue nc y of o cc ur re nc e (% ) 0 10 20 30 40 50 60 70 80 90 100 Cu m ul at ive fr eq ue nc y (% ) Dr um s d am ag ed Sc he du le c ha ng e Te mp era tur e t oo lo w Dr um s r us ted Or de rs wro ng Inc orr ec t la be ling Dr um s n ot se ale d Pa llet s p oo rly st ac ke d Do cu me nta tio n w ron g De lay – ba d w ea the r Individual Cumulative 21Chapter 1: Presenting and organizing data Table 1.9 Cross-classification or contingency table for professions in the United States.State Engineering Teaching Banking Government Agriculture Total California 20 19 12 23 23 97 Texas 34 62 15 51 65 227 Colorado 42 32 23 42 26 165 New York 43 40 23 35 54 195 Vermont 12 51 37 25 46 171 Michigan 24 16 15 16 35 106 South Dakota 34 35 12 24 25 130 Utah 61 25 19 29 61 195 Nevada 12 32 18 31 23 116 North Carolina 6 62 14 41 25 148 Total 288 374 188 317 383 1,550 Figure 1.18 Stacked histogram by state in the United States.0 20 40 60 80 100 120 140 160 180 200 220 240 California Texas Colorado New York Vermont Michigan South Dakota Utah Nevada North Carolina State N um be r i n sa m pl e Agriculture Government Banking Teaching Engineering 22 Statistics for Business Figure 1.

19 Stacked histogram by profession in the United States.0 50 100 150 200 250 300 350 400 450 Engineering Teaching Banking Government Agriculture Profession N um be r i n sa m pl e North Carolina Nevada Utah South Dakota Michigan Vermont New York Colorado Texas California Figure 1.The value of your money today The value of your money tomorrow Pictograms A pictogram is a picture, icon, or sketch that repr- esents quantitative data but in a categorical, quali- tative, or comparative manner.For example, a coin might be shown divided into sections indicating that portion of sales revenues that go to taxes, operating costs, profits, and capital expendi- tures.

Magazines such as Business Week, Time, or Newsweek make heavy use of pictograms.Pictograph is another term often employed for pictogram.20 gives an example of how inflation might be represented by showing a large sack of money for today, and a smaller sack for tomorrow.Attention must be made when using pictograms as they can easily distort the real facts of the data.

For example in the figure given, has our money been reduced by a factor of 50%, 100%, or 200%? We cannot say clearly.Pictograms are not covered further in this textbook.23Chapter 1: Presenting and organizing data This chapter has presented several tools useful for presenting data in a concise manner with the objective of clearly getting your message across to an audience.The chapter is divided into dis- cussing numerical and categorical data.Numerical data Numerical data is most often univariate, or data with a single variable, or bivariate which is information that has two related variables.

Univariate data can be converted into a frequency distribution that groups the data into classes according to the frequency of occurrence of val- ues within a given class.A frequency distribution can be simply in tabular form, or alternatively, it can be presented graphically as an absolute, or relative frequency, histogram.The advantage of a graphical display is that you see clearly the quantity, or proportion of information, that appear in defined classes.This can illustrate key information such as the level of your best, or worst, revenues, costs, or profits.A histogram can be converted into a frequency polygon which links the midrange of each of the classes.

The polygon, either in absolute or relative form, gives the pattern of the data in a continuous form showing where major frequencies occur.An exten- sion of the frequency distribution is the less than, or greater than ogive.The usefulness of ogive presentations is that it is visually apparent the amount, or percentage, that is more or less than certain values and may be indicators of performance.A stem-and-leaf display, a tool in EDA, is a frequency distribution where all data values are displayed according to stems, or leading val- ues, and leaves, or trailing values of the data.The commonly used line graph is a graphical pre- sentation of bivariate data correlating the x variable with its y variable.

Although we use the term line graph, the display does not have to be a straight line but it can be curvilinear or simply a line that is not straight! Categorical data Categorical data is information that includes qualitative or non-quantitative groupings.Numerical data can be represented in a categorical form where parts of the numerical values are put into a category such as product type or geographic location.In statistical analysis a common tool using categorical responses is the questionnaire, where respondents are asked C hapter Sum m ary 24 Statistics for Business opinions about a subject.If we give the categorical response a numerical score, a questionnaire can be easily analysed.A pie chart is a common visual representation of categorical data.

The pie chart is a circle where portions of the “pie” are usually named categories and a percentage of the complete data.A vertical histogram can also be used to illustrate categorical data where the x scale has a name, or label, and the y-axis is the amount or proportion of the data within that label.The vertical histogram can also be shown as a paral- lel or side-by-side histogram where now each label contains data say for two or more periods.

In this way a comparison of changes can be made within named categories.

The vertical histo- gram can be shown as a horizontal bar chart where it is now the y-axis that has the name, or label, and the x-axis the amount or proportion of data within that label.Similarly the horizon- tal bar chart can be shown as a parallel bar chart where now each label contains data say for two or more periods.Whether to use a vertical histogram or a horizontal bar chart is really a matter of personal preference.A visual tool often used in auditing or quality control is the Pareto diagram.This is a combination of vertical bars showing the frequency of occurrence of data according to given categories and a line graph indicating the accumulation of the data to 100%.

When data falls into several categories the information can be represented in a cross- classification or contingency table.This table indicates the amount of data within defined cat- egories.The cross-classification table can be converted into a stacked histogram according the desired categories, which is a useful graphical presentation of the various classifications.Finally, this chapter mentions pictograms, which are pictorial representations of information.These are often used in newspapers and magazines to represent situations but they are difficult to rigorously analyse and can lead to misrepresentation of information.

No further discussion of pictograms is given in this textbook.25Chapter 1: Presenting and organizing data 1.Buyout – Part I Situation Carrefour, France, is considering purchasing the total 50 retail stores belonging to Hardway, a grocery chain in the Greater London area of the United Kingdom.The prof- its from these 50 stores, for one particular month, in £’000s, are as follows.Illustrate this information as a closed-ended absolute frequency histogram using class ranges of £1,000 and logical minimum and maximum values for the data rounded to the nearest thousand pounds.Convert the absolute frequency histogram developed in Question 1 into a relative fre- quency histogram.Convert the relative frequency histogram developed in Question 2 into a relative fre- quency polygon.

Develop a stem-and-leaf display for the data using the thousands for the stem and the hundreds for the leaf.Compare this to the absolute frequency histogram.Illustrate this data as a greater than and a less than ogive using both absolute and rel- ative frequency values.

After examining the data presented in the figure from Question No.1, Carrefour management decides that it will purchase only those stores showing profits greater than £12,500.On this basis, determine from the appropriate ogive how many of the Hardway stores Carrefour would purchase? 2.Closure Situation A major United States consulting company has 60 offices worldwide.

The following are the revenues, in million dollars, for each of the offices for the last fiscal year.5 EXERCISE PROBLEMS 26 Statistics for Business annual operating cost per office for these, including salaries and all operating expenses, is $36 million.As a result of intense competition from other consulting firms and declining markets, management is considering closing those offices whose annual revenues are less than the average operating cost.Required In order to present the data to management, so they can understand the impact of their proposed decision, develop the following information.Present the revenue data as a closed-end absolute frequency distribution using log- ical lower and upper limits rounded to the nearest multiple $10 million and a class limit range of $5 million.What is the average margin per office for the consulting firm before any closure? 3.Present on the appropriate frequency distribution (ogive), the number of offices having less than certain revenues.To construct the distribution use the following criterion: ● Minimum on the revenue distribution is rounded to the closet multiple of $10 million.

● Maximum on the revenue distribution is rounded to the closest multiple of $10 million.From the distribution you have developed in Question 3, how many offices have rev- enues lower than $36 million and thus risk being closed? 5.If management makes the decision to close that number of offices determined in Question 3 above, estimate the new average margin per office.

564 27Chapter 1: Presenting and organizing data 3.Swimming pool Situation A local community has a heated swimming pool, which is open to the public each year from May 17 until September 13.The community is considering building a restaurant facility in the swimming pool area but before a final decision is made, it wants to have assurance that the receipts from the attendance at the swimming pool will help finance the construction and operation of the restaurant.In order to give some justification to its decision the community noted the attendance each day for one particular year and this information is given below.Develop an absolute value closed-limit frequency distribution table using a data range of 50 attendances and, to the nearest hundred, a logical lower and upper limit for the data.Convert this data into an absolute value histogram.Convert the absolute frequency histogram into a relative frequency histogram.Plot the relative frequency distribution histogram as a polygon.What are your obser- vations about this polygon? 4.Convert the relative frequency distribution into a greater than and less than ogive and plot these two line graphs on the same axis.What is the proportion of the attendance at the swimming pool that is �750 and �800 people? 6.

Develop a stem-and-leaf display for the data using the hundreds for the stem and the tens for the leaves.The community leaders came up with the following three alternatives regarding pro- viding the capital investment for the restaurant.(a) If the probability of more than 900 people coming to the swimming pool was at least 10% or the probability of less than 600 people coming to the swimming 869 755 729 926 821 709 1,088 785 830 709 678 1,019 825 843 940 826 750 835 956 743 835 630 791 795 903 790 931 869 878 808 845 692 830 794 993 847 901 837 755 810 791 609 878 778 761 763 726 745 874 728 870 798 507 763 764 779 678 690 1,004 792 848 823 769 773 919 682 672 829 915 883 699 650 780 743 861 610 582 748 744 680 930 776 871 759 580 669 716 980 724 880 669 712 732 968 620 852 749 860 811 748 822 651 539 658 796 825 685 707 895 806 609 952 565 869 560 751 790 907 621 619 28 Statistics for Business pool was not less than 10%.

Under these criteria would the community fund the restaurant? Quantify your answer both in terms of the 10% limits and the attend- ance values.(b) If the probability of more than 900 people coming to the swimming pool was at least 10% and the probability of less than 600 people coming to the swimming pool was not less than 10%.Under these criteria would the community fund the restaurant? Quantify your answer both in terms of the 10% limits and the attend- ance values.(c) If the probability of between 600 and 900 people coming to the swimming pool was at least 80%.Rhine river Situation On a certain lock gate on the Rhine river there is a toll charge for all boats over 20 m in length.00/m for every metre above the minimum value of 20 m.In a certain period the following were the lengths of boats passing through the lock gate.

Show this information in a stem-and-leaf display.Draw the ogives for this data using a logical maximum and minimum value for the limits to the nearest even number of metres.From the appropriate ogive approximately what proportion of the boats will not have to pay any toll fee? 4.Approximately what proportion of the time will the canal authorities be collecting at least €105 from boats passing through the canal? 5.Purchasing expenditures Situation The complete daily purchasing expenditures for a large resort hotel for the last 200 days in Euros are given in the table below.The purchases include all food, beverages, and non- food items for the five restaurants in the complex.It also includes energy, water for the 22.

00 29Chapter 1: Presenting and organizing data three swimming pools, laundry, which is a purchased service, gasoline for the courtesy vehicles, gardening and landscaping services.

Develop an absolute frequency histogram for this data using the maximum value, rounded up to the nearest €10,000, to give the upper limit of the data, and the mini- mum value, rounded down to the nearest €10,000, to give the lower limit.Use an interval or class width of €20,000.This histogram will be a closed-limit absolute fre- quency distribution.From the absolute frequency information develop a relative frequency distribution of sales.What is the percentage of purchasing expenditures in the range €180,000 to €200,000? 4.Develop an absolute frequency polygon of the data.This is a line graph connecting the midpoints of each class in the dataset.

What is the quantity of data in the highest frequency? 5.Develop an absolute frequency “more than” and “less than” ogive from the dataset.Develop a relative frequency “more than” and “less than” ogive from the dataset.From these ogives, what is an estimate of the percentage of purchasing expenditures less than €250,000? 8.From these ogives, 70% of the purchasing expenditures are greater than what amount? 63,680 307,024 188,973 242,746 217,724 194,157 230,211 192,285 235,015 195,577 197,613 332,923 173,876 219,573 113,864 295,731 175,622 297,536 205,173 224,937 195,651 165,355 217,076 86,157 293,373 151,135 187,173 110,336 188,977 332,212 161,275 288,466 99,886 274,856 167,175 102,382 273,411 159,262 298,256 161,075 153,862 116,240 187,173 147,564 248,146 228,577 185,377 210,573 81,340 237,524 132,476 291,411 238,840 217,177 122,211 157,775 106,155 187,124 224,276 303,466 172,613 94,957 206,973 112,676 262,773 179,377 137,860 204,462 144,826 194,157 197,741 183,409 144,283 141,476 156,213 175,612 246,571 161,741 173,187 295,173 150,651 136,609 177,766 241,124 134,811 68,141 163,240 115,540 194,157 223,124 190,777 168,898 106,155 185,375 185,377 260,973 182,696 182,336 187,124 128,860 106,787 218,626 147,956 108,230 155,875 165,215 102,415 203,137 97,430 274,777 179,998 141,412 198,880 156,523 179,075 238,624 242,977 137,860 244,256 213,577 163,076 282,568 157,849 212,211 154,138 188,276 139,777 190,777 141,221 269,212 124,157 90,230 191,876 114,476 222,415 86,211 180,531 108,230 254,336 152,276 180,533 139,496 140,141 242,802 142,978 181,186 171,880 221,324 201,415 233,215 128,624 159,833 198,466 130,676 253,076 225,880 125,251 161,372 127,076 168,977 203,377 223,011 118,525 231,651 120,415 148,426 241,171 177,226 275,936 157,077 130,162 146,621 224,741 182,677 132,424 249,651 134,249 246,524 208,615 257,373 215,377 173,866 119,876 146,682 251,251 148,421 270,536 192,346 124,101 220,777 126,880 170,257 154,755 249,475 175,496 259,173 166,480 263,320 152,266 125,773 30 Statistics for Business 6.Exchange rates Situation The table on next page gives the exchange rates in currency units per $US for two periods in 2004 and 2005.Construct a parallel bar chart for this data.

(Note in order to obtain a graph which is more equitable, divide the data for Japan by 100 and those for Denmark and Sweden by 10.What are your conclusions from this bar chart? 7.European sales Situation The table below gives the monthly profits in Euros for restaurants of a certain chain in Europe.Country Profits ($) Denmark 985,789 England 1,274,659 Germany 225,481 Ireland 136,598 Netherlands 325,697 Norway 123,657 Poland 429,857 Portugal 256,987 Czech Republic 102,654 Spain 995,796 16 November 2005 16 November 2004 Australia 1.

17 2 Economic and financial indicators, The Economist, 19 November 2005, p.31Chapter 1: Presenting and organizing data Required 1.Develop a pie chart for this information.Develop a histogram for this information in terms of absolute profits and percentage profits.Develop a bar chart for this information in terms of absolute profits and percentage profits.What are the three best performing countries and what is their total contribution to the total profits given? 5.

Which are the three countries that have the lowest contribution to profits and what is their total contribution? 8.Nuclear power Situation The table below gives the nuclear reactors in use or in construction according to country.of nuclear reactors Region Argentina 3 South America Armenia 1 Eastern Europe Belgium 7 Western Europe Brazil 2 South America Britain 27 Western Europe Bulgaria 4 Eastern Europe Canada 16 North America China 11 Far East Czech Republic 6 Eastern Europe Finland 4 Western Europe France 59 Western Europe Germany 18 Western Europe Hungary 4 Eastern Europe India 22 ME and South Asia Iran 2 ME and South Asia Japan 56 Far East Lithuania 2 Eastern Europe Mexico 2 North America Netherlands 1 Western Europe North Korea 1 Far East Pakistan 2 ME and South Asia Romania 2 Eastern Europe Russia 33 Eastern Europe Slovakia 8 Eastern Europe Slovenia 1 Eastern Europe South Africa 2 Africa 3 International Herald Tribune, 18 October 2004.(Continued) 32 Statistics for Business Required 1.

Develop a bar chart for this information by country sorted by the number of reactors.Develop a pie chart for this information according to the region.Develop a pie chart for this information according to country for the region that has the highest proportion of nuclear reactors.

Which three countries have the highest number of nuclear reactors? 5.Which region has the highest proportion of nuclear reactors and dominated by which country? 9.Textbook sales Situation The sales of an author’s textbook in one particular year were according to the following table.Country Sales (units) Country Sales (units) Australia 660 Mexico 10 Austria 4 Northern Ireland 69 Belgium 61 Netherlands 43 Botswana 3 New Zealand 28 Canada 147 Nigeria 3 China 5 Norway 78 Denmark 189 Pakistan 10 Egypt 10 Poland 4 Eire 25 Romania 3 England 1,632 South Africa 62 Finland 11 South Korea 1 France 523 Saudi Arabia 1 Germany 28 Scotland 10 Greece 5 Serbia 1 Hong Kong 2 Singapore 362 India 17 Slovenia 4 Iran 17 Spain 16 Israel 4 Sri Lanka 2 Italy 26 Sweden 162 Country No.

of nuclear reactors Region South Korea 20 Far East Spain 9 Western Europe Sweden 11 Western Europe Switzerland 5 Western Europe Ukraine 17 Eastern Europe United States 104 North America ME: Middle East.33Chapter 1: Presenting and organizing data Required 1.Develop a histogram for this data by country and by units sold, sorting the data from the country in which the units sold were the highest to the lowest.What is your criti- cism of this visual presentation? 2.Develop a pie chart for book sales by continent.

Which continent has the highest per- centage of sales? Which continent has the lowest book sales? 3.Develop a histogram for absolute book sales by continent from the highest to the lowest.Develop a pie chart for book sales by countries in the European Union.Which coun- try has the highest book sales as a proportion of total in Europe? Which country has the lowest sales? 5.

Develop a histogram for absolute book sales by countries in the European Union from the highest to the lowest.What are your comments about this data? 10.Textile wages Situation The table below gives the wage rates by country, converted to $US, for persons working in textile manufacturing.The wage rate includes all the mandatory charges which have to be paid by the employer for the employees benefit.

This includes social charges, med- ical benefits, vacation, and the like.4 Country Wage rate ($US/hour) Bulgaria 1.78 Country Sales (units) Country Sales (units) Japan 21 Switzerland 59 Jordan 3 Taiwan 938 Latvia 1 Thailand 2 Lebanon 123 UAE 2 Lithuania 1 Wales 135 Luxemburg 69 Zimbabwe 3 Malaysia 2 4 Wall street Journal Europe, 27 September 2005, p.

Develop a bar chart for this information.Determine the wage rate of a country relative to the wage rate in China.Plot on a combined histogram and line graph the sorted wage rate of the country as a histogram and a line graph for the data that you have calculated in Question 2.What are your conclusions from this data that you have presented? 11.

Immigration to Britain Situation Nearly a year and a half after the expansion of the European Union, hundreds of East Europeans have moved to Britain to work.Poles, Lithuanians Latvians and others are arriving at an average rate of 16,000 a month, as a result of Britain’s decision to allow unlimited access to the citizens of the eight East Europeans that joined the European Union in 2004.The immigrants work as bus drivers, farmhands, dentists, waitresses, builders, and sales persons.The following table gives the statistics for those new arrivals from Eastern Europe since May 2004.5 Nationality of applicant Registered to work Czech Republic 14,610 Estonia 3,480 Hungary 6,900 Latvia 16,625 Lithuania 33,755 Poland 131,290 Slovakia 24,470 Slovenia 250 Age range of applicant Percentage in range 18–24 42.

applied to work (May 2004–June 2005) Administration, business, and management 62,000 Agriculture 30,400 Construction 9,000 Entertainment and leisure 4,000 5 Fuller, T., Europe’s great migration: Britain absorbing influx from the East, International Herald Tribune, 21 October 2005, pp.35Chapter 1: Presenting and organizing data Required 1.Develop a bar chart of the nationality of the immigrant and the number who have registered to work.

Transpose the information from Question 1 into a pie chart.Develop a pie chart for the age range of the applicant and the percentage in this range.Develop a bar chart for the employment sector of the immigrant and those registered for employment in this sector.What are your conclusions from the charts that you have developed? 12.Pill popping Situation The table below gives the number of pills taken per 1,000 people in certain selected countries.Develop a bar chart for the data in the given alphabetical order.Develop a pie chart for the data and show on this the country and the percentage of pill consumption based on the information provided.Country Pills consumed per 1,000 people Canada 66 France 78 Italy 40 Japan 40 Spain 64 United Kingdom 36 USA 53 Employment sector of applicant No.applied to work (May 2004–June 2005) Food processing 11,000 Health care 10,000 Hospitality and catering 53,200 Manufacturing 19,000 Retail 9,500 Transport 7,500 Others 9,500 6 Wall Street Journal Europe, 25 February 2004.

Which country consumes the highest percentage of pills and what is this percentage amount to the nearest whole number? 4.How would you describe the consumption of pills in France compared to that in the United Kingdom? 13.Electoral College Situation In the United States for the presidential elections, people vote for a president in their state of residency.Each state has a certain number of electoral college votes according to the population of the state and it is the tally of these electoral college votes which deter- mines who will be the next United States president.

The following gives the electoral col- lege votes for each of the 50 states of the United States plus the District of Colombia.7 Also included is how the state voted in the 2004 United States presidential elections.8 State Electoral college votes Voted to Alabama 9 Bush Alaska 3 Bush Arizona 10 Bush Arkansas 6 Bush California 55 Kerry Colorado 9 Bush Connecticut 7 Kerry Delaware 3 Kerry District of Columbia 3 Kerry Florida 27 Bush Georgia 15 Bush Hawaii 4 Kerry Idaho 4 Bush Illinois 21 Kerry Indiana 11 Bush Iowa 7 Bush Kansas 6 Bush Kentucky 8 Bush Louisiana 9 Bush Maine 4 Kerry Maryland 10 Kerry Massachusetts 12 Kerry Michigan 17 Kerry Minnesota 10 Kerry Mississippi 6 Bush 7 Wall Street Journal Europe, 2 November 2004, p.37Chapter 1: Presenting and organizing data Required 1.Develop a pie chart of the percentage of electoral college votes for each state.Develop a histogram of the percentage of electoral college votes for each state.

How were the electoral college votes divided between Bush and Kerry? Show this on a pie chart.Which state has the highest percentage of electoral votes and what is the percentage of the total electoral college votes? 5.

What is the percentage of states including the District of Columbia that voted for Kerry? 14.

Chemical delivery Situation A chemical company is concerned about the quality of its chemical products that are delivered in drums to its clients.Over a 6-month period it used a student intern to State Electoral college votes Voted to Missouri 11 Bush Montana 3 Bush Nebraska 5 Bush Nevada 5 Bush New Hampshire 4 Kerry New Jersey 15 Kerry New Mexico 5 Bush New York 31 Kerry North Carolina 15 Bush North Dakota 3 Bush Ohio 20 Bush Oklahoma 7 Bush Oregon 7 Kerry Pennsylvania 21 Kerry Rhode Island 4 Kerry South Carolina 8 Bush South Dakota 3 Bush Tennessee 11 Bush Texas 34 Bush Utah 5 Bush Vermont 3 Kerry Virginia 13 Bush Washington 11 Kerry West Virginia 5 Bush Wisconsin 10 Kerry Wyoming 3 Bush 38 Statistics for Business measure quantitatively the number of problems that occurred in the delivery process.The following table gives the recorded information over the 6-month period.The column “reason” in the table is considered exhaustive.Construct a Pareto curve for this information.What is the problem that happens most often and what is the percentage of occur- rence? This is the problem area that you would probably tackle first.Which are the four problem areas that constitute almost 80% of the quality problems in delivery? 15.

Fruit distribution Situation A fruit wholesaler was receiving complaints from retail outlets on the quality of fresh fruit that was delivered.In order to monitor the situation the wholesaler employed a stu- dent to rigorously take note of the problem areas and to record the number of times these problems occurred over a 3-month period.The following table gives the recorded information over the 3-month period.The column “reasons” in the table is considered exhaustive.of occurrences in 3 months Bacteria on some fruit 9 Boxes badly loaded 62 Boxes damaged 17 Client documentation incorrect 23 Fruit not clean 25 Fruit squashed 74 Fruit too ripe 14 Reason No.of occurrences in 6-months Delay – bad weather 70 Documentation wrong 100 Drums damaged 150 Drums incorrectly sealed 3 Drums rusted 22 Incorrect labelling 7 Orders wrong 11 Pallets poorly stacked 50 Schedule change 35 Temperature too low 18 39Chapter 1: Presenting and organizing data Required 1.Construct a Pareto curve for this information.What is the problem that happens most often and what is the percentage of occur- rence? Is this the problem area that you would tackle first? 3.

What are the problem areas that cumulatively constitute about 80% of the quality problems in delivery of the fresh fruit? 16.Case: Soccer Situation When an exhausted Chris Powell trudged off the Millennium stadium pitch on the afternoon of 30 May 2005, he could not have been forgiven for feeling pleased with himself.Not only had he helped West Ham claw their way back into the Premiership league for the 2005–2006 season, but the left back had featured in 42 league cup and play off matches since reluctantly leaving Charlton Athletic the previous September.It had been a good season since opposition right-wingers had been vanquished and Powell and Mathew Etherington had formed a formidable left-sided partnership.If you did not know better, you might have suspected the engaging 35-year old was a decade younger.

9 For many people in England, and in fact, for most of Europe, football or soccer is their passion.Every Saturday many people, the young and the not-so-young, faithfully go and see their home team play.According to the accountants, Deloitte and Touche, the 20 clubs that make up the Barclay’s Bank spon- sored English Premiership league, the most watched and profitable league in Europe, had total revenues of almost £2 billion ($3.The best players command salaries of £100,000 a week excluding endorsements.10 In addition, at the end of the season, the clubs themselves are awarded price money depending on their position on the league tables at the end of the year.These prize amounts are indi- cated in Table 1 for the 2004–2005 season.The game results are given in Table 2 and the final league results in Table 3 and from these you can determine the amount that was awarded to each club.11 Labelling wrong 11 Orders not conforming 6 Route directions poor 30 9 Aizlewood, J.

, Powell back at happy valley, The Sunday Times, 28 August 2005, p., Business and the Beautiful Game, Kogan Page, International Herald Tribune, 1–2 October 2005, p.

11 News of the World Football Annual 2005–2006, Invincible Press, an imprint of Harper Collins, 2005.40 Statistics for Business Required These three tables give a lot of information on the premier leaguer football results for the 2004–2005 season.How could you put this in a visual form to present the information to a broad audience? Table 2 Club Games Home Away played Win Draw Lost For Away Win Draw Lost For Away Arsenal 38 13 5 1 54 19 12 3 4 33 17 Aston Villa 38 8 6 5 26 17 5 5 10 19 35 Birmingham City 38 8 6 5 24 15 4 6 10 16 31 Blackburn Rovers 38 5 8 6 24 22 3 7 8 11 21 Bolton Wanderers 38 9 5 5 25 18 7 5 7 24 26 Charlton Athletic 38 8 4 7 29 29 4 6 9 13 29 Chelsea 38 14 5 0 35 6 15 3 1 37 9 Crystal Palace 38 6 5 8 21 19 1 7 11 20 43 Everton 38 12 2 5 24 15 6 5 8 21 31 Fulham 38 8 4 7 29 26 3 4 11 23 34 Liverpool 38 12 4 3 31 15 5 3 11 21 26 Manchester City 38 8 6 5 24 14 5 7 7 23 25 Manchester United 38 12 6 1 31 12 10 5 4 27 14 Middlesbrough 38 9 6 4 29 19 5 7 7 24 27 Position Prize money (£) Position Prize money (£) 1 9,500,000 11 4,750,000 2 9,020,000 12 4,270,000 3 8,550,000 13 3,800,000 4 8,070,000 14 3,320,000 5 7,600,000 15 2,850,000 6 7,120,000 16 2,370,000 7 6,650,000 17 1,900,000 8 6,170,000 18 1,420,000 9 5,700,000 19 950,000 10 5,220,000 20 475,000 Table 1 41Chapter 1: Presenting and organizing data Club Games Home Away played Win Draw Lost For Away Win Draw Lost For Away Newcastle United 38 7 7 5 25 25 4 7 9 22 32 Norwich City 38 7 5 7 29 32 0 7 12 13 45 Portsmouth 38 8 4 7 30 26 4 5 12 13 33 Southampton 38 5 9 5 30 30 1 5 13 15 36 Tottenham 38 9 5 5 36 22 5 5 9 11 19 WBA 38 5 8 6 17 24 2 8 10 19 37 42 Statistics for Business Table 3 Arsenal – – – 3 – 1 3 – 0 3 – 0 2 – 2 3 – 0 2 – 2 5 – 1 7 – 0 Aston Villa 1 – 3 – – – 1 – 2 1 – 0 1 – 1 4 – 1 0 – 0 1 – 1 1 – 3 Birmingham 2 – 1 2 – 0 – – – 2 – 1 1 – 2 5 – 2 0 – 1 0 – 1 0 – 1 City Blackburn 0 – 1 2 – 2 3 – 3 – – – 0 – 1 6 – 3 0 – 1 1 – 0 0 – 0 Rovers Bolton 1 – 0 1 – 2 1 – 1 0 – 1 – – – 0 – 1 0 – 2 1 – 0 3 – 2 Wanderers Charlton 1 – 3 3 – 0 3 – 1 1 – 0 1 – 2 – – – 0 – 4 2 2 2 – 0 Athletic Chelsea 0 – 0 1 – 0 1 – 1 4 – 0 2 – 2 4 – 0 – – – 4 – 1 1 – 0 Crystal 1 – 1 2 – 0 2 – 0 0 – 0 0 – 1 0 – 0 0 – 2 – – – 1 – 3 Palace Everton 1 – 4 1 – 1 1 – 1 0 – 1 3 – 2 0 – 1 0 – 1 4 – 0 – – – Fulham 0 – 3 1 – 1 2 – 3 0 – 2 2 – 0 0 – 2 1 – 4 3 – 1 2 – 0 Liverpool 2 – 1 2 – 1 0 – 1 0 – 0 1 – 0 0 – 0 0 – 1 3 – 2 2 – 1 Manchester 0 – 1 2 – 0 3 – 0 1 – 1 0 – 1 1 – 1 1 – 0 3 – 1 0 – 1 City Manchester 2 – 0 3 – 1 2 – 0 0 – 0 2 – 0 0 – 0 1 – 3 5 – 2 0 – 0 United Middlesbrough 0 – 1 3 – 0 2 – 1 1 – 0 1 – 1 1 – 0 0 – 1 2 – 1 1 – 1 Newcastle 0 – 1 0 – 3 2 – 1 3 – 0 2 – 1 3 – 0 1 – 1 0 – 0 1 – 1 United Norwich City 1 – 4 0 – 0 1 – 0 1 – 1 3 – 2 1 – 1 1 – 3 1 – 1 2 – 3 Portsmouth 0 – 1 1 – 2 1 – 1 0 – 1 1 – 1 0 – 1 0 – 2 3 – 1 0 – 1 Southampton 1 – 1 2 – 3 0 – 0 3 – 2 1 – 2 3 – 2 1 – 3 2 – 2 2 – 2 Tottenham 4 – 5 5 – 1 1 – 0 0 – 0 1 – 2 0 – 0 0 – 2 1 – 1 5 – 2 WBA 0 – 2 1 – 1 2 – 0 1 – 1 2 – 1 1 – 1 1 – 4 2 – 2 1 – 0 Ev er to n Cr ys ta l P al ac e Ch el se a Ch ar lto n At hl et ic Bo lto n W an de re rs Bl ac kb ur n Ro ve rs Bi rm in gh am C ity As to n Vi lla Ar se na l 43Chapter 1: Presenting and organizing data 2 – 0 3 – 1 1 – 1 2 – 4 5 – 3 1 – 0 4 – 1 3 – 0 2 – 2 1 – 0 1 – 1 2 – 0 1 – 1 1 – 2 0 – 1 2 – 0 4 – 2 3 – 0 3 – 0 2 – 0 1 – 0 1 – 1 1 – 2 2 – 0 1 – 0 0 – 0 2 – 0 2 – 2 1 – 1 0 – 0 2 – 1 1 – 1 4 – 0 1 – 3 2 – 2 0 – 0 1 – 1 0 – 4 2 – 2 3 – 0 1 – 0 3 – 0 0 – 1 1 – 1 3 – 1 1 – 0 0 – 1 2 – 2 0 – 0 2 – 1 1 – 0 0 – 1 1 – 1 3 – 1 1 – 1 2 – 1 1 – 2 2 – 2 0 – 4 1 – 2 1 – 1 4 – 0 2 – 1 0 – 0 2 – 0 1 – 4 2 – 1 1 – 0 0 – 0 1 – 0 2 – 0 4 – 0 4 – 0 3 – 0 2 – 1 0 – 0 1 – 0 2 – 0 1 – 0 1 – 2 0 – 0 0 – 1 0 – 2 3 – 3 0 – 1 2 – 2 3 – 0 3 – 0 1 – 0 1 – 0 2 – 1 1 – 0 1 – 0 2 – 0 1 – 0 2 – 1 1 – 0 0 – 1 2 – 1 – – – 2 – 4 1 – 1 1 – 1 0 – 2 1 – 3 6 – 0 3 – 1 1 – 0 2 – 0 1 – 0 3 – 1 – – – 2 – 1 0 – 1 1 – 1 3 – 1 3 – 0 1 – 1 1 – 0 2 – 2 3 – 0 1 – 1 1 – 0 – – – 0 – 2 1 – 1 1 – 1 1 – 1 2 – 0 2 – 1 0 – 1 1 – 1 1 – 0 2 – 1 0 – 0 – – – 1 – 1 2 – 1 2 – 1 2 – 1 3 – 0 0 – 0 1 – 1 1 – 1 2 – 0 3 – 2 0 – 2 – – – 2 – 2 2 – 0 1 – 1 1 – 3 1 – 0 4 – 0 1 – 4 1 – 0 4 – 3 1 – 3 0 – 0 – – – 2 – 2 1 – 1 2 – 1 0 – 1 3 – 1 0 – 1 1 – 2 2 – 3 2 – 0 4 – 4 2 – 1 – – – 2 – 2 2 – 1 0 – 2 3 – 2 4 – 3 1 – 2 1 – 3 2 – 0 2 – 1 1 – 1 1 – 1 – – – 4 – 1 1 – 0 3 – 2 3 – 3 2 – 0 0 – 0 1 – 2 2 – 2 1 – 2 4 – 3 2 – 1 – – – 1 – 0 2 – 2 2 – 0 1 – 1 2 – 1 0 – 1 2 – 0 1 – 0 0 – 0 3 – 1 5 – 1 – – – 1 – 1 1 – 1 0 – 5 2 – 0 0 – 3 1 – 2 0 – 0 0 – 0 2 – 0 0 – 0 1 – 1 – – – W BA To tt en ha m So ut ha m pt on Po rt sm ou th N or w ic h Ci ty N ew ca st le U ni te d M id dl es br ou gh M an ch es te r Un ite d M an ch es te r Ci ty Li ve rp oo l Fu lh am This page intentionally left blank 2Characterizing and defining data Fast food and currencies How do you compare the cost of living worldwide? An innovative way is to look at the prices of a McDonald’s Big Mac in various countries as The Economist has being doing since 1986.1 From this information you might conclude that the Euro is overvalued by 17% against the $US; that the cost of living in Switzerland is the highest; and that it is cheaper to live in Malaysia.Alternatively you would know that worldwide the average price of a Big Mac is $2.51; that half of the Big Macs are less than $2.

40; and that the range of the prices of Big Macs is $3.These are some of the characteristics of the prices of data for Big Macs.These are some of the properties of statistical data that are covered in this chapter.1 “The Economist’s Big Mac index: Fast food and strong currencies”, The Economist, 11 June 2005.

46 Statistics for Business Country Price ($US) Argentina 1.13 Country Price ($US) It is useful to characterize data as these charac- teristics or properties of data can be compared or benchmarked with other datasets.

In this way decisions can be made about business situ- ations and certain conclusions drawn.The two common general data characteristics are central tendency and dispersion.The clustering of data around a central or a mid- dle value is referred to as the central tendency.The central tendency that we are most familiar with is average or mean value but there are others.Arithmetic mean The arithmetic mean or most often known as the mean or average value, and written by x–, is the most common measure of central tendency.It is determined by the sum of the all the values of the observations, x, divided by the number of elements in the observations, N.The equation is, 2(i) For example, assume the salaries in Euros of five people working in the same department are as in Table 2.The total of these five values is €172,000 and 172,000/5 gives a mean value of €34,400.

(On a grander scale, Goldman Sachs, the world’s leading investment bank reports that the average pay-packet of its 24,000 staff in 2005 was $520,000 and that included a lot of assistants and secretaries!2) The arithmetic mean is easy to understand, and every dataset has a mean value.x x N � ∑ Central Tendency of Data 47Chapter 2: Characterizing and defining data After you have studied this chapter you will be able to determine the properties of statistical data, to describe clearly their meaning, to compare datasets, and to apply these properties in decision- making.Specifically, you will learn the following characteristics.✔ Central tendency of data • Arithmetic mean • Weighted average • Median value • Mode • Midrange • Geometric mean ✔ Dispersion of data • Range • Variance and standard deviation • Expression for the variance • Expression for the standard deviation • Determining the variance and the standard deviation • Deviations about the mean • Coefficient of variation and the standard deviation ✔ Quartiles • Boundary limit of quartiles • Properties of quartiles • Box and whisker plot • Drawing the box and whisker plot with Excel ✔ Percentiles • Development of percentiles • Division of data L e a r n i n g o b j e c t i v e s Eric Susan John Helen Robert 40,000 50,000 35,000 20,000 27,000 Table 2.

2 “On top of the world – In its taste for risk, the world’s leading investment bank epitomises the modern financial system”, The Economist, 29 April 2006, p.Note that the arithmetic mean can be influ- enced by extreme values or outliers.In the above situation, John has an annual salary of €35,000 and his salary is currently above the average.

Now, assume that Susan has her salary increased to €75,000 per year.In recalculating the mean the average salary of the five increases to €39,400 per year.Nothing has happened to John’s situation but his salary is now below average.Is John now at a disadvantage? What is the reason that Susan received the increase? Thus, in using average values for analysis, you need to understand if it includes outliers and the circumstance for which the mean value is being used.The number of values does not necessarily influence the arithmetic mean.

In the above example, using the original data, suppose now that Brian and Delphine join the department at respective annual salaries of €34,000 and €34,800 as shown in Table 2.Weighted average The weighted average is a measure of central ten- dency and is a mean value that takes into account the importance, or weighting of each value in the overall total.For example in Chapter 1 we intro- duced a questionnaire as a method of evaluat- ing customer satisfaction.

4 is the type of questionnaire used for evaluating customer satisfaction.Here the questionnaire has the responses of 15 students regarding satisfaction of a course programme.(Students are the customers of the professors!) The X in each cell is the response of each student and the total responses for each category are in the last line.The weighted average of the student response is given by, Weighted average From the table we have, Weighted average Thus using the criterion of the weighted average the central tendency of the evaluation of the university programme is 3.

80, which translates into saying the programme is between satisfac- tory and good and closer to being good.Another use of weighted averages is in product costing.Assume that a manufacturing organ- ization uses three types of labour in the manu- facture of Product 1 and Product 2 as shown in Table 2.In making the finished product the semi-finished components must pass through the activities of drilling, forming, and assembly before it is completed.Note that in these differ- ent activities the hourly wage rate is different.Thus to calculate the correct average cost of � � � � � � 2 1 1 2 1 3 5 4 6 5 15 3 80 * * * * * .� Number of responses * score Total responses∑ 48 Statistics for Business Eric Susan John Helen Robert Brian Delphine 40,000 50,000 35,000 20,000 27,000 34,000 34,800 Table 2.3 Arithmetic mean not necessarily affected by the number values.

labour per finished unit, weighted averages are used as follows: Product A, labour cost, $/unit is $10.44 Product B, labour cost, $/unit is $10.94 If simply the average hourly wage rate was used, the hourly labour cost would be: Then if we use this hourly labour cost to deter- mine unit product cost we would have, Product A 12.

88/unit This is an incorrect way to determine the unit cost since we must use the contribution of each activity to determine the correct amount.Median value The median is another measure of central ten- dency that divides information or data into two equal parts.We come across the median when we talk about the median of a road.This is the white line that divides the road into two parts such that there is the same number of lanes on 10 50 12 75 3 12 50 .25 hour 49Chapter 2: Characterizing and defining data Table 2.Category Very Poor Satisfactory Good Very good Total poor responses Score 1 2 3 4 5 Student 1 X 1 Student 2 X 1 Student 3 X 1 Student 4 X 1 Student 5 X 1 1 Student 6 X 1 Student 7 X 1 Student 8 X 1 Student 9 X 1 Student 10 X 1 Student 11 X 1 Student 12 X 1 Student 13 X 1 Student 14 X 1 Student 15 X 1 Total responses 2 1 1 5 6 15 Labour Hourly Labour Labour operation wage rate hours/unit hours/unit Product A Product B Drilling $10.

When we have quan- titative data it is the middle value of the data array or the ordered set of data.To determine the median value it must first be rearranged in ascending (or descending) order.In ascending order this is as in Table 2.Since there are seven pieces of data, the middle, or the median, value is the 4th number, which in this case is 11.

The median value is of interest as indicates that half of the data lies above the median, and half below.For example, if the median, price of a house in a certain region is $200,000 then this indicates that half of the number of houses is above $200,000 and the other half is below.When n, the number of values in a data array is odd, the median is given by, 2(ii) Thus, if there are seven values in the dataset, then the median is (7 � 1)/2 or the 4th value as in the above example.When n, the number of values, is even, the median value is the average of the values deter- mined from the following relationship: 2(iii) When there are 6 values in a set of data, the median is the value of 6/2 and (6 � 2)/2 or the linear average of the 3rd and 4th value.The value of the median is unaffected by extreme values.

Consider again the salary situ- ation of the five people in John’s department as in Table 2.Again, if Susan’s salary is increased to €75,000 then the revised information is as in Table 2.John still has the median salary and so that on this basis, nothing has changed for John.However, when we used the average value as above, there was a change.

The number of values affects the median.

Assume Stan joins the department in the example above at the same original salary as Susan.There is an even number of values in the dataset and now the median is (35,000 � 40,000)/2 or €37,500.Again, nothing has happened n n 2 2 2 and ( )� n �1 2 50 Statistics for Business 9 13 12 7 6 11 12 Table 2.Eric Susan John Helen Robert 40,000 50,000 35,000 20,000 27,000 Table 2.

Helen Robert John Eric Susan 20,000 27,000 35,000 40,000 50,000 Table 2.Helen Robert John Eric Susan 20,000 27,000 35,000 40,000 75,000 Table 2.10 Median value – salaries unaffected by extreme values.

We do not have to order the data or even to take into account whether there is an even or odd number of values as Excel automatically takes this into consideration.For example, if we determine the median value of the sales data given in Table 1.For this dataset the median value is 100,296.

Mode The mode is another measure of central ten- dency and is that value that occurs most fre- quently in a dataset.It is of interest because that value that occurs most frequently is probably a response that deserves further investigation.12 are the monthly sales in $millions for the last year.Thus in forecasting future sales we might conclude that there is a higher probability that sales will be $12 million in any given month.The mode is unaffected by extreme values.For example, if the sales in January were $100 million instead of $10 million, the mode is still 12.However, the number of values might affect the mode.For example, if we use the following sales data in Table 2.

13 over the last 15 months, the modal value is now $14 million since it occurs 4 times.Unlike the mean and median, the mode can be used for qualitative as well as for quantitative data.For example, in a questionnaire, people were asked to give their favourite colour.The responses were according to Table 2.The modal value is blue since this response occurred 3 times.This type of information is useful say in the textile business when a firm is planning the preparation of new fabric or the automobile industry when the company is planning to put 51Chapter 2: Characterizing and defining data January 10 February 12 March 11 April 14 May 12 June 14 July 12 August 16 September 9 October 19 November 10 December 13 January 14 February 10 March 14 Table 2.13 Mode – might be affected by the number of values.Helen Robert John Eric Susan Stan 20,000 27,000 35,000 40,000 50,000 50,000 Table 2.11 Median value – number of values affects the median.

January 10 February 12 March 11 April 14 May 12 June 14 July 12 August 16 September 9 October 19 November 10 December 13 Table 2.12 Mode – that value that occurs most frequently.A dataset might be multi-modal when there are data values that occur equally frequently.

For example, bi-modal is when there are two values in a dataset that occur most frequently.15 is bi-modal as both the values 9 and 13 occur twice.When a dataset is bi-modal that indicates that there are two pieces of data that are of particular interest.

meaning that there are three, four, or more values that occur most frequently.Midrange The midrange is also a measure of central ten- dency and is the average of the smallest and largest observation in a dataset.16, the midrange is, The midrange is of interest to know where data sits compared to the midrange.17, The midrange is (50,000 � 20,000)/2 or 35,000 and so John’s salary is exactly at the midrange.Again assume Susan’s salary is increased to €75,000 to give the information in Table 2.Then the midrange is (20,000 � 75,000)/2 or €47,500 and John’s salary is now below the midrange.Thus, the midrange can be distorted by extreme values.

Geometric mean The geometric mean is a measure of central ten- dency used when data is changing over time.Examples might be the growth of investments, the inflation rate, or the change of the gross national product.For example, consider the growth of an initial investment of $1,000 in a savings account that is deposited for a period of 5 years.The inter- est rate, which is accumulated annually, is differ- ent for each year.19 gives the interest and the growth of the investment.The average growth rate, or geometric mean, is calculated by the relationship: n��(product����of �growth���rates)�� 2(iv) 13 6 2 19 2 9 5 � � � .52 Statistics for Business Yellow Green Purple Blue Red Blue Brown Pink Green Violet Rose Blue Table 2.Helen Robert John Eric Susan 20,000 27,000 35,000 40,000 50,000 Table 2.Helen Robert John Eric Susan 20,000 27,000 35,000 40,000 75,000 Table 2.18 Midrange – affected by extreme values.In this case the geometric mean is, This is an average growth rate of 6.Thus, the value of the $1,000 at the end of 5 years will be, The same value as calculated in Table 2.If the arithmetic average of the growth rates was used, the mean growth rate would be: or a growth rate slightly less of 6.

Using this mean interest rate, the value of the initial deposit at the end of 5 years would be, This is less than the amount calculated using the geometric mean.The difference here is small but in cases where interest rates are widely fluc- tuating, and deposit amounts are large, the dif- ference can be significant.1 000 1 0690 1 396 015 � 1 060 1 075 1 082 1 079 1 051 5 1 0690 .1 000 1 0693 1 398 195 � 1 060 1 075 1 082 1 079 1 051 1 0693 5 .

� 53Chapter 2: Characterizing and defining data Eric Susan John Helen Robert 40,000 50,000 35,000 20,000 27,000 Table 2.Dispersion is how much data is separated, spread out, or varies from other data values.It is import- ant to know the amount of dispersion, variation, or spread, as data that is more dispersed or separ- ated is less reliable for analytical purposes.Datasets can have different measures of disper- sion or variation but may have the same measure of central tendency.

In many situations, we may be more interested in the variation, than in the central value, since variation can be a measure of inconsistency.The following are the common measures of the dispersion of data.Range The range is the difference between the max- imum and the minimum value in a dataset.We have seen the use of the range in Chapter 1 in the development of frequency distributions.Another illustration is represented in Table 2.

20 which is the salary data presented earlier in Table 2.Here the range is the difference of the salaries for Susan and Helen, or €50,000 � €20,000 � €30,000.The range is affected by extreme values.For example, if we include in the dataset the salary of Francis, the manager of the department, who has a salary of €125,000 we then have Table 2.

Here the range is €125,000 � €20,000 � €105,000.The number of values does not necessarily affect the range.For example let us say that we Dispersion of Data Year Interest Growth Value rate (%) factor year-end 1 6.add the salary of Julie at €37,000 to the dataset in Table 2.Then the range is unchanged at €105,000.The larger the range in a dataset, then the greater is the dispersion, and thus the uncer- tainty of the information for analytical purposes.

Although we often talk about the range of data, the major drawback in using the range as a measure of dispersion is that it only considers two pieces of information in the dataset.In this case, any extreme, or outlying values, can distort the measure of dispersion as is illustrated by the information in Tables 2.Variance and standard deviation The variance and the related measure, standard deviation, overcome the drawback of using the range as a measure of dispersion as in their cal- culation every value in the dataset is con- sidered.

Although both the variance and standard deviation are affected by extreme values, the impact is not as great as using the range since an aggregate of all the values in the dataset are considered.The variance and particularly the standard deviation are the most often used meas- ures of dispersion in statistics.The variance is in squared units and measures the dispersion of a dataset around the mean value.The standard deviation has the same units of the data under consideration and is the square root of the vari- ation.We use the term “standard” in standard deviation as it represents the typical deviation for that particular dataset.

Expression for the variance There is a variance and standard deviation both for a population and a sample.The population variance, denoted by , is the sum of the squared difference between each observation, x, and the mean value, , divided by the number of data observations, N, or as follows: 2(v) ● For each observation of x, the mean value x is subtracted.This indicates how far this observation is from the mean, or the range of this observation from the mean.● By squaring each of the differences obtained, the negative signs are removed.● By dividing by N gives an average value.

The expression for the sample variance, s2, is analogous to the population variance and is, 2(vi) In the sample variance, x–, or x-bar the average of the values of x replaces x of the population vari- ance and (n � 1) replaces N the population size.One of the principal uses of statistics is to take a sample from the population and make estimates of the population parameters based only on the sample measurements.By convention when we use the symbol n it means we have taken a sample of size n from the population of size N.Using (n � 1) in the denominator reflects the fact that we have used x– in the formula and so we have lost one degree of freedom in our calculation.For example, consider you have a sum of $1,000 to s x x n 2 2 1 � � � ( ) ( ) ∑ x xx N 2 2 � ( )−∑ x 2 54 Statistics for Business Eric Susan John Francis Helen Robert 40,000 50,000 35,000 125,000 20,000 27,000 Table 2.

Eric Susan Julie John Francis Helen Robert 40,000 50,000 37,000 35,000 125,000 20,000 27,000 Table 2.22 Range is not necessarily affected by the number of values.distribute to your six co-workers based on certain criteria.To the first five you have the freedom to give any amount say $200, $150, $75, $210, $260.

To the sixth co-worker you have no degree of freedom of the amount to give which has to be the amount remaining from the original $1,000, which in this case is $105.When we are perform- ing sampling experiments to estimate the popula- tion parameter with (n � 1) in the denominator of the sample variance formula we have an unbiased estimate of the true population variance.If the sample size, n, is large, then using n or (n � 1) will give results that are close.Expression for the standard deviation The standard deviation is the square root of the variance and thus has the same units as the data used in the measurement.It is the most often used measure of dispersion in analytical work.

The population standard deviation, x, is given by, 2(vii) The sample standard deviation, s, is as follows: 2(viii) For any dataset, the closer the value of the stand- ard deviation is to zero, then the smaller is the dispersion which means that the data values are closer to the mean value of the dataset and that the data would be more reliable for subsequent analytical purposes.Note that the expression is sometimes used to denote the population standard deviation rather than x.Similarly is used to denote the mean value rather than x.That is the subscript x is dropped, the logic being that it is understood that the values are calculated using the random variable x and so it is not necessary to show them with the mean and standard deviation symbols! s s x x n � � � � 2 2 1 ( ) ( ) ∑ x x xx N � � � 2 2( )∑ 55Chapter 2: Characterizing and defining data Determining the variance and the standard deviation Let us consider the dataset given in Table 2.If we use equations 2(v) through to 2(viii) then we will obtain the population variance, the popu- lation standard deviation, the sample variance, and the sample standard deviation.These values and the calculation steps are shown in Table 2.Note that for any given dataset when you calcu- late the population variance it is always smaller than the sample variance since the denomin- ator, N, in the population variance is greater than the value N � 1.

Similarly for the same dataset the population standard deviation is always less than the calculated sample standard deviation.25, which is a summary of the final results of Table 2.Deviation about the mean The deviation about the mean of all observa- tions, x, about the mean value, x–, is zero or mathematically, 2(ix)( )x x� � 0∑ 9 13 12 7 6 11 12 Table 2.

And the deviation of the data around the mean value of 10 is as follows: (9 �10) � (13 �10) � (12 �10) � (7 �10) � (6 �10) � (11 �10) � (12 �10) � 0 This is perhaps a logical conclusion since the mean value is calculated from all the dataset values.Coefficient of variation and the standard deviation The standard deviation as a measure of dispersion on its own is not easy to interpret.

In general terms a small value for the standard deviation indicates that the dispersion of the data is low and conversely the dispersion is large for a high value of the standard deviation.However the magnitude of these values depends on what you are analysing.Further, how small is small and what about the units? If you say that the stand- ard deviation of the total travel time, including waiting, to fly from London to Vladivostok is 2 hours, the number 2 is small.However, if you convert that to minutes the value is 120, and a high 7,200, if you use seconds.

But in any event, the standard deviation has not changed! A way to overcome the difficulty in interpret- ing the standard deviation is to include the value of the mean of the dataset and use the coefficient of variation.

The coefficient of vari- ation is a relative measure of the standard devi- ation of a distribution, , to its mean, .The 56 Statistics for Business x (x � ) (x � )2 2 �1 1 13 3 9 12 2 4 7 �3 9 6 �4 16 11 1 1 12 2 4 Number of values, N 7 Total of values 70 44 Mean value, 10 N � 1 6 Table 2.7080 Measure of dispersion Value Population variance 6.coefficient of variation can be either expressed as a proportion or a percentage of the mean.It is defined as follows: 2(x) As an illustration, say that a machine is cutting steel rods used in automobile manufacturing where the average length is 1.5 m, and the stand- ard deviation of the length of the rods that are cut is 0.In this case the coef- ficient of variation is 0.25/150 (keeping all units in cm), which is 0.This value is small and perhaps would be acceptable from a quality control point of view.

However, say that the standard deviation is 6 cm or 0.06 is a small number but it gives a coeffi- cient of variation of 0.This value is probably unacceptable for precision engineering in automobile manufacturing.The coefficient of variation is also a useful measure to compare two sets of data.For example, in a manufacturing operation two operators are working on each of two machines.

Operator A produces an average of 45 units/day, with a standard deviation of the number of pieces pro- duced of 8 units.Operator B completes on aver- age 125 units/day with a standard deviation of 14 units.Which operator is the most consistent in the activity? If we just examine the standard deviation, it appears that Operator B has more variability or dispersion than Operator A, and thus might be considered more erratic.However if we compare the coefficient of variations, the value for Operator A is 8/45 or 17.78% and for Operator B it is 14/125 or 11.

On this com- parative basis, the variability for Operator B is less than for Operator A because the mean output for Operator B is more.The term / is strictly for the population distribution.

However, in absence of the values for the population, sample values of s/x-bar will give you an estimation of the coefficient of variation.In the earlier section on Central Tendency of Data we introduced the median or the value that divides ordered data into two equal parts.Another divider of data is the quartiles or those values that divide ordered data into four equal parts, or four equal quarters.With this division of data the positioning of information within the quartiles is also a measure of dispersion.

Quartiles are useful to indicate where data such as student’s grades, a person’s weight, or sales’ revenues are positioned relative to standardized data.

Boundary limits of quartiles The lower limit of the quartiles is the minimum value of the dataset, denoted as Q0, and the upper limit is the maximum value Q4.Between these two values is contained 100% of the dataset.There are then three quartiles within these outer limits.The 1st quartile is Q1, the 2nd quartile Q2, and the 3rd quartile Q3.We then have the boundary limits of the quartiles which are those values that divide the dataset into four equal parts such that within each of these boundaries there is 25% of the data.

Q0 Q1 Q2 Q3 Q4 Quartiles Coefficient of variation � 57Chapter 2: Characterizing and defining data Mean Standard Coefficient of output deviation variation / (%) Operator A 45 8 17.28, which gives the five quartile boundary limits plus additional properties related to the quartiles.Also indicated is the inter-quartile range, or mid-spread, which is the difference between the 3rd and the 1st quartile in a dataset or (Q3 � Q1).It measures the range of the middle 50% of the data.

One half of the inter-quartile range, (Q3 � Q1)/2, is the quartile deviation and this measures the average range of one half of the data.The smaller the quartile deviation, the greater is the concentration of the middle half of the observations in the dataset.The mid-hinge, or (Q3 � Q1)/2, is a measure of central tendency and is analogous to the midrange.Although like the range, these additional quartile properties only use two values in their calculation, distortion from extreme values is limited as the quartile values are taken from an ordered set of data.35,378 170,569 104,985 134,859 120,958 107,865 127,895 106,825 130,564 108,654 109,785 184,957 96,598 121,985 63,258 164,295 97,568 165,298 113,985 124,965 108,695 91,864 120,598 47,865 162,985 83,964 103,985 61,298 104,987 184,562 89,597 160,259 55,492 152,698 92,875 56,879 151,895 88,479 165,698 89,486 85,479 64,578 103,985 81,980 137,859 126,987 102,987 116,985 45,189 131,958 73,598 161,895 132,689 120,654 67,895 87,653 58,975 103,958 124,598 168,592 95,896 52,754 114,985 62,598 145,985 99,654 76,589 113,590 80,459 111,489 109,856 101,894 80,157 78,598 86,785 97,562 136,984 89,856 96,215 163,985 83,695 75,894 98,759 133,958 74,895 37,856 90,689 64,189 107,865 123,958 105,987 93,832 58,975 102,986 102,987 144,985 101,498 101,298 103,958 71,589 59,326 121,459 82,198 60,128 86,597 91,786 56,897 112,854 54,128 152,654 99,999 78,562 110,489 86,957 99,486 132,569 134,987 76,589 135,698 118,654 90,598 156,982 87,694 117,895 85,632 104,598 77,654 105,987 78,456 149,562 68,976 50,128 106,598 63,598 123,564 47,895 100,295 60,128 141,298 84,598 100,296 77,498 77,856 134,890 79,432 100,659 95,489 122,958 111,897 129,564 71,458 88,796 110,259 72,598 140,598 125,489 69,584 89,651 70,598 93,876 112,987 123,895 65,847 128,695 66,897 82,459 133,984 98,459 153,298 87,265 72,312 81,456 124,856 101,487 73,569 138,695 74,583 136,958 115,897 142,985 119,654 96,592 66,598 81,490 139,584 82,456 150,298 106,859 68,945 122,654 70,489 94,587 85,975 138,597 97,498 143,985 92,489 146,289 84,592 69,874 Quartile Position Value Q0 0 35,378 Q1 1 79,976 Q2 2 100,296 Q3 3 123,911 Q4 4 184,957 Q3 � Q1 Mid-spread 43,935 (Q3 � Q1)/2 Quartile deviation 21,968 (Q3 � Q1)/2 Mid-hinge 101,943 Mean 102,667 Box and whisker plot A useful visual presentation of the quartile val- ues is a box and whisker plot (from the face of a cat – if you use your imagination!) or sometimes referred to as a box plot.The box and whisker plot for the sales data is shown in Figure 2.Here, the middle half of the values of the dataset or the 50% of the values that lie in the inter- quartile range are shown as a box.

The vertical line making the left-hand side of the box is the 1st quartile, and the vertical line of the right-hand side of the box is the 3rd quartile.The 25% of the values that lie to the left of the box and the 25% of the values to the right of the box, or the other 50% of the dataset, are shown as two horizontal lines, or whiskers.The extreme left part of the first whisker is the minimum value, Q0, and the extreme right part of the second whisker is the maximum value, Q4.The larger the width of the box relative to the two whiskers indicates that the data is clustered around the middle 50% of the values.The box and whisker plot is symmetrical if the distances from Q0 to the median, Q2, and the distance from Q2 to Q4 are the same.

In addition, the distance from Q0 to Q1 equals the distance from Q3 to Q4 and the distance from Q1 to Q2 equals the distance from the Q2 to Q3 and fur- ther the mean and the median values are equal.The box and whisker plot is right-skewed if the distance from Q2 to Q4 is greater than the dis- tance from Q0 to Q2 and the distance from Q3 to Q4 is greater than the distance from Q0 to Q1.Also, the mean value is greater than the median.This means that the data values to the right of the median are more dispersed than those to the 59Chapter 2: Characterizing and defining data Figure 2.1 Box and whisker plot for the sales revenues.

Q4 � 184,957 Q2 � 100,296 Q3 � 123,911 Q0 � 35,378 Q1 � 79,976 0 20,000 40,000 60,000 80,000 100,000 120,000 140,000 160,000 180,000 200,000 Sales ($) left of the median.Conversely, the box and whisker plot is left-skewed if the distance from Q2 to Q4 is less than the distance from Q0 to Q2 and the distance from Q3 to Q4 is less than the dis- tance from Q0 to Q1.Also, the mean value is less than the median.This means that the data values to the left of the median are more dispersed than those to the right.There is further discussion on the skewed properties of data in Chapter 5 in the paragraph entitled Asymmetrical Data.Drawing the box and whiskerplot with Excel If you do not have add-on functions with Microsoft Excel one way to draw the box and whisker plot is to develop a horizontal and verti- cal line graph.The x-axis is the quartile values and the y-axis has the arbitrary values 1, 2, and 3.As the box and whisker plot has only three horizontal lines the lower part of the box has the arbitrary y-value of 1; the whiskers and the centre part of the box have the arbitrary value of 2; and the upper part of the box has the arbi- trary value of 3.

The procedure for drawing the box and whisker plot is as follows.Determine the five quartile boundary values Q0, Q1, Q2, Q3, and Q4 using the Excel quartile function.Set the coordinates for the box and whisker plot in two columns using the format in Table 2.For the 2nd column you enter the corresponding quartile value.

The reason that there are 13 coordinates is that when Excel creates the graph it connects every coordinate with a horizontal or vertical straight line to arrive at the box plot including going over some coordinates more than once.Say once we have drawn the box and whisker plot, the sales data from which it is constructed is considered our reference or benchmark.We now ask the question, where would we position Region A which has sales of $60,000, Region B which has sales of $90,000, Region C which has sales of $120,000, and Region D which has sales of $150,000? From the box and whisker plot of Figure 2.1 an amount of $60,000 is within the 1st quartile and not a great perform- ance; $90,000 is within the 2nd quartile or within the box or the middle 50% of sales.An amount of $120,000 is within the 3rd quartile and within the box or the middle 50% of sales and is a better performance.Finally, an amount of $150,000 is within the 4th quartile and a super- ior sales performance.As mentioned in Chapter 1, a box and whisker plot is another technique in exploratory data analysis (EDA) that covers methods to give an initial understanding of the characteristics of data being analysed.The percentiles divide data into 100 equal parts and thus give a more precise positioning of where information stands compared to the quar- tiles.For example, paediatricians will measure Percentiles 60 Statistics for Business Point No.

X Y 1 Q0 2 2 Q1 2 3 Q1 3 4 Q2 3 5 Q2 1 6 Q1 1 7 Q1 3 8 Q3 3 9 Q3 1 10 Q2 1 11 Q3 1 12 Q3 2 13 Q4 2 Table 2.29 Coordinates for a box and whisker plot.61Chapter 2: Characterizing and defining data Table 2.Percentile Value Percentile Value Percentile Value Percentile Value Percentile Value (%) ($) (%) ($) (%) ($) (%) ($) (%) ($) 0 35,378 1 45,116 21 76,589 41 92,717 61 107,865 81 132,592 2 47,894 22 77,620 42 93,858 62 108,670 82 133,963 3 52,675 23 78,318 43 95,101 63 109,811 83 134,864 4 55,437 24 78,589 44 96,075 64 110,342 84 135,101 5 56,896 25 79,976 45 96,595 65 111,632 85 136,962 6 58,975 26 81,197 46 97,533 66 112,899 86 137,962 7 60,072 27 81,848 47 98,040 67 113,720 87 138,811 8 61,204 28 82,384 48 99,137 68 115,277 88 140,682 9 63,199 29 83,337 49 99,830 69 117,267 89 143,095 10 64,130 30 84,404 50 100,296 70 118,954 90 145,085 11 65,707 31 85,206 51 100,972 71 120,614 91 146,584 12 66,861 32 85,865 52 101,492 72 121,098 92 150,426 13 68,809 33 86,723 53 102,407 73 122,166 93 152,657 14 69,499 34 87,160 54 102,987 74 123,116 94 153,519 15 70,397 35 87,680 55 103,958 75 123,911 95 160,341 16 71,320 36 88,682 56 103,985 76 124,660 96 163,025 17 72,189 37 89,556 57 104,764 77 125,086 97 164,325 18 73,394 38 89,778 58 105,407 78 127,187 98 165,756 19 74,396 39 90,654 59 106,238 79 128,877 99 170,709 20 75,694 40 91,833 60 106,839 80 130,843 100 184,957 the height and weight of small children and indicate how the child compares with others in the same age range using a percentile measure- ment.

For example assume the paediatrician says that for your child’s height he is in the 10th per- centile.This means that only 10% of all children in the same age range have a height less than your child, and 90% have a height greater than that of your child.This information can be used as an indicator of the growth pattern of the child.Another use of percentiles is in examination grading to determine in what percentile, a stu- dent’s grade falls.When you call up this func- tion you are asked to enter the dataset and the value of the kth percentile where k is to indicate the 1st, 2nd, 3rd percentile, etc.When you enter the value of k it has to be a decimal representation or a percentage of 100.For example the 15th per- centile has to be written as 0.As for the quartiles, you do not have to sort the data – Excel does this for you.

Using the same sales revenue information that we used for the quartiles, Table 2.30 gives the percentiles for this information using the percentage to indicate the percentile.For example a percentile of 15% is the 15th percentile or a percentile of 23% is the 23rd per- centile.Using this data we have developed Figure 2.2, which shows the percentiles as a histogram.

Say once again as we did for the quartiles, we ask the question, where would we position Region A which has sales of $60,000, Region B which has sales of $90,000, Region C which has sales of $120,000, and Region D which has sales of $150,000? From either Table 2.2 then we can say that for $60,000 this is at about the 7% percentile, which means that 93% of the sales are greater than this region and 7% are less – a poor performance.For $90,000 this is roughly the 39% percentile, which means that 61% of the sales are greater than this region and 39% are less – not a good performance.At the $120,000 level this is about the 71% percentile, which means that 29% of the sales are greater than this region and 71% are less – a reasonable performance.

Finally $150,000 is at roughly the 92% percentile which signifies that only 8% of the sales are greater than this region and 92% are less – a good performance.By describing the data using percentiles rather than using quartiles we have been able to be more precise as to where the region sales data are positioned.Division of data We can divide up data by using the median – two equal parts, by using the quartiles – four equal parts, or using the percentiles – 100 equal parts.In this case the median value equals the 3rd quar- tile which also equals the 50th percentile.For the raw sales data given in Table 1.

1 the median value is 100,296 (indicated at the end of paragraph median of this chapter), the value of the 2nd quartile, Q2, given in Table 2.28, is also 100,296 and the value of the 50th percentile, given in Table 2.0 20,000 40,000 60,000 80,000 100,000 120,000 140,000 160,000 180,000 200,000 0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 10 0 Percentile (%) Sa le s re ve nu es ($ ) 63Chapter 2: Characterizing and defining data This chapter has detailed the meaning and calculation of properties of statistical data, which we have classified by central tendency, dispersion, quartiles, and percentiles.Central tendency of data Central tendency is the clustering of data around a central or a middle value.If we know the central tendency this gives us a benchmark to situate a dataset and use this central value to compare one dataset with another.The most common measure of central tendency is the mean or average value, which is the sum of the data divided by the number of data points.

The mean value can be distorted by extreme value or outliers.

We also have the median, or that value that divides data into two halves.The median is not affected by extreme values but may be affected by the number of values.The mode is a measure of central tendency and is that value that occurs most often.The mode can be used for qualitative responses such as the colour that is preferred.There is the midrange, which is the average of the highest and lowest value in the dataset and this is very much dependent on extreme values.

We might use the weighted average when certain values are more important than others.If data are changing over time, as for example interest rates each year, then we would use the geometric mean as the measure of central tendency.Dispersion of data The dispersion is the way that data is spread out.If we know how data is dispersed, it gives us an indicator of its reliability for analytical purposes.Data that is highly dispersed is unreliable com- pared to data that is little dispersed.

The range is an often used measure of dispersion but it is not a good property as it is affected by extreme values.The most meaningful measures of disper- sion are the variance and the standard deviation, both of which take into consideration every value in the dataset.Mathematically the standard deviation is the square root of the variance and it is more commonly used than the variance since it has the same units of the dataset from which it is derived.For a given dataset, the standard deviation of the sample is always more than the standard deviation of the population since it uses the value of the sample size less one in its denominator whereas the population standard deviation uses in the denominator the number of data values.

A simple way to compare the relative dis- persion of datasets is to use the coefficient of variation, which is the ratio of the standard devi- ation to its mean value.Quartiles The quartiles are those values that divide ordered data into four equal values.Although, there are really just three quartiles, Q1 – the first, Q2 – the second, and Q3 – the third, we also refer to Q0, which is the start value in the quartile framework and also the minimum value.We also have Q4, which is the last value in the dataset, or the maximum value.Thus there are five quar- tile boundary limits.

The value of the 2nd quartile, Q2, is also the median value as it divides the data into two halves.By developing quartiles we can position information within the quartile framework and this is an indicator of its importance in the dataset.From the quartiles we can C hapter Sum m ary 64 Statistics for Business develop a box and whisker plot, which is a visual display of the quartiles.The middle box represents the middle half, or 50% of the data.The left-hand whisker represents the first 25% of the data, and the right-hand whisker represents the last 25%.

The box and whisker plot is distorted to the right when the mean value is greater than the median and distorted to the left when the mean is less than the median.Analogous to the range, in quartiles, we have the inter-quartile range, which is the difference between the 3rd and 1st quartile values.Also, analogous to the midrange we have the mid-hinge which is the average of the 3rd and 1st quartile.Percentiles Percentiles are those values that divide ordered data into 100 equal parts.Percentiles are useful in that by positioning where a value occurs in a percentile framework you can compare the importance of this value.

For example, in the medical profession an infant’s height can be positioned on a standard percentile framework for children’s height of the same age group which can then be an estimation of the height range of this child when he/she reaches adulthood.The 50th percentile in a dataset is equal to the 2nd quartile both of which are equal to the median value.65Chapter 2: Characterizing and defining data 1.Billing rate Situation An engineering firm uses senior engineers, junior engineers, computing services, and assistants on its projects.The billing rate to the customer for these categories is given in the table below together with the hours used on a recent design project.

Category Senior Junior Computing Assistants engineers engineers services Billing rate ($/hour) 85.00 Project hours 23,000 37,000 19,000 9,500 Required 1.

If this data was used for quoting on future projects, what would be the correct aver- age billing rate used to price a project? 2.If the estimate for performing a future job were 110,000 hours, what would be the billing amount to the customer? 3.What would be the billing rate if the straight arithmetic average were used? 2.Delivery Situation A delivery company prices its services according to the weight of the packages in certain weight ranges.This information together with the number of packages delivered last year is given in the table below.

Weight category Less than From 1 to From 5 to From 10 to Greater 1 kg 5 kg 10 kg 50 kg than 50 kg Price ($/package) 10.50 Number of packages 120,000 90,500 82,545 32,500 950 Required 1.What is the average price paid per package? 2.If next year it was estimated that 400,000 packages would be delivered, what would be an estimate of revenues? EXERCISE PROBLEMS 66 Statistics for Business 3.Investment Situation Antoine has $1,000 to invest.He has been promised two options of investing his money if he leaves it invested over a period of 10 years with interest calculated annually.

The interest rates for the following two options are in the tables below.Option 1 Option 2 Year Interest Year Interest rate (%) rate (%) 1 6.What is the average annual growth rate, geometric mean for Option 1? 2.What is the average annual growth rate, geometric mean for Option 2? 3.What would be the value of his investment at the end of 10 years if he invested in Option 1? 4.

What would be the value of his investment at the end of 10 years if he invested in Option 2? 5.What would need to be the interest rate in the 10th year for Option 2 in order that the value of his asset at the end of 10 years for Option 2 is the same as for Option 1? 4.Production Situation A custom-made small furniture company has produced the following units of furniture over the past 5 years.Year 2000 2001 2002 2003 2004 Production (units) 13,250 14,650 15,890 15,950 16,980 67Chapter 2: Characterizing and defining data Required 1.

What is the average percentage growth in this period? 2.If this average growth rate is maintained, what would be the production level in 2008? 5.Euro prices Situation The table below gives the prices in Euros for various items in the European Union.3 Milk Renault Big Stamp for Compact Can of (1 l) M gane Mac postcard disc Coke Austria 0.Determine the maximum, minimum, range, average, midrange, median, sample stand- ard deviation, and the estimated coefficient of variation using the sample values for all of the items indicated.

What observations might you draw from these characteristics? 6.Students Situation A business school has recorded the following student enrolment over the last 5 years Year 1997 1998 1999 2000 2001 Students 3,275 3,500 3,450 3,600 3,800 Required 1.What is the average percentage increase in this period? 2.If this rate of percentage increase is maintained, what would be the student popula- tion in 2005? 3 International Herald Tribune, 5/6 January 2002, p.

Construction Situation A firm purchases certain components for its construction projects.The price of these components over the last 5 years has been as follows.Year 1996 1997 1998 1999 2000 Price ($/unit) 105.

What is the average percentage price increase in this period? 2.If this rate of price increase is maintained, what would be the price in 2003? 8.Net worth Situation A small firm has shown the following changes in net worth over a 5-year period.Year 2000 2001 2002 2003 2004 Growth (%) 6.What is the average change in net worth over this period? 9.

Trains Situation A sample of the number of late trains each week, on a privatized rail line in the United Kingdom, was recorded over a period as follows.25 13 20 32 3 15 42 25 25 38 20 39 15 30 7 17 45 36 25 10 42 35 7 15 25 Required 1.From this information, what is the average number of trains late? 2.From this information, what is the median value of the number of trains late? 3.From this information, what is the mode value of the number of trains late? How many times does this modal value occur? 4.

From this information, what is the range? 5.From this information, what is the midrange? 6.From this information, what is the sample variance? 7.From this information, what is the sample standard deviation? 8.From this sample information, what is an estimate of the coefficient of variation? 9.

What can you say about the distribution of the data? 69Chapter 2: Characterizing and defining data 10.Summer Olympics 2004 Situation The table below gives the final medal count for the Summer Olympics 2004 held in Athens, Greece.4 Country Gold Silver Bronze Country Gold Silver Bronze Argentina 2 0 4 Japan 16 9 12 Australia 17 16 16 Kazakhstan 1 4 3 Austria 2 4 1 Kenya 1 4 2 Azerbaijan 1 0 4 Latvia 0 4 0 Bahamas 1 0 1 Lithuania 1 2 0 Belarus 2 6 7 Mexico 0 3 1 Belgium 1 0 2 Mongolia 0 0 1 Brazil 4 3 3 Morocco 2 1 0 Britain 9 9 12 The Netherlands 4 9 9 Bulgaria 2 1 9 New Zealand 3 2 0 Cameroon 1 0 0 Nigeria 0 0 2 Canada 3 6 3 North Korea 0 4 1 Chile 2 0 1 Norway 5 0 1 China 32 17 14 Paraguay 0 1 0 Columbia 0 0 1 Poland 3 2 5 Croatia 1 2 2 Portugal 0 2 1 Cuba 9 7 11 Romania 8 5 6 Czech Republic 1 3 4 Russia 27 27 38 Denmark 2 0 6 Serbia-Montenegro 0 2 0 Dominican Republic 1 0 0 Slovakia 2 2 2 Egypt 1 1 3 Slovenia 0 1 3 Eritrea 0 0 1 South Africa 1 3 2 Estonia 0 1 2 South Korea 9 12 9 Ethiopia 2 3 2 Spain 3 11 5 Finland 0 2 0 Sweden 4 1 2 France 11 9 13 Switzerland 1 1 3 Georgia 2 2 0 Syria 0 0 1 Germany 14 16 18 Taiwan 2 2 1 Greece 6 6 4 Thailand 3 1 4 Hong Kong 0 1 0 Trinidad and Tobago 0 0 1 Hungary 8 6 3 Turkey 3 3 4 India 0 1 0 Ukraine 9 5 9 Indonesia 1 1 2 United Arab Emirates 1 0 0 Iran 2 2 2 United States 35 39 29 Ireland 1 0 0 Uzbekistan 2 1 2 Israel 1 0 1 Venezuela 0 0 2 Italy 10 11 11 Zimbabwe 1 1 1 Jamaica 2 1 2 4 International Herald Tribune, 31 August 2004, p.If the total number of medals won is the criterion for rating countries, which coun- tries in order are in the first 10? 2.If the number of gold medals won is the criterion for rating countries, which coun- tries in order are in the first 10? 3.If there are three points for a gold medal, two points for a silver medal, and one point for a bronze medal, which countries in order are in the first 10? Indicate the weighted average for these 10 countries.What is the average medal count per country for those who competed in the Summer Olympics? 5.

Develop a histogram for the percentage of gold medals by country for those who won a gold medal.Which three countries have the highest percentage of gold medals out of all the gold medals awarded? 11.Printing Situation A small printing firm has the following wage rates and production time in the final sec- tion of its printing operation.Operation Binding Trimming Packing Wages ($/hour) 14.

For product costing purposes, what is the correct average rate per hour for 100 units for this part of the printing operation? 2.If we added in printing, where the wages are $25.00 hour and the production time is 45 minutes per 100 units, then what would be the new correct average wage rate for the operation? 12.Big Mac Situation The table below gives the price a Big Mac Hamburger in various countries converted to the $US.5 (This is the information presented in the Box Opener.

71Chapter 2: Characterizing and defining data Country Price ($US) Country Price ($US) Argentina 1.Determine the following characteristics of this data: (a) Maximum (b) Minimum (c) Average value (d) Median (e) Range (f) Midrange (g) Mode and how many modal values are there? (h) Sample standard deviation (i) Coefficient of variation using the sample standard deviation 2.

Illustrate the price of a Big Mac on a horizontal bar chart sorted according to price.What are the boundary limits of the quartiles? 4.Where in the quartile distribution do the prices of the Big Mac occur in Indonesia, Singapore, Hungary, and Denmark? What initial conclusions could you draw from this information? 6.

Draw a box and whisker plot for this data.Purchasing expenditures – Part II Situation The complete daily purchasing expenditures for a large resort hotel for the last 200 days in Euros are given in the table below.The purchases include all food and non-food items, 72 Statistics for Business and wine for the five restaurants in the complex, energy including water for the three swimming pools, laundry which is a purchased service, gasoline for the courtesy vehicles, gardening and landscaping services.63,680 307,024 188,973 242,746 217,724 194,157 230,211 192,285 235,015 195,577 197,613 332,923 173,876 219,573 113,864 295,731 175,622 297,536 205,173 224,937 195,651 165,355 217,076 86,157 293,373 151,135 187,173 110,336 188,977 332,212 161,275 288,466 99,886 274,856 167,175 102,382 273,411 159,262 298,256 161,075 153,862 116,240 187,173 147,564 248,146 228,577 185,377 210,573 81,340 237,524 132,476 291,411 238,840 217,177 122,211 157,775 106,155 187,124 224,276 303,466 172,613 94,957 206,973 112,676 262,773 179,377 137,860 204,462 144,826 194,157 197,741 183,409 144,283 141,476 156,213 175,612 246,571 161,741 173,187 295,173 150,651 136,609 177,766 241,124 134,811 68,141 163,240 115,540 194,157 223,124 190,777 168,898 106,155 185,375 185,377 260,973 182,696 182,336 187,124 128,860 106,787 218,626 147,956 108,230 155,875 165,215 102,415 203,137 97,430 274,777 179,998 141,412 198,880 156,523 179,075 238,624 242,977 137,860 244,256 213,577 163,076 282,568 157,849 212,211 154,138 188,276 139,777 190,777 141,221 269,212 124,157 90,230 191,876 114,476 222,415 86,211 180,531 108,230 254,336 152,276 180,533 139,496 140,141 242,802 142,978 181,186 171,880 221,324 201,415 233,215 128,624 159,833 198,466 130,676 253,076 225,880 125,251 161,372 127,076 168,977 203,377 223,011 118,525 231,651 120,415 148,426 241,171 177,226 275,936 157,077 130,162 146,621 224,741 182,677 132,424 249,651 134,249 246,524 208,615 257,373 215,377 173,866 119,876 146,682 251,251 148,421 270,536 192,346 124,101 220,777 126,880 170,257 154,755 249,475 175,496 259,173 166,480 263,320 152,266 125,773 Required 1.

Using the raw data determine the following data characteristics: (a) Maximum value (you may have done this in the exercise from the previous chapter) (b) Minimum value (you may have done this in the exercise from the previous chapter) (c) Range (d) Midrange (e) Average value (f) Median value (g) Mode and indicate the number of modal values (h) Sample variance (i) Standard deviation (assuming a sample) (j) Coefficient of variation on the basis of a sample 2.Determine the boundary limits for the quartile values for this data.What can you say about the distribution of this data? 5.Determine the percentile values for this data.Plot this information on a histogram with the x-axis being the percentile value, and the y-axis the dollar value of the retail sales.Verify that the median value, 2nd quartile, and the 50th quartile are the same.73Chapter 2: Characterizing and defining data 14.

Swimming pool – Part II Situation A local community has a heated swimming pool, which is open to the public each year from May 17 until September 13.The community is considering building a restaurant facility in the swimming pool area but before a final decision is made, it wants to have assurance that the receipts from the attendance at the swimming pool will help finance the construction and operation of the restaurant.In order to give some justification to its decision the com- munity noted the attendance for one particular year and this information is given below.869 755 729 926 821 709 1,088 785 830 709 678 1,019 825 843 940 826 750 835 956 743 835 630 791 795 903 790 931 869 878 808 845 692 830 794 993 847 901 837 755 810 791 609 878 778 761 763 726 745 874 728 870 798 507 763 764 779 678 690 1,004 792 848 823 769 773 919 682 672 829 915 883 699 650 780 743 861 610 582 748 744 680 930 776 871 759 580 669 716 980 724 880 669 712 732 968 620 852 749 860 811 748 822 651 539 658 796 825 685 707 895 806 609 952 565 869 560 751 790 907 621 619 Required 1.From this information determine the following properties of the data: (a) The sample size (b) Maximum value (c) Minimum value (d) Range (e) Midrange (f) Average value (g) Median value (h) Modal value and how many times does this value occur? (i) Standard deviation if the data were considered a sample (which it is) (j) Standard deviation if the data were considered a population (k) Coefficient of variation (l) The quartile values (m) The inter-quartile range (n) The mid-hinge 2.

Using the quartile values develop a box and whisker plot.What are your observations about the box plot? 4.Determine the percentiles for this data and plot them on a histogram.Buyout – Part II Situation Carrefour, France, is considering purchasing the total 50 retail stores belonging to Hardway, a grocery chain in the Greater London area of the United Kingdom.The profits from these 50 stores, for one particular month, in £’000s, are as follows.Using the raw data determine the following data characteristics: (a) Maximum value (this will have been done in the previous chapter) (b) Minimum value (this will have been done in the previous chapter) (c) Range (d) Midrange (e) Average value (f) Median value (g) Modal value and indicate the order of modality (single, bi, tri, etc.) (h) Standard deviation assuming the data was a sample (i) Standard deviation taking the data correctly as the population 2.

Determine the quartile values for the data and use these to develop a box and whisker plot.Determine the percentile values for the data and plot these on a histogram.Case: Starting salaries Situation A United States manufacturing company in Chicago has several subsidiaries in the 27 countries of the European Union including Calabas, Spain; Watford, United Kingdom; Bonn, Germany and Louny, Czech Republic.

It is planning to hire new engineers to work in these subsidiaries and needs to decide on the starting salary to offer these new hires.These new engineers will be hired from their country of origin to work in their home country.The human resource department of the parent firm in Chicago, who is not too 75Chapter 2: Characterizing and defining data familiar with the employment practices in Europe, has the option to purchase a database of annual starting salaries for engineers in the European Union from a consulting firm in Paris.This database, with values converted to Euros, is given in the table below.It was compiled from European engineers working in the automobile, aeronautic, chemicals, pharmaceutical, textiles, food, and oil refining sectors.

At the present time, the Chicago firm is considering hiring Markus Schroeder, offering a starting salary of €36,700, Xavier Perez offering a salary of €30,500, Joan Smith a salary of €32,700 and Jitka Sikorova a starting salary of €28,900.All these starting salaries include all social bene- fits and mandatory employer charges which have to be paid for the employee.Required Assume that you work with the human resource department in Chicago.Use the infor- mation from this current chapter, and also from Chapter 1 to present in detail the salary database prepared by the Paris consulting firm.Then in using your results, describe the characteristics of the four starting salaries that have been offered and give you comments.

34,756 30,196 29,164 37,022 32,842 28,356 43,504 31,380 31,030 28,366 25,700 40,750 33,012 33,726 37,594 33,038 30,004 33,388 38,224 29,728 33,400 35,450 31,658 35,044 36,104 27,894 37,224 34,754 33,122 32,310 33,800 27,662 33,208 31,752 39,724 33,866 36,032 34,132 30,180 32,380 31,634 24,370 35,136 33,858 30,456 30,538 29,052 29,796 34,978 29,110 34,786 33,936 33,586 30,530 30,568 31,178 24,652 27,580 40,160 33,926 33,928 32,932 30,774 30,914 36,750 27,280 26,886 33,152 36,580 35,324 27,956 26,016 25,802 29,722 34,454 35,964 23,282 29,908 29,772 27,200 37,198 31,056 34,852 30,370 30,828 26,776 28,650 33,958 28,974 35,204 26,752 28,478 29,264 37,898 32,724 34,082 29,948 34,410 32,456 29,928 32,884 25,974 21,566 26,310 31,836 35,898 27,396 28,292 35,784 32,220 24,342 36,302 32,184 34,788 32,098 30,044 31,610 36,282 24,842 24,742 29,514 35,566 27,556 39,886 31,468 33,302 37,980 38,174 35,644 37,370 35,072 27,400 35,838 29,858 24,062 28,606 28,012 34,442 35,018 31,638 26,154 29,706 33,850 28,668 31,870 25,572 26,576 28,758 32,580 24,114 34,878 31,860 31,216 31,668 37,490 42,072 29,242 28,086 25,054 33,248 33,654 25,892 34,902 33,294 31,712 32,312 27,546 31,472 34,020 32,704 40,202 27,252 28,870 36,414 29,586 28,906 35,434 34,332 33,564 34,268 24,246 35,214 33,102 29,274 29,454 34,126 29,412 31,588 30,766 31,052 34,614 31,630 31,024 29,242 30,924 35,032 37,334 26,660 36,616 25,342 30,076 31,902 35,114 32,348 41,184 28,972 34,588 27,312 30,404 31,478 26,422 31,648 33,078 33,640 34,240 25,632 24,528 31,188 30,006 34,650 28,466 27,616 33,994 35,368 33,804 27,050 34,070 36,012 36,410 31,840 39,782 28,378 29,328 36,144 33,010 28,592 32,782 36,774 39,144 36,902 28,662 27,522 29,200 31,992 30,564 30,762 25,860 35,620 25,192 41,490 34,250 25,212 35,678 24,912 35,648 36,622 36,884 29,488 29,060 38,692 29,052 33,884 35,202 38,824 33,376 36,488 31,982 34,902 33,068 34,518 25,146 27,834 38,990 34,944 23,394 30,276 37,124 26,756 32,142 30,388 27,624 28,718 36,828 26,528 29,168 31,612 28,822 29,296 28,374 29,990 (Continued) 76 Statistics for Business 30,914 34,812 37,508 30,446 35,390 38,916 33,842 25,442 28,088 28,234 34,652 34,286 26,474 35,394 36,636 34,596 36,196 32,412 31,272 31,822 29,696 33,552 28,900 30,384 32,274 22,320 28,934 35,738 36,010 39,038 22,044 34,022 37,750 27,146 34,570 29,514 31,042 30,672 33,482 34,774 36,518 29,638 28,976 31,146 38,434 27,468 39,570 28,502 31,762 38,600 27,134 32,334 27,928 31,150 31,858 31,544 27,254 27,716 41,482 27,082 32,470 28,128 28,584 33,120 36,764 35,450 32,854 31,848 33,474 26,842 33,396 27,358 33,832 38,088 35,074 29,114 26,380 31,256 37,080 29,622 33,060 25,426 31,616 31,876 35,838 29,376 36,654 30,398 36,030 34,196 29,732 29,744 30,544 31,854 30,884 23,768 31,520 30,336 27,442 29,796 39,302 25,424 28,924 32,072 29,204 34,906 32,434 23,710 31,964 33,328 39,956 33,386 38,184 35,326 34,468 37,616 35,588 37,312 32,484 29,522 31,332 35,136 35,186 32,964 31,962 34,070 41,396 28,170 35,352 31,300 24,190 37,292 33,146 32,972 30,260 31,178 26,772 39,376 31,860 37,080 32,568 27,990 32,378 22,508 32,644 27,158 31,868 33,050 29,624 32,368 28,176 27,664 31,840 26,800 33,252 32,622 35,966 29,264 31,546 26,292 31,116 30,834 30,254 30,690 23,930 31,202 32,166 30,396 33,698 29,704 31,496 33,730 31,714 34,046 29,756 38,372 35,666 31,344 35,976 33,036 27,500 33,882 40,496 32,218 30,110 36,168 31,654 28,880 27,502 29,082 36,524 30,512 33,882 34,350 39,062 24,674 33,384 27,472 21,954 27,934 33,346 33,426 31,722 29,566 31,000 30,522 33,942 32,490 35,134 29,644 38,754 32,214 31,220 32,604 23,588 29,648 32,470 38,824 30,820 34,294 22,856 29,514 28,000 35,398 31,934 27,104 34,994 25,006 31,186 35,164 28,352 26,626 36,052 31,134 34,064 32,186 29,724 36,968 32,558 34,596 34,646 29,832 33,784 36,346 33,692 41,182 29,374 36,574 26,868 37,596 25,132 30,618 23,684 33,918 35,336 26,862 35,756 31,754 28,090 28,236 30,780 32,894 26,608 30,890 33,530 34,210 31,072 36,742 32,982 41,776 33,250 36,836 32,390 29,626 38,642 29,406 27,086 27,902 36,370 30,522 29,572 30,944 33,000 34,314 31,148 35,300 24,016 27,878 38,818 33,910 26,838 34,214 33,470 31,070 32,100 28,982 27,632 28,432 31,854 32,852 32,596 25,810 36,426 33,452 31,704 34,938 30,704 35,736 34,682 36,700 34,238 27,012 34,812 30,624 30,418 34,730 33,134 30,692 32,142 34,450 34,766 31,824 29,126 34,594 31,088 39,328 27,676 34,518 30,296 35,742 22,830 33,332 33,486 31,544 32,932 29,596 28,628 35,662 27,524 35,074 37,378 33,426 30,336 29,462 32,180 35,530 36,288 32,148 27,738 30,110 29,610 29,252 36,378 33,632 28,574 26,076 33,118 28,660 35,970 35,806 30,698 32,620 28,642 35,738 34,744 34,828 29,520 23,676 32,424 32,538 27,782 29,062 27,266 30,916 29,868 29,746 35,976 32,204 30,992 35,100 38,164 30,698 31,002 34,276 30,846 29,952 27,972 35,484 31,812 32,620 31,974 31,932 33,348 27,468 33,736 27,100 31,120 30,492 39,210 28,310 27,216 31,428 33,268 29,196 29,868 35,784 31,938 33,570 27,300 37,214 28,758 30,968 33,402 36,310 37,372 35,490 35,254 23,456 29,628 29,966 32,102 41,046 31,504 26,562 33,400 28,768 32,270 27,726 32,422 30,504 27,662 29,844 30,178 33,942 32,794 27,536 27,354 29,754 31,814 29,426 31,498 39,300 26,742 40,572 36,102 32,950 20,376 35,892 29,254 36,222 30,880 30,040 25,360 36,004 28,592 29,334 26,960 25,978 27,216 32,292 33,090 27,826 31,052 31,774 32,562 32,112 26,386 37,556 28,554 23,048 36,176 27,392 37,216 24,314 24,410 36,304 29,568 33,214 31,284 37,264 25,518 33,618 36,218 31,148 26,620 31,178 31,490 28,338 26,770 31,498 77Chapter 2: Characterizing and defining data 31,404 32,206 34,552 34,842 26,664 24,960 32,798 22,856 33,082 32,514 26,856 29,672 33,786 30,502 31,766 31,854 35,450 29,188 32,692 29,830 28,858 36,308 30,292 30,298 32,124 31,730 33,534 35,440 28,990 29,606 29,554 25,062 28,502 34,388 31,052 34,826 34,024 33,926 32,330 33,460 32,216 32,160 40,642 27,986 33,040 36,398 36,084 25,664 29,852 37,400 29,674 33,100 32,048 30,606 34,902 34,538 32,438 28,844 30,502 29,178 26,656 26,730 26,690 31,236 35,788 29,438 33,088 28,930 27,342 32,070 36,686 30,786 27,364 35,570 39,390 28,258 35,902 33,858 27,742 31,358 39,762 33,386 37,550 30,652 24,938 33,852 30,508 30,422 34,022 29,790 33,316 26,600 29,916 31,562 22,092 32,998 34,746 35,340 30,336 32,256 30,336 31,462 31,918 31,994 25,040 30,986 32,220 26,830 28,882 29,426 33,048 31,510 33,382 32,680 35,802 36,704 29,836 31,160 33,318 25,824 37,688 34,382 34,504 31,868 30,872 36,156 42,592 33,636 38,870 25,470 34,658 30,430 36,060 37,306 39,048 35,334 28,598 32,664 34,958 39,414 42,786 43,258 35,260 35,068 30,454 30,880 34,776 29,942 26,144 26,432 25,936 31,368 26,992 26,452 28,084 28,036 28,780 36,382 35,248 32,926 36,662 27,056 27,762 28,616 34,842 28,582 37,860 31,134 36,704 29,992 37,560 32,108 29,358 27,562 29,490 31,316 35,590 33,520 30,462 28,802 35,772 34,220 34,490 29,224 37,310 30,246 27,920 30,000 35,144 29,814 28,584 32,202 35,650 24,874 36,094 34,774 38,626 30,520 24,750 28,578 39,180 30,170 30,220 29,564 34,306 33,834 34,368 27,344 32,548 32,702 30,792 23,460 31,302 29,472 25,530 29,028 34,350 33,748 35,530 31,732 This page intentionally left blank 3Basic probability and counting rules The wheel of fortune For many, gambling casinos are exciting establishments.The one-arm- bandits are colourful machines with flashing lights, which require no intelligence to operate.When there is a “win” coins drop noisily into aluminium receiving tray and blinking lights indicate to the world the amount that has been won.The gaming rooms for poker, or blackjack, and the roulette wheel have an air of mystery about them.

The dealers and servers are beautiful people, smartly dressed, who say very little and give an aura of superiority.

Throughout the casinos there are no clocks or windows so you do not see the time passing.Drinks are cheap, or maybe free, so having “a few” encourages you to take risk.The carpet patterns are busy so that you look at where the action is rather than looking at the floor.When you want to go to the toilet you have to pass by rows of slot machines and perhaps on the way you try your luck! Gambling used to be a by-word for racketeering.Now it has cleaned up its act and is more profitable than ever.

Today the gambling industry is run by respectable corporations instead of by the Mob and it is confident of winning public acceptance.1 million people, or more than one-quarter of all American adults visited a casino, on average 6 times each.Poker is a particular growth area and some 18% of Americans played poker in 2004, which was a 50% increase over 2003.Together, the United States’ 445 commercial casinos, that means excluding those 80 Statistics for Business owned by Indian tribes, had revenues in 2004 of nearly $29 billion.

Further, it paid state gaming taxes of $4.74 billion or almost 10% more than in 2003.A survey of 201 elected officials and civic leaders, not including any from gambling dependent Nevada and New Jersey, found that 79% believed casinos had had a positive impact on their communities.The company Partouche owns and operates very successful casinos in Belgium, France, Switzerland, Spain, Morocco, and Tunisia.

And, let us not forget the famed casino in Monte Carlo.Just about all casinos are associated with hotels and restaurants and many others include resort settings and spas.This makes the whole combination, gambling casinos, hotels, resorts, and spas a significant part of the service industry.1,2 1 The gambling industry, The Economist, 24 September 2005.In statistical analysis the outcome of certain situ- ations can be reliably estimated, as there are mathematical relationships and rules that gov- ern choices available.This is useful in decision- making since we can use these relationships to make probability estimates of certain outcomes and at the same time reduce risk.A principal objective of statistics is inferential statistics, which is to infer or make logical deci- sions about situations or populations simply by taking and measuring the data from a sample.

This sample is taken from a population, which is the entire group in which we are interested.We use the information from this sample to infer conclusions about the population.For example, we are interested to know how people will vote in a certain election.We sample the opinion of 5,500 of the electorate and we use this result to estimate the opinion of the population of 35 million.Since we are extending our sample results beyond the data that we have measured, this means that there is no guarantee but only a probability of being correct or of making the right decision.

The corollary to this is that there is a probability or risk of being incorrect.Probability The concept of probability is the chance that something happens or will not happen.In sta- tistics it is denoted by the capital letter P and is measured on an inclusive numerical scale of 0 to 1.If we are using percentages, then the scale is from 0% to 100%.If the probability is 0% then there is absolutely no chance that an out- come will occur.

Under present law, if you live in the United States, but you were born in Austria, the probability of you becoming president is 0% – in 2006, the current governor of California! At the top end of the probability scale is 100%, which means that it is certain the outcome will occur.The probability is 100% that someday you will die – though hopefully at an age way above the statistical average! Between the two extremes of 0 and 1 something might occur or might not occur.The meteorological office may announce that there is a 30% chance of rain Basic Probability Rules 81Chapter 3: Basic probability and counting rules After you have studied this chapter you will understand basic probability rules, risk in system reli- ability, and counting rules.You will then be able to apply these concepts to practical situations.The following are the specific topics to be covered.

✔ Basic probability rules • Probability • Risk • An event in probability • Subjective probability • Relative frequency probability • Classical probability • Addition rules in classical probability • Joint probability • Conditional probabilities under statistical dependence • Bayes’ Theorem • Venn diagram • Application of a Venn diagram and probability in services: Hospitality management • Application of probability rules in manufacturing: A bottling machine • Gambling, odds, and probability.✔ System reliability and probability • Series or parallel arrangements • Series systems • Parallel or backup systems • Application of series and parallel systems: Assembly operation.✔ Counting rules • A single type of event: Rule No.2 • Arrangement of different objects: Rule No.

L e a r n i n g o b j e c t i v e s today, which also means that there is a 70% chance that it will not.

The opposite of probabil- ity is deterministic where the outcome is certain on the assumption that the input data is reli- able.

For example if revenues are £10,000 and costs are £7,000 then it is sure that the gross profit is £3,000 (£10,000 � £7,000).With probability something happens or it does not happen, that is the situation is binomial, or there are only two possible outcomes.However that does not mean that there is a 50/50 chance of being right or wrong or a 50/50 chance of winning.If you toss a fair-sided coin, one that has not been “fixed”, you have a 50% chance of obtaining heads or 50% chance of throwing tails.If you buy one ticket in a fund raising raffle then you will either win or lose.

However, if there are 2,000 tickets that have been sold you have only a 1/2,000 or 0.05% chance of winning and a 1,999/2,000 or a 99.95% chance of losing! Risk An extension of probability, often encountered is business situations, but also in our personal life, is risk.Here, when we extend probability to risk we are putting a value on the outcomes.In business we might invest in new technology and say that there is a 70% probability of increasing market share but this also might mean that there is a risk of losing $100 million.

To insurance companies, the probability of an automobile driver aged between 18 and 25 years having an accident is considered greater than for people in higher age groups.Thus, to the insurance company young people present a high risk and so their premiums are higher than normal.If you drink and drive the probability of you having an accident is high.In this case you risk having an accident, or perhaps the risk of killing yourself.In this case the “value” on the outcome is more than monetary.

An event in probability In probability we talk about an event.An event is the result of an activity or experiment that has been carried out.If you obtain heads on the tossing of a coin, then “obtaining heads” would be an event.If you draw the King of Hearts from a pack of cards, then “drawing the King of Hearts” would be an event.If you select a light bulb from a production lot and it is defective then the “selection of a defective light bulb” would be an event.

If you obtain an A grade on an examination, then “obtaining an A grade” would be an event.If Susan wins a lottery, “Susan winning the lottery” would be an event.If Jim wins a slalom ski competition, “Jim win- ning the slalom” would be an event.Subjective probability One type of probability is subjective probability, which is qualitative, sometimes emotional, and simply based on the belief or the “gut” feeling of the person making the judgment.For example, you ask Michael, a single 22-year-old student what is the probability of him getting married next year? His response is 0%.

You ask his friend John, what he thinks is the probability of Michael getting married next year and his response is 50%.There are no numbers involved, and this particular situ- ation has never occurred before.For exam- ple, Salesperson A says that he is 80% certain of making a sale with a certain client, as he knows the client well.However, Salesperson B may give only a 50% probability level of making that sale.Both are basing their arguments on subjective probability.A manager who knows his employees well may be able to give a subjec- tive probability of his department succeeding in a particular project.This probability might differ from that of an outsider assessing the probability of success.

Very often, the subjective probability of people who are prepared to take risks, or risk takers, is higher than those persons who are risk averse, or afraid to take risks, since 82 Statistics for Business the former are more optimistic or gung ho individuals.Relative frequency probability A probability based on information or data collected from situations that have occurred previously is relative frequency probability.We have already seen this in Chapter 1, when we developed a relative frequency histogram for the sales data given in Figure 1.Here, if we assume that future conditions are similar to past events, then from this Figure 1.

2 we could say that there is a 15% probability that future sales will lie in the range of £95,000 to £105,000.Relative frequency probabilities have use in many business situations.For example, data taken from a certain country indicate that in a sample of 3,000 married couples under study, one-third were divorced within 10 years of mar- riage.Again, on the assumption that future conditions will be similar to past conditions, we can say that in this country, the probability of being divorced before 10 years of marriage is 1/3 or 33.This demographic information can then be extended to estimate needs of such things as legal services, new homes, and child- care.In collecting data for determining relative frequency probabilities, the reliability is higher if the conditions from which the data has been collected are stable and a large amount of data has been measured.Relative frequency prob- ability is also called empirical probability as it is based on previous experimental work.Also, the data collected is sometimes referred to as histor- ical data as the information after it has been col- lected is history.

Classical probability A probability measure that is also the basis for gambling or betting, and thus useful if you fre- quent casinos, is classical probability.

Classical probability is also known as simple probability or marginal probability and is defined by the follow- ing ratio: 3(i) In order for this expression to be valid, the prob- ability of the outcomes, as defined by the numerator (upper part of the ratio) must be equally likely.For example, let us consider a full pack of 52 playing cards, which is composed of the individual cards according to Table 3.The total number of possible outcomes is 52, the number of cards in the pack.We know in advance that the probability of drawing an Ace of Spades, or in fact any one single card, is 1/52 or 1.

Similarly in the throwing of one die there are six possible outcomes, the numbers 1, 2, 3, 4, 5, or 6.Thus, we know in advance that the prob- ability of throwing a 5 or any other number is Classical probability Number of outcomes w � hhere the event occurs Total number of posssible outcomes 83Chapter 3: Basic probability and counting rules Table 3.1 Composition of a pack of cards with no jokers.Suit Total Hearts Ace 1 2 3 4 5 6 7 8 9 10 Jack Queen King 13 Clubs Ace 1 2 3 4 5 6 7 8 9 10 Jack Queen King 13 Spades Ace 1 2 3 4 5 6 7 8 9 10 Jack Queen King 13 Diamonds Ace 1 2 3 4 5 6 7 8 9 10 Jack Queen King 13 Total 4 4 4 4 4 4 4 4 4 4 4 4 4 4 52 1/6 or 16.

In the tossing of a coin there are only two possible outcomes, heads or tails.Thus the probability of obtaining heads or tails is 1⁄2 or 50%.These illustrations of classical probability are also referred to as a priori probability since we know the probability of an event in advance with- out the need to perform any experiments or trials.Addition rules in classical probability In probability situations we might have a mutu- ally exclusive event.

A mutually exclusive event means that there is no connection between one event and another.They exhibit statistical inde- pendence.For example, obtaining heads on the tossing of a coin is mutually exclusive from obtaining tails since you can have either heads, or tails, but not both.Further, if you obtain heads on one toss of a coin this event will have no impact of the following event when you toss the coin again.In many chance situations, such as the tossing of coin, each time you make the experiment, every- thing resets itself back to zero.

My Canadian cousins had three girls and they really wanted a boy.They tried again thinking after three girls there must be a higher probability of getting a boy.This time they had twins – two girls! The fact that they had three girls previously had no bear- ing on the gender of the baby on the 4th trial.When two events are mutually exclusive then the probability of A or B occurring can be expressed by the following addition rule for mutually exclusive events P(A, or B) � P(A) � P(B) 3(ii) For example, in a pack of cards, the probability of drawing the Ace of Spades, AS, or the Queen of Hearts, QH, with replacement after the first draw, is by equation 3(ii).Replacement means that we draw a card, note its face value, and then put it back into the pack If we do not replace the first card that is with- drawn, and this first card is neither the Ace of Spades, or the Queen of Hearts then the prob- ability is given by the expression, That is, a slightly higher probability than in the case with replacement.

If two events are non-mutually exclusive, this means that it is possible for both events to occur.If we consider for example, the probability of draw- ing either an Ace or a Spade from a deck of cards, then the event Ace and Spade can occur together since it is possible that the Ace of Spades could be drawn.Thus an Ace and a Spade are not mutually exclusive events.In this case, equation 3(ii) for mutually exclusive events must be adjusted to avoid double accounting, or to reduce the proba- bility of drawing an Ace, or a Spade, by the chance we could draw both of them together, that is, the Ace of Spades.Thus, equation 3(ii) is adjusted to become the following addition rule for non- mutually exclusive events P(A, or B) � P(A) � P(B) � P(AB) 3(iii) Here P(AB) is the probability of A and B hap- pening together.

Thus from equation 3(iii) the probability of drawing an Ace or a Spade is, Or we can look at it another way: P P (Ace) (Spade) � � � � 4 52 7 69 13 52 25 00 .% P(Ace or Spade) � � � � 4 52 13 52 4 52 13 52 17 52 1 52 − − * 116 52 30 77� .A or Q ) % S S � � � � � 1 52 1 51 0 0192 0 0196 3 88 P( Q .%A or )S S � � � � 1 52 1 52 1 26 3 85 84 Statistics for Business P(Ace or a Spade) � 7.Joint probability The probability of two or more independent events occurring together or in succession is joint probability.This is calculated by the prod- uct of the individual marginal probabilities P(AB) � P(A) * P(B) 3(iv) Here P(AB) is the joint probability of events A and B occurring together or in succession.P(A) is the marginal probability of A occurring and P(B) is the marginal probability of B occurring.

The joint probability is always less than the marginal probability since we are determining the probability of more than one event occur- ring together in our experiment.Consider for example again in gambling where we are using one pack of cards.The clas- sical or marginal probability of drawing the Ace of Spades from a pack is 1/52 or 1.The probability of drawing the Ace of Spades both times in two successive draws with replacement is as follows: Here the value of 0.

037% for drawing the Ace of Spades twice in two draws is much less than the marginal productivity of 1.92% of drawing the Ace of Spades once in a single drawing.Assume in another gambling game, two dice are thrown together, and the total number obtained is counted.In order for the total count to be 7, the various combinations that must come up together on the dice are as given in Table 3.From classical probability we know that the chance of throwing a 1 and a 6 together, the combination in the 1st column, is from joint probability The chance of throwing a 2 and a 5 together, the combination in the 2nd column, is from joint probability Similarly, the joint probability for throwing a 3 and 4 together, a 4 and 3, a 5 and 2, and a 6 and 1 together is always 2.Thus, the probabil- ity that all 6 can occur is determined as follows from the addition rule 2.67% This is the same result using the criteria of clas- sical or marginal probability of equation 3(i), Here, the number of possible outcomes where the number 7 occurs is six.

The total number of possible outcomes are 36 by the joint probabil- ity of 6 * 6.Thus, the probability of obtaining a 7 on the throw of two dice is 6/36 � 16.67% In order to obtain the number 5, the combin- ations that must come up together are accord- ing to Table 3.Number of outcomes where the event occurs Tottal number of possible outcomes 1 6 1 6 1 36 2 78* .

% 85Chapter 3: Basic probability and counting rules Table 3.2 Possible combinations for obtaining 7 on the throw of two dice.

1st die 1 2 3 4 5 6 2nd die 6 5 4 3 2 1 Total throw 7 7 7 7 7 7 The probability that all four can occur is then from the addition rule, 2.11% if we round at the end of the calculation) Again from marginal probabilities this is 4/36 � 11.Thus again this is a priori probability since in the throwing of two dice, we know in advance that the probability of obtaining a 5 is 4/36 or 11.11% (see also the following section counting rules).

In gambling with slot machines or a one- arm-bandit, often the winning situation is obtaining three identical objects on the pull of a lever according to Figure 3.The probability of winning is joint probability and is given by, P(A1 A2 A3) � P(A1) * P(A2) * P(A3) 3(v) If there are six different objects on each wheel, but each wheel has the same objects, then the marginal probability of obtaining one object is 1/6 � 16.Then the joint probability of obtaining all three objects together is thus, 0.

46% If there are 10 objects on each wheel then the marginal probability for each wheel is 1/10 � 0.

In this case the joint probability is 0.This low value explains why in the long run, most gamblers lose! Conditional probabilities under statistical dependence The concept of statistical dependence implies that the probability of a certain event is depend- ent on the occurrence of another event.Consider the lot of 10 cubes given in Figure 3.

One cube is dark green and dotted; two cubes are light green and striped; three cubes are dark green and striped; and four cubes are light green and dotted.As there are 10 cubes, there are 10 possible events and the probability of selecting any one cube at random from the lot is 10%.The possible out- comes are shown in Table 3.

4 according to the configuration of each cube.Alternatively, this information can be pre- sented in a two by two cross-classification or contingency table as in Table 3.This shows that we have one cube that is dark green and dotted, three cubes that are dark green and striped, four cubes that are light green and dotted, and two cubes that are light green and striped.These formats are also shown in Figure 3.

Assume that we select a cube at random from the lot.Random means that each cube has an equally chance of being chosen.3 Possible combinations for obtaining 5 on the throw of two dice.

1st die 1 2 3 4 2nd die 4 3 2 1 Total throw 5 5 5 5 Probability of obtaining the same three fruits on a one-arm-bandit where there are 10 different fruits on each of the three wheels.P (ABC) � P (A ) * P (B ) * P (C ) Probability � 0.● The probability of the cube being light green is 6/10 or 60%.● The probability of the cube being dark green is 4/10 or 40%.

● The probability of the cube being striped is 5/10 or 50%.87Chapter 3: Basic probability and counting rules Figure 3.2 Probabilities under statistical dependence.10 cubes of the following format Table 3.4 Possible outcomes of selecting a coloured cube.

Event Probability (%) Colour Design 1 10 Dark green Dotted 2 10 Dark green Striped 3 10 Dark green Striped 4 10 Dark green Striped 5 10 Light green Striped 6 10 Light green Striped 7 10 Light green Dotted 8 10 Light green Dotted 9 10 Light green Dotted 10 10 Light green Dotted Table 3.5 Cross-classification table for coloured cubes.Dark green Light green Total Dotted 1 4 5 Striped 3 2 5 Total 4 6 10 Probability of occurrence with the total � 100% Light green and dotted 40% Dark green and dotted Dark green and striped Light green and striped 10% 30% 20% Figure 3.3 Probabilities under statistical dependence.● The probability of the cube being dotted is 5/10 or 50%.

● The probability of the cube being dark green and striped is 3/10 or 30%.● The probability of the cube being light green and striped is 2/10 or 20%.● The probability of the cube being dark green and dotted is 1/10 or 10% ● The probability of the cube being light green and dotted is 4/10 or 40%.Now, if we select a light green cube from the lot, what is the probability of it being dotted? The condition is that we have selected a light green cube.There are six light green cubes and of these, four are dotted, and so the probability is 4/6 or 66.

If we select a striped cube from the lot what is the probability of it being light green? The condition is that we have selected a striped cube.There are five striped cubes and of these two are light green, thus the probability is 2/5 or 40%.This conditional probability under statistical dependence can be written by the relationship, 3(vi) This is interpreted as saying that the probability of B occurring, on the condition that A has occurred, is equal to the joint probability of B and A happening together, or in succession, divided by the marginal probability of A.

Using the relationship from equation 3(vi) and referring to Table 3.

5, The relationship, The relationship, Bayes’ Theorem The relationship given in equation 3(vi) for con- ditional probability under statistical dependence is attributed to the Englishman, The Reverend Thomas Bayes (1702–1761) and is also referred to as Bayesian decision-making.It illustrates that if you have additional information, or based on the fact that something has occurred, certain prob- abilities may be revised to give posterior probabil- ities (post meaning afterwards).Consider that you are a supporter of Newcastle United Football team.Based on last year’s per- formance you believe that there is a high probabil- ity they have a chance of moving to the top of the league this year.However, as the current season moves on Newcastle loses many of the games even on their home turf.

In addition, two of their P P (striped, given dark green) (dotted, give� nn dark green) � � � 3 4 1 4 1 00.P P (striped, given light green) (dotted, giv� een light green) � � � 4 6 2 6 1 00.P P(dark green, given dotted) (dark gr � eeen and dotted) (dotted) / / P � � � 1 10 5 10 1 5 20 00.%% P P(light green, given striped) (light � green and striped) (striped) / / P � � � 2 10 5 10 2 5 440 00.% P P(dotted, given light green) (dotted and li � gght green) (light green) / / P � � � 4 10 6 10 2 3 33 33.

%% (striped, given light green) (striped andP P � light green) (light green) / / P � � � 2 10 6 10 1 3 13.%33 P B A P BA P A ( | ) ( ) � ( ) 88 Statistics for Business best players have to withdraw because of injuries.Thus, based on these new events, the probability of Newcastle United moving to the top of the league has to be revised downwards.Take into account another situation where insurance companies have actuary tables for the life expectancy of indi- viduals.Assume that your 18-year-old son is con- sidered for a life insurance.

However, as time moves on, your son starts smoking heavily.With this new infor- mation, your son’s life expectancy drops as the risk of contracting life-threatening diseases such as lung cancer increases.Thus, based on this pos- terior information, the probabilities are again revised downwards.Thus, if Bayes’ rule is correctly used it implies that it maybe unnecessary to collect vast amounts of data over time in order to make the best deci- sions based on probabilities.

Or, another way of looking at Bayes’ posterior rule is applying it to the often-used phrase of Hamlet, “he who hesitates is lost”.The phrase implies that we should quickly make a decision based on the information we have at hand – buy stock in Company A, purchase the house you visited, or take the high-paying job you were offered in Algiers, Algeria.3 However, if we wait until new information comes along – Company A’s financial accounts turns out are inflated, the house you thought about buying turns out is on the path of the construction of a new auto route, or new elections in Algeria make the political situation in the country unstable with a security risk for the population.In these cases, procrastination may be the best approach and, “he who hesitates comes out ahead”.Venn diagram A Venn diagram, named after John Venn an English mathematician (1834–1923), is a useful way to visually demonstrate the concept of mutually exclusive and non-mutually exclusive events.

A surface area such as a circle or rect- angle represents an entire sample space, and a particular outcome of an event is represented by part of this surface.If two events, A and B, are mutually exclusive, their areas will not overlap as shown in Figure 3.This is a visual representation for a pack of cards using a rectangle for the surface.Here the number of boxes is 52, which is entire sample space, or 100%.

Each card occupies 1 box and when we are considering two cards, the sum of occupied areas is 2 boxes or 2/52 � 3.If two events, are not mutually exclusive their areas would overlap as shown in Figure 3.Here again the number of boxes is 52, which is the entire sample space.

Each of the cards, 13 Spades and 4 Aces would normally occupy 1 box or a total of 17 boxes.However, one card is common to both events and so the sum of occupied areas is 17 � 1 boxes or 16/52 � 30.Application of a Venn diagram and probability in services: Hospitality management A business school has in its curriculum a hospitality management programme.This programme covers hotel management, the food industry, tourism, casino operation, and health spa management.

The programme includes a specialization in hotel management and tourist management and for these pro- grammes the students spend an additional year of enrolment.In one particular year there are 80 students enrolled in the programme.Of these 80 students, 15 elect to specialize in tourist management, 28 in hotel management, and 5 specializing in both tourist and hotel manage- ment.This information is representative of the general profile of the hospitality management programme.89Chapter 3: Basic probability and counting rules 3 Based on a real situation for the Author in the 1980s.

4 Venn diagram: mutually exclusive events.2nd Card 1st Card Number of boxes � 52 which is entire sample space � 100% Each card occupies 1 box Sum of occupied areas � 2 boxes or 2/52 � 3.85% Ace C Ace D KQJ1098765432AceS Ace H Number of boxes � 52 which is entire sample space Each card would normally occupy 1 box � 17 boxes However, one card is common to both events Sum of occupied areas � 17 � 1 boxes or 16/52 � 30.

5 Venn diagram: non-mutually exclusive events.Illustrate this situation on a Venn diagram The Venn diagram is shown in Figure 3.There are (23 � 5) in hotel management shown in the circle (actually an ellipse) on the left.

There are (10 � 5) in tourist man- agement in the circle on the right.The two circles overlap indicating the 5 students who are specializing in both hotel and tourist management.The rectangle is the total sample space of 80 students, which leaves (80 � 23 � 5 � 10) � 42 students as indi- cated not specializing.What is the probability that a random selected student is in tourist management? From the Venn diagram this is the total in tourist management divided by total sample space of 80 students or, 3.

What is the probability that a random selected student is in hotel management? From the Venn diagram this is the total in hotel management divided by total sample space of 80 students or, 4.What is the probability that a random selected student is in hotel or tourist management? From the Venn diagram this is, This can also be expressed by the counting rule equation 3(iii): 5.What is the probability that a random selected student is in hotel and tourist management? From the Venn diagram this is P(both H and T) � 5/80 � 6.Given a student is specializing in hotel manage- ment, what is the probability that a random selected student is also specializing in tourist management? This is expressed as P(T|H), and from the Venn diagram this is 5/28 � 17.

From equation 3(vi), this is also written as, P P P (T|H) (TH) (H) / / � � � � 5 80 28 80 5 28 17 86.% P P P P(H or T) (H) (T) (HT)� � � � � � � 28 80 15 80 5 80 47.% 91Chapter 3: Basic probability and counting rules Figure 3.6 Venn diagram for a hospitality management programme.Hotel management Tourist management 23 5 10 42 7.Given a student is specializing in tourist man- agement, what is the probability that a random selected student is also specializing in hotel man- agement? This is expressed as P(H|T), and from the Venn diagram this is 5/15 � 33.

From equation 3(vi), this is also written as, Application of probability rules in manufacturing: A bottling machine On an automatic combined beer bottling and capping machine, two major problems that occur are overfilling and caps not fitting cor- rectly on the bottle top.From past data it is known that 2% of the bottles are overfilled.Further past data shows that if a bottle is over- filled then 25% of the bottles are faulty capped as the pressure differential between the bottle and the capping machine is too low.Even if a bottle is filled correctly, then still 1% of the bot- tles are not properly capped.

What are the four simple events in this situation? The four simple events are: ● An overfilled bottle ● A normally filled bottle ● An incorrectly capped bottle ● A correctly capped bottle.What are joint events for this situation? There are four joint events: ● An overfilled bottle and correctly capped ● An overfilled bottle and incorrectly capped ● A normally filled bottle and correctly capped ● A normally filled bottle and incorrectly capped.What is the percentage of bottles that will be faulty capped and thus have to be rejected before final packing? Here there are two conditions where a bottle is rejected before packing.A bottle normally filled but faulty capped.● Joint probability of a bottle being over- filled and faulty capped is 0.5% ● Joint probability of a bottle filled normally and faulty capped is (1 � 0.98% ● By the addition rule, a bottle is faulty capped if it is overfilled and faulty capped or normally filled and faulty capped � 0.If the analysis were made looking at a sample of 10,000 bottles, how would this information appear in a cross-classification table? The cross-classification table is shown in Table 3.● Sample size is 10,000 bottles ● There are 2% bottles overfilled or 10,000 * 2% � 200 ● There are 98% of bottles filled correctly or 10,000 * 98% � 9,800 ● Of the bottles overfilled, 25% are faulty capped or 200 * 25% � 50 ● Thus bottles overfilled but correctly capped is 200 � 50 � 150 ● Bottles filled correctly but 1% are faulty capped or 9,800 * 1% � 98 ● Thus filled correctly and correctly capped is 9,800 � 98 � 9,702 ● Thus, bottles correctly capped is 9,702 � 150 � 9,852 ● Thus, all bottles incorrectly capped is 10,000 � 9,852 � 148 � 1.Gambling, odds, and probability Up to this point in the chapter you might argue that much of the previous analysis is related to gambling and then you might say, “but the business P P P (H|T) (HT) (T) / / � � � � 5 80 15 80 5 15 33.33% 92 Statistics for Business world is not just gambling”.That is true but do not put gambling aside.

Our capitalistic society is based on risk, and as a corollary, gambling, as is indicated by the Box Opener.We are confronted daily with gambling through government organ- ized lotteries, buying and selling stock, and gambling casinos.This service-related activity represents a non-negligible part of our economy! In risk, gambling, or betting we refer to the odds of wining.Although the odds are related to probability they are a way of looking at risk.The probability is the number of favourable out- comes divided by the total number of possible outcomes.

The odds of winning are the ratio of the chances of losing to the chances of win- ning.Earlier we illustrated that the probability of obtaining the number 7 in the tossing of two dice was 6 out of 36 throws, or 1 out of 6.Thus the probability of not obtaining the number 7 is 30 out of 36 throws or 5 out of 6.Thus the odds of obtaining the number 7 are 5 to 1.This can be expressed mathematically as, The odds of drawing the Ace of Spades from a full pack of cards are 51 to 1.

Although the odds depend on probability, it is the odds that matter when you are placing a bet or taking a risk! Probability concepts as we have just discussed can be used to evaluate system reliability.A system includes all the interacting components or activ- ities needed for arriving at an end result or prod- uct.In the system the reliability is the confidence that we have in a product, process, service, work team, or individual, such that we can operate under prescribed conditions without failure, or stopping, in order to produce the required output.In the supply chain of a firm for example, reliabil- ity might be applied to whether the trucks deliver- ing raw materials arrive on time, whether the suppliers produce quality components, whether the operators turn up for work, or whether the packing machines operate without breaking down.Generally, the more components or activi- ties in a product or a process, then the more com- plex is the system and in this case the greater is the risk of failure, or unreliability.

Series or parallel arrangement A product or a process might be organized in a series arrangement or parallel arrangement as illustrated schematically in Figure 3.This is a general structure, which contains n compon- ents in the case of a product, or n activities for processes.The value n can take on any integer value.The upper scheme shows a purely series arrangement and the lower a parallel arrange- ment.

Alternatively a system may be a combin- ation of both series and parallel arrangements.Series systems In the series arrangement, shown in the upper diagram of Figure 3.7, it means that for a system to operate we have to pass in sequence through Component 1, Component 2, Component 3, and eventually to Component n.System Reliability and Probability 5 6 1 6 5 1 / / � 93Chapter 3: Basic probability and counting rules Table 3.6 Cross-classification table for bottling machine.

Volume Capping Total Number Number that that fit does not fit Right 9,702 98 9,800 amount Overfilled 150 50 200 Total 9,852 148 10,000 For example, when an electric heater is operat- ing the electrical current comes from the main power supply (Component 1), through a cable (Component 2), to a resistor (Component 3), from which heat is generated.The reliability of a series system, RS, is the joint probability of the number of interact- ing components, n, according to the following relationship: RS � R1 * R2 * R3 * R4 * … Rn 3(vii) Here R1, R2, R3, etc.represent the reliability of the individual components expressed as a frac- tion or percentage.The relationship in equation 3(vii) assumes that each component is independ- ent of the other and that the reliability of one does not depend on the reliability of the other.In the electric heater example, the main power sup- ply, the electric cable, and the resistor are all inde- pendent of each other.

However, the complete electric heating system does depend on all the components functioning, or in the system they are interdependent.If one component fails, then the system fails.For the electric heater, if the power supply fails, or the cable is cut, or the resis- tor is broken then the heater will not function.The reliability, or the value of R, will be �100% (nothing is perfect) and may have a value of say 99%.This means that a component will perform as specified 99% of the time, or it will fail 1% of the time (100 � 99).

This is a binomial relation- ship since the component either works or it does not.Binomial means there are only two possible outcomes such as yes or no, true of false.Consider the system between point X and Y in the series scheme of Figure 3.Assume that component R1 has a relia- bility of 99%, R2 a reliability of 98%, and R3 a reliability of 97%.

The system reliability is then: RS � R1 * R2 * R3 � 0.11% 94 Statistics for Business Figure 3.7 Reliability: Series and parallel systems.Component 1 Component 2 Component 3 Component n X Y System connected in series System connected in parallel (backup) Component 1 Component 2X YComponent nComponent 3 In a situation where the components have the same reliability then the system reliability is given by the following general equation, where n is the number of components RS � Rn 3(viii) Note that as already mentioned for joint prob- ability, the system reliability RS is always less than the reliability of the individual components.Further, the reliability of the system, in a series arrangement of multiple components, decreases rapidly with the number of components.For example assume that we have a system where the average reliability of each component is 98%, then as shown in Table 3.

Further, to give a more complete picture, Figure 3.8 gives a family of curves showing the system reliability, for vari- ous values of the individual component reliability from 100% to 95%.

These curves illustrate the rapid decline in the system reliability as the num- ber of components increases.Parallel or backup systems The parallel arrangement is illustrated in the lower diagram of Figure 3.This illustrates that in order for equipment to operate we can pass through Component 1, Component 2, Component 3, or eventually Component n.Assume that we have two components in a parallel system, R1 the main component and R2 the backup or auxiliary component.

The reliabil- ity of a parallel system, RS, is then given by the relationship, RS � Probability of R1 working � Probability of R2 working * Probability of needing R2 The probability of needing R2 is when R1 is not working or (1 � R1).Thus, RS � R1 � R2(1 � R1) 3(ix) Reorganizing equation 3(ix) RS � R1 � R2 � R2 * R1 RS � 1 � R1 � R2 � R2 * R1 � 1 RS � 1 � (1 � R1 � R2 � R2 * R1) RS � 1 � (1 � R1)(1 � R2) 3(x) If there are n components in a parallel arrange- ment then the system reliability becomes RS � 1 � (1 � R1)(1 � R2)(1 � R3) (1 � R4) … (1 � Rn) 3(xi) where R1, R2, …, Rn represent the reliability of the individual components.The equation can be interpreted as saying that the more the number of backup units, then the greater is the system reli- ability.However, this increase of reliability comes at an increased cost since we are adding backup which may not be used for any length of time.When the backup components of quantity, n, have an equal reliability, then the system reli- ability is given by the relationship, RS � 1 � (1 � R)n 3(xii) Consider the three component system in the lower scheme of Figure 3.

7 between point X and Y with the principal component R1 having a 95Chapter 3: Basic probability and counting rules Table 3.7 System reliability for a series arrangement.Number of components 1 3 5 10 25 50 100 200 System reliability (%) 98.76 reliability of 99%, R2 the first backup com- ponent having a reliability of 98%, and R3 the second backup component having a reliability of 97% (the same values as used in the series arrangement).The system reliability is then from equation 3(xi), RS � 1 � (1 � R1)(1 � R2)(1 � R3) RS � 1 � (1 � 0.994% That is, a system reliability greater than with using a single generator.If we only had the first backup unit, R2 then the system reliability is, RS � 1 � (1 � R1)(1 � R2) RS � 1 � (1 � 0.98% Again, this is a reliability greater than the reliabil- ity of the individual components.Since the com- ponents are in parallel they are called backup units.The more the number of backup units, then the greater is the system reliability as illus- trated in Figure 3.Here the curves give the reli- ability with no backups (n � 1) to three backup components (n � 4).

Of course, ideally, we would always want close to 100% reliability, however, with greater reliability, the greater is the cost.Hospitals have back up energy systems in case of failure of the principal power supply.Most banks and other firms have backup computer systems containing client data should one system 96 Statistics for Business Figure 3.8 System reliability in series according to number of components, n.0 Individual component reliability (same for each component) (%) n � 1 n � 5 n � 3 n � 10 n � 25 n � 50 n � 100 n � 200 fail.The IKEA distribution platform in South- eastern France has a backup computer in case its main computer malfunctions.Without such a system, IKEA would be unable to organize delivery of its products to its retail stores in France, Spain, and Portugal.4 Aeroplanes have backup units in their design such that in the eventual failure of one component or subsys- tem there is recourse to a backup.

For example a Boeing 747 can fly on one engine, although at a much reduced efficiency.In August 2004, my wife and I were in a motor home in St.Petersburg Florida when hurricane Charlie was about to land.We were told of four possible escape routes to get out of the path.

The emergency services had designated several backup exit routes – thankfully! When backup systems are in place this implies redundancy since the backup units are not normally operational.The following is an application example of mixed series and parallel systems.Application of series and parallel systems: Assembly operation In an assembly operation of a certain product there are four components A, B, C, and D that have an individual reliability of 98%, 95%, 90%, and 85%, respectively.The possible ways 97Chapter 3: Basic probability and counting rules Figure 3.9 System reliability of a parallel or backup system according to number of components, n.

0 Reliability of component (same for each) (%) Sy st em re lia bi lity (% ) n � 1 n � 2 n � 3 n � 4 4 After a visit to the IKEA distribution platform in St.Quentin Falavvier, Near Lyon, France, 18 November 2005.of assembly the four components are given in Figures 3.Determine the system reli- ability of the four arrangements.1 Here this is completely a series arrangement and the system reliability is given by the joint probability of the individual reliabilities: ● Reliability is 0.● Probability of system failure is (1 � 0.2 Here this is a series arrangement in the top row in parallel with an assembly in the bottom row.The system reliability is calculated by first the joint probability of the individual reliabilities in the top row, in parallel with the reliability in the second row.● Probability of system failure is (1 � 0.98 Statistics for Business 85%90%95%98% DCBA Figure 3.3 Here we have four units in parallel and thus the system reliability is, ● 1 � (1 � 0.● Probability of system failure is (1 � 0.4 Here we have two units each in series and then the combination in parallel.● Probability of system failure is (1 � 0.In summary, when systems are connected in parallel, the reliability is the highest and the probability of system failure is the lowest.Counting rules are the mathematical relation- ships that describe the possible outcomes, or results, of various types of experiments, or trials.The counting rules are in a way a priori since you have the required information before you perform the analysis.

However, there is no probability involved.The usefulness of counting rules is that they can give you a precise answer to many basic design or analytical situations.The following gives five different counting rules.1 If the number of events is k, and the number of trials, or experiments is n, then the total possible outcomes of single types events are given by kn.

Suppose for example that a coin is tossed 3 times.Then the number of trials, n, is 3 and the number of events, k, is 2 since only heads or tails are the two possible events.The events, obtaining heads or tails are mutually exclusive since you can only have heads or tails in one throw of a coin.The col- lectively exhaustive outcome is 23, or 8.8 gives the possible outcomes of the coin toss experiment.1 in the three tosses of the coin, heads could be obtained each time.6 the first two tosses could be tails, and then the third heads.In tossing a coin just 3 times it is impossible to say what will be the possible outcomes.However, if there are many tosses say a 1,000 times, we can reasonably estimate that we will obtain approximately 500 heads and 500 tails.That is, the larger the number of trials, or exper- iments, the closer the result will be to the characteristic probability.In this case the char- acteristic probability, P(x) is 50% since there is Counting Rules 99Chapter 3: Basic probability and counting rules Table 3.

8 Possible outcomes of the tossing of a coin 8 times.Outcome 1 2 3 4 5 6 7 8 First toss Heads Heads Heads Tails Tails Tails Tails Heads Second toss Heads Heads Tails Heads Tails Tails Heads Tails Third toss Heads Tails Heads Heads Tails Heads Tails Tails an equal chance of obtaining either heads or tails.Thus the outcome is n * P(x) or 1,000 * 50% � 500.This idea is further elaborated in the law of averages in Chapter 4.2 If there are k1 possible events on the 1st trial or experiment, k2 possible events on the 2nd trial, k3 possible events on the 3rd trial, and kn possi- ble events on the nth trial, then the total possible outcomes of different events are calculated by the following relationship: k1 * k2 * k3 … kn 3(xiii) Suppose in gambling, two dice are used.The possible events from throwing the first die are six since we could obtain the number 1, 2, 3, 4, 5, or 6.Similarly, the possible events from throwing the second die are also six.Then the total possible different outcomes are 6 * 6 or 36.The relative frequency histogram of all the possible outcomes is shown in Figure 3.Note, that the number 7 has the highest possibility of occurring at 6 times or a probabil- ity of 16.This is the same value we found in the previous section on joint proba- bilities.Consider another example to determine the total different licence plate registrations that a country or community can possibly issue.Assume that the format for a licence plate is 212TPV.(This was the licence plate number of my first car, an Austin A40, in England, that I owned as a student in the 1960s the time of the Beatles – la belle poque!) In this format there are three numbers, followed by three letters.For numbers, there are 10 possible outcomes, the numbers from 0 to 9.

For letters, there are 26 possible outcomes, the letters A to Z.Thus the first digit of the licence plate can be the number 0 to 9, the same for the second, and the third.Similarly, the first letter can be any letter from A to Z, the same for the second letter, and the same for the third.Thus the total possible different combinations, or the number of licence plates is 17,566,000 on the assumption that 0 is pos- sible in the first place 100 Statistics for Business Table 3.9 Possible outcomes of the tossing of two dice.

1 2 3 4 5 6 7 8 9 10 11 12 1st die 1 2 3 4 5 6 1 2 3 4 5 6 2nd die 1 1 1 1 1 1 2 2 2 2 2 2 Total of both dice 2 3 4 5 6 7 3 4 5 6 7 8 Throw No.13 14 15 16 17 18 19 20 21 22 23 24 1st die 1 2 3 4 5 6 1 2 3 4 5 6 2nd die 3 3 3 3 3 3 4 4 4 4 4 4 Total of both dice 4 5 5 7 8 9 5 6 7 8 9 10 Throw No.25 26 27 28 29 30 31 32 33 34 35 36 1st die 1 2 3 4 5 6 1 2 3 4 5 6 2nd die 5 5 5 5 5 5 6 6 6 6 6 6 Total of both dice 6 7 8 9 10 11 7 8 9 10 11 12 10 * 10 * 10 * 26 * 26 * 26 � 17,576,000 If zero is not permitted in the first place, then the number possible is 15,818,000 9 * 10 * 10 * 26 * 26 * 26 � 15,818,000 Arrangement of different objects: Rule No.3 In order to determine the number of ways that we can arrange n objects is n!, or n factorial, where, n! � n(n � 1)(n � 2)(n � 3) … 1 3(xiv) This is the factorial rule.

Note, the last term in equation 3(xiv) is really (n � n) or 0, but in the factorial relationship, 0! � 1.For example, the number of ways that the three colours, red, yellow, and blue can be arranged is, 3! � 3 * 2 * 1 � 6 Table 3.10 gives these six possible arrange- ments.4 A permutation is a combination of data arranged in a particular order.The number 101Chapter 3: Basic probability and counting rules Figure 3.14 Frequency histogram of the outcomes of throwing two dice.78 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 2 4 53 6 7 8 9 10 11 12 Total value of the sum of the two dice Fr eq ue nc y of o cc ur re nc e( %) Table 3.

10 Possible arrangement of three different colours.Here the same two people can be together, providing they have different pos- itions.

For example in the 1st choice, Dan is the president and Sue is the secretary.In the 6th choice their positions are reversed.Sue is the president and Dan is the secretary.5 A combination is a selection of distinct items regardless of order.

The number of ways, or combinations, of arranging x objects, regard- less of order, from n objects is given by, 3(xvi) Again, assume that there are four candidates for two positions in an operating committee: Dan, Sue, Jim, and Ann.The number of ways a presi- dent and secretary can be chosen now without the same two people working together, regard- less of position is by equation 3(xvi) Table 3.5, combinations, by the value of x! in the denominator.For a given set of items the number of permutations will always be more than the number of combinations because with permutations the order of the data is impor- tant, whereas it is unimportant for combinations.4 2 4 2 4 2 6C � � � ! !( )! n xC n x n x � � ! !( )! 4 2 4 4 2 12P � � ! ( )!− n xP n n x � � ! ( )! 102 Statistics for Business Table 3.11 Permutations in organizing an operating committee.

Choice 1 2 3 4 5 6 7 8 9 10 11 12 President Dan Dan Dan Sue Sue Sue Jim Jim Jim Ann Ann Ann Secretary Sue Jim Ann Jim Ann Dan Ann Dan Sue Dan Sue Jim Table 3.12 Combinations for organizing an operating committee.Choice 1 2 3 4 5 6 President Dan Dan Dan Sue Sue Jim Vice president Sue Jim Ann Jim Ann Ann 103Chapter 3: Basic probability and counting rules This chapter has introduced rules governing basic probability and then applied these to reli- ability of system design.The last part of the chapter has dealt with mathematical counting rules.Basic probability rules Probability is the chance that something happens, or does not happen.

An extension of prob- ability is risk, where we can put a monetary value on the outcome of a particular action.In probability we talk about an event, which is the outcome of an experiment that has been under- taken.Probability may be subjective and this is the “gut” feeling or emotional response of the individual making the judgment.Relative frequency probability is derived from collected data and is thus also called empirical probability.A third is classical or marginal probability, which is the ratio of the number of desired outcomes to the total number of possible outcomes.

Classical probability is also a priori probability because before any action occurs we know in advance all possible outcomes.Gambling involving dice, cards, or roulette wheels are examples of classical probability since before playing we know in advance that there are six faces on a die, 52 cards in a pack.(We do not know in advance the number of slots on the roulette wheel – but the casino does!).Within classical probability, the addition rule gives the chance that two or more events occur, which can be modified to avoid double accounting.

To determine the probability of two or more events occurring together, or in succession, we use joint probability.

When one event has already occurred then this gives posterior probability meaning the new chance based on the condition that another event has already happened.Posterior probability is Bayes’ Theorem.To visually demonstrate relationships in classical probabilities we can use Venn diagrams where a surface area, such as a circle, represents an entire sample space, and a particular outcome of an event is shown by part of this surface.In gambling, particularly in horse racing, we refer to the odds of something happening.Odds are related to probability but odds are the ratio of the chances of losing to the chances of winning.

System reliability and probability A system is a combination of components in a product or many of the process activities that makes a business function.We often refer to the system reliability, which is the confidence that we have in the product or process operating under prescribed conditions without failure.If a system is made up of series components then we must rely on all these series components work- ing.If one component fails, then the system fails.To determine the system reliability, or system failure, we use joint probability.

When the probability of failure, even though small, can be cata- strophic such as for an airplane in flight, the power system in a hospital, or a bank’s computer- based information system, components are connected in parallel.The probability of failure of parallel systems is always less than the probability of failure for series systems for given individual component probabilities.However, on the downside, the cost is always higher for a parallel arrangement since we have a backup that (we hope) will hardly, or never, be used.C hapter Sum m ary 104 Statistics for Business Counting rules Counting rules do not involve probabilities.

However, they are a sort of a priori conditions, as we know in advance, with given criteria, exactly the number of combinations, arrangements, or outcomes that are possible.The first rule is that for a fixed number of possible events, k, then for an experiment with a sample of size, n, the possible arrangements is given by kn.If we throw a single die 4 times then the possible arrangements are 64 or 1,296.The second rule is if we have events of different types say k1, k2, k3 and k4 then the possible arrangements are k1 * k2 * k3 * k4.This rule will indicate, for example, the number of licence plate combinations that are possible when using a mix of numbers and letters.

The third rule uses the factorial relationship, n! for the number of different ways of organizing n objects.The fourth and fifth rules are permuta- tions and combinations, respectively.Permutations gives the number of possible ways of organ- izing x objects from a sample of n when the order is important.Combinations determine the number of ways of organizing x objects from a sample of n when the order is irrelevant.For given values of n and x the value using permutations is always higher than for combinations.

105Chapter 3: Basic probability and counting rules 1.Gardeners’ gloves Situation A landscape gardener employs several students to help him with his work.One morning they come to work and take their gloves from a communal box.This box contains only five left-handed gloves and eight right-handed gloves.If two gloves are selected at random from the box, without replacement, what is the probability that both gloves selected will be right handed? 2.If two gloves are selected at random from the box, without replacement, what is the probability that a pair of gloves will be selected? (One glove is right handed and one glove is left handed.If three gloves are selected at random from the box, with replacement, what is the probability that all three are left handed? 4.If two gloves are selected at random from the box, with replacement, what is the probability that both gloves selected will be right handed? 5.

If two gloves are selected at random from the box, with replacement, what is the probability that a correct pair of gloves will be selected? 2.Market Survey Situation A business publication in Europe does a survey or some of its readers and classifies the survey responses according to the person’s country of origin and their type of work.This information according to the number or respondents is given in the following contin- gency table.What is the probability that a survey response taken at random comes from a reader in Italy? Country Consultancy Engineering Investment Product Architecture banking marketing Denmark 852 232 541 452 385 France 254 365 842 865 974 Spain 865 751 695 358 845 Italy 458 759 654 587 698 Germany 598 768 258 698 568 EXERCISE PROBLEMS 106 Statistics for Business 2.

What is the probability that a survey response taken at random comes from a reader in Italy and who is working in engineering? 3.What is the probability that a survey response taken at random comes from a reader who works in consultancy? 4.What is the probability that a survey response taken at random comes from a reader who works in consultancy and is from Germany? 5.What is the probability that a survey response taken at random from those who work in investment banking comes from a reader who lives in France? 6.What is the probability that a survey response taken at random from those who live in France is working in investment banking? 7.

What is the probability that a survey response taken at random from those who live in France is working in engineering or architecture? 3.Getting to work Situation George is an engineer in a design company.When the weather is nice he walks to work and sometimes he cycles.In bad weather he takes the bus or he drives.

Based on past habits there is a 10% probability that George walks, 30% he uses his bike, 20% he drives, and 40% of the time he takes the bus.

If George walks, there is a 15% probability of being late to the office, if he cycles there is a 10% chance of being late, a 55% chance of being late if he drives, and a 20% chance of being late if he takes the bus.On any given day, what is the probability of George being late to work? 2.Given that George is late 1 day, what is the probability that he drove? 3.Given that George is on time for work 1 day, what is the probability that he walked? 4.

Given that George takes the bus 1 day, what is the probability that he will arrive on time? 5.Given that George walks to work 1 day, what is the probability that he will arrive on time? 4.Packing machines Situation Four packing machines used for putting automobile components in plastics packs operate independently of one another.The utilization of the four machines is given below.Packing machine A Packing machine B Packing machine C Packing machine D 30.

00% 107Chapter 3: Basic probability and counting rules Required 1.What is the probability at any instant that both packing machine A and B are not being used? 2.

What is the probability at any instant that all machines will be idle? 3.What is the probability at any instant that all machines will be operating? 4.What is the probability at any instant of packing machine A and C being used, and packing machines B and D being idle? 5.Study Groups Situation In an MBA programme there are three study groups each of four people.One study group has three ladies and one man.

One has two ladies and two men and the third has one lady and three men.One person is selected at random from each of the three groups in order to make a presentation in front of the class.What is the probability that this presentation group will be composed of one lady and two men? 6.Roulette Situation A hotel has in its complex a gambling casino.

In the casino the roulette wheel has the following configuration.1 2 3 4 5 6 7 8 9 9 8 7 6 5 4 3 2 1 108 Statistics for Business There are two games that can be played: Game No.1 Here a player bets on any single number.If this number turns up then the player gets back 7 times the bet.There is always only one ball in play on the roulette wheel.

2 Here a player bets on a simple chance such as the colours white or dark green, or an odd or even number.If this chance occurs then the player doubles his/her bet.If the number 5 turns up, then all players lose their bets.There is always only one ball in play on the roulette wheel.

What is the probability of the player receiving back £175? What is the probability that the player loses his/her bet? 2.1 a player places £25 on number 3 and £25 on number 4.What is the probability of the player winning? What is the probability that the player loses his/her bet? If the player wins how much money will he/she win? 3.1 if a player places £25 on each of several different numbers, then what is the maximum numbers on which he/she should bet in order to have a chance of winning? What is this probability of winning? In this case, if the player wins how much will he/she win? What is the probability that the player loses his entire bet? How much would be lost? 4.2 a player places £25 on the colour dark green.What is the probability of the player doubling the bet? What is the probability of the player losing his/her bet? 5.2 a player places £25 on obtaining the colour dark green and also £25 on obtaining the colour white.

In this case what is the probability a player will win some money? What is the probability of the player losing both bets? 6.

2 a player places £25 on an even number.What is the probability of the player doubling the bet? What is the probability of the player losing his/her bet? 7.What is the probability of the player doubling the bet? What is the probability of the player losing his/her bet? 7.Sourcing agents Situation A large international retailer has sourcing agents worldwide to search out suppliers of prod- ucts according the best quality/price ratio for products that it sells in its stores in the United States.The retailer has a total of 131 sourcing agents internationally.Of these 51 specialize in textiles, 32 in footwear, and 17 in both textiles and footwear.The remainder are general sourcing agents with no particular specialization.

All the sourcing agents are in a general database with a common E-mail address.When a purchasing manager from any of the retail stores needs information on its sourced products they send an E-mail to the general database address.Anyone of the 131 sourcing agents is able to respond to the E-mail.109Chapter 3: Basic probability and counting rules 1.Illustrate the category of the specialization of the sourcing agents on a Venn diagram.

What is the probability that at any time an E-mail is sent it will be received by a sour- cing agent specializing in textiles? 3.What is the probability that at any time an E-mail is sent it will be received by a sour- cing agent specializing in both textiles and footwear? 4.What is the probability that at any time an E-mail is sent it will be received by a sour- cing agent with no specialty? 5.Given that the E-mail is received by a sourcing agent specializing in textiles what is the probability that the agent also has a specialty in footwear? 6.

Given that the E-mail is received by a sourcing agent specializing in footwear what is the probability that the agent also has a specialty in textiles? 8.Subassemblies Situation A subassembly is made up of three components A, B, and C.A large batch of these units are supplied to the production site and the proportion of defective units in these is 5% of the component A, 10% of the component B, and 4% of the component C.What proportion of the finished subassemblies will contain no defective components? 2.

What proportion of the finished subassemblies will contain exactly one defective component? 3.What proportion of the finished subassemblies will contain at least one defective component? 4.What proportion of the finished subassemblies will contain more than one defective component? 5.What proportion of the finished subassemblies will contain all three defective components? 9.Workshop Situation In a workshop there are the four operating posts with their average utilization as given in the following table.

Each operating post is independent of the other.Operating post Utilization (%) Drilling 50 Lathe 40 Milling 70 Grinding 80 110 Statistics for Business Required 1.What is the probability of both the drilling and lathe work post not being used at any time? 2.What is the probability of all work posts being idle? 3.What is the probability of all the work posts operating? 4.

What is the probability of the drilling and the lathe work post operating and the milling and grinding not operating? 10.Assembly Situation In an assembly operation of a certain product there are four components A, B, C, and D which have an individual reliability of 98%, 95%, 90%, and 85%, respectively.The possible ways of assembly the four components making certain adjustments, are as follows.Method 1 Method 2 Method 3 Method 4 A B C D 99.00% 111Chapter 3: Basic probability and counting rules Required 1.

Determine the system reliability of each of the four possible ways of assembling the components.Determine the probability of system failure for each of the four schemes.Bicycle gears Situation The speeds on a bicycle are determined by a combination of the number of sprocket wheels on the pedal sprocket and the rear wheel sprocket.

The sprockets are toothed wheels over which the bicycle chain is engaged and the combination is operated by a derailleur system.To change gears you move a lever, or turn a control on the handle- bars, which derails the chain onto another sprocket.A bicycle manufacturer assembles customer made bicycles according to the number of speeds desired by clients.Using the counting rules, complete the following table regarding the number of sprockets and the number of gears available on certain options of bicycles.

Film festival Situation The city of Cannes in France is planning its next film festival.The festival will last 5 days and there will be seven films shown each day.The festival committee has selected the 35 films which they plan to show.How many different ways can the festival committee organize the films on the first day? Bicycle model Pedal sprocket Rear wheel sprocket Number of gears A 1 1 B 2 2 C 2 6 D 2 4 E 5 10 F 3 12 G 3 7 H 7 28 I 4 32 J 4 9 112 Statistics for Business 2.If the order of showing is important, how many different ways can the committee organize the showing of their films on the first day? (Often the order of showing films is important as it can have an impact on the voting results.How many different ways can the festival committee organize the films on (a) the second, (b) the third, (c) the fourth, (d) and the fifth and last day? 4.With the conditions according the Question No.

3, and again the order of showing the films is important, how many different ways are possible on (a) the second, (b) the third, (c) the fourth, (d) and the fifth and last day? 13.Flag flying Situation The Hilton Hotel Corporation has just built two large new hotels, one in London, England and the other in New York, United States.The hotel manager wants to fly appro- priate flags in front of the hotel main entrance.If the hotel in London wants to fly the flag of every members of the European Union, how many possible ways can the hotel organize the flags? 2.

If the hotel in London wants to fly the flag of 10 members of the European Union, how many possible ways can the flags be organized assuming that the hotel will con- sider all the flags of members of the European Union? 3.If the hotel in London wants to fly the flag of just five members of the European Union, how many possible ways can the flags be organized assuming that the hotel will consider all the flags of members of the European Union? 4.If the hotel in New York wants to fly the flag of all of the states of the United States how many possible ways can the flags be organized? 5.If the hotel in New York wants to fly the flag of all of the states of the United States in alphabetical order by state how many possible ways can the flags be organized? 14.Model agency Situation A dress designer has 21 evening gowns, which he would like to present at a fashion show.

However at the fashion show there are only 15 suitable models to present the dresses and the designer is told that the models can only present one dress, as time does not permit the presentation of more than 15 designs.How many different ways can the 21 different dress designs be presented by the 15 models? 113Chapter 3: Basic probability and counting rules 2.Once the 15 different dress designs have been selected for the available models in how many different orders can the models parade these on the podium if they all walk together in a single file? 3.Assume there was time to present all the 21 dresses.

Each time a presentation is made the 15 models come onto the podium in a single file.In this case how many permuta- tions are possible in presenting the dresses? 15.Thalassoth rapie Situation Thalassoth rapie is a type of health spar that uses seawater as a base of the therapy treat- ment (thalassa from the Greek meaning sea).The thalassoth rapie centres are located in coastal areas in Morocco, Tunisia, and France, and are always adjacent or physical attached to a hotel such that clients will typically stay say a week at the hotel and be cared for by (usually) female therapists at the thalassoth rapie centre.A week stay at a hotel with breakfast and dinner, and the use of the health spar may cost some £6,000 for two people.

A particular thalassoth rapie centre offers the following eight choices of individ- ual treatments.Bath and seawater massage (bain hydromassant).This is a treatment that lasts 20 minutes where the client lies in a bath of seawater at 37°C where mineral salts have been added.

In the bath there are multiple water jets that play all along the back and legs, which help to relax the muscles and improve blood circulation.

In this treatment the client lies face down while a fine warm seawater rain oscillates across the back and legs giving the client a relax- ing and sedative water massage (duration 20 minutes).Massage under a water spray (massage sous affusion).

This treatment is an individual massage by a therapist over the whole body under a fine shower of seawater.Oils are used during the massage to give a tonic rejuvenation to the complete frame (duration 20 minutes).Here the client is sprayed with a high-pressure water jet at a distance by a therapist who directs the jet over the soles of the feet, the calf muscles, and the back.

This treatment tones up the muscles, has an anti-cramp effect and increases the blood circulation (duration 10 minutes).Envelopment in seaweed (enveloppement d’algues).In this treatment the naked body is first covered with a warm seaweed emulsion.The client is then completely wrapped from the neck down in a heavy heated mattress.

This treatment causes the client to perspire eliminating toxins and recharges the body with iodine and other trace ele- ments from the seaweed (duration 30 minutes).5 Based on the Thalassoth rapie centre (Thalazur), avenue du Parc, 33120 Arcachon, France July 2005.Application of seawater mud (application de boue marine).This treatment is very simi- lar to the envelopment in seaweed except that mud from the bottom of the sea is used instead of seaweed.

Further, attention is made to apply the mud to the joints as this treatment serves to ease the pain from rheumatism and arthritis (duration 30 minutes).In this treatment the client lies on their back on the bare plastic top of a water bed maintained at 37°C.High-pressure water jets within the bed pound the legs and back giving a dry tonic massage (duration 15 minutes).

This is a massage by a therapist where oils are rubbed slowly into the body toning up the muscles and circulation system (duration 30 minutes).In addition to the individual treatments, there are also the following four treatments that are available in groups or which can be used at any time: 1.This is a group therapy where the participants have a gym session consisting of muscle stretching, breathing, and mental reflection (duration 30 minutes).Gymnastic in a seawater swimming pool (Aquagym).This is a group therapy where the participants have a gym session of running, walking, and jumping in a swim- ming pool (duration 30 minutes).The steam bath originated in North Africa and is where clients sits or lies in a marble covered room in which hot steam is pumped.This creates a humid atmosphere where the client perspires to clean the pores of the skin (maximum recommended duration, 15 minutes).The sauna originated in Finland and is a room of exotic wood panelling into which hot dry air is circulated.The temperature of a sauna can reach around 100°C and the dryness of the air can be tempered by pouring water over hot stones that add some humidity (maximum recommended duration, 10 minutes).Considering just the eight individual treatments, how many different ways can these be sequentially organized? 2.Considering just the four non-individual treatments, how many different ways can these be sequentially organized? 3.

Considering all the 12 treatments, how many different ways can these be sequen- tially organized? 4.One of the programmes offered by the thalassoth rapie centre is 6 days for five of the individual treatments alternating between the morning and afternoon.The morning session starts at 09:00 hours and finishes at 12:30 hours and the afternoon session starts at 14:00 hours and finishes at 17:00 hours.In this case, how many possible ways can a programme be put together without any treatment appearing twice on the same day? Show a possible weekly schedule.

115Chapter 3: Basic probability and counting rules 16.

Case: Supply chain management class Situation A professor at a Business School in Europe teaches a popular programme in supply chain management.In one particular semester there are 80 participants signed up for the class.When the participants register they are asked to complete a questionnaire regarding their sex, age, country of origin, area of experience, marital status, and the number of children.This information helps the professor organize study groups, which are balanced in terms of the participant’s background.This information is contained in the table below.

The profes- sor teaches the whole group of 80 together and there is always 100% attendance.The professor likes to have an interactive class and he always asks questions during his class.Required When you have a database with this type of information, there are many ways to analyse the information depending on your needs.The following gives some suggestions, but there are several ways of interpretation.What is the probability that if the professor chooses a participant at random then that person will be: (a) From Britain? (b) From Portugal? (c) From the United States? (d) Have experience in Finance? (e) Have experience in Marketing? (f) Be from Italy? (g) Have three children? (h) Be female? (i) Is greater than 30 years in age? (j) Are aged 25 years? (k) Be from Britain, have experience in engineering, and be single? (l) From Europe? (m) Be from the Americas? (n) Be single? 2.Given that a participant is from Britain then, what is the probability that that the per- son will: (a) Have experience in engineering? (b) Have experience in purchasing? 3.Given that a participant is interested in finance, then what is the probability that per- son is from an Asian country? 4.Given that a participant has experience in marketing, then what is the probability that person is from Denmark? 5.What is the average number of children per participant? 116 Statistics for Business Number Sex Age Country Experience Marital Children status 1 M 21 United States Engineering Married 0 2 F 25 Mexico Marketing Single 2 3 F 27 Denmark Marketing Married 0 4 F 31 Spain Engineering Married 2 5 F 23 France Production Married 0 6 M 26 France Production Single 3 7 M 25 Germany Engineering Single 0 8 F 29 Canada Production Single 3 9 M 32 Britain Engineering Married 2 10 F 21 Britain Finance Single 1 11 M 26 Spain Engineering Married 2 12 M 28 United States Finance Single 0 13 F 27 China Engineering Married 3 14 M 35 Germany Production Married 0 15 F 21 France Engineering Married 2 16 F 26 Germany Marketing Married 3 17 F 25 Britain Production Married 3 18 F 31 China Production Single 4 19 M 22 Britain Production Married 2 20 M 20 Britain Marketing Single 3 21 F 26 Germany Engineering Married 2 22 M 28 Portugal Engineering Single 1 23 M 29 Germany Engineering Single 0 24 M 35 Luxembourg Production Married 0 25 M 41 Germany Finance Married 3 26 F 25 Britain Marketing Single 0 27 M 23 Britain Engineering Married 3 28 F 23 Denmark Production Single 3 29 M 25 Denmark Marketing Single 2 30 F 26 Norway Finance Married 3 31 F 22 France Marketing Single 2 32 F 26 Portugal Engineering Married 3 33 F 28 Spain Engineering Single 3 34 M 24 Germany Production Married 2 35 M 23 Britain Engineering Single 1 36 M 25 United States Production Married 0 37 M 26 Canada Engineering Married 0 38 F 24 Canada Marketing Single 2 39 F 25 Denmark Marketing Single 0 40 M 28 Norway Engineering Married 3 41 F 31 France Finance Married 5 42 M 32 Britain Engineering Married 2 43 F 26 Britain Finance Single 3 44 M 21 Luxembourg Marketing Single 2 45 M 25 China Marketing Married 5 46 M 24 Japan Production Married 2 117Chapter 3: Basic probability and counting rules Number Sex Age Country Experience Marital Children status 47 F 25 France Marketing Single 0 48 F 26 Britain Marketing Married 3 49 M 24 Germany Production Single 2 50 F 21 Taiwan Engineering Married 1 51 F 31 China Engineering Single 3 52 F 35 Britain Marketing Married 0 53 M 38 United States Marketing Married 5 54 F 39 China Engineering Single 2 55 M 23 Portugal Purchasing Married 3 56 F 25 Indonesia Engineering Married 2 57 M 26 Portugal Purchasing Married 2 58 M 23 Britain Marketing Single 0 59 M 25 China Purchasing Married 3 60 M 26 Canada Engineering Single 0 61 F 24 Mexico Purchasing Married 3 62 M 25 China Engineering Single 0 63 F 28 France Production Married 1 64 M 31 United States Marketing Single 2 65 F 32 Britain Marketing Married 3 66 F 25 Germany Engineering Single 0 67 M 25 Spain Purchasing Married 2 68 M 25 Portugal Engineering Single 1 69 M 26 Luxembourg Production Single 3 70 F 24 Taiwan Marketing Single 0 71 M 25 Luxembourg Production Married 1 72 F 26 Britain Engineering Married 2 73 M 28 United States Engineering Single 3 74 F 25 France Engineering Married 0 75 M 26 France Production Single 0 76 F 31 Germany Marketing Single 0 77 M 40 France Engineering Married 3 78 F 25 Spain Marketing Single 2 79 M 26 Portugal Purchasing Married 1 80 M 23 Taiwan Production Single 1 This page intentionally left blank 4Probability analysis for discrete data The shopping mall How often do you go to the shopping mall – every day, once a week, or perhaps just once a month? When do you go? Perhaps after work, after dinner, in the morning when you think you can beat the crowds, or on the weekends? Why do you go? It might be that you have nothing else better to do, it is a grey, dreary day and it is always bright and cheerful in the mall, you need a new pair of shoes, you need a new coat, you fancy buying a couple of CDs, you are going to meet some friends, you want to see a film in the evening so you go to the mall a little early and just have a look around.

All these variables of when and why people go to the mall represent a complex random pattern of potential customers.How does the retailer manage this randomness? Further, when these potential customers are at the mall they behave in a binomial fashion – either they buy or they do not buy.Perhaps in the shopping mall there is a supermarket.It is Saturday, and the supermarket is full of people buying groceries.How to manage the waiting line or the queue at the cashier desk? This chapter covers some of these concepts.

Discrete data are statistical information com- posed of integer values, or whole numbers.They originate from the counting process.For example, we could say that 9 machines are shutdown, 29 bottles have been sold, 8 units are defective, 5 hotel rooms are vacant, or 3 students are absent.It makes little sense to say 91⁄2 machines are shutdown, 293⁄4 bottles have been sold, 81⁄2 units are defective, 51⁄2 hotel rooms are empty, or 31⁄4 students are absent.With discrete data there is a clear segregation and the data does not progress from one class to another.

If the values of discrete data occur in no special order, and there is no explanation of their con- figuration or distribution, then they are con- sidered discrete random variables.This means that, within the range of the possible values of the data, every value has an equal chance of occur- ring.In our gambling situation, discussed in Chapter 3, the value obtained by throwing a single die is random and the drawing of a card from a full pack is random.Besides gambling, there are many situations in the environment that occur randomly and often we need to understand the pattern of randomness in order to make appropriate decisions.

For example as illustrated in the Box Opener “The shopping mall”, the number of people arriving at a shop- ping mall in any particular day is random.If we knew the pattern it would help to better plan staff needs.The number of cars on a particular stretch of road on any given day is random and knowing the pattern would help us to decide on the appropriateness of installing stop signs, or traffic signals for example.The number of people seeking medical help at a hospital emer- gency centre is random and again understand- ing the pattern helps in scheduling medical staff and equipment.It is true that in some cases of randomness, factors like the weather, the day of the week, or the hour of the day, do influence the magnitude of the data but often even if we know these factors the data are still random.

Characteristics of a random variable Random variables have a mean value and a standard deviation.The mean value of random data is the weighted average of all the possible 120 Statistics for Business After you have studied this chapter you will learn the application of discrete random variables, and how to use the binomial and the Poisson distributions.If we assume that this particular pattern of randomness might be repeated we call this mean also the expected value of the random variable, or E(x).The variance of a distribution of a discrete ran- dom variable is given by the expression, 4(ii) This is similar to the calculation of the variance of a population given in Chapter 2, except that instead of dividing by the number of data values, which gives a straight average, here we are mul- tiplying by P(x) to give a weighted average.

The standard deviation of a random variable is the square root of the variance or, 4(iii) The following demonstrates the application of analysing the random variable in the throwing of two dice.Expected value of rolling two dice In Chapter 3, we used combined probabilities to determine that the chance of obtaining the Number 7 on the throw of two dice was 16.Let us turn this situation around and ask the question, “What is the expected value obtained in the throwing two dice, A and B?” We can use equation 4(i) to answer this question.1 gives the possible 36 combinations that can be obtained on the throw of two dice.As this table shows of the 36 combinations, there are just 11 different possible total values (2, 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12) by adding the numbers from the two dice.The number of possible ways that these 11 totals can be achieved is summarized in Column 2 of Table 4.2 and the probability P(x) of obtaining these totals is in Column 3 of the same table.

Using equation 4(i) we can calculate the expected or mean value of throwing two dice and the calculation and the individual results are in Columns 4 and 5.

The total in the last line of Column 4 indicates the probability of obtain- ing these eleven values as 36/36 or 100%.The expected value of throwing two dice is 7 as shown in the last line of Column 5.2 gives the calculation for the variation of obtaining the Number 7 using equation 4(ii).Finally, from equation 4(iii) the standard deviation is, �40.

Another way that we can determine the aver- age value of the number obtained by throwing two dice is by using equation 2(i) for the mean value given in Chapter 2: 2(i) From Column 1 of Table 4.2 the total value of the possible throws is, ∑x � 2 � 3 � 4 � 5 � 6 � 7 � 8 � 9 � 10 � 11 � 12 � 77 The value N, or the number of possible throws to give this total value is 11.Thus, The following is a business-related application of using the random variable.

Application of the random variable: Selling of wine Assume that a distributor sells wine by the case and that each case generates €6.Sales data for the last 200 days is given in Table 4.If we consider that this data is representative of future sales, then the frequency of occurrence of sales can be used to estimate the expected or x x N � � � ∑ 77 11 7 x x N � ∑ Standard deviation, � �∑( ) ( )x P xx 2 Variance, 2 2� �∑( ) ( )x P xx 121Chapter 4: Probability analysis for discrete data 122 Statistics for Business Table 4.2 Expected value of the outcome of the throwing of two dice.Value of Number of Probability x * P(x) Weighted (x � ) (x � )2 (x � )2 P(x) throw (x) possible P(x) value of x ways 2 1 1/36 2 * (1/36) 0.1 2 3 4 5 6 7 8 9 10 11 12 Die A 1 2 3 4 5 6 1 2 3 4 5 6 Die B 1 1 1 1 1 1 2 2 2 2 2 2 Total 2 3 4 5 6 7 3 4 5 6 7 8 Throw No.13 14 15 16 17 18 19 20 21 22 23 24 Die A 1 2 3 4 5 6 1 2 3 4 5 6 Die B 3 3 3 3 3 3 4 4 4 4 4 4 Total 4 5 6 7 8 9 5 6 7 8 9 10 Throw No.25 26 27 28 29 30 31 32 33 34 35 36 Die A 1 2 3 4 5 6 1 2 3 4 5 6 Die B 5 5 5 5 5 5 6 6 6 6 6 6 Total 6 7 8 9 10 11 7 8 9 10 11 12 Table 4.

1 Possible outcomes on the throw of two dice.Here, the values, “days this amount of wine is sold” are used to calculate the probability of future sales using the relationship, 4(iv) For example, from equation 4(iv) Probability of selling 12 cases is 80/200 � 40.00% The complete probability distribution is given in Table 4.

4, and the histogram of this frequency dis- tribution of the probability of sale is in Figure 4.

Using equation 4(i) to calculate the mean value, we have, x � 10 * 15% � 11 * 20% � 12 * 40% � 13 * 25% � 11.75 cases From this, an estimate of future profits is €6.Using equation 4(ii) to calculate the variance, 2 � (10 � 11.9875 cases2 Using equation 4(iii) to calculate the standard deviation we have, � �0.9937 Probability of selling amount days amount x � of sold total days considered in analysi x ss 123Chapter 4: Probability analysis for discrete data Cases sold per day 10 11 12 13 Total Days this amount 30 40 80 50 200 of wine is sold Probability of 15 20 40 25 100 selling this amount (%) Table 4.4 Cases of wine sold over the last 200 days.

1 Frequency distribution of the sale of wine.00 0 5 10 15 20 25 30 35 40 45 10 11 12 13 Cases of wine sold Fr eq ue nc y o f th is nu m be r s ol d ( %) Cases of wine 10 11 12 13 Total days sold per day Days this amount 30 40 80 50 200 of wine is sold Table 4.3 Cases of wine sold over the last 200 days.These calculations give a plausible approach of estimating average long-term future activity on the condition that the past is representative of the future.Covariance of random variables Covariance is an application of the distribution of random variables and is useful to analyse the risk associated with financial investments.

If we consider two datasets then the covariance, xy, between two discrete random variables x and y in each of the datasets is, xy � ∑(x � x)(y � y)P(xy) 4(v) Here x is a discrete random variable in the first dataset and y is a discrete random variable in the second dataset.The terms x and y are the mean or expected values of the corresponding datasets and P(xy) is the probability of each occurrence.The expected value of the sum of two ran- dom variables is, E(x � y) � E(x) � E(y) � x � y 4(vi) The variance of the sum of two random vari- ables is, Variance (x � y) � 2(x�y) � 2x � 2y � 2 xy 4(vii) The standard deviation is the square root of the variance or Standard deviation (x � y) � � 2(x�y) 4(viii) Covariance and portfolio risk An extension of random variables is covariance, which can be used to analyse portfolio risk.Assume that you are considering investing in two types of investments.One is a high growth fund, X, and the other is essentially a bond fund, Y.

An estimate of future returns, per $1,000 invested, according to expectations of the future outlook of the macro economy is given in Table 4.Using equation 4(i) to calculate the mean or expected values, we have, x � 20% * �$100 � 35% * $125 � 45% * $300 � $158.75 y � 20% * �$250 � 35% * $100 � 45% * $10 � $89.50 Using equation 4(ii) to calculate the variance, we have 2x � (�100 � 158.

75 Using equation 4(iii) to calculate the standard deviation, x � �22,767.64 The high growth fund, X, has a higher expected value than the bond fund, Y.However, the standard deviation of the high growth fund is higher and this is an indicator that the investment risk is greater.Using equation 4(v) to calculate the covariance, xy � (�100 � 158.

13 The covariance between the two investments is negative.This implies that the returns on the 124 Statistics for Business Economic Contracting Stable Expanding change Probability of economic change (%) 20 35 45 High growth fund (X ) �$100 $125 $300 Bond fund (Y ) $250 $100 $10 Table 4.investments are moving in the opposite direc- tion, or when the return on one is increasing, the other is decreasing and vice versa.From equation 4(vi) the expected value of the sum of the two investments is, x � y � $158.

25 From equation 4(vii) the variance of the sum of the two investments is, 2(x�y) � 22,767.69 From equation 4(viii) the standard deviation of the sum of the two investments is, � 2(x�y) � �$3,835.93 The standard deviation of the two funds is less than standard deviation of the individual funds because there is a negative covariance between the two investments.This implies that there is less risk with the joint investment than just with an individual investment.

If is the assigned weighting to the asset X, then since there are only two assets the situ- ation is binomial and thus the weighting for the other asset is (1 � ).The portfolio expected return for an investment of two assets, E(P), is, E(P) � p � x � (1 � ) y 4(ix) The risk associated with a portfolio is given by: � 2 2x � (1 � )2 2y � 2 (1 � ) xy 4(x) Assume that we have 40% of our investment in the high-risk fund, which means there is 60% in the bond fund.Then from equation 4(ix) the portfolio expected return is, p � x � (1 � ) y � 40% * $158.20 From equation 4(x) the risk associated with this portfolio is � 2 2x � (1 � )2 2y � 2 (1 � ) xy Thus in summary, the portfolio has an expected return of $117.20, or since this amount is based on an investment of $1,000, there is a return of 11.Further for every $1,000 invested there is a risk of $7.2 gives a graph of the expected return according to the associ- ated risk.This shows that the minimum risk is when there is 40% in the high growth fund and 60% in the bond fund.Although there is a higher expected return when the weighting in the high growth fund is more, there is a higher risk.Expected values and the law of averages When we talk about the mean, or expected value in probability situations, this is not the value that will occur next, or even tomorrow.

It is the value that is expected to be obtained in the long run.In the short term we really do not know what will happen.In gambling for example, when you play the slot machines, or one-armed bandits, you may win a few games.In fact, quite a lot of the money put into slot machines does flow out as jackpots but about 6% rests with the house.1 Thus if you continue playing, then in the long run you will lose because the gambling casinos have set their machines so that the casino will be the long-term winner.

If, not they would go out of business! With probability, it is the law of averages that governs.This law says that the average value obtained in the long term will be close to the expected value, which is the weighted outcome based on each of the probability of occurrence.The long-term result corresponding to the law of averages can be explained by Figure 4.This illustrates the tossing of a coin 1,000 times where we have a 50% probability of obtaining � � � 0.

96 125Chapter 4: Probability analysis for discrete data 1 Henriques, D., On bases, problem gamblers battle the odds, International Herald Tribune, 20 October 2005, p.

00 0 100 200 300 400 500 600 700 800 900 1,000 1,100 Number of coin tosses Pe rc en ta ge o f h ea ds o bt ai ne d (% ) Figure 4.

2 Portfolio analysis: expected value and risk.0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 0 10 20 30 40 50 60 70 80 90 100 Proportion in the high risk investment (%) Ex pe ct ed v al ue , a nd ri sk ($ ) Expected value, $ Risk, $ heads and a 50% probability of obtaining tails.The y-axis of the graph is the cumulative fre- quency of obtaining heads and the x-axis is the number of times the coin is tossed.In the early throws, as we toss the coin, the cumulative number of heads obtained may be more than the cumulative number of tails as illustrated.However, as we continue tossing the coin, the law of averages comes into play, and the cumu- lative number of heads obtained approaches the cumulative number of tails obtained.

After 1,000 throws we will have approximately 500 heads and 500 tails.This illustration supports the Rule 1 of the counting process given in Chapter 2.You can perhaps apply the law of averages on a non-quantitative basis to the behaviour in society.We are educated to be honest, respect- ful, and ethical.This is the norm, or the average, of society’s behaviour.

There are a few people who might cheat, steal, be corrupt, or be violent.In the short term these people may get away with it.However, often in the long run the law of averages catches up with them.They get caught, lose face, are punished or may be removed from society! In statistics, binomial means there are only two possible outcomes from each trial of an experi- ment.The tossing of a coin is binomial since the only possible outcomes are heads or tails.

In qual- ity control for the manufacture of light bulbs the principle test is whether the bulb illuminates or does not.If in a market survey, a respondent is asked if she likes a product, then the alternative response must be that she does not.If we know beforehand that a situation exhibits a binomial pattern then we can use the knowledge of statistics to better understand probabilities of occurrence and make suitable decisions.

We first develop a binomial distribution, which is a table or a graph showing all the possible outcomes of performing many times, the binomial-type experiment.Conditions for a binomial distribution to be valid In order for the binomial distribution to be valid we consider that each observation is selected from an infinite population, or one of a very large size usually without replacement.Alternatively, if the population is finite, such as a pack of 52 cards, then the selection has to be with replace- ment.Since there are only two possible outcomes, if we say that the probability of obtaining one outcome, or “success” is p, then the probability of obtaining the other, or “failure,” is q.

The value of q must be equal to (1 � p).The idea of failure here simply means the opposite of what you are testing or expecting.6 gives some various qualitative outcomes using p and q.

Other criteria for the binomial distribution are that the probability, p, of obtaining an out- come must be fixed over time and that the out- come of any result must be independent of a previous result.

For example, in the tossing of a coin, the probability of obtaining heads or tails 127Chapter 4: Probability analysis for discrete data Binomial Distribution Probability, p Success Win Works Good Present Pass Open Odd Yes Probability, q � (1 � p) Failure Lose Defective Bad Absent Fail Shut Even No Table 4.6 Qualitative outcomes for a binomial occurrence.remains always at 50% and obtaining a head on one toss has no effect on what face is obtained on subsequent tosses.In the throwing a die, an odd or even number can be thrown, again with a probability outcome of 50%.For each result one throw has no bearing on another throw.

In the drawing of a card from a full pack, the prob- ability of obtaining a black card (spade or clubs) or obtaining a red card (heart or diamond) is again 50%.If a card is replaced after the draw- ing, and the pack shuffled, the results of subse- quent drawings are not influenced by previous drawings.In these three illustrations we have the following relationship: Probability, p � (1 � p) � q � 0.Probability of x successes, in n trials 4(xii) ● p is the characteristic probability, or the prob- ability of success, ● q � (1 � p) or the probability of failure, ● x � number of successes desired, ● n � number of trials undertaken, or the sample size.The binomial random variable x can have any integer value ranging from 0 to n, the number of trials undertaken.Again, if p � 50%, then q is 50% and the resulting binomial distribution is symmetrical regardless of the sample size, n.This is the case in the coin toss experiment, obtaining an even or odd number on throwing a die, or selecting a black and red card from a pack.When p is not equal to 50% the distribution is skewed.

In the binomial function, the expression, px * q(n�x) 4(xiii) is the probability of obtaining exactly x successes out of n observations in a particular sequence.The relationship, 4(xiv) is how many combinations of the x successes, out of n observations are possible.We have already presented this expression in the count- ing process of Chapter 3.The expected value of the binomial distribution E(x) or the mean value, x, is the product of the number of trials and the characteristic probability.x � E(x) � n * p 4(xv) For example, if we tossed a coin 40 times then the mean or expected value would be, 40 * 0.

5 � 20 The variance of the binomial distribution is the product of the number of trials, the characteris- tic probability of success, and the characteristic probability of failure.2 � n * p * q 4(xvi) The standard deviation of the binomial distribu- tion is the square root of the variance, � � 2 � �(n * p * q) 4(xvii) Again for tossing a coin 40 times, Variance � 2 � n * p * q � 40 * 0.00 Standard deviation, � � 2 � �(n * p * q) � �10 � 3.

16 n x n x ! ! ( )!� � � �n x n x p qx n x ! ! !( ) ( )⋅ ⋅ 128 Statistics for Business Application of the binomial distribution: Having children Assume that Brad and Delphine are newly mar- ried and wish to have seven children.In the genetic makeup of both Brad and Delphine the chance of having a boy and a girl is equally pos- sible and in their family history there is no inci- dence of twins or other multiple births.What is the probability of Delphine giving birth to exactly two boys? For this situation, ● p � q � 50% ● x, the random variable can take on the val- ues, 0, 1, 2, 3, 4, 5, 6, and 7.● n, the sample size is 7 For this particular question, x � 2 and from equation 4(xii), 2.

Develop a complete binomial distribution for this situation and interpret its meaning.7 for the individual and cumulative values.The his- togram corresponding to this data is shown in Figure 4.

We interpret this information as follows: ● Probability of having exactly two boys � 16.● Probability of having more than two boys (3, 4, 5, 6, or 7 boys) � 77.● Probability of having at least two boys (2, 3, 4, 5, 6, or 7 boys) � 93.● Probability of having less than two boys (0 or 1 boy) � 6.50 boys (though not a feasible value) Standard deviation is �(n * p * q) � �(7 * 0.1890 Deviations from the binomial validity Many business-related situations may appear to follow a binomial situation meaning that the probability outcome is fixed over time, and the result of one outcome has no bearing on another.

However, in practice these two conditions might be violated.Consider for example, a manager is interviewing in succession 20 candidates for one position in his firm.Each candidate represents dis- crete information where their experience and ability are independent of each other.Thus, the interview process is binomial – either a particular p x p x ( ) ! ! ! ( ) � � � � � �2 7 2 7 2 0 50 0 50 2 5 040 2 7 2 ( ) .

, ( ) 22 120 0 25 0 0313 2 21 0 25 0 1313 16 * .%41 129Chapter 4: Probability analysis for discrete data Sample size (n) 7 Probability (p) 50.00% Random Probability of Probability of variable obtaining obtaining this (X) exactly cumulative this value of value of x (%) x (%) 0 0.7 Probability distribution of giving birth to a boy or a girl.As the manager continues the interviewing process he makes a subliminal comparison of competing candi- dates, in that if one candidate that is rated posi- tively this results perhaps in a less positive rating of another candidate.Thus, the evaluation is not entirely independent.Further, as the day goes on, if no candidate has been selected, the interviewer gets tired and may be inclined to offer the post to say perhaps one of the last few remaining candidates out of shear desperation! In another situation, consider you drive your car to work each morning.

When you get into the car, either it starts, or it does not.This is binomial and your expectation is that your car will start every time.The fact that your car started on Tuesday morning should have no effect on whether it starts on Wednesday and should not have been influenced on the fact that it started on Monday morning.However, over time, mechanical, electrical, and even elec- tronic components wear.Thus, on one day you turn the ignition in your car and it does not start! The Poisson distribution, named after the French- man, Denis Poisson (1781–1840), is another discrete probability distribution to describe events that occur usually during a given time interval.

Illustrations might be the number of cars arriv- ing at a tollbooth in an hour, the number of patients arriving at the emergency centre of a hospital in one day, or the number of airplanes waiting in a holding pattern to land at a major airport in a given 4-hour period, or the number of customers waiting in line at the cash checkout 130 Statistics for Business Figure 4.4 Probability histogram of giving birth to a boy (or girl).0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 0 2 3 4 6 Pr ob ab ilit y of g ivi ng b irt h to e xa ct ly th is nu m be r ( %) 1 5 7 Number of boys (or girls) 16.34 Poisson Distribution as highlighted in the Box Opener “The shopping mall”.Mathematical expression for the Poisson distribution The equation describing the Poisson probability of occurrence, P(x) is, 4(xviii) ● (lambda the Greek letter l) is the mean number of occurrences; ● e is the base of the natural logarithm, or 2.71828; ● x is the Poisson random variable; ● P(x) is the probability of exactly x occurrences.The standard deviation of the Poisson distribution is given by the square root of the mean number of occurrences or, � ���( ) 4(xix) In applying the Poisson distribution the assump- tions are that the mean value can be estimated from past data.Further, if we divide the time period into seconds then the following applies: ● The probability of exactly one occurrence per second is a small number and is constant for every one-second interval.

● The probability of two or more occurrences within a one-second interval is small and can be considered zero.● The number of occurrences in a given one- second interval is independent of the time at which that one-second interval occurs dur- ing the overall prescribe time period.● The number of occurrences in any one-second interval is independent on the number of occurrences in any other one-second interval.Application of the Poisson distribution: Coffee shop A small coffee shop on a certain stretch of high- way knows that on average nine people per hour come in for service.Sometimes the only waitress on the shop is very busy, and some- times there are only a few customers.

The owner has decided that if there is greater than a 10% chance that there will be at least 13 clients coming into the coffee shop in a given hour, the manager will hire another waitress.Develop the information to help the manager make a decision.To determine the probability of there being exactly 13 customers coming into the coffee shop in a given hour we can use equation 4(xviii) where in this case x is 13 and is 9.This distribu- tion is shown in Table 4.Column 2 gives the probability of obtaining exactly the random number, and Column 3 gives the cumulative values.5 gives the distribution his- togram for Column 2, the probability of obtaining the exact random variable.

This dis- tribution is interpreted as follows: ● Probability of exactly 13 customers enter- ing in a given hour � 5.● Probability of more than 13 customers entering in a given hour � (100 � 92.● Probability of at least 13 customers enter- ing in a given hour � (100 � 87.● Probability of less than 13 customers entering in a given hour � 87.Since the probability of at least 13 customers entering in a given hour is 12.42% or greater than 10% the manager should decide to hire another waitress.13 13 2 541 865 828 329 0 000123 6 13 9 � � � ,, , , .% 227 020 800 5 04� P x x x ( ) ! � � e 131Chapter 4: Probability analysis for discrete data Poisson approximated by the binomial relationship When the value of the sample size n is large, and the characteristic probability of occurrence, p, is small, we can use the Poisson distribution as a reasonable approximation of the binomial dis- tribution.

The criteria most often applied to make this approximation is when n is greater, or equal to 20, and p is less than, or equal to 0.If this requirement is met then the mean of the binomial distribution, which is given by the product n * p, can be substituted for the mean of the Poisson distribution, .The probability rela- tionship from equation 4(xviii) then becomes, 4(xx) The Poisson random variable, x in theory ranges from 0 to �.However, when the distribution is used as an approximation of the binomial distribution, the number of successes out of n observations can- not be greater than the sample size n.

From equa- tion 4(xx) the probability of observing a large number of successes becomes small and tends to zero very quickly when n is large and p is small.The following illustrates this approximation.Application of the Poisson–binomial approximation: Fenwick’s A distribution centre has a fleet of 25 Fenwick trolleys, which it uses every day for unloading and putting into storage products it receives on pallets from its suppliers.The same Fenwick’s are used as needed to take products out of storage and transfer them to the loading area.These 25 Fenwick’s are battery driven and at the end of the day they are plugged into the electric supply for recharging.

From past data it is known that on a daily basis on average one Fenwick will not be properly recharged and thus not available for use.From this table the probability of three Fenwick’s being out of service on any given day is 6.

Now if we use the binomial approxi- mation, then the characteristic probability P x np e x x np ( ) ( ) ! ( ) � � 132 Statistics for Business Mean value ( ) 9 Random Probability of Probability of variable (x) obtaining obtaining this exactly (%) cumulative value of x (%) 0 0.8 Poisson distribution for the coffee shop.133Chapter 4: Probability analysis for discrete data Figure 4.5 Poisson probability histogram for the coffee shop.000 2 4 6 8 10 12 14 0 3 7 10 11 12 13 14 15 16 17 18 19 20 21 22 23 Number of customers arriving in a given hour Fr eq ue nc y of th is oc cu rre nc e (% ) 9865421 Number of Fenwick’s 25 1 p 4.00% Random variable Poisson (%) Binomial (%) Random variable Poisson (%) Binomial (%) X Exact Exact 13 0.

9 Poisson and binomial distributions for Fenwick’s.

The sample size n is 25, the number of Fenwick’s.Then applying the binomial relationship of equation 4(xii) by generating the distribution using func- tion BINOMDIST we have the binomial distribution in Column 3 and Column 6 of Table 4.This indicates that on any given day, the probability of three Fenwick’s being out of service is 5.This is about the same result as using the Poisson relationship.9 we have given the probabilities to four decimal places to be able to compare values that are very close.

You can also notice that the probability of observ- ing a large number of “successes” tails off very quickly to zero.In this case it is for values of x beyond the Number 5.134 Statistics for Business This chapter has dealt with discrete random variables, their corresponding distribution, and the binomial and Poisson distribution.Distribution for discrete random variables When integer or whole number data appear in no special order they are considered discrete ran- dom variables.

This means that for a given range of values, any number is likely to appear.

The number of people in a shopping mall, the number of passengers waiting for the Tube, or the number of cars using the motorway is relatively random.The mean or the expected value of the random variable is the weighted outcome of all the possible outcomes.The variance is cal- culated by the sum, of the square of the difference between a given random variable and the mean of data multiplied by the probability of occurrence.As always, the standard deviation is the square root of the variance.When we have the expected value and the dispersion or spread of the data, these relationships can be useful in estimating long-term profits, costs, or budget figures.

An extension of the random variable is covariance analysis that can be used to estimate portfolio risk.The law of averages in life is underscored by the expected value in random vari- able analysis.We will never know exactly what will happen tomorrow, or even the day after, however over time or in the long range we can expect the mean value, or the norm, to approach the expected value.Binomial distribution The binomial concept was developed by Jacques Bernoulli a Swiss/French mathematician and as such is sometimes referred to as the Bernoulli process.Binomial means that there are only two possible outcomes, yes or no, right or wrong, works or does not work, etc.

For the binomial distribution to be valid the characteristic probability must be fixed over time, and the outcome of an activity must be independent of another.The mean value in a binomial distribution is the product of the sample size and the characteristic probability.The standard deviation is the square root of the product of the sample size, the characteristic probability, and the characteris- tic failure.If we know that data follows a binomial pattern, and we have the characteristic prob- ability of occurrence, then for a given sample size we can determine, for example, the probability C ha pt er S um m ar y 135Chapter 4: Probability analysis for discrete data of a quantity of products being good, the probability of a process operating in a given time period, or the probability outcome of a certain action.Although many activities may at first appear to be binomial in nature, over time the binomial relationship may be violated.

Poisson distribution The Poisson distribution, named after the Frenchman, Denis Poisson, is another discrete distri- bution often used to describe patterns of data that occur during given time intervals in waiting lines or queuing situations.In order to determine the Poisson probabilities you need to know the average number of occurrences, lambda, which are considered fixed for the experiment in ques- tion.In an experiment when the sample size is at least 20, and the characteristic probability is less then 5%, then the Poisson distribution can be approximated using the binomial relationship.When these conditions apply, the probability outcomes using either the Poisson or the binomial distributions are very close.

HIV virus Situation The Pasteur Institute in Paris has a clinic that tests men for the HIV virus.The testing is performed anonymously and the clinic has no way of knowing how many patients will arrive each day to be tested.Thus tomorrow’s number of patients is a random variable.Past daily records, for the last 200 days, indicate that from 300 to 315 patients per day are tested.

Thus the random variable is the number of patients per day – a discrete ran- dom variable.The Director of the clinic, Professor Michel is preparing his annual budget.The total direct and indirect cost for testing each patient is €50 and the clinic is open 250 days per year.Using the data in Table 1, what will be a reasonable estimated cost for this particular operation in this budget year? Assume that the records for the past 200 days are rep- resentative of the clinic’s operation.If the historical data for the testing is according to Table 2, what effect would this have on your budget? 3.Use the coefficient of variation (ratio of standard deviation to mean value or / ) to compare the data.EXERCISE PROBLEMS Men tested Days this level tested 300 2 301 7 302 10 303 12 304 12 305 14 306 18 307 20 308 24 309 22 310 18 311 16 312 12 313 5 314 4 315 4 Men tested Days this level tested 300 1 301 1 302 1 303 1 304 10 305 16 306 30 307 40 308 40 309 30 310 16 311 10 312 1 313 1 314 1 315 1 Table 1 Table 2 137Chapter 4: Probability analysis for discrete data 4.

Illustrate the distributions given by the two tables as histograms.Do the shapes of the distributions corroborate the information obtained in Question 3? Which of the data is the most reliable for future analysis, and why? 2.Rental cars Situation Roland Ryan operates a car leasing business in Wyoming, United States with 10 outlets in this state.He is developing his budgets for the following year and is proposing to use historical data for estimating his profits for the coming year.For the previous year he has accumulated data from two of his agencies one in Cheyenne, and the other in Laramie.

This data shown below gives the number of cars leased, and the number of days at which this level of cars are leased during 250 days per year when the leasing agencies are opened.Using the data from the Cheyenne agency, what is a reasonable estimate of the aver- age number of cars leased per day during the year the analysis was made? 2.If each car leased generates $22 in profit, using the average value from the Cheyenne data, what is a reasonable estimate of annual profit for the coming year for each agency? Cheyenne Cars leased Days at this level 20 2 21 9 22 12 23 14 24 14 25 18 26 24 27 26 28 29 29 27 30 25 31 20 32 15 33 8 34 6 35 1 Laramie Cars leased Days at this level 20 1 21 1 22 2 23 2 24 12 25 20 26 38 27 49 28 50 29 37 30 19 31 13 32 2 33 2 34 1 35 1 138 Statistics for Business 3.If the data from Laramie was used, how would this change the response to Question 1 for the average number of cars leased per day during the year the analysis was made? 4 If the data from Laramie was used, how would this change the response to Question 2 of a reasonable estimate of annual profit for the coming year for all 10 agencies? 5.

For estimating future activity for the leasing agency, which of the data from Cheyenne or Laramie would be the most reliable? Justify your response visually and quantitatively.Road accidents Situation In a certain city in England, the council was disturbed by the number of road accidents that occurred, and the cost to the city.Some of these accidents were minor just involving damage to the vehicles involved, others involved injury, and in a few cases, death to those persons involved.

These costs and injuries were obviously important but also the council wanted to know what were the costs for the services of the police and fire services.

When an accident occurred, on average two members of the police force were dispatched together with three members of the fire service.The estimated cost of the police was £35 per hour per person and £47 per hour per person for the fire service.The higher cost for the fire service was because the higher cost of the equipment employed.On average each accident took 3 hours to investigate.This including getting to the scene, doing whatever was necessary at the accident scene, and then writing a report.

The council conducted a survey of the number of accidents that occurred and this is in the table below.Plot a relative frequency probability for this data for the number of accidents that occurred.of days (x) occurred 0 7 1 35 2 34 3 46 4 6 5 2 6 31 7 33 8 29 9 31 10 47 11 34 12 30 139Chapter 4: Probability analysis for discrete data 2.Using this data, what is a reasonable estimate of the daily number of accidents that occur in this city? 3.What is the standard deviation for this information? 4.Do you think that there is a large variation for this data? 5.What is an estimated cost for the annual services of the police services? 6.

What is an estimated cost for the annual services of the fire services? 7.What is an estimated cost for the annual services of the police and fire services? 4.Express delivery Situation An express delivery company in a certain country in Europe offers a 48-hour delivery serv- ice to all regions of the United States for packages weighing less then 1 kg.If the firm is unable to delivery within this time frame it refunds to the client twice the fixed charge of €42.The following table gives the number of packages of less than one kilogram, each month, which were not delivered within the promised time-frame over the last three years.Plot a relative frequency probability for this data for the number of packages that were not delivered within the 48-hour time period.What is the highest frequency of occurrence for not meeting the promised time delivery? 3.

What is a reasonable estimate of the average number of packages that are not delivered within the promised time frame? 4.What is the standard deviation of the number of packages that are not delivered within the promised time frame? 5.If the firm sets an annual target of not paying to the client more than €4,500, based on the above data, would it meet the target? 6.What qualitative comments can you make about this data that might in part explain the frequency of occurrence of not meeting the time delivery? Month 2003 2004 2005 January 6 4 10 February 4 6 7 March 5 2 3 April 3 0 4 May 0 5 4 June 1 6 5 July 10 7 9 August 2 9 3 September 2 10 3 October 2 1 6 November 3 1 4 December 11 3 8 140 Statistics for Business 5.Bookcases Situation Jack Sprat produces handmade bookcases in Devon, United Kingdom.

Normally he operates all year-round but this year, 2005, because he is unable to get replacement help, he decides to close down his workshop in August and make no further bookcases.However, he will leave the store open for sales of those bookcases in stock.At the end of July 2005, Jack had 19 finished bookcases in his store/workshop.Sales for the previous 2 years were as follows: Required 1.Based on the above historical data, what is the expected number of bookcases sold per month? 2.

What is the highest probability of selling bookcases, and what is this quantity? 3.If the average sale price of a bookcase were £250.00 using the expected value, what would be the expected financial situation for Jack? 4.What are your comments about the answer to Question 3? 6.Investing Situation Sophie, a shrewd investor, wants to analyse her investment in two types of portfolios.

One is a high growth fund that invests in blue chip stocks of major companies, plus selected tech- nology companies.The other fund is a bond fund, which is a mixture of United States and European funds backed by the corresponding governments.Using her knowledge of finance and economics Sophie established the following regarding probability and financial returns per $1,000 of investment.Month 2003 2004 January 17 18 February 21 24 March 22 17 April 21 21 May 23 22 June 19 23 July 22 22 August 21 19 September 20 21 October 16 18 November 22 22 December 20 15 Economic change Contracting Stable Expanding Probability of economic change (%) 15 45 40 High growth fund, change ($/$1,000) �50 100 250 Bond fund change ($/$1,000) 200 50 10 141Chapter 4: Probability analysis for discrete data Required 1.Determine the expected values of the high growth fund, and the bond fund.

Determine the standard deviation of the high growth fund, and the bond fund.Determine the covariance of the two funds.

What is the expected value of the sum of the two investments? 5.What is the expected value of the portfolio? 6.What is the expected percentage return of the portfolio and what is the risk? 7.Gift store Situation Madame Charban owns a gift shop in La Ciotat.Last year she evaluated that the prob- ability of a customer who says they are just browsing, buys something, is 30%.

Suppose that on a particular day this year 15 customers browse in the store each hour.Required Assuming a binomial distribution, respond to the following questions 1.Develop the individual probability distribution histogram for all the possible outcomes.What is the probability that at least one customer, who says they are browsing, will buy something during a specified hour? 3.

What is the probability that at least four customers, who say they are browsing, will buy something during a specified hour? 4.What is the probability that no customers, who say they are browsing, will buy some- thing during a specified hour? 5.What is the probability that no more than four customers, who say they are brows- ing, will buy something during a specified hour? 8.European Business School Situation A European business school has a 1-year exchange programme with international uni- versities in Argentina, Australia, China, Japan, Mexico, and the United States.There is a strong demand for this programme and selection is based on language ability for the country in question, motivation, and previous examination scores.

Records show that in the 70% of the candidates that apply are accepted.The acceptance for the programme follows a Bernoulli process.Develop a table showing all the possible exact probabilities of acceptance if 20 candi- dates apply for this programme.Develop a table showing all the possible cumulative probabilities of acceptance if 20 candidates apply for this programme.Illustrate, on a histogram, all the possible exact probabilities of acceptance if 20 can- didates apply for this programme.If 20 candidates apply, what is the probability that exactly 10 candidates will be accepted? 5.

If 20 candidates apply, what is the probability that exactly 15 candidates will be accepted? 6.If 20 candidates apply, what is the probability that at least 15 candidates will be accepted? 7.If 20 candidates apply, what is the probability that no more than 15 candidates will be accepted? 8.If 20 candidates apply, what is the probability that fewer than 15 candidates will be accepted? 9.Clocks Situation The Chime Company manufactures circuit boards for use in electric clocks.

Much of the soldering work on the circuit boards is performed by hand and there are a proportion of the boards that during the final testing are found to be defective.Historical data indicates that of the defective boards, 40% can be corrected by redoing the soldering.The distribu- tion of defective boards follows a binomial distribution.Illustrate on a probability distribution histogram all of the possible individual out- comes of the correction possibilities from a batch of eight defective circuit boards.

What is the probability that in the batch of eight defective boards, none can be corrected? 3.What is the probability that in the batch of eight defective boards, exactly five can be corrected? 4.What is the probability that in the batch of eight defective boards, at least five can be corrected? 5.What is the probability that in the batch of eight defective boards, no more than five can be corrected? 6.

What is the probability that in the batch of eight defective boards, fewer than five can be corrected? 10.Computer printer Situation Based on past operating experience, the main printer in a university computer centre, which is connected to the local network, is operating 90% of the time.The head of Information Systems makes a random sample of 10 inspections.143Chapter 4: Probability analysis for discrete data Required 1.Develop the probability distribution histogram for all the possible outcomes of the operation of the computer printer.

In the random sample of 10 inspections, what is the probability that the computer printer is operating in exactly 9 of the inspections? 3.In the random sample of 10 inspections, what is the probability that the computer printer is operating in at least 9 of the inspections? 4.In the random sample of 10 inspections, what is the probability that the computer printer is operating in at most 9 of the inspections? 5.

In the random sample of 10 inspections, what is the probability that the computer printer is operating in more than 9 of the inspections? 6.

In the random sample of 10 inspections, what is the probability that the computer printer is operating in fewer than 9 of the inspections? 7.In how many inspections can the computer printer be expected to operate? 11.Bank credit Situation A branch of BNP-Paribas has an attractive credit programme.Customers meeting cer- tain requirements can obtain a credit card called “BNP Wunder”.Local merchants in surrounding communities accept this card.

The advantage is that with this card, goods can be purchased at a 2% discount and further, there is no annual cost for the card.Past data indicates that 35% of all card applicants are rejected because of unsatisfac- tory credit.Assuming that credit acceptance, or rejection, is a Bernoulli process, and samples of 15 applicants are made.Develop a probability histogram for this situation.

What is the probability that exactly three applicants will be rejected? 3.What is the probability that at least three applicants will be rejected? 4.What is the probability that more than three applicants will be rejected? 5.What is the probability that exactly seven applicants will be rejected? 6.

What is the probability that at least seven applicants will be rejected? 7.What is the probability that more than seven applicants will be rejected? 12.Biscuits Situation The Betin Biscuit Company every August offers discount coupons in the Rh ne-Alps Region, France for the purchase of their products.Historical data at Betin’s marketing 144 Statistics for Business department indicates that 80% of consumers buying their biscuits do not use the coupons.One day eight customers enter into a store to buy biscuits.

Develop an individual binomial distribution for the data.Plot this data as a relative frequency distribution.What is the probability that exactly six customers do not use the coupons for the Betin biscuits? 3.

What is the probability that exactly seven customers do not use the coupons? 4.What is the probability that more than four customers do not use the coupons for the Betin biscuits? 5.What is the probability that less than eight customers do not use the coupons? 6.What is the probability that no more than three customers do not use the coupons? 13.Bottled water Situation A food company processes sparkling water into 1.

The speed of the bot- tling line is very high and historical data indicates that after filling, 0.This filling and ejection operation is considered to follow a Poisson distribution.For 2,000 bottles, develop a probability histogram from zero to 15 bottles falling from the line.What is the probability that for 2,000 bottles, none are ejected from the line? 3.What is the probability that for 2,000 bottles, exactly four are ejected from the line? 4.What is the probability that for 2,000 bottles, at least four are ejected from the line? 5.

What is the probability that for 2,000 bottles, less than four are ejected from the line? 6.What is the probability that for 2,000 bottles, no more than four are ejected from the line? 14.Cash for gas Situation A service station, attached to a hypermarket, has two options for gasoline or diesel pur- chases.Customers either using a credit card that they insert into the pump, serve them- selves with fuel such that payment is automatic.This is the most usual form of purchase.

The other option is the cash-for-gas utilization area.Here the customers fill their tank and then drive to the exit and pay cash, to one of two attendants at the exit kiosk.This form of distribution is more costly to the operator principally because of the salaries of the attendants in the kiosk.The owner of this service station wants some assurance that 145Chapter 4: Probability analysis for discrete data there is a probability of greater than 90% that 12 or more customers in any hour use the automatic pump.

Past data indicates that on average 15 customers per hour use the automatic pump.

The Poisson relationship will be used for evaluation.Develop a Poisson distribution for the cash-for-gas utilization area.Should the service station owner be satisfied with the cash-for-gas utilization, based on the criteria given? 3.

From the information obtained in Question 2 what might you propose for the owner of the service station? 15.Cashiers Situation A supermarket store has 30 cashiers full time for its operation.From past data, the absenteeism due to illness is 4.Develop an individual Poisson distribution for the data.Plot this data as a relative fre- quency distribution? 2.Using the Poisson distribution, what is the probability that on any given day exactly three cashiers do not show up for work? 3.Using the Poisson distribution, what is the probability that less than three cashiers do not show up for work? 4.Using the Poisson distribution, what is the probability that more than three cashiers do not show up for work? 5.

Develop an individual binomial distribution for the data.Plot this data as a relative frequency distribution.Using the binomial distribution, what is the probability that on any given day exactly three cashiers do not show up for work? 7.Using the binomial distribution, what is the probability that less than three cashiers do not show up for work? 8.

Using the binomial distribution, what is the probability that more than three cashiers do not show up for work? 9.What are your comments about the two frequency distribution that you have developed, and the probability values that you have determined? 16.Case: Oil well Situation In an oil well area of Texas are three automatic pumping units that bring the crude oil from the ground.These pumps are installed to operate continuously, 24 hours per day, 365 days 146 Statistics for Business per year.Each pump delivers 156 barrels per day of oil when operating normally and the oil is sold at a current price of $42 per barrel.

There are times when the pumps stop because of blockages in the feed pipes and the severe weather conditions.When this occurs, the auto- matic controller at the pump wellhead sends an alarm to a maintenance centre.Here there is always a crew on-call 24 hours a day.When a maintenance crew is called in there is always a three-person team and they bill the oil company for a fixed 10-hour day at a rate of $62 per hour, per crewmember.The data below gives the operating performance of these three pumps in a particular year, for each day of a 365-day year.

In the table, “1” indicates the pump is operating, “0” indicates the pump is down (not operating).Required Describe this situation in probability and financial terms.1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 Pump No.2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 0 1 1 Pump No.

3 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 0 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 147Chapter 4: Probability analysis for discrete data Pump No.1 1 1 1 1 1 1 0 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 Pump No.2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 Pump No.3 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 0 1 1 1 0 1 1 0 1 This page intentionally left blank 5Probability analysis in the normal distribution Your can of beer or your bar of chocolate When you buy a can of beer written 33 cl on the label, you have exactly a volume of 33 cl in the can, right? You are almost certainly wrong as this implies a volume of 33.When you buy a bar of black chocolate it is stamped on the label, net weight 100 g.Again, it is highly unlikely that you have 100.In operations, where the target, or the machine setting is to obtain a certain value it is just about impossible to always obtain this value.Some values will be higher, and some will be lower, just because of the variation of the filling process for the cans of beer or the moulding operation for the chocolate bars.

The volume of the beer in the can, or the weight of the bar of chocolate, should not be consistently high since over time this would cost the producing firm too much money.Conversely, the volume or weight cannot be always too low as the firm will not be respecting the information given on the label and clearly this would be unethical.These measurement anomalies can be explained by the normal distribution.The normal distribution is developed from continu- ous random variables that unlike discrete random variables, are not whole numbers, but take frac- tional or decimal values.

As we have illustrated in the box opener “Your can of beer or your bar of chocolate”, the nominal volume of beer in a can, or that amount indicated on the label, is 33 cl.

However, the actual volume when measured may be in fact 32.The nominal weight of a bar of chocolate is 100 g but the actual weight when measured may be in fact 99.We may note that the runner completed the Santa Barbara marathon in 3 hours and 4 minutes and 32 sec- onds.

For all these values of volume, weight, and time there is no distinct cut-off point between the data values and they can overlap into other class ranges.A normal distribution is the most important prob- ability distribution, or frequency of occurrence, to describe a continuous random variable.It is widely used in statistical analysis.The concept was developed by the German, Karl Friedrich Gauss (1777–1855) and is thus it is also known as the Gaussian distribution.It is valuable to understand the characteristics of the normal distribution as this can provide information about probability outcomes in the business environment and can be a vital aid in decision-making.

Characteristics The shape of the normal distribution is illus- trated in Figure 5.The x-axis is the value of the random variable, and the y-axis is the fre- quency of occurrence of this random variable.As we mentioned in Chapter 3, if the frequency of occurrence can represent future outcomes, then the normal distribution can be used as a measure of probability.The following are the basic characteristics of the distribution: ● It is a continuous distribution.

● It is bell, mound, or humped shaped and it is symmetrical around this hump.When it is Describing the Normal Distribution 150 Statistics for Business After you have studied this chapter you will understand and be able to apply the most widely used tool in statistics, the normal distribution.symmetrical it means that the left side is a mirror image of the right side.

● The central point, or the hump of the distri- bution, is at the same time the mean, median, mode, and midrange.● The left and right extremities, or the two tails of the normal distribution, may extend far from the central point implying that the associated random variable, x, has a range, �� � x � ��.● The inter-quartile range is equal to 1.Regarding the tails of the distributions most real-life situations do not extend indefinitely in both directions.In addition, negative val- ues or extremely high positive values would not be possible.However, for these situations the normal distribution is still a reasonable approximation.Mathematical expression The mathematical expression for the normal distribution, and from which the continuous curve is developed, is given by the normal distri- bution density function, 5(i) ● f(x) is the probability density function.● e is the base of the natural logarithm equal to 2.● x is the value of the random variable.

● x is the mean value of the distribution.f x x x x x( ) ( / ) ( )/ � � � 1 2 1 2 2 e Empirical rule for the normal distribution There is an empirical rule for the normal distribu- tion that states the following: ● No matter the values of the mean or the stand- ard deviation, the area under the curve is always unity.This means that the area under the curve represents all or 100% of the data.26%) of all the data falls within �1 standard deviations from the mean.

This means that the boundary limits of this 68% of the data are � .44%) of all the data falls within �2 standard deviations from the mean.This means that the bound- ary limits of this 95% of the data are � 2 .73%) of all the data falls within �3 standard devia- tions from the mean.This means that the boundary limits of this almost 100% of the data are � 3 .Effect of different means and/or different standard deviations The mean measures the central tendency of the data, and the standard deviation measures its spread or dispersion.Datasets in a normal distri- bution may have the following configurations: ● The same mean, but different standard devi- ations as illustrated in Figure 5.

Here there are three distributions with the same mean but with standard deviations of 2.The smaller the standard deviation, here 2.

50, the curve is narrower and the data congregates around the mean.The larger the standard deviation, here 10.0, the flatter is the curve and the deviation around the mean is greater.● Different means but the same standard devi- ation as illustrated in Figure 5.00 for the three curves and their shape is identical.However their means 152 Statistics for Business m s � 10.2 Normal distribution: the same mean but different standard deviations.are �10, 0, and 20 so that they have different positions on the x-axis.● Different means and also different standard deviations are illustrated in Figure 5.

Here the flatter curve has a mean of �10.The middle curve has a mean of 0 and a standard deviation of 5.

In conclusion, the shape of the normal distribu- tion is determined by its standard deviation, and the mean value establishes its position on the x-axis.

As such, there is an infinite combination of curves according to their respective mean and standard deviation.However, a set of data can be uniquely defined by its mean and stand- ard deviation.Kurtosis in frequency distributions Since continuous distributions may have the same mean, but different standard deviations, the different standard deviations alter the sharpness or hump of the peak of the curve as illustrated by the three normal distributions given in Figure 5.This difference in shape is the kurtosis, or the char- acteristic of the peak of a frequency distribution curve.

The curve that has a small standard deviation, � 2.5 is leptokurtic after the Greek word lepto meaning slender.The peak is sharp, and as shown in Figure 5.The curve that has a standard deviation, � 10.0 is platykurtic after the Greek word platy meaning broad, or flat, and this flatness can be seen also in Figure 5.153Chapter 5: Probability analysis in the normal distribution �60 �50 �40 �30 �20 �10 0 10 20 30 40 50 60 s � 10: m � �10 s � 10: m � 20 s � 10: m � 0 Figure 5.3 Normal distribution: the same standard deviation but different means.The intermediate curve where the standard devi- ation � 5.0 is called mesokurtic since the peak of the curve is in between the two others.In statistics, recording the kurtosis value of data gives a measure of the sharpness of the peak and as a corollary a measure of its disper- sion.The kurtosis value of a relatively flat peak is negative, whereas for a sharp peak it is positive and becomes increasingly so with the sharpness.The importance of knowing these shapes is that a curve that is leptokurtic is more reliable for ana- lytical purposes.

Transformation of a normal distribution Continuous datasets might be for example, the volume of beer in cans, the weight of chocolate bars, or the distance travelled by an automobile tyre.In the normal distribution the units for these measurements for the mean and the standard deviation are different.There are centilitres for the beer, grams for the chocolate, or kilometres for the tyres.However, all these datasets can be transformed into a standard normal distribution using the following normal distribution transform- ation relationship: 5(ii) ● x is the value of the random variable.

● x is the mean of the distribution of the random variables.● x is the standard deviation of the distribution.● z is the number of standard deviations from x to the mean of this distribution.Since the numerator and the denominator (top and bottom parts of the equation) have the z x x x � � 154 Statistics for Business �60 �50 �40 �30 �20 �10 0 10 20 30 40 50 60 s � 10: m � �10 s � 5: m � 0 s � 2.4 Normal distribution: different means and different standard deviations.same units, there are no units for the value of z.Further, since the value of x can be more, or less, than the mean value, then z can be either plus or minus.For example, for a certain format the mean value of beer in a can is 33 cl and from past data we know that the standard devi- ation of the bottling process is 0.Assume that a single can of beer is taken at random from the bottling line and its volume is 33.In this case using equation 5(ii), Alternatively, the mean value of a certain size chocolate bar is 100 g and from past data we know that the standard deviation of a produc- tion lot of these chocolate bars is 0.Assume one slab of chocolate is taken at ran- dom from the production line and its weight is 100.

In this case using equation 5(ii), Again assume that the mean value of the life of a certain model tyre is 35,000 km and from past data we know that the standard deviation of the life of a tyre is 1,500 km.Then suppose that one tyre is taken at random from the production line and tested on a rolling machine.Then using equation 5(ii), Thus in each case we have the same number of standard deviations, z.

This is as opposed to the value of the standard deviation, , in using three different situations each with different units.We have converted the data to a standard normal distribution.This is how the normal fre- quency distribution can be used to estimate the probability of occurrence of certain situations.The standard normal distribution A standard normal distribution has a mean value, , of zero.The area under the curve to the left of the mean is 50.

00% and the area to the right of the mean is also 50.00 the area under the curve represents 99.

00, then the area under the curve represents 95.

00 the area under the curve represents 68.These relationships are illus- trated in Figure 5.

These areas of the curve are indicated with the appropriate values of z on the x-axis.Also, indicated on the x-axis are the val- ues of the random variable, x, for the case of a bar of chocolate of a nominal weight of 100.00 g, and a population standard deviation of 0.These values of x are deter- mined as follows.Reorganizing equation 5(ii) to make x the subject, we have, x � x � z x 5(iii) Thus, when z is 2 the value of x from equation 5(iii) is, x � 100.80 Alternatively, when z is �3 the value of x from equation 5(iii) is, x � 100.

80 The other values of x are calculated in a similar manner.Note the value of z is not necessarily a whole number but it can take on any numerical value such as �0.35, which give areas under the curve from the left-hand tail to the z-value of 32.When z is negative it means that the area under the curve from the left is less than 50% and when z is positive it means that the area from the z x x x � � � � � � 37 250 35 000 1 500 2 250 1 500 1 50 , , , , , .z x x x � � � � � � 100 60 100 00 0 40 0 60 0 40 1 50 .z x x x � � � � � � 33 75 33 00 0 50 0 75 0 50 1 50 .155Chapter 5: Probability analysis in the normal distribution left of the curve is greater than 50%.These area values can also be interpreted as probabilities.

Thus for any data of any continuous units such as weight, volume, speed, length, etc.all inter- vals containing the same number of standard devi- ations, z from the mean, will contain the same proportion of the total area under the curve for any normal probability distribution.These tables give the area of the curve either to the right or the left side of the mean and from these tables probabilities can be estimated.The logic of the z-values in Excel is that the area of the curve increases from 0% at the left to 100% as we move to the right of the curve.The following four useful normal distribution func- tions are found in Excel.156 Statistics for Business 0 Fr eq ue nc y, o r p ro ba bi lity o f o cc u rr e n ce 1 2 3�1�2�3z x 100.5 Areas under a standard normal distribution.It is not necessary to learn by heart which func- tion to use because, as for all the Excel func- tions, when they are selected, it indicates what values to insert to obtain the result.

Application of the normal distribution: Light bulbs General Electric Company has past data concern- ing the life of a particular 100-Watt light bulb that shows that on average it will last 2,500 hours before it fails.The standard deviation of this data is 725 hours and the illumination time of a light bulb is considered to follow a normal distribution.Thus for this situation, the mean value, , is con- sidered a constant at 2,500 hours and the stan- dard deviation, , is also a constant with a value of 725 hours.

Thus we can say that the probability of a single light bulb taken from the production line has an 84.

95% probability of lasting not more than 3,250 hours.This concept is shown on the normal distribution in Figure 5.What is the probability that a light bulb of this kind selected at random from the production line will last at least 3,250 hours? Here we are interested in the area of the curve on the right where x is at least 3,250 hours.

05% z � � � � 3 250 2 500 725 750 725 1 0345 , , .

157Chapter 5: Probability analysis in the normal distribution Life of a light bulb, hours Fr eq ue nc y o f o cc ur re nc e 2,500 3,250 Area � 84.6 Probability that the life of a light bulb lasts no more than 3,250 hours.probability that a single light bulb taken from the production line has a 15.05% probability of lasting at least 3,250 hours.

This is shown on the normal distribution in Figure 5.What is the probability that a light bulb of this kind selected at random will last no more than 2,000 hours? Using equation 5(ii), where the random vari- able, x, is now 2,000 hours, The fact that z has a negative value implies that the random variable lies to the left of the mean; which it does since 2,000 hour is less than 2,500 hours.52% probability that a single light bulb taken from the production line will last no more than 2,000 hours.

This is shown on the normal distribution curve in Figure 5.

What is the probability that a light bulb of this kind selected at random will last between 2,000 and 3,250 hours? In this case we are interested in the area of the curve between 2,000 hours and 3,250 hours where 2,000 hours is to the left of the mean and 3,250 is greater than the mean.We can determine this probability by several methods.Method 1 ● Area of the curve 2,000 hours and below is 24.

● Area of the curve 3,250 hours and above is 15.Thus, area between 2,000 and 3,250 hours is (100.Method 2 Since the normal distribution is symmetrical, the area of the curve to the left of the mean is 50.00% and also the area of the curve to the right of the mean is 50.

Thus, ● Area of the curve between 2,000 and 2,500 hours is (50.● Area of the curve between 2,500 and 3,250 hours is (50.z � � �� �� 2 000 2 500 725 500 725 0 6897 , , .

158 Statistics for Business Life of a light bulb, hours Fr eq ue nc y o f o cc ur re nc e 2,500 3,250 Area � 15.

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7 Probability that the life of a light bulb lasts at least 3,250 hours.Thus, area of the curve between 2,000 and 3,250 hours is (25.Method 3 ● Area of the curve at 3,250 hours and below is 84 .Method 3 ● Area of the curve at 3,250 hours and below is 84.

● Area of the curve at 2,000 hours and below is 24.

Thus, area of the curve between 2,000 and 3,250 hours is (84 **qty.;top titles;ISBN;hyperlinks;NEWS icon;last name of 1st author;authors without affiliation;title;subtitle;series;ed.;year;pages arabic;cover;medium type;bibliography;MRW/KBL;price status ;US $ net;main subject;subject 1;bic code 1;subject 2;bic code 2;bookstore location;product category;publisher;availability status **.Thus, area of the curve between 2,000 and 3,250 hours is (84.This situation is shown on the normal distribution curve in Figure 5 **qty.;top titles;ISBN;hyperlinks;NEWS icon;last name of 1st author;authors without affiliation;title;subtitle;series;ed.;year;pages arabic;cover;medium type;bibliography;MRW/KBL;price status ;US $ net;main subject;subject 1;bic code 1;subject 2;bic code 2;bookstore location;product category;publisher;availability status **.This situation is shown on the normal distribution curve in Figure 5.159Chapter 5: Probability analysis in the normal distribution Life of a light bulb, hours Fr eq ue nc y o f o cc ur re nc e 2,5002,000 Area � 24.8 Probability that the life of a light bulb lasts no more than 2,000 hours best websites to order custom macro economics presentation Business single spaced Rewriting.

8 Probability that the life of a light bulb lasts no more than 2,000 hours.

Fr eq ue nc y o f o cc ur re nc e 2,5002,000 Area � 60.43% 3,250 Life of a light bulb, hours Figure 5.9 Probability that the light bulb lasts between 2,000 and 3,250 hours.What are the lower and upper limits in hours, symmetrically distributed, at which 75% of the light bulbs will last? In this case we are interested in 75% of the middle area of the curve.

The area of the curve outside this value is (100.Since the normal distri- bution is symmetrical, the area on the left side of the limit, or the left tail, is 25/2 or 12.

Similarly, the area on the right of the limit, or the right tail, is also 12.From the normal probability functions in Excel, given the value of 12.

Again, since the curve is symmetrical the value of z on the left side is �1.From equation 5(iii) where z at the upper limit is 1.1503, x � 2,500 and x is 725, x (upper limit) � 2,500 � 1.1503 * 725 � 3,334 hours At the lower limit z is �1.1503 * 725 � 1,666 hours These values are also shown on the normal distribution curve in Figure 5.

If General Electric has 50,000 of this particu- lar light bulb in stock, how many bulbs would be expected to fail at 3,250 hours or less? In this case we simply multiply the population N, or 50,000 by the area under the curve by the answer determined in Question No.24 or 42,477 light bulbs rounded to the nearest whole number.If General Electric has 50,000 of this particular light bulb in stock, how many bulbs would be expected to fail between 2,000 and 3,250 hours? Again, we multiply the population N, or 50,000, by the area under the curve deter- mined by the answer to Question No.96 or 30,217 light bulbs rounded to the nearest whole number.160 Statistics for Business Life of a light bulb, hours Fr eq ue nc y o f o cc ur re nc e 2,5001,666 Area � 75.10 Symmetrical limits between which 75% of the light bulbs will last.161Chapter 5: Probability analysis in the normal distribution Frequency polygon Box-and-whisker plot Figure 5.11 Sales revenue: comparison of the frequency polygon and its box-and- whisker plot.In all these calculations we have determined the appropriate value by first determining the value of z.A quicker route in Excel is to use the func- tion NORMDIST where the mean, standard devi- ation, and the value of x are entered.

It is a matter of preference which of the functions to use.I like to calculate z, since with this value it is easy to position the situ- ation on the normal distribution curve.A lot of data follows a normal distribution par- ticularly when derived from an operation set to a nominal value.The weight of a nominal 100-g chocolate bar, the volume of liquid in a nom- inal 33-cl beverage can, or the life of a tyre mentioned earlier follow a normal distribution.

Some of the units examined will have values greater than the nominal figure and some less.However, there may be cases when other data may not follow a normal distribution and so if you apply the normal distribution assumptions erroneous conclusions may be made.Verification of normality To verify that data reasonably follows a normal distribution you can make a visual comparison.For small datasets a stem-and-leaf display as pre- sented in Chapter 1, will show if the data appears normal.

For larger datasets a frequency polygon also developed in Chapter 1 or a box-and-whisker plot, introduced in Chapter 2, can be developed to see if their profiles look normal.

11 shows a frequency polygon and the box-and-whisker plot for the sales revenue data presented in Chapters 1 and 2.Another verification of the normal assumption is to determine the properties of the dataset to see if they correspond to the normal distribution Demonstrating That Data Follow a Normal Distribution criteria.If they do then the following relation- ships should be close.● The mean is equal to the median value.

● The inter-quartile range is equal to 1.● The range of the data is equal to six times the standard deviation.● About 68% of the data lies between �1 stand- ard deviations of the mean.● About 95% of the data lies between �2 stand- ard deviations of the mean.

● About 100% of the data lies between �3 standard deviations of the mean.1 gives the proper- ties for the 200 pieces of sales data presented in Chapter 1.The percentage values are calculated by using the equation 5(iii) first to find the limits for a given value of z using the mean and stan- dard deviation of the data.Then the amount of data between these limits is determined and this 162 Statistics for Business 35,378 170,569 104,985 134,859 120,958 107,865 127,895 106,825 130,564 108,654 109,785 184,957 96,598 121,985 63,258 164,295 97,568 165,298 113,985 124,965 108,695 91,864 120,598 47,865 162,985 83,964 103,985 61,298 104,987 184,562 89,597 160,259 55,492 152,698 92,875 56,879 151,895 88,479 165,698 89,486 85,479 64,578 103,985 81,980 137,859 126,987 102,987 116,985 45,189 131,958 73,598 161,895 132,689 120,654 67,895 87,653 58,975 103,958 124,598 168,592 95,896 52,754 114,985 62,598 145,985 99,654 76,589 113,590 80,459 111,489 109,856 101,894 80,157 78,598 86,785 97,562 136,984 89,856 96,215 163,985 83,695 75,894 98,759 133,958 74,895 37,856 90,689 64,189 107,865 123,958 105,987 93,832 58,975 102,986 102,987 144,985 101,498 101,298 103,958 71,589 59,326 121,459 82,198 60,128 86,597 91,786 56,897 112,854 54,128 152,654 99,999 78,562 110,489 86,957 99,486 132,569 134,987 76,589 135,698 118,654 90,598 156,982 87,694 117,895 85,632 104,598 77,654 105,987 78,456 149,562 68,976 50,128 106,598 63,598 123,564 47,895 100,295 60,128 141,298 84,598 100,296 77,498 77,856 134,890 79,432 100,659 95,489 122,958 111,897 129,564 71,458 88,796 110,259 72,598 140,598 125,489 69,584 89,651 70,598 93,876 112,987 123,895 65,847 128,695 66,897 82,459 133,984 98,459 153,298 87,265 72,312 81,456 124,856 101,487 73,569 138,695 74,583 136,958 115,897 142,985 119,654 96,592 66,598 81,490 139,584 82,456 150,298 106,859 68,945 122,654 70,489 94,587 85,975 138,597 97,498 143,985 92,489 146,289 84,592 69,874 Table 5.

1 Sales revenues: properties compared to normal assumptions.The fol- lowing gives an example of the calculation.x (for z � �1) � 102,667 � 30,880 � 71,787 x (for z � �1) � 102,667 � 30,880 � 133,547 Using Excel, there are 129 pieces of data between these limits and so 129/200 � 64.50% x (for z � �2) � 102,667 � 2 * 30,880 � 40,907 x (for z ��2) � 102,667 � 2 * 30,880 � 164,427 Using Excel, there are 192 pieces of data between these limits and 192/200 � 96.00% x (for z � �3) � 102,667 � 3 * 30,880 � 10,027 x (for z � �3) � 102,667 � 3 * 30,880 � 19,5307 Using Excel, there are 200 pieces of data between these limits and 200/200 � 100.00% Thus from the visual displays, and the proper- ties of the sales data, the normal assumption seems reasonable.

As a proof of this, if we go back to Chapter 1 from the ogives for this sales data we showed that, ● From the greater than ogive, 80.00% of the sales revenues are at least $75,000.00% of the revenues are no more than $145,000.If we assume a normal distribution then at least 80% of the sales revenue will appear in the area of the curve as illustrated in Figure 5.

Using this and the mean, and standard deviation values for the sales data using equation 5(iii) we have, x � 102,667 � (�0.

8416) * 30,880 � $76,678 This value is only 2.

2% greater than the value of $75,000 determined from the ogive.163Chapter 5: Probability analysis in the normal distribution Fr eq ue nc y o f o cc ur re nc e m Area � 80.12 Area of the normal distribution containing at least 80% of the data.Similarly, if we assume a normal distribution then 90% of the sales revenue will appear in the area of the curve as illustrated in Figure 5.

Using this and the mean, and standard deviation values for the sales data using equation 5(iii) we have, x � 102,667 � 1.2816 * 30,880 � $142,243 This value is only 1.

9% less than the value of $145,000 determined from the ogive.Asymmetrical data In a dataset when the mean and median are sig- nificantly different then the probability distribu- tion is not normal but is asymmetrical or skewed.A distribution is skewed because values in the frequency plot are concentrated at either the low (left side) or the high end (right side) of the x-axis.When the mean value of the dataset is greater than the median value then the distri- bution of the data is positively or right-skewed where the curve tails off to the right.This is because it is the mean that is the most affected by extreme values and is pulled over to the right.

Here the distribution of the data has its mode, the hump, or the highest frequency of occur- rence, at the left end of the x-axis where there is a higher proportion of relatively low values and a lower proportion of high values.The median is the middle value and lies between the mode and the mean.If the mean value is less than the median, then the data is negatively or left-skewed such that the curve tails off to the left.This is because it is the mean that is the most affected by extreme values and is pulled back to the left.Here the distribution of the data has its mode, the hump, or the highest frequency of occur- rence, at the right end of the x-axis where there is a higher proportion of large values and lower proportion of relatively small values.

Again, the median is the middle value and lies between the mode and the mean.This concept of symmetry and asymmetry is illustrated by the following three situations.For a certain consulting Firm A, the monthly salaries of 1,000 of its worldwide staff are shown by the frequency polygon and its associ- ated box-and-whisker plot in Figure 5.Here 164 Statistics for Business Fr eq ue nc y o f o cc ur re nc e m x Area � 90.

13 Area of the normal distribution giving upper limit of 90% of the data.the data is essentially symmetrically distributed.The mean value is $15,893 and the median value is $15,907 or the mean is just 0.The maximum salary is $21,752 and the minimum is $10,036.Thus, 500, or 50% of the staff have a monthly salary between $10,036 and $15,907 and 500, or the other 50%, have a salary between $15,907 and $21,752.From the graph the mode is about $15,800 with the frequency at about 19.2% or essentially the mean, mode, and median are approximately the same.Here the frequency polygon and the box-and-whisker plot are right-skewed.The mean value is now $12,964 and the median value is $12,179 or the mean is 6.The maximum salary is still $21,752 and the min- imum $10,036.

Now, 500, or 50%, of the staff have a monthly salary between $10,036 and $12,179 and 500, or the other 50%, have a salary between $12,179 and $21,752 or a larger range of smaller values than in the case of the symmetrical distribution, which explains the lower average value.From the graph the mode is about $11,500 with the frequency at about 24.Thus in ascending order, we have the mode ($11,500), median ($12,179), and mean ($12,964).Here the frequency polygon and the box-and-whisker plot are left-skewed.The mean value is now $18,207 and the median value is $19,001 or the mean is 4.The maximum salary is still $21,752 and the minimum $10,036.

Now, 500, or 50%, of the staff have a monthly salary between $10,036 and $19,001 and 500, or the other 50%, have a salary between 165Chapter 5: Probability analysis in the normal distribution 0% 2% 4% 6% 8% 10% 12% 14% 16% 18% 20% 22% 8,000 9,000 10,000 11,000 12,000 13,000 14,000 15,000 16,000 17,000 18,000 19,000 20,000 21,000 22,000 23,000 24,000 8,000 9,000 10,000 11,000 12,000 13,000 14,000 15,000 16,000 17,000 18,000 19,000 20,000 21,000 22,000 23,000 24,000 Monthly salary, $ Figure 5.14 Frequency polygon and its box-and-whisker plot for symmetrical data.166 Statistics for Business 0% 2% 4% 6% 8% 10% 12% 14% 16% 18% 20% 22% 24% 26% Monthly salary, $ 8,000 9,000 10,000 11,000 12,000 13,000 14,000 15,000 16,000 17,000 18,000 19,000 20,000 21,000 22,000 23,000 24,000 8,000 9,000 10,000 11,000 12,000 13,000 14,000 15,000 16,000 17,000 18,000 19,000 20,000 21,000 22,000 23,000 24,000 0% 2% 4% 6% 8% 10% 12% 14% 16% 18% 20% 22% 24% 26% Monthly salary, $ 8,000 9,000 10,000 11,000 12,000 13,000 14,000 15,000 16,000 17,000 18,000 19,000 20,000 21,000 22,000 23,000 24,000 8,000 9,000 10,000 11,000 12,000 13,000 14,000 15,000 16,000 17,000 18,000 19,000 20,000 21,000 22,000 23,000 24,000 Figure 5.15 Frequency polygon and its box-and-whisker plot for right-skewed data.16 Frequency polygon and its box-and-whisker plot for left-skewed data.167Chapter 5: Probability analysis in the normal distribution Data Area to left of No.of standard point data point (%) deviations at data point 1 5.$19,001 and $21,752 or a smaller range of upper values compared to the symmetrical distri- bution, which explains the higher mean value.From the graph the mode is about $20,500 with the frequency at about 24.

Thus in ascend- ing order we have the mean ($18,207), median ($19,001), and the mode ($20,500).Testing symmetry by a normal probability plot Another way to establish the symmetry of data is to construct a normal probability plot.This procedure is as follows: ● Organize the data into an ordered data array.● For each of the data points determine the area under the curve on the assumption that the data follows a normal distribution.For exam- ple, if there are 19 data points in the array then the curve has 20 portions.

(To divide a segment into n portions you need (n � 1) limits.For example, for 19 data values Table 5.2 gives the area under the curve and the corresponding value of z.Note that the value of z has the same numerical values moving from left to right and at the median, z is 0 since this is a standardized nor- mal distribution.

● Plot the data values on the y-axis against the z-values on the x-axis.If the graph is essentially a straight line with a positive slope then the data follows a normal distribution.If the graph is non-linear of a concave format then the data is right-skewed.If the graph has a convex format then the data is left-skewed.

The three normal probability plots that show clearly the profiles for the normal, right-, and left-skewed datasets for the consulting data of Figures 5.Percentiles and the number of standard deviations In Chapter 2, we used percentiles to divide up the raw sales data originally presented in Figure 1.

1 and then to position regional sales informa- tion according to its percentile value.Using the concept from the immediate previous para- graph, “Testing symmetry by a normal probability plot”, we can relate the percentile value and the number of standard deviations.3, in the column “z”, we show the number of stand- ard deviations going from �3.The next column, “percentile” gives the area to the left of this number of stand- ard deviations, which is also the percentile value on the basis the data follows a normal distribution, which we have demonstrated in the paragraph “Demonstrating that data follow a normal distribution” in this chapter.The third 168 Statistics for Business 8,000 10,000 12,000 14,000 16,000 18,000 20,000 22,000 24,000 �4.0000 Number of standard deviations Sa la ry Normal distribution Right-skewed Left-skewed Figure 5.z Percentile (%) Value ($) z Percentile (%) Value ($) z Percentile (%) Value ($) �3.3 Positioning of sales data, according to z and the percentile.column, “Value, $” is the sales amount corres- ponding to the number of standard deviations and also the percentile.1 the standard deviation, z � 1, for this sales data is $30,888.20 (let’s say $31 thousand) and the mean 102,666.Thus if sales are �1 standard deviations from the mean they would be approximately 103 � 31 � $134 thousand.3 the value is $135 thousand (rounding), or a negligible difference.Similarly a value of z � �1 puts the sales at 103 � 31 � $72 thousand.3 the value is $71 thousand which again is close.Thus, using the standard z-values we have a measure of the dispersion of the data.

This is another way of looking at the spread of information.In Chapter 4, we presented the binomial distri- bution.Under certain conditions, the discrete binomial distribution can be approximated by the continuous normal distribution, enabling us to perform sampling experiments for discrete data but using the more convenient normal dis- tribution for analysis.This is particularly useful for example in statistical process control (SPC).Conditions for approximating the binomial distribution The conditions for approximating the binomial distribution are that the product of the sample size, n, and the probability of success, p, is greater, or equal to five and at the same time the product of the sample size and the probability of failure is also greater than or equal to five.

That is, np 5 5(iv) n(1 � p) 5 5(v) From Chapter 4, equation 4(xv), the mean or expected value of the binomial distribution is, x � E(x) � np And from equation 4(xvii) the standard devi- ation of the binomial distribution is given by, When the two normal approximation conditions apply, then from using equation 5(ii) substitut- ing for the mean and standard deviation we have the following normal–binomial approximation: 5(vi) The following illustrates this application.Application of the normal–binomial approximation: Ceramic plates A firm has a continuous production operation to mould, glaze, and fire ceramic plates.It knows from historical data that in the operation 3% of the plates are defective and have to be sold at a marked down price.The quality control manager takes a random sample of 500 of these plates and inspects them.

Can we use the normal distribution to approxi- mate the normal distribution? The sample size n is 500, and the probability p is 3%.Using equations 5(iv) and 5(v) np � 500 * 0.03 � 15 or a value �5 n(1 � p) � 500 * (1 � 0.97 � 485 and again a value �5 Thus both conditions are satisfied and so we can correctly use the normal distribution as an approximation of the binomial distribution.

This gives the probability of exactly 20 plates being defective of 4.z x x np npq x np np p x x � � � � � � � ( )1 � � � � 2 1( ( )np p npq( )) Using a Normal Distribution to Approximate a Binomial Distribution 169Chapter 5: Probability analysis in the normal distribution 3.

Using the normal–binomial approximation what is the probability that 20 of the plates are defective? From equation 4(xv), the mean value of the binomial distribution is, x � np � 500 * 0.This gives the probability of exactly 20 plates being defective of 4.This is a value not much different from 4.(Note if we had used a cumulative value � 1 this would give the area from the left of the normal distribution curve to the value of x.) Continuity correction factor Now, the normal distribution is continuous, and is shown by a line graph, whereas the bino- mial distribution is discrete illustrated by a histogram.Another way to make the normal– binomial approximation is to apply a continuity correction factor so that we encompass the range of the discrete value recognizing that we are superimposing a histogram to a continuous curve.

In the previous ceramic plate example, if we apply a correction factor of 0.5–20, the ran- dom variable x then on the lower side we have x1 � 19.Using equation 5(vi) for these two values of x gives z x np np p 1 1 1 19 5 500 0 03 500 0 03 1 0 0 � � � � � �( ) .)500 0 003 0 997 170 Statistics for Business Fr eq ue nc y of o cc ur re nc e 15 x19.1797 gives the area under the curve from the left to x � 19.The difference between these two areas is 4.This value is again close to those values obtained in the worked example for the ceramic plates.Sample size to approximate the normal distribution The conditions that equations 5(iv) and 5(v) are met depend on the values of n and p.When p is large then for a given value of n the product np is large; conversely n(1 � p) is small.The mini- mum sample size possible to apply the normal– binomial approximation is 10.In this case the probability, p, must be equal to 50% as for exam- ple in the coin toss experiment.

As the probabil- ity p increases in value, (1 � p) decreases and so for the two conditions to be valid the sample size n has to be larger.If for example p is 99%, then the minimum sample size in order to apply the normal distribution assumption is 500 illus- trated as follows: p � 99% and thus, np � 500 * 99% � 495 (1 � p) � 1% and thus, n(1 � p) � 500 * 1% � 5 Figure 5.19 gives the relationship of the min- imum values of the sample size, n, for values of p from 10% to 90% in order to satisfy both equa- tions 5(iv) and 5(v).z x np np p 2 2 1 20 5 500 0 03 500 0 03 1 0 0 � � � � � �( ) .� � � � 171Chapter 5: Probability analysis in the normal distribution 0 5 10 15 20 25 30 35 40 45 50 55 0% 5% 10% 15% 20% 25% 30% 35% 40% 45% 50% 55% 60% 65% 70% 75% 80% 85% 90% 95% Probability, p Sa m pl e siz e, n (u nit s) Figure 5.19 Minimum sample size in a binomial situation to be able to apply the normal distribution assumption.172 Statistics for Business This chapter has been entirely devoted to the normal distribution.

Describing the normal distribution The normal distribution is the most widely used analytical tool in statistics and presents graphically the profile of a continuous random variable.Situations which might follow a normal distribution are those processes that are set to produce products according to a target or a mean value such as a bottle filling operation, the filling of yogurt pots, or the pouring of liquid chocolate into a mould.Simply because of the nature, or randomness of these operations, we will find volume or weight val- ues below and above the set target value.Visually, a normal distribution is bell or hump shaped and is symmetrical around this hump such that the left side is a mirror image of the right side.The central point of the hump is at the same time the mean, median, mode, and midrange.

The left and right extremities, or the two tails of the normal distribution, may extend far from the central point.No matter the value of the mean or the standard deviation, the area under the curve of the normal distribution is always unity.26% of all the data falls within �1 standard deviations from the mean, 95.44% of the data falls within �2 standard deviations from the mean, and 99.

73% of data is �3 standard deviations from the mean.These empirical relationships allow the normal distribu- tion to be used to determine probability outcomes of many situations.Data in a normal distribution can be uniquely defined by its mean value and standard deviation and these values define the shape or kurtosis of the distribution.A distribution that has a small standard deviation relative to its mean has a sharp peak or is leptokurtic.A distribution that has a large standard deviation relative to its mean has a flat peak and is platykurtic.

A distribution between these two extremes is mesokurtic.The importance of knowing these shapes is that a curve that is leptokurtic is more reliable for analytical purposes.When we know the values of the mean value, , the standard deviation, , and the random variable, x, of a dataset we can trans- form the absolute values of the dataset into standard values.This then gives us a standard normal distribution which has a mean value of 0 and plus or minus values of z, the number of standard deviations from the mean corresponding to the area under the curve.Demonstrating that data follow a normal distribution To verify that data follows a normal distribution there are several tests.

We can develop a stem- and-leaf display if the dataset is small.For larger datasets we can draw a box-and-whisker plot, or plot a frequency polygon, and see if these displays are symmetrical.Additionally, we can determine the properties of the data to see if the mean is about equal to the median, that the inter-quartile range is equal to 1.33 times the standard deviation, that the data range is about six times the stand- ard deviation, and that the empirical rules governing the number of standard deviations and the area under the curve are respected.

If the mean and median value in a dataset are significantly dif- ferent then the data is asymmetric or skewed.

When the mean is greater than the median the distribution is positively or right-skewed, and when the mean is less than the median the distri- bution is negatively or left-skewed.A more rigorous test of symmetry involves developing a normal C ha pt er S um m ar y 173Chapter 5: Probability analysis in the normal distribution probability plot which involves organizing the data into an ordered array and determining the val- ues of z for defined equal portions of the data.If the normal probability plot is essentially linear with a positive slope, then the data is normal.If the plot is non-linear and concave then the data is right-skewed, and if it is convex then the data is left-skewed.Since we have divided data into defined portions, the normal probability plot is related to the data percentiles.

A normal distribution to approximate a binomial distribution When both the product of sample size, n, and probability, p, of success and the product of sample size and probability of failure (1 � p) are greater or equal to five then we can use a normal distribu- tion to approximate a binomial distribution.This condition applies for a minimum sample size of 10 when the probability of success is 50%.For other probability values the sample size will be larger.This normal–binomial approximation has practicality in sampling experiments such as statistical process control.Renault trucks Situation Renault Trucks, a division of Volvo Sweden, is a manufacturer of heavy vehicles.It is interested in the performance of its Magnum trucks that it sells throughout Europe to both large and smaller trucking companies.Based on service data throughout the Renault agencies in Europe it knows that on an annual basis the distance travelled by its trucks, before a major overhaul is necessary, is 150,000 km with a standard deviation of 35,000 km.The data is essentially normally distributed, and there were 62,000 trucks in the analysis.What proportion of trucks can be expected to travel between 82,000 and 150,000 km per year? 2.What is the probability that a randomly selected truck travels between 72,000 and 140,000 km per year? 3.What percentage of trucks can be expected to travel no more than 50,000 km per year and at least 190,000 km per year? 4.How many of the trucks in the analysis, are expected to travel between 125,000, and 200,000 km in the year? 5.In order to satisfy its maintenance and quality objectives Renault Trucks desires that at least 75% of its trucks travel at least 125,000 km.

Does Renault Trucks reach this objective? Justify your answer by giving the distance at which at least 75% of the trucks travel.90% of the trucks are expected to travel? 7.For analytical purposes for management, develop a greater than ogive based on the data points developed from Questions 1–6.

Telephone calls Situation An analysis of 1,000 long distance telephone calls made from a large business office indicates that the length of these calls is normally distributed, with an average time of 240 seconds, and a standard deviation of 40 seconds.What percentage of these calls lasted no more than 180 seconds? 2.What is the probability that a particular call lasted between 180 and 300 seconds? 3.

How many calls lasted no more than 180 seconds and at least 300 seconds? 4.What percentage of these calls lasted between 110 and 180 seconds? 5.What is the length of a particular call, such that only 1% of all calls are shorter? EXERCISE PROBLEMS 175Chapter 5: Probability analysis in the normal distribution 3.Training programme Situation An automobile company has installed an enterprise resource planning (ERP) system to better manage the firm’s supply chain.The human resource department has been instructed to develop a training programme for the employees to fully understand how the new system functions.

This training programme has a fixed lecture period and at the end of the programme there is a self-paced on-line practical examination that the partici- pants have to pass before they are considered competent with the new ERP system.If they fail the examination they are able to retake it as many times as they wish in order to pass.When the employee passes the examination they are considered competent with the ERP system and they immediately receive a 2% salary increase.During the last several months, average completion of the programme, which includes passing the examination, has been 56 days, with a standard deviation of 14 days.The time taken to pass the exam- ination is considered to follow a normal distribution.

What is the probability that an employee will successfully complete the programme between 40 and 51 days? 2.What is the probability an employee will successfully complete programme in 35 days or less? 3.What is the combined probability that an employee will successfully complete the programme in no more than 34 days or more than 84 days? 4.What is the probability that an employee will take at least 75 days to complete the training programme? 5.

What are the upper and lower limits in days within which 80% of the employees will successfully complete the programme? 4.Cashew nuts Situation Salted cashew nuts sold in a store are indicated on the packaging to have a nominal net weight of 125 g.Tests at the production site indicate that the average weight in a pack- age is 126.If you buy a packet of these cashew nuts at a store, what is the probability that your packet will contain more than 127 g? 2.If you buy a packet of these cashew nuts at a store, what is the probability that your packet will contain less than the nominal indicated weight of 125 g? 3.What is the minimum and maximum weight of a packet of cashew nuts in the mid- dle 99% of the cashew nuts? 4.In the packets of cashew nuts, 95% will contain at least how much in weight? 176 Statistics for Business 5.

Publishing Situation Cathy Peck is the publishing manager of a large textbook publishing house in England.Based on passed information she knows that it requires on average, 10.5 months to pub- lish a book from receipt of manuscript from the author to getting the book on the mar- ket.She also knows that from past publishing data a normal distribution represents the distribution time for publication, and that the standard deviation for the total process from review, publication, to distribution is 3.In a certain year she is told that she will receive 19 manuscripts for publication.From the manuscripts she is promised to receive this year for publication, approxi- mately how many can Cathy expect to publish within the first quarter? 2.From the manuscripts she is promised to receive this year for publication, approxi- mately how many can Cathy expect to publish within the first 6 months? 3.From the manuscripts she is promised to receive this year for publication, approxi- mately how many can Cathy expect to publish within the third quarter? 4.

From the manuscripts she is promised to receive this year for publication, approxi- mately how many can Cathy expect to publish within the year? 5.If by the introduction of new technology, the publishing house can reduce the aver- age publishing time and the standard deviation by 30%, how many of the 19 manu- scripts could be published within the year? 6.Gasoline station Situation A gasoline service sells, on average 5,000 litre of diesel oil per day.The standard devia- tion of this sale is 105 litre per day.The assumption is that the sale of diesel oil follows a normal distribution.

What is the probability that on a given day, the gas station sells at least 5,180 litre? 2.What is the probability that on a given day, the gas station sells no more than 4,850 litre? 3.What is the probability that on a given day, the gas station sells between 4,700 and 5,200 litre? 4.What is the volume of diesel oil sales at which the sales are 80% more? 5.

The gasoline station is open 7 days a week and diesel oil deliveries are made once a week on Monday morning.To what level should diesel oil stocks be replenished if the owner wants to be 95% certain of not running out of diesel oil before the next deliv- ery? Daily demand of diesel oil is considered reasonably steady.177Chapter 5: Probability analysis in the normal distribution 7.Ping-pong balls Situation In the production of ping-pong balls the mean diameter is 370 mm and their standard deviation is 0.The size distribution of the production of ping-pong balls is con- sidered to follow a normal distribution.What percentage of ping-pong balls can be expected to have a diameter that is between 369 and 370 mm? 2.What is the probability that the diameter of a randomly selected ping-pong ball is between 372 and 369 mm? 3.What is the combined percentage of ping-pong balls can be expected to have a diam- eter that is no more than 368 mm or is at least 371 mm? 4.

If there are 25,000 ping-pong balls in a production lot how many of them would have a diameter between 368 and 371 mm? 5.What is the diameter of a ping-pong ball above which 75% are greater than this diameter? 6.What are the symmetrical limits of the diameters between which 90% of the ping- pong balls would lie? 7.What can you say about the shape of the normal distribution for the production of ping-pong balls? 8.Marmalade Situation The nominal net weight of marmalade indicated on the jars is 340 g.

The filling machines are set to the nominal weight and the standard deviation of the filling operation is 3.What percentage of jars of marmalade can be expected to have a net weight between 335 and 340 g? 2.What percentage of jars of marmalade can be expected to have a net weight between 335 and 343 g? 3.

What is the combined percentage of jars of marmalade that can be expected to have a net weight that is no more than 333 g and at least 343 g? 4.If there are 40,000 jars of marmalade in a production lot how many of them would have a net weight between 338 and 345 g? 5.What is the net weight of jars of marmalade above which 85% are greater than this net weight? 6.What are the symmetrical limits of the net weight between which 99% of the jars of marmalade lie? 7.

The jars of marmalade are packed in cases of one dozen jars per case.

What propor- tion of cases will be above 4.1 kg in net weight? 178 Statistics for Business 9.Restaurant service Situation The profitability of a restaurant depends on how many customers can be served and the price paid for a meal.Thus, a restaurant should service the customers as quickly as pos- sible but at the same time providing them quality service in a relaxed atmosphere.A cer- tain restaurant in New York, in a 3-month study, had the following data regarding the time taken to service clients.

It believed that it was reasonable to assume that the time taken to service a customer, from showing to the table and seating, to clearing the table after the client had been serviced, could be approximated by a normal distribution.What is the average time and standard deviation to serve a customer such that the restaurant can then receive another client? 2.What is the probability that a customer can be serviced between 90 and 125 minutes? 3.What is the probability that a customer can be serviced between 70 and 140 minutes? 4.

What is the combined probability that a customer can be serviced in 70 minutes or less and at least 140 minutes? 5.If in the next month it is estimated that 1,200 customers will come to the restaurant, to the nearest whole number, what is a reasonable estimate of the number of customers that can be serviced between 70 and 140 minutes? 6.Again, on the basis that 1,200 customers will come to the restaurant in the next month, 85% of the customers will be serviced in a minimum of how many minutes? 10.Yoghurt Situation The Candy Corporation has developed a new yogurt and is considering various prices for the product.Marketing developed an initial daily sales estimate of 2,400 cartons, with a standard deviation of 45.

Prices for the yogurt were then determined based on that forecast.A later revised estimate from marketing was that average daily sales would be 2,350 cartons.Activity Average time (minutes) Variance Showing to table, and seating client 4.7744 179Chapter 5: Probability analysis in the normal distribution Required 1.According to the revised estimate, what is the probability that a day’s sale will still be over 2,400 given that the standard deviation remains the same? 2.According to the revised estimate, what is the probability that a day’s sale will be at least 98% of 2,400? 11.Motors Situation The IBB Company has just received a large order to produce precision electric motors for a French manufacturing company.

To fit properly, the drive shaft must have a diameter of 4.The production manager indicates that in inventory there is a large quan- tity of steel rods with a mean diameter of 4.What is the probability of a steel rod from this inventory stock, meeting the drive shaft specifications? 12.Doors Situation A historic church site wishes to add a door to the crypt.The door opening for the crypt is small and the church officials want to enlarge the opening such that 95% of visitors can pass through without stooping.

Statistics indicate that the adult height is normally distributed, with a mean of 1.Based on the design criterion, what height should the doors be made to the nearest cm? 2.

If after consideration, the officials decided to make the door 2 cm higher than the value obtained in Question 1, what proportion of the visitors would have to stoop when going through the door? 13.

Machine repair Situation The following are the three stages involved in the servicing of a machine.Activity Mean time (minutes) Standard deviation (minutes) Dismantling 20 4 Testing and adjusting 30 7 Reassembly 15 3 180 Statistics for Business Required 1.What is the probability that the dismantling time will take more than 28 minutes? 2.What is the probability that the testing and adjusting activity alone will take less than 27 minutes? 3.What is the probability that the reassembly activity alone will take between 13 and 18 minutes? 4.

What is the probability that an allowed time of 75 minutes will be sufficient to complete the servicing of the machine including dismantling, testing and adjusting, and assembly? 14.Savings Situation A financial institution is interested in the life of its regular savings accounts opened at its branch.This information is of interest as it can be used as an indicator of funds available for automobile loans.An analysis of past data indicates that the life of a regular savings account, maintained at its branch, averages 17 months, with a standard deviation of 171 days.For calculation purposes 30 days/month is used.

The distribution of this past data was found to be approximately normal.If a depositor opens an account with this savings institution, what is the probability that there will still be money in that account in 20 months? 2.What is the probability that the account will have been closed within 2 years? 3.What is the probability that the account will still be open in 2.

What is the chance an account will be open in 3 years? 15.Buyout – Part III Situation Carrefour, France, is considering purchasing the total 50 retail stores belonging to Hardway, a grocery chain in the Greater London area of the United Kingdom.The profits from these 50 stores, for one particular month, in £ ’000s, are as follows.(This is the same information as provided in Chapters 1 and 2.

5 181Chapter 5: Probability Analysis in the Normal Distribution Required 1.Carrefour management decides that it will purchase only those stores showing profits greater than £12,500.On the basis that the data follow a normal distribution, calculate how many of the Hardway stores Carrefour would purchase? (You have already calcu- lated the mean and the standard deviation in the Exercise Buyout – Part II in Chapter 2.How does the answer to Question 1 compare to the answer to Question 6 of buyout in Chapter 1 that you determined from the ogive? 3.

What are your conclusions from the answers determined from both methods.Case: Cadbury’s chocolate Situation One of the production lines of Cadbury Ltd turns out 100-g bars of milk chocolate at a rate of 20,000/hour.The start of this production line is a stainless steel feeding pipe that delivers the molten chocolate, at about 80°C, to a battery of 10 injection nozzles.These nozzles are set to inject a little over 100 g of chocolate into flat trays which pass under- neath the nozzles.

Afterwards these trays move along a conveyer belt during which the chocolate cools and hardens taking the shape of the mould.In this cooling process some of the water in the chocolate evaporates in order that the net weight of the chocolate comes down to the target value of 100 g.At about the middle of the conveyer line, the moulds are turned upside down through a reverse system on the belt after which the belt vibrates slightly such that the chocolate bars are ejected from the mould.The next pro- duction stage is the packing process where the bars are first wrapped in silver foil then wrapped in waxed paper onto which is printed the product type and the net weight.The final part of this production line is where the individual bars of chocolate are packed in cardboard cartons.

Immediately upstream of the start of the packing process, the bars of chocolate pass over an automatic weighing machine that measures at random the indi- vidual weights.A printout of the weights for a sample of 1,000 bars, from a production run of 115,000 units is given in the table below.The production cost for these 100 g chocolate bars is £0.Required From the statistical sample data presented, how would you describe this operation? What are your opinions and comments? 109.31 183Chapter 5: Probability analysis in the normal distribution 84.42 This page intentionally left blank 6Theory and methods of statistical sampling The sampling experiment was badly designed! A well-designed sample survey can give pretty accurate predictions of the requirements, desires, or needs of a population.However, the accuracy of the survey lies in the phrase “well-designed”.A classic illustration of sampling gone wrong was in 1948 during the presidential election campaign when the two candidates were Harry Truman, the Democratic incumbent and Governor Dewey of New York, the Republican candidate.The Chicago Tribune was “so sure” of the outcome that the headlines in their morning daily paper of 3 November 1948 as illustrated in Figure 6.

In fact Harry Truman won by a narrow but decisive victory of 49.5% of the popular vote to Dewey’s 45% and with an electoral margin of 303 to 189.The Chicago Tribune had egg on their face; something went wrong with the design of their sample experiment!1,2 1 Chicago Daily Tribune, 3 November 1948.

(1982), America in the Twentieth Century, 5th edition, McGraw Hill, New York, pp.1 Harold Truman holding aloft a copy of the November 3rd 1948 morning edition of the Chicago Tribune.

In business, and even in our personal life, we often make decisions based on limited data.What we do is take a sample from a population and then make an inference about the population charac- teristics, based entirely on the analysis of this sample.For example, when you order a bottle of wine in a restaurant, the waiter pours a small quantity in your glass to taste.Based on that small quantity of wine you accept or reject the bottle of wine as drinkable.The waiter would hardly let you drink the whole bottle before you decide it is no good! The United States Dow Jones Industrial Average consists of just 30 stocks but this sample average is used as a measure of eco- nomic changes when in reality there are hun- dreds of stocks in the United States market where millions of dollars change hands daily.

In political elections, samples of people’s voting intentions are made and based on the propor- tion that prefer a particular candidate, the expected outcome of the nation’s election may be presented beforehand.In manufacturing, lots of materials, assemblies, or finished products are sampled at random to see if pieces conform to appropriate specifications.If they do, the assump- tion is that the entire population, the production line or the lot from where these samples are taken, meet the desired specifications and so all the units can be put onto the market.And, how many months do we date our future spouse before we decide to spend the rest of our life together! The usual purpose of taking and analysing a sample is to make an estimate of the population parameter.As the sample size is smaller than the population we have no guarantee of the population param- eter that we are trying to measure, but from the sample analysis, we draw conclusions.If we really wanted to guarantee our conclusion we would have to analyse the whole population but 187Chapter 6: Theory and methods of statistical sampling Statistical Relationships in Sampling for the Mean After you have studied this chapter you will understand the theory, application, and practical methods of sampling, an important application of statistical analysis.The topics are broken down according to the following themes: ✔ Statistical relations in sampling for the mean • Sample size and population • Central limit theory • Sample size and shape of the sampling distribution of the means • Variability and sample size • Sample mean and the standard error.✔ Sampling for the means for an infinite population • Modifying the normal transformation relationship • Application of sampling from an infinite normal population: Safety valves ✔ Sampling for the means from a finite population • Modification of the standard error • Application of sampling from a finite population: Work week ✔ Sampling distribution of the proportion • Measuring the sample proportion • Sampling distribution of the proportion • Binomial concept in sampling for the proportion • Application of sampling for proportions: Part-time workers ✔ Sampling methods • Bias in sampling • Randomness in your sample experiment • Excel and random sampling • Systematic sampling • Stratified sampling • Several strata of interest • Cluster sampling • Quota sampling • Consumer surveys • Primary and secondary data L e a r n i n g o b j e c t i v e s in most cases this is impractical, too costly, takes too long, or is clearly impossible.An alternative to inferential statistics is descriptive statistics which involves the collection and analysis of the dataset in order to characterize just the sampled dataset.

Sample size and population A question that arises in our sampling work to infer information about the population is what should be the size of the sample in order to make a reliable conclusion? Clearly the larger the sample size, the greater is the probability of being close to estimating the correct population parameter, or alternatively, the smaller is the risk of making an inappropriate estimate.To demonstrate the impact of the sample size, consider an experiment where there is a population of seven steel rods, as shown in Figure 6.The number of the rod and its length in centimetres is indicated in Table 6.The total length of these seven rods is 35 cm (2 � 3 � 4 � 5 � 6 � 6 � 9).This translates into a mean value of the length of the rods of 5 cm (35/7).If we take samples of these rods from the population, with replacement, then from the counting relations in Chapter 3, the possible combinations of rods that can be taken, the same rod not appearing twice in the sample, is given by the relationship, 3(xvi) Here, n is the size of the population, or in this case 7 and x is the size of the sample.For example, if we select a sample of size of 3, the number of possible different combinations from equation 3(xvi) is, If we increase the sample sizes from one to seven rods, then from equation 3(xvi) the total possible number of different samples is as given in Table 6.Thus, we sample from the popul- ation first with a sample size of one, then two, three, etc.Each time we select a sample we determine the sample mean value of the length of rods selected.For example, if the sample size is 3 and rods of length 2, 4, and 6 cm are selected, then the mean length, x–, of the sample would be, 2 4 6 3 4 00 � � � .cm Combinations 7! 3!(7 )! 7! 3! 4! 7 6 5 4 3 2 � � � � 3 * * * * * * ** * * * * * * 1 35 3 2 1 4 3 2 1 � Combinations ! !( )! � � n x n x 188 Statistics for Business Figure 6.

2 Seven steel rods and their length in centimetres.9cm 6cm 6cm 5cm 4cm 3cm 2cm Rod 1 2 3 4 5 6 7 number Rod 2.of possible 7 21 35 35 21 7 7 different samples Table 6.2 Number of samples from a population of seven steel rods.The possible combinations of rod sizes for the seven different samples are given in Table 6.

(Note that there are two rods of length 6 cm.) For a particular sample size, the sum of all the sample means is then divided by the number of samples withdrawn to give the mean value of the samples or, x=.For example, for a sample of size 3, the sum of the sample means is 175 and 189Chapter 6: Theory and methods of statistical sampling No.Size � 1 Size � 2 Size � 3 Size � 4 Size � 5 Size � 6 Size � 7 Mean Mean Mean Mean Mean Mean Mean 1 2 2.3 Samples of size 1 to 7 taken from a population of size 7.this number divided by the sample number of 35 gives 5.These values are given at the bottom of Table 6.What we conclude is that the sam- ple means are always equal to 5 cm, or exactly the same as the population mean.

Next, for each sample size, a frequency distri- bution of mean length is determined.The left-hand column gives the sample mean and the other columns give the number of occurrences within a class limit according to the sample size.For example, for a sample size of four there are four sample means greater than 4.

This data is now plotted as a frequency histogram in Figures 6.9 where each of the seven histograms have the same scale on the x-axis.

9 we can see that as the sample size increases from one to seven, the dis- persion about the mean value of 5 cm becomes smaller or alternatively more sample means lie closer to the population mean.For the sample size of seven, or the whole population, the dis- persion is zero.The mean of the sample means, x=, is always equal to the population mean of 5 or they have the same central tendency.

This experiment demonstrates the concept of the central limit theory explained in the following section.Central limit theory The foundation of sampling is based on the cen- tral limit theory, which is the criterion by which information about a population parameter can be inferred from a sample.The central limit theory states that in sampling, as the size of the sample increases, there becomes a point when the dis- tribution of the sample means, , can be approxi- mated by the normal distribution.This is so even though the distribution of the population itself may not necessarily be normal.The distribution of the sample means, also called sampling distribution of the means, is a probability distribution of all the possible means of samples taken from a population.

This concept of sampling and sampling means is illustrated by the information in Table 6.Here the produc- tion line is producing 500,000 chocolate bars, and this is the population value, N.The mould- ing for the chocolate is set such that the weight of each chocolate bar should be 100 g.This is the nominal weight of the chocolate bar and is x 190 Statistics for Business Sample Sample size mean x= 1 2 3 4 5 6 7 2.

00 1 0 0 0 0 0 0 Total 7 21 35 35 21 7 1 Table 6.4 Frequency distribution within sample means for different sample sizes.

191Chapter 6: Theory and methods of statistical sampling Figure 6.3 Samples of size 1 taken from a population of size 7.0 0 Mean length of rod (cm) Fr eq ue nc y o f t hi s l en gt h Figure 6.4 Samples of size 2 taken from a population of size 7.0 0 Mean length of rod (cm) Fr eq ue nc y o f t hi s l en gt h 192 Statistics for Business Figure 6.

5 Samples of size 3 taken from a population of size 7.0 0 Mean length of rod (cm) Fr eq ue nc y o f t hi s l en gt h Figure 6.6 Samples of size 4 taken from a population of size 7.0 0 Mean length of rod (cm) Fr eq ue nc y o f t hi s l en gt h 193Chapter 6: Theory and methods of statistical sampling Figure 6.7 Samples of size 5 taken from a population of size 7.

0 0 Mean length of rod (cm) Fr eq ue nc y o f t hi s l en gt h Figure 6.8 Samples of size 6 taken from a population of size 7.0 0 Mean length of rod (cm) Fr eq ue nc y o f t hi s l en gt h the population mean, .For quality control pur- poses an inspector takes 10 random samples from the production line in order to verify that the weight of the chocolate is according to speci- fications.Each bar in the sample is weighed and these individual weights, and the mean weight of each sample, are recorded.48 g, and the weight of the 15th bar is 98.

The mean weight of this first sample, x–1, is 99.The mean weight of the 10th sample, x–10, is 100.The mean value of the means of all the 10 samples, x= is 99.The values of x– plotted in a frequency distribution would give a sampling distribution of the means (though only 10 values are insufficient to show a correct distribution).Sample size and shape of the sampling distribution of the means We might ask, what is the shape of the sampling distribution of the means? From statistical experi- ments the following has been demonstrated: ● For most population distributions, regardless of their shape, the sampling distribution of the means of samples taken at random from the population will be approximately normally distributed if samples of at least a size of 30 units each are withdrawn.● If the population distribution is symmetrical, the sampling distribution of the means of samples taken at random from the population will be approximately normal if samples of at least 15 units are withdrawn.

● If the population is normally distributed, the sampling distribution of the means of samples 194 Statistics for Business Figure 6.9 Samples of size 7 taken from a population of size 7.0 0 Mean length of rod (cm) Fr eq ue nc y o f t hi s l en gt h taken at random from the population will be normally distributed regardless of the sample size withdrawn.The practicality of these relationships with the central limit theory is that by sampling, either from non-normal populations or normal populations, inferences can be made about the population parameters without having information about the shape of the population distribution other than the information obtained from the sample.Variability and sample size Consider a large organization such as a govern- ment unit that has over 100,000 employees.

This is a large enough number so that it can be considered infinite.Assume that the distribution of the employee salaries is considered normal with an average salary of $40,000.Sampling of individual salaries is made using random com- puter selection: ● Assume a random sample of just one salary value is selected that happens to be $90,000.This value is a long way from the mean value of $40,000.

● Assume now that random samples of two salaries are taken which happen to be $60,000 and $90,000.

The average of these is $75,000 (60,000 � 90,000)/2 .This is still far from $40,000 but closer than in the case of a single sample.195Chapter 6: Theory and methods of statistical sampling ● Company is producing a lot (population) of 500,000 chocolate bars ● Nominal weight of each chocolate bar is 100 g ● To verify the weight of the population, an inspector takes 10 random samples from production ● Each sample contains 15 slabs of chocolate ● Mean value of each sample is determined.This is x– ● Mean value of all the x– is x= or 99.85 g ● A distribution can be plotted with x– on the x-axis.

Sample number 1 2 3 4 5 6 7 8 9 10 1 100.● If now random samples of five salaries $60,000, $90,000, $45,000, $15,000, and $20,000 come up, the mean value of these is $46,000 or closer to the population aver- age of $40,000.Thus, by taking larger samples there is a higher probability of making an estimate close to the population parameter.

Alternatively, increasing the sample size reduces the spread or variability of the average value of the samples taken.Sample mean and the standard error The mean of a sample is x– and the mean of all possible samples withdrawn from the popu- lation is x=.From the central limit theory, the mean of the entire sample means taken from the population can be considered equal to the population mean, x: =x � x 6(i) And because of this relationship in equation 6(i), the arithmetic mean of the sample is said to be an unbiased estimator of the population mean.By the central limit theory, the standard devi- ation of the sampling distribution, x-, is related to the population standard deviation, x, and the sample size, n, by the following relationship: 6(ii) This indicates that as the size of the sample increases, the standard deviation of the sam- pling distribution decreases.The standard devi- ation of the sampling distribution is more usually referred to as the standard error of the sample means, or more simply the standard error as it represents the error in our sampling experiment.

For example, going back to our illustration of the salaries of the government employees, if we take a series of samples from the employees and measure each time, the x– value of salaries, we will almost certainly have different values each time simply because the chances are that our salary numbers in our sample will be different.That is, the difference between each sample, among the several samples, and the population causes variability in our analysis.This variabil- ity, as measured by the standard error of equa- tion 6(ii), is due to the chance or sampling error in our analysis between the samples we took and the population.The standard error indi- cates the magnitude of the chance error that has been made, and also the accuracy when using a sample statistic to estimate the population parameter.A distribution of sample means that has less variability, or less spread out, as evi- denced by a small value of the standard error, is a better estimator of the population parameter than a distribution of sample means that is widely dispersed with a larger standard error.

As a comparison to the standard error, we have a standard deviation of a population.This is not an error but a deviation that is to be expected since by their very nature, populations show variation.There are variations in the age of people, variations in the volumes of liquid in cans of soft drinks, variations in the weights of a nominal chocolate bar, variations in the per capita income of individuals, etc.These com- parisons are illustrated in Figure 6.10, which shows the shape of a normal distribution with its standard deviation, and the corresponding profile of the sample distribution of the means with its standard error.

An infinite population is a collection of data that has such a large size that sampling from an infinite population involving removing or destroying some of the data elements does not significantly impact the population that remains.x x n � 196 Statistics for Business Sampling for the Means from an Infinite Population Modifying the normal transformation relationship In Chapter 5, we have shown that the standard relationship between the mean, x, the standard deviation, x, and the random variable, x, in a normal distribution is as follows: 5(ii) An analogous relationship holds for the sampling distribution as shown in the lower distribution of Figure 6.10 where now: ● the random variable x is replaced by the sam- ple mean x–; ● the mean value x is replaced by the sample mean x=; ● the standard deviation of the normal distribu- tion, x, is replaced by the standard deviation of the sample distribution or the sample error, x-.The standard equation for the sampling distri- bution of the means now becomes, 6(iii) Substituting from equations 6(i) to 6(iii), the stan- dard equation then becomes, 6(iv) This relationship can be used using the four normal Excel functions already presented in Chapter 4, except that now the mean value of the sample mean, x–, replaces the random vari- able, x, of the population distribution, and the standard error of the sampling distribution replaces the standard deviation of the population.The following application illustrates the use of this relationship.

x n z x x x x nx x x x x � � � � � � z x x xx x x � � � � z x x x � � 197Chapter 6: Theory and methods of statistical sampling Figure 6.10 Population distribution and the sampling distribution.Population distribution Sampling distribution Mean � �x Standard deviation � �x x x Mean � x � �x � �x n � x Application of sampling from an infinite normal population: Safety valves A manufacturer produces safety pressure valves that are used on domestic water heaters.In the production process, the valves are automatically preset so that they open and release a flow of water when the upstream pressure in a heater exceeds 7 bars.In the manufacturing process there is a tolerance in the setting of the valves and the release pressure of the valves follows a normal distribution with a standard deviation of 0.

What proportion of randomly selected valves has a release pressure between 6.1 bars? Here we are only considering a single valve, or a sample of size 1, from the population between 6.

Thus, the probability that a randomly selected valve has a release pressure between 6.If many random samples of size eight were taken, what proportion of sample means would have a release pressure between 6.1 bars? Here now we are sampling from the normal population with a sample size of 8.Using equation 6(ii) the standard error is, Using this value in equation 6(iv) when x– � 6.Thus, the propor- tion of sample means that would have a release pressure between 6.

If many random samples of size 20 were taken, what proportion of sample means would have a release pressure between 6.1 bars? Here now we are sampling from the popu- lation with a sample size of 20.Using equation 6(ii) the standard error is, Using this value in equation 6(iv) when x– � 6.

x x n � � � � 0 3 20 0 3 4 4721 0 0671 .z x n x x � � � � � � 7 1 7 0 0 1061 0 1 0 1061 0 9425 .z x n x x � � � � �� �� 6 8 7 0 0 1061 0 2 0 1061 1 8850 .x x n � � � � 0 3 8 0 3 2 8284 0 1061 .z x x x � � � � � � 7 1 7 0 0 3 0 1 0 3 0 3333 .z x x x � � � � �� �� 6 8 7 0 0 3 0 2 0 3 0 6667 .198 Statistics for Business gives the area under the curve from the left of 0.Thus, the proportion of sample means that would have a release pressure between 6.If many random samples of size 50 were taken, what proportion of sample means would have a release pressure between 6.1 bars? Here now we are sampling from the popu- lation with a sample size of 50.Using equation 6(ii) the standard error is, Using this value in equation 6(iv) when x– � 6.Thus, the proportion of sample means that would have a release pressure between 6.To summarize this situation we have the results in Table 6.

6 and the concept is illustrated in the dis- tributions of Figure 6.What we observe is that not only the standard error decreases as the sam- ple size increases but there is a larger proportion between the values of 6.That is a larger cluster around the mean or target value of 7.Alternatively, as the sample size increases there is a smaller dispersion of the val- ues.For example, in the case of a sample size of 1 there is 37.81% of the data clustered around the values of 6.

In the case of a sample size of 50 there is 99.

08% clustered around the mean and only 0.Note, in applying these calcula- tions the assumption is that the sampling distribu- tions of the mean follow a normal distribution, and the relation of the central limit theory applies.A finite population is a collection of data that has a stated, limited, or small size.It implies that if one piece of the data from the population is z x n x x � � � � � � 7 1 7 0 0 0424 0 1 0 0424 2 3585 .z x n x x � � � � �� �� 6 8 7 0 0 0424 0 2 0 0424 4 714 .x x n � � � � 0 3 50 0 3 7 0711 0 0424 .z x n x x � � � � � � 7 1 7 0 0 0671 0 1 0 0671 1 4903 .199Chapter 6: Theory and methods of statistical sampling Sample size 1 8 20 50 Standard error, 0.Sampling for the Means from a Finite Population destroyed, or removed, there would be a signifi- cant impact on the data that remains.Modification of the standard error If the population is considered finite, that is the size is relatively small and there is sampling with replacement (after each item is sampled it is put back into the population), then we can use the equation for the standard error already presented, 6(ii) x x n � 200 Statistics for Business Figure 6.71% However, if we are sampling without replace- ment, the standard error of the mean is modi- fied by the relationship, 6(v) Here the term, 6(vi) is the finite population multiplier, where N is the population size, and n is the size of the sample.

This correction is applied when the ratio of n/N is greater than 5%.In this case, equation 6(iv) now becomes, 6(vii) The application of the finite population multi- plier is illustrated in the following application exercise.Application of sampling from a finite population: Work week A firm has 290 employees and records that they work an average of 35 hours/week with a stand- ard deviation of 8 hours/week.What is the probability that an employee selected at random will be working between �2 hours/week of the population mean? In this case, again we have a single unit (an employee) taken from the population where the standard deviation x is 8 hours/week.

34% or less than 5% and so the population multiplier is not needed.We know that the difference between the random variable and the population, (x � x) is equal to �2.For a value of (x � x) � �2 we have again from equation 6(ii), Or we could have simply concluded that z is �0.2500 since the assumption is that the curve follows a normal distribution, and a normal distribution is, by definition, sym- metrical.Thus, the probability that an employee selected at random will be working between �2 hours/week is, 59.If a sample size of 19 employees is taken, what is the probability that the sample means lies between �2 hours/week of the population mean? In this case, again we have a sample, n, of size 19 taken from a population, N, of size 290.The ratio n/N is, This ratio is greater than 5% and so we use the finite population multiplier in order to calculate the standard error.From equa- tion 6(vi), n N � � 19 290 0 0655 6 55.

%or of the population z x x x � � �� �� 2 8 0 2500.z x n x n N n N x x x x � � � � � � 1 N n N � �1 x x n N n N � � �1 201Chapter 6: Theory and methods of statistical sampling From equation 6(v) the corrected standard error of the distribution of the mean is, From equation 6(vii) where now x– � x � �2.Thus, the probability that the sample means lie between �2 hours/week is, 86.

74%, obtained for a sample of size 1, because as we increase the sample size, the sampling distribution of the means is clustered around the population mean.This concept is illus- trated in Figure 6.z x n N n N x x � � � � � �� 1 2 1 7773 1 1253 .

z x n N n N x x � � � � � � 1 2 1 7773 1 1253 .x x n N n N � � � � � � 1 8 19 0 9684 8 0 9684 4 3589 1 77 * .773 N n N � � � � � � � � 1 290 19 290 1 271 289 0 9377 0 9684 ( ) ( ) .96% Sample size � 19 �2 35 In sampling we may not be interested in an absolute value but in a proportion of the popu- lation.For example, what proportion of the popu- lation will vote conservative in the next United Kingdom elections? What proportion of the popu- lation in Paris, France has a salary more than €60,000 per year? What proportion of the houses in Los Angeles country in United States has a mar- ket value more than $500,000? In these cases, we have established a binomial situation.

In the United Kingdom elections either a person votes conserva- tive or he or she do not.In Paris either an individ- ual earns a salary more than €60,000/year, or they do not.In Los Angeles country, either the houses have a market value greater than $500,000 or they do not.In these types of situations we use sampling for proportions.Measuring the sample proportion When we are interested in the proportion of the population, the procedure is to sample from the population and then again use inferential statis- tics to draw conclusions about the population proportion.

The sample proportion, p–, is the ratio of that quantity, x, taken from the sample having the desired characteristic divided by the sample size, n, or, 6(viii) For example, assume we are interested in people’s opinion of gun control.We sample 2,000 people from the State of California and 1,450 say they are for gun control.The proportion in the sam- ple that says they are for gun control is thus 72.We might extend this sample experiment further and say that 72.

50% of the population of California is for gun control or even go further and conclude that 72.50% of the United States population is for gun control.However, these would be very uncertain conclu- sions since the 2,000-sample size may be nei- ther representative of California, and probably not of the United States.This experiment is binomial because either a person is for gun con- trol or is not.

Thus, the proportion in the sample that is against gun control is 27.

Sampling distribution of the proportion In our sampling process for the proportion, assume that we take a random sample and meas- ure the proportion having the desired character- istic and this is p–1.We then take another sample from the population and we have a new value p–2.If we repeat this process then we possibly will have different values of p–.

The probability distribution of all possible values of the sample proportion, p–, is the sampling distribution of the proportion.This is analogous to the sampling distribution of the means, x–, discussed in the previous section.Binomial concept in sampling for the proportion If there are only two possibilities in an outcome then this is binomial.In the binomial distribu- tion the mean number of successes, , for a sam- ple size, n, with a characteristic probability of success, p, is given by the relationship presented in Chapter 4: � np 4(xv) Dividing both sides of this equation by the sam- ple size, n, we have, 6(ix) The ratio /n is now the mean proportion of successes written as p–.Thus, 6(x) p p� n np n p� � p x n � 203Chapter 6: Theory and methods of statistical sampling Sampling Distribution of the Proportion Again from Chapter 4, the standard deviation of binomial distribution is given by the relationship, 4(xvii) where the value q � 1 � p And again dividing by n, 6(xi) where the ratio /n is the standard error of the proportion, p-, and thus, 6(xii) From equation 6(iv) we have the relationship, 6(iv) From Chapter 5, we can use the normal distri- bution to approximate the binomial distribution when the following two conditions apply: np 5 5(iv) n(1 � p) 5 5(v) That is, the products np and n(1 � p) are both greater or equal to 5.

Thus, if these criteria apply then by substituting in equation 6(iv) as follows, x–, the sample mean by the average sample proportion, p– x, the population mean by the population proportion, p x-, the standard error of the sample means by p-, the standard error of the proportion and using the relationship developed in equation 6(iii), we have, 6(xiii) Since from equation 6(xii), 6(xiv) Then, 6(xv) Alternatively, we can say that the difference between the sample proportion p– and the popu- lation proportion p is, 6(xvi) The application of this relationship is illustrated as follows.Application of sampling for proportions: Part-time workers The incidence of part-time working varies widely across Organization for Cooperation and Development (OECD) countries .The clear leader is the Netherlands where part-time employment accounts for 33% of all jobs.If a sample of 100 people of the work force were taken in the Netherlands, what proportion between 25% and 35%, in the sample, would be part-time workers? Now, the sample size is 100 and so we need to test again whether we can use the normal probability assumption by using equations 5(iv) and 5(v).

33 and n is 100, thus from equation 5(iv), np � 100 * 0.33 � 33 or greater than 5 From equation 5(v), n(1 � p) � 100(1 � 0.33) � 67 or again greater than 5 p p z p p n � � �( )1 z p p p p n � � �( )1 p pq n p p n � � �( )1 z x x x p p x x x p � � � � � � z x x x x x x � � � � p pq n p p n � � �( )1 n pqn n pqn n pq n p p n � � � � � 2 1( ) � � �npq np p( )1 204 Statistics for Business 3 Economic and financial indicators, The Economist, 20 July 2002, p.Thus, we can apply the normal probability assumption.The population proportion p is 33%, or 0.33, and thus from equation 6(xiv) the standard error of the proportion is, The lower sample proportion, p–, is 25%, or 0.The upper sample proportion, p–, is 35%, or 0.Thus, the proportion between 25% and 35%, in the sample, that would be part-time workers is, 66.If a sample of 200 people of the work force were taken in the Netherlands, what proportion between 25% and 35%, in the sample, would be part-time workers? First, we need to test whether we can use the normal probability assumption by using equations 5(iv) and 5(v).33 and n is 200, thus from equation 5(iv), np � 200 * 0.33 � 66 or greater than 5 From equation 5(v) n(1 � p) �200(1 � 0.33) � 134 or again greater than 5 Thus, we can apply the normal probabil- ity assumption.

The population proportion p is 33%, or 0.33, and thus from equation 6(xiv) the standard error of the proportion is, The lower sample proportion, p–, is 25%, or 0.The upper sample proportion, p–, is 35%, or 0.Thus, the proportion between 25% and 35% in a sample size of 200 that would be part-time workers is, 72.

7182 Note that again this value is larger than in the first situation since the sample size was 100 rather than 200.As for the mean, as the sample size increases the values will cluster around the mean value of the popu- lation.

Here the mean value of the propor- tion for the population is 33% and the z p p p � � � � �� � 0 35 0 33 0 0332 0 02 0 0332 0 6015 .z p p p � � � � �� �� 0 25 0 33 0 0332 0 0800 0 0332 2 40 .661 p � � � � � 0 33 1 0 33 200 0 33 0 67 200 0 0011 0 03 .332 z p p p � � � � �� � 0 35 0 33 0 0469 0 02 0 0469 0 4264 .z p p p � � � � �� �� 0 25 0 33 0 0469 0 0800 0 0469 1 70 .

558 p � � � � � 0 33 1 0 33 100 0 33 0 67 100 0 0022 0 04 .669 205Chapter 6: Theory and methods of statistical sampling sample proportions tested were 25% and 35% or both on the opposite side of the mean value of the proportion.This concept is illustrated in Figure 6.The purpose of sampling is to make reliable esti- mates about a population.

It is usually impossi- ble, and too expensive, to sample the whole population so that when a sampling experiment is developed it should as closely as possible paral- lel the population conditions.As the box opener “The sampling experiment was badly designed!” indicates, the sampling experiment to determine voter intentions was obviously badly designed.This section gives considerations when undertak- ing sampling experiments.Bias in sampling When you sample to make estimates of a popu- lation you must avoid bias in the sampling experiment.

Bias is favouritism, purposely or unknowingly, present in the sample data that gives lopsided, misleading, or unrepresentative results.

For example, you wish to obtain the voting intentions of the people in the United Kingdom and you sample people who live in the West End of London.This would be biased as the West End is pretty affluent and the voters sampled are more likely to vote Tory (Conservative).To measure the average intelligence quotient (IQ) of all the 18- year-old students in a country you take a sample of students from a private school.This would be biased because private school students often come from high-income families and their education level is higher.To measure the average income of residents of Los Angeles, California you take 206 Statistics for Business Figure 6.

62% Sample size � 200 33% Sampling Methods a sample of people who live in Santa Monica.This would be biased as people who live in Santa Monica are wealthy.

Randomness in your sample experiment A random sample is one where each item in the population has an equal chance of being selected.Assume a farmer wishes to determine the average weight of his 200 pigs.He samples the first 12 who come when he calls.They are probably the fittest – thus thinner than the rest! Or, a hotel manager wishes to determine the quality of the maid service in his 90-room hotel.If the maid works in order, then the first 15 probably were more thoroughly cleaned than the rest – the maid was less tired! These sampling experiments are not random and probably they are not representative of the population.In order to perform random sampling, you need a framework for your sampling experiment.For example, as an auditor you might wish to analyse 10% of the financial accounts of the firm to see if they conform to acceptable accounting practices.A business might want to sample 15% of its clients to obtain the level of customer satisfaction.A hotel might want to sample 12% of the condition of its hotel rooms to obtain a quality level of its operation.

You first create a ran- dom number in a cell and copy this to other cells.Suppose that as an auditor you have 630 accounts in your population and you wish to examine 10% of these accounts or 63.You then generate 63 random numbers between 1 and 630 and you examine those accounts whose numbers correspond to the numbers generated by the ran- dom number function.7 shows 63 random numbers within the range 1 to 630.Thus, you would examine those accounts corresponding to those numbers.The same procedure would apply to the farmer and his pigs.

Each pig would have identi- fication, either a tag, tattoo, or embedded chip giving a numerical indication from 1 to 200.The farmer would generate a list of 12 random numbers between 1 and 200 as indicated in Table 6.8, and weigh those 12 pigs that corres- pond to those numbers.Systematic sampling When a population is relatively homogeneous and you have a listing of the items of interest such as invoices, a fleet of company cars, physical units such as products coming off a production 207Chapter 6: Theory and methods of statistical sampling 389 386 309 75 174 350 314 70 219 380 473 249 56 323 270 147 605 426 440 285 353 339 173 583 620 624 331 84 219 78 560 272 347 171 476 589 396 285 306 557 300 183 406 114 485 105 161 528 438 510 288 437 374 368 512 49 368 25 75 36 415 251 308 Table 6.7 Table of 63 random numbers between 1 and 630.

142 26 178 146 72 7 156 95 176 144 113 194 Table 6.8 Table of 12 random numbers between 1 and 200.line, inventory going into storage, a stretch of road, or a row of houses, then systematic sample may be appropriate.You first decide at what fre- quency you need to take a sample.For example, if you want a 4% sample you analyse every 25th unit � 4% of 100 is 25.

If you want a 5% sample you analyse every 20th unit � 5% of 100 is 20.5% sample you analyse every 200 units � 0.Care must be taken in using systematic sampling that no bias occurs where the interval you choose corresponds to a pattern in the oper- ation.

For example, you use systematic sampling to examine the filling operation of soft drink machine.It so happens that there are 25 filling nozzles on the machine.In this case, you will be sampling a can that has been filled from the same nozzle.

The United States population census, undertaken every 10 years, is a form of systematic sample where although every household receives a sur- vey datasheet to complete, every 10th household receives a more detailed survey form to complete.

Stratified sampling The technique of stratified sampling is useful when the population can be divided into relatively homogeneous groups, or strata, and random sampling is made only on the strata of interest.For example, the strata may be students, people of a certain age range, male or female, married or single households, socio-economic levels, affiliated with the labour or conservative party, etc.Stratified sampling is used because it more accurately reflects the characteristics of the tar- get population.Single people of a certain socio- economic class are more likely to buy a sports car; people in the 20–25 have a different prefer- ence of music and different needs of portable phones than say those in the 50–55-age range.Stratified sampling is used when there is a small variation within each group, but a wide variation among groups.

For example, teenagers in the age range 13 to 19 and their parents in the age range 40 to 50 differ very much in their tastes and ideas! Several strata of interest In a given population you may have several well-defined strata and perhaps you wish to take a representative sample from this population.Consider for example, the 1st row of Table 6.9 which gives the number of employees by func- tion in a manufacturing company.Each func- tion is a stratum since it defines a specific activity.Suppose we wish to obtain the employ- ees’ preference of changing from the current 8 hours/day, 5 days a week to a proposed 10 hours/day, 4 days/week.

In order to limit the cost and the time of the sampling experiment we decide to only survey 60 of the employees.There are a total of 1,200 employees in the firm and so 60 represents 5% of the total workforce (60/1,200).Thus, we would take a random sam- ple of 5% of the employees from each of the departments or strata such that the sampling experiment parallels the population.The number that we would survey is given in the 2nd row of Table 6.208 Statistics for Business Department Administration Operations Design R&D Sales Accounting Information Total Systems Employees 160 300 200 80 260 60 140 1,200 Sample size 8 15 10 4 13 3 7 60 Table 6.Cluster sampling In cluster sample the population is divided into groups, or clusters, and each cluster is then sampled at random.For example, assume Birmingham is targeted for preference of a cer- tain consumer product.The city is divided into clusters using a city map and an appropriate number of clusters are selected for analysis.

Cluster sampling is used when there is consider- able variation in each group or cluster, but groups are essentially similar.Cluster sampling, if properly designed, can provide more accurate results than simple random sampling from the population.Quota sampling In market research, or market surveys, inter- viewers carrying out the experiment may use quota sampling where they have a specific target quantity to review.In this type of sampling often the population is stratified according to some criteria so that the interviewer’s quota is based within these strata.For example, the inter- viewer may be interested to obtain information regarding a ladies fashion magazine.

The inter- viewer conducts her survey in a busy shopping area such as London’s Oxford Street.Using quota sampling, in her survey she would only interview females, perhaps less than 40, and who are elegantly dressed.This stratification should give a reasonable probability that the selected candidates have some interest, and thus an opinion, regarding the fashion magazine in question.If you are in an area where surveys are being carried out, it could be that you do not fit the strata desired by the interviewer.For example, you are male and the interviewer is targeting females, you appear to be over 50 and the inter- viewer is targeting the age group under 40, you are white and the interviewer is targeting other ethnic groups, etc.

Consumer surveys If your sampling experiment involves opinions say concerning a product, a concept, or a situa- tion, then you might use a consumer survey, where responses are solicited from individuals who are targeted according to a well-defined sampling plan.The sampling plan would use one, or a combination of the methods above – simple random sampling, systematic, stratified, cluster, or quota sampling.The survey informa- tion is prepared on questionnaires, which might be sent through the mail, completed by tele- phone, sent by electronic mail, or requested in person.In the latter case this may be either going door-to-door, or soliciting the information in areas frequented by potential consumers such as shopping malls or busy pedestrian areas.The collected survey data, or sample, is then analysed and used to forecast or make estimates for the population from which the survey data was taken.

Surveys are often used to obtain ideas about a new product, because required data is unavailable from other sources.When you develop a consumer survey remem- ber that it is perhaps you who have to analyse it afterwards.Thus, you should structure it so that this task is straightforward with responses that are easy to organize.For example, rather than asking the question “How old are you?” give the respondent age cate- gories as for example in Table 6.

Here these cat- egories are all encompassing.Alternatively, if you want to know the job of the respondent rather than asking, “What is you job?” ask the question, “Which of the following best describes your pro- fessional activity?” as for example in Table 6.209Chapter 6: Theory and methods of statistical sampling Under Over 25 25–34 35–44 45–54 55–65 65 Table 6.

This is not all encompassing but there is a cate- gory “Other” for activities that may have been overlooked.Soliciting information from consumers is not easy, “everyone is too busy”.Postal responses have a very low response and their use has declined.

Those people who do respond may not be representative in the sample.

Telephone sur- veys give a higher return because voice contact has been obtained.However, again the sample obtained may not be representative as those contacted may be the unemployed, retirees or elderly people, or non-employed individuals who are more likely to be at home when the telephone call is made.The other segment of the population, usually larger, is not available because they are working.Though if you have access to portable phone numbers this may not apply.Electronic mail surveys give a reasonable response, as it is very quick to send the survey back.

However, the questionnaire only reaches those who have E-mail, and then those who care to respond.Person-to-person contact gives a much higher response for consumer surveys since if you are stopped in the street, a relatively large proportion of people will accept to be questioned.There is the cost of designing the question- naire such that it is able to solicit the correct response.There is the operating side of collect- ing the data, and then the subsequent analysis.

Often businesses use outside consulting firms specialized in developing consumer surveys.Primary and secondary data In sampling if we are responsible for carrying out the analysis, or at least responsible for designing the consumer surveys, then the data is considered primary data.If the sample experi- ment is well designed then this primary data can provide very useful information.The disad- vantage with primary data is the time, and the associated cost, of designing the survey and the subsequent analysis.In some instances it may be possible to use secondary data in analytical work.

Secondary data is information that has been developed by someone else but is used in your analytical work.Secondary data might be demographic information, economic trends, or consumer patterns, which is often available through the Internet.The advantage with sec- ondary data, provided that it is in the public domain, is that it costs less or at best is free.The disadvantage is that the secondary data may not contain all the information you require, the format may not be ideal, and/or it may be not be up to date.Thus, there may be a trade-off between using less costly, but perhaps less accurate, sec- ondary data, and more expensive but more reli- able, primary data.

210 Statistics for Business Construction Consulting Design Education Energy Financial services Government Health care Hospitality Insurance Legal Logistics Manufacturing Media communications Research Retail Telecommunications Tourism Other (please describe) Table 6.11 Which of the following best describes your professional activity? 211Chapter 6: Theory and methods of statistical sampling This chapter has looked at sampling covering specifically, basic relationships, sampling for the mean in infinite and finite populations, sampling for proportions, and sampling methods.Statistical relations in sampling for the mean Inferential statistics is the estimate of population characteristics based on the analysis of a sam- ple.The larger the sample size, the more reliable is our estimate of the population parameter.It is the central limit theory that governs the reliability of sampling.

This theory states that as the size of the sample increases, there becomes a point when the distribution of the sample means can be approximated by the normal distribution.In this case, the mean of all sample means withdrawn from the population is equal to the population mean.Further, the standard error of the estimate in a sampling distribution is equal to the population standard deviation divided by the square root of the sample size.Sampling for the means for an infinite population An infinite population is a collection of data of a large size such that by removing or destroying some data elements the population that remains is not significantly affected.Here we can modify the transformation relationship that apply to the normal distribution and determine the num- ber of standard deviations, z, as the sample mean less the population mean divided by the stand- ard error.

When we use this relationship we find that the larger the sample size, n, the more the sample data clusters around the population mean implying that there is less variability.Sampling for the means from a finite population A finite population in sampling is defined such that the ratio of the sample size to the population size is greater than 5%.This means that the sample size is large relative to the population size.When we have a finite population we modify the standard error by multiplying it by a finite popu- lation multiplier, which is the square root of the ratio of the population size minus the sample size to the population size minus one.When we have done this, we can use this modified relationship to infer the characteristics of the population parameter.

Again as before, the larger the sample size, the more the data clusters around the population mean and there is less variability in the data.Sampling distribution of the proportion A sample proportion is the ratio of those values that have the desired characteristics divided by the sample size.The binomial relationship governs proportions, since either values in the sample have the desired characteristics or they do not.Using the binomial relationships for the mean and the standard deviation, we can develop the standard error of the proportion.With this standard error of the proportion, and the value of the sample proportion, we can make an estimate of the population proportion in a similar manner to making an estimate of the population mean.

Again, the larger the sample size, the closer is our estimate to the population proportion.C hapter Sum m ary 212 Statistics for Business Sampling methods The key to correct sampling is to avoid bias, that is not taking a sample that gives lopsided results, and to ensure that the sample is random.If we have a relatively homogeneous population we can use systematic sampling, which is taking samples at predetermined intervals according to the desired sample size.Stratified sampling can be used when we are interested in a well-defined strata or group.

Stratified sampling can be extended when there are several strata of interest within a population.Cluster sampling is another way of making a sampling experiment when the population is divided up into manageable clusters that represent the population, and then sampling an appropriate quantity within a cluster.Quota sampling is when an interviewer has a certain quota, or number of units to analyse that may be according to a defined strata.Consumer surveys are part of sampling where respondents complete questionnaires that are sent through the post, by E-mail, completed over the phone, or face to face contact.

When you construct a questionnaire for a consumer surveys, avoid having open-ended questions as these are more difficult to analyse.

In sampling there is pri- mary data, or that collected by the researcher, and secondary data that maybe in the public domain.Primary data is normally the most useful but is usually more costly to develop.213Chapter 6: Theory and methods of statistical sampling 1.Credit card Situation From past data, a large bank knows that the average monthly credit card account bal- ance is £225 with a standard deviation of £98.What is the probability that in an account chosen at random, the average monthly balance will lie between £180 and £250? 2.What is the probability that in 10 accounts chosen at random, the sample average monthly balance will lie between £180 and £250? 3.What is the probability that in 25 accounts chosen at random, the sample average monthly balance will lie between £180 and £250? 4.What assumptions are made in determining these estimations? 2.Food bags Situation A paper company in Finland manufactures treated double-strength bags used for hold- ing up to 20 kg of dry dog or cat food.These bags have a nominal breaking strength of 8 kg/cm2 with a production standard deviation of 0.The manufacturing process of these food bags follows a normal distribution.

What percentage of the bags produced has a breaking strength between 8.What percentage of the bags produced has a breaking strength between 6.

What proportion of the sample means of size 10 will have breaking strength between 8.What proportion of the sample means of size 10 will have breaking strength between 6.Compare the answers of Questions 1 and 3, and 2 and 4.What distribution would the sample means follow for samples of size 10? 3.Telephone calls Situation It is known that long distance telephone calls are normally distributed with the mean time of 8 minutes, and the standard deviation of 2 minutes.EXERCISE PROBLEMS 214 Statistics for Business Required 1.What is the probability that a call taken at random will last between 7.What is the probability that a call taken at random will last between 7.If random samples of 25 calls are selected, what is the probability that a call taken at random will last between 7.

If random samples of 25 calls are selected, what is the probability that a call taken at random will last between 7.If random samples of 100 calls are selected, what is the probability that a call taken at random will last between 7.If random samples of 100 calls are selected, what is the probability that a call taken at random will last between 7.

Soft drink machine Situation A soft drinks machine is regulated so that the amount dispensed into the drinking cups is on average 33 cl.The filling operation is normally distributed and the standard deviation is 1 cl no matter the setting of the mean value.

What is the volume that is dispensed such that only 5% of cups contain this amount or less? 2.If the machine is regulated such that only 5% of the cups contained 30 cl or less, by how much could the nominal value of the machine setting be reduced? In this case, on average a customer would be receiving what percentage less of beverage? 3.With a nominal machine setting of 33 cl, if samples of 10 cups are taken, what is the volume that will be exceeded by 95% of sample means? 4.There is a maintenance rule such that if the sample average content of 10 cups falls below 32.

50 cl, a technician will be called out to check the machine settings.In this case, how often would this happen at a nominal machine setting of 33 cl? 5.What should be the nominal machine setting to ensure that no more than 2% main- tenance calls are made? In this case, on average customers will be receiving how much more beverage? 5.Baking bread Situation A hypermarket has its own bakery where it prepares and sells bread from 08:00 to 20:00 hours.One extremely popular bread, called “pave supreme”, is made and sold continu- ously throughout the day.

This bread, which is a nominal 500 g loaf, is individually kneaded, left for 3 hours to rise before being baked in the oven.During the kneading and 215Chapter 6: Theory and methods of statistical sampling baking process moisture is lost but from past experience it is known that the standard deviation of the finished bread is 17 g.If you go to the store and take at random one pave supreme, what is the probability that it will weigh more than 520 g? 2.You are planning a dinner party and so you go to the store and take at random four pave supremes, what is the probability that the average weight of the four weigh more than 520 g? 3.

Say that you are planning a larger dinner party and you go to the store and take at random eight pave supremes, what is the probability that the average weight of the eight breads weigh more than 520 g? 4.If you go to the store and take at random one pave supreme, what is the probability that it will weigh between 480 and 520 g? 5.If you go to the store and take at random four pave supremes, what is the probability that the average weight of the loaves will be between 480 and 520 g? 6.If you go to the store and take at random eight pave supremes, what is the probability that the average weight of the loaves will be between 480 and 520 g? 7.Explain the differences between Questions 1 to 3.

Explain the differences between Questions 4 to 6.Why is the progression the reverse of what you see for Questions 1 to 3? 6.Financial advisor Situation The amount of time a financial advisor spends with each client has a population mean of 35 minutes, and a standard deviation of 11 minutes.If a random client is selected, what is the probability that the time spent with the client will be at least 37 minutes? 2.If a random client is selected, there is a 35% chance that the time the financial advi- sor spends with the client will be below how many minutes? 3.If random sample of 16 clients are selected, what is the probability that the average time spent per client will be at least 37 minutes? 4.If a random sample of 16 clients is selected, there is a 35% chance that the sample mean will be below how many minutes? 5.If random sample of 25 clients are selected, what is the probability that the average time spent per client will be at least 37 minutes? 6.

If a random sample of 25 clients is selected, there is a 35% chance that the sample mean will be below how many minutes? 7.Explain the differences between Questions 1, 3, and 5.What assumptions do you make in responding to these questions? 216 Statistics for Business 7.Height of adult males Situation In a certain country, the height of adult males is normally distributed, with a mean of 176 cm and a variance of 225 cm2.

If one adult male is selected at random, what is the probability that he will be over 2 m? 2.What are the upper and lower limits of height between which 90% will lie for the population of adult males? 3.If samples of four men are taken, what percentage of such samples will have average heights over 2 m? 4.What are the upper and lower limits between which 90% of the sample averages will lie for samples of size four? 5.

If samples of nine men are taken, what percentage of such samples will have average heights over 2 m? 6.What are the upper and lower limits between which 90% of the sample averages will lie for samples of size nine? 7.Wal-Mart Situation Wal-Mart of the United States, after buying ASDA in Great Britain, is now looking to move into France.

It has targeted 220 supermarket stores in that country and the present owner of these said that profits from these supermarkets follows a normal distribution, have the same mean, with a standard deviation of €37,500.Financial information is on a monthly basis.If Wal-Mart selects a store at random what is the probability that the profit from this store will lie within €5,400 of the mean? 2.If Wal-Mart management selects 50 stores at random, what is the probability that the sample mean of profits for these 50 stores will lie within €5,400 of the mean? 9.

Automobile salvage Situation Joe and three colleagues have created a small automobile salvage company.Their work consists of visiting sites that have automobile wrecks and recovering those parts that can be resold.Often from these wrecks they recoup engine parts, computers from the electrical systems, scrap metal, and batteries.From past work, salvaged components on an average generate €198 per car with a standard deviation of €55.Joe and his three col- leagues pay themselves €15 each per hour and they work 40 hours/week.

Between them 217Chapter 6: Theory and methods of statistical sampling they are able to complete the salvage work on four cars per day.One particular period they carry out salvage work at a site near Hamburg, Germany where there are 72 wrecked cars.What is the correct standard error for this situation? 2.What is the probability that after one weeks work the team will have collected enough parts to generate total revenue of €4,200? 3.

On the assumption that the probability outcome in Question No.2 is achieved, what would be the net income to each team member at the end of 1 week? 10.Education and demographics Situation According to a survey in 2000, the population of the United States in the age range 25 to 64 years, 72% were white.Further in this same year, 16% of the total population in the same age range were high school dropouts and 27% had at least a bachelor’s degree.If random samples of 200 people in the age range 25 to 64 are selected, what proportion of the samples between 69% and 75% will be white? 2.If random samples of 400 people in the age range 25 to 64 are selected, what proportion of the samples between 69% and 75% will be white? 3.If random samples of 200 people in the age range 25 to 64 are selected, what proportion of the samples between 13% and 19% will be high school dropouts? 4.If random samples of 400 people in the age range 25 to 64 are selected, what propor- tion of the samples between 13% and 19% will be high school dropouts? 5.If random samples of 200 people in the age range 25 to 64 are selected, what proportion of the samples between 24% and 30% will have at least a bachelors degree? 6.

If random samples of 400 people in the age range 25 to 64 are selected, what proportion of the samples between 24% and 30% will have at least a bachelors degree? 7.Explain the difference between each paired question of 1 and 2; 3 and 4; and 5 and 6.World Trade Organization Situation The World Trade Organization talks, part of the Doha Round, took place in Hong Kong between 13 and 18 December 2005.According to data, the average percentage tariff imposed on all imported tangible goods and services in certain selected countries is as follows5: 4 Losing ground, Business Week, 21 November 2005, p.

5 US, EU walk fine line at heart of trade impasse, The Wall Street Journal, 13 December 2005, p.218 Statistics for Business Required 1.If a random sample of 200 imported tangible goods or service into the United States were selected, what is the probability that the average proportion of the tariffs for this sample would be between 1% and 4%? 2.

If a random sample of 200 imported tangible goods or service into Burkina Faso were selected, what is the probability that the average proportion of the tariffs for this sam- ple would be between 10% and 14%? 3.If a random sample of 200 imported tangible goods or service into India were selected, what is the probability that the average proportion of the tariffs for this sam- ple would be between 25% and 32%? 4.If a random sample of 400 imported tangible goods or service into the United States were selected, what is the probability that the average proportion of the tariffs for this sample would be between 1% and 4%? 5.If a random sample of 400 imported tangible goods or service into Burkina Faso were selected, what is the probability that the average proportion of the tariffs for this sam- ple would be between 10% and 14%? 6.If a random sample of 400 imported tangible goods or service into India were selected, what is the probability that the average proportion of the tariffs for this sample would be between 25% and 32%? 7.

Explain the difference between each paired question of 1 and 2; 3 and 4; and 5 and 6.Female illiteracy Situation In a survey conducted in three candidate countries for the European Union – Turkey, Romania, and Croatia and three member countries – Greece, Malta, and Slovakia Europe in 2003, the female illiteracy of those over 15 was reported as follows6: Required 1.If random samples of 250 females over 15 were taken in Turkey in 2003, what pro- portion between 12% and 22% would be illiterate? 2.If random samples of 500 females over 15 were taken in Turkey in 2003, what pro- portion between 12% and 22% would be illiterate? Turkey Greece Malta Romania Croatia Slovakia 19% 12% 11% 4% 3% 0.

5% United States India European Union Burkina Faso Brazil 3.4% 6 Too soon for Turkish delight, The Economist, 1 October 2005, p.219Chapter 6: Theory and methods of statistical sampling 3.If random samples of 250 females over 15 were taken in Malta in 2003, what propor- tion between 9% and 13% would be illiterate? 4.If random samples of 500 females over 15 were taken in Malta in 2003, what proportion between 9% and 13% would be illiterate? 5.

If random samples of 250 females over 15 were taken in Slovakia in 2003, what proportion between 0.If random samples of 500 females over 15 were taken in Slovakia in 2003, what proportion between 0.What is your explanation for the difference between each paired question of 1 and 2; 3 and 4; and 5 and 6? 8.If you took a sample of 200 females over 15 from Istanbul and the proportion of those females illiterate was 0.Unemployment Situation According to published statistics for 2005, the unemployment rate among people under 25 in France was 21.

These numbers in part are considered to be reasons for the riots that occurred in France in 2005.

If random samples of 100 people under 25 were taken in France in 2005, what pro- portion between 12% and 15% would be unemployed? 2.If random samples of 200 people under 25 were taken in France in 2005, what pro- portion between 12% and 15% would be unemployed? 3.If random samples of 100 people under 25 were taken in Germany in 2005, what proportion between 12% and 15% would be unemployed? 4.If random samples of 200 people under 25 were taken in Germany in 2005, what proportion between 12% and 15% would be unemployed? 5.

If random samples of 100 people under 25 were taken in Britain in 2005, what pro- portion between 12% and 15% would be unemployed? 6.If random samples of 200 people under 25 were taken in Britain in 2005, what pro- portion between 12% and 15% would be unemployed? 7.If random samples of 100 people under 25 were taken in the United States in 2005, what proportion between 12% and 15% would be unemployed? 8.If random samples of 200 people under 25 were taken in the United States in 2005, what proportion between 12% and 15% would be unemployed? 7 France’s young and jobless, Business Week, 21 November 2005, p.What is you explanation for the difference between each paired question of 3 and 4; 5 and 6; and 7 and 8? 10.Why do the data for France in Questions 1 and 2 not follow the same trend as for the questions for the other three countries? 14.Manufacturing employment Situation According to a recent survey by the OECD in 2005 employment in manufacturing as a percent of total employment, has fallen dramatically since 1970.The following table gives the information for OECD countries8: Required 1.

If random samples of 200 people of the working population were taken from Germany in 2005, what proportion between 20% and 26% would be in manufacturing? 2.If random samples of 400 people of the working population were taken from Germany in 2005, what proportion between 20% and 26% would be in manufacturing? 3.If random samples of 200 people of the working population were taken from Britain in 2005, what proportion between 13% and 15% would be in manufacturing? 4.If random samples of 400 people of the working population were taken from Britain in 2005, what proportion between 13% and 15% would be in manufacturing? 5.If random samples of 200 people of the working population were taken from the United States in 2005, what proportion between 6% and 10% would be in manufacturing? 6.

If random samples of 400 people of the working population were taken from the United States in 2005, what proportion between 6% and 10% would be in manufacturing? 7.What is your explanation for the difference between each paired question of 1 and 2; 3 and 4; and 5 and 6.If a sample of 100 people was taken in Germany in 2005 and the proportion of the people in manufacturing was 32%, what conclusions might you draw? Country Germany Italy Japan France Britain Canada United States 1970 40% 28% 27% 28% 35% 23% 25% 2005 23% 22% 18% 16% 14% 14% 10% 8 Industrial metamorphosis, The Economist, 1 October 2005, p.221Chapter 6: Theory and methods of statistical sampling 15.Homicide Situation In December 2005, Steve Harvey, an internationally known AIDS outreach worker was abducted at gunpoint from his home in Jamaica and murdered.9 According to the statis- tics of 2005, Jamaica is one the world’s worst country for homicide.How it compares with some other countries according to the number of homicides per 100,000 people is given in the table below10: Required 1.

If you lived in Jamaica what is the probability that some day you would be a homicide statistic? 2.

If you lived in Britain what is the probability that some day you would be a homicide statistic? Compare this probability with the previous question? What is another way of expressing this probability between the two countries? 3.If random samples of 1,000 people were selected in Jamaica what is the proportion between 0.If random samples of 2,000 people were selected in Jamaica what is the proportion between 0.

Explain the difference between Questions 3 and 4.Humanitarian agency Situation A subdivision of the humanitarian organization, doctors without borders, based in Paris has 248 personnel in its database according to the table below.

This database gives in alphabetical order the name of the staff members, gender, age at last birthday, years with the organization, the country where the staff member is based, and their training in the medical field.You wish to get information about the whole population included in this database including criteria such as job satisfaction, safety concerns in the country of work, human relationships in the country, and other qualitative factors.For budget rea- sons you are limited to interviewing a total of 40 people and some of these will be done by telephone but others will be personal interviews in the country of operation.Britain United States Zimbabwe Argentina Russia Brazil S.Africa Columbia Jamaica 2 6 8 14 21 25 44 47 59 9 A murder in Jamaica, International Herald Tribune, 14 December 2005, p.

10 Less crime, more fear, The Economist, 1 October 2005, p.222 Statistics for Business Required Develop a sampling plan to select the 40 people.Consider total random sampling, cluster sampling, and strata sampling.

Of the plans that you select draw conclusions.Which do you believe is the best experiment? Explain your reasoning: No.Name Gender Age Years Country Medical with where training agency based 1 Abissa, Yasmina Murielle F 26 2 Chile Nurse 2 Adekalom, Maily F 45 17 Brazil General medicine 3 Adjei, Abena F 41 16 Chile Nurse 4 Ahihounkpe, Ericka F 29 5 Kenya Physiotherapy 5 Akintayo, Funmilayo F 46 12 Brazil Nurse 6 Alexandre, Ga lle F 46 19 Kenya Nurse 7 Alibizzata, Myl ne F 31 2 Chile Radiographer 8 Ama, Eric M 30 2 Brazil Nurse 9 Angue Assoumou, M lodie F 47 18 Chile Nurse 10 Arfort, Sabrina F 47 12 Cambodia Nurse 11 Aubert, Nicolas M 50 20 Costa Rica Nurse 12 Aubery, Olivia F 34 12 Thailand Nurse 13 Aulombard, Audrey F 49 18 Brazil Physiotherapy 14 Awitor, Euloge M 36 12 Kenya Nurse 15 Ba, Oumy F 27 1 Chile Nurse 16 Bakouan, Aminata F 24 3 Vietnam Nurse 17 Banguebe, Sandrine F 41 1 Costa Rica Nurse 18 Baque, Nicolas M 42 14 Kenya Nurse 19 Batina, C dric M 32 2 Kenya Nurse 20 Batty-Ample, Agn s F 31 10 Chile Nurse 21 Baud, Maxime F 44 18 Costa Rica Nurse 22 Belkora, Youssef M 46 2 Brazil Radiographer 23 Berard, Emmanuelle F 41 17 Vietnam Nurse 24 Bernard, Eloise F 40 3 Vietnam Surgeon 25 Berton, Alexandra M 46 22 Ivory Coast Nurse 26 Besenwald, Laetitia F 28 2 Brazil Nurse 27 Beyschlag, Natalie F 34 8 Brazil Nurse 28 Black, Kimberley F 32 8 Kenya Nurse 29 Blanchon, Paul M 23 1 Vietnam Nurse 30 Blondet, Thomas M 34 1 Kenya Nurse 31 Bomboh, Patrick M 31 11 Chile Nurse 32 Bordenave, Bertrand M 32 10 Brazil Physiotherapy 33 Bossekota, Ariane F 37 9 Kenya Nurse 34 Boulay, Gr gory M 36 7 Kenya Nurse 35 Bouziat, Lucas M 53 26 Kenya Nurse 36 Briatte, Pierre-Edouard M 48 28 Kenya Nurse 223Chapter 6: Theory and methods of statistical sampling No.Name Gender Age Years Country Medical with where training agency based 37 Brunel, Laurence F 27 3 Ivory Coast General medicine 38 Bruntsch-Lesba, Natascha F 55 30 Cambodia Nurse 39 Buzingo, Patrick M 46 5 Thailand Nurse 40 Cablova, Dagmar F 53 14 Kenya Nurse 42 Chabanel, Gael F 31 2 Ivory Coast Nurse 43 Chabanier, Maud F 27 1 Brazil Nurse 44 Chahboun, Zineb F 53 23 Brazil Physiotherapy 45 Chahed, Samy M 46 12 Thailand Nurse 46 Chappon, Romain F 40 18 Costa Rica Nurse 47 Chartier, Henri M 28 8 Vietnam Nurse 48 Chaudagne, Stanislas M 45 10 Chile Radiographer 49 Coffy, Robin M 48 24 Ivory Coast Nurse 50 Coissard, Alexandre M 36 8 Chile Nurse 51 Collomb, Fanny F 54 18 Ivory Coast Nurse 52 Coradetti, Louise F 36 11 Cambodia Nurse 53 Cordier, Yan M 43 1 Brazil Surgeon 54 Crombe, Jean-Michel M 27 7 Brazil Nurse 55 Croute, Benjamin M 42 17 Vietnam Nurse 56 Cusset, Johannson M 42 12 Cambodia Nurse 57 Czajkowski, Mathieu M 51 10 Brazil Nurse 58 Dadzie, Kelly M 34 10 Chile Nurse 59 Dandjouma, Ainaou F 50 2 Nigeria Nurse 60 Dansou, Joel M 54 27 Brazil Physiotherapy 61 De Messe Zinsou, Thierry M 38 2 Ivory Coast Nurse 62 De Zelicourt, Gonzague M 55 4 Cambodia Nurse 63 Debaille, Camille F 50 26 Kenya Nurse 64 Declippeleir, Olivier M 31 9 Chile Nurse 65 Delahaye, Benjamin M 47 11 Ivory Coast Nurse 66 Delegue, H loise F 31 6 Brazil Radiographer 67 Delobel, Delphine F 23 3 Kenya Nurse 68 Demange, Aude F 30 1 Vietnam Nurse 69 Deplano, Guillaume M 54 33 Thailand Nurse 70 Desplanches, Isabelle F 34 10 Thailand Nurse 71 Destombes, H l ne F 31 7 Brazil Nurse 72 Diallo, Ralou Maimouna F 50 25 Ivory Coast General medicine 73 Diehl, Pierre M 45 11 Brazil Nurse 74 Diop, Mohamed M 25 5 Chile Nurse 75 Dobeli, Nathalie F 33 10 Chile Physiotherapy 76 Doe-Bruce, Othalia Ayele E F 53 19 Cambodia Nurse 77 Donnat, M lanie F 51 16 Thailand Nurse 78 Douenne, Fran ois-Xavier M 37 15 Ivory Coast Surgeon 79 Du Mesnil Du Buisson, Edouard M 52 21 Vietnam Nurse 80 Dubourg, Jonathan M 44 3 Cambodia Nurse 81 Ducret, Camille F 50 16 Thailand Nurse (Continued) 224 Statistics for Business No.

Name Gender Age Years Country Medical with where training agency based 82 Dufau, Guillaume M 45 25 Chile Nurse 83 Dufaud, Charly M 28 5 Costa Rica Radiographer 84 Dujardin, Agathe F 36 15 Kenya Nurse 85 Dutel, S bastien M 50 11 Cambodia Nurse 86 Dutraive, Benjamin M 33 2 Brazil Nurse 87 Eberhardt, Nadine F 26 6 Cambodia Physiotherapy 88 Ebibie N’ze, Yannick M 28 8 Chile Nurse 89 Errai, Skander M 47 5 Thailand Nurse 90 Erulin, Caroline F 42 18 Ivory Coast General medicine 91 Escarboutel, Christel F 52 3 Ivory Coast Nurse 92 Etien, St phanie F 54 12 Brazil Nurse 93 Felio, S bastien M 32 4 Kenya Nurse 94 Fernandes, Claudio M 29 5 Vietnam Surgeon 95 Fillioux, St phanie F 32 9 Brazil Nurse 96 Flandrois, Nicolas M 31 10 Ivory Coast Nurse 97 Gaillardet, Marion F 23 3 Brazil Nurse 98 Garapon, Sophie F 31 3 Kenya Nurse 99 Garnier, Charles M 27 6 Costa Rica Nurse 100 Garraud, Charlotte F 43 11 Brazil Radiographer 101 Gassier, Vivienne F 33 13 Brazil Nurse 102 Gava, Mathilde F 26 4 Ivory Coast Nurse 103 Gerard, Vincent M 50 5 Thailand Physiotherapy 104 Germany, Julie F 29 9 Kenya Nurse 105 Gesrel, Valentin M 40 11 Nigeria Nurse 106 Ginet-Kauders, David M 54 33 Costa Rica Nurse 107 Gobber, Aur lie F 32 1 Costa Rica Nurse 108 Grangeon, Baptiste M 33 13 Chile Nurse 109 Gremmel, Antoine M 31 3 Brazil Nurse 110 Gueit, Delphine F 46 2 Thailand Nurse 111 Guerite, Camille M 45 8 Cambodia Nurse 112 Guillot, Nicholas M 33 5 Brazil Nurse 113 Hardy, Gilles M 30 3 Chile Nurse 114 Hazard, Guillaume M 38 9 Cambodia Nurse 115 Honnegger, Doroth e F 45 2 Vietnam Nurse 116 Houdin, Julia F 49 7 Thailand Physiotherapy 117 Huang, Shan-Shan F 35 14 Costa Rica Nurse 118 Jacquel, H l ne F 47 16 Vietnam Nurse 119 Jiguet-Jiglairaz, S bastien M 55 35 Chile Nurse 120 Jomard, Sam M 34 7 Kenya Surgeon 121 Julien, Lo c F 35 2 Brazil Nurse 122 Kacou, Joeata F 48 3 Vietnam Nurse 123 Kasalica, Aneta F 51 13 Brazil Nurse 124 Kasalica, Darko M 24 4 Brazil Nurse 125 Kassab, Philippe M 29 1 Brazil Radiographer 126 Kervaon, Nathalie F 45 24 Brazil Nurse 225Chapter 6: Theory and methods of statistical sampling No.Name Gender Age Years Country Medical with where training agency based 127 Kimbakala-Koumba, Madeleine F 44 11 Costa Rica Nurse 128 Kolow, Alexandre M 40 7 Chile Nurse 129 Latini, St phane F 50 23 Chile Nurse 130 Lauvaure, Julien M 42 14 Brazil Nurse 132 Legris, Baptiste M 38 18 Ivory Coast Nurse 133 Lehot, Julien M 37 16 Vietnam Physiotherapy 134 Lestangt, Aur lie F 29 8 Cambodia Nurse 135 Li, Si Si F 32 5 Chile Nurse 136 Liubinskas, Ricardas M 25 4 Cambodia Nurse 137 Loyer, Julien M 34 10 Vietnam Nurse 138 Lu Shan Shan F 31 8 Chile Nurse 139 Marchal, Arthur M 45 12 Nigeria Nurse 140 Marganne, Richard M 25 4 Chile Nurse 141 Marone, Lati F 33 4 Brazil Nurse 142 Martin, Cyrielle F 42 5 Kenya Physiotherapy 143 Martin, St phanie F 46 16 Brazil Nurse 144 Martinez, St phanie F 25 5 Thailand Nurse 145 Maskey, Lilly F 23 1 Vietnam Nurse 146 Masson, C dric M 29 8 Brazil Nurse 147 Mathisen, M linda F 48 12 Cambodia Nurse 148 Mermet, Alexandra F 25 1 Brazil Nurse 149 Mermet, Florence F 27 1 Brazil Radiographer 150 Michel, Doroth e F 54 24 Brazil Nurse 151 Miribel, Julien M 53 6 Vietnam Nurse 152 Monnot, Julien F 40 5 Chile Nurse 153 Montfort, Laura F 53 16 Nigeria Nurse 154 Murgue, Fran ois M 32 2 Kenya Nurse 155 Nauwelaers, Emmanuel F 55 24 Thailand Nurse 156 Nddalla-Ella, Claude F 35 14 Thailand Surgeon 157 Ndiaye, Baye Mor M 50 23 Vietnam Nurse 158 Neulat, Jean-Philippe M 37 17 Cambodia Physiotherapy 159 Neves, Christophe M 28 2 Brazil Nurse 160 Nicot, Guillaume M 29 8 Brazil Nurse 161 Oculy, Fr deric M 45 12 Chile Nurse 162 Okewole, Maxine M 51 21 Kenya Nurse 163 Omba, Nguie M 47 24 Ivory Coast Nurse 164 Ostler, Emilie F 28 1 Brazil Nurse 165 Owiti, Brenda F 25 1 Kenya Surgeon 166 Ozkan, Selda F 43 21 Nigeria Nurse 167 Paillet, Ma t F 55 28 Kenya Nurse 168 Penillard, Clo F 38 5 Ivory Coast Nurse 169 Perera, William M 43 17 Nigeria Nurse 170 Perrenot, Christophe M 30 3 Kenya Nurse 171 Pesenti, Johan M 47 3 Costa Rica Radiographer (Continued) 226 Statistics for Business No.Name Gender Age Years Country Medical with where training agency based 172 Petit, Dominique F 48 17 Thailand Nurse 173 Pfeiffer, C line F 39 7 Ivory Coast Nurse 174 Philetas, Ludovic M 24 3 Thailand Nurse 175 Portmann, Kevin M 45 21 Chile Nurse 176 Pourrier, Jennifer F 41 15 Thailand Physiotherapy 177 Prou, Vincent M 42 7 Chile Nurse 178 Raffaele, Gr gory M 55 26 Cambodia Nurse 179 Ramanoelisoa, Eliane Goretti F 49 18 Vietnam Nurse 180 Rambaud, Philippe M 27 1 Costa Rica Nurse 181 Ranjatoelina, Andrew M 43 5 Brazil Nurse 182 Ravets, Emmanuelle F 30 8 Nigeria Nurse 183 Ribieras, Alexandre M 45 14 Cambodia Nurse 184 Richard, Damien M 23 1 Kenya Nurse 185 Rocourt, Nicolas M 41 9 Brazil Nurse 186 Rossi-Ferrari, S bastien M 37 2 Thailand Nurse 187 Rouviere, Gr gory M 51 31 Ivory Coast Nurse 188 Roux, Alexis M 23 1 Kenya Nurse 189 Roy, Marie-Charlotte F 51 14 Costa Rica Nurse 190 Rudkin, Steven M 41 21 Thailand General medicine 191 Ruget, Joffrey M 24 4 Brazil Nurse 192 Rutledge, Diana F 38 11 Thailand Nurse 193 Ruzibiza, Hubert M 35 12 Brazil Nurse 194 Ruzibiza, Oriane F 45 10 Brazil Nurse 195 Sadki, Khalid M 35 13 Vietnam Nurse 196 Saint-Quentin, Florent M 55 22 Brazil Physiotherapy 197 Salami, Mistoura F 45 20 Cambodia Nurse 198 Sambe, Mamadou F 31 5 Ivory Coast Nurse 199 Sanvee, Pascale F 51 23 Vietnam Radiographer 200 Saphores, Pierre-Jean M 32 2 Cambodia Nurse 201 Sassioui, Mohamed M 48 10 Chile Nurse 202 Savall, Arnaud M 47 18 Brazil Nurse 203 Savinas, Tamara F 54 34 Ivory Coast Nurse 204 Schadt, St phanie F 33 13 Brazil Surgeon 205 Schmuck, C line F 54 9 Costa Rica Nurse 206 Schneider, Aur lie F 53 3 Costa Rica Nurse 207 Schulz, Amir M 39 10 Vietnam Nurse 208 Schwartz, Olivier M 46 21 Brazil Nurse 209 Seimbille, Alexandra M 47 1 Chile Nurse 210 Servage, Benjamin M 47 23 Ivory Coast Nurse 211 Sib, Brigitte F 51 13 Brazil Nurse 212 Sinistaj, Irena F 36 12 Brazil Nurse 213 Six, Martin M 34 3 Costa Rica Nurse 214 Sok, Steven M 26 1 Costa Rica Nurse 215 Souah, Steve M 50 23 Ivory Coast Nurse 227Chapter 6: Theory and methods of statistical sampling No.Name Gender Age Years Country Medical with where training agency based 216 Souchko, Edouard M 38 7 Nigeria Radiographer 217 Soumare, Anna F 52 22 Vietnam Nurse 218 Straub, Elodie F 25 2 Brazil Nurse 219 Sun, Wenjie F 31 4 Kenya Physiotherapy 220 SuperVielle Brouques, Claire F 40 7 Kenya Surgeon 221 Tahraoui, Davina F 31 1 Thailand Nurse 222 Tall, Kadiatou F 33 8 Thailand Nurse 223 Tarate, Romain M 49 3 Costa Rica Nurse 224 Tessaro, Laure F 39 18 Ivory Coast Nurse 225 Tillier, Pauline F 29 8 Kenya Nurse 226 Trenou, K mi M 44 19 Thailand Nurse 227 Triquere, Cyril M 23 1 Brazil Nurse 228 Tshitungi, Mesenga F 40 2 Ivory Coast Nurse 229 Vadivelou, Christophe M 55 17 Chile Physiotherapy 230 Vande-Vyre, Julien M 25 5 Brazil Nurse 231 Villemur, Claire F 41 17 Costa Rica Nurse 232 Villet, Diana F 33 8 Nigeria Nurse 233 Vincent, Marion F 36 2 Ivory Coast General medicine 234 Vorillon, Fabrice M 32 4 Chile Nurse 235 Wadagni, Imelda F 45 10 Vietnam Nurse 236 Wallays, Anne F 30 2 Vietnam Nurse 237 Wang, Jessica F 38 18 Brazil Nurse 238 Weigel, Samy M 34 13 Ivory Coast Nurse 239 Wernert, Lucile F 24 2 Kenya Nurse 240 Willot, Mathieu M 52 30 Brazil Nurse 241 Wlodyka, S bastien M 40 6 Kenya Nurse 242 Wurm, Debora F 46 15 Chile Nurse 243 Xheko, Eni F 28 1 Costa Rica Nurse 244 Xu, Ning F 48 13 Brazil Physiotherapy 245 Yuan, Zhiyi M 39 18 Brazil Surgeon 246 Zairi, Leila F 51 26 Vietnam Radiographer 247 Zeng, Li F 25 3 Ivory Coast Nurse 248 Zhao, Lizhu F 33 9 Thailand General medicine This page intentionally left blank 7Estimating population characteristics Turkey and the margin of error The European Union, after a very heated debate, agreed in early October 2005 to open membership talks to admit Turkey, a Muslim country of 70 million people.This agreement came only after a tense night-and-day discussion with Austria, one of the 25-member states, who strongly opposed Turkey’s membership.

Austria has not forgotten fighting back the invading Ottoman armies in the 16th and 17th centuries.Reservations to Turkey’s membership is also very strong in other countries as shown in Figure 7.1 where an estimated 70% or more of the population in each of Austria, Cyprus, Germany, France, and Greece are opposed to membership.This estimated information is based on a survey response of a sample of about 1,000 people in each of the 10 indicated countries.The survey was conducted in the period May–June 2005 with an indicated margin of error of �3% points.

This survey was made to estimate population characteristics, which is the essence of the material in this chapter., “Turkey gains EU approval to begin membership talks”, Wall Street Journal Europe, 4 October 2005, pp.1 Survey response of attitudes to Turkey joining the European Union.0% 10% 20% 30% 40% 50% 60% 70% 80% 90% 100% UK Turkey Sweden Poland Italy Greece Germany France Cyprus Austria Margin of error is � 3% In favor Undecided Against In Chapter 6, we discussed statistical sampling for the purpose of obtaining information about a population.This chapter expands upon this to use sampling to estimate, or infer, population parame- ters based entirely on the sample data.By its very nature, estimating is probabilistic as there is no cer- tainty of the result.

However, if the sample experi- ment is correctly designed then there should be a reasonable confidence about conclusions that are made.Thus from samples we might with confi- dence estimate the mean weight of airplane pas- sengers for fuel-loading purposes, the proportion of the population expected to vote Republican, or the mean value of inventory in a distribution centre.The mean or average value of data is the sum of all the data taken divided by the number of measurements taken.The units of measurement can be financial units, length, volume, weight, etc.

Point estimates In estimating, we could use a single value to estimate the true population mean.

For example, if the grade point average of a random sample of students is 3.75 then we might estimate that the population average of all students is also 3.Or, we might select at random 20 items of inventory from a distribution centre and calcu- late that their average value is £25.In this case we would estimate that the population average of the entire inventory is £25.Here we have used the sample mean x– as a point esti- mate or an unbiased estimate of the true popula- tion mean, x.The problem with one value or a point estimate is that they are presented as being exact and that unless we have a super Estimating the Mean Value 231Chapter 7: Estimating population characteristics After you have studied this chapter you will understand how sampling can be extended to make estimates of population parameters such as the mean and the proportion.To facilitate comprehen- sion the chapter is organized as follows: ✔ Estimating the mean value • Point estimates • Interval estimates • Confidence level and reliabil- ity • Confidence interval of the mean for an infinite population • Application of confidence inter- vals for an infinite population: Paper • Sample size for estimating the mean of an infinite population • Application for determining the sample size: Coffee • Confidence interval of the mean for a finite population • Application of the confidence interval for a finite population: Printing ✔ Estimating the mean using the Student-t distribution • The Student-t distribution • Degrees of freedom in the t-distribution • Profile of the Student-t distribution • Confidence intervals using a Student-t distribution • Excel and the Student-t distribution • Application of the Student-t distribution: Kiwi fruit • Sample size and the Student-t distribution • Re-look at the example kiwi fruit using the normal distribution ✔ Estimating and auditing • Estimating the population amount • Application of auditing for an infinite population: tee-shirts • Application of auditing for a finite population: paperback books ✔ Estimating the proportion • Interval estimate of the proportion for large samples • Sample size for the proportion for large samples • Application of estimation for proportions: Circuit boards ✔ Margin of error and levels of confidence • Explaining margin of error • Confidence levels L e a r n i n g o b j e c t i v e s crystal ball, the probability of them being pre- cisely the right value is low.

Point estimates are often inadequate as they are just a single value and thus, they are either right or wrong.In practice it is more meaningful to have an inter- val estimate and to quantify these intervals by probability levels that give an estimate of the error in the measurement.Interval estimates With an interval estimate we might describe situations as follows.The estimate for the proj- ect cost is between $11.9 million and I am 95% confident of these figures.The esti- mate for the sales of the new products is between 22,000 and 24,500 units in the first year and I am 90% confidence of these figures.The estimate of the price of a certain stock is between $75 and $90 but I am only 50% confi- dent of this information.The estimate of class enrolment for Business Statistics next academic year is between 220 and 260 students though I am not too confident about these figures.Thus the interval estimate is a range within which the population parameter is likely to fall.

Confidence level and reliability Suppose a subcontractor A makes refrigerator compressors for client B who assembles the final refrigerators.In order to establish the terms of the final customer warranty, the client needs information about the life of compressors since the compressor is the principal working compon- ent of the refrigerator.Assume that a random sample of 144 compressors is tested and that the mean life of the compressors, x–, is determined to be 6 years or 72 months.Using the concept of point estimates we could say that the mean life of all the compressors manufactured is 72 months.Here x– is the estimator of the population mean x and 72 months is the estimate of the population mean obtained from the sample.

However, this information says nothing about the reliability or confidence that we have in the estimate.The subcontractor has been making these compressors for a long time and knows from past data that the standard deviation of the working life of compressors is 15 months.Then since our sample size of 144 is large enough, the standard error of the mean can be calcu- lated by using equation 6(ii) from Chapter 6 from the central limit theory: This value of 1.25 months is one standard error of the mean, or it means that z � �1.If we assume that the life of a compressor follows a normal distribu- tion then we know from Chapter 5 that 68.26% of all values in the distribution lie within �1 standard deviations from the mean.From equa- tion 6(iv), or When z � �1 then the lower limit of the com- pressor life is, When z � �1 then the upper limit is, Thus we can say that, the mean life of the com- pressors is about 72 months and there is a 68.26% (about 68%) probability that the mean value will be between 70.Two standard errors of the mean, or when z � �2, is 2 * 1.� � � z x n x x / x x n � � � � 15 144 15 12 1 25.

months 232 Statistics for Business Chapter 5, if we assume a normal distribution, 95.44%, of all values in the distribution lie within �2 standard deviations from the mean.When z � �2 then using equation 6(iv), the lower limit of the compressor life is, When z � �2 then the upper limit is, Thus we can say that, the mean life of the com- pressor is about 72 months and there is a 95.44% (about 95%) probability that the mean value will be between 69.

Finally, three standard errors of the mean is 3 * 1.75 months and again from Chapter 5, assuming a normal distribution, 99.73%, of all values in the distribution lie within �3 standard deviations from the mean.

When z � �3 then using equation 6(iv), the lower limit of compressor life is, When z � �3 then the upper limit is, Thus we can say that the mean life of the com- pressor is about 72 months and there is almost a 99.73% (about 100%) probability that the mean value will be between 68.The best estimate is that the mean compres- sor life is 72 months and the manufacturer is about 68% confident that the compressor life is in the range 70.Here the confidence interval is between 70.The best estimate is that the mean compres- sor life is 72 months and the manufacturer is about 95% confident that the compressor life is in the range 69.Here the confidence interval is between 69.The best estimate is that the mean compres- sor life is 72 months and the manufacturer is about 100% confident that the compressor life is in the range 68.Here the confidence interval is between 68.

It is important to note that as our confidence level increases, going from 68% to 100%, the confi- dence interval increases, going from a range of 2.This is to be expected as we become more confident of our estimate, we give a broader range to cover uncertainties.Confidence interval of the mean for an infinite population The confidence interval is the range of the esti- mate being made.From the above compressor example, considering the �2 confidence inter- vals, we have 69.50 months as the respective lower and upper limits.Between these limits this is equivalent to 95.44% of the area under the normal curve, or about 95%.A 95% confidence interval estimate implies that if all possible samples were taken, about 95% of them would include the true population mean, , somewhere within their interval, whereas, about 5% of them would not.This concept is illustrated in Figure 7.

The �2 intervals for sample numbers 1, 2, 4, and 5 contain the population mean , whereas for samples 3 and 6 do not contain the population mean within their interval.The level of confidence is (1 � ), where is the total proportion in the tails of the distribu- tion outside of the confidence interval.Since the distribution is symmetrical, the area in each tail is /2 as shown in Figure 7.

As we have shown in the compressor situation, the x � � �72 3 1 25 75 75* .months 233Chapter 7: Estimating population characteristics confidence intervals for the population estimate for the mean value are thus, 7(i) This implies that the population mean lies in the range given by the relationship, 7(ii) Application of confidence intervals for an infinite population: Paper Inacopia, the Portuguese manufacturer of A4 paper commonly used in computer printers wants to be sure that its cutting machine is operating correctly.The width of A4 paper is expected to be 21.

00 cm and it is known that the standard deviation of the cutting machine is 0.The quality control inspector pulls a random sample of 60 sheets from the produc- tion line and the average width of this sample is 20.Determine the 95% confidence intervals of the mean width of all the A4 paper coming off the production line? x z n x z n x x x� � � � x z x z n x x� � � 234 Statistics for Business � 2 � 2 Confidence interval Mean Figure 7.3 Confidence interval and the area in the tails.x4x3 x1 x6x2 x5 x3 x1 x2 x6 x4 x5 �2sx Interval for sample No.

2 m We have the following information: Sample size, n, is 60 Sample mean, x–, is 20.9986 cm Population standard deviation, , is 0.0100 Standard error of the mean is, The area in the each tail for a 95% confi- dence limit is 2.Since the dis- tribution is symmetrical, the upper value is numerically the same at �1.50% � 95%) which is the area of the curve from the left to the upper value of z.) From equation 7(i) the confidence limits are, 20.0011 cm Thus we would say that our best estimate of the width of the computer paper is 20.9986 cm and we are 95% confident that the width is in the range 20.Since this interval contains the population expected mean value of 21.0000 cm, we can conclude that there seems to be no problem with the cutting machine.Determine the 99% confidence intervals of the mean width of all the A4 paper coming off the production line.

The area in each tail for a 99% confidence limit is 0.Since the distribution is symmet- rical, the upper value is �2.50% � 99%) which is the area of the curve from the left to the upper value of z.

) From equation 7(i) the confidence limits are, 20.0019 cm Thus we would say that our best estimate of the width of the computer paper is 20.9986 cm and we are 99% confident that the width is in the range 20.Again, since this interval contains the expected mean value of 21.

0000 cm, we can conclude that there seems to be no problem with the cutting machine.Note that the limits in Question 2 are wider than in Question 1 since we have a higher confidence level.Sample size for estimating the mean of an infinite population In sampling it is useful to know the size of the sample to take in order to estimate the popula- tion parameter for a given confidence level.We have to accept that unless the whole population is analysed there will always be a sampling error.If the sample size is small, the chances are that the error will be high.

If the sample size is large there may be only a marginal gain in reli- ability in the estimate of our population mean but what is certain is that the analytical experi- ment will be more expensive.Thus, what is an appropriate sample size, n, to take for a given confidence level? The confidence limits are related the sample size, n, by equation 6(iv) or, 6(iv) The range from the population mean, on the left side of the distribution when z is negative, is �(x– � x) or x � x– on the left side of the distri- bution, and x– � x on the right side of the � � � z x n x x / n � � 0 0100 60 0 0013 .235Chapter 7: Estimating population characteristics distribution curve.Reorganizing equation 6(iv) by making the sample size, n, the subject gives, 7(iii) The term, x– � x, is the sample error and if we denote this by e, then the sample size is given by, 7(iv) Thus for a given confidence level, which then gives the value of z, and a given confidence limit the required sample size can be determined.

Note in equation 7(iv) since n is given by squared value it does not matter if we use a negative or positive value for z.The following worked example illus- trates the concept of confidence intervals and sample size for an infinite population.Application for determining sample size: Coffee The quality control inspector of the filling machine for coffee wants to estimate the mean weight of coffee in its 200 gram jars to within �0.It is known that the standard devi- ation of the coffee filling machine is 2 g.

What sample size should the inspector take to be 95% confidence of the estimate? Using equation 7(iv), The area in the each tail for a 95% confidence limit is 2.Since the distribution is symmetrical, the upper value is numerically the same at �1.50% � 95%) which is the area of the curve from the left to the upper value of z.9600 (it does not matter whether we use plus or minus since we square the value) x is 2 g e is �0.50 g Thus the quality control inspector should take a sample size of 62 (61 would be just slightly too small).

Confidence interval of the mean for a finite population As discussed in Chapter 6 (equation 6(vi)), if the population is considered finite, that is the ratio n/N is greater than 5%, then the standard error should be modified by the finite population multiplier according to the expression, 6(vi) In this case the confidence limits for the pop- ulation estimation from equation 7(i) are modi- fied as follows: 7(v) Application of the confidence interval for a finite population: Printing A printing firm runs off the first edition of a textbook of 496 pages.

After the book is printed, the quality control inspector looks at 45 ran- dom pages selected from the book and finds that the average number of errors in these pages is 2.These include printing errors of colour and alignment, but also typing errors which originate from the author and the editor.The x z x z n N n Nx x� � � � � ( ) ( )1 x x n N n N � � � ⋅ 1 n � � � 1 9600 2 00 0 50 61 463 62 2 .( ⎛ ⎝ ⎜⎜⎜ ⎞ ⎠ ⎟⎟⎟⎟ rouunded up) n z e x� ⎛ ⎝ ⎜⎜⎜⎜ ⎞ ⎠ ⎟⎟⎟⎟ 2 n z e x� ⎛ ⎝ ⎜⎜⎜⎜ ⎞ ⎠ ⎟⎟⎟⎟ 2 n x x x z � � ⎛ ⎝ ⎜⎜⎜⎜ ⎞ ⎠ ⎟⎟⎟⎟⎟ 2 236 Statistics for Business inspector knows that based on past contracts for a first edition of a book the standard deviation of the number of errors per page is 0.What is a 95% confidence interval for the mean number of errors in the book? Sample size, n, is 45 Population size, N, is 496 Sample mean, x–, errors per page is 2.70 Population standard deviation, , is 0.07% This value is greater than 5%, thus, we must use the finite population multiplier: Uncorrected standard error of the mean is Corrected standard error of the mean, � 0.0711 Confidence level is 95%, thus area in each tail is 2.

Since the distribution is symmet- rical, the upper value is numerically the same at �1.Thus from equation 7(v) the lower confi- dence limit is, 2.56 Thus from equation 7(v) the upper confi- dence limit is, 2.

84 Thus we could say that the best estimate of the errors in the book is 2.70 per page and that we are 95% confident that the errors lie between 2.

There may be situations in estimating when we do not know the population standard deviation and that we have small sample sizes.In this case there is an alternative distribution that we apply called the Student-t distribution, or more simply the t-distribution.The Student-t distribution In Chapter 6, in the paragraph entitled, “Sample size and shape of the sampling distribution of the means”, we indicated that the sample size taken has an influence on the shape of the sampling distributions of the means.

If we sample from population distributions that are normal, such that we know the standard devia- tion, , any sample size will give a sampling distribution of the means that are approxi- mately normal.However, if we sample from populations that are not normal, we are obliged to increase our sampling size to at least 30 units in order that the sampling distribution of the means will be approximately normally distributed.Thus, what do we do when we have small sample sizes that are less than 30 units? To be correct, we should use a Student-t distribution.The Student-t distribution, like the normal distribution, is a continuous distribution for small amounts of data.It was developed by William Gossett of the Guinness Brewery, in Dublin, Ireland in 1908 (presumably when he had time between beer production!) and pub- lished under the pseudonym “student” as the Guinness company would not allow him to put his own name to the development.

The Student-t distributions are a family of distributions each one having a different shape and characterized by a parameter called the degrees of freedom.The density function, from which the Student-t Estimating the Mean Using the Student-t Distribution x x n N n N � � � � 1 0 0745 0 9545.N n N � � � � � � � 1 496 45 496 1 451 495 0 9545.237Chapter 7: Estimating population characteristics distribution is drawn, has the following rela- tionship: 7(vi) Here, is the degree of freedom, is the value of 3.1416, and t is the value on the x-axis similar to the z-value of a normal distribution.Degrees of freedom in the Student-t distribution Literally, the degrees of freedom means the choices that you have regarding taking certain actions.

For example, what is the degree of free- dom that you have in manoeuvring your car into a parking slot? What is the degree of free- dom that you have in contract or price negoti- ations? What is the degree of freedom that you have in negotiating a black run on the ski slopes? In the context of statistics the degrees of free- dom in a Student-t distribution are given by (n � 1) where n is the sample size.

This then implies that there is a degree of freedom for every sample size.To understand quantitatively the degrees of freedom consider the following.There are five variables v, w, x, y, and z that are related by the following equation: 7(vii) Since there are five variables we have a choice, or the degree of freedom, to select four of the five.After that, the value of the fifth variable is auto- matically fixed.For example, assume that we give v, w, x, and y the values 14, 16, 12, and 18, respectively.

Then from equation 7(vii) we have, Thus automatically the fifth variable, z, is fixed at a value of 5 in order to retain the validity of the equation.Here we had five variables to give a degree of freedom of four.In general terms, for a sample size of n units, the degree of freedom is the value determined by (n � 1).Profile of the Student-t distribution Three Student-t distributions, for sample size n of 6, 12, and 22, or sample sizes less than 30, are illustrated in Figure 7.The degrees of free- dom for these curves, using (n � 1) are respect- ively 5, 11, and 21.These three curves have a profile similar to the normal distribution but if we superimposed a normal distribution on a Student-t distribution as shown in Figure 7.5, we see that the normal distribution is higher at the peak and the tails are closer to the x-axis, compared to the Student-t distribution.The Student-t distribution is flatter and you have to go further out on either side of the mean value before you are close to the x-axis indicating greater variability in the sample data.This is the penalty you pay for small sample sizes and where the sampling is taken from a non-normal population.

As the sample size increases the profile of the Student-t distribution approaches that of the normal distribution and as is illus- trated in Figure 7.4 the curve for a sample size of 22 has a smaller variation and is higher at the peak.Confidence intervals using a Student-t distribution When we have a normal distribution the confi- dence intervals of estimating the mean value of the population are as given in equation 7(i): 7(i)x z n x� z � � � � � � � � 5 13 14 16 12 18 65 60 5 * ( ) 14 16 12 18 5 13 � � � � � z v w x y z� � � � � 5 13 f t t ( ) / ! / � � � � 1 2 2 2 1 2( )⎡⎣⎢ ⎤⎦⎥ ( )⎡⎣⎢ ⎤⎦⎥ ⎡ ⎣ ⎢ ⎢ ⎤⎤ ⎦ ⎥ ⎥ � �( )/ 1 2 238 Statistics for Business When we are using a Student-t distribution, Equation 7(i) is modified to give the following: 7(viii) Here the value of t has replaced z, and has replaced , the population standard deviation.This new term, , means an estimate of the population standard deviation.Numerically it is equal to s, the sample standard deviation by the relationship, 7(ix) We could avoid writing , as some texts do, and simply write s since they are numerically the same.

However, by putting it is clear that our only alternative to estimate our confidence ( ) ( ) � � � � s x x n 2 1 x t n x� 239Chapter 7: Estimating population characteristics Figure 7.4 Three Student-t distributions for different sample sizes.Mean value Sample size � 6 Sample size � 12 Sample size � 22 � Student-t distribution Normal distribution Figure 7.limits is to use an estimate of the population stan- dard deviation as measured from the sample.

Excel and the Student-t distribution There are two functions in Excel for the Student-t distribution.When we use the t-distribution in estimating, the number of tails is always two – that is, one on the left and one on the right.(This is not necessarily the case for hypothesis testing that is discussed in Chapter 8.(Note the difference in the way you enter the variables for the Student-t and the normal distri- bution.For the Student-t you enter the area in the tails, whereas for the normal distribution you enter the area of the curve from the extreme left to a value on the x-axis.) Application of the Student-t distribution: Kiwi fruit Sheila Hope, the Agricultural inspector at Los Angeles, California wants to know in milligrams, the level of vitamin C in a boat load of kiwi fruits imported from New Zealand, in order to compare this information with kiwi fruits grown in the Central Valley, California.Sheila took a random sample of 25 kiwis from the ship’s hold and measured the vitamin C content.1 gives the results in milligrams per kiwi sampled.Estimate the average level of vitamin C in the imported kiwi fruits and give a 95% confidence level of this estimate.Since we have no information about the popu- lation standard deviation, and the sample size of 25 is less than 30, we use a Student-t distribution.Estimate of the population standard devia- tion, � s � 12.Standard error of the sample distribution, Required confidence level (given) is 95%.

Area outside of confidence interval, , (100% � 95%) is 5%.

From equation 7(viii), Thus the estimate of the average level of vita- min C in all the imported kiwis is 100.24 mg with a 95% confidence that the lower level of our estimate is 95.

01 mg and the upper level Upper confidence level, x t n x� � � .100 24 2 06339 2 5346 100 24 5 2312 105 47 * .� � � Lower confidence level, x t n x� � � .100 24 2 06339 2 5346 100 24 5 2312 95 01 * .x n � � 12 6731 5 00 2 5346 n , 240 Statistics for Business Table 7.1 Milligrams of vitamins per kiwi sampled.109 88 91 136 93 101 89 97 115 92 114 106 94 109 110 97 89 117 105 92 83 79 107 100 93 is 105.This information is illustrated on the Student-t distribution in Figure 7.

Sample size and the Student-t distribution We have said that the Student-t distribution should be used when the sample size is less than 30 and the population standard deviation is unknown.Some analysts are more rigid and use a sample size of 120 as the cut-off point.What should we use, a sample size of 30 or a sample size of 120? The movement of the value of t relative to the value of z is illustrated by the data in Table 7.2 and the corresponding graph in 241Chapter 7: Estimating population characteristics 95.

2 Values of t and z with different sample sizes.50% Sample Upper Upper Sample Upper Upper size, n Student-t z size, n Student-t z 5 2.Here we have the Student-t value for a confidence level of 95% for sample sizes ran- ging from 5 to 200.In the column (t � z)/z we see that the difference between t and z is 4.When the sample size increases to 120 then the difference is just 1.

Is this difference signifi- cant? It really depends on what you are sam- pling.We have to remember that we are making estimates so that we must expect errors.In the medical field small differences may be import- ant but in the business world perhaps less so.Let us take another look at the kiwi fruit example from above using z rather than t values.

Re-look at the example Kiwi fruit using the normal distribution Here all the provided data and the calculations are the same as previously but we are going to assume that we can use the normal distribution for our analysis.Required confidence level (given) is 95%.Area outside of confidence interval, , (100% � 95%) is 5%, which means that there is an area of 2.5% in both tails for a symmetrical 242 Statistics for Business Figure 7.7 As the sample size increases the value of t approaches z.

8500 0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180 190 200 Sample size, n Up pe r z o r t v al ue Upper t value Upper z value distribution.From equation 7(i), The corresponding values that we obtained by using the Student-t distribution were 95.

Since in reality we would report probability of our confidence for the vitamin level of kiwis between 95 and 105 mg, the difference between using z and t in this case in insignificant.Auditing is the methodical examination of financial accounts, inventory items, or operating processes to verify that they confirm with stand- ard practices or targeted budget levels.

Estimating the population amount We can use the concepts that we have developed in this chapter to estimate the total value of goods such as, for example, inventory held in a distribution centre when, for example, it is impossible or very time consuming to make an audit of the population.In this case we first take a random and representative sample and deter- mine the mean financial value x–.If N is the total number of units, then the point estimate for the population total is the size of the population, N, multiplied by the sample mean, or, Total � N x– 7(x) It is unlikely we know the standard deviation of the large population of inventory and so we would estimate the value from the sample.If the sample size is less than 30 we use the Student-t distribution and the confidence intervals are given as follows by multiplying both terms in equation 7(viii) to give, 7(xi) Alternatively, if the population is considered finite, that is the ratio of n/N 5%, then the standard error has to be modified by the esti- mated finite population multiplier to give, 7(xii) Thus the confidence intervals when the stand- ard deviation is unknown, the sample size is less than 30, and the population is finite, are, 7(xiii) The following two applications illustrate the use of estimating the total population amount for auditing purposes.Application of auditing for an infinite population: tee-shirts A store on Duval Street in Key West Florida, wishes to estimate the total retail value of its tee-shirts, tank tops, and sweaters that it has in its store.

The inventory records indicate that there are 4,500 of these clothing articles on the shelves.The owner takes a random sample of 29 items and Table 7.3 gives the prices in dollars indicated on the articles.Confidence intervals: N x Nt n N n N � � � 1 Estimated standard error: n N n N � �1 Confidence intervals: N x Nt n � Estimating and Auditing Upper confidence level, x z n x� � � .100 00 1 96000 2 5346 100 24 4 9678 105 21 * .� � � Lower confidence level, x z n x� � � .

100 24 1 96000 2 5346 100 24 4 9678 95 27 * .� � � 243Chapter 7: Estimating population characteristics 1.Estimate the total retail value of the clothing items within a 99% confidence limit.Population size, N, is 4,500 Estimated total retail value is N x– � 4,500 * 25.

Since this value is less than 5% we do not need to use the finite population multiplier.Estimated population standard deviation, , is $11.Estimated standard error of the sample distribution, is 2.Since we do not know the population stand- ard deviation, and the sample size is less than 30 we use the Student-t distribution.From equation 7(xi) the lower confidence limit for the total value is, and the upper confidence limit is, Thus the owner estimates the average, or point estimate, of the total retail value of the clothing items in his Key West store as $113,897 (rounded) and he is 99% confi- dent that the value lies between $88,303.Application of auditing for a finite population: paperback books A newspaper and bookstore at Waterloo Station wants to estimate the value of paper backed books it has in its store.

The owner takes a ran- dom sample of 28 books and determines that the average retail value is £4.57 with a sample standard deviation of 53 pence.There are 12 shelves of books and the owner estimates that there are 45 books per shelf.Estimate the total retail value of the books within a 95% confidence limit.

Estimated population amount of books, N, is 12 * 45 or 540.Estimated total retail value is N x– � 540 * 4.Since this value is greater than 5% we use the finite population multiplier Finite population multiplier .

Estimated population standard deviation, , is £0.From equation 7(xii) the estimated standard error is, .

n N n N � � � � 1 0 53 28 0 9746 0 0976 540 28 540 1 512 539 0 9746 � � � � .113 896 55 4 500 2 7633 2 0582 or $ , .113 896 55 4 500 2 7633 2 0582 or $ , ./ , x n/ �11 0836 29 244 Statistics for Business Table 7.From equation 7(xiii) the lower confidence limit is, From equation 7(xiii) the upper confidence limit is, Thus the owner estimates the average, or point estimate, of the total retail value of the paper back books in the store as £2,467.80 (£2,468 rounded) and that she is 95% confident that the value lies between £2,359.Rather than making an estimate of the mean value of the population, we might be interested to estimate the proportion in the population.For example, we take a sample and say that our point estimate of the proportion expected to vote conservative in the next United Kingdom election is 37% and that we are 90% confident that the proportion will be in the range of 34% and 40%.When dealing with proportions then the sample proportion, p–, is a point estimate of the population proportion p.The value p– is determined by taking a sample of size n and measuring the proportion of successes.

Interval estimate of the proportion for large samples When analysing the proportions of a popula- tion then from Chapter 6 we developed the fol- lowing equation 6(xi) for the standard error of the proportion, : 6(xi) where n is the sample size and p is the popula- tion proportion of successes and q is the popula- tion proportion of failures equal to (1 � p).Further, from equation 6(xv), Reorganizing this equation we have the following expression for the confidence intervals for the estimate of the population proportion as follows: 7(xiv) Thus, analogous to the estimation for the means, this implies that the confidence intervals for an estimate of the population proportion lie in the range given by the following expression: 7(xv) If we do not know the population proportion, p, then the standard error of the proportion can be estimated from the following equation by replacing p with p–: 7(xvi) In this case, is the estimated standard error of the proportion and p– is the sample proportion of successes.If we do this then equation 7(xv) is modified to give the expression, 7(xvii)p z p p n p p z p p n � � � � � �( ) ( )1 1 p ( ) p p p n � �1 p z p p n p p z p p n � � � � � �( ) ( )1 1 p p z p p n � � �( )1 z p p p p n �� � �( )1 p pq n p p n � � �( )1 p Estimating the Proportion N x Nt n N n N � � � � � � , .1 2 467 80 540 2 0518 0 0976£ ££2 575 96, .1 2 467 80 540 2 0518 0 0976£ ££2 359 64, .

245Chapter 7: Estimating population characteristics Sample size for the proportion for large samples In a similar way for the mean, we can determine the sample size to take in order to estimate the population proportion for a given confidence level.From the relationship of 7(xiv) the inter- vals for the estimate of the population propor- tion are, 7(xviii) Squaring both sides of the equation we have, Making n, the sample size the subject of the equation gives, 7(xix) If we denote the sample error, (p– � p) by e then the sample size is given by the relationship, 7(xx) While using this equation, a question arises as to what value to use for the true population propor- tion, p, when this is actually the value that we are trying to estimate! One possible approach is to use the value of p– if this is available.Alternatively, we can use a value of p equal to 0.5 or 50% as this will give the most conserva- tive sample size.This is because for a given value of the confidence level say 95% which defines z, and the required sample error, e, then a value of p of 0.

4 and illustrated by the graph in Figure 7.

The following is an application of the estimation for proportions including an esti- mation of the sample size.Application of estimation for proportions: Circuit boards In the manufacture of electronic circuit boards a sample of 500 is taken from a production line and of these 15 are defective.What is a 90% confidence interval for the pro- portion of all the defective circuit boards pro- duced in this manufacturing process? Proportion defective, p–, is 15/500 � 0.030 or also From equation 7(xvi) the estimate of the stan- dard error of the proportion is, When we have a 90% confidence interval, and assuming a normal distribution, then the area of the distribution up to the lower confidence level is (100% � 90%)/2 � 5% ( ) .p p p n � � � � � 1 0 03 0 97 500 0 0291 500 0 0076 500 15 500 0 97 � � .n z p p e � �2 2 1( ) n z p p p p � � � 2 2 1( ) ( ) ( ( ) p p z p p n � � � )2 2 1 p p z p p n � �� �( )1 246 Statistics for Business Table 7.4 Conservative value of p for sample size.0000 and the area of the curve up to the upper confidence level is 5% � 90% � 95%.From equation 7(xvii) the lower confidence limit is, From equation 7(xvii) the upper confidence limit is, Thus we can say that from our analysis, the proportion of all the manufactured circuit boards which are defective is 0.Further, we are 90% confident that this pro- portion lies in the range of 0.

If we required our estimate of the proportion of all the defective manufactured circuit boards to be within a margin of error of �0.01 at a 98% confidence level, then what size of sample should we take? When we have a 98% confidence interval, and assuming a normal distribution, then the area of the distribution up to the lower confidence level is (100% � 98%)/2 � 1% and the area of the curve up to the upper confidence level is 1% � 98% � 99%.0 03 1 6449 0 0076 0 03 0 0125 0 04255 p z p� � � � � � .0 03 1 6449 0 0076 0 03 0 0125 0 01775 247Chapter 7: Estimating population characteristics Figure 7.8 Relation of the product, p(1 � p) with the proportion, p.

The sample proportion p– is used for the pop- ulation proportion p or 0.Using equation 7(xx), It does not matter which value of z we use, �2.

3263, since we are squaring z and the negative value becomes positive.Thus the sample size to estimate the popula- tion proportion of the number of defective cir- cuits within an error of margin of error of �0.An alternative, more conservative approach is to use a value of p � 0.

In this case the sam- ple size to use is, This value of 2,500 is significantly higher than 1,575 and would certainly add to the cost of the sampling experiment with not necessarily a sig- nificant gain in the accuracy of the results.When we make estimates the question arises (or at least it should) “How good is your estimate?” That is to say, what is the margin of error? In addition, we might ask, “Why don’t we always use a high confidence level of say 99% as this would signify a high degree of accuracy?” These two issues are related and are discussed below.Explaining margin of error When we analyse our sample we are trying to esti- mate the population parameter, either the mean value or the proportion.When we do this, there will be a margin of error.

This is not to say that we have made a calculation error, although this can occur, but the margin of error measures the max- imum amount that our estimate is expected to dif- fer from the actual population parameter.The margin of error is a plus or minus value added to the sample result that tells us how good is our estimate.If we are estimating the mean value then, 7(xxi) This is the same as the confidence limits from equation 7(i).In the worked example paper, at a confidence level of 95%, the margin of error is �1.Thus, another way of reporting our results is to say that we estimate that the width of all the com- puter paper from the production line is 20.9986 cm and we have a margin of error of �0.Now if we look at equation 7(xxi), when we have a given standard deviation and a given confidence level the only term that can change is the sample size n.Thus we might say, let us analyse a bigger sample in order to obtain a smaller margin of error.This is true, but as can be seen from Figure 7.9, which gives the ratio of as a percentage according to the sample size in units, there is a diminishing return.Increasing the sample size does reduce the margin of error but at a decreasing rate.

If we double the sample size from 60 to 120 units the ratio of changes from 12.13% or a difference 1/ n 1/ n Margin of error is � z n x Margin of Error and Levels of Confidence n z p p e � � � 2 2 1 2 3263 2 3263 0 50 0 50 0 01 0 ( ) ., n z p p e � � � 2 2 1 2 3263 2 3263 0 03 0 97 0 01 0 ( ) .From a sample size of 120 to 180 the value of changes from 9.68% or, if we go from a sample size of 360 to 420 units the value of goes from 5.With the increasing sample size the cost of testing of course increases and so there has to be a balance between the size of the sam- ple and the cost.If we are estimating for proportions then the margin of error is from equation 7(xvii) the value, 7(xxii) Since for proportions we are trying to estimate the percentage for a situation then the margin of error is a plus or minus percentage.In the worked example circuit boards the margin of error at a 90% level of confidence is, This means that our estimate could be 1.

25% less than our estimated propor- tion or a range of 2.The margin of error quoted in a sampling situation is important as it can give uncertainty to our conclusions.1, for example, we see that 52% of the Italian population is against Turkey join- ing the European Union.Based on just this information we might conclude that the major- ity of the Italians are against Turkey’s member- ship.However, if we then bring in the �3% margin of error then this means that we can ( ) .p z p p n �� � �� �� �� 1 1 6449 0 03 0 97 500 0 0125 11 25.% � � z p p n ( )1 1/ n 1/ n 249Chapter 7: Estimating population characteristics Figure 7.9 The change of with increase of sample size.1/ n 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 0 60 120 180 240 300 360 420 480 540 600 660 720 780 840 900 960 Sample size, n units 1/ �n (% ) have 49% against Turkey joining the Union (52 � 3), which is not now the majority of the population.

Our conclusions are reversed and in cases like these we might hear the term for the media “the results are too close to call”.Thus, the margin of error must be taken into account when surveys are made because the result could change.If the margin of error was included in the survey result of the Dewey/Truman election race, as presented in the Box Opener of Chapter 6, the Chicago Tribune may not have been so quick to publish their morning paper! Confidence levels If we have a confidence level that is high say at 99% the immediate impression is to think that we have a high accuracy in our sampling and esti- mating process.However this is not the case since in order to have high confidence levels we need to have large confidence intervals or a large margin of error.In this case the large intervals give very broad or fuzzy estimates.

This can be illustrated qualitatively as follows.Assume that you have contracted a new house to be built of 170 m2 living space on 2,500 m2 of land.You are concerned about the time taken to complete the project and you ask the constructor various questions concerning the time frame.These are given in the 1st col- umn of Table 7.

Possible indicated responses to these are given in the 2nd column and the 3rd and 4th columns, respectively, give the implied confidence interval and the implied confidence level.Thus, for a house to be finished in 10 years the constructor is almost certain because this is an inordinate amount of time and so we have put a confidence level of 99%.Again to ask the question for 5 years the confidence level is high at 95%.At 2 years there is a confidence level of 80% if everything goes better than planned.At 18 months there is a 50% confidence if there are, for example, ways to expedite the work.

At 6 months we are essentially saying it is impossible.(The time to completely construct a house varies with location but some 18 months to 2 years to build and completely finish all the landscaping is a reasonable time frame.Your question Constructor’s response Implied confidence interval Implied confidence level 1.

Will my house be I am certain �99% 10 years finished in 10 years? 2.Will my house be I am pretty sure �95% 5 years finished in 5 years? 3.Will my house be I think so �80% 2 years finished in 2 years? 4.Will my house be Probably not About 1% 0.50 years finished in 6 months? 251Chapter 7: Estimating population characteristics This chapter has covered estimating the mean value of a population using a normal distribu- tion and a Student-t distribution, using estimating for auditing purposes, estimating the popu- lation proportion, and discussed the margin of error and confidence intervals.Estimating the mean value We can estimate the population mean by using the average value taken from a random sample.However this single value is often insufficient as it is either right or wrong.

A more objective analysis is to give a range of the estimate and the probability, or the confidence, that we have in this estimate.When we do this in sampling from an infinite normal distribution we use the standard error.The standard error is the population standard deviation divided by the square root of the sample size.This is then multiplied by the number of standard deviations in order to determine the confidence intervals.The wider the confidence interval then the higher is our confidence and vice-versa.

If we wish to determine a required sample size, for a given confidence interval, this can be calculated from the interval equation since the num- ber of standard deviations, z, is set by our level of confidence.If we have a finite population we must modify the standard error by the finite population multiplier.Estimating the mean using the Student-t distribution When we have a sample size that is less than 30, and we do not know the population standard devi- ation, to be correct we must use a Student-t distribution.The degree of freedom is the sample size less one.

When we do not know the population stan- dard deviation we must use the sample standard deviation as an estimate of the population standard deviation in order to calculate the confidence intervals.As we increase the size of the sam- ple the value of the Student-t approaches the value z and so in this case we can use the normal dis- tribution relationship.Estimating and auditing The activity of estimating can be extended to auditing financial accounts or values of inventory.To do this we multiply both the average value obtained from our sample, and the confidence interval, by the total value of the population.Since it is unlikely that we know the population standard deviation in our audit experiment we use a Student-t distribution and use the sample standard deviation in order to estimate our population standard deviation.

When our popula- tion is finite, we correct our standard error by multiplying by the finite population multiplier.Estimating the proportion If we are interested in making an estimate of the population proportion we first determine the standard error of the proportion by using the population value, and then multiply this by the num- ber of standard deviations to give our confidence limits.If we do not have a value of the population C hapter Sum m ary 252 Statistics for Business proportion then we use the sample value of the proportion to estimate our standard error.We can determine the sample size for a required confidence level by reorganizing the confidence level equation to make the sample size the subject of the equation.The most conservative sample size will be when the value of the proportion p has a value of 0.

Margin of error and levels of confidence In estimating both the mean and the proportion of a population the margin of error is the max- imum amount of difference between the value of the population and our estimated amount.The larger the sample size then the smaller is the margin of error.However, as we increase the size of the sample the cost of our sampling experiment increases and there is a diminishing return on the margin of error with sample size.Although at first it might appear that a high confidence level of say close to 100% indicates a high level of accuracy, this is not the case.

In order to have a high confidence level we need to have broader confidence limits and this leads to rather vague or fuzzy estimates.253Chapter 7: Estimating population characteristics 1.Ketchup Situation A firm manufactures and bottles tomato ketchup that it then sells to retail firms under a private label brand.One of its production lines is for filling 500 g squeeze bottles, which after being filled are fed automatically into packing cases of 20 bottles per case.

In the filling operation the firm knows that the standard deviation of the filling operation is 8 g.

In a randomly selected case, what would be the 95% confidence intervals for the mean weight of ketchup in a case? 2.In a randomly selected case what would be the 99% confidence intervals for the mean weight of ketchup in a case? 3.Explain the differences between the answers to Questions 1 and 2.About how many cases would have to be selected such that you would be within �2 g of the population mean value? 5.What are your comments about this sampling experiment from the point-of-view of randomness? 2.Light bulbs Situation A subsidiary of GE manufactures incandescent light bulbs.The manufacturer sampled 13 bulbs from a lot and burned them continuously until they failed.The number of hours each burned before failure is given below.

342 426 317 545 264 451 1,049 631 512 266 492 562 298 Required 1.Determine the 80% confidence intervals for the mean length of the life of light bulbs.How would you explain the concept illustrated by Question 1? 3.Determine the 90%, confidence intervals for the mean length of the life of light bulbs.

Determine the 99% confidence intervals for the mean length of the life of light bulbs.Explain the differences between Questions 1, 3, and 4.Ski magazine Situation The publisher of a ski magazine in France is interested to know something about the average annual income of the people who purchase their magazine.Over a period of EXERCISE PROBLEMS 254 Statistics for Business three weeks they take a sample and from a return of 758 subscribers, they determine that the average income is €39,845 and the standard deviation of this sample is €8,542.Determine the 90% confidence intervals of the mean income of all the magazine readers of this ski magazine? 2.Determine the 99% confidence intervals of the mean income of all the magazine readers of this ski magazine? 3.

How would you explain the difference between the answers to Questions 1 and 2? 4.Households Situation A random sample of 121 households indicated they spent on average £12 on take-away restaurant foods.The standard deviation of this sample was £3.Calculate a 90% confidence interval for the average amount spent by all households in the population.

Calculate a 95% confidence interval for the average amount spent by all households in the population.Calculate a 98% confidence interval for the average amount spent by all households in the population.Explain the differences between the answers to Questions 1–3.Taxes Situation To estimate the total annual revenues to be collected for the State of California in a cer- tain year, the Tax Commissioner took a random sample of 15 tax returns.The taxes paid in $US according to these returns were as follows: $34,000 $2,000 $12,000 $39,000 $16,000 $7,000 $9,000 $72,000 $23,000 $15,000 $0 $19,000 $6,000 $12,000 $43,000 Required 1.Determine the 80%, 95%, and 99% confidence intervals for the mean tax returns.

Using for example the 95% confidence interval, how would you present your analysis to your superior? 255Chapter 7: Estimating population characteristics 3.How do you explain the differences in these intervals and what does it say about con- fidence in decision-making? 6.Vines Situation In the Beaujolais wine region north of Lyon, France, a farmer is interested to estimate the yield from his 5,200 grape vines.He samples at random 75 of the grape vines and finds that there is a mean of 15 grape bunches per vine, with a sample standard deviation of 6.

Construct a 95% confidence limit for the bunch of grapes for the total of 5,200 grape vines.How would you express the values determined in the previous question? 3.

Would your answer change if you used a Student-t distribution rather than a normal distribution? 7.

Floor tiles Situation A hardware store purchases a truckload of white ceramic floor tiles from a supplier know- ing that many of the tiles are imperfect.Imperfect means that the colour may not be uni- form, there may be surface hairline cracks, or there may be air pockets on the surface finish.The store will sell these at a marked-down price and it knows from past experience that it will have no problem selling these tiles as customers purchase these for tiling a base- ment or garage where slight imperfections are not critical.A store employee takes a ran- dom sample of 25 tiles from the storage area and counts the number of imperfections.This information is given in the table below.

7 4 1 2 3 4 3 3 2 8 5 5 2 3 1 3 1 6 7 5 8 2 3 4 8 Required 1.To the nearest whole number, what is an estimate of the mean number of imper- fections on the lot of white tiles? This would be a point estimate.What is an estimate of the standard error of the number of imperfections on the tiles? 3.Determine a 90% confidence interval for the mean amount of imperfections on the floor tiles.

This would mean that you would be 90% confident that the mean amount of imperfections lies within this range.Determine a 99% confidence interval for the mean amount of imperfections on the floor tiles.This would mean that you would be 99% confident that the mean amount of imperfections lies within this range.What is your explanation of the difference between the limits obtained in Questions 3 and 4? 8.World’s largest companies Situation Every year Fortune magazine publishes information on the world’s 500 largest com- panies.This information includes revenues, profits, assets, stock holders equity, number of employees, and the headquarters of the firm.The following table gives a random sam- ple of the revenues of 35 of those 500 firms for 2006, generated using the random func- tion in Excel.2 Company Revenues ($millions) Country Royal Mail Holdings 16,153.

0 United States Liberty Mutual Insurance 25,520.9 South Korea Archer Daniels Midland 36,596.7 Japan Matsushita Electric Industrial 77,871.0 United States Magna International 24,180.4 Japan 2 The World’s Largest Corporations, Fortune, Europe Edition, 156(2), 23 July 2007, p.257Chapter 7: Estimating population characteristics Company Revenues ($millions) Country Mediceo Paltac Holdings 18,524.

Using the complete sample data, what is an estimate for the average value of rev- enues for the world’s 500 largest companies? 2.Using the complete sample data, what is an estimate for the standard error? 3.Using the complete sample data, determine a 95% confidence interval for the mean value of revenues for the world’s 500 largest companies.

This would mean that you would be 95% confident that the average revenues lie within this range.Using the complete sample data, determine a 99% confidence interval for the mean value of revenues for the world’s 500 largest companies.This would mean that you would be 95% confident that the average revenue lies within this range.Explain the difference between the answers obtained in Questions 3 and 4.Using the first 15 pieces of data, give an estimate for the average value of revenues for the world’s 500 largest companies? 7.Using the first 15 pieces of data, what is an estimate for the standard error? 8.Using the first 15 pieces of data, determine a 95% confidence interval for the mean value of revenues for the world’s 500 largest companies.

This would mean that you would be 95% confident that the average revenue lies within this range.Using the first 15 pieces of data, determine a 99% confidence interval for the mean value of revenues for the world’s 500 largest companies.This would mean that you would be 95% confident that the average revenue lies within this range.Explain the difference between in the answers obtained in Questions 8 and 9.Explain the differences between the results in Questions 1 through 4 and those in Questions 6 through 9 and justify how you have arrived at your results.Hotel accounts Situation A 125-room hotel noted that in the morning when clients check out there are often questions and complaints about the amount of the bill.

These complaints included over- charging on items taken from the refrigerator in the room, wrong billing of restaurant meals consumed, and incorrect accounts of laundry items.On a particular day the hotel 258 Statistics for Business is full and the night manager analyses a random sample of 19 accounts and finds that there is an average of 2.Based on passed analysis the night manager believes that the population standard deviation is 0.From this sample experiment, what is the correct value of the standard error? 2.What are the confidence intervals for a 90% confidence level? 3.What are the confidence intervals for a 95% confidence level? 4.What are the confidence intervals for a 99% confidence level? 5.

Explain the differences between Questions 2, 3, and 4? 10.Automobile tyres Situation An automobile repair company has an inventory of 2,500 different sizes, and different makes of tyres.It wishes to estimate the value of this inventory and so it takes a random sample of 30 tyres and records their cost price.This sample information in Euros is given in the table below.44 34 66 48 42 36 88 76 68 34 89 73 69 55 72 88 60 74 80 75 57 36 95 50 61 41 32 62 91 65 Required 1.

What is an estimation of the cost price of the total amount of tyres in inventory? 2.Determine a 95% confidence interval for the cost price of the automobile tyres in inventory.How would you express the answers to Questions 1 and 2 to management? 4.

Determine a 99% confidence interval for the cost price of the automobile tyres in inventory.

Explain the differences between Questions 2 and 4? 6.How would you suggest a random sample of tyres should be taken from inventory? What other comments do you have? 11.Stuffed animals Situation A toy store in New York estimates that it has 270 stuffed animals in its store at the end of the week.An assistant takes a random sample of 19 of these stuffed animals and determines that the average retail price of these animals is $13.

259Chapter 7: Estimating population characteristics Required 1.What is the correct value of the standard error of the sample? 2.What is an estimate of the total value of the stuffed animals in the store? 3.

Give a 95% confidence limit of the total retail value of all the stuffed animals in inventory.Give a 99% confidence limit of the total retail value of all the stuffed animals in inventory.Explain the difference between Questions 3 and 4.

Shampoo bottles Situation A production operation produces plastic shampoo bottles for Procter and Gamble.At the end of the production operation the bottles pass through an optical quality control detector.Any bottle that the detector finds defective is automatically ejected from the line.In 1,500 bottles that passed the optical detector, 17 were ejected.

What is a point estimate of the proportion of shampoo bottles that are defective in the production operation? 2.Obtain 90% confidence intervals for the proportion of defective bottles produced in production.Obtain 98% confidence intervals for the proportion of defective bottles produced in production.

If an estimate of the proportion of defectives to within a margin of error of �0.005 of the population proportion at 90% confidence were required, and you wanted to be conservative in you analysis, how many bottles should pass through the optical detector? No information is available from past data.If an estimate of the proportion of defectives to within a margin of error of �0.

005 of the population proportion at 98% confidence were required, and you wanted to be conservative in you analysis, how many bottles should pass through the optical detector? No information is available from past data.What are your comments about the answer obtained in Question 4 and 5 and in gen- eral terms for this sampling process.Night shift Situation The management of a large factory, where there are 10,000 employees, is considering the introduction of a night shift.

The human resource department took a random sample of 800 employees and found that there were 240 who were not in favour of a night shift.260 Statistics for Business Required 1.What is the proportion of employees who are in favour of a night shift? 2.What are the 95% confidence limits for the population who are not in favour? 3.What are the 95% confidence limits for the proportion who are in favour of a night shift? 4.

What are the 98% confidence limits for the population who are not in favour? 5.What are the 98% confidence limits for the proportion who are in favour of a night shift? 6.What is your explanation of the difference between Questions 3 and 5? 14.Ski trip Situation The Student Bureau of a certain business school plans to organize a ski trip in the French Alps.The bureau selects a random sam- ple of 40 students and of these 24 say they will be coming skiing.What is an estimate of the proportion of students who say they will not be coming skiing? 2.Obtain 90% confidence intervals for the proportion of students who will be coming skiing.

Obtain 98% confidence intervals for the proportion of students who will be coming skiing.How would you explain the different between the answers to Questions 2 and 3? 5.What would be the conservative value of the sample size in order that the Student Bureau can estimate the true proportion of those coming skiing within plus or minus 0.02 at a confidence level of 90%? No other sample information has been taken.

What would be the conservative value of the sample size in order that the Student Bureau can estimate the true proportion of those coming skiing within plus or minus 0.02 at a confidence level of 98%? No other sample information has been taken.Hilton hotels Situation Hilton hotels, based in Watford, England, agreed in December 2005 to sell the inter- national Hilton properties for £3.

3 billion to United States-based Hilton group.This transaction will create a worldwide empire of 2,800 hotels stretching from the Waldorf- Astoria in New York to the Phuket Arcadia Resort in Thailand.3 The objective of this new 3 Timmons, H., “Hilton sets the stage for global expansion”, International Herald Tribune, 30 December 2005, p.261Chapter 7: Estimating population characteristics chain is to have an average occupancy, or a yield rate, of all the hotels at least 90%.In order to test whether the objectives are able to be met, a member of the finance depart- ment takes a random sample of 49 hotels worldwide and finds that in a 3-month test period, 32 of these had an occupancy rate of at least 90%.What is an estimate of the proportion or percentage of the population of hotels that meet the objectives of the chain? 2.What is a 90% confidence interval for the proportion of hotels who meet the object- ives of the chain? 3.

What is a 98% confidence interval for the proportion of hotels who meet the object- ives of the chain? 4.How would you explain the difference between the answers to Questions 2 and 3? 5.What would be the conservative value of the sample size that should be taken in order that the hotel chain can estimate the true proportion of those meeting the objectives is within plus or minus 10% of the true proportion at a confidence level of 90%? No earlier sample information is available.What would be the conservative value of the sample size that should be taken in order that the hotel chain can estimate the true proportion of those meeting the objectives is within plus or minus 10% of the true proportion at a confidence level of 98%? No earlier sample information is available.

What are your comments about this sample experiment that might explain inconsist- encies? 16.Case: Oak manufacturing Situation Oak manufacturing company produces kitchen appliances, which it sells on the European market.One of its new products, for which it has not yet decided to go into full commercial- ization, is a new computerized food processor.The company made a test market, during the first 3 months that this product was on sale.

Six stores were chosen for this study in the European cities of Milan, Italy; Hamburg, Germany; Limoges, France; Birmingham, United Kingdom; Bergen, Norway; and Barcelona, Spain.The weekly test market sales for these out- lets are given in the table below.Oak had developed this survey, because their Accounting Department had indicated that at least 130,000 of this food processor need to be sold in the first year of commercialization to break-even.They reasonably assumed that daily sales were independent from country to country, store to store, and from day to day.Management wanted to use a confidence level of 90% in its analysis.

For the first year of commercializa- tion after the “go” decision, the food processor is to be sold in a total of 100 stores in the six countries where the test market had been carried out.262 Statistics for Business Milan, Hamburg, Limoges, Birmingham, Bergen, Barcelona, Italy Germany France United Kingdom Norway Spain 3 29 15 34 25 21 8 29 16 22 19 0 20 13 32 31 25 5 8 22 31 28 35 14 17 23 32 23 25 16 11 20 15 20 20 9 12 29 16 26 34 13 3 17 46 39 29 11 6 22 27 24 24 3 13 26 20 35 33 16 12 19 28 37 36 4 13 21 2 20 39 1 15 47 28 27 38 15 0 31 29 30 12 18 15 33 36 34 33 6 5 42 33 25 26 18 2 32 18 21 35 14 17 13 33 26 30 21 19 19 28 16 28 14 18 23 27 31 34 20 17 20 34 23 20 19 12 20 16 25 29 9 17 17 30 12 20 12 6 34 32 22 36 1 Required Based on this information what would be your recommendations to the management of Oak manufacturing? 8Hypothesis testing of a single population You need to be objective The government in a certain country says that radiation levels in the area surrounding a nuclear power plant are well below levels considered harmful.Three people in the area died of leukaemia.The local people immediately put the blame on the radioactive fallout.Does the death of three people make us assume that the government is wrong with its information and that we make the assumption, or hypothesis, that radiation levels in the area are abnormally high? Alternatively, do we accept that the deaths from leukaemia are random and are not related to the nuclear power facility? You should not accept, or reject, a hypothesis about a population parameter – in this case the radiation levels in the surrounding area of the nuclear power plant, simply by intuition.

You need to be objective in decision-making.For this situation an appropriate action would be to take representative samples of the incidence of leukaemia cases over a reasonable time period and use these to test the hypothesis.This is the purpose of this chapter (and the following chapter) to find out how to use hypothesis testing to determine whether a claim is valid.There are many instances when published claims are not backed up by solid statistical evidence.A hypothesis is a judgment about a situation, outcome, or population parameter based simply on an assumption or intuition with no concrete backup information or analysis.

Hypothesis test- ing is to take sample data and make on objective decision based on the results of the test within an appropriate significance level.Thus like esti- mating, hypothesis testing is an extension of the use of sampling presented in Chapter 6.Significance level When we make quantitative judgments, or hypotheses, about situations, we are either right, or wrong.However, if we are wrong we may not be far from the real figure or that is our judgment is not significantly different.

Consider the following: ● A contractor says that it will take 9 months to construct a house for a client.The house is finished in 9 months and 1 week.The com- pletion time is not 9 months however it is not significantly different from the estimated time construction period of 9 months.● The local authorities estimate that there are 20,000 people at an open air rock concert.Ticket receipts indicate there are 42,000 attendees.

This number of 42,000 is signifi- cantly different from 20,000.● A financial advisor estimates that a client will make $15,000 on a certain investment.The number $14,900 is not $15,000 but it is not significantly different from $15,000 and the client really does not have a strong reason to complain.However, if the client made only $8,500 he would probably say that this is significantly different from the estimated $15,000 and has a justified reason to say that he was given bad advice.

Thus in hypothesis testing, we need to decide what we consider is the significance level or the level of importance in our evaluation.This significance level is giving a ceiling level usually in terms of Concept of Hypothesis Testing 264 Statistics for Business After you have studied this chapter you will understand the concept of hypothesis testing, how to test for the mean and proportion and be aware of the risks in testing.The topics of these themes are as follows: ✔ Concept of hypothesis testing • Significance level • Null and alternative hypothesis ✔ Hypothesis testing for the mean value • A two-tail test • One-tail, right-hand test • One-tail, left-hand test • Acceptance or rejection • Test statistics • Application when the standard deviation of the population is known: Filling machine • Application when the standard deviation of the population is unknown: Taxes ✔ Hypothesis testing for proportions • Testing for proportions from large samples • Application of hypothesis testing for proportions: Seaworthiness of ships ✔ The probability value in testing hypothesis • p-value of testing hypothesis • Application of the p-value approach: Filling machine • Application of the p-value approach: Taxes Application of the p-value approach: Seaworthiness of ships • Interpretation of the p-value ✔ Risks in hypothesis testing • Errors in hypothesis testing • Cost of making an error • Power of a test L e a r n i n g o b j e c t i v e s percentages such as 1%, 5%, 10%, etc.To a certain extent this is the subjective part of hypothesis test- ing since one person might have a different crite- rion than another individual on what is considered significant.However in accepting, or rejecting a hypothesis in decision-making, we have to agree on the level of significance.

This significance value, which is denoted as alpha, , then gives us the critical value for testing.Null and alternative hypothesis In hypothesis testing there are two defining statements premised on the binomial concept.One is the null hypothesis, which is that value considered correct within the given level of sig- nificance.The other is the alternative hypothesis, which is that the hypothesized value is not cor- rect at the given level of significance.The alter- native hypothesis as a value is also known as the research hypothesis since it is a value that has been obtained from a sampling experiment.

For example, the hypothesis is that the aver- age age of the population in a certain country is 35.The alter- native to the null hypothesis is that the average age of the population is not 35 but is some other value.In hypothesis testing there are three pos- sibilities.The first is that there is evidence that the value is significantly different from the hypothe- sized value.

The second is that there is evidence that the value is significantly greater than the hypothesized value.The third is that there is evi- dence that the value is significantly less than the hypothesized value.Note, that in these sen- tences we say there is evidence because as always in statistics there is no guarantee of the result but we are basing our analysis of the population based only on sampling and of course our sam- ple experiment may not yield the correct result.These three possibilities lead to using a two-tail hypothesis test, a right-tail hypothesis test, and a left-tail hypothesis test as explained in the next section.In hypothesis testing for the mean, an assump- tion is made about the mean or average value of the population.

Then we take a sample from this population, determine the sample mean value, and measure the difference between this sample mean and the hypothesized population value.If the difference between the sample mean and the hypothesized population mean is small, then the higher is the probability that our hypothe- sized population mean value is correct.If the difference is large then the smaller is the prob- ability that our hypothesized value is correct.A two-tail test A two-tail test is used when we are testing to see if a value is significantly different from our hypothesized value.For example in the above population situation, the null hypothesis is that the average age of the population is 35 years and this is written as follows: Null hypothesis: H0: x � 35 8(i) In the two-tail test we are asking, is there evi- dence of a difference.

In this case the alternative to the null hypothesis is that the average age is not 35 years.This is written as, Alternative hypothesis: H1: x � 35 8(ii) When we ask the question is there evidence of a difference, this means that the alternative value can be significantly lower or higher than the hypothesized value.For example, if we took a sample from our population and the average age of the sample was 36.2 years we might say that the average age of the population is not signifi- cantly different from 35.In this case we would accept the null hypothesis as being correct.

However, if in our sample the average age was Hypothesis Testing for the Mean Value 265Chapter 8: Hypothesis testing of a single population 52.7 years then we may conclude that the aver- age age of the population is significantly different from 35 years since it is much higher.Alterna- tively, if in our sample the average age was 21.2 years then we may also conclude that the aver- age age of the population is significantly different from 35 years since it is much lower.In both of these cases we would reject the null hypothesis and accept the alternative hypothesis.

Since this is a binomial concept, when we reject the null hypothesis we are accepting the alternative hypothesis.Conceptually the two-tailed test is illustrated in Figure 8.Here we say that there is a 10% level of significance and in this case for a two-tail test there is 5% in each tail.

One-tail, right-hand test A one-tail, right-hand test is used to test if there is evidence that the value is significantly greater than our hypothesized value.

For example in the above population situation, the null hypothesis is that the average age of the population is equal to or less than 35 years and this is written as follows: Null hypothesis: H0: x � 35 8(iii) The alternative hypothesis is that the average age is greater than 35 years and this is written as, Alternative hypothesis: H1: x � 35 8(iv) Thus, if we took a sample from our population and the average age of the sample was say 36.2 years we would probably say that the average age of the population is not significantly greater than 35 years and we would accept the null hypoth- esis.Alternatively, if in our sample the average age was 21.2 years then although this is significantly less than 35, it is not greater than 35.Again we would accept the null hypothesis.

However, if in 266 Statistics for Business Figure 8.Question being asked, “Is there evidence of a difference?” “Is there evidence that the average age is not 35?” H0:�x � 35 H1:�x � 35 Reject null hypothesis if sample mean falls in either of these regions 35 At a 10% significance level, there is 5% of the area in each tail If sample means falls in this region we accept the null hypothesis our sample the average age was 52.7 years then we may conclude that the average age of the popu- lation is significantly greater than 35 years and we would reject the null hypothesis and accept the alternative hypothesis.Note that for this situ- ation we are not concerned with values that are significantly less than the hypothesized value but only those that are significantly greater.

Again, since this is a binomial concept, when we reject the null hypothesis we accept the alternative hypothesis.Conceptually the one-tail, right-hand test is illustrated in Figure 8.Again we say that there is a 10% level of significance, but in this case for a one-tail test, all the 10% area is in the right-hand tail.One-tail, left-hand test A one-tail, left-hand test is used to test if there is evidence that the value is significantly less than our hypothesized value.

For example again let us consider the above population situation.The null hypothesis, H0: x, is that the average age of the population is equal to or more than 35 years and this is written as follows: H0: x 35 8(v) The alternative hypothesis, H1: x, is that the average age is less than 35 years.This is written, H1: x � 35 8(vi) Thus, if we took a sample from our population and the average age of the sample was say 36.2 years we would say that there is no evidence that the average age of the population is significantly less than 35 years and we would accept the null hypothesis.Or, if in our sample the average age was 52.

7 years then although this is significantly greater than 35 it is not less than 35 and we would accept the null hypothesis.However, if in our sample the average age was 21.2 years then we may conclude that the average age of the 267Chapter 8: Hypothesis testing of a single population Figure 8.2 One-tailed hypothesis test (right hand).Question being asked, “Is there evidence of something being greater?” “Is there evidence that the average age is greater than 35?” H0:�x � 35 H1:�x � 35 If sample means falls in this region we accept the null hypothesis Reject null hypothesis if sample mean falls in this region 35 At a 10% significance level, all the area is in the right tail population is significantly less than 35 years and we would reject the null hypothesis and accept the alternative hypothesis.

Note that for this situation we are not concerned with values that are signifi- cantly greater than the hypothesized value but only those that are significantly less than the hypothesized value.Again, since this is a binomial concept, when we reject the null hypothesis we accept the alternative hypothesis.Conceptually the one-tail, left-hand test is illustrated in Figure 8.With the 10% level of significance shown means that for this one-tail test all the 10% area is in the left-hand tail.

Acceptance or rejection The purpose of hypothesis testing is not to ques- tion the calculated value of the sample statistic, but to make an objective judgment regarding the difference between the sample mean and the hypothesized population mean.If we test at the 10% significance level this means that the null hypothesis would be rejected if the differ- ence between the sample mean and the hypoth- esized population mean is so large than it, or a larger difference would occur, on average, 10 or fewer times in every 100 samples when the hypothesized population parameter is correct.Assuming the hypothesis is correct, then the significance level indicates the percentage of sample means that are outside certain limits.Even if a sample statistic does fall in the area of acceptance, this does not prove that the null hypothesis H0 is true but there simply is no sta- tistical evidence to reject the null hypothesis.Acceptance or rejection is related to the values of the test statistic that are unlikely to occur if the null hypothesis is true.

However, they are not so unlikely to occur if the null hypothesis is false.3 One-tailed hypothesis test (left hand).Question being asked, “Is there evidence of something being less than?” “Is there evidence that the average age is less than 35?” If sample means falls in this region we accept the null hypothesis Reject null hypothesis if sample mean falls in this region 35 At a 10% significance level, all the area is in the left tail H0:�x � 35 H1:�x � 35 Test statistics We have two possible relationships to use that are analogous to those used in Chapter 7.If the population standard deviation is known, then using the central limit theorem for sampling, the test statistic, or the critical value is, test statistics, 8(vii) Where, ● H0 is the hypothesized population mean.

● The numerator, x– � x, measures how far, the observed mean is from the hypothesized mean.● x is the population standard deviation.● the denominator in the equation, is the standard error.

● z, is how many standard errors, the observed sample mean is from the hypothesized mean.If the population standard deviation is unknown then the only standard deviation we can determine is the sample standard deviation, s.This value of s can be considered an estimate of the population standard deviation sometimes written as x.If the sample size is less than 30 then we use the Student-t distribution, pre- sented in Chapter 7, with (n � 1) degrees of freedom making the assumption that the popu- lation from which this sample is drawn is nor- mally distributed.

In this case, the test statistic can be calculated by, 8(viii) Where, ● H0 is again the hypothesized population mean.

● The numerator, x– � H0, measures how far, the observed mean is from the hypothesized mean.● x is the estimate of the population standard deviation and is equal to the sample standard deviation, s.● the denominator in the equation, is the estimated standard error.

● t, is how many standard errors, the observed sample mean is from the hypothesized mean.The following applications illustrate the proced- ures for hypothesis testing.Application when the standard deviation of the population is known: Filling machine A filling line of a brewery is for 0.50 litre cans where it is known that the standard deviation of the filling machine process is 0.The qual- ity control inspector performs an analysis on the line to test whether the process is operating according to specifications.If the volume of liq- uid in the cans is higher than the specification limits then this costs the firm too much money.If the volume is lower than the specifications then this can cause a problem with the external inspectors.A sample of 25 cans is taken and the average of the sample volume is 0.At a significance level, , of 5% is there evidence that the volume of beer in the cans from this bot- tling line is different than the target volume of 0.50 litre? Here we are asking the question if there is there evidence of a difference so this means it is a two-tail test.The null and alternative hypotheses are written as follows: Null hypothesis: H0: x � 0., x n t x nx � � H0 x n , z x nx � � H0 269Chapter 8: Hypothesis testing of a single population 270 Statistics for Business Figure 8.89 Critical valueCritical value Test statistic 2.0% of area And, since we know the population standard deviation we can use equation 8(vii) where, ● H0 is the hypothesized population mean, or 0.

● x is the population standard deviation, or 0.Thus, the standard error of the sample is 0.The test statistic from equation 8(vii) is, At a significance level of 5% for the test of a difference there is 2.

Since the value of the test statistic or 1.96, or alterna- tively within the boundaries of �1.96 then there is no statistical evidence that the volume of beer in the cans is significantly different than 0.Thus we would accept the null hypoth- esis.These relationships are shown in Figure 8.

z x nx � � � � H0 0 0189 0 01 1 8900 .At a significance level, , of 5% is there evidence that the volume of beer in the cans from this bottling line is greater than the target volume of 0.50 litre? Here we are asking the question if there is evi- dence of the value being greater than the tar- get value and so this is a one-tail, right-hand test.The null and alternative hypotheses are as follows: Null hypothesis: H0: x � 0.Nothing has changed regarding the test statis- tic and it remains 1.However for a one-tail test, at a significance level of 5% for the test there is 5% in the right tail.The area of the curve for the upper level is 100% � 5.

Since now the value of the test statistic or 1.

89 is greater than the critical value of 1.64 then there is evidence that the volume of beer in all of the cans is significantly greater than 0.Conceptually this situ- ation is shown on the normal distribution curve in Figure 8.271Chapter 8: Hypothesis testing of a single population Application when the standard deviation of the population is unknown: Taxes A certain state in the United States has made its budget on the bases that the average individual average tax payments for the year will be $30,000.The financial controller takes a ran- dom sample of annual tax returns and these amounts in United States dollars are as follows.this can be taken as an estimate of the popu- lation standard deviation, x.Estimate of the standard error is, From equation 8(vii) the sample statistic is, Since the sample statistic, �1.8523, is not less than the test statistic of �2.

1315, there is no reason to reject the null hypothesis and so we accept that there is no evidence that the average of all the tax receipts will be signifi- cantly different from $30,000.Note in this situation, as the test statistic is negative we are on the left side of the curve and so we only make an evaluation with the negative values of t.Another way of making the analysis, when we are looking to see if there is a differ- ence, is to see whether the sample statistic of �1.8523 lies within the critical boundary values of t � �2.t x nx � � �� �� H0 8 500 4 453 93 1 8523 , , .

x n � � 17 815 72 16 4 453 93 34,000 12,000 16,000 10,000 2,000 39,000 7,000 72,000 24,000 15,000 19,000 12,000 23,000 14,000 6,000 43,000 1.At a significance level, , of 5% is there evidence that the average tax returns of the state will be different than the budget level of $30,000 in this year? The null and alternative hypotheses are as follows: Null hypothesis: H0: x � $30,000.Alternative hypothesis: H1: x � $30,000.

Since we have no information of the popula- tion standard deviation, and the sample size is less than 30, we use a Student-t distribution.Note, that since this is a two-tail test there is 2.5% of the area in each of the tails and t has a plus or minus value.Mean value of this sample data, x–, is $21,750.8523 Critical valueCritical value Test statistic 2.

5% of area 272 Statistics for Business 2.At a significance level, , of 5% is there evidence that the tax returns of the state will be less than the budget level of $30,000 in this year? This is a left-hand, one-tail test and the null and alternative hypothesis are as follows: Null hypothesis: H0: x $30,000.Alternative hypothesis: H1: x � $30,000.Again, since we have no information of the population standard deviation, and the sam- ple size is less than 30, we use a Student-t distribution.

Here we have a one-tail test and thus all of the value of , or 5%, lies in one tail.The value of the sample statistic t remains unchanged at �1.8523 is less than the test statistic, �1.

8523, then there is reason to reject the null hypothesis and to accept the alternative hypothesis that there is evidence that the average value of all the tax receipts is significantly less than $30,000.Note that in this situation we are on the left side of the curve and so we are only interested in the negative value of t.This situation is conceptually shown on the Student-t distribution curve of Figure 8.In hypothesis testing for the proportion we test the assumption about the value of the popula- tion proportion.

In the same way for the mean, we take a sample from this population, deter- mine the sample proportion, and measure the difference between this proportion and the hypothesized population value.If the difference between the sample proportion and the hypoth- esized population proportion is small, then the higher is the probability that our hypothesized population proportion value is correct.If the dif- ference is large then the probability that our hypothesized value is correct is low.Hypothesis testing for proportions from large samples In Chapter 6, we developed the relationship from the binomial distribution between the popula- tion proportion, p, and the sample proportion p–.

On the assumption that we can use the normal distribution as our test reference then from equation 6(xii) we have the value of z as follows: 6(xii) In hypothesis testing for proportions we use an analogy as for the mean where p is now the z p p p p p p np � � � � � ( )1 Hypothesis Testing for Proportions Figure 8.

0% of area 273Chapter 8: Hypothesis testing of a single population hypothesized value of the proportion and may be written as pH0.

Thus, equation 6(xii) becomes, 8(ix) The application of the hypothesis testing for proportions is illustrated below.Application of hypothesis testing for proportions: Seaworthiness of ships On a worldwide basis, governments say that 0.80, or 80%, of merchant ships are seaworthy.Greenpeace, the environmental group, takes a random sample of 150 ships and the analysis indicates that from this sample, 111 ships prove to be seaworthy.At a 5% significance level, is there evidence to suggest that the seaworthiness of ships is differ- ent than the hypothesized 80% value? Since we are asking the question is there a difference then this is a two-tail test with 2.The hypothesis test is written as follows: H0: p � 0.The proportion of ships that are seaworthy is equal to 0.The proportion of ships that are not seaworthy is different from 0.Sample proportion p– that is seaworthy is 111/150 � 0.From the sample, the number of ships that are not seaworthy is 39 (150 � 111).Sample proportion q– � (1 � p–) that is not seaworthy is 39/150 � 0.z p p p p p p np � � � � � H H H H 0 0 0 0 1( ) The standard error of the proportion, or the denominator in equation 8(ix).Thus the sample test statistic from equation 6(xii) is, Since the test statistic of �1.9600 then we accept the null hypothesis and say that at a 5% significance level there is no evidence of a significance dif- ference between the 80% of seaworthy ships postulated.Conceptually this situation is shown in Figure 8.At a 5% significance level, is there evidence to suggest that the seaworthiness of ships is less than the 80% indicated? This now becomes a one-tail, left-hand test where we are asking is there evidence that z p p p � � �� �� 0 06 0 0327 1 8349 .p p p n � � � � � H H0 0 1 0 80 0 20 150 0 16 150 0 0327 ( ) .

8349 Critical valueCritical value Test statistic 2.

5% of area 274 Statistics for Business the proportion is less than 80% The hypoth- esis test is thus written as, H0: p 0.The proportion of ships is not less than 0.In this situation the value of the sample statistics remains unchanged at �1.

8349, but the critical value of z is different.Now we reject the null hypothesis because the value of the test statistic, �1.8349 is less than the critical value of �1.

Thus our conclusion is that there is evidence that the proportion of ships that are not sea- worthy is significantly less than 0.Conceptually this situation is shown on the distribution in Figure 8.Up to this point our method of analysis has been to select a significance level for the hypothesis, The Probability Value in Testing Hypothesis which then translates into a critical value of z or t, and then test to see whether the sample statistic lies within the boundaries of the critical value.If the test statistic falls within the boundaries then we accept the null hypothesis.If the test statistic falls outside, then we reject the null hypothesis and accept the alternative hypothesis.Thus we have created a binomial “yes” or “no” situation by examining whether there is suffi- cient statistical evidence to accept or reject the null hypothesis.p-value of testing hypothesis An alternative approach to hypothesis testing is to ask, what is the minimum probable level that we will tolerate in order to accept the null hypothesis of the mean or the proportion? This level is called the p-value or the observed level of significance from the sample data.

It answers the question that, “If H0 is true, what is the probability of obtaining a value of x–, (or p–, in the case of propor- tions) this far or more from H0.If the p-value, as determined from the sample, is to the null hypothesis is accepted.Alternatively, if the p-value is less than then the null hypothesis is rejected and the alternative hypothesis is accepted.The use of the p-value approach is illustrated by re-examining the previous applications, Filling machine, Taxes, and Seaworthiness of ships.Application of the p-value approach: Filling machine 1.

At a significance level, , of 5% is there evidence that the volume of beer in the cans from this bottling line is different than the target volume of 0.50 litre? As before, a sample of 25 cans is taken and the average of the sample volume is 0.The test statistic, z x nx � � � � H0 0 0189 0 01 1 8900 .8900 area of the curve from the left is 97.

Thus the area in the right-hand tail is 100% � 97.Since this is a two-tail test the area in the left tail is also 2.Since we have a two-tail test, the area in each of the tail set by the significance level is 2.50% then we accept the null hypothesis and conclude that the volume of beer in the cans is not different from 0.

At a significance level, , of 5% is there evidence that the volume of beer in the cans from this bottling line is greater than the target volume of 0.50 litre? The value of the test statistic of 1.

8900 gives an area in the right-hand tail of 2.We now have a one-tail, right-hand test when the significance level is 5%.00% we reject the null hypothesis and accept the alternative hypoth- esis and conclude that there is evidence that the volume of beer in the cans is greater than 0.Application of the p-value approach: Taxes 1.At a significance level, , of 5% is there evidence that the average tax returns of the state will be different than the budget level of $30,000 in this year? The sample statistic gives a t-value which is equal to �1.

8523, for a two-tail test, indi- cates a probability of 8.00% we accept the null hypothesis and conclude that there is no evi- dence to indicate that the average tax receipts are significantly less than $30,000.At a significance level, , of 5% is there evidence that the tax returns of the state will be less than the budget level of $30,000 in this year? The sample statistic gives a Student-t value which is equal to �1.00% we reject the null hypothesis and conclude that there is evi- dence to indicate that the average tax receipts are significantly less than $30,000.Application of the p-value approach: Seaworthiness of ships 1.As this is a two-tail test then there is 2.50% we accept the null hypothesis and conclude that there is no evidence to indicate that the seaworthiness of ships is different from the hypothesized value of 80%.At a 5% significance level, is there evidence to suggest that the seaworthiness of ships is less than the 80% indicated? As this is a one-tail, left-hand test then there is 5% in the tail.

00% we reject the null hypothesis and conclude that there is evidence to indicate that the seaworthiness of ships is less than the hypothesized value of 80%.z p p p � � �� �� 0 06 0 0327 1 8349 .

275Chapter 8: Hypothesis testing of a single population 276 Statistics for Business Sample Test statistic p-value mean x-bar z % 0.1 Sample mean and a corresponding z and p-value.

Interpretation of the p-value In hypothesis testing we are making inferences about a population based only on sampling.The sampling distribution permits us to make probabil- ity statements about a sample statistic on the basis of the knowledge of the population parameter.In the case of the filling machine for example where we are asking is there evidence that the volume of beer in the can is greater than 0.The probability of obtaining a sample mean of 0.5189 litre from a population whose mean is 0.Thus we have observed an unlikely event or an event so unlikely that we should doubt our assumptions about the population mean in the first place.

Note, that in order to calculate the value of the test statistic we assumed that the null hypothesis is true and thus we have reason to reject the null hypothesis and accept the alternative.The p-value provides useful information as it measures the amount of statistical evidence that supports the alternative hypothesis.1, which gives values of the sample mean, the value of the test statistic, and the correspond- ing p-value for the filling machine situation.As the sample mean gets larger, or moves further away from the hypothesized population mean of 0.

5000 litre tend to indicate that the alternative hypothesis is true or the smaller the p-value, the more the statistical evidence there is to support the alternative hypothesis.Remember that the p-value is not to be inter- preted by saying that it is the probability that the null hypothesis is true.

You cannot make a prob- ability assumption about the population param- eter 0.

5000 litre as this is not a random variable.In hypothesis testing there are risks when you sample and then make an assumption about the population parameter.This is to be expected since statistical analysis gives no guarantee of the result but you hope that the risk of making a wrong decision is low.Errors in hypothesis testing The higher the value of the significance level, used for hypothesis testing then the higher is the percentage of the distribution in the tails.In this case, when is high, the greater is the probabil- ity of rejecting a null hypothesis.

Since the null hypothesis is true, or is not true, then as increases there is a greater probability of reject- ing the null hypothesis when in fact it is true.Looking at it another way, with a high signifi- cance level, that is a high value of , it is unlikely we would accept a null hypothesis when it is in fact not true.This relationship is illustrated in the normal distributions of Figure 8.At the 1% significance level, the probability of accept- ing the hypothesis, when it is false is greater than at a significance level of 50%.

Alternatively, the risk of rejecting a null hypothesis when it is in fact true is greater at a 50% significance level, than at a 1% significance level.These errors in hypothesis testing are referred to as Type I or Type II errors.Risks in Hypothesis Testing A Type I error occurs if the null hypothesis is rejected when in fact it is true.The probability of a Type I error is called where is also the level of significance.A Type II error is accepting a null hypothesis when it is not true.

The probability of a Type II error is called .When the acceptance region is small, or is large, it is unlikely we would accept a null hypothesis when it is false.However, at a risk of being this sure, we will often reject a null hypothesis when it is in fact true.The level of significance to use depends on the cost of the error as illustrated as follows.Cost of making an error Consider that a pharmaceutical firm makes a certain drug.

A quality inspector tests a sample of the product from the reaction vessel where the drug is being made.He makes a Type I error in his analysis.That is he rejects a null hypoth- esis when it is true or concludes from the sample that the drug does not conform to quality speci- fications when in fact it really does.As a result, all the production quantity in the reaction ves- sel is dumped and the firm starts the production all over again.In reality the batch was good and could have been accepted.

In this case, the firm incurs all the additional costs of repeating the production operation.Alternatively, suppose the quality inspector makes a Type II error, or accepts a null hypothesis when it is in fact false.In this case the produced pharmaceutical product is accepted and commercialized but it does not conform to quality specifications.This may mean 277Chapter 8: Hypothesis testing of a single population The higher the significance level for testing the hypothesis, the greater is the probability of rejecting a null hypothesis when it is true.However, we would rarely accept a null hypothesis when it is not true.

5% of area 5% of area 5% of area x x x 99% 90% Significance level is the total area in the tail(s) 50% 25% of area 25% of area Significance level of 10% Significance level of 50% Figure 8.that users of the drug could become sick, or at worse die.

The “cost” of this error would be very high.In this situation, a pharmaceutical firm would prefer to make a Type I error, or destroy- ing the production lot, rather than take the risk of poisoning the users.This implies having a high value of such as 50% as illustrated in Figure 8.Suppose in another situation, a manufactur- ing firm is making a mechanical component that is used in the assembly of washing machines.

An inspector takes a sample of this component from the production line and measures the appropri- ate properties.He makes a Type I error in the analysis.He rejects the null hypothesis that the component conforms to specifications, when in fact the null hypothesis is true.In this case to correct this conclusion would involve an expen- sive disassembly operation of many components on the shop floor that have already been pro- duced.On the other hand if the inspector had made a Type II error, or accepting a null hypoth- esis when it is in fact false, this might involve a less expensive warranty repairs by the dealers when the washing machines are commercial- ized.

In this latter case, the cost of the error is rela- tively low and manufacturer is more likely to prefer a Type II error even though the marketing image may be damaged.In this case, the manu- facturer will set low levels for such as 10% as illustrated in Figure 8.The cost of an error in some situations might be infinite and irreparable.

Under Anglo-Saxon law the null hypothesis, is that a person if charged with mur- der is considered innocent of the crime and the court has to prove guilt.In this case, the jury would prefer to commit a Type II error or accept- ing a null hypothesis that the person is innocent, when it is in fact not true, and thus let the guilty person go free.The alternative would be to accept a Type I error or rejecting the null hypothesis that the person is innocent, when it is in fact true.In this case the person would be found guilty and risk the death penalty (at least in the United States) for a crime that they did not commit.Power of a test In any analytical work we would like the proba- bility of making an error to be small.

Thus, in hypothesis testing we would like the probability of making a Type I error, , or the probability of making a Type II error to be small.Thus, if a null hypothesis is false then we would like the hypothesis test to reject this conclusion every time.However, hypothesis tests are not perfect and when a null hypothesis is false, a test may not reject it and consequently a Type II error, , is made or that is accepting a null hypothesis when it is false.When the null hypothesis is false this implies that the true population value, does not equal the hypothesized population value but instead equals some other value.For each possible value for which the alternative hypothesis is true, or the null hypothesis is false, there is a different probability, of accepting the null hypothesis when it is false.

We would like this value of to be as small as possible.Alternatively, we would like (1 � ) the probability of rejecting a null hypothesis when it is false, to be as large as pos- sible.Rejecting a null hypothesis when it is false is exactly what a good hypothesis test ought to do.Alternatively, a low value of (1 � ) approaching zero means that the test is not working well and the test is not rejecting the null hypothesis when it is false.The value of (1 � ), the measure of how well the test is doing, is called the power of the test.2 summarizes the four possibilities that can occur in hypothesis testing and what type of errors might be incurred.Again, as in all statisti- cal work, in order to avoid errors in hypothesis testing, utmost care must be made to ensure that the sample taken is a true representation of the population.

278 Statistics for Business 279Chapter 8: Hypothesis testing of a single population Decision you make In reality for the population In reality for the population – null hypothesis, H0 is true – null hypothesis, H0 is false – what your test indicates – what your test indicates Null hypothesis, • Test statistic falls in the region • Test statistic falls in the region H0 is accepted (1 � ) (1 � ) • Decision is correct • Decision is incorrect • No error is made • A Type II error, is made Null hypothesis, • Test statistic falls in the region • Test statistic falls in the region H0 is rejected • Decision is incorrect • Decision is correct • A Type I error, is made • No error is made • Power of test is (1 � ) Table 8.2 Sample mean and a corresponding z and p-value.This chapter has dealt with hypothesis testing or making objective decisions based on sample data.The chapter opened with describing the concept of hypothesis testing, then presented hypothesis testing for the mean, hypothesis testing for proportions, the probability value in test- ing hypothesis, and finally summarized the risks in hypothesis testing.Concept of hypothesis testing Hypothesis testing is to sample from a population and decide whether there is sufficient evidence to conclude that the hypothesis appears correct.

In testing we need to decide on a significance level, which is the level of importance in the difference between values before we accept an alternative hypothesis.The significance level establishes a critical value, which is the barrier beyond which decisions will change.The concept of hypothesis testing is binomial.There is the null hypothesis denoted by H0, which is the announced value.Then there is the alternative hypothesis, H1 which is the other situation we accept should we reject the null hypothesis.

When we reject the null hypothesis we automatically accept the alternative hypothesis.Hypothesis testing for the mean value In hypothesis testing for the mean we are trying to establish if there is statistical evidence to accept a hypothesized average value.The first is to estab- lish if there is a significant difference from the hypothesize mean.Another is to test to see if there is evidence that a value is significantly greater than the hypothe- sized amount.This gives rise to a one-tail, right-hand test.The third is a left-hand test that decides if a value is significantly less than a hypothesized value.In all of these tests the first step is to determine a sample test value, either z, or t, depending on our knowledge of the population.We C hapter Sum m ary 280 Statistics for Business then compare this test value to our critical value, which is a direct consequence of our signifi- cance level.

If our test value is within the limits of the critical value, we accept the null hypoth- esis.Otherwise we reject the null hypothesis and accept the alternative hypothesis.Hypothesis testing for proportions The hypothesis test for proportions is similar to the test for the mean value but here we are trying to see if there is sufficient statistical evidence to accept or reject a hypothesized population proportion.The criterion is that we can assume the normal distribution in our analytical procedure.As for the mean, we can have a two-tail test, a one-tail, left-hand test, or a one-tail, right-hand test.

We estab- lish a significance level and this sets our critical value of z.We than determine the value of our sam- ple statistic and compare this to the critical value determined from our significance level.If the test statistic is within our boundary limits we accept the null hypothesis, otherwise we reject it.The probability value in testing hypothesis The probability, or p-value, for hypothesis testing is an alternative approach to the critical value method for testing assumptions about the population mean or the population proportion.

The p-value is the minimum probability that we will tolerate before we reject the null hypothesis.

When the p-value is less than , our level of significance, we reject the null hypothesis and accept the alternative hypothesis.Risks in hypothesis testing As in all statistical methods there are risks when hypothesis testing is carried out.If we select a high level of significance, which means a large value of the greater is the risk of rejecting a null hypothesis when it is in fact true.However if we have a high value of , the risk of accepting a null hypothesis when it is false is low.

A Type II error called occurs if we accept a null hypothesis when it is in fact false.The value of (1 � ), is a measure of how well the test is doing and is called the power of the test.The closer the value of (1 � ) is to unity implies that the test is working quite well.281Chapter 8: Hypothesis testing of a single population 1.Sugar Situation One of the processing plants of B ghin Say, the sugar producer, has problems controlling the filling operation for its 1 kg net weight bags of white sugar.

The quality control inspector takes a random sample of 22 bags of sugar and finds that the weight of this sample is 1,006 g.It is known from experience that the standard deviation of the filling operation is 15 g.At a significance level of 5% for analysis, using the critical value method, is there evi- dence that the net weight of the bags of sugar is different than 1 kg? 2.If you use the p-value for testing are you able to verify your conclusions in Question 1? Explain your reasoning.

What are the confidence limits corresponding to a significance level of 5%.How do these values corroborate your conclusions for Questions 1 and 2? 4.At a significance level of 10% for analysis, using the critical value method, is there evidence that the net weight of the bags of sugar is different than 1 kg? 5.If you use the p-value for testing are you able to verify your conclusions in Question 4? Explain your reasoning.

What are the confidence limits corresponding to a significance level of 10%.How do these values corroborate your conclusions for Questions 4 and 5? 7.Why is it necessary to use a difference test? Why should this processing plant be con- cerned with the results? 2.Neon lights Situation A firm plans to purchase a large quantity of neon light bulbs from a subsidiary of GE for a new distribution centre that it is building.

The subsidiary claims that the life of the light bulbs is 2,500 hours, with a standard deviation of 40 hours.Before the firm finalizes the purchase it takes a random sample of 20 neon bulbs and tests them until they burn out.The average life of the sample of these bulbs is 2,485 hours.(Note, the firm has a spe- cial simulator that tests the bulb and in practice it does not require that the bulbs have to be tested for 2,500 hours.Using the critical value approach, at a 5% significance level, is there evidence to sug- gest that the life of the light bulbs is different than 2,500 hours? 2.If you use the p-value for testing are you able to verify your conclusions in Question 1? Explain your reasoning.EXERCISE PROBLEMS 282 Statistics for Business 3.Using the critical value approach, at a 5% significance level, is there evidence to sug- gest that the life of the light bulbs is less than 2,500 hours? 4.If you use the p-value for testing are you able to verify your conclusions in Question 3? Explain your reasoning.

If the results from the Questions 3 and 4 what options are open to the purchasing firm? 3.Graphite lead Situation A company is selecting a new supplier for graphite leads which it uses for its Pentel-type pencils.The supplier claims that the average diameter of its leads is 0.The company wishes to verify this claim because if the lead is significantly too thin it will break.If it is significantly too thick it will jam in the pencil.It takes a sample of 30 of these leads and measures the diameter with a micrometer gauge.The diameter of the samples is given in the table below.

At a 5% significance level, using the critical value concept, is there evidence to sug- gest that the diameter of the lead is different from the supplier’s claim? 2.At a 5% significance level, using the p-value concept, verify your answer obtained in Question 1.What are the confidence limits corresponding to a significance level of 5%.How do these values corroborate your conclusions for Questions 1 and 2? 4.At a 10% significance level, using the critical value concept, is there evidence to sug- gest that the diameter of the lead is different from the supplier’s claim? 5.At a 10% significance level, using the p-value concept, verify your answer obtained in Question 1.What are the confidence limits corresponding to a significance level of 10%.How do these values corroborate your conclusions for Questions 4 and 5? 7.The mean of the sample data is an indicator whether the lead is too thin or too thick.If you applied the appropriate one-tail test what conclusions would you draw? Explain your logic.

Industrial pumps Situation Pumpet Corporation manufactures electric motors for many different types of industrial pumps.One of the parts is the drive shaft that attaches to the pump.7600 283Chapter 8: Hypothesis testing of a single population for the drive shafts is that they should not be below a certain diameter.If this is the case, then when in use, the shaft vibrates, and eventually breaks.

In the way that the drive shafts are machined, there are never problems of the shafts being oversized.For one par- ticular model, MT 2501, the specification calls for a nominal diameter of the drive shaft of 100 mm.The company took a sample of 120 drive shafts from a large manufactured lot and measured their diameter.The results were as follows: Required 1.

Pumpet normally uses a significance level of 5% for its analysis.

In this case, using the critical value method, is there evidence that the shaft diameter of model MT 2501 is significantly below 100 mm? If so there would be cause to reject the lot.If you use the p-value for testing are you able to verify your conclusions in Question 1? Explain your reasoning.A particular client of Pumpet insists that a significance level of 10% be used for analy- sis as they have stricter quality control limits.Using this level, and again making the test using the critical value criteria, is there evidence that the drive shaft diameter is significantly below 100 mm causing the lot to be rejected? Explain your reasoning.If you use the p-value for testing are you able to verify your conclusions in Question 3? Explain your reasoning.If instead of using the whole sample size indicated in the table you used just the data in the first three columns, how would your conclusions from the Questions 1 to 4 change? 6.From your answer to Question 5, what might you recommend? 100.Automatic teller machines (ATMs) Situation Banks in France are closed for 2.5 days from Saturday afternoon to Tuesday morning.In this case banks need to have a reasonable estimate of how much cash to make available in their ATMs.

For BNP-Paribas in its branches in the Rhone region in the Southeast of France it estimates that for this 2.5-day period the demand from its customers from those branches with a single ATM machine is €3,200 with a population standard devia- tion of €105.A random sample of the withdrawal from 36 of its branches indicate a sample average withdrawal of €3,235.Using the concept of critical values, then at the 5% significance level does this data indicate that the mean withdrawal from the machines is different from €3,200? 2.

Re-examine Question 1 using the p-value approach.Are your conclusions the same? Explain your conclusions? 3.What are the confidence limits at 5% significance? How do these values corroborate your answers to Questions 1 and 2? 4.Using the concept of critical values then at the 1% significance level does this data indicate that the mean life of the population of the batteries is different from €3,200? 5.Re-examine Question 4 using the p-value approach.

Are your conclusions the same? Explain your conclusions.What are the confidence limits at 1% significance? How do these values corroborate your answers to Questions 4 and 5? 7.Here we have used the test for a difference.

Why is the bank interested in the differ- ence rather than a one-tail test, either left or right hand? 6.

Bar stools Situation A supplier firm to IKEA makes wooden bar stools of various styles.In the production process of the bar stools the pieces are cut before shaping and assembling.The specifi- cations require that the length of the legs of the bar stools is 70 cm.If the length is more than 70 cm they can be shaved down to the required length.However if pieces are significantly less than 70 cm they cannot be used for bar stools and are sent to another production area where they are re-cut to use in the assembly of standard chair legs.

In the production of the legs for the bar stools it is known that the standard deviation of the process is 2.In a production lot of legs for bar stools the quality 285Chapter 8: Hypothesis testing of a single population control inspector takes a random sample and the length of these is according to the following table.At a 5% significance level, using the concept of critical value testing, does this sample data indicate that the length of the legs is less than 70 cm? 2.

At the 5% significance level, using the p-value concept, does your answer corroborate the conclusion of Question 1? Give your reasoning.At a 10% significance level, using the concept of critical value testing, does this sam- ple data indicate that the length of the legs is less than 70 cm? 4.At the 10% significance level, using the p-value concept, does your answer corroborate the conclusion of Question 3? Give your reasoning.Since we know the standard deviation we are correct to use the normal distribution for this hypothesis test.Assume that we did not know the process standard deviation and as the sample size of 32 is close to a cut-off point of 30, we used the Student-t distribu- tion.In this case, would our analysis change the conclusions of Questions 1to 4? 7.Salad dressing Situation Amora salad dressing is made in Dijon in France.One of their products, made with wine, indicates on the label that the nominal volume of the salad dressing is 1,000 ml.

In the filling process the firm knows that the standard deviation is 5.The quality control inspector takes a random sample of 25 of the bottles from the production line and meas- ures their volumes, which are given in the following table.At the 5% significance level, using the concept of critical value testing, does this sam- ple data indicate that the volume of salad dressing in the bottles is different than the volume indicated on the label? 65 74 69 68 68 72 72 73 71 75 74 68 67 69 67 68 67 70 68 68 68 71 67 70 69 69 69 67 72 66 67 72 993.

At the 5% significance level, using the p-value concept, does your answer corrobor- ate the conclusion of Question 1? Give your reasoning.At the 5% significance level, what are the confidence intervals when the test is asking for a difference in the volume? How do these intervals confirm your answers to Questions 1 and 2? 4.At the 5% significance level, using the concept of critical value testing, does this data indicate that the volume of salad dressing in the bottles is less than the volume indicated on the label? 5.At the 5% significance level, using the p-value concept, does your answer corrob- orate the conclusion of Question 4? Give your reasoning.

Why is the test mentioned in Question 4 important? 7.What can you say about the sensitivity of this sampling experiment? 8.Apples Situation In an effort to reduce obesity among children, a firm that has many vending machines in schools is replacing chocolate bars with apples in its machines.Unlike chocolate bars that are processed and thus the average weight is easy to control, apples vary enormously in weight.

The vending firm asks its supplier of apples to sort them before they are delivered as it wants the average weight to be 200 g.The criterion for this is that the vending firm wants to be reasonably sure that each child who purchases an apple is getting one of equivalent weight.A truck load of apples arrives at the vendor’s depot and an inspector takes a ran- dom sample of 25 apples.The following is the weight of each apple in the sample.At the 5% significance level, using the concept of critical value testing, does this sample data indicate that the weight of the truck load of apples is different than the desired 200 g? 2.At the 5% significance level, using the p-value concept, does your answer corrob- orate the conclusion of Question 1? Give your reasoning.At the 5% significance level what are the confidence intervals when the test is asking for a difference in the volume.How do these intervals confirm your answers to Questions 1 and 2? 4.

At the 5% significance level, using the concept of critical value testing, does this sample data indicate that the weight of the truck load of apples is less than the desired 200 g? 5.At the 5% significance level, using the p-value concept, does your answer corrob- orate the conclusion of Question 4? Give your reasoning.198 201 202 186 199 199 208 196 196 196 207 195 187 199 189 195 190 195 197 209 199 205 203 190 199 287Chapter 8: Hypothesis testing of a single population 9.Batteries Situation A supplier of batteries claimed that for a certain type of battery the average life was 500 hours.The quality control inspector of a potential buying company took a random sam- ple of 15 of these batteries from a lot and tested them until they died.

The life of these batteries in hours is given in the table.Using the concept of critical values, then at the 5% significance level does this data indicate that the mean life of the population of the batteries is different from the hypothesized value? 2.Re-examine Question 1 using the p-value approach.Are your conclusions the same? Explain your reasoning? 3.

Using the concept of critical values then at the 5% significance level does this data indicate that the mean life of the population of the batteries is greater than the hypothesized value? 4.Re-examine Question 3 using the p-value approach.Are your conclusions the same? Explain your reasoning? 5.Explain the rationale for the differences in the answers to Questions 1 and 3, and the differences in the answers to Questions 2 and 4.Hospital emergency Situation A hospital emergency service must respond rapidly to sick or injured patients in order to increase rate of survival.A certain city hospital has an objective that as soon as it receives an emergency call an ambulance is on the scene within 10 minutes.The regional director wanted to see if the hospital objectives were being met.Thus during a weekend (the busiest time for hospital emergencies) a random sample of the time taken to respond to emergency calls were taken and this information, in minutes, is in the table below.350 925 796 689 501 485 546 551 512 589 489 568 685 578 398 8 14 15 20 7 12 7 8 21 13 9 17 22 10 9 288 Statistics for Business Required 1.

At the 5% significance level, using the concept of critical value testing, does this sam- ple data indicate that the response time is different from 10 minutes? 2.At the 5% significance level, using the p-value concept, does your answer corro- borate the conclusion of Question 1? Give your reasoning.At the 5% significance level what are the confidence intervals when the test is asking for a difference? How do these intervals confirm your answers to Questions 1 and 2? 4.At the 5% significance level, using the concept of critical value testing, does this data indicate that the response time for an emergency call is greater than 10 minutes? 5.

At the 5% significance level, using the p-value concept, does your answer corroborate the conclusion of Question 4? Give your reasoning.Which of these two tests is the most important? 11.Equality for women Situation According to Jenny Watson, the commission’s chair of the United Kingdom Sex Discrimination Act (SDA), there continues to be an unacceptable pay gap of 45% between male and female full time workers in the private sector.

1 A sample of 72 women is taken and of these 22 had salaries less than their male counterparts for the same type of work.

Using the critical value approach for a 1% significance level, is there evidence to sug- gest that the salaries of women is different than the announced amount of 45%? 2.Using the p-value approach are you able to corroborate your conclusions from Question 1.What are the confidence limits at the 1% level? How do they agree with your conclu- sions of Questions 1 and 2? 4.Using the critical value approach for a 5% significance level, is there evidence to sug- gest that the salaries of women is different than the announced amount of 45%? 5.Using the p-value approach are you able to corroborate your conclusions from Question 3.What are the confidence limits at the 5% level? How do they agree with your conclu- sions of Questions 1 and 2? 7.How would you interpret these results? 12.Gas from Russia Situation Europe is very dependent on natural gas supplies from Russia.In January 2006, after a bit- ter dispute with Ukraine, Russia cut off gas supplies to Ukraine but this also affected other 1 Overell, S., Act One in the play for equality, Financial Times, 5 January 2006, p.

289Chapter 8: Hypothesis testing of a single population European countries’ gas supplies.This event jolted European countries to take a re-look at their energy policies.Based on 2004 data the quantity of imported natural gas of some major European importers and the amount from Russia in billions of cubic metres was according to the table below.2 The amounts from Russia were on a contractual basis and did not necessarily correspond to physical flows.

Industrial users have gas flow monitors at the inlet to their facilities according to the source of the natural gas.Samples from 35 industrial users were taken from both Italy and Poland and of these 7 industrial users in Italy and 31 in Poland were using gas imported from Russia.Using the critical value approach at a 5% significance level, is there evidence to sug- gest that the proportion of natural gas Italy imports from Russia is different than the amount indicated in the table? 2.Using the p-value approach are you able to corroborate your conclusions from Question 1.

What are the confidence limits at the 5% level? How do they agree with your conclu- sions of Questions 1 and 2? 4.Using the critical value approach for a 10% significance level, is there evidence to suggest that the proportion of natural gas Italy imports from Russia is different than the amount indicated in the table? 5.Using the p-value approach are you able to corroborate your conclusions from Question 4.

What are the confidence limits at the 10% level? How do they agree with your conclusions of Questions 4 and 5? Country Total imports Imports from Russia (m3 billions) (m3 billions) Germany 91., “Russia blinks in gas fight as crisis rattles Europe”, The Wall Street Journal, 3 January 2005, pp.

Using the critical value approach at a 5% significance level, is there evidence to suggest that the proportion of natural gas Poland imports from Russia is different than the amount indicated in the table? 8.Using the p-value approach are you able to corroborate your conclusions from Question 7.What are the confidence limits at the 5% level? How do they agree with your con- clusions of Questions 7 and 8? 10.Using the critical value approach for a 10% significance level, is there evidence to suggest that the proportion of natural gas Poland imports from Russia is different than the amount indicated in the table? 11.Using the p-value approach are you able to corroborate your conclusions from Question 10.What are the confidence limits at the 10% level? How do they agree with your con- clusions of Questions 10 and 11? 13.How would you interpret these results of all these questions? 13.International education Situation Foreign students are most visible in Australian and Swiss universities, where they make up more than 17% of all students.Although the United States attracts more than a quar- ter of the world’s foreign students, they account for only some 3.

Almost half of all foreign students come from Asia, particularly China and India.Social sciences, business, and law are the fields of study most popular with overseas scholars.The table below gives information for selected countries for 2003.3 Country Foreign students Country Foreign students as % of total as % of total Australia 19.

0 3 Economic and financial indicators, The Economist, 17 September 2005, p.291Chapter 8: Hypothesis testing of a single population Random samples of 45 students were selected in Australia and in Britain.

Of those in Australia, 14 were foreign, and 10 of those in Britain were foreign.Using the critical value approach, at a 1% significance level, is there evidence to suggest that the proportion of foreign students in Australia is different from that indicated in the table? 2.Using the p-value approach are you able to corroborate your conclusions from Question 1.What are the confidence limits at the 1% level? How do they agree with your con- clusions of Questions 1 and 2? 4.Using the critical value approach, at a 5% significance level, is there evidence to suggest that the proportion of foreign students in Australia is different from that indicated in the table? 5.Using the p-value approach are you able to corroborate your conclusions from Question 4.What are the confidence limits at the 5% level? How do they agree with your con- clusions of Questions 1 and 2? 7.Using the critical value approach, at a 1% significance level, is there evidence to suggest that the proportion of foreign students in Britain is different from that indi- cated in the table? 8.Using the p-value approach are you able to corroborate your conclusions from Question 7.What are the confidence limits at the 1% level? How do they agree with your con- clusions of Questions 7 and 8? 10.Using the critical value approach, at a 5% significance level, is there evidence to suggest that the proportion of foreign students in Britain is different from that indi- cated in the table? 11.Using the p-value approach are you able to corroborate your conclusions from Question 10.What are the confidence limits at the 5% level? How do they agree with your con- clusions of Questions 10 and 11? 14.United States employment Situation According to the United States labour department the jobless rate in the United States fell to 4.It was reported that 108,000 jobs were created in December and 305,000 in November.

Taken together, these new jobs created over the past 2 months allowed the United States to end the year with about 2 million more jobs than it had 4 Andrews, E.S,” International Herald Tribune, 7/8 January 2006, p.

292 Statistics for Business 12 months ago.4 Random samples of 83 people were taken in both Palo Alto, California and Detroit, Michigan.Of those from Palo Alto, 4 said they were unemployed and 8 in Detroit said they were unemployed.Using the critical value approach, at a 5% significance level, is there evidence to suggest that the unemployment rate in Palo Alto is different from the national unemployment rate? 2.Using the p-value approach are you able to corroborate your conclusions from Question 1.What are the confidence limits at the 5% level? How do they agree with your con- clusions of Questions 1 and 2? 4.

Using the critical value approach, at a 10% significance level, is there evidence to suggest that the unemployment rate in Palo Alto is different from the national unemployment rate? 5.Using the p-value approach are you able to corroborate your conclusions from Question 4.What are the confidence limits at the 10% level? How do they agree with your con- clusions of Questions 4 and 5? 7.

Using the critical value approach, at a 5% significance level, is there evidence to suggest that the unemployment rate in Detroit is different from the national unemployment rate? 8.Using the p-value approach are you able to corroborate your conclusions from Question 7.What are the confidence limits at the 5% level? How do they agree with your con- clusions of Questions 7 and 8? 10.

Using the critical value approach, at a 10% significance level, is there evidence to suggest that the unemployment rate in Detroit is different from the national unemployment rate? 11.Using the p-value approach are you able to corroborate your conclusions from Question 10.What are the confidence limits at the 10% level? How do they agree with your con- clusions of Questions 10 and 11? 13.

Explain your results for Palo Alto and Detroit.Mexico and the United States Situation On 30 December 2005 a United States border patrol agent shot dead an 18-year-old Mexican as he tried to cross the border near San Diego, California.The patrol said the shooting was in self-defence and that the dead man was a coyote, or people smuggler.In 293Chapter 8: Hypothesis testing of a single population 2005, out of an estimated 400,000 Mexicans who crossed illegally into the United States, more than 400 died in the attempt.

Illegal immigration into the United States has long been a problem and to control the movement there are plans to construct a fence along more than a third of the 3,100 km border.According to data for 2004, there are some 10.5 million Mexicans in the United States, which represents some 31% of the foreign-born United States population.The recorded Mexicans in the United States of America is equiv- alent to 9% of Mexico’s total population.In addition, it is estimated that there are some 10 million undocumented immigrants in the United States of which 60% are considered to be Mexican.

5 A random sample of 57 foreign-born people were taken in the United States and of these 11 said they were Mexican and of those 11, two said they were illegal.What is the probability that a Mexican who is considering to cross the United States border will die or be killed in the attempt? 2.Using the critical value approach, at a 5% significance level, is there evidence to suggest that the proportion of Mexicans, as foreign-born people, living in the United States is different from the indicated data? 3.Using the p-value approach are you able to corroborate your conclusions from Question 2.

What are the confidence limits at the 5% level? How do they agree with your con- clusions of Questions 1 and 2? 5.Using the critical value approach, at a 10% significance level, is there evidence to suggest that the proportion of Mexicans, as foreign-born people, living in the United States is different from the indicated data? 6.Using the p-value approach are you able to corroborate your conclusions from Question 5.

What are the confidence limits at the 10% level? How do they agree with your con- clusions of Questions 5 and 6? 8.Using the critical value approach, at a 5% significance level, is there evidence to suggest that the number of undocumented Mexicans living in the United States is different from the indicated data? 9.Using the p-value approach are you able to corroborate your conclusions from Question 8.

What are the confidence limits at the 5% level? How do they agree with your con- clusions of Questions 8 and 9? 11.Using the critical value approach, at a 10% significance level, is there evidence to suggest that the number of undocumented Mexicans living in the United States is different from the indicated data? 12.Using the p-value approach are you able to corroborate your conclusions from Question 11.

5 “Shots across the border,” The Economist, 14 January 2006, p.What are the confidence limits at the 10% level? How do they agree with your conclusions of Questions 11 and 12? 14.

What are your comments about the difficulty in carrying out this hypothesis test? 16.Case: Socrates and Erasmus Situation The Socrates II European programme supports cooperation in education in eight areas, from school to higher education, from new technologies, to adult learners.Within Socrates II is the programme Erasmus that was established in 1987 with the objective to facilitate the mobility of higher education students within European universities.The programme is named after the philosopher, theologian, and humanist, Erasmus of Rotterdam (1465–1536).Erasmus lived and worked in several parts of Europe in quest of knowledge and experience believing such contacts with different cultures could only furnish a broad knowledge.

He left his fortune to the University of Basel and became a precursor of mobility grants.The Erasmus programme has 31 participating countries that include the 25 member states of the European Union, the three European Economic area countries of Iceland, Liechtenstein, and Norway, and the current three candidate countries – Romania, Bulgaria, and Turkey.The programme is open to universities for all higher education programmes including doctoral courses.In between the academic years 1987–1988 to 2003–2004 more than 1 million university students had spent an Erasmus period abroad and there are 2,199 higher education institutions participating in the pro- gramme.

The European Union budget for 2000–2006 is €950 million of which about is €750 million is for student grants.

In the academic year 2003–2004, the Erasmus stu- dents according to their country of origin and their country of study, or host country is given in the cross-classification Table 1 and the field of study for these students accord- ing to their home country is given in Table 2.It is the target of the Erasmus programme to have a balance in the gender mix and the programme administrators felt that the profile for subsequent academic years would be similar to the profile for the academic year 2003–2004.6 Required A sample of random data for the Erasmus programme for the academic year 2005–2006 was provided by the registrar’s office and this is given in Table 3.Does this information bear out the programme administrator’s beliefs if this is tested at the 1%, 5%, and 10% significance level for a difference? 6 :comm/eduation/programmes/socrates/erasmus/ 295 Chapter 8: Hypothesis testing of a single population Subject AT BE BG CY CZ DK EE FI FR DE GR HU IS IE IT LV Agricultural sciences 37 156 51 0 187 18 6 64 398 181 81 136 3 3 317 14 Architecture, Planning 128 163 32 0 168 54 12 30 519 762 149 75 0 30 877 9 Art and design 193 209 42 0 182 60 47 326 651 906 143 114 24 90 756 31 Business studies 1,117 1,089 97 7 584 364 47 1,383 6,573 5,023 306 450 56 593 1,963 88 Education, Teacher training 260 414 12 24 228 74 2 100 320 535 81 126 22 24 267 27 Engineering, Technology 248 384 133 3 481 112 22 487 2,833 1,376 143 147 20 52 1,545 10 Geography, Geology 32 28 12 0 90 27 9 33 259 433 46 66 3 12 206 14 Humanities 147 105 14 0 148 141 9 136 598 1,048 131 64 13 51 1,144 13 Languages, Philological sciences 505 603 73 15 464 346 51 316 3,321 3,528 327 248 47 305 3,346 21 Law 231 357 37 0 185 103 28 117 1,449 1,474 191 159 7 142 1,455 7 Mathematics, Informatics 146 139 86 0 123 20 4 108 570 803 104 64 4 45 392 13 Medical sciences 144 349 60 12 222 115 12 291 399 1,021 172 125 4 46 1,045 8 Natural sciences 143 51 33 0 113 33 4 93 843 879 87 29 3 62 453 6 Social sciences 250 500 48 3 309 171 32 307 1,787 2,067 343 200 15 210 2,220 38 Communication and information 112 212 19 0 14 44 12 100 295 425 38 23 0 32 723 5 science Other areas 28 30 2 0 91 4 8 60 166 227 43 32 0 8 120 4 Total 3,721 4,789 751 64 3,589 1,686 305 3,951 20,981 20,688 2,385 2,058 221 1,705 16,829 308 Table 1 Students by field of study 2003–2004 according to home country.296 Statistics for Business Subject LI LT LU MT NL NO PL PT RO SK SI ES SE UK EUI Total Agricultural sciences 0 48 0 0 80 27 112 69 61 37 23 566 19 23 0 2,717 Architecture, Planning 9 37 4 2 109 19 321 264 64 18 24 854 64 96 0 4,893 Art and design 0 63 4 3 145 69 232 205 87 34 38 905 90 489 0 6,138 Business studies 10 241 15 6 1,089 275 1,342 386 290 169 146 3,244 902 1,332 0 29,187 Education, Teacher training 0 56 43 11 354 92 126 215 47 15 17 602 69 163 0 4,326 Engineering, Technology 0 189 6 9 224 112 752 479 604 106 35 3,109 424 269 0 14,314 Geography, Geology 0 25 8 2 84 5 158 66 147 10 6 450 31 88 0 2,350 Humanities 0 33 2 1 81 39 171 60 116 22 12 654 48 206 8 5,215 Languages, Philological sciences 0 92 14 7 253 84 675 334 451 84 97 2,568 121 2,875 0 21,171 Law 0 87 6 31 303 77 429 190 98 25 51 1,413 195 754 1 9,602 Mathematics, Informatics 0 65 0 1 55 35 301 87 176 23 3 674 46 92 0 4,179 Medical sciences 0 85 8 32 219 142 247 407 209 71 6 1,211 176 232 0 7,070 Natural sciences 0 43 7 4 51 22 361 216 206 29 2 1,062 84 220 0 5,139 Social sciences 0 97 19 5 992 137 928 487 355 29 65 1,701 313 585 1 14,214 Communication and information 0 17 1 5 264 10 68 155 54 3 19 800 56 83 0 3,589 science Other areas 0 16 1 0 85 11 53 162 40 7 2 221 29 32 0 1,482 Total 19 1,194 138 119 4,388 1,156 6,276 3,782 3,005 682 546 20,034 2,667 7,539 10 135,586 Table 1 (Continued).

297 Chapter 8: Hypothesis testing of a single population Home Country Code AT BE BG CY CZ DK EE FI FR DE GR HU IS IE IT LV Austria AT 79 3 5 51 104 7 227 528 262 30 30 15 132 461 5 Belgium BE 105 11 1 51 84 5 218 768 306 75 28 3 121 467 4 Bulgaria BG 52 46 14 16 136 227 62 6 39 Cyprus CY 1 0 0 2 14 9 4 13 0 3 Czech Republic CZ 211 134 103 241 510 931 78 43 180 Denmark DK 70 44 2 19 2 5 260 302 13 3 12 36 111 Estonia EE 16 10 19 47 42 59 6 2 26 Finland FI 229 148 5 9 126 37 35 413 654 72 162 14 111 190 9 France FR 361 420 9 10 206 500 21 727 2,804 218 169 23 1,081 1,550 3 Germany DE 387 330 17 7 207 410 25 918 3,997 165 171 47 926 1,755 23 Greece GR 71 140 6 8 63 45 1 116 420 356 20 2 27 248 1 Hungary HU 110 98 44 201 276 566 42 15 227 Iceland IS 10 4 54 1 26 40 3 2 16 Ireland IE 35 47 6 1 26 30 2 40 557 292 12 5 109 Italy IT 339 633 8 7 86 357 28 367 2,859 1,994 180 129 29 230 4 Latvia LV 8 27 13 42 18 111 2 2 9 Liechtenstein LI 0 0 2 3 1 1 Lithuania LT 49 70 145 180 77 294 18 10 67 Luxembourg LU 17 1 0 0 2 2 0 1 27 39 0 0 0 9 0 Malta MT 4 5 0 0 0 2 0 6 3 6 0 6 52 Netherlands NL 98 184 1 0 44 158 7 275 543 391 42 49 11 88 256 6 Norway NO 50 29 0 0 0 53 0 15 156 190 15 0 0 17 85 0 Poland PL 159 358 362 310 855 1,870 122 74 481 Portugal PT 53 250 8 8 103 63 3 95 325 295 53 59 4 19 713 5 Rumania RO 38 163 29 33 1,125 457 87 21 448 Slovakia SK 44 50 11 52 80 191 24 2 58 Slovenia SI 59 30 19 24 62 125 6 1 56 Spain ES 298 1,054 11 0 169 573 12 501 3,412 2,553 178 67 21 513 4,250 1 Sweden SE 142 42 0 0 38 25 10 24 484 426 17 28 9 80 137 3 United Kingdom UK 143 117 5 4 107 136 8 233 2,303 1,127 60 31 9 21 740 1 EUI* EUR 2 4 1 Total 3,161 4,513 90 62 1,298 3,396 166 4,932 20,275 16,874 1,593 951 199 3,587 12,743 65 Table 2 Erasmus students 2003–2007 by home country and host country.298 Statistics for Business Home Country Code LI LT LU MT NL NO PL PT RO SK SI ES SE UK Total Austria AT 1 12 0 14 215 82 22 60 8 6 16 631 305 410 3,721 Belgium BE 0 7 3 13 377 40 69 207 30 10 9 1,287 149 341 4,789 Bulgaria BG 23 34 43 9 44 751 Cyprus CY 2 3 5 8 64 Czech Republic CZ 203 189 286 163 317 3,589 Denmark DK 3 4 117 27 12 15 5 5 259 30 330 1,686 Estonia EE 10 4 30 26 8 305 Finland FI 15 16 377 15 60 58 13 22 29 479 101 552 3,951 France FR 25 6 43 891 246 314 288 167 30 40 5,115 1,062 4,652 20,981 Germany DE 8 49 1 28 862 463 395 283 27 26 24 4,325 1,653 3,159 20,688 Greece GR 1 1 5 106 17 14 90 3 0 2 374 109 139 2,385 Hungary HU 145 42 125 58 109 2,058 Iceland IS 13 1 36 2 13 221 Ireland IE 4 5 110 8 10 18 3 291 57 37 1,705 Italy IT 1 28 71 607 156 174 766 129 29 20 5,688 399 1,511 16,829 Latvia LV 24 4 9 32 7 308 Liechtenstein LI 4 2 1 5 19 Lithuania LT 30 51 61 120 22 1,194 Luxembourg LU 0 0 0 0 0 1 6 0 0 0 14 3 16 138 Malta MT 7 2 3 1 22 119 Netherlands NL 0 10 0 18 140 21 93 14 3 5 907 389 635 4,388 Norway NO 0 0 0 78 0 36 0 0 0 231 42 159 1,156 Poland PL 294 222 546 286 337 6,276 Portugal PT 1 26 0 4 250 38 125 68 7 14 920 95 178 3,782 Rumania RO 72 119 285 42 86 3,005 Slovakia SK 3 29 30 59 17 32 682 Slovenia SI 25 30 63 17 29 546 Spain ES 0 24 0 9 1,263 200 176 992 59 32 22 670 2,974 20,034 Sweden SE 0 11 0 11 236 22 24 25 3 0 6 370 494 2,667 United Kingdom UK 0 3 0 12 365 69 42 97 10 16 6 1,636 238 7,539 EUI* EUR 1 2 10 Total 11 218 14 253 6,733 1,523 1,459 3,766 536 181 201 24,076 6,082 16,628 135,586 * European University Institute, Florence.299Chapter 8: Hypothesis testing of a single population Family name First name Home country Study area Gender Algard Erik Norway Business studies M Alinei Gratian Rumania Business studies M Andersen Birgitte Brix Denmark Engineering, Technology F Bay Hilde Norway Social sciences F Bednarczyk Tomasz Poland Law M Berberich R mi Germany Engineering, Technology M Berculo Ruwan Netherlands Business studies M Engler Dorothea Germany Geography, Geology F Ernst Folker Germany Business studies M Fouche Elie France Education, Teacher training M Garcia Miguel Spain Communication and information M science Guenin Aur lie France Humanities F Johannessen Sanne Lyng Denmark Business studies F Justnes Petter Norway Languages, Philological sciences M Kauffeldt Ane Katrine Denmark Business studies F Keddie Nikki United Kingdom Mathematics, Informatics F Lorenz Jan Sebastian Germany Business studies M Mallet Guillaume France Business studies M Manzo Margherita Italy Business studies F Margineanu Florin Rumania Agricultural sciences M Miechowka Anne Sophie France Engineering, Technology F Mynborg Astrid Denmark Humanities F Napolitano Silvia Italy Architecture, Planning F Neilson Alison United Kingdom Business studies F Ou Kalvin France Education, Teacher training M Rachbauer Thomas Austria Engineering, Technology M Savreux Margaux France Mathematics, Informatics F Seda Jiri Czech Republic Agricultural sciences M Semoradova Petra Czech Republic Natural sciences F Torres Maria Teresa Spain Humanities F Ungerstedt Malin Sweden Law F Ververken Alexander Belgium Languages, Philological sciences M Viscardi Alessandra Italy Business studies F Zawisza Katarzyna Poland Business studies F Table 3 Sample of Erasmus student enrollments for the academic year 2005–2006.This page intentionally left blank 9Hypothesis testing for different populations Women still earn less than men On 27 February 2006 the Women and Work Commission (WWC) published its report on the causes of the “gender pay gap” or the difference between men’s and women’s hourly pay.

According to the report, British women in full-time work currently earn 17% less per hour than men.Also in February, the European Commission brought out its own report on the pay gap across the whole European Union.Its findings were similar in that on an hourly basis, women earn 15% less than men for the same work.In the United States, the difference in median pay between men and women is around 20%.According to the WWC report the gender pay gap opens early.

Boys and girls study different subjects in school, and boy’s subjects lead to more lucrative careers.They then work in different sorts of jobs.As a result, average hourly pay for a woman at the start of her working life is only 91% of a man’s, even though nowadays she is probably better qualified.1 How do we compile this type of statistical information? We can use hypothesis testing for more than one type of population – the subject of this chapter.1 “Women’s pay: The hand that rocks the cradle”, The Economist, 4 March 2006, p.

In Chapter 8, we presented by sampling from a single population, how we could test a hypoth- esis or an assumption about the parameter of this single population.In this chapter we look at hypothesis testing when there is more than one population involved in the analysis.The difference between the mean of two inde- pendent populations is a hypothesis test to sample in order to see if there is a significant difference between the parameters of two independent popu- lations, as for example the following: ● A human resource manager wants to know if there is a significant difference between the salaries of men and the salaries of women in his multinational firm.● A professor of Business Statistics is interested to know if there is a significant difference between the grade level of students in her morning class and in a similar class in the afternoon.

● A company wants to know if there is a sig- nificant difference in the productivity of the employees in one country and another country.● A firm wishes to know if there is a difference in the absentee rate of employees in the morn- ing shift and the night shift.● A company wishes to know if sales volume of a certain product in one store is different from another store in a different location.In these cases we are not necessarily interested in the specific value of a population parameter but more to understand something about the relation between the two parameters from the populations.That is, are they essentially the same, or is there a significant difference? Difference Between the Mean of Two Independent Populations 302 Statistics for Business After you have studied this chapter you will understand how to extend hypothesis testing for two populations and to use the chi-square hypothesis test for more than two populations.

The subtopics of these themes are as follows: ✔ Difference between the mean of two independent populations • Difference of the means for large samples • The test statistic for large samples • Application of the differences in large sam- ples: Wages of men and women • Testing the difference of the means for small sample sizes • Application of the differences in small samples: Production output ✔ Differences of the means between dependent or paired populations • Application of the dif- ferences of the means between dependent samples: Health spa ✔ Difference between the proportions of two populations with large samples • Standard error of the difference between two proportions • Application of the differences of the proportions between two populations: Commuting ✔ Chi-square test for dependency • Contingency table and chi-square application: Work schedule preference • Chi-square distribution • Degrees of freedom • Chi-square distribution as a test of independence • Determining the value of chi-square • Excel and chi-square functions • Testing the chi-square hypothesis for work preference • Using the p-value approach for the hypothesis test • Changing the significance level L e a r n i n g o b j e c t i v e s 303Chapter 9: Hypothesis testing for different populations Difference of the means for large samples The hypothesis testing concept between two popu- lation means is illustrated in Figure 9.The fig- ure on the left gives the normal distribution for Population No.1 and the figure on the right gives the normal distribution for Population No.Underneath the respective distributions are the sampling distributions of the means taken from that population.From the data another distri- bution can be constructed, which is then the difference between the values of sample means taken from the respective populations.Assume, for example, that we take a random sample from Population 1, which gives a sample mean of x–1.Similarly we take a random sample from Popula- tion 2 and this gives a sample mean of x–2.The difference between the values of the sample means is then given as, 9(i) When the value of x–1 is greater than x – 2 then the result of equation 9(i) is positive.

When the value of x–1 is less than x – 2 then the result of equation 9(i) value is negative.If we construct a distribution of the difference of the entire sample means then we will obtain a sampling distribu- tion of the differences of all the possible sample means as shown in Figure 9.The mean of the sample distribution of the differences of the mean is written as, 9(ii) When the mean of the two populations are equal then x–1 � x–2 � 0.1 f (x) Sampling distribution from population No.

1 Sampling distribution from population No.

2 Mean � m1 Standard deviation � s1 Mean � m2 Standard deviation � s2 mx1 � m1 mx2 � m2 f (x) From Chapter 6, using the central limit theory we developed the following relationship for the standard error of the sample mean: 6(ii) Extending this relationship for sampling from two populations, the standard deviation of the distribution of the difference between the sample means, as given in Figure 9.2, is determined from the following relationship: 9(iii) where 21 and 22 are respectively the variance of Population 1 and Population 2, 1 and 2 are the standard deviations and n1 and n2 are the sample sizes taken from these two populations.This relationship is also the standard error of the difference between two means.If we do not know the population standard deviations, then we use the sample standard deviation, s, as an estimate of the population standard deviationIn this case the estimated standard deviation of the dis- tribution of the difference between the sample means is, 9(iv) The test statistic for large samples From Chapter 6, when we have just one popula- tion, the test statistic z for large samples, that is greater than 30, is given by the relationship, 6(iv) When we test the difference between the means of two populations then the equation for the test statistic becomes, 9(v) Alternatively, if we do not know the popu- lation standard deviation, then equation 9(v) becomes, 9(vi) In this equation, (x–1 � x – 2) is the difference between the sample means taken from the population and ( 1 � 2)H0 is the difference of the hypothesized means of the population.The following application example illustrates this concept.z x x n n � � � � � ( ) ( ) 1 2 1 2 1 2 1 2 2 2 0 H z x x n n � � � � � ( ) ( )1 2 1 2 1 2 1 2 2 2 0 H z x n x x � � x x n n1 2 1 2 1 2 2 2 � � � x x n n1 2 1 2 1 2 2 2 � � � x x n � 304 Statistics for Business Standard error of difference �x1�x2 Mean �x1�x2 Distribution of all possible values of X1 � X2 Figure 9.2 Distribution of all possible values of difference between two means.305Chapter 9: Hypothesis testing for different populations Application of the differences in large samples: Wages of men and women A large firm in the United States wants to know, the relationship between the wages of men and women employed at the firm.

Sampling the employees gave the information in $US in Table 9.At a 10% significance level, is there evidence of a difference between the wages of men and women? At a 10% significance level we are asking the question is there a difference, which means to say that values can be greater or less than.The null and alternative hypotheses are as follows: ● Null hypothesis, H0: 1 � 2 is that there is no significant difference in the wages.● Alternative hypothesis, H1: 1 � 2 is that there is a significant difference in the wages.

Since we have only a measure of the sample standard deviation s and not the population standard deviation , we use equation 9(vi) to determine the test or sample statistic z: Here, x–1 � x – 2 � 28.50 and 1 � 2 � 0 since the null hypothesis is that there is no difference between the population means.The standard error of the difference between the means is from equation 9(iv): Thus, Since the sample, or test statistic, of �1.

8886 is less than the critical value of �1.6449 we reject the null hypothesis and conclude that there is evidence to indicate that the wages of women are significantly different from that of men.As discussed in Chapter 8 we can also use the p-value approach to test the hypothesis.In this example the sample value of z � �1.95% is less than 5% we reject the null hypothesis.This is the same conclusion as previously.The representation of this worked example is illustrated in Figure 9.

Testing the difference of the means for small sample sizes When the sample size is small, or less than 30 units, then to be correct we must use the z �� �� 0 50 0 2648 1 8886 .x x n n1 2 1 2 1 2 2 2 2 22 40 130 1 90 140 0 2 � � � � � � 6648 z x x n n � � � � � ( ) ( ) 1 2 1 2 1 2 1 2 2 2 0 H Sample mean x– ($) Sample standard deviation, s ($) Sample size, n Population 1, women 28.1 Difference in the wages of men and women.When we use the Student-t distribution the population standard deviation is unknown.

Thus to estimate the standard error of the difference between the two means we use equation 9(iv): 9(iv) However, a difference from the hypothesis test- ing of large samples is that here we make the assumption that the variance of Population 1, 21 is equal to the variance of Population 2, 22, or 21 � 22.This then enables us to use a pooled variance such that the sample variance, s21, taken from Population 1 can be pooled, or com- bined, with s22, to give a value s2p.This value of the pooled estimate s2p is given by the relationship, 9(vii) This value of s2p is now the best estimate of the variance common to both populations 2, on the assumption that the two population variances are equal.Note, that the denominator in equa- tion 9(vii), can be rewritten as, (n1 � 1) � (n2 � 1) � (n1 � n2 � 2) 9(viii) This is so because we now have two samples and thus two degrees of freedom.Note that in Chapter 8 when we took one sample of size n in order to use the Student-t distribution we had (n � 1) degrees of freedom.

Combining equations 9(iv) and 9(vii) the rela- tionship for the estimated standard error of the difference between two sample means, when there are small samples on the assumption that the population variances are equal, is given by, 9(ix) Then by analogy with equation 9(vi) the value of the Student-t distribution is given by, 9(x) If we take samples of equal size from each of the populations, then since n1 � n2, equation 9(vii) becomes as follows: 9(xi) Further, the relationship in the denominator of equation 9(x) can be rewritten as, 9(xii)1 1 1 1 2 1 2 1 1n n n n � � � � ⎛ ⎝ ⎜⎜⎜⎜ ⎞ ⎠ ⎟⎟⎟⎟⎟ ⎛ ⎝ ⎜⎜⎜⎜ ⎞ ⎠ ⎟⎟⎟⎟⎟ nn1 s n s n s n n n p 2 1 1 2 2 2 2 1 2 1 1 1 1 1 1 � � � � � � � � � ( ) ( ) ( ) ( ) ( )ss n s n n n s s 1 2 1 2 2 1 1 1 1 2 2 2 1 1 1 1 � � � � � � � � ( ) ( ) ( ) ( )( ) (( )( ) ( ) n s s 1 1 2 2 2 1 1 1 2� � � � t x x s n n � � � � � ( ) ( )1 2 1 2 2 1 2 0 1 1 H p ⎛ ⎝ ⎜⎜⎜⎜ ⎞ ⎠ ⎟⎟⎟⎟⎟ x x s n n1 2 1 1 1 2 � � �p s n s n s n np 2 1 1 2 2 2 2 1 2 1 1 1 1 � � � � � � � ( ) ( ) ( ) ( ) x x n n1 2 1 2 1 2 2 2 � � � 306 Statistics for Business z �1.3 Difference in wages between men and women.307Chapter 9: Hypothesis testing for different populations Thus equation 9(x) can be rewritten as, 9(xiii) The use of the Student-t distribution for small samples is illustrated by the following example.Application of the differences in small samples: Production output One part of a car production firm is the assembly line of the automobile engines.In this area of the plant, the firm employs three shifts: morn- ing 07:00–15:00 hours, evening 15:00–23:00 hours, and the night shift 23:00–07:00 hours.The manager of the assembly line believes that the production output on the morning shift is greater than that on the night shift.

Before the manager takes any action he first records the output on 16 days for the morning shift, and 13 days for the night shift.At a 1% significance level, is there evidence that the output of engines on the morning shift is greater than that on the evening shift? At a 1% significance level we are asking the question is there evidence of the output on the morning shift being greater than the output on the night shift.

This is then a one-tail test with 1% in the upper tail.● The null hypothesis is that there is no dif- ference in output, H0: M � N ● The alternative hypothesis is that the out- put on the morning shift is greater than that on the night shift, H1: M � N.From the sample data we have the informa- tion given in Table 9.

From equation 9(vii), s n s n s n np 2 1 1 2 2 2 2 1 2 1 1 1 1 16 1 � � � � � � � � � ( ) ( ) ( ) ( ) ( ) ** .2 3910 13 1 3 4548 16 1 13 1 8 480 2 2� � � � � � 88 t x x s s n � � � � � ( ) ( )1 2 1 2 1 2 2 2 1 0 H ⎛ ⎝ ⎜⎜⎜⎜ ⎞ ⎠ ⎟⎟⎟⎟⎟ Morning (m) 29 24 28 29 31 27 29 28 26 23 25 28 27 27 30 23 Night (n) 22 23 21 25 31 22 28 30 20 22 23 25 26 Table 9.

2 Production output between morning and night shifts.Sample mean x– Sample standard deviation, s Sample size, n Population 1, morning 27.3 Production output between morning and night shifts.From equation 9(x) the sample or test value of the Student-t value is, Since the sample test value of t of 2.4494 is less than the critical value t of 2.4727 we conclude that there is no significant differ- ence between the production output in the morning and night shifts.

00% and so our conclusion is the same in that we accept the null hypothesis.The concept of this worked example is illustrated in Figure 9.

How would your conclusions change if a 5% level of significance were used? In this situation nothing happens to the sample or test value of the Student-t which remains at 2.7033 we concluded that at a 5% level the production output in the morning shift is significantly greater than that in the night shift.00% and so our conclusion is the same that we reject the null hypothesis.This new concept is illustrated in Figure 9.t x x s n n � � � � � � ( ) ( )1 2 1 2 2 1 2 1 1 H p 0 ⎛ ⎝ ⎜⎜⎜⎜ ⎞ ⎠ ⎟⎟⎟⎟⎟ 227 1250 24 4615 0 8 4808 1 16 1 13 .4 Production output between the morning and night shifts.

5 Production output between the morning and night shifts.309Chapter 9: Hypothesis testing for different populations In the previous section we discussed analysis on populations that were essentially independent of each other.In the wage example we chose sam- ples from a population of men and a population of women.In the production output example we looked at the population of the night shift and the morning shift.Sometimes in sampling experiments we are interested in the differences of paired samples or those that are dependent or related, often in a before and after situation.

Examples might be weight loss of individuals after a diet programme, productivity improve- ment after an employee training programme, or sales increases of a certain product after an advertising campaign.The purpose of these tests is to see if improvements have been achieved as a result of a new action.When we make this type of analysis we remove the effect of other variables or extraneous factors in our analysis.The analytical procedure is to consider statistical analysis on the difference of the values since there is a direct rela- tionship rather than the values before and after.Application of the differences of the means between dependent samples: Health spa A health spa in the centre of Brussels, Belgium advertises a combined fitness and diet programme where it guarantees that participants who are overweight will lose at least 10 kg in 6 months if they scrupulously follow the course.The weights of all participants in the programme are recorded each time they come to the spa.The authorities are somewhat sceptical of the advertising claim so they select at random 13 of the regular par- ticipants and their recorded weights in kilograms before and after 6 months in the programme are given in Table 9.At a 5% significance level, is there evidence that the weight loss of participants in this programme is greater than 10 kg? Here the null hypothesis is that the weight loss is not more than 10 kg or H0: � 10 kg.The alternative hypothesis is that the weight loss is more than 10 kg, or H1: � 10 kg.We are interested not in the weights before and after but in the difference of the weights and thus we can extend Table 9.The test is now very similar to hypothesis testing for a single population since we are making our analysis just on the difference.At a significance level of 5% all of the area lies in the right-hand tail.3999 Estimated standard error of the mean is x n� � � / .4 3999 13 1 2203 Differences of the Means Between Dependent or Paired Populations Before, kg (1) 120 95 118 92 132 102 87 92 115 98 109 110 95 After, kg (2) 101 87 97 82 121 87 74 84 109 87 100 101 82 Table 9.

Sample, or test value of Student-t is, Since this sample value of t of 1.4498 is less than the critical value of t of 1.7823 we accept the null hypothesis and conclude that based on our sampling experiment that the weight loss in this programme over a 6-month period is not more than 10 kg.00% and so our conclusion is the same in that we accept the null hypothesis.The concept for this is illustrated in Figure 9.The sample or test value of the Student-t remains unchanged at 1.3562 and thus we reject the null hypothesis and conclude that the publicity for the programme is correct and that the average weight loss is greater than 10 kg.00% and so our conclusion is the same in that we reject the null hypothesis.This concept is illustrated in Figure 9.11 7692 10 1 2203 1 7692 1 2203 1 44498 310 Statistics for Business t 1.

Before, kg (1) 120 95 118 92 132 102 87 92 115 98 109 110 95 After, kg (2) 101 87 97 82 121 87 74 84 109 87 100 101 82 Difference, kg 19 8 21 10 11 15 13 8 6 11 9 9 13 311Chapter 9: Hypothesis testing for different populations Again as in all hypotheses testing, remem- ber that the conclusions are sensitive to the level of significance used in the test.

There are situations we might be interested to know if there is a significant difference between the proportion or percentage of some criterion of two different populations.

For example, is there a significant difference between the percentage output of one firm’s production site and the other? Is there a difference between the health of British people and Americans? (The answer is yes, according to a study in the Journal of the American Medical Association.2) Is there a signifi- cant difference between the percentage effec- tiveness of one drug and another drug for the same ailment? In these situations we take sam- ples from each of the two groups and test for the percentage difference in the two populations.The procedure behind the test work is similar to the testing of the differences in means except rather than looking at the difference in numerical values we have the differences in percentages.Standard error of the difference between two proportions In Chapter 6 (equation 6(xi)) we developed the following equation for the standard error of the proportion, p–: 6(xi) where n is the sample size, p is the population proportion of successes, and q is the population proportion of failures equal to (1 � p).Then by analogy with equation 9(iii) for the difference in the standard error for the means we have the equation for the standard error of the difference between two proportions as, 9(xiv) where p1, q1 are respectively the proportion of success and failure and n1 is the sample size taken from the first population and p2, q2, and n2 are the corresponding values for the second popula- tion.

If we do not know the population propor- tions then the estimated standard error of the difference between two proportions is, 9(xv) Here, p–1, q – 1, p – 2, q – 2 are the values of the propor- tion of successes and failures taken from the sample.In Chapter 8 we developed that the number of standard deviations, z, in hypothesizing for a sin- gle population proportion as, 8(ix) By analogy, the value of z for the difference in the hypothesis for two population proportions is, 9(xvi) The use of these relationships is illustrated in the following worked example.Application of the differences of the proportions between two populations: Commuting A study was made to see if there was a signifi- cance difference between the commuting time z p p p p p p � � � � � ( ) ( ) 1 2 1 2 1 2 H0 z p p p � � H0 p p p q n p q n1 2 1 1 1 2 2 2 � � � p p p q n p q n1 2 1 1 1 2 2 2 � � � p pq n p p n � � �( )1 Difference Between the Proportions of Two Populations with Large Samples 2 “Compared with Americans, the British are the picture of health”, International Herald Tribune, 22 May 2006, p.of people working in downtown Los Angeles in Southern California and the commuting time of people working in downtown San Francisco in Northern California.

The benchmark for com- muting time was at least 2 hours per day.A ran- dom sample of 302 people was selected from Los Angeles and 178 said that they had a daily commute of at least 2 hours.A random sample of 250 people was selected in San Francisco and 127 replied that they had a commute of at least 2 hours.At a 5% significance level, is there evidence to suggest that the proportion of people com- muting Los Angeles is different from that of San Francisco? Sample proportion of people commuting at least 2 hours to Los Angeles is, p1 � 178/302 � 0.

4106 Sample proportion of people commuting at least 2 hours to San Francisco is, p2 � 127/250 � 0.4920 This is a two-tail test since we are asking the question is there a difference? ● Null hypothesis is that there is no differ- ence or H0: p1 � p2 ● Alternative hypothesis is that there is a dif- ference or, H1: p1 � p2 From equation 9(xv) the estimated standard error of the difference between two propor- tions is, From equation 9(xvi) the sample value of z is, This is a two-tail test at 5% significance, so there is 2.Using func- tion NORMSINV gives a critical value of z of �1.9600 we accept the null hypothesis and conclude that there is no significant difference between commuting time in Los Angeles and San Francisco.We obtain the same conclusion when we use the p-value for making the hypothesis test.50% the critical value, and so again we accept the null hypothesis.This concept is illustrated in Figure 9.

At a 5% significance level, is there evidence to suggest that the proportion of people commuting z p p p p p p � � � � � � � � ( ) ( ) ( .) 1 2 1 2 1 2 0 5894 0 5080 H0 00 0 0424 1 9181 .

p p p q n p q n 1 2 1 1 1 2 2 2 0 5894 0 4106 302 0 50 � � � � � 550 0 4920 250 0 0424 * .

313Chapter 9: Hypothesis testing for different populations Los Angeles is greater than those working in San Francisco? This is a one-tail test since we are asking the question, is one population greater than the other? Here all the 5% is in the upper tail.● Null hypothesis is that there is a popula- tion not greater or H0: p1 � p2 ● Alternative hypothesis is that a population is greater than or, H1: p1 � p2 Here we use � since less than or equal is not greater than and so thus satisfies the null hypothesis.The sample test value of z remains unchanged at 1.

However, using func- tion NORMSDIST the 5% in the upper tail corresponds to a critical z-value of 1.6449 we reject the null hypothesis and conclude that there is statistical evidence that the commuting time for Los Angeles people is significantly greater than for those persons in San Francisco.Using the p-value approach, the area in the upper tail corresponding to a sample test value of 1.Now this value is less than the 5% significant value and so the conclusion is the same that there is evidence to suggest that the commuting time for those in Los Angeles is greater than for those in San Francisco.

This new situa- tion is illustrated in Figure 9.In testing samples from two different popula- tions we examined the difference between either two means, or alternatively, two proportions.If we have sample data which give proportions from more than two populations then a chi-square test can be used to draw conclusions about the populations.The chi-square test enables us to decide whether the differences among several sample proportions is significant, or that the dif- ference is only due to chance.

Suppose, for example, that a sample survey on the proportion of people in certain states of the United States who exercise regularly was found to be 51% in California, 34% in Ohio, 45% in New York, and 29% in South Dakota.If this dif- ference is considered significant then a conclu- sion may be that location affects the way people behave.If it is not significant, then the differ- ence is just due to chance.Thus, assuming a firm is considering marketing a new type of jog- ging shoe then if there is a significant difference between states, its marketing efforts should be weighted more on the state with a higher level of physical fitness.The chi-square test will be demonstrated as follows using a situation on work schedule preference.

Contingency table and chi-square application: Work schedule preference We have already presented a contingency or cross-classification table, in Chapter 2.This table Chi-Square Test for Dependency z 1.presents data by cross-classifying variables according to certain criteria of interest such that the cross-classification accounts for all con- tingencies in the sampling data.Assume that a large multinational company samples its employ- ees in the United States, Germany, Italy, and England using a questionnaire to discover their preference towards the current 8-hour/day, 5-day/week work schedule and a proposed 10-hour/day, 4-day/week work schedule.The sample data collected using an employee questionnaire is in Table 9.

In this contingency table, the columns give preference according to country and the rows give the preference according to the work schedule criteria.These sample values are the observed frequencies of occurrence, fo.This is a 2 4 contingency table as there are two rows and four columns.Neither the row totals, nor the column totals are con- sidered in determining the dimension of the table.

In order to test whether preference for a certain work schedule depends on the location, or there is simply no dependency, we test using a chi- square distribution.Chi-square distribution The chi-square distribution is a continuous prob- ability distribution and like the Student-t distribu- tion there is a different curve for each degree of freedom, .The x-axis is the value of chi-square, written 2 where the symbol is the Greek letter c.Since we are dealing with 2, or to the power of two, the values on the x-axis are always positive and extend from zero to infinity.

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10 gives three chi-square distribu- tions for degrees of freedom, , of respectively 4, 8, and 12.For small values of the curves are positively or right skewed.As the value of increases the curve takes on a form similar to a normal distribution **By T.S. Jain 2015 UPKAR PRAKASHAN, AGRA-2 Introducing Direct Shopping Now you can purchase from our vast range of books and magazines at your convenience : «▻ Pay by Credit Card/Debit Which Power Point view displays each slide of the presentation as a thumbnail and is useful for rearranging slides ?**.As the value of increases the curve takes on a form similar to a normal distribution.

The mode or the peak of the curve is equal to the degrees of freedom less two.

For example, for the three curves illus- trated, the peak of each curve is for values of 2 equal to 2, 6, and 10, respectively.

Degrees of freedom The degrees of freedom in a cross-classification table are calculated by the relationship, (Number of rows �1) * (Number of columns �1) 9(xviii) Consider Table 9 **Help me do an corruption powerpoint presentation original 24 hours Standard British Academic**.Degrees of freedom The degrees of freedom in a cross-classification table are calculated by the relationship, (Number of rows �1) * (Number of columns �1) 9(xviii) Consider Table 9.7 which is a 3 4 contin- gency table as there are three rows and four columns.R1 through R3 indicate the rows and C1 through C4 indicates the columns.The row totals are given by TR1 through TR3 and the col- umn totals by TC1 through TC4.The value of the row totals and the column totals are fixed and the “yes” or “no” in the cells indicate whether or not we have the freedom to choose a value in this cell.

For example, in the column designated f ( ) ( / ) ! ( ) / ( / ) / 2 2 2 2 11 2 1 1 2 2 2 � � � �e 314 Statistics for Business Preference United States Germany Italy England Total 8 hours/day 227 213 158 218 816 10 hours/day 93 102 97 92 384 Total 320 315 255 310 1,200 Table 9.6 Work preference sample data or observed frequencies, fo.315Chapter 9: Hypothesis testing for different populations by C1 we have only the freedom to choose two values, the third value is automatically fixed by the total of that column.In this table we have the freedom to choose only six values or the same as determined from equation 9(x).

Degrees of freedom � (3 � 1) * (4 � 1) � 2 * 3 � 6 Chi-square distribution as a test of independence Going back to our cross-classification on work preferences in Table 9.6 let as say that, pU is the proportion in the United States who prefer the present work schedule pG is the proportion in Germany who prefer the present work schedule Figure 9.10 Chi-square distribution for three different degrees of freedom.0% 2% 4% 6% 8% 10% 12% 14% 16% 18% 20% 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 f( 2 ) 2 df � 4 df � 8 df � 12 C1 C2 C3 C4 Total rows R1 yes yes yes no TR1 R2 yes yes yes no TR2 R3 no no no no TR3 Total columns TC1 TC2 TC3 TC4 TOTAL Table 9.pI is the proportion in the Italy who prefer the present work schedule pE is the proportion in England who prefer the present work schedule The null hypothesis H0 is that the population pro- portion favouring the current work schedule is not significantly different from country to country and thus we can write the null hypothesis situ- ation as follows: H0: pU � pG � pI � pE 9(xix) This is also saying that for the null hypothesis of the employee preference of work schedule is inde- pendent of the country of work.Thus, the chi- square test is also known as a test of independence.The alternative hypothesis is that population proportions are not the same and that the pref- erence for the work schedule is dependent on the country of work.In this case, the alternative hypothesis H1 is written as, H1: pU � pG � pI � pE 9(xx) Thus in hypothesis testing using the chi-square distribution we are trying to determine if the population proportions are independent or dependent according to a certain criterion, in this case the country of employment.This test determines frequency values as follows.

Determining the value of chi-square From Table 9.6 if the null hypothesis is correct and that there is no difference in the preference for the work schedule, then from the sample data.● Population proportion who prefer the 8-hour/day schedule is 816/1,200 � 0.6800 ● Population proportion who prefer the 10-hour/day schedule is 384/1,200 � 0.3200 We then use these proportions on the sample data to estimate the population proportion that pre- fer the 8-hour/day or the 10-hour/day schedule.

For example, the sample size for the United States is 320 and so assuming the null hypothesis, the estimated number that prefers the 8-hour/day schedule is 0.The esti- mated number that prefers the 10-hour/day schedule is 0.40 since the choice is one schedule or the other.Thus the complete expected data, on the assumption that the null hypothesis is correct is as in Table 9.

These are then considered expected frequencies, fe.Another way of calculating the expected fre- quency is from the relationship, 9(xxi) TRo and TCo are the total values for the rows and columns for a particular observed frequency fo in a sample of size n.6 let us consider the cell that gives the observed frequency for Germany for a preference of an 8-hour/day schedule.

f ne o oTR TC� * 316 Statistics for Business Preference United States Germany Italy England Total 8 hours/day 217.8 Work preference – expected frequencies, fe.317Chapter 9: Hypothesis testing for different populations TRo � 816 TCo � 315 n � 1,200 Thus, The value of chi-square, 2, is given by the relationship, 9(xxii) where fo is the frequency of the observed data and fe is the frequency of the expected or theoretical data.9 the value of the sample chi-square as shown is, Note in order to verify that your calculations are correct, the total amount in the fo column must equal to total in the fe column and also the total ( fo � fe) must be equal to zero.Excel and chi-square functions In Microsoft Excel there are three functions that are used for chi-square testing.Testing the chi-square hypothesis for work preference As for all hypothesis tests we have to decide on a significance level to test our assumption.Let us say for the work preference situation that we consider 5% significance.In addition, for the chi- square test we also need the degrees of freedom.f f f o e e ∑ 2 2 � �( )f f f o e e ∑ f ne o oTR TC� � � * * , .816 315 1 200 214 20 fo fe fo � fe (fo � fe)2 227 217.

9 Work preference – observed and expected frequencies.The positions of this critical value and the value of the sample or test chi-square value are shown in Figure 9.Since the value of the sample chi-square statistic, 6.

3325, is less than the crit- ical value of 7.8147 at the 5% significance level given, we accept the null hypothesis and say that there is no statistical evidence to conclude that the preference for the work schedule is sig- nificantly different from country to country.We can avoid performing the calculations shown in Table 9.6 and the expected frequency values fe as given in Table 9.65% which is the area in the chi- square distribution for the observed data.65% and the degrees of freedom to give the sample chi-square value of 6.65%, which is the area in the chi-square dis- tribution.This is also the p-value for the observed data.

00% the significance level we accept the null hypothesis or the same conclusion as before.This concept is illustrated in Figure 9.8147 Critical value at 5% significance Figure 9.11 Chi-square distribution for work preferences.8147 Critical value at 5% significance Figure 9.12 Chi-square distribution for work preferences.2514 Critical value at 10% significance 6.13 Chi-square distribution for work preferences.

319Chapter 9: Hypothesis testing for different populations Changing the significance level For the work preference situation we made the hypothesis test at 5% significance.What if we increased the significance level to 10%? In this case nothing happens to our sampling data and we still have the following information that we have already generated.● Area under the chi-square distribution represented by the sampling data is 9.2514 (using chi-square values), alternatively 9.We reject the null hypothesis and conclude that the country of employment has some bear- ing on the preference for a certain work schedule.This new relationship is illustrated in Figure 9.

This chapter has dealt with extending hypothesis testing to the difference in the means of two independent populations and the difference in the means of two dependent or paired populations.It also looks at hypothesis testing for the differences in the proportions of two populations.The last part of the chapter presented the chi-square test for examining the dependency of more than two populations.In all cases we propose a null hypothesis H0 and an alternative hypothesis H1 and test to see if there is statistical evidence whether we should accept, or reject, the null hypothesis.

Difference between the mean of two independent populations The difference between the mean of two independent populations is a test to see if there is a signifi- cant difference between the two population parameters such as the wages between men and women, employee productivity in one country and another, the grade point average of students in one class or another, etc.In these cases we may not be interested in the mean values of one popu- lation but in the difference of the mean values of both populations.We develop first a probability distribution of the difference in the sample means.From this we determine the standard deviation of the distribution by combining the standard deviation of each sample using either the popul- ation standard deviations, if these are known, or if they are not known, using estimates of the pop- ulation standard deviation measured from the samples.From the sample test data we determine the sample z-value and compare this to the z-value dictated by the given significance level .

Alternatively, we can make the hypothesis test using the p-value approach and the conclusion will be the same.When we have small sample sizes our analytical approach is similar except that we use a pooled sampled variance and the Student-t distribution for our analytical tool.Differences of the means between dependent or paired populations This hypothesis test of the differences between paired samples has the objective to see if there are measured benefits gained by the introduction of new programmes such as employee train- ing to improve productivity or to increase sales, fitness programmes to reduce weight or increase stamina, coaching courses to increase student grades, etc.In these type of hypothesis C hapter Sum m ary 320 Statistics for Business test we are dealing with the same population in a before and after situation.In this case we measure the difference of the sample means and this becomes our new sampling distribution.

The hypothesis test is then analogous to that for a single population.For large samples we use a z-value for our critical test and a Student-t distribution for small sample sizes.Difference between the proportions of two populations with large samples This hypothesis test is to see if there is a significant difference between the proportion or per- centage of some criterion of two different populations.The test procedure is similar to the differ- ences in means except rather than measuring the difference in numerical values we measure the differences in percentages.We calculate the standard error of the difference between two proportions using a combination of data taken from the two samples based on the proportion of successes from each sample, the proportion of failures taken from each sample, and the respect- ive sample sizes.

We then determine whether the sample z-value is greater or lesser than the critical z-value.If we use the p-value approach we test to see whether the area in the tail or tails of the distribution is greater or smaller than the significance level .Chi-square test for dependency The chi-square hypothesis test is used when there are more than two populations and tests whether data is dependent on some criterion.The first step is to develop a cross-classification table based on the sample data.This information gives the observed frequency of occurrence, fo.

Assuming that the null hypothesis is correct we calculate an expected value of the frequency of occurrence, fe, using the sample proportion of successes as our benchmark.To perform the chi- square test we need to know the degrees of freedom of the cross-classification table of our sam- ple data.This is (number of rows � 1) * (number of columns � 1).The hypothesis test is based on the chi-square frequency distribution, which has a y-axis of frequency and a positive x-axis 2 extending from zero.There is a chi-square distribution for each degree of freedom of the cross-classification table.

The test procedure is to see whether the sample test value 2 is greater or lesser than the critical value 2.Alternatively we use the p-value approach and see whether the area under the curve determined from the sample data is greater or smaller than the signifi- cance level, .321Chapter 9: Hypothesis testing for different populations 1.Gasoline prices Situation A survey of 102 gasoline stations in France in January 2006 indicated that the average price of unleaded 95 octane gasoline was €1.

381 per litre with a standard deviation of €0.

Another sample survey taken 6 months later at 97 gasoline stations indicated that the average price of gasoline was €1.4270 per litre with a standard deviation of €0.Indicate an appropriate null and alternative hypotheses for this situation if we wanted to know if there is a significant difference in the price of gasoline.Using the critical value method, at a 2% significance level, does this data indicate that there has been a significant increase in the price of gasoline in France? 3.Confirm your conclusions to Question 2 using the p-value approach.Using the critical value method would your conclusions change at a 5% significance level? 5.Confirm your conclusions to Question 4 using the p-value approach.What do you think explains these results? 2.Tee shirts Situation A European men’s clothing store wants to test if there was a difference in the price of a cer- tain brand of tee shirts sold in its stores in Spain and Italy.

It took a sample of 41 stores in Spain and found that the average price of the tee shirts was €27.It took a sample of 49 stores in Italy and found that the average price of the tee shirts was €26.Indicate an appropriate null and alternative hypotheses for this situation if we wanted to know if there is a significant difference in the price of tee shirts in the two countries.Using the critical value method, at a 1% significance level, does the data indicate that there is a significant difference in the price of tee shirts in the two countries? 3.

Confirm your conclusions to Question 2 using the p-value approach.Using the critical value method would your conclusions change at a 5% significance level? 5.Confirm your conclusions to Question 4 using the p-value approach.Indicate an appropriate null and alternative hypothesis for this situation if we wanted to test if the price of tee shirts is significantly greater in Spain than in Italy? EXERCISE PROBLEMS 322 Statistics for Business 7.Using the critical value method, at a 1% significance level, does the data indicate that the price of tee shirts is greater in Spain than in Italy? 8.Confirm your conclusions to Question 7 using the p-value criterion? 3.Inventory levels Situation A large retail chain in the United Kingdom wanted to know if there was a significant difference between the level of inventory kept by its stores that are able to order on-line through the Internet with the distribution centre and those that must use FAX.The headquarters of the chain collected the following sample data from 12 stores that used direct FAX and 13 that used Internet connections for the same non-perishable items in terms of the number of days’ coverage of inventory.

For example, the first value for a store using FAX has a value of 14.This means that the store has on average 14 days supply of products to satisfy estimated sales until the next delivery arrives from the distribution centre.Stores FAX 14 11 13 14 15 11 15 17 16 14 22 16 Stores internet 12 8 14 11 6 3 15 8 7 22 19 3 4 Required 1.Indicate an appropriate null and alternative hypotheses for this situation if we wanted to show if those stores ordering by FAX kept a higher inventory level than those that used Internet.Using the critical value method, at a 1% significance level, does this data indicate that those stores using FAX keep a higher level of inventory than those using Internet? 3.Confirm your conclusions to Question 2 using the p-value approach.Using the critical value method, at a 5% significance level, does this data indicate that those stores using FAX keep a higher level of inventory than those using Internet? 5.Confirm your conclusions to Question 4 using the p-value approach.

How might you explain the conclusions obtained from Questions 4 and 5? 4.Restaurant ordering Situation A large franchise restaurant operator in the United States wanted to know if there was a difference between the number of customers that could be served if the person taking the order used a database ordering system and those that used the standard hand- written order method.In the database system when an order is taken from a customer 323Chapter 9: Hypothesis testing for different populations it is transmitted via the database system directly to the kitchen.

When orders are made by hand the waiter or waitress has to go to the kitchen and give the order to the chef.

The franchise operator believed that up to 25% more customers per hour could be served if the restaurants were equipped with a database ordering system.The following sample data was taken from some of the many restaurants within the franchise of the average number of customers served per hour per waiter or waitress.Standard (S) 23 20 34 6 25 25 31 22 30 Using database (D) 30 38 43 37 67 43 42 34 50 34 45 Required 1.What is an appropriate null and alternative hypotheses for this situation? 2.

Using the critical value method, at a 1% significance level, does the data support the belief of the franchise operator? 3.Confirm your conclusions to Question 2 using the p-value approach.Using the same 1% significance level how could you rewrite the null and alternative hypothesis to show that the data indicates better the franchise operator’s belief? 5.Test your relationship in Question 4 using the critical value method.

Confirm your conclusions to Question 5 using the p-value approach.What do you think are reasons that some of the franchise restaurants do not have a database ordering system? 5.Sales revenues Situation A Spanish-based ladies clothing store with outlets in England is concerned about the low store sales revenues.

In an attempt to reverse this trend it decides to conduct a pilot pro- gramme to improve the sales training of its staff.It selects 11 of its key stores in the Birmingham and Coventry area and sends these sales staff progressively to a training programme in London.This training programme includes how to improve customer contact, techniques of how to spend more time on the high-revenue products, and gen- erally how to improve team work within the store.The firm decided that it would extend the training programme to its other stores in England if the training programme increased revenues by more than 10% of revenues in its pilot stores before the programme.The table below gives the average monthly sales in £ ’000s before and after the training programme.

The before data is based on a consecutive 6-month period.The after data is based on a consecutive 3-month period after the training programme had been completed for all pilot stores.324 Statistics for Business Store number 1 2 3 4 5 6 7 8 9 10 11 Average sales before (£ ’000s) 256 202 203 189 302 275 259 358 249 265 302 Average sales after (£ ’000s) 302 289 345 259 357 299 368 402 258 267 391 Required 1.What is the benchmark of sales revenues on which the hypothesis test programme is based? 2.Indicate the null and alternative hypotheses for this situation if we wanted to know if the training programme has reached its objective? 3.

Using the critical value approach at a 1% significance level, does it appear that the objectives of the training programme have been reached? 4.Verify your conclusion to Question 3 by using the p-value approach.Using the critical value approach at a 5% significance level, does it appear that the training programme has reached its objective? 6.Verify your conclusion to Question 5 by using the p-value approach.

What are your comments on this test programme? 6.Hotel yield rate Situation A hotel chain is disturbed about the low yield rate of its hotels.It decides to see if improvements could be made by extensive advertising and reducing prices.It selects nine of its hotels and measures the average yield rate (rooms occupied/rooms available) in a 3-month period before the advertising, and a 3-month period after advertising for the same hotels.

The data collected is given in the following table.Hotel number 1 2 3 4 5 6 7 8 9 Yield rate before (1) 52% 47% 62% 65% 71% 59% 81% 72% 91% Yield rate after (2) 72% 66% 75% 78% 77% 82% 89% 79% 96% Required 1.Indicate the null and alternative hypotheses for this situation if we wanted to know if the advertising programme has reached an objective to increase the yield rate by more than 10%? 2.Using the critical value approach at a 1% significance level, does it appear that the objectives of the advertising programme have been reached? 3.Verify your conclusion to Question 2 by using the p-value approach.

Using the critical value approach at a 15% significance level, does it appear that the objectives of the advertising programme have been reached? 5.Verify your conclusion to Question 4 by using the p-value approach.Should management be satisfied with the results obtained? 325Chapter 9: Hypothesis testing for different populations 7.

Migraine headaches Situation Migraine headaches are not uncommon.They begin with blurred vision either in one or both eyes and then are often followed by severe headaches.There are medicines available but their efficiency is often questioned.Studies have indicated that migraine is caused by stress, drinking too much coffee, or consuming too much sugar.

A study was made on 10 volunteer patients who were known to be migraine sufferers.

These patients were first asked to record over a 6-month period the number of migraine headaches they experienced.This was then calculated into the average number per month.Then they were asked to stop drinking coffee for 3 months and record again the number of migraine attacks they experienced.This again was reduced to a monthly basis.The com- plete data is in the table below.

Patient 1 2 3 4 5 6 7 8 9 10 Average number per month before (1) 23 27 24 18 31 24 23 27 19 28 Average number per month after (2) 12 18 14 5 12 12 15 12 6 14 Required 1.Indicate the null and alternative hypothesis for this situation if we wanted to show that the complete elimination of coffee in a diet reduced the impact of migraine headaches by 50%.Using the critical value approach at a 1% significance level, does it appear that elimin- ating coffee the objectives of the reduction in migraine headaches has been reached? 3.Verify your conclusion to Question 2 by using the p-value approach.

Using the critical value approach at a 10% significance level, does it appear that elimin- ating coffee the objectives of the reduction in migraine headaches has been reached? 5.Verify your conclusion to Question 4 by using the p-value approach.At a 1% significance level, approximately what reduction in the average number of headaches has to be experienced before we can say that eliminating coffee is effective? 7.

What are your comments about this experiment? 8.Hotel customers Situation A hotel chain was reviewing its 5-year strategic plan for hotel construction and in par- ticular whether to include a fitness room in the new hotels that it was planning to build.It had made a survey in 2001 on customers’ needs and in a questionnaire of 408 people surveyed, 192 said that they would prefer to make a reservation with a hotel that had a fitness room.A similar survey was made in 2006 and out of 397 persons who returned 326 Statistics for Business the questionnaire, 210 said that a hotel with a fitness room would influence booking decision.Indicate an appropriate null and alternative hypotheses for this situation.Using the critical value approach at a 5% significance level, does it appear that there is a significant difference between customer needs for a fitness room in 2006 than in 2001? 3.Verify your conclusion to Question 2 by using the p-value approach.Indicate the null and alternative hypotheses for this situation if we wanted to see if the customer needs for a fitness room in 2006 is greater than that in 2001.Using the critical value approach at a 5% significance level, does it appear that cus- tomer needs in 2006 are greater than in 2001? 6.Verify your conclusion to Question 5 by using the p-value approach.What are your comments about the results? 9.Flight delays Situation A study was made at major European airports to see if there had been a significant dif- ference in flight delays in the 10-year period between 1996 and 2005.A flight was con- sidered delayed, either on takeoff or landing, if the difference was greater than 20 minutes of the scheduled time.In 2005, in a sample of 508 flights, 310 were delayed more than 20 minutes.In 1996 out of a sample of 456 flights, 242 were delayed.

Indicate an appropriate null and alternative hypothesis for this situation.Using the critical value approach at a 1% significance level, does it appear that there is a significant difference between flight delays in 2005 and 1996? 3.Verify your conclusion to Question 2 by using the p-value approach.

Using the critical value approach at a 5% significance level, does it appear that there is a significant difference in flight delays in 2005 and 1996? 5.Verify your conclusion to Question 4 by using the p-value approach.Indicate an appropriate null and alternative hypotheses for this situation to respond to the question has there been a significant increase in flight delays between 1996 and 2005? 7.

From the relationship in Question 6 and using the critical value approach, what are your conclusions if you test at a significance level of 1%? 8.What has to be the significance level in order for your conclusions in Question 7 to be different? 9.What are your comments about the sample experiment? 327Chapter 9: Hypothesis testing for different populations 10.World Cup Situation The soccer World Cup tournament is held every 4 years.

In 2002 it was in Korea and Japan, and in June 1998 it was in France.A survey was taken to see if people’s interest in the World Cup had changed in Europe between 1998 and 2006.A random sample of 99 people was taken in Europe in early June 1998 and 67 said that they were interested in the World Cup.In 2006 out of a sample of 112 people taken in early June, 92 said that they were interested in the World Cup.Indicate an appropriate null and alternative hypotheses for this situation to test whether people’s interest in the World Cup has changed between 1998 and 2006.Using the critical value approach at a 1% significance level, does it appear that there is a difference between people’s interest in the World Cup between 1998 and 2006? 3.Verify your conclusion to Question 2 by using the p-value approach.Using the critical value approach at a 5% significance level, does it appear that there is a difference between people’s interest in the World Cup between 1998 and 2006? 5.Verify your conclusion to Question 4 by using the p-value approach.Indicate an appropriate null and alternative hypotheses to test whether there has been a significant increase in interest in the World Cup between 1998 and 2006? 7.From the relationship in Question 6 and using the critical value approach, what are your conclusions if you test at a significance level of 1%? 9.

Confirm your conclusions to Question 7 using the p-value criterion.What are your comments about the sample experiment? 11.Travel time and stress Situation A large company located in London observes that many of its staff are periodically absent from work or are very grouchy even when at the office.Casual remarks indicate that they are stressed by the travel time into the City as their trains are crowded, or often late.

As a result of these comments the human resource department of the firm sent out 200 questionnaires to its employees asking the simple question what is your commuting time to work and how do you feel your stress level on a scale of high, moderate, and low.The table below summarizes the results that it received.Travel time High stress level Moderate stress level Low stress level Less than 30 minutes 16 12 19 30 minutes to 1 hour 23 21 31 Over 1 hour 27 25 12 328 Statistics for Business Required 1.Indicate the appropriate null hypothesis and alternative hypothesis for this situation if we wanted to test to see if stress level is dependent on travel time.Using the critical value approach of the chi-square test at a 1% significance level, does it appear that there is a relationship between stress level and travel time? 3.Verify your conclusion to Question 2 by using the p-value approach of the chi-square test.Using the critical value approach of the chi-square test at a 5% significance level, does it appear that there is a relationship between stress level and travel time? 5.Corroborate your conclusion to Question 4 by using the p-value approach of the chi-square test.

Would you say based on the returns received that the analysis is a good representation of the conditions at the firm? 7.What additional factors need to be considered when we are analysing stress (a much overused word today!)? 12.Investing in stocks Situation A financial investment firm wishes to know if there is a relationship between the coun- try of residence and an individual’s saving strategy regarding whether or not they invest in stocks.This information would be useful as to increase the firm’s presence in countries other than the United States.

The following information was collected by simple tele- phone contact on the number of people in those listed countries on whether or not they used the stock market as their investment strategy.Savings strategy United States Germany Italy England Invest in stocks 206 121 147 151 Do not invest in stocks 128 118 143 141 Required 1.Show the appropriate null hypothesis and alternative hypothesis for this situation if we wanted to test if there is a dependency between savings strategy and country of residence.Using the critical value approach of the chi-square test at a 1% significance level, does it appear that there is a relationship between investing in stocks and the country of residence? 3.

Verify your conclusion to Question 1 by using the p-value approach of the chi-square test.329Chapter 9: Hypothesis testing for different populations 4.Using the critical value approach of the chi-square test at a 3% significance level, does it appear that there is a relationship between investing in stocks and the country of residence? 5.Verify your conclusion to Question 3 by using the p-value approach of the chi-square test.What are your observations from the sample data and what is a probable explanation? 13.Automobile preference Situation A market research firm in Europe made a survey to see if there was any correlation between a person’s nationality and their preference in the make of automobile they pur- chase.The sample information obtained is in the table below.Germany France England Italy Spain Volkswagen 44 27 26 19 48 Renault 27 32 24 17 32 Peugeot 22 33 22 24 27 Ford 37 16 37 25 36 Fiat 25 15 30 31 19 Required 1.

Indicate the appropriate null and alternative hypotheses to test if the make of auto- mobile purchased is dependent on an individual’s nationality.

Using the critical value approach of the chi-square test at a 1% significance level, does it appear that there is a relationship between automobile purchase and nation- ality? 3.Verify your results to Question 2 by using the p-value approach of the chi-square test.What has to be the significance level in order that there appears a breakeven situ- ation between a dependency of nationality and automobile preference? 5.

What are your comments about the results? 14.Newspaper reading Situation A cooperation of newspaper publishers in Europe wanted to see if there was a relation- ship between salary levels and the reading of a morning newspaper.A survey was made in Italy, Spain, Germany, and France and the sample information obtained is given in the table below.330 Statistics for Business Salary bracket �€16,000 �€16,000 to �€50,000 to �€75,000 to �€100,000 �€50,000 �€75,000 �€100,000 Salary category 1 2 3 4 5 Always read 36 55 65 65 62 Sometimes 44 40 47 47 52 Never read 30 28 19 19 22 Required 1.Indicate the appropriate null and alternative hypotheses to test if reading a news- paper is dependent on an individual’s salary.

Using the critical value approach of the chi-square test at a 5% significance level, does it appear that there is a relationship between reading a newspaper and salary? 3.Verify your results to Question 2 by using the p-value approach of the chi-square test.Using the critical value approach of the chi-square test at a 10% significance level, does it appear that there is a relationship between reading a newspaper and salary? 5.

Verify your results to Question 4 by using the p-value approach of the chi-square test.What are your comments about the sample experiment? 15.Wine consumption Situation A South African producer is planning to increase its export of red wine.Before it makes any decision it wants to know if a particular country, and thus the culture, has any bear- ing on the amount of wine consumed.

Using a market research firm it obtains the fol- lowing sample information on the quantity of red wine consumed per day.Amount consumed England France Italy Sweden United States Never drink 20 10 15 8 12 One glass or less 72 77 70 62 68 Between one and two 85 65 95 95 48 More than two 85 79 77 85 79 Required 1.Show the appropriate null hypothesis and alternative hypothesis for this situation if we wanted to test if there is a dependency between wine consumption and country of residence.331Chapter 9: Hypothesis testing for different populations 2.Using the critical value approach of the chi-square test at a 1% significance level, does it appear that there is a relationship between wine consumption and the coun- try of residence? 3.

Verify your conclusion to Question 2 by using the p-value approach.To the nearest whole number, what has to be the minimum significance level in order to change the conclusion to Question 1? This is the p-value.What is the chi-square value for the significance level of Question 4? 6.

Based on your understanding of business, what is the trend in wine consumption today? 16.Case: Salaries in France and Germany Situation Business students in Europe wish to know if there is a difference between salaries offered in France and those offered in Germany.An analysis was made by taking random samples from alumni groups in the 24–26 age group.This information is given in the table below.France 52,134 45,294 43,746 55,533 49,263 42,534 65,256 47,070 46,545 42,549 38,550 61,125 49,518 50,589 56,391 49,557 45,006 50,082 57,336 44,592 50,100 53,175 47,487 52,566 54,156 41,841 55,836 52,131 49,683 48,465 50,700 41,493 49,812 47,628 59,586 50,799 54,048 51,198 45,270 48,570 47,451 36,555 52,704 50,787 45,684 45,807 43,578 44,694 52,467 43,665 52,179 50,904 50,379 45,795 45,852 46,767 36,978 41,370 60,240 50,889 50,892 49,398 46,161 46,371 55,125 40,920 40,329 49,728 54,870 52,986 41,934 39,024 38,703 44,583 51,681 53,946 34,923 44,862 44,658 40,800 55,797 46,584 52,278 45,555 46,242 40,164 42,975 50,937 43,461 52,806 40,128 42,717 43,896 56,847 49,086 51,123 44,922 51,615 48,684 44,892 49,326 38,961 32,349 39,465 47,754 53,847 41,094 42,438 53,676 48,330 36,513 54,453 48,276 52,182 48,147 45,066 47,415 54,423 37,263 37,113 44,271 53,349 41,334 59,829 47,202 49,953 56,970 57,261 53,466 56,055 52,608 41,100 53,757 44,787 36,093 42,909 42,018 51,663 52,527 47,457 39,231 44,559 50,775 43,002 47,805 38,358 39,864 43,137 48,870 36,171 52,317 47,790 46,824 47,502 56,235 63,108 43,863 42,129 37,581 49,872 50,481 38,838 52,353 49,941 47,568 48,468 41,319 47,208 51,030 49,056 60,303 40,878 43,305 54,621 44,379 43,359 53,151 51,498 50,346 51,402 36,369 52,821 49,653 43,911 44,181 51,189 44,118 47,382 46,149 46,578 51,921 47,445 46,536 43,863 46,386 52,548 56,001 39,990 54,924 38,013 Germany 45,716 48,491 53,373 49,169 62,600 44,037 52,574 41,514 46,214 47,847 40,161 48,105 50,279 51,133 52,045 38,961 37,283 47,406 45,609 52,668 43,268 41,976 51,671 53,759 51,382 41,116 51,786 54,738 55,343 48,397 60,469 43,135 44,579 54,939 50,175 43,460 49,829 55,896 59,499 56,091 43,566 41,833 44,384 48,628 46,457 46,758 39,307 54,142 38,292 63,065 332 Statistics for Business 52,060 38,322 54,231 37,866 54,185 55,665 56,064 44,822 44,171 58,812 44,159 51,504 53,507 59,012 50,732 55,462 48,613 53,051 50,263 52,467 38,222 42,308 59,265 53,115 35,559 46,020 56,428 40,669 48,856 46,190 41,988 43,651 55,979 40,323 44,335 48,050 43,809 44,530 43,128 45,585 46,989 52,914 57,012 46,278 53,793 59,152 51,440 38,672 42,694 42,916 52,671 52,115 40,240 53,799 55,687 52,586 55,018 49,266 47,533 48,369 45,138 50,999 43,928 46,184 49,056 33,926 43,980 54,322 54,735 59,338 33,507 51,713 57,380 41,262 52,546 44,861 47,184 46,621 50,893 52,856 55,507 45,050 44,044 47,342 58,420 41,751 60,146 43,323 48,278 58,672 41,244 49,148 42,451 47,348 48,424 47,947 41,426 42,128 63,053 41,165 49,354 42,755 43,448 50,342 55,881 53,884 49,938 48,409 50,880 40,800 Required 1.

Using all of the concepts developed from Chapters 1 to 9 how might you interpret and compare this data from the two countries? 10Forecasting and estimating from correlated data Value of imported goods into the United States Forecasting customer demand is a key activity in business.Forecasts trigger strategic and operations planning.Forecasts are used to determine capital budgets, cash flow, hiring or termination of personnel, warehouse space, raw material quantities, inventory levels, transportation volumes, outsourcing requirements, and the like.If we make an optimistic forecast – estimating more than actual, we may be left with excess inventory, unused storage space, or unwanted personnel.If we are pessimistic in our forecast – estimating less than actual, we may have stockouts, irritated or lost customers, or insufficient storage space.

Thus business must be accurate in forecasting.An often used approach is to use historical or collected data as the basis for forecasting on the assumption that past information is the bellwether for future activity.1, which is a time series analysis for the amount of goods imported into the United States each year from 1996 to 2006.

1 Consider for example that we are now in the year 1970.In this case, we would say that there has been a reasonable linear growth in imported goods in the last decade from 1960.Then if we used a linear relationship for this 1 US Census Bureau, Foreign Trade division, /foreign-trade/statistics/historical goods, 8 June 2007.334 Statistics for Business period to forecast the value of imported goods for 2006, we would arrive at a value of $131,050 million.

The actual value is $1,861,380 million or our forecast is low by an enormous factor of 14! As the data shows, as the years progress, there is an increasing or an almost exponential growth that is in part due to the growth of imported goods particularly from China, India, and other emerging countries many of which are destined for Wal-Mart! Thus, rather than using a linear relationship we should use a polynomial relationship on all the data or perhaps a linear regression relationship just for the period 2000–2005.

Quantitative forecasting methods are extremely useful statistical techniques but you must apply the appropriate model and understand the external environment.Forecasting concepts are the essence of this chapter.1 Value of imported goods into the United States, 1960–2006.0 200,000 400,000 600,000 800,000 1,000,000 1,200,000 1,400,000 1,600,000 1,800,000 2,000,000 19 60 19 65 19 70 19 75 19 80 19 85 19 90 19 95 20 00 20 05 20 10 $m illio ns Year A useful part of statistical analysis is correlation, or the measurement of the strength of a relation- ship between variables.

If there is a reasonable correlation, then regression analysis is a math- ematical technique to develop an equation that describes the relationship between the variables in question.The practical use of this part of statis- tical analysis is that correlation and regression can be used to forecast sales or to make other deci- sions when the developed relationship from past data can be considered to mimic future conditions.A time series is past data presented in regular time intervals such as weeks, months, or years to illustrate the movement of specified variables.Financial data such as revenues, profits, or costs can be presented in a time series.Operating data for example customer service level, capacity util- ization of a tourist resort, or quality levels can be similarly shown.

Macro-economic data such as Gross National Product, Consumer Price Index, or wage levels are typically illustrated by a time series.In a time series we are presenting one variable, such as revenues, against another variable, time, and this is called bivariate data.Scatter diagram A scatter diagram is the presentation of the time series data by dots on an x�y graph to see if there is a correlation between the two variables.The time, or independent variable, is presented on A Time Series and Correlation 335Chapter 10: Forecasting and estimating from correlated data After you have studied this chapter you will understand how to correlate bivariate data and use regression analysis to make forecasts and estimates for business decisions.The variable on the y-axis is considered the dependent variable since it is “dependent”, or a function, of the time.Time is always shown on the x-axis and considered the independent variable since whatever happens today – an earthquake, a flood, or a stock market crash, tomorrow will always come! Application of a scatter diagram and correlation: Sale of snowboards – Part I Consider the information in Table 10.1, which is a time series for the sales of snowboards in a sports shop in Italy since 1990.Using in Excel the graphical command XY(scatter), the scatter diagram for the data from Table 10.We 336 Statistics for Business Table10.Year x Sales, units y 1990 60 1991 90 1992 110 1993 320 1994 250 1995 525 1996 400 1997 800 1998 1,200 1999 985 2000 1,600 2001 1,550 2002 2,000 2003 2,500 2004 2,100 2005 2,400 Figure 10.2 Scatter diagram for the sale of snowboards.

0 200 400 600 800 1,000 1,200 1,400 1,600 1,800 2,000 2,200 2,400 2,600 2,800 19 89 19 90 19 91 19 92 19 93 19 94 19 95 19 96 19 97 19 98 19 99 20 00 20 01 20 02 20 03 20 04 20 05 20 06 Year Sn ow bo ar ds s ol d (un its ) can see that it appears there is a relationship, or correlation, between the sale of snowboards, and the year in that sales are increasing over time.(Note that in Appendix II you will find a guide of how to develop a scatter diagram in Excel.) Coding time series data Very often in presenting time series data we indi- cate the time period by using numerical codes starting from the number 1, rather than the actual period.This is especially the case when the time is mixed alphanumeric data since it is not always convenient to perform calculations with such data.For example, a 12-month period would appear coded as in Table 10.

With the snowboard sales data calculation is not a problem since the time in years is already numerical data.However, the x-values are large and these can be cumbersome in subsequent calculations.3 gives the scatter diagram using a coded value for x where 1 � 1990, 2 � 1991, 3 � 1992, etc.

The form of this scatter diagram in Figure 10.337Chapter 10: Forecasting and estimating from correlated data Table 10.Month Code January 1 February 2 March 3 April 4 May 5 June 6 July 7 August 8 September 9 October 10 November 11 December 12 Figure 10.3 Scatter diagram for the sale of snowboards using coded values for x.0 200 400 600 800 1,000 1,200 1,400 1,600 1,800 2,000 2,200 2,400 2,600 2,800 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 Year (1�1990, 2 � 1991, etc.) Sn ow bo ar ds s ol d (un its ) Coefficient of correlation Once we have developed a scatter diagram, a next step is to determine the strength or the importance of the relationship between the time or independent variable x, and the depend- ent variable y.One measure is the coefficient of correlation, r, which is defined by the rather horrendous-looking equation as follows: Coefficient of correlation, 10(i) Here n is the number of bivariate (x, y) values.

The value of r is either plus or minus and can take on any value between 0 and 1.If r is nega- tive, it means that for the range of data given the variable y decreases with x.If r is positive, it means that y increases with x.The closer the value of r is to unity, the stronger is the relation- ship between the variables x and y.

When r approaches zero it means that there is a very weak relationship between x and y.

The calculation steps using equation 10(i) are given in Table 10.3 using a coded value for the time period rather than using the numerical val- ues of the year.You simply enter the corresponding values for x and y where x can either be the indicated period (provided it is in numerical form) or the code value.It does not matter which, as the result is the same.

In the case of the snowboard sales given in the example, r � �0.0 and thus it indicates there is a strong correlation between x and y.Coefficient of determination The coefficient of determination, r2, is another measure of the strength of the relationship r n xy x y n x x n y y � � � � ∑∑∑ ∑∑ ∑∑( )⎡ ⎣ ⎢⎢ ⎤ ⎦ ⎥⎥ ( ) ⎡ ⎣ ⎢⎢ 2 2 2 2 ⎤⎤ ⎦ ⎥⎥ 338 Statistics for Business Table 10.3 Coefficients of correlation and determination for snowboards using coded values of x.n 16 n xy 3,251,200 x y 2,297,040 n x2 23,936 ( x)2 18,496 n y2 464,912,800 ( y)2 285,272,100 n xy � x y 954,160 n x2 � ( x)2 5,440 n y2 � ( y)2 179,640,700 r 0.9316 x (year) x (coded) y xy x2 y2 1990 1 60 60 1 3,600 1991 2 90 180 4 8,100 1992 3 110 330 9 12,100 1993 4 320 1,280 16 102,400 1994 5 250 1,250 25 62,500 1995 6 525 3,150 36 275,625 1996 7 400 2,800 49 160,000 1997 8 800 6,400 64 640,000 1998 9 1,200 10,800 81 1,440,000 1999 10 985 9,850 100 970,225 2000 11 1,600 17,600 121 2,560,000 2001 12 1,550 18,600 144 2,402,500 2002 13 2,000 26,000 169 4,000,000 2003 14 2,500 35,000 196 6,250,000 2004 15 2,100 31,500 225 4,410,000 2005 16 2,400 38,400 256 5,760,000 Total 136 16,890 203,200 1,496 29,057,050 between x and y.

Since it is the square of the coefficient of correlation, r, where r can be either negative or positive, the coefficient of determina- tion always has a positive value.Further, since r is always equal to, or less than 1.0, numerically the value of r2, the coefficient of determination, is always equal to or less than r, the coefficient of correlation.0 which means that there is a perfect correlation between x and y.For the snowboard sales the value of r2 is 0.Again for completeness, the calculation using equation 10(ii) is shown in Table 10.

How good is the correlation? Analysts vary on what is considered a good cor- relation between bivariate data.I say that if you have a value of r2 of at least 0.8944), then there is a reasonable relationship between the independent variable and the dependent variable.Once we have developed a scatter diagram for a time series data, and the strength of the rela- tionship between the dependent variable, y, and the independent time variable, x, is reasonably strong, then we can develop a linear regression equation to define this relationship.After that, we can subsequently use this equation to fore- cast beyond the time period given.Linear regression line The linear regression line is the best straight line that minimizes the error between the data points on the regression line and the correspond- ing actual data from which the regression line is developed.

The following equation represents the regression line: 10(iii) Here, ● a is a constant value and equal to the intercept on the y-axis; ● b is a constant value and equal to the slope of the regression line; ● x is the time and the independent variable value; ● y is the predicted, or forecast value, of the actual dependent variable, y.The values of the constants a and b can be cal- culated by the least squares method using the following two relationships: 10(iv) 10(v) Another approach is to calculate b and a using the average value of x or x–, and the aver- age value of y or y– using the two equations below.It does not matter which we use as the result is the same: 10(vi) a � y�bx� 10(vii) b xy nx y x n x � � � ∑ ∑ 2 ( )2 b n xy x y n x x � � � ∑∑∑ ∑∑ ( )2 2 a x y x xy n x x � � � 2 2 2 ∑ ∑ ∑∑ ∑∑ ( ) y a bx� � Linear Regression in a Time Series Data r n xy x y n x x n y y 2 2 2 2 2 2 � � � � ∑∑∑ ∑∑ ∑∑ ( ) ( )⎡ ⎣ ⎢⎢ ⎤ ⎦ ⎥⎥ ( ) ⎡⎡ ⎣ ⎢⎢ ⎤ ⎦ ⎥⎥ 339Chapter 10: Forecasting and estimating from correlated data The calculations using these four equations are given in Table 10.4 for the snowboard sales using the coded values for x.However, again it is not necessary to perform these calculations because all the relationships can be developed from Microsoft Excel as explained in the next section.

Application of developing the regression line using Excel: Sale of snowboards – Part II Once we have the scatter diagram for the bivari- ate data we can use Microsoft Excel to develop the regression line.To do this we first select the data points on the scatter diagram and then proceed as follows: ● In the menu select of Excel, select Chart ● Select Add trend line ● Select Type ● Select Linear ● Select Options and check Display equation on chart and Display R-squared value on chart This final window is shown in Figure E-7 of the Appendix II.The regression line using the coded values of x is shown in Figure 10.On the graph we have the regression line written as follows, which is a different form as represented by equation 10(iii).

2500 In the form of equation 10(iii) it would be reversed and written as, y � �435.

3971x 340 Statistics for Business Table 10.4 Regression constants for snowboards using coded value of x.n 16 x 136 y 16,890 x2 1,496 xy 203,200 n x2 23,936 ( x)2 18,496 n xy 3,251,200 a using equation 10(iv) �435.2500 x (year) x (coded) y xy x2 y2 1990 1 60 60 1 3,600 1991 2 90 180 4 8,100 1992 3 110 330 9 12,100 1993 4 320 1,280 16 102,400 1994 5 250 1,250 25 62,500 1995 6 525 3,150 36 275,625 1996 7 400 2,800 49 160,000 1997 8 800 6,400 64 640,000 1998 9 1,200 10,800 81 1,440,000 1999 10 985 9,850 100 970,225 2000 11 1,600 17,600 121 2,560,000 2001 12 1,550 18,600 144 2,402,500 2002 13 2,000 26,000 169 4,000,000 2003 14 2,500 35,000 196 6,250,000 2004 15 2,100 31,500 225 4,410,000 2005 16 2,400 38,400 256 5,760,000 Total 136 16,890 203,200 1,496 29,057,050 Average 8.6250 However, the regression information is the same where y is y , and the slope of the line, b, is 175.3971 and, a, the intercept on the y-axis is �435.These numbers are the same as calculated and presented in Figure 10.The slope of the line means that the sale of snowboards increases by 175.The intercept, a, means that when x is zero the sales are �432.25 units which has no meaning for this situation.9316, which appears on the graph, is the same as previously calculated though note that Microsoft Excel uses upper case R2 rather than the lower case r2.When the value of a is negative, but the slope of the curve is positive, it is normal to show the above equations for this example in the form y � 175.That is, avoid starting an equation with a negative value.The regression line using the actual values of the year is shown in Figure 10.This is because the values of x are the real values and not coded values.For example, assume that we want to forecast the sale of 341Chapter 10: Forecasting and estimating from correlated data Figure 10.4 Regression line for the sale of snowboards using coded value of x.

9316 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 0 200 400 600 800 1,000 1,200 1,400 1,600 1,800 2,000 2,200 2,400 2,600 2,800 Sn ow bo ar ds s ol d (un its ) Year snowboards for 2010.This gives a forecast value of y of 3,248 units.Alternatively, we can use the coded values of x that appear in the 2nd column of Table 10.2 and the corresponding actual data for y.If we do this, we must use a code value for the year 2010, which in this case is 21.

(Year 2005 has a code of 16, thus year 2010 � 16 � 5 � 21.) Note that in any forecasting using time series data, the assumption is that the pattern of past years will be repeated in future years, which may not necessarily be the case.Also, the fur- ther out we go in time, the less accurate will be the forecast.For example, a forecast of sales for next year may be reasonably reliable, whereas a forecast 20 years from now would not.

The variability of the estimate In Chapter 2, we presented the sample standard deviation, s, of data by the equation, Sample standard deviation, 2(viii) The standard deviation is a measure of the variability around the sample mean, x–, for each random variable x, in a given sample size, n.

Further, the deviation of all the observations, x, about the mean value x– is zero (equation 2(ix)), or, 2(ix)( )x x� �∑ 0 s s x x n � � � � 2 1 ( ) ( ) 2∑ 342 Statistics for Business Figure 10.5 Regression line for the sale of snowboards using actual year.9316 0 200 400 600 800 1,000 1,200 1,400 1,600 1,800 2,000 2,200 2,400 2,600 2,800 19 89 19 90 19 91 19 92 19 93 19 94 19 95 19 96 19 97 19 98 19 99 20 00 20 01 20 02 20 03 20 04 20 05 20 06 Year Sn ow bo ar ds s ol d (un its ) In a similar manner, a measure of the variabil- ity around the regression line is the standard error of the estimate, se, given by, Standard error of the estimate, 10(viii) Here n is the number of bivariate data points (x, y).The value of se has the same units of the dependant variable y.The denominator in this equation is (n � 2) or the number of degrees of freedom, rather than (n � 1) in equation 2(viii).In equation 10(viii) two degrees of freedom are lost because two statistics, a and b, are used in regression to compute the standard error of the estimate.Like the standard deviation, the closer to zero is the value of the standard error then there is less scatter or deviation around the regression line.

If this is the case, this translates into saying that the linear regression model is a good fit of the observed data, and we should have reasonable confidence in the estimate or forecast made.The regression equation, is determined so that the vertical distance between the observed, or data values, y, and the predicted values, y , balance out when all data are considered.To do this we select a cellblock of two columns by five rows and enter the given x- and y-values and input 1 both times for the constant data.The statistics for the regression line for the snowboard data are given in Table 10.The explanations are given to the left and to the right of each column.Those that we have discussed so far are highlighted and also note in this matrix that we have again the value of b, the slope of the line; the value a, the intercept on the y-axis, and the coefficient of determination, r2.We also have the degrees of freedom, or (n � 2).

The other statistics are not used here but their mean- ing in the appropriate format is indicated in Table E-3 of Appendix II.Confidence in a forecast In a similar manner to confidence limits in esti- mating presented in Chapter 7, we can deter- mine the confidence limits of a forecast.If we have a sample size greater than 30 then the confidence intervals are given by, y � zse 10(x)( )y y� � ∑ 0 s y y ne 2 � � � ( )∑ 2 343Chapter 10: Forecasting and estimating from correlated data Table 10.1764 se standard error of determination estimate 190.7380 14 degrees of freedom (n� 2) 10,459,803.1471 With sample sizes no more than 30, we use a Student-t relationship and the confidence limits are, y � tse 10(xi) For our snowboard sales situation we have a fore- cast of 3,248 units for 2010.To obtain a confi- dence level, we use a Student-t relationship since we have a sample size of 16.

Then using equation 10(xi) and the standard error from Table 10.5 the confidence limits are as follows: Lower limit is 3,248 � 1.

1764 � 3,361 Thus to better define our forecast we could say that our best estimate of snowboard sales in 2010 is 3,248 units and that we are 90% confi- dent that the sales will be between 2,836 and 3,361 units.Alternative approach to develop and verify the regression line Now that we have determined the statistical val- ues for the regression line, as presented in Table 10.

5, we can use these values to develop the spe- cific values of the regression points and further to verify the standard error of the estimate, se.The calculation steps are shown in Table 10.The col- umn y is calculated using equation 10(iii) and imputing the constant values of a and b from Table 10.

The total of (y � y ) in Column 5 verifies equation 10(ix).And, using the total value of (y � y )2 in Column 6, the last column of Table 10.6, and inserting this in equation 10(viii) veri- fies the value of the standard error of the estimate of Table 10.6 Calculating the standard estimate of the regression line using coded values of x.18 n 16 In the previous sections we discussed correlation and how a dependent variable changed with time.Another type of correlation is when one variable is dependent, or a function, not of time, on some other variable.In these situations we say that the move- ment of the dependent variable, y, is caused by the change of the dependent variable, x and the correlation can be used for causal forecasting or estimating.The analytical approach is very simi- lar to linear regression for a time series except that time is replaced by another variable.

The following example illustrates this.Application of causal forecasting: Surface area and house prices In a certain community in Southern France, a real estate agent has recorded the past sale of houses according to sales price and the square metres of living space.Develop a scatter diagram for this information.Does there appear to be a reasonable correlation between the price of homes, and the square metres of living space? Here this is a causal relationship where the price of the house is a function, or is “caused” by the square metres of living space.Thus, the square metres is the independent vari- able, x, and the house price is the dependent variable y.Using the same approach as for the previous snowboard example in a time series analysis, Figure 10.6 gives the scatter diagram for this causal relationship.

Visually it appears that within the range of the data given, the house prices generally increase lin- early with square metres of living space.Show the regression line and the coefficient of determination on the scatter diagram.What can you say about the coefficients of determination and cor- relation? What is the slope of the regression line and how is it interpreted? The regression line is shown in Figure 10.

7 together with the coefficient of determina- tion.The relationships are as follows: Regression equation, y � �1,263,749.6133x Coefficient of determination, r2 � 0.8623 Coefficient of correlation, r � �r2� � ��0.

9286 Since the coefficient of determination is greater than 0.8, and thus the coefficient of correlation is greater than 0.9 we can say that there is quite a strong correlation Linear Regression and Causal Forecasting 345Chapter 10: Forecasting and estimating from correlated data Table 10.Square metres, x Price (€) y 100 260,000 180 425,000 190 600,000 250 921,000 360 2,200,000 200 760,500 195 680,250 110 690,250 120 182,500 370 2,945,500 280 1,252,500 450 5,280,250 425 3,652,000 390 3,825,240 60 140,250 125 280,125 346 Statistics for Business Figure 10.6 Scatter diagram for surface area and house prices.0 1,000,000 2,000,000 3,000,000 4,000,000 5,000,000 6,000,000 0 50 100 150 200 250 300 350 400 450 500 Area (m2) Pr ic e (€ ) Figure 10.7 Regression line for surface area and house prices.8623 0 1,000,000 2,000,000 3,000,000 4,000,000 5,000,000 6,000,000 0 50 100 150 200 250 300 350 400 450 500 Area (m2) Pr ic e (€ ) between house prices and square metres of living space.The slope of the regression line is 11,646.

6133 (say 11,650); this means to say that for every square metre in living space, the price of the house increases by about €11,650 within the range of the data given.

If a house on the market has a living space of 310 m2, what would be a reasonable estimate of the price? Give the 85% confidence intervals for this price.8 the statistics for the regression line.

8, the value of t for a confidence level of 85% is 1.Using equation 10(xi), y � tse Lower limit of price estimate using the stand- ard error of the estimate from Table 10.5231 * 609,444 � €1,418,463 Upper limit is, €2,346,700 � 1.5231 * 609,444 � €3,274,938 Thus we could say that a reasonable estimate of the price of a house with 310 m2 living space is €2,346,700 and that we are 85% con- fident that the price lies in the range €1,418,463 (say €1,418,460) to €3,274,938 (say €3,274,940).The danger with making this estimate is that 800 m2 is outside of the limits of our observed data (it ranges from 60 to 450 m2).

Thus the assumption that the linear regres- sion equation is still valid for a living space area of 800 m2 may be erroneous.Thus you must be careful in using causal forecasting beyond the range of data collected.In the previous section on causal forecasting we considered the relationship between just one dependent variable and one independent variable.Forecasting Using Multiple Regression 347Chapter 10: Forecasting and estimating from correlated data Table 10.8 Regression statistics for surface area and house prices.

0004 se standard error of determination estimate 87.1999 * 1012 Multiple regression takes into account the rela- tionship of a dependent variable with more than one independent variable.

In business, sales revenues can be a func- tion of advertising expenditures, number of sales staff, number of branch offices, unit prices, num- ber of competing products on the market, etc.In this situation, the forecast estimate is a causal regression equation containing several independ- ent variables.Multiple independent variables The following is the equation that describes the multiple regression model: y � a � b1x1 � b2x2 � b3x3 � … � bkxk 10(xii) Here, ● a is a constant and the intercept on the y-plane; ● x1, x2, x3, and xk are the independent variables; ● b1, b2, b3 and bk are constants and slopes of the line corresponding to x1, x2, x3, and xk; ● y is the forecast or predicted value given by the best fit for the actual data; ● k is a value equal to the number of independ- ent variables in the model.

Also note that the more the number of independent variables in the rela- tionship then the more complex is the model, and possibly the more uncertain is the predicted value.Standard error of the estimate As for linear regression, there is a standard error of the estimate se that measures the degree of dispersion around the multiple regres- sion plane.It is as follows: 10(xiii) Here, ● y is the actual value of the dependant variable; ● y is the corresponding predicted value of dependant variable from the regression equation; ● n is the number of bivariate data points; ● k is the number of independent variables.This is similar to equation 10(viii) for linear regres- sion except that there is now a term k in the denominator where the value (n � k � 1) is the degrees of freedom.

As an illustration, if the num- ber of bivariate data points n is 16, and there are four independent variables then the degrees of freedom are 16 � 4 � 1 � 11.In linear regres- sion, with the same 16 bivariate data values, the number of independent variables, k, is 1, and so the degrees of freedom are 14 � 1 � 1 � 12 or the denominator as given by equation 10(xiii).As before, the smaller the value of the standard error of the estimate, the better is the fit of the regression equation.Coefficient of multiple determination Similar to linear regression there is a coefficient of multiple determination r2 that measures the strength of the relationship between all the independent variables and the dependent vari- able.

The calculation of this is illustrated in the following worked example.Application example of multiple regression: Supermarket A distributor of Nestl coffee to supermarkets in Scandinavia visits the stores periodically to s y y n ke 2( ) � � � � ∑ 1 348 Statistics for Business meet the store manager to negotiate shelf space and to discuss pricing and other sales-related activities.For one particular store the distribu- tor had gathered the data in Table 10.9 regard- ing the unit sales for a particular size of instant coffee, the number of visits made to that store, and the total shelf space that was allotted.

Determine the coefficient of determination.The difference is that we now select a virgin area of three rows and five columns and we enter two columns for the independent variable x, visits per month and the shelf space.

The statistics that we need from this table are in the shaded cells and are as follows: ● a, the intercept on the y-plane � 14,227.67; ● b1, the slope corresponding to x1, the vis- its per month � 4,827.01; ● b2, the slope corresponding to the shelf space, x2 � 9,997.

64; ● se, the standard error of the estimate � 5,938.51; ● Coefficient of determination, r2 � 0.Again, the other statistics in the non-shaded areas are not used here but their meaning, in the appropriate format, are indicated in Table E-4 of Appendix II.The equation, or model, that describes this relation is from equation 10(xii) for two independent variables: y � a � b1x1 � b2x2 y � 14,227.

64x2 As the coefficient of determination, 0.8 the strength of the rela- tionship is quite good.

349Chapter 10: Forecasting and estimating from correlated data Table 10.Unit sales/month, Visits/month, Shelf space, y x1 (m2) x2 90,150 9 3.10 Regression statistics for sales of Nestl coffee–two variables.Estimate the monthly unit sales if eight visits per month were made to the supermarket and the allotted shelf space was 3.What are the 85% confidence levels for this estimate? Here x1 is the estimate of sales of eight visits per month, and x2 is the shelf space of 3.

The monthly sales are determined from the regression equation: y � 14,227.10, the value of t for a confidence level of 85% is 1.The confidence limits of sales using the standard error of the estimate of 5,938.10 are, Lower confidence limit is 82,837 � 1.51 � 73,237 units Upper confidence limit is, 82,837 � 1.51 � 92,437 units Thus we can say that using this regression model our best estimate of monthly sales is 82,837 units and that we are 85% confident that the sales will between 73,237 and 92,437 units.

Assume now that for the sales data in Table 10.9 the distributor looks at the unit price of the coffee sold during the period that the analysis was made.This expanded information is in Table 10.11 showing now the variation in the unit price of a jar of coffee.

Determine the coefficient of determination.The statistics that we need from this table are: ● a, the intercept on the y-plane � 75,658.05; ● b1, the slope corresponding to x1 the vis- its per month � 2,984.28; ● b2, the slope corresponding to the shelf space, x2 � 4,661,82; 350 Statistics for Business Table 10.11 Sales of Nestl coffee with three variables.Sales, Visits/month, Shelf space (m2), Price (€/unit), y x1 x2 x3 90,150 9 3.

75 ● b3, the slope corresponding to the price, x3 � �18,591.

50; ● se, the standard error of the estimate � 5,491.60; ● Coefficient of determination, r2 � 0.The equation or model that describes this relation is from equation 10(xii) for three independent variables: y � a � b1x1 � b2x2 � b3x3 y � 75,658.50x3 As the coefficient of determination, 0.8 the strength of the rela- tionship is quite good.

Estimate the monthly unit sales if eight visits per month were made to the supermarket, the allotted shelf space was 3.00 m2, and the unit price of coffee was €2.What are the 85% confidence levels for this estimate? Here x1 is the estimate of sales of eight visits per month, x2 is the shelf space of 3.

00 m2, and x3 is the unit sales price of coffee of €2.Estimated monthly sales are deter- mined from the regression equation, y � 75,658.The degrees of freedom are 6 from Table 10.

12, the value of t for a confidence level of 85% is 1.The confidence limits of sales using the standard error of the estimate of 5,491.60 � 57,977 units Upper limit is, 67,039 � 1.60 � 76,101 units Thus we can say that using this regression model, the best estimate of monthly sales is 67,039 units and that we are 85% confident that the sales will between 57,977 and 76,101 units.Examples of these are: the sales of mobile phones from about 1995 to 2000; the increase of HIV contamination in Forecasting Using Non-linear Regression 351Chapter 10: Forecasting and estimating from correlated data Table 10.12 Regression statistics for coffee sales – three variables.32 #N/A #N/A Africa; and the increase in the sale of DVD play- ers.Curvilinear relationships can take on a var- iety of forms as discussed below.Once we have the scatter diagram for this bivariate data we can use Microsoft Excel to develop the regression line.

To do this we first select the data points on the graph and then from the Menu chart proceed sequentially as follows: ● Add trend line ● Type polynomial power ● Options ● Display equation on chart and Display R-squared value on chart.A second-degree or quadratic poly- nomial function, where x has a power of 2 for the surface area and house price data of Table 10.The regression equation and the corresponding coefficient of determina- tion are as follows: y � 41.The regression equation 352 Statistics for Business Figure 10.8 Second-degree polynomial for house prices.9653 0 1,000,000 2,000,000 3,000,000 4,000,000 5,000,000 6,000,000 0 50 100 150 200 250 300 350 400 450 500 Area (m2) Pr ic e (€ ) and the corresponding coefficient of determina- tion are as follows: y � �0.9729 We can see that as the power of x increases the closer is the coefficient of determination to unity or the better fit is the model.Note for this same data when we used linear regression, Figure 10.7, the coefficient of determination was 0.10 and the follow- ing is the equation with the corresponding coef- ficient of determination: y � 110,415.9298 In business, particularly when selling is involved, seasonal patterns often exist.For example, in the Northern hemisphere the sale of swimwear is higher in the spring and summer than in the autumn and winter.

The demand for heating oil is higher in the autumn and winter, and the sale of cold beverages is higher in the summer than in the winter.The linear regres- sion analysis for a time series analysis, discussed Seasonal Patterns in Forecasting 353Chapter 10: Forecasting and estimating from correlated data Figure 10.9729 0 1,000,000 2,000,000 3,000,000 4,000,000 5,000,000 6,000,000 0 50 100 150 200 250 300 350 400 450 500 Area (m2) Pr ic e (€ ) early in the chapter, can be modified to take into consideration seasonal effects.The following application illustrates one approach.Application of forecasting when there is a seasonal pattern: Soft drinks Table 10.13 gives the past data for the number of pallets of soft drinks that have been shipped from a distribution centre in Spain to various retail outlets on the Mediterranean coast.

Plot the actual data and see if a seasonal pattern exists The actual data is shown in Figure 10.

11 and from this it is clear that the data is seasonal.Note that for the x-axis we have used a coded value for each season starting with winter 2000 with a code value of 1.Determine a centred moving average A centred moving average is the average value around a designated centre point.Here we determine the average value around a partic- ular season for a 12-month period, or four quarters.

For example, the following relation- ship indicates how we calculate the centred moving average around the summer quarter (usually 15 August) for the current year n: For example if we considered the centre period as summer 2000 then the centred 0 5 1 0 1 0 1 .* ( ) winter spring summer n n n � � � .* ( ) 0 0 5 1 4 autumn winter n n� � 354 Statistics for Business Figure 10.9298 0 1,000,000 2,000,000 3,000,000 4,000,000 5,000,000 6,000,000 0 50 100 150 200 250 300 350 400 450 500 Area (m2) Pr ic e (€ ) 355Chapter 10: Forecasting and estimating from correlated data Table 10.Year Quarter Actual sales (pallets) 2000 Winter 14,844 Spring 15,730 Summer 16,665 Autumn 15,443 2001 Winter 15,823 Spring 16,688 Summer 17,948 Autumn 16,595 2002 Winter 16,480 Spring 17,683 Summer 18,707 Autumn 17,081 Year Quarter Actual sales (pallets) 2003 Winter 18,226 Spring 19,295 Summer 19,028 Autumn 17,769 2004 Winter 18,909 Spring 20,064 Summer 19,152 Autumn 18,503 2005 Winter 19,577 Spring 20,342 Summer 20,156 Autumn 19,031 Figure 10.11 Seasonal pattern for the sales of soft drinks.14,000 15,000 16,000 17,000 18,000 19,000 20,000 21,000 22,000 23,000 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 Quarter (1�winter 2000) Sa le s (pa lle ts) moving average around this quarter using the actual data from Table 10.

13 is as follows: We are determining a centred moving aver- age and so the next centre period is autumn 2000.For this quarter, we drop the data for winter 2000 and add spring 2001 and thus the centred moving average around autumn 2000 is as follows: Thus each time we move forward one quarter we drop the oldest piece of data and add the next quarter.The values for the centred mov- ing average for the complete period are in Column 5 of Table 10.Note that we 0 50 15 730 1 0 16 665 1 0 15 443 1 0 15 82 .

* , � � � 33 0 5 16 688 4 16 035 00 � � .0 5 14 844 1 0 15 730 1 0 16 665 1 0 15 443 .14 Sales of soft drinks – seasonal indexes and regression.1 2 3 4 5 6 7 8 9 Year Quarter Code Actual sales Centred moving SIp Seasonal Sales/SI Regression (pallets) average index SI forecast, y

74 cannot determine a centred moving average for winter and spring 2000 or for summer and autumn of 2005 since we do not have all the necessary information.The line graph for this centred moving average is in Figure 10.Divide the actual sales by the moving average to give a period seasonal index, SIp This is the ratio, This data is in Column 6 of Table 10.What we have done here is compared actual sales to the average for a 12-month period.It gives a specific seasonal index for each month.For example, if we consider 2004 the ratios, rounded to two decimal places, are as in Table 10.

We interpret this by saying that sales in the winter 2004 are 2% below the year (1 � 0.98), in the spring they are 3% above the year, 6% above the year for the summer, and 10% below the year for autumn 2004 (1 � 0.Determine an average seasonal index, SI, for the four quarters This is determined by taking the average of all the ratios, SIp for like seasons.For example,SIp (Actual recorded sales in a period) (Mov � iing average for the same period) 357Chapter 10: Forecasting and estimating from correlated data Figure 10.12 Centred moving average for the sale of soft drinks.15,000 16,000 17,000 18,000 19,000 20,000 21,000 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 Coded period (1�winter 2000) Ce nt re d m ov in g av er ag e of p al le ts Table 10.15 Sales of soft drinks – seasonal indexes.

90 the seasonal index for the summer is calcu- lated as follows: The seasonal indices for the four seasons are in Table 10.

Note that the average value of these indices must be very close to unity since they represent the movement for one year.These same indices, but rounded to two decimal places, are shown in Column 7 of Table 10.Note, for similar seasons, the values are the same.

Divide the actual sales by the seasonal index, SI This data is shown in Column 8.What we have done here is removed the seasonal effect of the sales, and just showed the trend in sales without any contribution from the sea- sonal period.Another way to say is that the sales are deseasonalized.The line graph for these deseasonalized sales is in Figure 10.

Develop the regression line for the desea- sonalized sales The regression line is shown in Figure 10.The regression equation and the 1 0552 1 0654 1 0565 1 0588 1 0646 5 1 0601 .

� � � � � 358 Statistics for Business Table 10.16 Sales of soft drinks – seasonal indexes.15,000 16,000 17,000 18,000 19,000 20,000 21,000 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 Coded period (1�winter 2000) Sa le s/ SI corresponding coefficient of determination are as follows: y � 230.

11 the x-values of Column 1 and the y-values from Column 8 to give the statistics in Table 10.Using the corresponding values of a and b we have developed the regression line values as shown in Column 9 of Table 10.From the regression line forecast desea- sonalized sales for the next four quarters This can be done in two ways.Either from the Excel table, continue the rows down for 2006 using the code values of 25 to 28 for the four seasons.

14 Deseasonalized sales and regression line for soft drinks.9673 15,000 16,000 17,000 18,000 19,000 20,000 21,000 22,000 23,000 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 Coded period (1�winter 2000) Sa le s/ SI Table 10.17 Sales of soft drinks – seasonal indexes.0335 se, standard error determination of estimate 650.0810 22 degrees of freedom (n � 2) 61,282,861 2,073,931 FORECAST where the x-values are the code values 25 to 28 and the actual values of x are the code values 1 to 24 and the actual values of y are the deseasonalized sales for these same coded periods.

These values are in Column 6 of Table 10.Multiply the forecast regression sales by the SI to forecast 2006 seasonal sales The forecast seasonal sales are shown in Column 4 of Table 10.What we have done is reversed our procedure by now multiplying the regression forecast by the SI.When we developed the data we divided by the SI to obtain a deseasonalized sale and used the regression analysis on this information.The actual and forecast sales are shown in Figure 10.Although at first the calculation procedure may seem laborious, it can be very quickly executed using an Excel spread sheet and the given functions.

We must remember that a forecast is just that – a forecast.Thus when we use statistical analysis to forecast future patterns we have to exercise caution when we interpret the results.The fol- lowing are some considerations.Time horizons Often in business, managers would like a forecast to extend as far into the future as possible.However, the longer the time period the more uncertain is the model because of the changing environment – What new technologies will come onto the market? What demographic changes will occur? How will interest rates move? An approach to recognize this is to develop forecast models for different time periods – say short, medium, and long-term.

The forecast model for the shorter time period would provide the most reliable information.Collected data Quantitative forecast models use collected or historical data to estimate future outcomes.In collecting data it is better to have detailed rather than aggregate information, as the latter might camouflage situations.For example, assume that you want to forecast sales of a certain prod- uct of which there are six different models.You could develop a model of revenues for all of the six models.

However, revenues can be distorted by market changes, price increases, or exchange Considerations in Statistical Forecasting 360 Statistics for Business Table 10.1 2 3 4 5 6 Year Quarter Code Forecast Seasonal Regression sales index SI forecast, y12 rates if exporting or importing is involved.It would be better first to develop a time series model on a unit basis according to product range.This base model would be useful for tracking inventory movements.It can then be extended to revenues simply by multiplying the data by unit price.

Coefficient of variation When past data is collected to make a forecast, the coefficient of variation of the data, or the ratio of the standard deviation to the mean ( / ), is an indicator of how reliable is a forecast model.For example, consider the time series data in Table 10.361Chapter 10: Forecasting and estimating from correlated data Figure 10.15 Actual and forecast sales for soft drinks.

14,000 15,000 16,000 17,000 18,000 19,000 20,000 21,000 22,000 23,000 24,000 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 Quarter (1�winter 2000) Sa le s (pa lle ts) Forecast Table 10.Period Product A Product B January 1,100 800 February 1,024 40 March 1,080 564 April 1,257 12 May 1,320 16 June 1,425 456 July 1,370 56 August 1,502 12 September 1,254 954 s (as a sample) 164.17 variation, / For product A the coefficient of variation is low meaning that the dispersion of the data relative to its mean is small.In this case a forecast model should be quite reliable.

On the other hand, for Product B the coefficient of variation is greater than one or the sample standard devi- ation is greater than the mean.Here a forecast model would not be as reliable.In situations like this perhaps there is a seasonal activity of the product and this should be taken into account in the selected forecast model.In using the coeffi- cient of variation as a guide, care should be taken as if there is a trend in the data that will of course impact the coefficient.As already discussed in the chapter, plotting the data on a scatter diagram would be a visual indicator of how good is the past data for forecasting purposes.

Note that in determining the coefficient of variation we have used the sample standard deviation, s, as an esti- mate of the population standard deviation, .Market changes Market changes should be anticipated in forecast- ing.For example, in the past, steel requirements might be correlated with the forecast sale of automobiles.However plastic and composite materials are rapidly replacing steel, so this factor would distort the forecast demand for steel if the old forecasting approach were used.

Alternatively, more and more uses are being found for plastics, so this element would need to be incorporated into a forecast for the demand for plastics.

These types of events may not affect short-term planning but certainly are important in long-range forecasting when capital appropriation for plant and equip- ment is a consideration.Models are dynamic A forecast model must be a dynamic working tool with the flexibility to be updated or modi- fied as soon as new data become available that might impact the outcome of the forecast.For example, an economic model for the German economy had to be modified with the fall of the Berlin Wall in 1989 and the fusion of the two Germanys.Similarly, models for the European Economy have been modified to take into account the impact of the Euro single currency.Model accuracy All managers want an accurate model.

The accuracy of the model, whether it is estimated at 10%, 20%, or say 50% can only be within a range bounded by the error in the collected data.Further, accuracy must be judged in light of control a firm has over resources and exter- nal events.Besides accuracy, also of interest in a forecast is when turning points in situations might be expected such as a marked increase (or decrease) in sales so that the firm can take advantage of the opportunities, or be prepared for the threats.Curvilinear or exponential models We must exercise caution in using curvilinear functions, where the predicted value y changes rapidly with x.Even though the actual collected data may exhibit a curvilinear relationship, an exponential growth often cannot be sustained in the future often because of economic, mar- ket, or demographic reasons.

In the classic life cycle curve in marketing, the growth period for successful new products often follows a curvi- linear or more precisely an exponential growth model but this profile is unlikely to be sustained as the product moves into the mature stage.In the worked example, surface area and house prices, we developed the following two- degree polynomial equation: y � 41.1408 Using this for a surface area of 1,000 m2 fore- casts a house price of €32.

3 million, which is 362 Statistics for Business beyond the affordable range for most people.Consider also the sale of snowboards worked example presented at the beginning of the chap- ter.Here we developed a linear regression model that gave a coefficient of determination of 0.9316 and the model forecast sales of 3,248 units for 2010.Now if we develop an exponen- tial relationship for this same data then this would appear as in Figure 10.

The equation describing this curve is, y � e0.2479x The data gives a respectable coefficient of deter- mination of 0.If we use this to make a forecast for sale of snowboards in 2010 we have a value of 2.

62 10216 which is totally unreasonable.Selecting the best model It is difficult to give hard and fast rules to select the best forecasting model.The activity may be a trial and error process selecting a model and test- ing it against actual data or opinions.If a quanti- tative forecast model is used there needs to be consideration of subjective input, and vice-versa.This model was needed to estimate financial returns from future oil exploration, drilling, refin- ery, and chemical plant operation.The model basis was a combined multiple regression and curvilin- ear relationships incorporating variables in the United States economy such as changes in the GNP, interest rates, energy consumption, chemical 363Chapter 10: Forecasting and estimating from correlated data Figure 10.9191 0 200 400 600 800 1,000 1,200 1,400 1,600 1,800 2,000 2,200 2,400 2,600 2,800 3,000 3,200 3,400 3,600 3,800 4,000 4,200 4,400 19 89 19 90 19 91 19 92 19 93 19 94 19 95 19 96 19 97 19 98 19 99 20 00 20 01 20 02 20 03 20 04 20 05 20 06 Year Sn ow bo ar ds s ol d (un its ) production and forecast chemical use, demo- graphic changes, taxation, capital expenditure, seasonal effects, and country political risk.Throughout the development, the model was tested against known situations.The model proved to be a reasonable forecast of future prices.

A series of forecast models have been devel- oped by a group of political scientists who study the United States elections.These models use com- bined factors such as public opinions in the pre- ceding summer, the strength of the economy, and the public’s assessment of its economic well- being.The models have been used in all the United States elections since 1948 and have proved highly accurate.2 In 2007 the world economy suf- fered a severe decline as a result of bank loans to low income homeowners.Jim Melcher, a money manager based in New York, using complex deriv- ative models forecast this downturn and pulled out of this risky market and saved his clients mil- lions of dollars.

3 364 Statistics for Business 2 Mathematically, Gore is a winner, International Herald Tribune, 1 September 2000.3 Warnings were missed in US loan meltdown, International Herald Tribune, 20 August 2007.This chapter covers forecasting using bivariate data and presents correlation, linear and mul- tiple regression, and seasonal patterns in data.A time series and correlation A time series is bivariate information of a dependent variable, y, such as sales with an independent variable x representing time.

Correlation is the strength between these variables and can be illus- trated by a scatter diagram.

If the correlation is reasonable, then regression analysis is the tech- nique to develop an equation that describes the relationship between the two variables.The coefficient of correlation, r, and the coefficient of determination, r2, are two numerical measures to record the strength of the linear relationship.Both of these coefficients have a value between 0 and 1.The closer either is to unity then the stronger is the correlation.The coefficient of correl- ation can be positive or negative whereas the coefficient of determination is always positive.

Linear regression in a time series data The linear regression line for a time series has the form, y � a � bx, where y is the predicted value of the dependent variable, a and b are constants, and x is the time.The regression equa- tion gives the best straight line that minimizes the error between the data points on the regres- sion line and the corresponding actual data from which the regression line is developed.To forecast using the regression equation, knowing a and b, we insert the time, x, into the regres- sion equation to give a forecast value y .The variability around the regression line is measured by the standard error of the estimate, se.We can use the standard error of the estimate to give the confidence in our forecast by using the relationship y � zse for large sample sizes and y � tse for sample sizes no more than 30.

C ha pt er S um m ar y 365Chapter 10: Forecasting and estimating from correlated data Linear regression and casual forecasting We can also use the linear regression relationship for causal forecasting.In causal forecasting all of the statistical relationships of correlation, predic- tion, variability, and confidence level of the forecast apply exactly as for a time series data.The only difference is that the value of the independent variable x is not time.Forecasting using multiple regression Multiple regression is when there is more than one independent variable x to give an equation of the form, y � a � b1x1 � b2x2 � b3x3 � … � bkxk.

A coefficient of multiple determination, r2, measures the strength of the relationship between the dependent variable y and the various independent variables x, and again there is a standard error of the estimate, se.Alternatively it may have an exponential relationship of the form y � aebx.Again with both these relationships we have a coefficient of determination that illustrates the strength between the dependent variable and the independent variable.

Seasonal patterns in forecasting Often in selling seasonal patterns exist.In this case we develop a forecast model by first removing the seasonal impact to calculate a seasonal index.If we divide the actual sales by the seasonal index we can then apply regression analysis on this smoothed data to obtain a regression forecast.When we multiply the regression forecast by the seasonal index we obtain a forecast by season.Considerations in statistical forecasting When we forecast using statistical data the longer the time horizon then the more inaccurate is the model.

Other considerations are that we should work with specific defined variables rather than aggregated data and that past data must be representative of the future environment for the model to be accurate.Further, care must be taken in using curvilinear models as though the coefficient of determination indicates a high degree of accuracy, the model may not follow market changes.Safety record Situation After the 1999 merger of Exxon with Mobil, the newly formed corporation, ExxonMobil implemented worldwide its Operations Integrity Management System (OIMS), a pro- gramme that Exxon itself had developed in 1992 in part as a result of the Valdez oil spill in Alaska in 1989.Since the implementation of OIMS the company has experienced fewer safety incidents and its operations have become more reliable.

These results are illustrated in the table below that shows the total incidents reported for every 200,000 hours worked since 1995.4 Year Incidents per 200,000 hours 1995 1.Develop the linear regression equation that best describes this data.Using the regression information, what is annual change in the number of safety incidents reported by ExxonMobil? 4.

What quantitative data indicates that there is a reasonable relationship over time with the safety incidents reported by ExxonMobil? 5.

Using the regression equation what is a forecast of the number of reported incidents in 2007? 6.Using the regression equation what is a forecast of the number of reported incidents in 2010? What are your comments about this result? 7.From the data, what might you conclude about the future safety record of ExxonMobil? EXERCISE PROBLEMS 4 Managing risk in a challenging business, The Lamp, ExxonMobil, 2007, (2), p.367Chapter 10: Forecasting and estimating from correlated data 2.

is a distributor of office supplies including agendas, diaries, computer paper, pens, pencils, paper clips, rubber bands, and the like.For a particular geographic region the company records over a 4-year period indicated the following monthly sales in pound sterling as follows.Using a coded value for the data with January 2003 equal to 1, develop a time series scatter diagram for this information.

What is an appropriate linear regression equation to describe the trend of this data? 3.What might be an explanation for the relative increase in sales for the months of August and September? 4.What can you say about the reliability of the regression model that you have created? Justify your reasoning.Month £ ‘000s January 2003 14 February 2003 18 March 2003 16 April 2003 21 May 2003 15 June 2003 19 July 2003 22 August 2003 31 September 2003 33 October 2003 28 November 2003 27 December 2003 29 January 2004 26 February 2004 28 March 2004 31 April 2004 33 May 2004 34 June 2004 35 July 2004 38 August 2004 41 September 2004 43 October 2004 37 November 2004 37 December 2004 41 Month £ ‘000s January 2005 42 February 2005 43 March 2005 42 April 2005 41 May 2005 41 June 2005 42 July 2005 43 August 2005 49 September 2005 52 October 2005 47 November 2005 48 December 2005 49 January 2006 51 February 2006 50 March 2006 52 April 2006 54 May 2006 57 June 2006 54 July 2006 48 August 2006 59 September 2006 61 October 2006 57 November 2006 56 December 2006 61 368 Statistics for Business 5.

What are the average quarterly sales as predicted by the regression equation? 6.What would be the forecast of sales for June 2007, December 2008, and December 2009? Which would be the most reliable? 7.What are your comments about the model you have created and its use as a forecast- ing tool? 3.Road deaths Situation The table below gives the number of people killed on French roads since 1980.Develop the linear regression equation that best describes this data.Is the linear equation a good forecasting tool for forecasting the future value of the road deaths? What quantitative piece of data justifies your response? 4.

Using the regression information, what is the yearly change of the number of road deaths in France? 5.Using the regression information, what is the forecast of road deaths in France in 2010? 6.Using the regression information, what is the forecast of road deaths in France in 2030? 7.What are your comments about the forecast data obtained in Questions 5 and 6? Year Deaths 1980 12,543 1981 12,400 1982 12,400 1983 11,833 1984 11,500 1985 10,300 1986 10,833 1987 9,855 1988 10,548 1989 10,333 1990 10,600 1991 9,967 Year Deaths 1992 9,083 1993 8,500 1994 8,333 1995 8,000 1996 8,067 1997 7,989 1998 8,333 1999 7,967 2000 7,580 2001 7,720 2002 7,242 5 Metro-France 16 May 2003, p.369Chapter 10: Forecasting and estimating from correlated data 4.Carbon dioxide Situation The data below gives the carbon dioxide emissions, CO2, for North America, in millions of metric tons carbon equivalent.Carbon dioxide is one of the gases widely believed to cause global warming.6 Year North America 1992 1,600 1993 1,625 1994 1,650 1995 1,660 1996 1,750 1997 1,790 1998 1,800 1999 1,825 2000 1,850 2001 1,800 Required 1.Plot the information on a time series scatter diagram and develop the linear regres- sion equation for the scatter diagram.

What are the indicators that demonstrate the strength of the relationship between carbon dioxide emission and time? What are your comments about these values? 3.What is the annual rate of increase of carbon dioxide emissions using the regression relationship? 4.Using the regression equation, forecast the carbon dioxide emissions in North America for 2010? 5.From the answer in Question 3, what is your 95% confidence limit for this forecast? 6.

Using the regression equation, forecast the carbon dioxide emissions in North America for 2020? 7.What are your comments about using this information for forecasting? 5.Restaurant serving Situation A restaurant has 55 full-time operating staff that includes kitchen staff and servers.Since the restaurant is open for lunch and dinner 7 days a week there are times that the restaurant does not have the full complement of staff.In addition, there are times when 6 Insurers weigh moves on global warming, Wall Street Journal Europe, 7 May 2003, p.

370 Statistics for Business staff are simply absent as they are sick.The restaurant manger conducted an audit to determine if there was a relationship between the number of staff absent and the aver- age time that a client had to wait for the main meal.This information is given in the table below.

Number of Average waiting staff absent time (minutes) 7 24 1 5 3 12 8 30 0 3 4 16 2 15 3 20 5 22 9 27 Required 1.

For the information given, develop a scatter diagram between number of staff absent and the average time that a client has to wait for the main meal.Using regression analysis, what is a quantitative measure that illustrates a reason- able relationship between the waiting time and the number of staff absent? 3.What is the linear regression equation that describes the relationship? 4.What is an estimate of the time delay per employee absent? 5.

When the restaurant has the full compliment of staff, to the nearest two decimal places, what is the average waiting time to obtain the main meal as predicted by the linear regression equation? 6.If there are six employees absent, estimate the average waiting time as predicted by the linear regression equation.If there are 20 employees absent, estimate the average waiting time as predicted by the linear regression equation.What are your comments about this result? 8.

What are some of the random occurrences that might explain variances in the wait- ing time? 6.Product sales Situation A hypermarket made a test to see if there was a correlation between the shelf space of a special brand of raison bread and the daily sales.The following is the data that was collected over a 1-month period.371Chapter 10: Forecasting and estimating from correlated data Shelf space Daily sales (m2) units 0.Illustrate the relationship between the sale of the bread and the allocated shelf space.Develop a linear regression equation for the daily sales and the allocated shelf space.50 m2, what is the estimated daily sale of this bread? 4.00 m2, what is the estimated daily sale of this bread? What are your comments about this forecast? 5.What does this sort of experiment indicate from a business perspective? 7.

German train usage Situation The German rail authority made an analysis of the number of train users on the net- work in the southern part of the country since 1993 covering the months for June, July, and August.The Transport Authority was interested to see if they could develop a rela- tionship between the number of users and another easily measurable variable.In this way they would have a forecasting tool.The variables they selected for developing their models were the unemployment rate in this region and the number of foreign tourists visiting Germany.The following is the data collected: Year Unemployment No.

of tourists Train users rate (%) (millions) (millions) 1993 11.4 4 11 (Continued) 372 Statistics for Business Year Unemployment No.

of tourists Train users rate (%) (millions) (millions) 1997 11.Illustrate the relationship between the number of train users and unemployment rate on a scatter diagram.

Using simple regression analysis, what are your conclusions about the correlation between the number of train users and the unemployment rate? 3.Illustrate the relationship between the number of train users and foreign tourists on a scatter diagram.Using simple regression analysis, what are your conclusions about the correlation between the number of train users and the number of foreign tourists? 5.

In any given year, if the number of foreign tourists were estimated to be 10 million, what would be a forecast for the number of train users? 6.If a polynomial correlation (to the power of 2) between train users and foreign tourists was used, what are your observations? 8.sells cosmetic products by simply advertising in throwaway newspapers and by ladies who organize Yam parties in order to sell directly the products.The table below gives data on a monthly basis for revenues, in pound sterling, for sales of cosmetics each month for the last year according to advertising budget and the equivalent number of people selling full time.

This data is to be analysed using multiple regression analysis.of revenues budget persons yam parties 721,200 47,200 542 101 770,000 54,712 521 67 580,000 25,512 328 82 910,000 94,985 622 75 315,400 13,000 122 57 373Chapter 10: Forecasting and estimating from correlated data Sales Advertising Sales No.of revenues budget persons yam parties 587,500 46,245 412 68 515,000 36,352 235 84 594,500 25,847 435 85 957,450 64,897 728 81 865,000 67,000 656 37 1,027,000 97,000 856 99 965,000 77,000 656 100 Required 1.Does the relationship appear strong? Quantify.From the answer developed in Question 1, assume for a particular month it is pro- posed to allocate a budget of £30,000 and there will be 250 sales persons available.In this case, what would be an estimate of the sales revenues for that month? 3.What are the 95% confidence intervals for Question 2? 4.

Does the relationship appear strong? Quantify.From the answer developed in Question 4, assume for a particular month it is pro- posed to allocate a budget of $US 4,000 to use 30 sales persons, with a target to make 21,000 sales contacts.Then what would be an estimate of the sales for that month? 6.

What are the 95% confidence intervals for Question 5? 9.Hotel revenues Situation A hotel franchise in the United States has collected the revenue data in the following table for the several hotels in its franchise.Year Revenues ($millions) 1996 35 1997 37 1998 44 1999 51 2000 50 2001 58 2002 59 (Continued) 374 Statistics for Business Year Revenues ($millions) 2003 82 2004 91 2005 104 Required 1.From the given information develop a linear regression model of the time period against revenues.What is the coefficient of determination for relationship developed in Question 1? 3.What is the annual revenue growth rate based on the given information? 4.From the relationship in Question 1, forecast the revenues in 2008 and give the 90% confidence limits.From the relationship in Question 1, forecast the revenues in 2020 and give the 90% confidence limits.

From the given information develop a two-degree polynomial regression model of the time period against revenues.What is the coefficient of determination for relationship developed in Question 6? 8.From the relationship in Question 6, forecast the revenues in 2008.

From the relationship in Question 6, forecast the revenues in 2020.What are your comments related to making a forecast for 2008 and 2020? 10.

Hershey Corporation Situation Dan Smith has in his investment portfolio shares of Hershey Company, Pennsylvania, United States of America, a Food Company well known for its chocolate.

Dan bought a round lot (100 shares) in September 1988 for $28.Since that date, Dan participated in Hershey’s reinvestment programme.That meant he reinvested all quar- terly dividends into the purchase of new shares.In addition, from time to time, he made optional cash investment for new shares.

The share price, and the number of shares held by Dan, at the end of each quarter since the time of the initial purchase, and the 1st quarter 2007, is given in Table 1.9584 375Chapter 10: Forecasting and estimating from correlated data Required 1.For the data given and using a coded value for the quarter starting at unity for September 1988, develop a line graph for the price per share.How might you explain the shape of the line graph? 2.For the data given and using a coded value for the quarter starting at unity for September 1988, develop a time series scatter diagram for the asset value (value of Table 1 (Continued).

5265 376 Statistics for Business the portfolio) of the Hershey stock.Show on the scatter diagram graph the linear regression line for the asset value.What is the equation that represents the linear regression line? 4.What information indicates quantitatively the accuracy of the asset value and time for this model? Would you say that the regression line could be used to reasonably forecast future values? 5.

From the linear regression equation, what is the annual average growth rate in dol- lars per year of the asset value of the portfolio? 6.Dan plans to retire at the end of December in 2020 (4th quarter 2020).Using the lin- ear regression equation, what is a forecast of the value of Dan’s assets in Hershey stock at this date? 7.At a 95% confidence level, what are the upper and lower values of assets at the end of December 2020? 8.What occurrences or events could affect the accuracy of forecasting the value of Hershey’s asset value in 2020? 9.

Qualitatively, would you think there is great risk for Dan in finding that the value of his assets is significantly reduced when he retires? Justify your response.Compact discs Situation The table below gives the sales by year of music compact discs by a selection of Virgin record stores.Year CD sales (millions) 1995 45 1996 52 1997 79 1998 72 1999 98 2000 99 2001 138 2002 132 2003 152 2004 203 Required 1.Develop the linear regression equation that best describes this data.Is the equation a good forecasting tool for CD record sales? What quantitative piece of data justifies your response? 377Chapter 10: Forecasting and estimating from correlated data 3.From the linear regression function, what is the forecast for CD sales in 2007? 4.From the linear regression function, what is the forecast for CD sales in 2020? 5.

Does a second-degree polynomial regression line have a better fit for this data? Why? 6.What would be the forecast for record sales calls in 2007 using the polynomial rela- tionship developed in Question 5? 7.What would be the forecast for record sales calls in 2020 using the polynomial rela- tionship developed in Question 5? 8.United States imports Situation The data in Table 1 is the amount of goods imported into the United States from 1960 until 2006.

7 (This is the same information presented in the Box Opener “Value of imported goods into the States” of this chapter.) 7 US Census Bureau, Foreign Trade division, /foreign-trade/statistics/historical goods, 8 June 2007.Table 1 Year Imported goods ($millions) 1960 14,758 1961 14,537 1962 16,260 1963 17,048 1964 18,700 1965 21,510 1966 25,493 1967 26,866 1968 32,991 1969 35,807 1970 39,866 1971 45,579 1972 55,797 1973 70,499 1974 103,811 1975 98,185 Year Imported goods ($millions) 1976 124,228 1977 151,907 1978 176,002 1979 212,007 1980 249,750 1981 265,067 1982 247,642 1983 268,901 1984 332,418 1985 338,088 1986 368,425 1987 409,765 1988 447,189 1989 477,665 1990 498,438 1991 491,020 Year Imported goods ($millions) 1992 536,528 1993 589,394 1994 668,690 1995 749,374 1996 803,113 1997 876,794 1998 918,637 1999 1,031,784 2000 1,226,684 2001 1,148,231 2002 1,167,377 2003 1,264,307 2004 1,477,094 2005 1,681,780 2006 1,861,380 378 Statistics for Business Required 1.Develop a time series scatter data for the complete data.From the scatter diagram developed in Question 1 develop linear regression equa- tions using just the following periods to develop the equation where x is the year.Also give the corresponding coefficient of determination: 1960–1964; 1965–1969; 1975–1979; 1985–1989; 1995–1999; 2002–2005.Using the relationships developed in Question 2, what would be the forecast values for 2006? 4.

Compare these forecast values obtained in Question 3 with the actual value for 2006.

Develop the linear regression equation and the corresponding coefficient of deter- mination for the complete data and show this information on the scatter diagram.Develop the exponential equation and the corresponding coefficient of determina- tion for the complete data and show this information on the scatter diagram.Develop the fourth power polynomial equation and the corresponding coefficient of determination for the complete data and show this information on the scatter diagram.Use the linear, exponential, and polynomial equations developed in Questions 5, 6, and 7 to forecast the value of imports to the United States for 2010.Use the equation for the period, 2002–2005, developed in Question 3 to forecast United States imports for 2010.

Discuss your observations and results for this exercise including the forecasts that you have developed.English pubs Situation The data below gives the consumption of beer in litres at a certain pub on the river Thames in London, United Kingdom between 2003 and 2006 on a monthly basis.Month 2003 2004 2005 2006 January 15,000 16,200 16,900 17,100 February 37,500 45,000 47,000 52,500 March 127,500 172,500 210,000 232,500 April 502,500 540,000 675,000 720,000 May 567,500 569,500 697,500 757,500 June 785,000 715,000 765,000 862,500 July 827,500 948,600 1,098,000 1,124,500 August 990,000 978,400 1,042,300 1,198,500 September 622,500 682,500 765,000 832,500 October 75,000 82,500 97,500 105,000 November 15,000 17,500 20,000 22,500 December 7,500 8,500 8,200 9,700 379Chapter 10: Forecasting and estimating from correlated data Required 1.

Develop a line graph on a quarterly basis for the data using coded values for the quar- ters.That is, winter 2003 has a coded value of 1.Plot a graph of the centred moving average for the data.What is the linear regression equation that describes the centred moving average? 3.

Determine the ratio of the actual sales to the centred moving average for each quar- ter.What is your interpretation of this information for 2004? 4.What are the seasonal indices for the four quarters using all the data? 5.What is the value of the coefficient of determination on the deseasonalized sales data? 6.What would be an estimate of the annual consumption of beer in 2010? What are your comments about this forecast? 14.Mersey Store Situation The Mersey Store in Arkansas, United States is a distributor of garden tools.The table below gives the sales by quarter since 1997.Show graphically that the sales for Mersey are seasonal.Use the multiplication model, predict sales by quarter for 2005.Show graphically the moving average, deseasonalized sales, regression line, and forecast.

Year Quarter Sales 1997 Winter 11,302 Spring 12,177 Summer 13,218 Autumn 11,948 1998 Winter 11,886 Spring 12,198 Summer 13,294 Autumn 11,785 1999 Winter 11,875 Spring 12,584 Summer 13,332 Autumn 12,354 2000 Winter 12,658 Spring 13,350 Summer 14,358 Autumn 13,276 Year Quarter Sales 2001 Winter 13,184 Spring 14,146 Summer 14,966 Autumn 13,665 2002 Winter 13,781 Spring 14,636 Summer 15,142 Autumn 13,415 2003 Winter 14,327 Spring 15,251 Summer 15,082 Autumn 14,002 2004 Winter 14,862 Spring 15,474 Summer 15,325 Autumn 14,425 380 Statistics for Business 15.Swimwear Situation The following table gives the sale of swimwear, in units per month, for a sports store in Redondo Beach, Southern California, United States of America during the period 2003 through 2006.Month 2003 2004 2005 2006 January 150 75 150 75 February 375 450 450 525 March 1,275 1,725 2,100 2,325 April 5,025 5,400 6,750 7,200 May 5,175 5,625 6,975 7,575 June 5,850 6,150 7,650 8,625 July 5,275 5,486 6,980 7,245 August 4,900 5,784 6,523 6,985 September 3,225 3,825 4,650 5,325 October 750 825 975 1,050 November 150 75 150 225 December 75 150 85 175 Required 1.Develop a line graph on a quarterly basis for the data using coded values for the quar- ters.That is, winter 2003 has a coded value of 1.

Plot a graph of the centred moving average for the data.What is the linear regression equation that describes the centred moving average? 3.Determine the ratio of the actual sales to the centred moving average for each quar- ter.What is your interpretation of this information for 2005? 4.

What are the seasonal indices for the four quarters using all the data? 5.Why are unit sales as presented preferable to sales on a dollar basis? 16.

Case: Saint Lucia Situation Saint Lucia is an overseas territory of the United Kingdom with a population in 2007 of 171,000.

It is an island of 616 square miles and counts as its neighbours Barbados, Saint Vincent, The Grenadines, and Martinique.It is an island with a growing tourist industry and offers the attraction of long sandy beaches, stunning nature trails, superb diving in deep blue waters, and relaxing spas.8 With increased tourism goes the demand for hotel and restaurants.Related to these two hospitality institutions is the volume of wine in thousand litres, sold per month during 8 Based on information from a Special Advertising Section of Fortune, 2 July 2007, p.381Chapter 10: Forecasting and estimating from correlated data 2005, 2006, and 2007.In addition, the local tourist bureau published data on the number of tourists visiting Santa Lucia for the same period.Required Use the data for forecasting purposes and develop and test an appropriate model.

Table 2 Month Tourist bookings 2005 2006 2007 January 28,700 29,800 30,800 February 23,200 25,200 28,000 March 29,000 28,000 31,000 April 23,500 26,000 28,400 May 21,900 25,000 27,500 June 25,300 31,000 32,000 July 26,000 25,550 31,000 August 20,100 23,200 22,000 September 22,300 24,100 26,000 October 25,100 25,100 27,000 November 22,600 27,000 28,000 December 27,000 31,900 30,200 Table 1 Month Unit wine sales (1,000 litres) 2005 2006 2007 January 530 535 578 February 436 477 507 March 522 530 562 April 448 482 533 May 422 498 516 June 499 563 580 July 478 488 537 August 400 428 440 September 444 430 511 October 486 486 480 November 437 502 499 December 501 547 542 This page intentionally left blank 11Indexing as a method for data analysis Metal prices Metal prices continued to soar in early 2006 as illustrated in Figure 11.1, which gives the index value for various metals for the first half of 2006 based on an index of 100 at the beginning of the year.The price of silver has risen by some 65%, gold by 32%, and platinum by 21%.Aluminium, copper, lead, nickel, and zinc are included in The Economist metals index curve and here the price of copper has increased by 60% and nickel by 45%.1 Indexing is another way to present statistical data and this is the subject of this chapter.

1 Metal prices, economic and financial indicators, The Economist, 6 May 2006, p.90 100 110 120 130 140 150 160 170 180 1 J an ua ry 20 06 1 F eb rua ry 20 06 1 M arc h 2 00 6 1 A pri l 2 00 6 1 M ay 20 06 In de x Silver Economist metals index Gold Platinum In Chapter 10, we introduced bivariate time-series data showing how past data can be used to fore- cast or estimate future conditions.

There may be situations when we are more interested not in the absolute values of information but how data compare with other values.For example, we might want to know how prices have changed each year or how the productivity of a manu- facturing operation has increased over time.For these situations we use an index number or index value.The index number is the ratio of a certain value to a base value usually multiplied by 100.When the base value equals 100 then the meas- ured values are a percentage of the base value as illustrated in the box opener “Metal prices”.

Perhaps the most common indices are quantity and price indexes.In their simplest form they measure the relative change in time respective to a given base value.Quantity index number with a fixed base As an example of a quantity index consider the information in Table 11.The 1st column is the time period in years and the 2nd column is the absolute values of enrolment in an MBA programme for a certain business school over the last 10 years from 1995.

Here the data for 1995 is considered the index base value.The 3rd column gives the ratio of a particular year to the base value.The 4th column is the ratio for each year multiplied by 100.The index number for the base period is 100 and this is obtained by the ratio (95/95) * 100.

If we consider the year 2000, the enrolment for the MBA programme is 125 candidates.This gives a ratio to the 1995 data of 125/95 or 1.Relative Time-Based Indexes 385Chapter 11: Indexing as a method for data analysis After studying this chapter you will learn how to present and analyse statistical data using index values.The subjects treated are as follows: ✔ Relative time-based indexes • Quantity index number with a fixed base • Price index number with a fixed base • Rolling index number with a moving base • Changing the index base • Comparing index numbers • Consumer price index (CPI) and the value of goods and services • Time series deflation.

✔ Relative regional indexes (RRIs) • Selecting the base value • Illustration by comparing the cost of labour.✔ Weighting the index number • Unweighted index number • Laspeyres weighted price index • Paasche weighted price index • Average quantity-weighted price index.L e a r n i n g o b j e c t i v e s Year Enrolment Ratio to Index base value number 1995 95 1.We can interpret this information by saying that enrolment in 2000 is 132% of the enrolment in 1995, or alternatively an increase of 32%.In 2004 the enrolment is only 74% of the 1995 enrolment or 26% less (100% � 74%).The general equation for this index, IQ, which is called the relative quantity index, is, 11(i) Here Q0 is the quantity at the base period, and Qn is the quantity at another period.This other period might be at a future date or after the base period.

Alternatively, it could be a past period or before the base period.Price index number with a fixed base Another common index, calculated in a similar way to the quantity index, is the price index, which compares the level of prices from one period to another.The most common price index is the consumer price index, that is used as a measure of inflation by comparing the general price level for specific goods and services in the economy.The data is collected and compiled by government agencies such as Bureau of Labour Statistics in the United Kingdom and a similar department in the United States.In the European Union the organization concerned is Eurostat.

2 which gives the average price of unleaded regular petrol in the United States for the 12-month period from January 2004.2 (For comparison the price is also given $ per litre where 1 gallon equals 3.) In this table, we can see that the price of gasoline has increased 28% in the month of June com- pared to the base month of January.

In a similar manner to the quantity index, the general equation for this index, IP, called the relative price index is, 11(ii) Here P0 is the price at the base period, and Pn is the price at another period.I P P n P � 0 100* I Q Q n Q � 0 100* 386 Statistics for Business Table 11.2 Average price of unleaded gasoline in the United States in 2004.Month $/gallon $/litre Ratio to base Index value number January 1.18 118 2 US Department of Labor Statistics, .

Rolling index number with a moving base We may be more interested to know how data changes periodically, rather than how it changes according to a fixed base.In this case, we would use a rolling index number.3 which is the same enrolment MBA data from Table 11.

In the last column we have an index showing the change relative to the previous year.For example, the rolling index for 1999 is given by (64/56) * 100 � 114.This means that in 1999 there was a 14% increase in student enrolment compared to 1998.In 2002 the index compared to 2001 is calculated by (54/102) * 100 � 53.

This means that enrol- ment is down 47% (100 � 53) in 2002 com- pared to 2001, the previous year.Again the value of the index has been rounded to the nearest whole number.Changing the index base When the base point of data is too far in the past the index values may be getting too high to be meaningful and so we may want to use a more recent index so that our base point corresponds more to current periods.4 where the 2nd column shows the relative sales for a retail store based on an index of 100 in 1980.

The 3rd column shows the index on a basis of 1995 equal to 100.The index value for 1995, for example, is (295/295) * 100 � 100.The index value for 1998 is (322/295) * 100 � 109.The index values for the other years are determined in the same manner.By transposing the data in this manner we have brought our index information closer to our current year.

Comparing index numbers Another interest that we might have is to com- pare index data to see if there is a relationship between one index number and another.5 which is index data for the number of new driving licences issued and the number of recorded automobile accidents in a certain community.The 2nd col- umn, for the number of driving licences issued, gives information relative to a base period of 1960 equal to 100.The 3rd column gives the number of recorded automobile accidents but in this case the base period of 100 is for the year 387Chapter 11: Indexing as a method for data analysis Year Enrolment Ratio to Annual immediate change previous Rolling period index 1995 95 1996 97 1.

Year Sales index Sales index 1980� 100 1995� 100 1995 295 100 1996 286 97 1997 301 102 1998 322 109 1999 329 112 2000 345 117 2001 352 119 2002 362 123 2003 359 122 2004 395 134 Table 11.It is inappropriate to compare data of dif- ferent base periods and what we have done is converted the number of driving licences issued to a base period of the year 2000 equal to 100.

In this case, in 2000 the index is (406/ 406) * 100 � 100.

Then for example, the index in 1995 is (307/406) * 100 � 76 and in 2004 the index is (469/406) * 100 � 116.In both cases, the indices are rounded to the nearest whole number.Now that we have the indices on a common base it is easier to compare the data.For example, we can see that there appears to be a relationship between the number of new driving licenses issued and the recorded automobile accidents.More specifically in the period 1995–2000, the index for automobile accidents went from 62 to 100 or a more rapid increase than for the issue of driving licences which went from 76 to 100.

However, in the period 2000–2004, the increase was not as pronounced going from 100 to 112 compared to the number of licenses issued going from 100 to 116.This could have been perhaps because of better police surveillance, a better road infrastructure, or other reasons.2 gives a graph of the data where we can see clearly the changes.Comparing index numbers has a similarity to causal regression analysis presented in Chapter 10, where we determined if the change in one variable was caused by the change in another variable.

CPI and the value of goods and services The CPI is a measure of how prices have changed over time.It is determined by measuring the value of a “basket” of goods in one base period and then comparing the value of the same basket of goods at a later period.The change is most often presented on a ratio measurement scale.This bas- ket of goods can include all items such as food, consumer goods, housing costs, mortgage inter- est payments, indirect taxes, etc.Alternatively, the CPI can be determined by excluding some of these items.

When there is a significant increase in the CPI then this indicates an inflationary period.6 gives the CPI in the United Kingdom for 1990 for all items.3 For this period the CPI has increased by 9.(Note that we have included the CPI for December 1989, in order to determine the annual change for 1990.

) Say now, for example, your annual salary at the end of 1989 was £50,000 and then at the end of 1990 it was increased to £54,000.Your salary has increased by an amount of 8% (£54,000 � 50,000)/50,000 and your man- ager might expect you to be satisfied.However, if you measure your salary increase to the CPI of 9.34% the “real” value or “worth” of your salary has in fact gone down.You have less spending power than you did at the end of 1989 and would not unreasonably be dissatisfied.

7 which is the CPI in the United Kingdom for 2001 for all items.For this period the CPI has increased by only 0.70% 388 Statistics for Business Year Driving Automobile Driving licenses accidents licenses issued 2000� 100 issued 1960� 100 2000� 100 1995 307 62 76 1996 325 71 80 1997 335 79 83 1998 376 83 93 1999 411 98 101 2000 406 100 100 2001 413 105 102 2002 421 108 104 2003 459 110 113 2004 469 112 116 Table 11.5 Automobile accidents and driving licenses issued.

389Chapter 11: Indexing as a method for data analysis Figure 11.2 Automobile accidents and driving licences issued.50 60 70 80 90 100 110 120 1994 1995 1996 1997 1998 1999 2000 2001 2002 2003 2004 2005 In de x nu m be r: 2 00 0� 10 0 Automobile accidents Issued automobile licences Year Month Index December 1989 118.Say now a person’s annual salary at the end of 2000 was £50,000 and then at the end of 2001 it was £54,000.The salary increase is 8% as before (£54,000 � 50,000)/50,000 .This person should be satisfied as compared to the CPI increase of 0.

70% there has been a real increase in the salary and thus in the spending power of the individual.Time series deflation In order to determine the real value in the change of a commodity, in this case salary from the pre- vious section, we can use time series deflation.Time series deflation is illustrated as follows using first the information from Table 11.6: Base value of the salary at the end of 1989 is £50,000/year At the end of 1989, the base salary index is, At the end of 1990, the salary index to the base period is, Ratio of the CPI at the base period to the new period is, Multiply the salary index in 1990 by the CPI ratio to give the real value index (RVI) or, 108 * 0.77 This means to say that the real value of the salary has in fact declined by 1.If we do the same calculation using the CPI for 2001 using Table 11.

7 then we have the following: Base value of the salary at the end of 2000 is £50,000/year At the end of 2001, the base salary index is, At the end of 2001, the salary index to the base period is, Ratio of the CPI at the base period to the new period is Multiply the salary index in 2001 by the CPI ratio to give the RVI or, 108 * 0.25 This means to say that the real value of the salary has increased by 7.In summary, if you have a time series, x-values of a commodity and an index series, I-values, over the same period, n, then the RVI of a com- modity for this period is, 11(iii) If we substitute in equation 11(iii) the salary and CPI information for 1990 we have the following: This means a real decrease of 1.

Similarly, if we substitute in equation 11(iii) the salary and CPI information for 2000 we have the following: This means a real increase of 7.RVI � � 54 000 50 000 172 2 173 4 100 107 25 , , * .RVI � � 54 000 50 000 118 8 129 9 100 98 77 , , * .RVI � x x I I n n0 0 100* * RVI Current value of commodity Base value of � commodity Base indicator Current indicator * ** 100 172 2 173 4 0 9931 .

� 54 000 50 000 100 108 , , * � 50 000 50 000 100 100 , , * � 118 8 129 9 0 9145 .� 54 000 50 000 100 108 , , * � 50 000 50 000 100 100 , , * � 390 Statistics for Business Notice that the commodity ratio and the indi- cator ratio are in the reverse order since we are deflating the value of the commodity according to the increase in the consumer price.Index numbers may be used to compare data between one region and another.For example, we might be interested to compare the cost of liv- ing in London to that of New York, Paris, Tokyo, and Los Angeles or the productivity of one pro- duction site to others.When we use indexes in this manner the time variable is not included.Selecting the base value When we use indexes to compare regions to others, we first decide what our base point is for comparison and then develop the relative regional index (RRI), from this base value.

Again, we multiply the ratio by 100 so that the calculated index values represent a percentage change.As an illustration, when I was an engi- neer in Los Angeles our firm was looking to open a design office in Europe.One of the crite- ria for selection was the cost of labour in various selected European countries compared to the United States.This type of comparison is illus- trated in the following example.Illustration by comparing the cost of labour In Table 11.

8 are data on the cost of labour in various countries in terms of the statutory 391Chapter 11: Indexing as a method for data analysis Relative Regional Indexes Table 11.Country Minimum wage plus Index, Index, Index, social security contributions United States� 100 Britain� 100 France� 100 as percent of labour cost of average worker (%) Australia 46 139 107 85 Belgium 40 121 93 74 Britain 43 130 100 80 Canada 36 109 84 67 Czech Republic 33 100 77 61 France 54 164 126 100 Greece 51 155 119 94 Ireland 49 148 114 91 Japan 32 97 74 59 Luxembourg 50 152 116 93 New Zealand 42 127 98 78 Poland 35 106 81 65 Portugal 50 152 116 93 Slovakia 44 133 102 81 South Korea 25 76 58 46 Spain 37 112 86 69 United States 33 100 77 61 � Value at other region Value at base region * 1100 1000� V Vb * Relative regional index minimum wage plus the mandatory social secu- rity contributions as a percentage of the labour costs of the average worker in that country.4 In Column 3, we have converted the labour cost value into an index using the United States as the base value of 100.This is determined by the calculation (33%/33%) * 100.

The base values of the other countries are then determined by the ratio of that country’s value to that of the United States.For example, the index for Australia is 139, (46%/33%) * 100 for South Korea it is 76, (25%/33%) * 100 and for Britain it is 130, (43%/33%) * 100 .We inter- pret this index data by saying that the cost of labour in Australia is 39% more than in the United States; 24% less in South Korea than in the United States (100% � 76%); and 30% more in Britain than in the United States.8 gives comparisons using Britain as the base country such that the base value for Britain is 100 (43%/43%) * 100 .

We interpret this data in Column 4, for example, by saying that compared to Britain, the labour cost in Australia is 7% more, 16% more in Portugal and 16% less in Canada.Column 5 gives similar index information using France as the base coun- try with an index of 100 (54%/54%) * 100) .Here, for example, the cost of labour in Australia is 15% less than in France, in Britain it is 20% less, and in South Korea it is a whopping 54% less than in France.In fact from Column 5 we see that France is the most expensive country in terms of the cost of labour and this in part explains why labour intensive industries, particularly manufac- turing, relocate to lower cost regions.Index numbers may be unweighted or weighted according to certain criteria.

The unweighted index number means that each item in arriving at the index value is considered of equal impor- tance.In the weighted index number, emphasis or weighting is put onto factors such as quan- tity or expenditure in order to calculate the index.Unweighted index number Consider the information in Table 11.9 that gives the price of a certain 11 products bought in a hypermarket in £UK for the years 2000 and 2005.If we use equation 11(ii) then the price index is, To the nearest whole number this is 129, which indicates that in using the items given, prices rose 29% in the period 2000 to 2005.

Now, for example, assume that an additional item, a lap- top computer is added to Table 11.9 to give the I P P n P � � � 0 100 96 16 74 50 100 129 07* .392 Statistics for Business Weighting the Index Number Item and unit size 2000, P0 2005, Pn (weight, volume, (£/unit) (£/unit) or unit) Bread, loaf 1.

9 Eleven products purchased in a hypermarket.

4 Economic and financial indicators, The Economist, 2 April 2005, p.Again using equation 11(ii) the price index is, This indicates that prices have declined by 51% (100 � 49) in the period 2000 to 2005.

We know intuitively that this is not the case.In determining these price indexes using equation 11(ii), we have used an unweighted aggregate index meaning that in the calculation each item in the index is of equal importance.In a similar manner we can use equation 11(i) to calculate an unweighted quantity index.This is a major disadvantage of an unweighted index as it neither attaches importance or weight to the quantity of each of the goods pur- chased nor to price changes of high volume purchased items and to low volume purchased items.For example, a family may purchase 200 lettuces/year but probably would only pur- chase a laptop computer say every 5 years.

Thus, to be more meaningful we should use a weighted price index.The concept of weighting or putting importance on items of data was first intro- duced in Chapter 2.Laspeyres weighted price index The Laspeyres weighted price index, after its originator, is determined by the following relationship: 11(iv) Here, ● Pn is the price in the current period.● Q0 is the quantity consumed in the base period.

Note that with this method, the quantities in the base period, Q0 are used in both the numera- tor and the denominator of the equation.In addition, the value of the denominator ∑P0Q0 remains constant for each index and this makes comparison of successive indexes simpler where the index for the first period is 100.11 gives the calculation procedure for the Laspeyres price index for the items in Table 11.

9 with the addition that here the quantities con- sumed in the base period 2000 are also indi- cated.Here we have assumed that the quantity of laptop computers consumed is 1⁄6 or 0.17 for the 6-year period between 2000 and 2005 Thus, from equation 11(iv), Laspeyres price index in 2000 is, P Q P Q n 0 0 0 100 7 466 50 7 466 50 100 100 00 ∑ ∑ * , .� � oor 100 Laspeyres weighted price index � P Q P Q n 0 0 0 1∑ ∑ * 000 I P P n P or a � � � 0 100 1 447 51 2 925 60 100 49 48 * , .nn index of 49 393Chapter 11: Indexing as a method for data analysis Item and unit size 2000, P0 2005, Pn (weight, volume, (£/unit) (£/unit) or unit) Bread, loaf 1.70 Laptop computer 2,850,00 1,350,00 TOTAL 2,925.10 Twelve products purchased in a hypermarket.Laspeyres price index in 2000 is, Thus, if we have selected a representative sam- ple of goods we conclude that the price index for 2005 is 134 based on a 2000 index of 100.

This is the same as saying that in this period prices have increased by 34%.With the Laspeyres method we can compare index changes each year when we have the new prices.For example, if we had prices in 2003 for the same items, and since we are using the quantities for the base year, we can determine a new index for 2003.A disadvantage with this method is that it does not take into account the change in consumption patterns from year to year.For example, we may purchase less of a particular item in 2005 than we purchased in 2000.

Paasche weighted price index The Paasche price index, again after its originator, is calculated in a similar manner to the Laspeyres index except that now current quantities in period n are used rather than quantities in the base period.The Paasche equation is, 11(v) Here, ● Pn is the price in the current period.● Qn is the quantity consumed in the current period n.Thus, in the Paasche weighted price index, unlike, the Laspeyres weighted price index, the value of the denominator ∑P0Qn changes according to the period with the value of Qn.

The Paasche price index is illustrated by Table 11.12, which has the same prices for the base period but Paasche price index � P Q P Q n n n ∑ ∑ 0 100* P Q P Q n 0 0 0 100 9 986 00 7 466 50 100 133 76 ∑ ∑ * , .134 394 Statistics for Business Table 11.Item and unit size 2000, 2005, Quantity (units) P0*Q0 Pn*Q0 (weight, volume, or unit) P0 Pn consumed in (£/unit) (£/unit) 2000, Q0 Bread, loaf 1.00 the quantities are for the current consumption period.These revised quantities show that per- haps the family is becoming more health con- scious, in that the consumption of bread, wine, coffee, cheese, and petrol (family members walk) is down whereas the consumption of lettuce, apples, fish, and chicken (white meat) is up.Thus, using equation 11(v), Paasche price index in 2000 is, Paasche price index in 2005 is, Thus, with the Paasche index using revised con- sumption patterns it indicates that the prices have increased 30% in the period 2000 to 2005.Average quantity-weighted price index In the Laspeyres method we used quantities consumed in early periods and in the Paasche method quantities consumed in later periods.

12 there were changes in consumption patterns so that we might say that the index does not fairly represent the period in question.An alternative approach to the Laspeyres and Paasches meth- ods is to use fixed quantity values that are considered representative of the consumption patterns within the time periods considered.

These fixed quantities can be the average quan- tities consumed within the time periods consid- ered or some other appropriate fixed values.

In this case, we have an average quantity weighted price index as follows: Average quantity-weighted price index 11(vi)� P Q P Q n a a ∑ ∑ 0 100* P Q P Q n n n ∑ ∑ 0 100 10 891 05 8 400 50 100 129 65 * , .P Q P Q n n n ∑ ∑ 0 100 8 400 50 8 400 50 100 100 00 * , .

� � oor 100 395Chapter 11: Indexing as a method for data analysis Table 11.Item and unit size 2000, 2005, Quantity P0 * Qn Pn * Qn (weight, volume, or unit) P0 Pn consumed in (£/unit) (£/unit) 2005, Qn Bread, loaf 1.

05 Here, ● Pn is the price in the current period.● Qa is the average quantity consumed in the total period in consideration.From equation 11(vi) using this information, Average quantity weighted price index in 2000 is, Average quantity weighted price index in 2005 is, Rounding up this indicates that prices have increased 32% in the period.This average quan- tity consumed is in fact a fixed quantity and so this approach is sometimes referred to as a fixed weight aggregate price index.

The usefulness of this index is that we have the flexibility to choose the base price P0 and the fixed weight Qa.Here we have used an average weight but this fixed quantity can be some other value that we con- sider more appropriate.P Q P Q n a a ∑ ∑ 0 100 10 438 53 7 933 50 100 131 58 * , .P Q P Q n a a ∑ ∑ 0 100 7 933 50 7 933 50 100 100 00 * , .� � oor 100 396 Statistics for Business Table 11.

13 Average price index Item and unit size 2000, 2005, Average quantity P0*Qa Pn*Qa (weight, volume, P0 Pn consumed between or unit) (£/unit) (£/unit) 2000 and 2005, Qa Bread, loaf 1.53 397Chapter 11: Indexing as a method for data analysis This chapter has introduced relative time-based indexes, RRIs, and weighted indexes as a way to present and analyse statistical data.Relative time-based indexes The most common relative time-based indexes are the quantity and price index.In their most common form these indexes measure the relative change over time respective to a given fixed base value.The base value is converted to 100 so that the relative values show a percentage change.

An often used price index is the CPI which indicates the change in prices over time and thus is a relative measure of inflation.

Rather than having a fixed base we can have rolling index where the base value is the previous period so that the change we measure is relative to the pre- vious period.This is how we would record annual or monthly changes.When the index base is too far in the past the index values may become too high to be meaningful.In this case, we con- vert the historical sales index to 100 by dividing this value by itself and multiplying by 100.The new relative index values are then the old values divided by the historical index value.

Relative index values can be compared to others to see if there is a relationship between one index and another.This is analogous to causal regression analysis where we establish whether the change in one variable is caused by the change in another variable.A useful comparison of indexes is to compare the index of wage or salary changes to see if they are in line with the change in the CPI.To do this we use a time series deflation which determines the real value in the change of a commodity.Relative regional indexes The goal of relative regional indexes (RRIs) is to compare the data values at one region to that of a base region.

Some RRIs might be the cost of living in other locations compared to say New York; the price of housing in major cities compared to say London; or as illustrated in the chapter, the cost of labour compared to France.There can be many RRIs depending on the values that we wish to compare.Weighting the index An unweighted index is one where each element used to calculate the index is considered to have equal value.A weighted price index is where different weights are put onto the index to indicate their importance in calculating the index.The Laspeyres price index is where the index is weighted by multiplying the price in the current period, by the quantity of that item consumed in the base period, and dividing the total value by the sum of the product of the price in the base period and the consumption in the base period.

A criticism of this index is that if the time period is long it does not take into account changing consumption patterns.An alternative to the Laspeyres index is the Paasche weighted price index, which is the ratio of total product of current consumption and current price, divided by the total product of current consumption and base price.An alternative to both the Laspeyres and Paasche index is to use an average of the quantity consumed during the period consid- ered.In this way, the index is fairer and more representative of consumption patterns in the period.C hapter Sum m ary 398 Statistics for Business 1.

Backlog Situation Fluor is a California-based engineering and constructing company that designs and builds power plants, oil refineries, chemical plants, and other processing facilities.In the following table are the backlog revenues of the firm in billions of dollar since 1988.5 Backlog is the amount of work that the company has contracted but which has not yet been executed.Normally, the volume of work is calculated in terms of labour hours and material costs and this is then converted into estimated revenues.The backlog repre- sents the amount of work that will be completed in the future.

Develop the quantity index numbers for this data where 1988 has an index value of 100.How would you describe the backlog of the firm, based on 1988, in 1989, 2000, and 2005? 3.Develop the quantity index for this data where the year 2000 has an index value of 100.

How would you describe the backlog of the firm, based on 2000, in 1989, 1993, and 2005? 5.Why is an index number based on 2000 preferred to an index number of 1988? 6.Develop a rolling quantity index from 1988 based on the change from the previous period.Using the rolling quantity index, how would you describe the backlog of the firm, in 1990, 1994, 1998, and 2004? EXERCISE PROBLEMS Year Backlog Year Backlog Year Backlog ($billions) ($billions) ($billions) 1988 6.399Chapter 11: Indexing as a method for data analysis 2.

Gold Situation The following table gives average spot prices of gold in London since 1987.6 In 1969 the price of gold was some $50/ounce.In 1971 President Nixon allowed the $US to float by eliminating its convertibility into gold.Concerns over the economy and scarcity of natu- ral resources resulted in the gold price reaching $850/ounce in 1980 which coincided with peaking inflation rates.Develop the price index numbers for this data where 1987 has an index value of 100.How would you describe gold prices, based on 1987, in 1996, 2001, and 2005? 3.Develop the price index numbers for this data where the year 1996 has an index value of 100.

How would you describe gold prices, based on 1996, in 1987, 2001, and 2005? 5.Why is an index number based on 1996 preferred to an index number of 1987? 6.Develop a rolling price index from 1987 based on the change from the previous period.Using the rolling price index, which year saw the biggest annual decline in the price of gold? 8.Using the rolling price index, which year saw the biggest annual increase in the price of gold? 3.United States gasoline prices Situation The following table gives the mid-year price of regular gasoline in the United States in cents/gallon since 19907 and the average crude oil price for the same year in $/bbl.8 Year Gold price Year Gold price ($/ounce) ($/ounce) 1987 446 1997 331 1988 437 1998 294 1989 381 1999 279 1990 384 2000 279 1991 362 2001 271 1992 344 2002 310 1993 360 2003 364 1994 384 2004 410 1995 384 2005 517 1996 388 6 Newmont, 2005 Annual Report.7 US Department of Energy, (consulted July 2006).

400 Statistics for Business Required 1.Develop the price index for regular grade gasoline where 1990 has an index value of 100.How would you describe gasoline prices based on 1990, in 1993, 1998, and 2005? 3.

Develop the price index numbers for this data where 2000 has an index value of 100.How would you describe gasoline prices, based on 2000, in 1993, 1998, and 2005? 5.Why might an index number based on 2000 be preferred to an index number of 1990? 6.Develop a rolling price index from 1990 based on the change from the previous period.

Using the rolling price index, which year saw the biggest annual increase in the price of regular gasoline? 8.Develop the price index for crude oil prices where 1990 has an index value of 100.Plot the index values of the gasoline prices developed in Question 1 to the crude oil index values developed in Question 8.

What are your comments related to the graphs you developed in Question 9? 4.Coffee prices Situation The following table gives the imported price of coffee into the United Kingdom since 1975 in United States cents/pound.9 Year Price of regular Oil price ($/bbl) grade gasoline (cents/US gallon) 1990 119.80 62 9 International Coffee Organization, (consulted July 2006).401Chapter 11: Indexing as a method for data analysis Required 1.Develop the price index for the imported coffee prices where 1975 has an index value of 100.How would you describe coffee prices based on 1975, in 1985, 1995, and 2004? 3.

Develop the price index for the imported coffee prices where 1990 has an index value of 100.How would you describe coffee prices based on 1990, in 1985, 1995, and 2004? 5.Develop the price index for the imported coffee prices where 2000 has an index value of 100.How would you describe coffee prices based on 2000, in 1985, 1995, and 2004? 7.Which index base do you think is the most appropriate? 8.Develop a rolling price index from 1975 based on the change from the previous period.Using the rolling price index, which year and by what amount was the biggest annual increase in the price of imported coffee? 10.

Using the rolling price index, which year and by what amount, was the annual decrease in the price of imported coffee? 11.Why are coffee prices not a good measure of the change in the cost of living? 5.Boeing Situation The following table gives summary financial and operating data for the United States Aircraft Company Boeing.10 All the data is in $US millions except for the earnings per share.Year US cents/1b Year US cents/1b 1975 329.

94 10 The Boeing Company 2005 Annual Report.

402 Statistics for Business Required 1.Develop the index numbers for revenues using 2005 as the base.How would you describe the revenues for 2001 using the base developed in Question 1? 3.Develop the index numbers for earnings/share using 2001 as the base? 4.

How would you describe the earnings/share for 2005 using the base developed in Question 3? 5.Develop a rolling index for revenues since 2001.Use the index values developed in Question 5, how would you describe the progression of revenues? 6.Ford Motor Company Situation The following table gives selected financial data for the Ford Motor Company since 1992.

11 2005 2004 2003 2002 2001 Revenues 54,845 52,457 50,256 53,831 57,970 Net earnings 2,572 1,872 718 492 2,827 Earnings/share 3.20 Backlog 160,473 109,600 104,812 104,173 106,591 Year Revenues Net income Stock price, Stock price, Dividends Vehicle sales automotive total company high low ($/share) North America ($millions) ($millions) ($/share) ($/share) units 000s 1992 84,407 �7,835 8.40 11 Ford Motor Company Annual Reports, 2002 and 2005.403Chapter 11: Indexing as a method for data analysis Required 1.Develop the index numbers for revenues using 1992 as the base.

How would you describe the revenues for 2005 using the base developed in Question 1? 3.Develop the rolling index for revenues starting from 1992.Using the rolling index based on the previous period, in which years did the revenues decline, and by how much? 5.

Develop the index numbers for North American vehicle sales using 1992 as the base.Based on the index numbers developed in Question 5 which was the best comparative year for vehicle sales, and which was the worst? 7.From the information given, and from the data that you have developed, how would you describe the situation of the Ford Motor Company? 7.Drinking Situation In Europe, alcohol consumption rates are rising among the young.

The following table gives the percentage of 15- and 16-year olds who admitted to being drunk 3 times or more in a 30-day period in 2003.Using Britain as the base, develop a relative regional index for the percentage of 15- and 16-year olds who admitted to being drunk 3 times or more in a 30-day period.Using the index for Britain developed in Question 1, how would you describe the per- centage of 15- and 16-year olds who admitted to being drunk 3 times or more in a 30-day period in Ireland, Greece, and Germany? 3.

Using France as the base, develop a relative regional index for the percentage of 15- and 16-year olds who admitted to being drunk 3 times or more in a 30-day period.Using the index for France developed in Question 3, how would you describe the per- centage of 15- and 16-year olds who admitted to being drunk 3 times or more in a 30-day period in Ireland, Greece, and Germany? Country Percentage Britain 23.00 12 Europe at tipping point, International Herald Tribune, 26 June 2006, p.Using Denmark as the base, develop a relative regional index for the percentage of 15- and 16-year olds who admitted to being drunk 3 times or more in a 30-day period.Using the index for Denmark developed in Question 3, how would you describe the percentage of 15- and 16-year olds who admitted to being drunk 3 times or more in a 30-day period in Ireland, Greece, and Germany? 7.Based on the data what general conclusions can you draw? 8.Part-time work Situation The following table gives the people working part time in 2005 by country as a percent- age of total employment and also the percentage of those working part time who are women.

Part-time work is defined as working less than 30 hours/week.Using the United States as the base, develop a relative regional index for the percentage of people working part time.Using the index for the United States developed in Question 1, how would you describe the percentage of people working part time in Australia, Greece, and Switzerland? Country Working part time, percentage Percentage of part of total employment timers who are women Australia 27.

40 13 Economic and financial indicators, The Economist, 24 June 2006, p.405Chapter 11: Indexing as a method for data analysis 3.Using the Netherlands as the base, develop a relative regional index for the percentage of people working part time.Using the index for the Netherlands developed in Question 3, how would you describe the percentage of people working part time in Australia, Greece, and Switzerland? What can you say about the part-time employment situation in the Netherlands? 5.Using Britain as the base, develop a relative regional index for the percentage of people working part time who are women? 6.

Using the index for Britain as developed in Question 5, how would you describe the percentage of people working part time who are women for Australia, Greece, and Switzerland? 9.Cost of living Situation The following table gives the purchase price, at medium-priced establishments, of certain items and rental costs in major cities worldwide in 2006.14 These numbers are a measure of the cost of living.The exchange rates used in the tables are £1.City Rent of 2 Bus or Compact International Cup of Fast food bedroom subway disc (£) newspaper coffee hamburger unfurnished (£/ride) (£/copy) including meal (£) apartment service (£) (£/month) Amsterdam 926 1.58 14 Global/worldwide cost of living survey ranking, 2006, /costofliving.406 Statistics for Business Required 1.Using rental costs as the criterion, how do Amsterdam, Berlin, New York, Paris, Sydney, Tokyo, and Vancouver, compare to London? 2.Using rental costs as the criterion, how do Amsterdam, Berlin, New York, Paris, Sydney, Tokyo, and Vancouver, compare to Madrid? 3.Using rental costs as the criterion, how do Amsterdam, Berlin, New York, Paris, Sydney, Tokyo, and Vancouver, compare to Prague? 4.Using the sum of all the purchase items except rent as the criterion, how do Amsterdam, Berlin, New York, Paris, Sydney, Tokyo, and Vancouver, compare to London? 5.

Using the sum of all the purchase items except rent as the criterion, how do Amsterdam, Berlin, New York, Paris, Sydney, Tokyo, and Vancouver, compare to Madrid? 6.Using the sum of all the purchase items except rent as the criterion, how do Amsterdam, Berlin, New York, Paris, Sydney, Tokyo, and Vancouver, compare to Prague? 7.Using rental costs as the criterion, how does the most expensive city compare to the least expensive city? Identify the cities.Corruption Situation The Berlin-based organization, Transparency International, defines corruption as the abuse of public office for private gain, and measures the degree to which corruption is perceived to exist among a country’s public officials and politicians.

It is a composite index, drawing on 16 surveys from 10 independent institutions, which gather the opin- ions of business people and country analysts.Only 159 of the world’s 193 countries are included in the survey due to an absence of reliable data from the remaining countries.The scores range from 10 or squeaky clean, to zero, highly corrupt.A score of 5 is the number Transparency International considers the borderline figure distinguishing countries that do not have a serious corruption problem.The following table gives the corruption index for the first 50 countries in terms of being the least corrupt.

15 Country Index Country Index Australia 8.6 15 The 2005 Transparency International Corruption Perceptions Index, www.

407Chapter 11: Indexing as a method for data analysis Required 1.From the countries in the list which country is the least corrupt and which is the most corrupt? 2.

What is the percentage of countries that are above the borderline limit as defined by Transparency International, in not having a serious corruption problem? 3.

Compare Denmark, Finland, Germany, and England using Spain as the base.Compare Denmark, Finland, Germany, and England using Italy as the base.Compare Denmark, Finland, Germany, and England using Greece as the base.

Compare Denmark, Finland, Germany, and England using Portugal as the base.What conclusions might you draw from the responses to Questions 3 to 6? 11.Road traffic deaths Situation Every year over a million people die in road accidents and as many as 50 million are injured.

Over 80% of the deaths are in emerging countries.This dismal toll is likely to get much worse as road traffic increases in the developing world.The following table gives the annual road deaths per 100,000 of the population.9 16 Emerging market indicators, The Economist, 17 April 2004, p.

408 Statistics for Business Required 1.From the countries in the list, which country is the most dangerous to drive and which is the least dangerous? 2.How would you compare Belgium, The Dominican Republic, France, Latvia, Luxembourg, Mauritius, Russia, and Venezuela to Britain? 3.How would you compare Belgium, The Dominican Republic, France, Latvia, Luxembourg, Mauritius, Russia, and Venezuela to the United States? 4.

How would you compare Belgium, The Dominican Republic, France, Latvia, Luxembourg, Mauritius, Russia, and Venezuela to Kuwait? 5.How would you compare Belgium, The Dominican Republic, France, Latvia, Luxembourg, Mauritius, Russia, and Venezuela to New Zealand? 6.What are your overall conclusions and what do you think should be done to improve the statistics? 12.Family food consumption Situation The following table gives the 1st quarter 2003 and 1st quarter 2004 prices of a market basket of grocery items purchased by an American family.17 In the same table is the con- sumption of these items for the same period.

Country Deaths per Country Deaths per 100,000 people 100,000 people Belgium 16 Luxembourg 17 Britain 5 Mauritius 45 China 16 New Zealand 13 Columbia 18 Nicaragua 23 Costa Rica 19 Panama 18 Dominican Republic 39 Peru 18 Ecuador 18 Poland 12 El Salvador 42 Romania 11 France 4 Russia 20 Germany 6 Saint Lucia 14 Italy 13 Slovenia 14 Japan 8 South Korea 24 Kuwait 21 Thailand 21 Latvia 25 United States 15 Lithuania 22 Venezuela 24 17 World Food Prices, (consulted July 2006).409Chapter 11: Indexing as a method for data analysis Required 1.Calculate an unweighted price index for this data.Calculate an unweighted quantity index for this data.

Develop a Laspeyres weighted price index for this data.Develop a Paasche weighted price index using the 1st quarter 2003 for the base price.Develop an average quantity weighted price index using 2003 as the base price period and the average of the consumption between 2003 and 2004.Discuss the usefulness of these indexes.Meat Situation A meat wholesaler exports and imports New Zealand lamb, (frozen whole carcasses) United States beef, poultry, United States broiler cuts and frozen pork.

Table 1 gives the prices for these products in $US/ton for the period 2000 to 2005.18 Table 2 gives the quan- tities handled by the meat wholesaler in the same period 2000 to 2005.Product unit amount 1st quarter 1st quarter 1st quarter 2003 1st quarter 2004 2003 ($/unit) 2004 ($/unit) quantity (units) quantity (units) Ground chuck beef (1 lb) 2.87 98 182 Vegetable oil (32 oz bottle) 2.09 21 72 18 International Commodity Prices, /es/esc/prices/CIWPQueryServlet (con- sulted July 2006).

Table 1 Average annual price of meat product ($US/ton).Product 2000 2001 2002 2003 2004 2005 New Zealand Lamb 2,618.17 410 Statistics for Business Required 1.Develop a Laspeyres weighted price index using 2000 as the base period.Develop a Paasche weighted price index using 2005 as the base period.Develop an average quantity weighted price index using the average quantities con- sumed in the period and 2005 as the base period for price.Develop an average quantity weighted price index using as the base both the average quantity distributed in the period and the average price for the period.What are your observations about the data and the indexes obtained? 14.

Beverages Situation A wholesale distributor supplies sugar, coffee, tea, and cocoa to various coffee shops in the west coast of the United States.The distributor buys these four commodities from its supplier at the prices indicated in Table 1 for the period 2000 to 2005.19 Table 2 gives the quantities distributed by the wholesaler in the same period 2000 to 2005.19 International Commodity Prices, /es/esc/prices/CIWPQueryServlet (con- sulted July 2006).

Table 2 Amount handled each year (tons).

Product 2000 2001 2002 2003 2004 2005 New Zealand Lamb 54,000 67,575 72,165 79,125 85,124 95,135 Beef, United States 105,125 107,150 109,450 110,125 115,125 120,457 Poultry, United States 118,450 120,450 122,125 125,145 129,875 131,055 Pork, United States 41,254 42,584 45,894 47,254 49,857 51,254 Table 1 Average annual price of commodity.Commodity 2000 2001 2002 2003 2004 2005 Sugar (US cents/lb) 8.37 Table 2 Amount distributed each year (kg).Commodity 2000 2001 2002 2003 2004 2005 Sugar 75,860 80,589 85,197 94,904 104,759 112,311 Tea 29,840 34,441 39,310 47,887 50,966 59,632 Coffee 47,300 52,429 58,727 66,618 73,427 79,303 Cocoa 27,715 29,156 30,640 35,911 41,219 46,545 411Chapter 11: Indexing as a method for data analysis Required 1.Develop a Laspeyres weighted price index using 2000 as the base period.Develop a Paasche weighted price index using 2005 as the base period.

Develop an average quantity weighted price index using the average quantities con- sumed in the period and 2005 as the base period for price.Develop an average quantity weighted price index using as the base both the average quantity distributed in the period and the average price for the period.What are your observations about the data and the indexes obtained? 15.Non-ferrous metals Situation Table 1 gives the average price of non-ferrous metals in $US/ton in the period 2000 to 2005.20 Table 2 gives the consumption of these metals in tons for a manufacturing con- glomerate in the period 2000 to 2005.Develop a Laspeyres weighted price index using 2000 as the base period.

Develop a Paasche weighted price index using 2005 as the base period.Develop an average quantity weighted price index using the average quantities con- sumed in the period and 2005 as the base period for price.Develop an average quantity weighted price index using as the base both the average quantity consumed in the period and the average price for the period.What are your observations about the data and the indexes obtained? 20 London Metal Exchange, /dataprices (consulted July 2006).Metal 2000 2001 2002 2003 2004 2005 Aluminium 1,650 1,500 1,425 1,525 1,700 2,050 Copper 1,888 1,688 1,550 2,000 2,800 3,550 Tin 5,600 4,600 4,250 5,500 7,650 7,800 Zinc 1,100 900 800 900 1,150 1,600 Metal 2000 2001 2002 2003 2004 2005 Aluminium 53,772 100,041 86,443 63,470 126,646 102,563 Copper 75,000 93,570 106,786 112,678 79,345 126,502 Tin 18,415 13,302 14,919 22,130 21,916 18,535 Zinc 36,158 48,187 32,788 47,011 49,257 31,712 Table 1 Average metal price, $US/ton.

Case study: United States energy consumption Situation The following table gives the energy consumption by source in the United States since 1973 in million British Thermal Units (BTUs).21 Required Using the concept of indexing, describe the consumption pattern of energy in the United States.Year Coal Natural Petroleum Nuclear Hydroelectric Biomass Geothermal Solar Wind gas products 1973 12,971,490 22,512,399 34,839,926 910,177 2,861,448 1,529,068 42,605 1974 12,662,878 21,732,488 33,454,627 1,272,083 3,176,580 1,539,657 53,158 1975 12,662,786 19,947,883 32,730,587 1,899,798 3,154,607 1,498,734 70,153 1976 13,584,067 20,345,426 35,174,688 2,111,121 2,976,265 1,713,373 78,154 1977 13,922,103 19,930,513 37,122,168 2,701,762 2,333,252 1,838,332 77,418 1978 13,765,575 20,000,400 37,965,295 3,024,126 2,936,983 2,037,605 64,350 1979 15,039,586 20,665,817 37,123,381 2,775,827 2,930,686 2,151,906 83,788 1980 15,422,809 20,394,103 34,202,356 2,739,169 2,900,144 2,484,500 109,776 1981 15,907,526 19,927,763 31,931,050 3,007,589 2,757,968 2,589,563 123,043 1982 15,321,581 18,505,085 30,231,314 3,131,148 3,265,558 2,615,048 104,746 1983 15,894,442 17,356,794 30,053,921 3,202,549 3,527,260 2,831,271 129,339 28 1984 17,070,622 18,506,993 31,051,327 3,552,531 3,385,811 2,879,817 164,896 55 68 1985 17,478,428 17,833,933 30,922,149 4,075,563 2,970,192 2,864,082 198,282 111 60 1986 17,260,405 16,707,935 32,196,080 4,380,109 3,071,179 2,840,995 219,178 147 44 1987 18,008,451 17,744,344 32,865,053 4,753,933 2,634,508 2,823,159 229,119 109 37 1988 18,846,312 18,552,443 34,221,992 5,586,968 2,334,265 2,936,991 217,290 94 9 1989 19,069,762 19,711,690 34,211,114 5,602,161 2,837,263 3,062,458 317,163 55,291 22,033 1990 19,172,635 19,729,588 33,552,534 6,104,350 3,046,391 2,661,655 335,801 59,718 29,007 1991 18,991,670 20,148,929 32,845,361 6,422,132 3,015,943 2,702,412 346,247 62,688 30,796 1992 19,122,471 20,835,075 33,526,585 6,479,206 2,617,436 2,846,653 349,309 63,886 29,863 1993 19,835,148 21,351,168 33,841,477 6,410,499 2,891,613 2,803,184 363,716 66,458 30,987 1994 19,909,463 21,842,017 34,670,274 6,693,877 2,683,457 2,939,105 338,108 68,548 35,560 1995 20,088,727 22,784,268 34,553,468 7,075,436 3,205,307 3,067,573 293,893 69,857 32,630 1996 21,001,914 23,197,419 35,756,853 7,086,674 3,589,656 3,127,341 315,529 70,833 33,440 1997 21,445,411 23,328,423 36,265,647 6,596,992 3,640,458 3,005,919 324,959 70,237 33,581 1998 21,655,744 22,935,581 36,933,540 7,067,809 3,297,054 2,834,635 328,303 69,787 30,853 1999 21,622,544 23,010,090 37,959,645 7,610,256 3,267,575 2,885,449 330,919 68,793 45,894 2000 22,579,528 23,916,449 38,403,623 7,862,349 2,811,116 2,906,875 316,796 66,388 57,057 2001 21,914,268 22,905,783 38,333,150 8,032,697 2,241,858 2,639,717 311,264 65,454 69,617 2002 21,903,989 23,628,207 38,401,351 8,143,089 2,689,017 2,649,007 328,308 64,391 105,334 2003 22,320,928 22,967,073 39,047,308 7,958,858 2,824,533 2,811,514 330,554 63,620 114,571 2004 22,466,195 23,035,840 40,593,665 8,221,985 2,690,078 2,982,342 341,082 64,500 141,749 2005 22,830,007 22,607,562 40,441,180 8,133,222 2,714,661 2,780,755 351,671 64,467 149,490 21 Energy Information Administration, Monthly Energy Review, June 2006 (posted 27 June 2006), .Appendix I: Key terminology and formula in statistics A priori probability is being able to make an estimate of probability based on information already available.

Absolute in this textbook context implies pre- senting data according to the value collected.Absolute frequency histogram is a vertical bar chart on x-axis and y-axis.The x-axis is a numer- ical scale of the desired class width, and the y-axis gives the length of the bar which is proportional to the quantity of data in a given class.Addition rule for mutually exclusive events is the sum of the individual probabilities.Addition rule for non-mutually exclusive events is the sum of the individual probabilities less than the probability of the two events occurring together.

Alternative hypothesis is another value when the hypothesized value, or null hypothesis, is not correct at the given level of significance.Arithmetic mean is the sum of all the data val- ues divided by the amount of data.Asymmetrical data is numerical information that does not follow a normal distribution.Average quantity weighted price index is, where P0 and Pn are prices in the base and cur- rent period, respectively, and Qa is the average quantity consumed during the period under consideration.

This index is also referred to as a fixed weight aggregate price index.Average value is another term used for arith- metic mean.Backup is an auxiliary unit that can be used if the principal unit fails.In a parallel arrangement we have backup units.Bar chart is a type of histogram where the x-axis and y-axis have been reversed.

It can also be called a Gantt chart after the American engin- eer Henry Gantt.Bayesian decision-making implies that if you have additional information, or based on the fact that something has occurred, certain probabilities P Q P Q n a a ∑ ∑ 0 100* Expressions and formulas presented in bold letters in the textbook can be found in this section in alphabetical order.In this listing when there is another term in bold letters it means it is explained elsewhere in this Appendix I.At the end of this listing is an explanation of the symbols used in this equation.Further, if you want to know the English equivalent of those that are Greek symbols, you can find that in Appendix III.

may be revised to give posterior probabilities (post meaning afterwards).Bayes’ theorem gives the relationship for statis- tical probability under statistical dependence.Benchmark is the value of a piece of data which we use to compare other data.Bernoulli process is where in each trial there are only two possible outcomes, or binomial.

The probability of any outcome remains fixed over time and the trials are statistically independent.The concept comes from Jacques Bernoulli (1654–1705) a Swiss/French mathematician.Bias in sampling is favouritism, purposely or unknowingly, present in sample data that gives lopsided, misleading, false, or unrepresentative results.Bi-modal means that there are two values that occur most frequently in a dataset.Binomial means that there are only two pos- sible outcomes of an event such as yes or no, right or wrong, good or bad, etc.

Binomial distribution is a table or graph show- ing all the possible outcomes of an experiment for a discrete distribution resulting from a Bernoulli process.Bivariate data involves two variables, x and y.Any data that is in graphical form is bivariate since a value on the x-axis has a corresponding value on the y-axis.Boundary limits of quartiles are Q0, Q1, Q2, Q3, and Q4, where the indices indicate the quartile value going from the minimum value Q0 to the maximum value Q4.Box and whisker plot is a visual display of quartiles.

The box contains the middle 50% of the data.The 1st whisker on the left contains the first 25% of the data and the 2nd whisker on the right contains the last 25%.Box plot is an alternative name for the box and whisker plot.Category is a distinct class into which infor- mation or entities belong.Categorical data is information that includes a qualitative response according to a name, label, or category such as the categories of Asia, Europe, and the United States or the categories of men and women.

With categorical information there may be no quantitative data.Causal forecasting is when the movement of the dependent variable, y, is caused or impacted by the change in value of the independent variable, x.Categories are the groups into which data is organized.Central limit theory in sampling states that as the size of the sample increases, there becomes a point when the distribution of the sample means, x–, can be approximated by the normal distribution.

If the sample size taken is greater than 30, then the sample distribution of the means can be considered to follow a normal distribution even though the population is not normal.

Central moving average in seasonal forecast- ing is the linear average of four quarters around a given central time period.As we move forwards in time the average changes by eliminating the oldest quarter and adding the most recent.Central tendency is how data clusters around a central measure such as the mean value.Characteristic probability is that which is to be expected or that which is the most common in a statistical experiment.Chi-square distribution is a continuous probabi- lity distribution used in this text to test a hypothesis associated with more than two populations.

414 Statistics for Business Chi-square test is a method to determine if there is a dependency on some criterion between the proportions of more than two populations.Class is a grouping into which data is arranged.The age groups, 20–29; 30–39; 40–49; 50–59 years are four classes that can be group- ings used in market surveys.Class range is the breadth or span of a given class.Class width is an alternative description of the class range.

Classical probability is the ratio of the number of favourable outcomes of an event divided by the total possible outcomes.Classical probabil- ity is also known as marginal probability or sim- ple probability.Closed-ended frequency distribution is one where all data in the distribution is contained within the limits.Cluster sampling is where the population is divided into groups, or clusters, and each cluster is then sampled at random.Coefficient of correlation, r is a measure of the strength of the relation between the inde- pendent variable x and the dependent variable y.

The value of r can take any value between �1.00 and the sign is the same as the slope of the regression line.Coefficient of determination, r2 is another measure of the strength of the relation between the variables x and y.The value of r2 is always positive and less than or equal to the coefficient of correlation, r.

Coefficient of variation of a dataset is the ratio of the standard deviation to the mean value, / .Collectively exhaustive gives all the possible outcomes of an experiment.Combination is the arrangement of distinct items regardless of their order.The number of combinations is calculated by the expression, Conditional probability is the chance of an event occurring given that another event has already occurred.Confidence interval is the range of the estimate at the prescribed confidence level.

Confidence level is the probability value for the estimate, such as a 95%.Confidence level may also be referred to as the level of confidence.Confidence limits of a forecast are given by, y � zse, when we have a sample size greater than 30 and by y � tse, for sample sizes less than 30.The values of z and t are determined by the desired level of confidence.Constant value is one that does not change with a change in conditions.

The beginning let- ters of the alphabet, a, b, c, d, e, f, etc., either lower or upper case, are typically used to represent a constant.Consumer price index is a measure of the change of prices.Consumer surveys are telephone, written, elec- tronic, or verbal consumer responses concerning a given issue or product.

Continuity correction factor is applied to a random variable when we wish to use the normal- binomial approximation.Continuous data has no distinct cut-off point and continues from one class to another.The volume of beer in a can may have a nominal value of 33 cl but the actual volume could be 32.Continuous probability distribution is a table or graph where the variable x can take any value within a defined range.n xC n x n x � � ! !( )! 415Appendix 1: Key terminology and formula in statistics Contingency table indicates data relationships when there are several categories present.

It is also referred to as a cross-classification table.Continuous random variables can take on any value within a defined range.Correlation is the measurement of the strength of the relationship between variables.Counting rules are the mathematical relation- ships that describe the possible outcomes, or results, of various types of experiments, or trials.

Covariance of random variables is an applica- tion of the distribution of random variables often used to analyse the risk associated with financial investments.

Critical value in hypothesis testing is that value outside of which the null hypothesis should be rejected.Cross-classification table indicates data rela- tionships when there are several categories pre- sent.It is also referred to as a contingency table.Cumulative frequency distribution is a display of dataset values cumulated from the minimum to the maximum.

In graphical form this it is called an ogive.It is useful for indicating how many observations lie above or below certain values.Data array is raw data that has been sorted in either ascending or descending order.

Data characteristics are the units of measure- ment that describe data such as the weight, length, volume, etc.Data point is a single observation in a dataset.Dataset is a collection of data either unsorted or sorted.Degrees of freedom means the choices that you have taken regarding certain actions.Degrees of freedom in a cross-classification table are (No.

Degrees of freedom in a Student-t distribution are given by (n � 1), where n is the sample size.Dependent variable is that value that is a func- tion or is dependent on another variable.Graphically it is positioned on the y-axis.

Descriptive statistics is the analysis of sample data in order to describe the characteristics of that particular sample.Deterministic is where outcomes or decisions made are based on data that are accepted and can be considered reliable or certain.For example, if sales for one month are $50,000 and costs $40,000 then it is certain that net income is $10,000 ($50,000 � $40,000).Deviation about the mean of all observations, x, about the mean value x–, is zero.Discrete data is information that has a distinct cut-off point such as 10 students, 4 machines, and 144 computers.

Discrete data come from the counting process and the data are whole numbers or integer values.Discrete random variables are those integer values, or whole numbers, that follow no particular pattern.Dispersion is the spread or the variability in a dataset.Distribution of the sample means is the same as the sampling distribution of the means.Empirical probability is the same as relative frequency probability.

Empirical rule for the normal distribution states that no matter the value of the mean or the standard deviation, the area under the curve is always unity.26% of all data 416 Statistics for Business falls within �1 standard deviations from the mean, 95.44% falls within �2 standard devi- ations from the mean, and 99.73% of all data falls within �3 standard deviations from the mean.

Estimate in statistical analysis is that value judged to be equal to the population value.Estimated standard error of the proportion is, where p– is the sample proportion and n is the sample size.Estimated standard error of the difference between two proportions is, Estimated standard deviation of the distribution of the difference between the sample means is, Estimating is forecasting or making a judgment about a future situation using entirely, or in part, quantitative information.Estimator is that statistic used to estimate the population value.Event is the outcome of an activity or experi- ment that has been carried out.

Expected value of the binomial distribution E(x) or the mean value, x, is the product of the number of trials and the characteristic probabil- ity, or x � E(x) � np.Expected value of the random variable is the weighted average of the outcomes of an experi- ment.It is the same as the mean value of the random variable and is given by the relation- ship, x � xP(x) � E(x).Experiment is the activity, such as a sampling process, that produces an event.Exploratory data analysis (EDA) covers those techniques that give analysts a sense about data that is being examined.A stem-and-leaf display and a box and whisker plot are methods in EDA.Factorial rule for the arrangement of n different objects is n! � n(n � 1)(n � 2)(n � 3) … (n � n), where 0! � 1.Finite population is a collection of data that has a stated, limited, or a small size.

The number of playing cards (52) in a pack is considered finite.

Finite population multiplier for a population of size N and a sample of size n is, Fixed weight aggregate price index is the same as the average quantity weighted price index.Fractiles divide data into specified fractions or portions.Frequency distribution groups data into defined classes.The distribution can be a table, polygon, or histogram.We can have an absolute frequency distribution or a relative frequency distribution.

Frequency polygon is a line graph connecting the midpoints of the class ranges.Functions in the context of this textbook are those built-in macros in Microsoft Excel.In this book, it is principally the statistical functions that are employed.However, Microsoft Excel contains financial, logic, database, and other functions.Gaussian distribution is another name for the normal distribution after its German originator, Karl Friedrich Gauss (1777–1855).

N n N � �1 x x n n1 2 1 2 1 2 2 2 � � � p p p q n p q n1 2 1 1 1 2 2 2 � � � ( ) p p p n � �1 417Appendix 1: Key terminology and formula in statistics Geometric mean is used when data is chan- ging over time.It is calculated by the nth root of the growth rates for each year, where n is the number of years.Graphs are visual displays of data such as line graphs, histograms, or pie charts.Greater than ogive is a cumulative frequency distribution that illustrates data above certain values.It has a negative slope, where the y-values decrease from left to right.

Groups are the units or ranges into which data is organized.Histogram is a vertical bar chart showing data according to a named category or a quantitative class range.Historical data is information that has occurred, or has been collected in the past.Horizontal bar chart is a bar chart in a horizontal form where the y-axis is the class and the x-axis is the proportion of data in a given class.Hypothesis is a judgment about a situation, outcome, or population parameter based simply on an assumption or intuition with initially no concrete backup information or analysis.

Hypothesis testing is to test sample data and make on objective decision based on the results of the test using an appropriate significance level for the hypothesis test.Graphically the independent variable is always positioned on the x-axis.Index base value is the real value of a piece of data which is used as the reference point to determine the index number.Index number is the ratio of a certain value to a base value usually multiplied by 100.

When the base value equals 100 then the measured values are a percentage of the base.The index number may be called as the index value.Index value is an alternative for the index number.Inferential statistics is the analysis of sample data for the purpose of describing the character- istics of the population parameter from which that sample is taken.Infinite population is a collection of data that has such a large size so that by removing or destroying some of the data elements it does not significantly impact the population that remains.

Integer values are whole numbers originating from the counting process.Interval estimate gives a range for the estimate of the population parameter.Inter-quartile range is the difference between the values of the 3rd and the 1st quartile in the dataset.It measures the range of the middle half of an ordered dataset.Joint probability is the chance of two events occurring together or in succession.

Kurtosis is the characteristic of the peak of the distribution curve.Laspeyres weighted price index is, where Pn is the price in the current period, P0 is the price in the base period and Q0 is the quan- tity consumed in the base period.Law of averages implies that the average value of an activity obtained in the long run will be close to the expected value, or the weighted out- come based on each probability of occurrence.Least square method is a calculation technique in regression analysis that determines the best P Q P Q n 0 0 ∑ ∑ 0 100* 418 Statistics for Business straight line for a series of data that minimizes the error between the actual and forecast data.Leaves are the trailing digits in a stem-and-leaf display.

Left-skewed data is when the mean of a dataset is less than the median value, and the curve of the distribution tails off to the left side of the x-axis.Left-tail hypothesis test is used when we are asking the question, “Is there evidence that a value is less than?” Leptokurtic is when the peak of a distribution is sharp, quantified by a small standard deviation.Less than ogive is a cumulative frequency distri- bution that indicates the amount of data below certain limits.As a graph it has a positive slope such that the y-values increase from left to right.Level of confidence in estimating is (1 � ), where is the proportion in the tails of the dis- tribution, or that area outside of the confidence interval.

Line graph shows bivariate data on x-axis and y-axis.If time is included in the data this is always indicated on the x-axis.Linear regression line takes the form y � a � bx.It is the equation of the best straight line for the data that minimizes the error between the data points on the regression line and the correspon- ding actual data from which the regression line is developed.

Margin of error is the range of the estimate from the true population value.

Marginal probability is the ratio of the number of favourable outcomes of an event divided by the total possible outcomes.Marginal probability is also known as classical probability or simple probability.Mean value is another way of referring to the arithmetic mean.Mean value of random data is the weighted average of all the possible outcomes of the ran- dom variable.

Median is the middle value of an ordered set of data.It divides the data into two equal halves.The 2nd quartile and the 50th percentile are also the median value.Mesokurtic describes the curve of a distribu- tion when it is intermediate between a sharp peak, leptokurtic and a relatively flat peak, or platykurtic.Mid-hinge in quartiles is the average of the 3rd and 1st quartile.

Midpoint of a class range is the maximum plus the minimum value divided by 2.Midrange is the average of the smallest and the largest observations in a dataset.Mid-spread range is another term for the inter- quartile range.Mode is that value that occurs most frequently in a dataset.It can be represented by an equation of the general form, y � a � b1x1 � b2x2 � b3x3 � … � bkxk.Mutually exclusive events are those that can- not occur together.Normal-binomial approximation is applied when np 5 and n(1 � p) 5.In this case, substituting for the mean value and the stand- ard deviation of the binomial distribution in the normal distribution transformation relationship we have, z x x np npq x np np p � � � � � � � ( )1 419Appendix 1: Key terminology and formula in statistics Normal distribution, or the Gaussian distribu- tion, is a continuous distribution of a random variable.It is symmetrical, has a single hump, and the mean, median and mode are equal.

The tails of the distribution may not immediately cut the x-axis.Normal distribution density function, which describes the shape of the normal distribution is, Normal distribution transformation relation- ship is, where z is the number of standard deviations, x is the value of the random variable, x is the mean value of the dataset and x is the standard deviation of the dataset.Non-linear regression is when the dependent variable is represented by an equation where the power of some or all the independent vari- ables is at least two.These powers of x are usu- ally integer values.Non-mutually exclusive events are those that can occur together.

Null hypothesis is that value that is considered correct in the experiment.Numerical codes are used to transpose quali- tative or label data into numbers.For example, if the time period is January, February, March, etc.Odds are the chance of winning and are the ratio of the probability of losing to the chances of winning.Ogive is a frequency distribution that shows data cumulatively.A less than ogive indicates data less than certain values and a greater than ogive shows data more than certain values.An ogive can illustrate absolute data or relative data.One-arm-bandit is the slang term for the slot machines that you find in gambling casinos.

The game of chance is where you put in a coin or chip, pull a lever and hope that you win a lucky combination! One-tail hypothesis test is used when we are interested to know if something is less than or greater than a stipulated value.If we ask the ques- tion, “Is there evidence that the value is greater than?” then this would be a right-tail hypothesis test.Alternatively, if we ask the question, “Is there evidence that the value is less than?” then this would be a left-tail hypothesis test.Ordered dataset is one where the values have been arranged in either increasing or decreas- ing order.Outcomes of a single type of event are kn, where k is the number of possible events, and n is the number of trials.

Outcomes of different types of events are k1 * k2 * k3 * * kn, where k1, k2, , kn are the number of possible events.Outliers are those numerical values that are either much higher or much lower than other values in a dataset and can distort the value of the central tendency, such as the average, and the value of the dispersion such as the range or standard deviation.P in upper case or capitals is often the abbrevi- ation used for probability.Paired samples are those that are dependent or related, often in a before and after situation.Examples are the weight loss of individuals after a diet programme or productivity improvement after a training programme.

Pareto diagram is a combined histogram and line graph.The frequency of occurrence of the data is indicated according to categories on the z x x x � � f x x x x x( ) ( / ) ( )/ � � � 1 2 1 2 2 e 420 Statistics for Business histogram and the line graph shows the cumu- lated data up to 100%.Parallel bar chart is similar to a parallel his- togram but the x-axis and y-axis have been reversed.

Parallel arrangement in design systems is such that the components are connected giving a choice to use one path or another.

Which ever path is chosen the system continues to function.Parallel histogram is a vertical bar chart show- ing the data according to a category and within a given category there are sub-categories such as different periods.A parallel histogram is also referred to a side-by-side histogram.Parameter describes the characteristic of a population such as the weight, height, or length.Percentiles are fractiles that divide ordered data into 100 equal parts.Permutation is a combination of data arranged in a particular order.The number of ways, or permutations, of arranging x objects selected in order from a total of n objects is, Pictogram is a diagram, picture, or icon that shows data in a relative form.Pictograph is an alternative name for the pictogram.Pie chart is a circle graph showing the percent- age of the data according to certain categories.

The circle, or pie, contains 100% of the data.Platykurtic is when the curve of a distribution has a flat peak.Numerically this is shown by a larger value of the coefficient of variation, / .p-value in hypothesis testing is the observed level of significance from the sample data or the minimum probable level that we will tolerate in order to accept the null hypothesis of the mean or the proportion.Point estimate is a single value used to estimate the population parameter.

Poisson distribution describes events that occur during a given time interval and whose average value in that time period is known.Population is all of the elements under study and about which we are trying to draw conclusions.Population standard deviation is the square root of the population variance.Population variance is given by, where N is the amount of data, x is the particu- lar data value, and x is the mean value of the dataset.

Portfolio risk measures the exposure associ- ated with financial investments.Posterior probability is one that has been revised after additional information has been received.Power of a hypothesis test is a measure of how well the test is performing.Primary data is that collected directly from the source.Probability is a quantitative measure, expressed as a decimal or percentage value, indicating the likelihood of an event occurring.

The value 2 2 � �( )x N x∑ P x x x ( ) ! � � e n xP n n x � � ! ( )! 421Appendix 1: Key terminology and formula in statistics 1 � P(x) is the likelihood of the event not occurring.Probabilistic is where there is a degree of uncertainty, or probability of occurrence from the supplied data.Quad-modal is when there are four values in a dataset that occur most frequently.Qualitative data is information that has no numerical response and cannot immediately be analysed.Quantitative data is information that has a numerical response.

Quartiles are those three values which divide ordered data into four equal parts.Quartile deviation is one half of the inter- quartile range, or (Q3 � Q1)/2.Questionnaires are evaluation sheets used to ascertain people’s opinions of a subject or a product.Quota sampling in market research is where each interviewer in the sampling experiment has a given quota or number of units to analyse.Random implies that any occurrence or value is possible.

Random sample is where each item of data in the sample has an equal chance of being selected.Random variable is one that will have different values as a result of the outcome of a random experiment.Range is the numerical difference between the highest and lowest value in a dataset.Ratio measurement scale is where the difference between measurements is based on starting from a base point to give a ratio.The consumer price index is usually presented on a ratio measurement scale.

Raw data is collected information that has not been organized.Real value index (RVI) of a commodity for a period is, Regression analysis is a mathematical technique to develop an equation describing the relation- ship of variables.It can be used for forecasting and estimating.Relative in this textbook context is presenting data compared to the total amount collected.It can be expressed either as a percentage or fraction.

Relative frequency histogram has vertical bars that show the percentage of data that appears in defined class ranges.Relative frequency distribution shows the per- centage of data that appears in defined class ranges.Relative frequency probability is based on information or experiments that have previ- ously occurred.It is also known as empirical probability.

Relative price index where P0 is the price at the base period, and Pn is the price at another period.

Relative quantity index where Q0 is the quantity at the base period and Qn is the quantity at another period.Relative regional index (RRI) compares the value of a parameter at one region to a selected base region.It is given by, Value at other region Value at base region * 1000 1000� V Vb * I Q QQ n� ( / )0 100* I P PP n� ( / )0 100* , RVI Current value of commodity Base value of � commodity Base indicator Current indicator * ** 100 422 Statistics for Business Reliability is the confidence we have in a prod- uct, process, service, work team, individual, etc.to operate under prescribed conditions without failure.Reliability of a series system, RS is the product of the reliability of all the components in the system, or RS � R1 * R2 * R3 * R4 * * Rn.

The value of Rs is less than the reliability of a single component.Reliability of a parallel system, RS is one minus the product of all the parallel components not working, or RS � 1 � (1 � R1)(1 � R2)(1 � R3) (1 � R4) (1 � Rn).The value of RS is greater than the reliability of an individual component.Replacement is when we take an element from a population, note its value, and then return this element back into the population.Representative sample is one that contains the relevant characteristics of the population and which occur in the same proportion as in the population.

Research hypothesis is the same as the alter- native hypothesis and is a value that has been obtained from a sampling experiment.Right-skewed data is when the mean of a dataset is greater than the median value, and the curve of the distribution tails off to the right side of the x-axis.Right-tail hypothesis test is used when we are asking the question, Is there evidence that a value is greater than? Risk is the loss, often financial, that may be incurred when an activity or experiment is undertaken.Rolling index number is the index value com- pared to a moving base value often used to show the change of data each period.Sample is the collection of a portion of the population data elements.

Sampling is the analytical procedure with the objective to estimate population parameters.Sampling distribution of the means is a distri- bution of all the means of samples withdrawn from a population.Sampling distribution of the proportion is a probability distribution of all possible values of the sample proportion, p–.Sampling error is the inaccuracy in a sam- pling experiment.Sample space gives all the possible outcomes of an experiment.

Sample standard deviation, s is the square root of the sample variation, ���s2.Sample variance, s2 is given by, where n is the amount of data, x is the particu- lar data value, and x– is the mean value of the dataset.Sampling from an infinite population means that even if the sample were not replaced, then the probability outcome for a subsequent sample would not significantly change.Sampling with replacement is taking a sample from a population, and after analysis, the sam- ple is returned to the population.Sampling without replacement is taking a sample from a population, and after analysis not returning the sample to the population.

Scatter diagram is the presentation of time series data in the form of dots on x-axis and y-axis to illustrate the relationship between the x and y variables.Score is a quantitative value for a subjective response often used in evaluating questionnaires.Secondary data is the published information collected by a third party.Series arrangement is when in system, compon- ents are connected sequentially so that you have to pass through all the components in order that the system functions.

Shape of the sampling distribution of the means is about normal if random samples of at least size 30 are taken from a non-normal popu- lation; if samples of at least 15 are withdrawn from a symmetrical distribution; or samples of any size are taken from a normal population.Side-by-side bar chart is where the data is shown as horizontal bars and within a given cat- egory there are sub-categories such as different periods.Side-by-side histogram is a vertical bar chart showing the data according to a category and within a given category there are sub-categories such as different periods.A side-by-side histo- gram is also referred to as a parallel histogram.Significantly different means that in compar- ing data there is an important difference between two values.

Significantly greater means that a value is considerably greater than a hypothesized value.Significantly less means that a value is consid- erably smaller than a hypothesized value.Significance level in hypothesis testing is how large, or important, is the difference before we say that a null hypothesis is invalid.It is denoted by , the area outside the distribution.Simple probability is an alternative for marginal or classical probability.

Simple random sampling is where each item in the population has an equal chance of being selected.Skewed means that data is not symmetrical.Stacked histogram shows data according to categories and within each category there are sub-categories.It is developed from a cross- classification or contingency table.Standard deviation of a random variable is the square root of the variance or, Standard deviation of the binomial distribution is the square root of the variance, or � ��� 2 ��(npq).

Standard error of the difference between two proportions is, Standard deviation of the distribution of the dif- ference between sample means is, Standard deviation of the Poisson distribution is the square root of the mean number of occur- rences or, � ���( ).Standard deviation of the sampling distribution, x–, is related to the population standard devia- tion, x, and sample size, n, from the central limit theory, by the relationship, Standard error of the estimate of the linear regression line is, s y y ne � � � ( )2 2 ∑ x x n � x x n n1 2 1 2 1 2 2 2 � � � p p p q n p q n1 2 1 1 1 2 2 2 � � � � �( ) ( )x P xx 2∑ 424 Statistics for Business Standard error of the difference between two means is, Standard error of the proportion, p– is, Standard error of the sample means, or more simply the standard error is the error in a sam- pling experiment.It is the relationship, Standard error of the estimate in forecasting is a measure of the variability of the actual data around the regression line.Standard normal distribution is one which has a mean value of zero and a standard deviation of unity.

Statistic describes the characteristic of a sam- ple, taken from a population, such as the weight, volume length, etc.

Statistical dependence is the condition when the outcome of one event impacts the outcome of another event.Statistical independence is the condition when the outcome of one event has no bearing on the outcome of another event, such as in the tossing of a fair coin.Stems are the principal data values in a stem- and-leaf display.Stem-and-leaf display is a frequency distribu- tion where the data has a stem of principal values, and a leaf of minor values.In this display, all data values are evident.

Stratified sampling is when the population is divided into homogeneous groups or strata and random sampling is made on the strata of interest.Student-t distribution is used for small sample sizes when the population standard deviation is unknown.Subjective probability is based on the belief, emotion or “gut” feeling of the person making the judgment.Symmetrical in a box and whisker plot is when the distances from Q0 to the median Q2, and the distance from Q2 to Q4, are the same; the dis- tance from Q0, to Q1 equals the distance from Q3 to Q4 and the distance from Q1 to Q2 equals the distance from the Q2 to Q3; and the mean and the median value are equal.Symmetrical distribution is when one half of the distribution is a mirror image of the other half.

System is the total of all components, pieces, or processes in an arrangement.Purchasing, trans- formation, and distribution are the processes of the supply chain system.Systematic sampling is taking samples from a homogeneous population at a regular space, time or interval.Time series is historical data, which illustrate the progression of variables over time.Time series deflation is a way to determine the real value in the change of a commodity using the consumer price index.

Transformation relationship is the same as the normal distribution transformation relationship.Tri-modal is when there are three values in a dataset that occur most frequently.Type I error occurs if the null hypothesis is rejected when in fact the null hypothesis is true.x x n � p pq n p p n � � �( )1 x x n n1 2 1 2 1 2 2 2 � � � 425Appendix 1: Key terminology and formula in statistics Type II error is accepting a null hypothesis when the null hypothesis is not true.Two-tail hypothesis test is used when we are asking the question, “Is there evidence of a dif- ference?” Unbiased estimate is one that on an average will equal to the parameter that is being estimated.

Univariate data is composed of individual val- ues that represent just one random variable, x.Unreliability is when a system or component is unable to perform as specified.Unweighted aggregate index is one that in the calculation each item in the index is given equal importance.Variable value is one that changes according to certain conditions.The ending letters of the alphabet, u, v, w, x, y, and z, either upper or lower case, are typically used to denote variables.

Variance of a distribution of a discrete random variable is given by the expression, Variance of the binomial distribution is the product of the number of trials n, the character- istic probability, p, of success, and the character- istic probability, q, of failure, or 2 � npq.Venn diagram is a representation of probabil- ity outcomes where the sample space gives all possible outcomes and a portion of the sample space represents an event.Vertical histogram is a graphical presentation of vertical bars where the x-axis gives a defined class and the y-axis gives data according to the frequency of occurrence in a class.Weighted average is the mean value taking into account the importance or weighting of each value in the overall total.The total weight- ings must add up to 1 or 100%.

Weighted mean is an alternative for the weighted average.Weighted price index is when different weights or importance is given to the items used to cal- culate the index.What if is the question asking, “What will be the outcome with different information?” Wholes numbers are those with no decimal or fractional components.2 2� �( ) ( )x P xx∑ 426 Statistics for Business 427Appendix 1: Key terminology and formula in statistics Symbol Meaning Mean number of occurrences used in a Poisson distribution Mean value of population n Sample size in units N Population size in units p Probability of success, fraction or percentage q Probability of failure � (1 � p), fraction or percentage Q Quartile value r Coefficient of correlation r 2 Coefficient of determination s Standard deviation of sample Standard deviation of population Estimate of the standard deviation of the population se Standard error of the regression line t Number of standard deviations in a Student distribution x Value of the random variable.The independent variable in the regression line x– Average value of x y Value of the dependent variable y– Average value of y y Value of the predicted value of the dependent variable z Number of standard deviations in a normal distribution Note: Subscripts or indices 0, 1, 2, 3, etc.

indicate several data values in the same series.Symbols used in the equations This page intentionally left blank Appendix II: Guide for using Microsoft Excel in this textbook (Based on version 2003) The most often used tools in this statistics textbook are the development of graphs and the built-in functions of the Microsoft Excel pro- gram.The following sections give more information on their use.Note in these sections the words shown in italics correspond exactly to the headings used in the Excel screens but these may not always be the same terms as used in this textbook.

For example, Excel refers to chart type, whereas in the text I call them graphs.When you click on the graph icon as shown in Figure E-1 you will obtain the screen that is illustrated in Figure E-2.Here in the tab Standard Types you have a selection of the Chart type or graphs that you can produce.The key ones that are used in this text are the first five in the list – Column (histogram), Bar, Line, Pie, and XY (Scatter).When you click on any of these options you will have a selection of the various formats that are available.

For example, Figure E-2 illustrates the Chart sub-type for the Column options and Figure E-3 illustrates the Chart sub- types for the XY (Scatter) option.Assume, for example, you wish to draw a line graph for the data given in Table E-1 that is con- tained in an Excel spreadsheet.You first select (highlight) this data and then choose the graph option XY (Scatter).You then click on Next and this will illustrate the graph you have formed as shown in Figure E-4.

This Step 2 of 4 of the chart wizard as shown at the top of the window.

If you click on the tab, Series at the top of the screen, you can make modifications to the input data.If you then click on Next again you will have Step 3 of 4, which gives the various Chart options for presenting your graph.Finally, when you again click on Next you will have Chart Location according to the screen shown in Generating Excel Graphs Figure E.431Appendix II: Guide for using microsoft excel in this textbook x 1 2 3 4 5 y 5 9 14 12 21 Table E-1 x, y data.

This gives you a choice of making a graph As new sheet, that is as a new file for your graph or As object in, which is the graph in your spread sheet.For organizing my data I always prefer to create a new sheet for my graphs, but the choice is yours! Regardless of what type of graph you decide to make, the procedure is the same as indicated in the previous paragraph.One word of caution is the choice in the Standard Types between using Line and XY (Scatter).For any line graph I always use XY (Scatter) rather than Line as with this pres- entation the x and y data are always correlated.In Chapter 10, we discussed in detail linear regression or the development of a straight line that is the best fit for the data given.

Figure E-7 shows the screen for developing this linear regression line.433Appendix II: Guide for using microsoft excel in this textbook If you click on the fx object in the tool bar as shown in Figure E-1, and select, All, in the command, Or select a category, you will have the screen as shown in Figure E-8.This gives a list- ing of all the functions that are available in Excel in alphabetical order.Table E-2 gives those functions that are used in this textbook and their use.Each func- tion indicated can be found in appropriate chap- ters of this textbook.(Note, for those living south of the Isle of Wight, you have the equiva- lent functions in French!) Using the Excel Functions Figure E.Table E-2 Excel functions used in this book.

English French For determining ABS ABS Gives the absolute value of a number.That is the negative numbers are ignored AVERAGE MOYENNE Mean value of a dataset EXPONDIST NTIELLE Cumulative distribution exponential function, given the value of the ransom variable x and the mean value .Use a value of cumulative � 1 CEILING Rounds up a number to the nearest integer value 434 Statistics for Business Table E-2 Excel functions used in this book.(Continued) English French For determining CHIDIST X Gives the area in the chi-distribution when you enter the chi-square value and the degrees of freedom CHIINV E Gives the chi-square value when you enter the area in the chi-square distribution and the degrees of freedom CHITEST X Gives the area in the chi-square distribution when you enter the observed and expected frequency values COMBIN COMBIN Gives the number of combinations of arranging x objects from a total sample of n objects CONFIDENCE NCE Returns the confidence interval for a population mean CORREL COEFFICIENT;CORRELATION Determines the coefficient of correlation for a bivariate dataset COUNT NBVAL The number of values in a dataset CHIINV E Returns the inverse of the one-tailed probability of the chi-squared distribution BINOMDIST ALE Binomial distribution given the random variable, x, and characteristic probability, p.If cumulative � 0, the individual value is determined.

If cumulative � 1, the cumulative values are determined IF SI Evaluates a condition and returns either true or false based on the stated condition FACT FACT Returns the factorial value n! of a number FLOOR Rounds down a number to the nearest integer value FORECAST PREVISION Gives a future value of a dependent variable, y, from known variables x and y, data assuming a linear relationship between the two FREQUENCY FREQUENCE Determines how often values occur in a dataset GEOMEAN RIQUE Gives the geometric mean growth rate from the annual growth rates data.The percentage of geometric mean is the geometric mean growth rate less than 1 GOAL SEEK VALEUR CIBLE Gives a value according to a given criteria.If you use cumulative � 0 this gives a point value for exactly x occurring NORMINV E Value of the random variable x given probability, p, mean value, , standard deviation, , and cumulative � 1 435Appendix II: Guide for using microsoft excel in this textbook Table E-2 Excel functions used in this book.English French For determining NORMSDIST RD The probability, p, given the number of standard deviations z NORMSINV RD.

Determines the number of standard deviations, z, INVERSE given the value of the probability, p OFFSET DECALER Repeats a cell reference to another line or column according to the offset required PEARSON PEARSON Determines the Pearson product moment correlation, or the coefficient of correlation, r PERCENTILE CENTILE Gives the percentile value of a dataset.Select the data and enter the percentile, 0.02, etc PERMUT PERMUTATION Gives the number of permutations of organising x objects from a total sample of n objects POISSON N Poisson distribution given the random variable, x, and the mean value, .

If cumulative � 0, the individual value is determined.

If cumulative � 1, the cumulative values are determined POWER PUISSANCE Returns the result of a number to a given power RAND ALEA Generates a random number between 0 and 1 RANDBETWEEN Generates a random number between the numbers you specify ROUND ARRONDI Rounds to the nearest whole number RSQ INATION Determines the coefficient of determination, r2 or gives the square of the Pearson product moment correlation coefficient IF SI Logical statement to test a specified condition SLOPE PENTE Determines the slope of a regression line SQRT RACINE Gives the square root of a given value STDEV ECARTYPE Determines the standard deviation of a dataset on the basis for a sample STDEVP ECARTYPEP Determines the standard deviation of a dataset on the basis for a population SUM SOMME Determines the total of a defined dataset SUMPRODUCT SOMMEPROD Returns the sum of two columns of data TDIST T Probability of a random variable, x, given the degrees of freedom, and the number of tails.If the number of tails � 1, the area to the right is determined.P Determines the variance of a dataset on the basis it is a population bk, bk�1, b2, b1, a, slope due to slope due to slope due to slope due to intercept variable xk variable xk�1 variable x2 variable x1 on y-axis sek, sek�1, se2, se1, sea, standard error for standard error for standard error for standard error for standard slope bk slope bk�1 slope b2 slope b1 error for intercept a r2, coefficient of se, determination standard error of estimate F-ratio df, degree of freedom SSreg, SSresid, sum of squares due sum of squares of to regression residual (explained variation) (unexplained variation) 436 Statistics for Business b Slope due to variable x a intercept on y-axis seb Standard error for slope, b sea standard error for intercept a r2 coefficient of determination se standard error of estimate F F-ratio for analysis of variance df degree of freedom (n � 2) SSreg sum of squares due to regression SSresid sum of squares of residual (explained variation) (unexplained variation) Table E-3 Microsoft Excel and the linear regression function.Table E-4 Microsoft Excel and the multiple regression function.

A virgin block of cells at least two columns by five rows are selected.When the y and x data are entered into the function, the various statis- tical data are returned in a format according to Table E-3.As for simple linear regression, multiple regres- sion functions can be solved with the Excel regres- sion function.Here now a virgin block of cells is selected such that the number of columns is at least equal to the number of variables plus one and the number of rows is equal to five.

When the y and x data are entered into the function, the various statistical data are returned in a for- mat according to Table E-4.Simple Linear Regression Multiple Regression Appendix III: Mathematical relationships Statistics involves numbers and the material in this textbook is based on many mathematical relationships, fundamental ideas, and conversion factors.Constants and variables A constant is a value which does not change under any circumstances.The straight line dis- tance from the centre of Trafalgar Square in London to the centre of the Eiffel Tower in Paris is constant.

However, the driving time from these two points is a variable as it depends on road, traffic, and weather conditions.By convention, constants are represented algebraically by the beginning letters of the alphabet either in lower or upper case.Lower case a, b, c, d, e, …… Upper case A, B, C, D, E, …… A variable is a number whose value can change according to various conditions.By convention variables are represented algebraically by the end- ing letters of the alphabet again either in lower or upper case.Lower case u, v, w, x, y, z Upper case U, V, W, X, Y, Z The variables denoted by the letters x and y are the most commonly encountered.

Where two- dimensional graphs occur, x is the abscissa or horizontal axis, and y is the ordinate or vertical axis.In three-dimensional graphs, the letter z is used to denote the third axis.In textbooks, articles, and other documents you will see constants and variables written in either upper case or lower case.There seems to be no recognized rule; however, I prefer to use the lower case.

Your memory of basic mathematical relationships may be rusty.The objective of this appendix is to give a detailed revision of arithmetic relationships, rules, and conversions.The following concepts are covered: Constants and variables • Equations • Integer and non-integer numbers • Arithmetic operating sym- bols and equation relationships • Sequence of arithmetic operations • Equivalence of algebraic expressions • Fractions • Decimals • The Imperial and United States measuring system • Temperature • Conversion between fractions and decimals • Percentages • Rules for arithmetic calculations for non-linear relationships • Sigma, • Mean value • Addition of two variables • Difference of two variables • Constant multiplied by a variable • Constant summed n times • Summation of a random variable around the mean • Binary numbering system • Greek alphabet S u b j e c t m a t t e r Equations An equation is a relationship where the values on the left of the equal sign are equal to the values on the right of the equal sign.Values in any part of an equation can be variables or constants.The following is a linear equation meaning that the power of the variables has the value of unity: y � a � bx This equation represents a straight line where the constant cutting the y-axis is equal to a and the slope of the curve is equal to b.

An equation might be non-linear meaning that the power of any one of the variables has a value other than unity as for example, y � a � bx3 � cx2 � d Integer and non-integer numbers An integer is a whole number such as 1, 2, 5, 19, 25, etc.In statistics an integer is also known as a discrete number or discrete variable if the number can take any different values.Non-integer numbers are those that are not whole numbers such as the fractions 1⁄2, 3⁄4, or 31⁄2, 73⁄4, etc; or decimals such as 2.Arithmetic operating symbols and equation relationships The following are arithmetic operating symbols and equation relationships: � Addition � Subtraction � Plus or minus � Equals � Not equal to � Divide / This means ratio but also divide.For example, 3⁄4 means the ratio of 3 to 4 but also 3 divided by 4.� Greater than � Less than Greater or equal to � Less than or equal to � Approximately equal to For multiplication we have several possibilities to illustrate the operation.When we multiply two algebraic terms a and b together this can be shown as: ab; a.

b; a b; or a * b With numbers, and before we had computers, the multiplication or product of two values was written using the symbol for multiplication: 6 4 � 24 With Excel the symbol * is used as the multi- plication sign and so the above relationship is written as: 6 * 4 � 24 It is for this reason that in this textbook, the symbol * is used for the multiplication sign rather than the historical symbol.Sequence of arithmetic operations When we have expressions related by operating symbols the rule for calculation is to start first to calculate the terms in the Brackets, then Division and/or Multiplication, and finally Addition and/or Subtraction (BDMAS) as shown in Table M-1.If there are no brackets in the expression and only addition and subtraction operating sym- bols then you work from left to write.438 Statistics for Business Symbol Term Evaluation sequence B Brackets 1st D Division 2nd M Multiplication 2nd A Addition Last S Subtraction Last Table M-1 Sequence of arithmetic operations.

439Appendix III: Mathematical relationships Equivalence of algebraic expressions Algebraic or numerical expressions can be writ- ten in various forms as Table M-3 illustrates.Fractions Fractions are units of measure expressed as one whole number divided by another whole num- ber.The common fraction has the numerator on the top and the denominator on the bottom: The common fraction is when the numerator is less than the denominator which means that the number is less than one as for example, 1⁄7, 3⁄4 and 5⁄12.The improper fraction is when the numerator Common fraction Numerator Denominator � Expression Answer Operation 25 � 11 � 7 21 Calculate from left to right 9 * 6 � 4 50 Multiplication before subtraction �22 * 4 �88 A minus times a plus is a minus �12 * �6 72 Minus times a minus equals a minus 6 � 9 * 5 � 3 48 Multiplication then addition and subtraction 7(9) 63 A bracket is equivalent to a multiplication operation 9(5 � 7) 108 Addition in the bracket then the multiplication (7 � 4)(12 � 3)�6 21 Expression in brackets, multiplication, then subtraction 20 * 3 � 10 � 11 17 Multiplication and divisions first then addition Table M-2 Calculation procedures for addition and subtraction.

Arithmetic rule Example a � b � b � a 6 � 7 � 7 � 6 � 13 a � (b � c) � a � b � c � (a � b) � c 9 � (7 � 3) � 9 � 7 � 3 � (9 � 7) � 3 � 19 a � b � �b � a 15 � 21 � �21 � 15 � �6 a * b � b * a 6 * 7 � 7 * 6 � 42 a * (b � c) � a * b � a * c 3 * (8 � 4) � 3 * 8 � 3 * 4 � 36 Table M-3 Algebraic and numerical expressions.

is greater than the denominator, which means that the number is greater than unity as for example , and .In this case these improper fractions can be reduced to a whole number and proper fractions to give 42⁄7, 57⁄9 and 61⁄3.The rules for adding, subtracting multiplying, and dividing fractions are given in Table M-4.Decimals A decimal number is a fraction, whose denomi- nator is any power of 10 so that it can be writ- ten using a decimal point as for example: 7/10 � 0.051 The metric system, used in continental Europe, is based on the decimal system and changes in units of 10.Tables M-5, M-6, M-7, and M-8, give 19 3 52 9 30 7 440 Statistics for Business Micrometre Millimetre Centimetre Decimetre Metre Decametre Hectometre Kilometre ( m) (mm) (cm) (dm) (m) (dam) (hm) (km) 109 1,000,000 100,000 10,000 1,000 100 10 1 108 100,000 10,000 1,000 100 10 1 0.000001 Table M-5 Length or linear measure.Square Square Square Square Square Are Hectare Square micrometre millimetre centimetre decimetre metre (a) (ha) kilometre ( m2) (mm2) (cm2) (dam2) (m2) (km2) 1018 1012 1010 108 1,000,000 10,000 100 1 1016 1010 108 1,000,000 10,000 100 1 0.

00000001 10�10 Table M-6 Surface or area measure.1 1 a b b a ab � � � 1 5 1 6 6 5 5 6 11 30 � � � � * a c b c a b c � � � 4 5 16 5 4 16 5 20 5 4� � � � � a c b c a b c � � � 4 7 2 7 4 2 7 2 7 � � � � a c b d a b c d * * * � 4 5 7 6 4 7 5 6 28 30 14 15 * * * � � � a c b d a d c b � * * 4 5 7 6 4 6 5 7 24 35 � � * * 441 Appendix III: M athem atical relationships Microlitre Millilitre Centilitre Decilitre Litre Decalitre Hectolitre Kilolitre Cubic centimetre Cubic decimetre Cubic metre ( l) (ml) (cl) (dl) (l) (dal) (hl) (kl) (cm3) (dm3) (m3) 109 106 100,000 10,000 1,000 100 10 1 106 1,000 1 108 100,000 10,000 1,000 100 10 1 0.001 10�6 10�9 Table M-7 Volume or capacity measure.442 Statistics for Business Microgram Milligram Centigram Decigram Gram Decagram Hectogram Kilogram Metric ton ( g) (mg) (cg) (dg) (g) (dag) (hg) (kg) (t) 1012 109 108 107 1,000,000 100,000 10,000 1,000 1 109 1,000,000 100,000 10,000 1,000 100 10 1 0.000001 10�7 10�8 10�9 10�12 Table M-8 Mass or weight measure.Inches Feet Yards Miles Millimetres Centimetres Metres Kilometres (in) (ft) (yd) (mi) (mm) (cm) (m) (km) 1 0.6214 10,000 100,000 1,000 1 Table M-9 Conversions for length or linear measurement.the relationships for length, surface, volume, and weight.The Imperial, US, and metric measuring system The Imperial measuring system is used in the United States and partly in England though there are efforts to change to the metric system.The Imperial numbering system is quirky with no apparent logic as compared to the metric system.Tables M-9, M-10, M-11, M-12, and M-13 give approximate conversion tables for key meas- urements.

The United States system is not always the same as the Imperial measuring system.Temperature In Europe, usually the Celsius system is used for recording temperature.Here the freezing point of water is measured at 0°C and the boiling point is 100°C.In the United States, and sometimes in the United Kingdom, the Fahrenheit system is used where the freezing point of water is given at 32°F and the boiling point as 212°F.443Appendix III: Mathematical relationships USA USA USA Imperial Imperial Imperial Cubic Litres gallon quart pint gallon quart pint inches (in3) (l) 1.

1365 Table M-11 Conversions for capacity or volume measure.Ounce (oz) Pound (lb) Short ton Long ton Grams (g) Kilograms (kg) Metric ton 1 0.0161 Table M-12 Conversions for mass or weight measure.2231 1 1 Table M-13 Conversion for pressure.

Square Square Square Square Are Square Square Area Hectare Square inch feet yard mile (a) centimetre metre (ha) kilometre (in2) (ft2) (yd2) (mi2) (cm2) (m2) (km2) 1 0.004 Table M-10 Conversions for surface or area measure.

The following is the relationship between the two scales: When F � 212 F then C �5 ⁄9(212�32) �5 ⁄9 * 180 � 100 C C F F C� � � �59 9 532 32( ) When C � 100 C then F �9⁄5C � 32 �9⁄5 * 100 � 32 � 212 F When C � 35 C then F �9⁄5C � 32 �9⁄5 * 35 � 32 � 95 F When F � 104 F then C �5 ⁄9 (104 � 32) �5 ⁄9 * 72 � 40 C.Conversion between fractions and decimals To convert from a fraction to a decimal representation you divide the numerator or upper value in the fraction, by the denominator or the lower value in the fraction.To convert from a two-decimal presentation to a fraction, you divide the numbers after the decimal point by 100 and simplify until both the numerator and denomi- nator are the smallest possible integer values.That is, you find the lowest common denominator.To convert from a three-decimal presentation to a fraction you divide to numbers after the deci- mal point by 1,000 etc.

Percentages The term percent means per 100 so that 50 per- cent, usually written, 50%, is 50 per 100.In order to change from A fraction to a percentage multiply by 100 3⁄4 � 75% (3⁄4 * 100 � 75%) A decimal to a percentage multiply by 100 0.12% A percentage to a fraction divide by 100 25% � 25/100 � 1⁄4 A percentage to a decimal move the decimal place two places to the left 25.2586 Rules for arithmetic calculations for non-linear relationships Table M-15 gives algebraic operations when the power of the variable, or the constant, is non-linear.Sigma, Very often in statistics we need to determine the sum of a set of data and to indicate this we use the Greek letter sigma, .If we have a set of n values of data of a variable x then the sum of these is written as: x � x1 � x2 � x3 � x4 � … � xn M(i) where x1, x2, x3, etc.

are the individual values in the dataset.If we have a dataset consisting of the values 7, 3, 2, 11, 21, and 9 then the sum of these values is: x � 7 � 3 � 2 � 11 � 21 � 9 � 53 Mean value The mean or average of a dataset is equal to the sum of the individual values, x divided by the number of observations, n: M(ii)x x n � ∑ 444 Statistics for Business Fraction 1/2 1/3 3/4 1/8 7/8 4(7⁄8) 6(2⁄3) 2(3⁄16) Decimal 0.1875 Table M-14 Conversion of fraction and decimals.

If x has values of 1, �5, 9, �2, 6, then n � 5 and, Addition of two variables The total of the addition of two variables is equal to the total of the individual sum of each variable: M(iii) Assume the five values of the dataset (x, y) are, (1, 6); (�5, �2); (9, �8); (�2, 5); (6, 4): Difference of two variables The sum of the difference of two variables is equal to the sum of the individual differences of each variable: M(iv) If we have the following five values, (1, 6), (�5, �2), (9, �8), (�2, 5), (6, 4) Constant multiplied by a variable The sum of a constant times a variable equals to the constant times the sum of the variables: M(v)( )kx k x∑ ∑� ( ) ( ) ( ) ( ) ( ) ( ) ( ) x y x y � � � � � � � � � � � � � � ∑ 1 6 5 2 9 8 2 5 6 4 ∑∑ ∑∑ �� � � � � � � � � � � � � � � � � 5 3 17 7 2 4 1 5 9 2 6 6 2 8 5 x y ( ) ( 44 9 5 4 ) x y� � � �∑∑ ( )x y x y−∑ ∑∑� � ( ) ( ) ( ) ( ) ( ) ( ) ( ) x y x y � � � � � � � � � � � � � � ∑ 1 6 5 2 9 8 2 5 6 4 ∑∑ ∑∑ � � � � � � � � � � � � � � � � � 7 7 1 3 10 14 1 5 9 2 6 6 2 8 5 x y ( ) ( 44 9 5 14 ) x y� � � �∑∑ ( )x y x y� � �∑ ∑∑ x � � � � � � � ( ) .1 5 9 2 6 5 9 5 1 80 445Appendix III: Mathematical relationships Arithmetic rule Example xaxb � x(a�b) 53 * 52 � 5(3�2) � 55 � 3,125 (xa)b � xab (53)2 � 56 � 15,625 1/xa � x�a 1/22 � 2�2 � 0.25 xa/xb � x(a�b) 53/52 � 5(3�2) � 51 � 5 xa/xa � x(a�a) � x0 � 1 62/62 � 6(2�2) � 60 � 1 64 144 64 144 8 12 2 3 0 6667� � � � .x y x y � 16 25 16 25 4 5 20* * *� � �x y x y* *� Table M-15 Algebraic operations involving powers.If k � 5, and x has values of 1, �5, 9, �2, 6, then, Constant summed n times A constant summed n times is equal to the n times the constant: M(vi) If n � 6, and k � 5, Summation of a random variable around the mean Summation rules can be used to demonstrate that the summation of a random variable around a mean is equal to zero.

Or, M(vii) From equation M(iv) equation M(vii) becomes, M(viii) For any fixed set of data, x–, is a constant and thus from equation M(vi), M(ix) Thus from equation M(viii) we have, M(x) From equation M(ii), Or, M(xi) Thus substituting in equation M(x) we have, M(xii) That is the sum of the random variables about the mean is equal to zero.Binary numbering system In the textbook, we introduced the binomial distri- bution.Related to this is the binary numbering system or binary code which is a system of arith- metic based on two digits, zero and one.The on/off system of most electrical appliances is based on the binary system where 0 � off and 1 � on.The arithmetic of computers is based on the binary code.

A digit, either the 0 or the 1 in the binary code is called the bit, or binary digit.When a binary digit is moved one space to the left, and a zero is placed after it, the resulting number is twice the original number.In the binary code, the actual value of 1 depends on the position of the 1 in a binary number, reading from right to left.A digit doubles its value each time it moves one place further to the left as shown in Table M-16.Table M-17 gives the equivalent between binary numbers, and decimal numbers from 1 to 10.

Note, we are reading from the right so that the 1st position is at the extreme right, the 2nd posi- tion to the immediate left of the 1st position, etc.Greek alphabet In statistics, letters of the Greek alphabet are sometimes used as abbreviations to denote various terms.Table M-18 gives the Greek letters, both in upper and lower case and their English equivalent, and some of the areas where they are often used.To write these Greek letters in Microsoft Word you write the English letter and then change the font to symbol.( )x x x x x nx x x � � � � � � � � ∑ ∑ ∑ ∑ ∑ ∑ 0 nx x�∑ x x n � ∑ ( )x x x x x nx� � � � � �∑ ∑ ∑ ∑ 0 x nx∑ � ( )x x x x� � � �∑ ∑∑ 0 ( )x x� �∑ 0 k � � � � � � � � 5 5 5 5 5 5 30 5 6∑ k n k� *∑ ( ) ( kx k x ∑ ∑ � � � � � � � � � � � 5 1 5 5 5 9 5 2 5 6 45 5 1 5 9 2 6)) � 45 446 Statistics for Business 0001 Value of 1 in 1st position is equal to 1 0010 Value of 1 in 2nd position is equal to 2 0100 Value of 1 in 3rd position is equal to 4 1000 Value of 1 in 4th position is equal to 8 Table M-16 Concept of binary numbering system.

447Appendix III: Mathematical relationships Upper Lower Name English Use case case equivalent A Alpha a Hypothesis testing, exponential smoothing B Beta b Hypothesis testing Gamma g Delta d Calculus as the derivative E Epsilon e Z Zeta z H Eta h Theta th (q) Angle of a triangle I Iota i K Kappa k Lambda l Poisson distribution, queuing theory M Mu m Mean value N Nu n Xi x O o Omicron o Pi p Circle constant P Rho r, rh Sigma s Total (upper case), standard deviation (lower case) T Tau t Y Upsilon u , Phi ph (f, j) X Chi ch (c) Chi-square test in hypothesis testing Psi ps (y) Omega o (w) Electricity Table M-18 Greek alphabet and English equivalent.Decimal Binary Explanation 0 0000 1 0001 Value of 1 in the 1st position � 1 2 0010 Value of 1 in the 2nd position � 2 3 0011 Value of 1 in the 1st position � value of 1 in the 2nd position � 3 4 0100 Value of 1 in the 3rd position � 4 5 0101 Value of 1 in the 3rd position � value of 1 in the 1st position � 5 6 0110 Value of 1 in 3rd position � value of 1 in 2nd position � 6 7 0111 Value of 1 in the 3rd position � value of 1 in 2nd position � value of 1 in 1st position � 7 8 1000 Value of 1 in the 4th position � 8 9 1001 Value of 1 in 4th position � value of 1 in 1st position � 9 10 1010 Value of 1 in 4th position � value of 1 in 2nd position � 10 Table M-17 Relationship between decimal and binary numbers.This page intentionally left blank Appendix IV: Answers to end of chapter exercises Chapter 1: Presenting and Organizing Data 1.Solutions – swimming pool 3 Polygon is symmetrical 5 20% 7 (a) From ogive, probability of more than 900 using the pool is 14.Probability of fewer than 600 people using the pool is 5.Thus, both the conditions are satisfied (b) In Part (a) we have indicated that although the criteria was one or the other, we have shown that both conditions apply, or and.Further at the 10% benchmark there will be an estimated 925 people – this is greater than 900.

Also, at the 10% level there will be an estimated to be 640 people – this number is greater than the 600 criterion (c) Probability of at least 600 is 95.

00% and the probability of at least 900 is 14.Alternatively, probability of less than 900 is 85.83% and the probability of less than 600 is 5.

83% Question Answer Question Answer Question Answer 2 For all countries, except Canada where there is no change, the US dollar has become stronger.(It takes more of the local currency to buy dollars in 2005 than it did in 2004) 1 Minimum value is €63,680 so use €60,000.Maximum value is €332,923 so use €334,000.In this way there is an even number of limits 3 15.

50% 4 31 7 85% 8 €151,000 Question Answer 2 Logical minimum value is 16 m and logical maximum is 38 m 3 22% 4 31% Question Answer 450 Statistics for Business 4.Solutions – European sales 4 Best three performing countries are England, Denmark, and Spain with over two-thirds of the profit (67.

04%) 5 Worst performing countries are Ireland, Norway, and Czech Republic with less than 10% of the profits (7.47%) 4 United States (104); France (59); Japan (56) 5 Western Europe (30.Solutions – nuclear power Question Answer Question Answer Question Answer 451Appendix IV: Answers to end of chapter exercises 1 Range of data from maximum to minimum is large so visual presentation of smaller values is not clear 2 Europe (56.

66%); Serbia, Latvia, and Lithuania (each with 0.03%) 6 No sales recorded for the United States.For some reason, the textbook cannot be sold in this country Question Answer 9.

Solutions – textbook sales 2 Times more than China, France (40.00) 4 Explains why the textile firms (and other industries) are losing out to China Question Answer 10.Solutions – textile wages 5 Largest proportion of these new immigrants is from Poland (56.

They are young with over 80% no more than 34-years old.Over a quarter are in administration, business, and management (27.Solutions – immigration to Britain 3 France 21% (20.

69%) 4 France consumes more than twice the amount of pills that is consumed in the United Kingdom Question Answer 12.Solutions – electoral college 452 Statistics for Business 2 Drums damaged (32.Solutions – chemical delivery 2 Fruit squashed (27.Not necessarily as bacteria on the fruit (3.

32%), even though less frequent, is a more serious problem 3 Fruit squashed (27.Solutions – fruit distribution 453Appendix IV: Answers to end of chapter exercises Chapter 2: Characterizing and Defining Data 1.Solutions – production 454 Statistics for Business 5.

Solutions – Euro prices 1 Milk (1l) Renault Big Mac Stamp for Compact disc Can of coke M gane postcard Maximum 1.82% 2 Finland has some of the highest prices.A Big Mac has the lowest deviation relative to the mean value.

Spain’s prices are all below the mean and the median value, which is an indicator of the lower cost of living Question Answer 1 3.Solutions – net worth 455Appendix IV: Answers to end of chapter exercises 1 24.This value occurs 5 times 4 42 5 24 6 147.09% 9 Data is pretty evenly distributed as the mean, median, and mode are quite close Question Answer 9.Solutions – summer Olympics 1 United States (103); Russia (92); China (63); Australia (49); Germany (48); Japan (37); France (33); Italy (32); Britain (30); South Korea (30) 2 United States (35); China (32); Russia (27); Australia (17); Japan (16); Germany (14); France (11); Italy (10); Britain (9); Cuba (9) 3 United States (35.Solutions – Big Mac (Continued ) 456 Statistics for Business 4 0.

92 5 The price of the Big Mac in Indonesia is within the 1st quartile indicating a country with a low cost of living; Singapore is within the 2nd quartile and the inter-quartile range – indicating a relative low cost of living.The price for Hungary is within the 3rd quartile and the inter-quartile range and has a relative high cost of living.The price for Denmark is within the 4th quartile and indicates a high cost of living country Question Answer 1 (a) 332,923; (b) 63,680; (c) 269,243; (d) 198,302; (e) 184,767.00; (g) 194,157; 3 values; (h) 3,105,927,434; (i) 55,730.

00 4 Mean is greater than the median and so the distribution is right-skewed 5 0% � 63,680.Solutions – purchasing expenditures – Part II 1 (a) 120; (b) 1,088; (c) 507; (d) 581; (e) 797.63 3 It is reasonably symmetrical Question Answer 14.Solutions – swimming pool – Part II (Continued ) 12.Solutions – (Continued ) 457Appendix IV: Answers to end of chapter exercises 4 Percentile Value Percentile Value Percentile Value Percentile Value (%) (%) (%) (%) 0 507.Solutions – (Continued ) 458 Statistics for Business Chapter 3: Basic Probability and Counting Rules 1.Solutions – packing machines (Continued ) 459Appendix IV: Answers to end of chapter exercises 3 8.Solutions – sub-assemblies (Continued ) 4.

Solutions – (Continued ) 460 Statistics for Business 4 1.Solutions – bicycle gears 1 Values for each of the three columns are as follows: (A) 1; 1; 1 (B) 1; 2; 2 (C) 2; 3; 6 (D) 2; 4; 8 (E) 2; 5; 10 (F) 3; 4; 12 (G) 3; 7; 21 (H) 4; 7; 28 (I) 4; 8; 32 (J) 4; 9; 36 Question Answer 12.Solutions – film festival 1 6,724,520 2 33,891,580,800 3 (a) 1,184,040; (b) 116,280; (c) 3,432; (d) 1 4 (a) 5,967,561,600; (b) 586,051,200; (c) 17,297,280; (d) 5,040 Question Answer 13.08889 1028 (There are 27 countries in the European Union as of 2007) 2 8,436,285 Question Answer (Continued ) 8.Solutions – (Continued ) 461Appendix IV: Answers to end of chapter exercises 3 80,730 4 3.Solutions – model agency 1 54,264 2 1,307,674,368,000 3 70,959,641,905,152,000 Question Answer 15.

Solutions – thalassoth rapie 1 40,320 2 24 3 479,001,600 4 56 Question Answer The following is a possible schedule for Question 4.

Note, here that the treatment, the enveloppement d’algues and application de boue marine are not put together.Even though they are different treat- ments, they have similarities regarding their benefits 13.bain hydromassant 462 Statistics for Business Chapter 4: Probability Analysis for Discrete Data 1.Solutions – HIV virus 1 $3,843,750 2 No change.Remains the same at €3,843,750 3 Table 1 � 1.

69% 4 Data from Table 2 as it shows less dispersion Question Answer 2.50 cars 2 $151,250 3 No change 4 No change 5 Laramie coefficient of variation at 7.64% which is high 5 £505,470 6 £1,018,161 7 £1,523,631 Question Answer 3.

67% for three packages not being delivered on-time 3 4.33 6 There are high occurrences of non-deliveries in the winter months, which might be explained by bad weather.Also high occurrences in the summer period perhaps because of high volume of air traffic Question Answer 463Appendix IV: Answers to end of chapter exercises 5.50 (opportunity cost) 4 Expected values are long-term and may not necessarily apply to the immediate following month Question Answer 1 $137.Solutions – European business school 464 Statistics for Business 9.Solutions – biscuits (Continued ) 465Appendix IV: Answers to end of chapter exercises 5 83.The owner should not be satisfied as the percentage of 12 or more using the pump is only 81.52% 3 Only have one attendant at the exit kiosk.This would reduce operating costs by about 50%.If this means a longer wait for customers they may prefer to use the pumps that take credit card purchase, or they may go elsewhere! Provide an incentive to use the cash-for-gas utilization, such as lower prices, a gift.

However, this would only be a suitable approach if overall more customers use the service station, and the method does not just siphon off customers who before used credit card for purchases Question Answer 2 10.42% 9 They both tail-off rapidly to the right.Since P � 5% and sample size is greater than 20 it is reasonable to use the Poisson–binomial approximation Question Answer 15.Solutions – (Continued ) 466 Statistics for Business Chapter 5: Probability Analysis in the Normal Distribution 1.87% 4 42,526 5 Yes; 126,393 km 6 258,159 km Question Answer 1 6.

74% 5 About 38 and 74 days Question Answer 3.69 gm Question Answer 467Appendix IV: Answers to end of chapter exercises 5.Solutions – publishing 1 Almost none (0.95% 4 4,912 litre 5 35,457 litre Question Answer 6.

23 mm 7 Slender, with a sharp peak or leptokurtic as the coefficient of variation is small, 0.36% Question Answer (Continued ) 468 Statistics for Business 4 About 26,755,27 5 336.78% Question Answer 1 Average service time is 125.Solutions – doors 1 196 cm 2 About 3% (3.

Solutions – (Continued ) 469Appendix IV: Answers to end of chapter exercises 1 2.45) 2 The values are very close 3 When data follows closely a normal distribution it is not necessary to construct the ogives for this type of information Question Answer 15.Solutions – buyout – Part III 470 Statistics for Business 1 27.

81% 4 Larger the sample size the data clusters around the mean 5 That the central limit theory applies and the sample means follow a normal distribution Question Answer Chapter 6: Theory and Methods of Statistical Sampling 1.19% 5 Since we have a sample of size 10 rather than just a size 1 from the population, data clusters around the mean.In Questions 1 and 3, the percentage increases as the limits are around the mean.In Questions 2 and 4, the percentage decreases as the limits are away from the mean 6 Normal as the population is normal Question Answer 2.38% 7 With larger sample sizes data lies closer to population mean Question Answer 3.Solutions – telephone calls 471Appendix IV: Answers to end of chapter exercises 4.91% 7 With larger sample sizes data lies closer to population mean 8 For Questions 1 to 3 limit values are on the right side of the mean value.For Questions 4 to 6 the limit values are on either side of the mean values Question Answer 5.15 minutes 7 As the sample size increases there are the probability of being beyond 37 minutes declines as more values are clustered around the mean value 8 The population from which the samples are drawn follows a normal distribution Question Answer 6.Solutions – financial advisor 472 Statistics for Business 1 5.

22 cm 7 With increase in sample size data lies closer to the mean value Question Answer 7.35% 7 The larger the sample size, the closer is the data clustered around the population proportion.Note, that in each case the percentages given are on either side of the population proportion Question Answer 10.Solutions – education and demographics 473Appendix IV: Answers to end of chapter exercises 1 56.37% 7 The larger the sample size, the closer is the data clustered around the population proportion.

Note, that in each case the percentages given are on either side of the population proportion Question Answer 11.Solutions – World Trade Organization 1 88.11% 7 Larger the sample size, the data clusters around the population mean 8 No.Istanbul is the commercial area of Turkey and the people are educated.The sample taken would be biased if it was meant to represent the whole of Turkey Question Answer 12.

Solutions – unemployment (Continued ) 474 Statistics for Business 7 29.02% 9 As the limits given are on either side of the population proportion the percentages increase with sample size 10 The limits for France are on the left side of the distribution and as the sample size increases the percentage here declines Question Answer 1 68.

23% 7 The larger the sample size, the more the data is clustered around the mean 8 You might say that the German data for the population is an underestimation.Perhaps a more palatable reason could be that the sample was taken from an industrial area of Germany and thus the sample is biased Question Answer 14.Solutions – manufacturing employment 1 0.002%; you have a 30 times more chance of being shot in Jamaica than in Britain 3 0.

93% 5 The larger the sample size, the closer the values are to the mean Question Answer 15.Solutions – (Continued ) 475Appendix IV: Answers to end of chapter exercises 1 496.32 g 3 The higher the confidence intervals, the broader are the limits 4 About 3 cases (between 61 and 62 cases) 5 A case of 20 bottles may not be random since they probably come off the production line in sequence Question Answer Chapter 7: Estimating Population Characteristics 1.

74 2 I estimate that the mean life of these light bulbs is about 473 hours (473.46) and I am 80% confident that the mean life will lie in the range 394 (394.02 5 Higher the certainty, broader the range Question Answer 2.

18 3 The higher the confidence level, the broader are the limits Question Answer 3.

Solutions – ski magazine 1 Lower limit £11.63 4 The higher the confidence level, the wider are the limits Question Answer 4.Solutions – households 476 Statistics for Business 1 At 80%, lower limit is 13,923.520 2 I estimate that the mean value of the tax returns for the year in question is $20,600,00 and I am 95% confident that they will lie between $9,954 and $31,246 3 More confident you are about information (closer to 100%) the wider is the interval Question Answer 5.36 2 I estimate that in my vineyard this year I will have an average of 15 grape bunches per vine and I am 95% confident that this number will lie between 13.75 4 The higher the confidence level, the broader are the limits Question Answer 7.

53 million 5 Higher the certainty, broader the range 6 $38,954.Solutions – world’s largest companies (Continued ) 477Appendix IV: Answers to end of chapter exercises 9 $19,337.59 million 10 Higher the certainty, broader the range 11 For Questions 1 to 4 we have used a larger sample size than for Questions 6 to 9 and by the central limit theory, the mean values determined are probably closer to the population value.For the first four questions we have a sample size of 35 and a population of 500, which gives a value of n/N of 7% so we should use the finite correction multiplier.

In Questions 6 to 9 n/N is 3% so the finite population multiplier is not necessary.However, we have used a Student-t distribution as the sample size is less than 30 Question Answer 1 0.1825 5 The higher the confidence level, the broader is the range Question Answer 9.45 using z 3 My best estimate of the total value of the tyres in inventory is €155,083 and I am 95% confident that the value lies between €136,532 and €173,635 4 €130,081.

74 using z 5 The higher the confidence level, the broader are the limits 6 Use random number related to the database of the tyres in inventory.Just selecting tyres from the racks at random may not result in a representative sample Question Answer 10.

Solutions – stuffed animals (Continued ) 8.Solutions – (Continued ) 478 Statistics for Business 3 $3,645.

78 5 Higher the confidence, broader the limits Question Answer 1 0.

0177 4 27,056 5 54,119 6 The conservative value requires a very high inspection quantity but since the inspection process is automatic perhaps this is not a problem.Note that this is not a random inspection as the device samples every bottle Question Answer 12.

77% 6 The higher the confidence limits, the broader is the range Question Answer 13.Solutions – (Continued ) 479Appendix IV: Answers to end of chapter exercises 4 The higher the level of confidence, the broader are the limits 5 1,691 6 3,383 Question Answer 1 0.13% 4 The higher the level of confidence, the broader are the limits 5 68 6 136 7 There will be hotels in the Southern and Northern hemisphere and since hotel occupancy is seasonal, the data that is taken can be distorted Question Answer 15.Solutions – (Continued ) 480 Statistics for Business 1 No.8762, which is within the critical boundaries.5% Alternatively, the total p-value is 6.06% which is less than the -value of 5% 3 Lower value is 999.

The target value of 1,000 g is contained within these limits 4 Yes.8762, which is outside the critical boundaries.06% which is less than the -value of 10% 6 Lower value is 1,000.The target value of 1,000 g is not contained within these limits 7 The firms should not be under filling the bags as this would not be abiding by the net weight value printed on the label.In this case it appears that the bags are overfilled and this is costing B ghin Say money.As an illustration assume that 1 million of these bags are sold per year and they in fact contain 6 g more of sugar.This is equivalent to 6,000 kg bags of sugar.50/bag this would be a loss of an estimated €3,000/year from this production line Question Answer Chapter 8: Hypothesis Testing for a Single Population 1.6771 which is inside the critical boundary limits.

35% which is greater than the value of 5.0% 5 The purchaser can refuse the purchase and go to another supplier.Alternatively the purchaser can negotiate a lower price Question Answer 2.Thus test value is within the boundaries of the critical value.5% in each tail Question Answer (Continued ) 481Appendix IV: Answers to end of chapter exercises 2 p-value is 3.5% we accept the null hypothesis 3 Lower limit is 0.

7000 mm is contained within these limits 4 Yes.8396 and the critical value of z is now �1.Test value is outside boundaries of the critical value 5 p-value is 3.29% and since this is less than the value of 5% we reject the null hypothesis.7000 mm is not within these limits 7 Since sample mean of 0.

7168 mm is greater than the hypothesized value of 0.7000 mm it implies that the diameter is on the high side.Thus we would use a right-hand, one-tail test.At both 5% and 10% limits the null hypothesis should be rejected as test statistic is greater.80%, which is greater than 5% but less than 10% 6 The hypothesis testing could be performed with a smaller sample size which would be less expensive.However note that at the 5% level the results between the test and critical value are close Question Answer 4.Alternatively we can say that the p-value is 4.00 is not contained within these limits Question Answer 5.Solutions – automatic teller machine (Continued ) 3.Solutions – (Continued ) 482 Statistics for Business 4 No.

The test value is within these boundaries.Alternatively, we can say the p-value is 4.The value of €3,200 is contained within these limits 7 If the bank has significantly less in the machines there is a risk of the machines running out of cash and this is poor customer service.Alternatively, if the bank puts significantly too much cash in the machines that is underutilized this money could be invested elsewhere to earn interest Question Answer 1 No.

This is a left-hand, one-tail test with 5.This is a left-hand, one-tail test with 10.0% 5 From the data, the sample standard deviation is 2.54 cm, which gives a sample statistic of �1.At a significance level of 5% the Student-t value is �1.

At a significance level of 10% the Student-t value is �1.Thus there is no change to the conclusions from Questions 1 to 4 Question Answer 6.The nominal value of 1,000 ml is contained within these intervals 4 Yes.0% 6 If as the test suggests there is significantly less than the 1,000 ml as indicated on the label, the manufacturer is not being honest to the customer and they could be in trouble with the control inspectors 7 Very sensitive as small changes in volume will swing the results either side of the barrier limits Question Answer 7.Solutions – (Continued ) 483Appendix IV: Answers to end of chapter exercises 1 No.The value of 200 g is contained within these intervals 4 Yes.Thus test value is within the boundaries of the critical value 2 p-value is 6.15% and since this is greater than the value of 5% we accept the null hypothesis 3 Yes.0330 and the critical value of t is now �1.

Thus test value is outside the boundaries of the critical value – it is higher 4 p-value is 3.07% and since this is less than the value of 5% we reject the null hypothesis 5 In Questions 1 and 2 we are testing to see if there is a difference thus the value of is 5% but there is only 2.5% in each tail which gives a relatively high value of t.In Questions 3 and 4, we are still testing at the 5% significance level but all of this area is the right tail.

Thus t is smaller than for the two-tail test and explains the shift from acceptance to rejection Question Answer 9.The value of 10 minutes is contained within these intervals 4 Yes.0% 6 The test for being greater than 10 minutes.If this is the case, the hospital is not meeting its objectives with the risk to the life of patients Question Answer 10.Solutions – hospital emergency 484 Statistics for Business 1 Accept the null hypothesis.5% in each tail 2 In each tail the p-value is 0.38% which is greater than the -value of 1% 3 Limits are 15.The value of 45% is within these limits so verifies that we should accept the null hypothesis 4 Reject the null hypothesis.

5% in each tail 5 In each tail the p-value is 0.

38% which is less than the -value of 5% 6 Limits are 19.The value of 45% is outside these limits and so verifies that we should reject the null hypothesis 7 At the significance level of 1% the test would support the conclusions.At the significance level of 5% indications are that the pay gap is more than 45% Question Answer 11.Solutions – equality for women 1 From the table proportion in Italy imports from Russia is 34.7711 which is inside the critical boundaries.5% in each tail 2 In each tail the p-value is 3.65% which is greater than the -value of 5% 3 Limits are 4.

20% is within these limits so verifies that we should accept the null hypothesis 4 Reject the null hypothesis.7711 which is outside the critical boundaries.0% in each tail 5 In each tail the p-value is 3.65% which is less than the -value of 10% 6 Limits are 6.02% is not within these limits so verifies that we should reject the null hypothesis 7 From the table proportion Poland imports from Russia is 86.3074 which is inside the critical boundaries.5% in each tail 8 In each tail the p-value is 37.85% which is greater than the -value of 5% Question Answer 12.Solutions – gas from Russia (Continued ) 485Appendix IV: Answers to end of chapter exercises 9 Limits are 77.

81% is within these limits so verifies that we should accept the null hypothesis 10 Accept the null hypothesis.3074 which is inside the critical boundaries.0% in each tail 11 In each tail the p-value is 37.85% which is greater than the -value of 10% 12 Limits are 79.81% is within these limits so verifies that we should accept the null hypothesis 13 From this data it appears that Italy is tending to use less than the contractual amount of natural gas from Russia.It perhaps can easily call on imports from say Libya or other North African countries.On the other hand Poland seems to be using close to or more than the contractual amount of natural gas from Russia.This is perhaps because historically and geographically Poland is closer to Russia and is not yet in a position to exploit other suppliers Question Answer 1 Accept the published value or the null hypothesis.0710 which is inside the critical boundaries.5% in each tail 2 In each tail the p-value is 1.84% which is greater than the -value of 1% 3 Limits are 16.0% is within these limits so verifies that we should accept the null hypothesis 4 Reject the null hypothesis.0710 which is outside the critical boundaries.

5% in each tail 5 In each tail the p-value is 1.84% which is less than the -value of 5% 6 Limits are 19.0% is not within these limits so verifies that we should reject the null hypothesis 7 Accept the published value or the null hypothesis.

2546 which is inside the critical boundaries.5% in each tail 8 In each tail the p-value is 1.

42% which is greater than the -value of 1% 9 Limits are 9.5% is within these limits so verifies that we should accept the null hypothesis Question Answer 13.Solutions – international education (Continued ) 12.Solutions – (Continued ) 486 Statistics for Business 10 Reject the null hypothesis.

2546 which is outside the critical boundaries.5% in each tail 11 In each tail the p-value is 1.

42% which is less than the -value of 5% 12 Limits are 12.5% is not within these limits so verifies that we should reject the null hypothesis Question Answer 1 Accept the published value or the null hypothesis.0341 which is inside the critical boundaries.5% in each tail 2 In each tail the p-value is 48.28% which is greater than the -value of 5% 3 Limits are 0.90% is within these limits so verifies that we should accept the null hypothesis 4 Accept the published value or the null hypothesis.0341 which is inside the critical boundaries.0% in each tail 5 In each tail the p-value is 48.28% which is greater than the -value of 10% 6 Limits are 0.

90% is within these limits so verifies that we should accept the null hypothesis 7 Reject the published value or the null hypothesis.

9998 which is outside the critical boundaries.5% in each tail 8 In each tail the p-value is 2.55% which is less than the -value of 5% 9 Limits are 4.90% is not within these limits so verifies that we should reject the null hypothesis 10 Reject the published value or the null hypothesis.9998 which is outside the critical boundaries.

0% in each tail 11 In each tail the p-value is 2.55% which is less than the -value of 10% Question Answer 14.Solutions – US employment (Continued ) 13.Solutions – (Continued ) 487Appendix IV: Answers to end of chapter exercises 12 Limits are 5.90% is not within these limits so verifies that we should reject the null hypothesis 13 The data for Palo Alto indicates low levels of unemployment.This region is a service economy principally in the hi-tech sector.People are young and are more able to move from one company to another.Detroit is manufacturing and this sector is in decline.

People are older and do not have flexibility to move from one industry to another Question Answer 15.Solutions – Mexico and the United States 1 0.10% 2 Accept the indicated value or the null hypothesis.9102 which is inside the critical boundaries.5% in each tail 3 In each tail the p-value is 2.61% which is greater than the -value of 5% 4 Limits are 7.00% is within these limits so verifies that we should accept the null hypothesis 5 Reject the indicated value or the null hypothesis.9102 which is outside the critical boundaries.

0% in each tail 6 In each tail the p-value is 2.61% which is less than the -value of 10% 7 Limits are 9.

00% is not within these limits so verifies that we should reject the null hypothesis 8 Reject the published value or the null hypothesis.7059 which is well outside the critical boundaries.5% in each tail 9 In each tail the p-value is 0% which is less than 2.Alternatively the p-value total is 0% which is less than the -value of 5% 10 Limits indicate �9.The value of 60% is far from being within these limits so verifies that we should soundly reject the null hypothesis 11 Reject the published value or the null hypothesis.7059 which is well outside the critical boundaries.0% in each tail Question Answer (Continued ) 14.Solutions – (Continued ) 488 Statistics for Business 12 In each tail the p-value is 0% which is less than 5.

Alternatively the p-value total is 0% which is less than the -value of 10% 13 Limits indicate �7.The value of 60% is not within these limits so verifies that we should reject the null hypothesis 14 Random sampling for verifying that an individual is from Mexico can only be done from census data.

Taking a random sample of foreign-born individuals would be biased.To an interviewer, and undocumented alien is unlikely to be truthful and nothing says that the random sample were documented, or undocumented.Exercise underscores difficulty of obtaining reliable information about illegal immigrants Question Answer 15.Solutions – (Continued ) 489Appendix IV: Answers to end of chapter exercises 1 Null hypothesis is H0: 1 � 2.Alternative hypothesis is H1: 1 � 2, where 1 is the mean value in January 2006 and 2 is the mean value in June 2006 2 Yes.

At 2% significance, the critical value of z is �2.3263 and the sample test statistic is �2.8816 3 Significance level is 2% and the p-value is 0.At 5% significance, the critical value of z now becomes �1.The sample test statistic remains at �2.8816 5 Significance level is 5% and the p-value is still 0.1979% 6 In 2006 the price of crude oil rose above $70/barrel.

Gasoline is a refined product from crude oil Question Answer Chapter 9: Hypothesis Testing for Different Populations 1.Solutions – gasoline prices 1 Null hypothesis is H0: S � I.Alternative hypothesis is H1: S � I, where S is the mean value for Spain and I is the mean value for Italy.At 1% significance, the critical value of z is �2.

Accept the null hypothesis 3 Significance level is 1% with 0.At 5% significance, the critical value of z is now �1.9600 and the sample test statistic is still 2.Reject the null hypothesis 5 Significance level is 5% with 2.Alternative hypothesis is H1: S � I, where S is the mean value for Spain and I is the mean value for Italy.

At 1% significance, the critical value of z is �2.Reject the null hypothesis 8 Significance level is 1% with 1.

Solutions – tee shirts 1 Null hypothesis is H0: F � I.Alternative hypothesis is H1: F � I, where F is the mean value for stores using FAX and I is the mean value for those using Internet.

At 1% significance, the critical value of Student-t is 2.4999 and the sample test statistic t is 2.Solutions – inventory levels (Continued ) 490 Statistics for Business 3 Significance level is 1% and the p-value is 1.

At 5% significance, the critical value of Student-t is 1.7139 and the sample test statistic t is still 2.4320 5 Significance level is 5% and the p-value is still 1.16% 6 The statistical evidence implies that when orders are made by FAX, the delivery is longer.

For this reason, those stores using FAX keep a higher coverage of inventory in order to minimize the risk of a stockout.Perhaps FAX orders sit in an in-tray before they are handled.However Internet orders are normally processed immediately as they would go into the supplier’s database Question Answer 1 Null hypothesis is H0: D � S.Alternative hypothesis is H0: D � S, where D is the mean value using database ordering and S is the mean value for standard method.Since we are asking is there a difference, this is a two-tail test 2 No.

At a 1% significance level, accept the null hypothesis.8784 and the sample test statistic t is 4.5% in the tail and the sample p-value is 0.Alternative hypothesis is H0: D � S � 1.

This is now a one-tail test 5 Using this new criterion we reject the null hypothesis.The critical value of Student-t is now 2.5524 and the sample test statistic t is still 2.This is a one-tail test 6 Significance level is 1% and the sample p-value is 0.

87% (a one-tail test) 7 The investment is too much to install a database system.It could be that they prefer the manual system as they can more easily camouflage real revenues for tax purposes! Question Answer 4.Solutions – restaurant ordering 1 Average store sales of those in the pilot programme before is £260,000/store.Thus benchmark is an increase of 10% or £26,000/store 2 Null hypothesis H0: 1 � 2 � £26,000.Alternative hypothesis H1: 1 � 2 � £26,000 3 No.

7638 4 Significance level is 1% and p-value is 1.Solutions – sales revenues (Continued ) 3.Solutions – (Continued ) 491Appendix IV: Answers to end of chapter exercises 6 Significance level is 5% and p-value is 1.04% 7 Even at 1% significance the results are very close to the critical test level.

Retail sales are very sensitive to seasons so we need to be sure that the sampling data was carried out in like periods Question Answer 1 Null hypothesis H0: 1 � 2 � 10%.Alternative hypothesis H1: 1 � 2 � 10% 2 No.8965 3 Significance level is 1% and p-value is 13.

1081 5 Significance level is 15% and p-value is still 13.30% 6 That depends if these new yield rate levels are sustainable and if the advertising cost and the reduction in price levels have led to a net income increase for the hotels.

Also, note that a high significance level is used before we can say that the objectives have been reached Question Answer 6.Solutions – hotel yield rate 1 Monthly average of the migraine attacks before is 24.8214 3 Significance level is 1% and sample p-value is 36.3830 5 Significance level is 10% and sample p-value is still 36.

69% 6 About a 60% reduction would be required.28% reduction this gives a mean value afterwards of 9.99% 7 Patients may experience less attacks knowing they are undertaking a test (physiological effect).Also eliminating coffee may reduce or increase stress or may reduce or increase consumption of sugar so all variables have not been isolated Question Answer 7.

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Solutions – (Continued ) 492 Statistics for Business 1 Null hypothesis is H0: p6 � p1.

The indices refer to the year the test was made.86% in each tail 4 Null hypothesis is H0: p6 � p1 **Should i purchase corruption powerpoint presentation confidentially US Letter Size Platinum single spaced Custom writing**.86% in each tail 4 Null hypothesis is H0: p6 � p1.

The indices refer to the year the test was made.This is a one-tail, right-hand test 5 Yes.86% in the tail 7 The data results are close **SMS Gateway Enterprise 3 12 15 with Product keys Programs Crack nbsp**.86% in the tail 7 The data results are close.The fitness craze seems to have leveled off in this decade.Other variables that need to be considered in the sampling are the gender of customers, and whether the customers at the hotel are for business or pleasure? Question Answer 8 elrubius.es/homework/should-i-purchase-custom-advertising-homework-american-premium-a4-british-european-professional.Other variables that need to be considered in the sampling are the gender of customers, and whether the customers at the hotel are for business or pleasure? Question Answer 8.Solutions – hotel customers 1 Null hypothesis is H0: p05 � p96.

Alternative hypothesis is H0: p05 � p96.The indices refer to the year the test was made.63% in each tail 6 Null hypothesis is H0: p05 � p96.Alternative hypothesis is H0: p05 � p96.The indices refer to the year the test was made.This is a one-tail test 7 Reject the null hypothesis.

63% or the p-value of the sampling experiment 9 The experiment could be defined better.

What is a major European airport? When was the sampling made – winter or summer? In winter, there are weather problems.

Certainly though airplane traffic has increased significantly and delays have increased.A 20-minute delay may be excessive for some passengers Question Answer 9.Solutions – World Cup 1 Null hypothesis is H0: p98 � p06.

Alternative hypothesis is H0: p98 � p06.The indices refer to the year the test was made.This is a two-tail test Question Answer (Continued ) 493Appendix IV: Answers to end of chapter exercises 2 No.74% in each tail 6 Null hypothesis is H0: p06 � p98.Alternative hypothesis is H0: p06 � p98.The indices refer to the year the test was made.Note, here we have reversed the position and put the year 2006 first.Alternative hypothesis is H0: p98 � p06 though this does not fit with the way the question is phrased 7 Reject the null hypothesis.What was the gender of the people sample? (Men are more likely to be enthusiastic about the World Cup than omen.) The country surveyed is also important.

Not all European countries have a team in the World Cup Question Answer 1 Null hypothesis H0: pHS � pMS � pLS.Alternative hypothesis H1: pHS � pMS � pI � pLS where the indices refer to the stress level 2 No.5604 3 Significance level is 1% and p-value is 4.

4877 and sample chi-square value is still 9.5604 5 Significance level is 5% and p-value is 4.Percentage returns are 93% (186/200) 7 Stress may not only be a result of travel but could be personal reasons or work itself Question Answer 11.Solutions – travel time and stress 1 Null hypothesis H0: pU � pG � pI � pE.Alternative hypothesis H1: pU � pG � pI � pE where the indices refer to the first letter of the country 2 No.It seems that the investment strategy of individuals is independent of the country of residence.

3449 3 Significance level is 1% and p-value is 1.Solutions – investing in stocks (Continued ) 10.Solutions – (Continued ) 494 Statistics for Business 4 Yes.

It seems that the investment strategy of individuals is dependent on the country of residence.9473 5 Significance level is 3% and p-value is still 1.22% 6 According to the sample, nearly 62% of individuals invest in stocks in the United States while in the European countries it is between 50% and 52%.

The higher level in the United States is to be expected as Americans are usually more risk takers and the social security programmes are not as strong as in Europe Question Answer 1 Null hypothesis H0: pG � pF � pE � pI � pS.Alternative H1: pG � pF � pE � pI � pS where the indices refer to the country 2 Yes.7764 3 Significance level is 1% and p-value is 0.

12% the p-value of the sample data 5 Results are not surprising.Even though we have the European Union people are still nationalistic about certain things including the origin of the products they buy Question Answer 13.Solutions – automobile preference 1 Null hypothesis H0: p1 � p2 � p3 � p4 � p5.Alternative H1: p1 � p2 � p3 � p4 � p5 where the indices refer to the salary category 2 No.

81% 6 There are many variables in this sample experiment for example the country of readership, the newspaper that is read, and the profession of the reader.

Depending on the purpose of this analysis another sampling experiment should be carried out to take into account these other variables Question Answer 14.Solutions – (Continued ) 495Appendix IV: Answers to end of chapter exercises 1 Null hypothesis H0: pE � pF � pI � pS � pU.Alternative H1: pE � pF � pI � pS � pU where the indices refer to the first letter of the country 2 No.It seems that the amount of wine consumed is independent of the country of residence.

2170 3 Significance level is 1% and p-value is 2.00% since the p-value of the sample is 2.

7418 6 In Europe wine consumption per capita is down but wine exporters such as South Africa, Chile, Australia, and California are up – at the expense of French wine! Question Answer 15.Solutions – wine consumption 496 Statistics for Business Chapter 10: Forecasting and Estimating from Correlated Data 1.895 incidents per 200,000 hours worked 4 The coefficient of determination, r2, which is 0.9391 or the coefficient of correlation of �0.

Note that the coefficient of correlation has the same negative sign as the slope of the regression line 5 0.11 reported incidents per 200,000 hours worked 6 �0.16 reported incidents per 200,000 hours worked.We cannot have negative reported incidents of safety 7 It suggests that the safety record of ExxonMobil has reached its minimum level and if anything will now decline in an exponential fashion.Using the linear regression equation to forecast for 2009 already shows a negative 0.

07 incidents per 200,000 hours worked Question Answer 2 y9002x 3 This might correspond to increased purchases due to back-to-school period 4 r2 is 0.9689 which indicate a good correlation so there is a reasonable reliability for the regression model 5 £2,701 increase per quarter 6 June 2007 is £66,057.This is most reliable as it is closer period to actual data.December 2008 is £82,261 and December 2009 is £93,064 7 Forecasting using revenues has a disadvantage in that it covers up the increases in prices over time.From an operating point of view it may be better to report unit sales in order to see those items which are the major revenue contributors Question Answer 2.59x � using the years as the x-values 3 The coefficient of determination, r 2.Since it is quite close to unity, the regression line seems to be a reasonable predictor for future values.9649 4 A reduction of about 246 deaths per year 5 4,983 Question Answer 3.Solutions – road deaths (Continued ) 497Appendix IV: Answers to end of chapter exercises 6 71 7 The value of 4,983 for Question No.5 is achievable with a lot of effort on the part of drivers and national and local authorities.

These values indicate the danger of making forecasts beyond the collected data and way into the future Question Answer 1 y0000 2 Coefficient of determination is 0.8822 and the coefficient of correlation is 0.For the data given they are greater than 0.8 and indicate a reasonable relationship 3 28.

54 millions of metric tons of carbon per year 4 2,120 millions of metric tons of carbon per year 5 Between 2,043 and 2,198 million metric tons of carbon equivalent.To be correct we use the Student-t distribution.6 2,406 millions of metric tons of carbon per year 7 Worldwide there is a lot of discussion and action on reducing carbon dioxide emissions, so that the rate of increase may decline and thus making forecasts using the given data may be erroneous Question Answer 4.Solutions – carbon dioxide 2 Coefficient of determination, r2 � 0.8627 or coefficient of correlation, r � �0.

A value of 20 for x is beyond the range of the collected data.

This would be an unacceptable internal operating situation (36% of employees are absent).

Further, externally many clients would not wait over an hour to get served 8 People order different meals, and different cooking conditions (Rare, medium, and well cooked steaks).Planning not optimum in the kitchen or in the dining room.Kitchen staff not polyvalent so bottlenecks for some dishes.Clients change their minds Question Answer 5.Solutions – (Continued ) 498 Statistics for Business 2 Regression equation is yThe relationship is strong 3 28 units/day 4 65 units/day.The shelf space of 5 m2 is beyond the range of the data collected.Is it realistic for the store? 5 The extent of product exposure has an impact on sales Question Answer 6.

Solutions – product sales 2 Regression equation is, y5054, relationship is not strong 4 Regression equation is, y

9156, relationship is strong 5 About 21 million (20.However this profile cannot be maintained indefinitely Question Answer 7.Coefficient of multiple determination r2 is 0.9640 which indicates a strong relationship 2 £485,649 3 Lower limit is £381,835 and upper limit is £589,463 4 yCoefficient of multiple determination r2 is 0.9650, a strong relationship 5 £477,485 6 Lower limit is £366,799 and upper limit is £588,171 Question Answer 8.Solutions – hotel revenues (Continued ) 499Appendix IV: Answers to end of chapter exercises 4 $116.

90 million 10 2008 is closer to the period when the data was measured and so is more reliable.The further away in the time horizon that forecasts are made, the more difficult it is to make a forecast Question Answer 1 The number of shares more than doubles in mid-1996 and mid-2004 indicating a stock split in those periods 3 y

7610 4 The coefficient of determination, r2.8 the regression line is a reasonable predictor for future values 5 $2,931/year 6 $90,090 7 $79,248,61 and $100,930,94 8 A severe economic downturn which could depress stock prices 9 Probably not significantly reduced.Hershey has been around for a long time.People will continue to eat chocolate and related foods.Hershey is expanding their market into Asia so that it should help to keep up their revenues Question Answer 10.

Solutions – compact discs (Continued ) 9.Solutions – (Continued ) 500 Statistics for Business 7 844.

Note the value obtained depends on how many decimal places are used for the coefficient.Answers to Question 6 and 7 use eight decimal places 8 Using polynomial relationships implies that we can reach infinite values.The answer to Question 7 is almost twice that of Question 4.In fact sales of CDs are now declining with Internet downloading of music and further, there is evidence that the CD, introduced in 1982, will soon be replaced by other technologies.

1 1 At 25, compact disc faces retirement, International Herald Tribune, 17 August 2007, p.9767 3 Using the actual data for 1960–1964; forecast for 2006 is $61,998.60 million Using the actual data for 1965–1969; forecast for 2006 is $169,292.20 million Using the actual data for 1975–1979; forecast for 2006 is $962,778.00 million Using the actual data for 1985–1989; forecast for 2006 is $1,088, 270.

60 million Using the actual data for 1995–1999; forecast for 2006 is $1,488,250.00 million Using the actual data for 2002–2005; forecast for 2006 is $1,836,638.00 million 4 The actual value for 2006 is $1,861,380 million.Even though the coefficients of regression determined in Question 2 show a strong correlation for the time series data, only using the data for 2002–2005 is a close agreement (a difference of some 1%).

This shows that the closer your actual data is to the forecast there is a better reliability.Also, you must know the business environment – in this case there has been a significant change in the volume of US imports since 1960 5 1960�2006; y8000; r2 � 0,9935 8 Linear: $1,345,457 million; exponential: 3,860,721; polynomial: 284,264,832 9 $2,573,675 million 10 As the years move on, the slope of the curve increases.Thus using linear regression based on earlier period data is not useful.From this we might conclude that the forecast developed from this relation for 2010 is the most reliable Question Answer 12.

Solutions – (Continued ) 501Appendix IV: Answers to end of chapter exercises 1 There is a seasonal variation with highest sales in the spring and summer which is to be expected.Also, sales are increasing over time 2 yThis means that for 2004 winter sales are 80.9031 6 Winter 328,017; spring 2,551,173; summer 3,580,963; fall 149,200 litre 7 8,402,514 litre in 2010.

There is a lot of concern about the amount of beer consumed by the English and the hope is that consumption will decline or at least not increase Question Answer 13.Solutions – English pubs 2 Q1: 14,919; Q2: 15,750; Q3: 16,544; Q4: 15,021 Question Answer 14.Solutions – Mersey Store 1 There is a seasonal variation and sales are increasing over time 2 y7796 � 364,7968x 3 Ratios for 2005 are winter 0.

This means that for 2005 winter sales are 74.

1120 5 Winter 3,302; spring 25,905; summer 22,598; fall 1,555 units 6 Unit sales are more appropriate for comparison.If dollar sales are used this masks the impact of price increases over time Question Answer 15.Solutions – swimwear 502 Statistics for Business 1 1988 � 100; 126; 144; 168; 221; 222; 211; 221; 237; 216; 190; 137; 2000 � 150; 173; 146; 159; 222; 224 2 In 1989 the backlog was 26% more than in 1988; in 2000 it was 50% more than in 1988; in 2005 it was 124% more than in 1988 3 1988 � 67; 84; 96; 112; 147; 148; 140; 147; 158; 144; 126; 91; 2000� 100; 115; 97; 106; 148; 149 4 In 1989 the backlog was 33% less than in 2000; in 1993 it was 48% more than in 2000; in 2005 it was 49% more than in 2000 5 1988 is a long way back and with a base of 1988 gives some index values greater than 200 which is sometimes difficult to interpret, 2000 is closer to the present time and all the index values are less than 200 making interpretation more understandable 6 1989 � 126; 114; 117; 132; 100; 95; 105; 107; 91; 88; 72; 2000 � 109; 115; 84; 109; 139; 101 7 In 1990 the backlog increased 14% over 1989.

In 1994 the backlog decreased by 5% compared to 1993.In 1998 the backlog decreased by 12% compared to 1997.In 2004 the backlog increased by 39% compared to 2003 Question Answer Chapter 11: Indexing as a Method of Data Analysis 1.Solutions – backlog 1 1987 � 100; 98; 85; 86; 81; 77; 81; 86; 86; 1996 � 87; 74; 66; 63; 63; 61; 70; 82; 92; 116 2 In 1996 the price was 13% less than in 1987.In 2005 it was 16% more than in 1987 3 1988 � 115; 113; 98; 99; 93; 89; 93; 99; 99; 1996 � 100; 85; 76; 72; 72; 70; 80; 94; 106; 133 4 In 1987 the price was 15% more than in 1996.In 2005 it was 33% more than in 1996 5 1987 is a long way back.A base of 1996 gives index values for the last 10 years 6 1988 � 98; 87; 101; 94; 95; 105; 107; 100; 101; 85; 89; 95; 100; 97; 114; 117; 113; 126 7 1997 when the price decreased by 15% from the previous year 8 2005 when the price increased by 26% from the previous year Question Answer 2.Solutions – gold 503Appendix IV: Answers to end of chapter exercises 1 1990 � 100.

5% more than in 1990 or more than double the price 3 1990 � 83.4% more than in 2000 5 1990 is in the last century.A base of 2000 gives the benchmark start for the 21st century 6 1991 � 94.

9% increase over the previous year 8 1990 � 100.0 10 There is an approximate correlation between the two in that when crude oil prices rise, then the price of refined gasoline rises.However, there is always a lag when gasoline prices change to correspond to crude oil price changes Question Answer 3.Solutions – US gasoline prices 1 1975 � 100; 138.Solutions – coffee prices (Continued ) 504 Statistics for Business 7 A base of 2000 is probably the best as it starts the 21st century.A base of 1975 is inappropriate as it is too far back and we are introducing percentage values more than 100% which is not always easy to grasp 8 1976 � 138.7 9 In 1977 when prices increased by 121.5% 10 In 1981 when prices decreased by 20.4% 11 Coffee beans are a commodity and the price is dependent on the weather and the quantity harvested.Also country unrest can have an impact such as in the Ivory Coast in 2005 Question Answer 1 100; 96; 92; 98; 106 2 They were 6% higher than in 2005 3 94; 66; 25; 84; 100 4 They were 6% less than in 2001 5 105; 104; 93; 93 6 There was a 7% annual decline in both 2002 and 2003 and a 4% annual increase in 2004 and a 5% annual increase in 2005 Question Answer 5.

Solutions – Ford Motor Company 1 1992 � 100.6% more vehicles were sold than in 1992 7 Not that great; stock price has fallen; dividends are half what they were in 1992; North American vehicle sales look to be on the decline Question Answer 4.Solutions – (Continued ) 505Appendix IV: Answers to end of chapter exercises 1 Britain � 100; 113; 70; 13; 43; 13; 113; 30; 13; Sweden � 39 2 In Ireland 13% more people than in Britain admit to have been drunk.In Greece the figure is 87% less than in Britain and in Germany it is 57% less 3 Britain � 767; 867; 533; France � 100; 333; 100; 867; 233; 100; Sweden � 300 4 In Ireland 767% more people than in France admit to have been drunk (almost 8 times as many).In Greece the figure is about the same as in France and in Germany it is 233% more or over 3 times as much 5 Britain � 88; Denmark � 100; 62; 12; 38; 12; 100; 27; 12; Sweden � 35 6 In Ireland the drinking level is about the same as in Denmark.In Greece it is about 88% less than in Denmark and in Germany it is 62% less than in Denmark 7 Southern Europeans seem to admit being drunk less than Northern Europeans.

In Southern Europe drinking wine with a meal is common and this perhaps suppresses the notion for “binge” drinking Question Answer 7.Solutions – drinking 1 Australia � 208; 123; 137; 188; 138; 136; 88; 108; 169; 46; Ireland � 138; 115; 200; 277; 169; 162; 77; 92; 112; 196; 42; United States � 100 2 In Australia there is 108% more people working part time than in the United States.In Greece there is 54% less people working part time.In Switzerland there is 96% more people working part time than in the United States 3 Australia � 75; 44; 49; 68; 50; 49; 32; 39; 61; 17; 50; 42; 72; Netherlands � 100; 61; 58; 28; 33; 40; 71; 15; United States � 36 4 In Australia there is 15% less people working part time than in the Netherlands.In Greece there is 39% less people working part time.

In Switzerland there is 29% less people working part time than in the Netherlands 5 Australia � 88; 108; 105; Britain � 100; 89; 83; 82; 102; 105; 90; 102; 101; 88; 99; 97; 97; 88; 101; 90; 107; 77; 88 6 In Australia the proportion of women working part time is 12% less than in Britain.In Greece the proportion of women working part time is 10% less than in Britain.In Switzerland the proportion of women working part time is 7% less than in Britain Question Answer 8.Solutions – part-time work 1 Amsterdam 46% less; Berlin 58% less; New York 18% more; Paris 23% less; Sydney 35% less; Tokyo 38% more; Vancouver 53% less Question Answer 9.Solutions – cost of living (Continued ) 506 Statistics for Business 2 Amsterdam 4% more; Berlin 19% more; New York 14% less; Paris 9% less; Sydney 16% less; Tokyo 13% less; Vancouver 10% less 3 Amsterdam 23% more; Berlin 6% less; New York 165% more; Paris 73% more; Sydney 46% more; Tokyo 212% more; Vancouver 16% less 4 Amsterdam 12% more; Berlin 6% less; New York 14% less; Paris 9% less; Sydney 16% less; Tokyo 13% less; Vancouver 16% less 5 Amsterdam 10% more; Berlin 8% less; New York 16% less; Paris 11% less; Sydney 18% less; Tokyo 14% less; Vancouver 18% less 6 Amsterdam 14% more; Berlin 4% less; New York 12% less; Paris 7% less; Sydney 15% less; Tokyo 11% less; Vancouver 15% less 7 Most expensive is Tokyo which is 425% more than the least expensive or over 5 times more expensive Question Answer 1 Iceland is the least corrupt; the countries of Czech Republic, Greece, Slovakia, and Namibia are equally the most 2 78% 3 Spain 100.

3 7 Of the selected eight countries from the European Union those in North appear to be less corrupt than in the South Question Answer 10.

Solutions – corruption 1 Mauritius is most dangerous and France is the least 2 Belgium is 220% more dangerous; Dominican Republic 680% more; France 20% less; Latvia 400% more; Luxembourg 220% more; Mauritius 800% more; Russia 300% more; Venezuela 380% more 3 Belgium is 6.7% more dangerous; Dominican Republic 160% more; France 73.3% more; Venezuela 60% more Question Answer 11.Solutions – road traffic deaths (Continued ) 9.Solutions – (Continued ) 507Appendix IV: Answers to end of chapter exercises 4 Belgium is 23.8% less dangerous; Dominican Republic 85.7% more; France 81% less; Latvia 19% more; Luxembourg 29% less; Mauritius 114.

1% more dangerous; Dominican Republic 200% more; France 69.6% more 6 Emerging countries are more dangerous to drive than developed countries.

Improved road conditions, better driving education, better traffic control, and severer penalties when traffic laws are infringed Question Answer 1 2003 � 100.3 6 The unweighted price index measures inflation.The unweighted quantity index measures consumption changes for this family.The other indexes take into account price and quantity.The values are close as although the numerator changes, the denominator changes by almost proportionality the same amount Question Answer 12.Solutions – family food consumption 1 2000 � 100.

0 5 The meat prices show a reasonable trend during the period and there is an increasing trend in the quantity of meat handled.This explains a reasonable trend of the indexes Question Answer 13.

Solutions – (Continued ) 508 Statistics for Business 1 2000 � 100.1 5 The commodity prices show no trend during the period but there is an increasing trend of the consumption of the commodity.This explains the fluctuation of the indexes Question Answer 14.6 5 There is neither a trend in the commodity price or the usage of the metals during the 6-year period.This helps to explain the fluctuation of the indexes Question Answer 15.Solutions – non-ferrous metals Bibliography Alreck, P.Survey Research Handbook, 3rd edition, McGraw Hill, New York.News of the World, Football Annual 2005–2006, Invincible Press, London, ISBN 0-00-720582-1.Basic Business Statistics, 10th edition, Pearson, Pren- tice Hall, New Jersey, ISBN 0-13-196869-6.

Against the Gods: The Remarkable Story of Risk, Wiley, New York, USA, ISBN 0-471- 12104-5.Stats Means Business: A Guide to Business Statistics, Butterworth-Heinemann, Oxford, ISBN 0-7506-5364-7.Economist (The), Editorial offices in major cities.Eurostat, Statistical information for Europe and else- where, Financial Times, Printed in London and other world cities.Daily newspaper published Monday through Friday on a distinguishing salmon coloured paper, Fortune, Published monthly in the Netherlands by Time Warner, Francis, A.Business Mathematics and Statistics, 6th edition, Thomson, London, ISBN 1-84480-128-4.International Herald Tribune, Published by the New York Times, edited in Paris and Hong Kong and printed in Paris, France.

Daily newspaper published Monday through Saturday, Levin, R.Statistics for Management, 7th edition, Prentice Hall, New Jersey, ISBN 0-13-606716-6.Organization for Economic and Development, www.The Visual Display of Quantitative Information, 2nd edition (May 2001), Graphics Press, Connecticut, USA, ISBN 0-961-39214-2.Statistical information on many aspects of the UK, www.United Nations Conference on Trade and Development, United States Government Statistics.Statistics pro- duced by more than 70 agencies of the United States Federal Government, Wall Street Journal, Europe, Printed in Belgium.

Daily newspaper published Monday through Friday, Waller, D.Operations Management: A Supply Chain Approach, 2nd edition, Thomson, Learning, London, ISBN 1-86152-803-5.There are many books related to statistics, and numerous sources of statistical information.

The following are some that I have used.This page intentionally left blank Index 80/20 rule 17 a priori probability, concepts 84–6, 99, 104 ABS function, Excel 433 absolute frequency histograms, concepts 5–6, 16, 17, 23 absolute frequency ogives 9–11 absolute frequency polygons 7–9, 23 acceptance/rejection issues, hypothesis testing 265–9, 276–80, 305–20 addition rules classical probability 84–5, 91–2, 103 equation 84 alcohol 348 algebraic expressions, concepts 439 alternative hypothesis see also hypothesis testing concepts 265–80, 305–20 answers to end of chapter exercises 449–508 appendices 413–508 answers to end of chapter exercises 449–508 Excel usage guide 429–36 key terminology/formulas in statistics 413–26, 427, 438, 446–7 mathematical relationships 437–47 arithmetic operating symbols list 438 rules for calculations 444–5 sequence of operations 438 arithmetic mean see also mean… coefficient of variation 56–7, 63, 361–2, 365 concepts 47–8, 54–7, 59–60, 63, 120–1, 151–73, 187–212, 231–43, 301–20, 444–6 definition 47, 120–1, 231 deviation about the mean 55–6 different populations hypothesis testing 301–32 distribution of the sample means 190, 194–6, 211 equation 47, 444–5 estimates 231–43 examples 47–8, 161–2, 164–9 hypothesis testing 265–72, 279–80, 302–20 median 59–60, 151–2, 161–2, 164–9, 172–3 normality of data 161–9, 172–3 sampling for the mean 187–212, 231–52 arrangement of different objects (rule no.5), counting rules 102, 104 commuting times 311–13 computer systems, backups 96–7 conditional probabilities equation 88 statistical dependence 86–8, 103 confidence intervals applications 234–5, 236–7, 243–5 concepts 231, 232–52, 343–4 definition 233–4 examples 233–7, 243–5 exercises and answers 253–62, 475–9 finite populations 231, 236–7, 244–5, 251–2 infinite populations 233–6, 243–5, 251–2 student-t distribution 231, 238–43, 251–2, 344, 350–1 confidence levels concepts 231, 232–52, 343–4, 350–1 definition 232–3 exercises and answers 253–62, 475–9 forecasts 343–4, 364 margin of error 248–52 regression analysis 343–4, 364 constants, concepts 339, 437, 445–6 consumer price index (CPI) see also indexing concepts 386, 388–91, 397 consumer surveys concepts 209–10, 212 definition 209 examples 209–10 contingency tables see cross-classification (contingency) tables continuity correction factor, normal–binomial approximation 170–1 continuous random variables concepts 150–73 definition 150 control-shift-enter keys 5 CORREL function, Excel 338, 434 correlation 335–81 see also forecasts causal forecasting 345–7, 388 concepts 335–9, 345–7, 364–5 definition 335, 338 equation 338 Excel 338 good correlation value 339 costs of errors, hypothesis testing 277–80 COUNT function, Excel 434 counting rules arrangement of different objects (rule no.3) 101, 104 combinations of objects (rule no.

5) 102, 104 concepts 81, 99–102, 104 definition 99 different types of events (rule no.2) 100–1, 104 examples 100–2 exercises and answers 105–17, 458–61 permutations of objects (rule no.4) 101–2, 104 single type of event (rule no.1) 99–100, 104 covariance concepts 120, 124–6, 134–5 definition 124 equation 124 examples 124–6 portfolio risk 124–6 critical levels see also significance levels concepts 264–80, 302–20 cross-classification (contingency) tables chi-square hypothesis test 313–20 concepts 20, 21, 24, 86–8, 92–3, 313–20 definition 20 degrees of freedom 314–15 examples 21, 87, 93, 313–14 cumulative distributions see ogives currencies 45–6 curvilinear functions 351–3, 362–5 see also non-linear regression 513Index data see also categorical…; numerical… asymmetrical data 60, 164–9, 172–3 characterizing data 45–77 concepts 3–24 defining data 45–77 indexing 383–412 normality of data 161–9, 172–3 presenting data 1–43, 449–52 primary/secondary data 210, 212 data arrays, concepts 50 Data menu bar 12 decimals concepts 439–42, 444, 446–7 fractions 444 defining data concepts 45–77 exercises and answers 65–77, 453–7 degrees of freedom chi-square hypothesis test 314–19 cross-classification (contingency) tables 314–15 student-t distribution 237–43, 251, 269–72, 306–11, 344, 347 dependent populations, hypothesis testing 309–11, 319–20 dependent variables, definition 336, 345 derivatives 364 descriptive statistics, definition 188 development of percentiles 61–2 see also percentiles deviation about the mean, concepts 55–6 Dewey, Governor 185–6 dice-tossing experiments 100, 104, 120, 121–2 different populations exercises and answers 321–32, 489–95 hypothesis testing 301–32 different types of events (rule no.73 EUR, Germany 10096,"MST Morphine sulphate tablets 100 mg =30$","pure12",29.14 EUR, United Kingdom 10097,"MORPHINE SULPHATE TABLETS (CONTINUS) 30$ =100 mg tablet","pure12",29.14 EUR, United Kingdom 10099," Learn The Basics of Ethical Hacking and Penetration Testing","junkiepig666",18.

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68 EUR, Philippines 11296,"115 Pills Ambien (Generic) 10 MG","Meds2Buy",186.89 EUR, Philippines 11297,"100 Pills Ambien (Generic) 10 MG","Meds2Buy",168.66 EUR, Philippines 11300,"75 Pills Ambien (Generic) 10 MG","Meds2Buy",136.75 EUR, Philippines 11301,"60 Pills Ambien (Generic) 10 MG","Meds2Buy",113.96 EUR, Philippines 11302,"45 Pills Ambien (Generic) 10 MG","Meds2Buy",95.

72 EUR, Philippines 11304,"30 Pills Ambien (Generic) 10 MG","Meds2Buy",77.49 EUR, Philippines 11305,"30 Pills Ritalin 10 MG","Meds2Buy",77.49 EUR, Philippines 11306,"45 Pills Ritalin 10 MG","Meds2Buy",95.72 EUR, Philippines 11307,"60 Pills Ritalin 10 MG","Meds2Buy",113.96 EUR, Philippines 11308,"75 Pills Ritalin 10 MG","Meds2Buy",136.

75 EUR, Philippines 11310,"100 Pills Ritalin 10 MG","Meds2Buy",173.22 EUR, Philippines 11311,"115 Pills Ritalin 10 MG","Meds2Buy",186.89 EUR, Philippines 11312,"130 Pills Ritalin 10 MG","Meds2Buy",209.68 EUR, Philippines 11314,"115 Pills Rivotril 2 MG","Meds2Buy",145.87 EUR, Philippines 11316,"100 Pills Rivotril 2 MG","Meds2Buy",132.

19 EUR, Philippines 11317,"90 Pills Rivotril 2 MG","Meds2Buy",123.07 EUR, Philippines 11318,"75 Pills Rivotril 2 MG","Meds2Buy",109.4 EUR, Philippines 11319,"60 Pills Rivotril 2 MG","Meds2Buy",95.72 EUR, Philippines 11320,"45 Pills Rivotril 2 MG","Meds2Buy",86.

61 EUR, Philippines 11335,"HACKING WORDLISTS!!!","junkiepig666",18.

22 EUR, Germany 11339,"Avast Internet Security 2015 With Serial Keys Valid Until 2017","junkiepig666",4.55 EUR, Germany 11354,"1 Liter of HoneyWanna 750mg THC","hashishmaster",410.25 EUR, Canada 11355,"500ml HoneyWanna ","hashishmaster",205.12 EUR, Canada 11356,"100ml HoneyWanna","hashishmaster",40.98 EUR, Canada 11392,"Speed 2g","THEFRENCHCONNECTION",13 EUR, Belgium 11434,"Heroin T 4, 5g","THEFRENCHCONNECTION",360 EUR, Belgium 11469,"Temgesic 0.

2mg Bubrenorfiini 10kpl* SILKKITIEN HALVIN*","Nappi Kauppa",50 EUR, Finland 11479,"Propral 40mg 10kpl","Nappi Kauppa",15 EUR, Finland 11587,"Ultimate Guide to PayPal Credit ","junkiepig666",4.55 EUR, Germany 11594,"LIST OF CARDABLE SITES ","junkiepig666",4.55 EUR, Germany 11596,"Udemy - How I Make Money By Building And Selling Blogs In 4 Hours","junkiepig666",2.73 EUR, Germany 11598,"4 Ways To Hack Facebook","junkiepig666",2.73 EUR, Germany 11698,"The Ultimate List of Onion Websites","junkiepig666",2.

73 EUR, Germany 11699,"Guide to Civilian Warfare and Sabotage","junkiepig666",2.73 EUR, Germany 11715,"80 x Kratom capsules","dionysos",32 EUR, Austria 11765,"►►► CC & PAYPAL to BITCOIN - 10 different methods +BONUS◄◄◄","CardBoss",2.73 EUR, United States 11766,"►►► NON VBV - Easy RESET - BIN LIST *WorldWide* UPDATE ","CardBoss",2.73 EUR, United States 11767,"►►► NON VBV - Easy RESET - BIN LIST *WorldWide* UPDATE ","CardBoss",2.73 EUR, United States 11780,"ULTRA SPAM Software - All you need 4 Spamming","CardBoss",2.

73 EUR, Finland 11787,"♕ ♕ ULTRA Hackers Collection OVER 10.73 EUR, United States 11792,"How To Become A Better Negotiator","CardBoss",0.9 EUR, United States 11805,"How to get the TRUTH out of anyone","CardBoss",0.9 EUR, United States 11837,"ULTRA HQ - FRANCE - FR - FULLZ (MMN/BILL/DOB)","CardBoss",9.

91 EUR, United Kingdom 12007,"PGP Tutorial For Newbs (Gpg4Win)","KingCarder",0.91 EUR, United Kingdom 12129,"Zahra AAA hashish, 10 grams","projeccao",66 EUR, France 12133,"e-liquide BHO, 2 ml","projeccao",30 EUR, France 12135,"MDMA, 1 gram","projeccao",20 EUR, France 12223,"10 Grams ORGANIC RIF Moroccan Hashish","danhash",66.09 EUR, Spain 12260,"1g Cocaine Grade AA++","lawpee911",148.6 EUR, United States 12261,"1g Cocaine Grade B","lawpee911",75.

59 EUR, United States 12269,"20 Grams ORGANIC AZLA Moroccan Hashish ","danhash",145.86 EUR, Spain 12365,"7g Weed - Bestaussiegear POSTAGE INCLUDED ","bestaussiegear",51.33 EUR, Australia 12367,"1g PREMIUM ICE SHARDS CHEAPEST AUSSIE PRICE INCLUDED POSTAGE","bestaussiegear",123.5g PREMIUM ICE SHARDS CHEAPEST AUSSIE PRICE INCLUDED POSTAGE","bestaussiegear",394.

31 EUR, Australia 12484,"Trim 1g ( 10g)","carlos lopez",3 EUR, Czech Republic 12579,"FUB-AMB 100g","Jimmy chemicals",282.73% Amphetamine (Speed) Paste ","love2xlr8",84 EUR, Netherlands 12584,"50 gr.73% Amphetamine (Speed) Paste ","love2xlr8",153 EUR, Netherlands 12615,"BEIN SPORTS FR ACCOUNT (lifetime)","TheTopDigital",10.94 EUR, France 12645,"LoL EUW ACCOUNT (lifetime)","TheTopDigital",17.

32 EUR, Finland 12778,"Tramadol 50mg x 20 caps","stealthmeds",20 EUR, United Kingdom 1279,"(PE+unknown) iti ruisku, 5ml READY","faceless",8.4 EUR, Finland 12911,"Norge: 5 gram Bubba Kush","hogstandard",131.55 EUR, Norway 12912,"Norge: 10 gram Bubba Kush","hogstandard",225.13 EUR, Norway 12914,"Norge: 5 gram Og Kush","hogstandard",104.03 EUR, Norway 12940,"electronic pocket scale","dionysos",23 EUR, Austria 12973,"14 GRAMS - HYDRO BUDS - **Free Express Shipping**","Utopia",124.

86 EUR, Australia 12974,"7 GRAMS - HYDRO BUDS - **Free Express Shipping**","Utopia",66.5 GRAMS - HYDRO BUDS - **Free Express Shipping**","Utopia",41.62 EUR, Australia 12977,"1 GRAM - HYDRO BUDS - **Free Express Shipping**","Utopia",17.34 EUR, Australia 13000," - LIFETIME PORN PREMIUM ACCOUNT ","Wooody",9.

11 EUR, Finland 13001," - LIFETIME PORN PREMIUM ACCOUNT ","Wooody",9.11 EUR, Finland 13003," - LIFETIME PORN PREMIUM ACCOUNT ","Wooody",9.11 EUR, Finland 13004," - LIFETIME PORN PREMIUM ACCOUNT ","Wooody",9.11 EUR, Finland 13005," - LIFETIME PORN PREMIUM ACCOUNT ","Wooody",9.

11 EUR, Finland 13006," - LIFETIME PORN PREMIUM ACCOUNT ","Wooody",9.

11 EUR, Finland 13007," - LIFETIME PORN PREMIUM ACCOUNT ","Wooody",9.11 EUR, Finland 13008," - LIFETIME PORN PREMIUM ACCOUNT ","Wooody",9.11 EUR, Finland 13009," - LIFETIME PORN PREMIUM ACCOUNT ","Wooody",9.11 EUR, Finland 13010," - LIFETIME PORN PREMIUM ACCOUNT ","Wooody",9.11 EUR, Finland 13011," - LIFETIME PORN PREMIUM ACCOUNT ","Wooody",9.

11 EUR, Finland 13012," - LIFETIME PORN PREMIUM ACCOUNT ","Wooody",9.11 EUR, Finland 13013," - LIFETIME PORN PREMIUM ACCOUNT ","Wooody",9.11 EUR, Finland 13015," - LIFETIME PORN PREMIUM ACCOUNT ","Wooody",9.11 EUR, Finland 13016," - LIFETIME PORN PREMIUM ACCOUNT ","Wooody",9.11 EUR, Finland 13017," - LIFETIME PORN PREMIUM ACCOUNT ","Wooody",9.

11 EUR, Finland 13019," - LIFETIME PORN PREMIUM ACCOUNT ","Wooody",9.11 EUR, Finland 13020," - LIFETIME PORN PREMIUM ACCOUNT ","Wooody",9.11 EUR, Finland 13021," - LIFETIME PORN PREMIUM ACCOUNT ","Wooody",9.11 EUR, Finland 13022," - LIFETIME PORN PREMIUM ACCOUNT ","Wooody",9.11 EUR, Finland 13023," - LIFETIME PORN PREMIUM ACCOUNT ","Wooody",9.

11 EUR, Finland 13024," - LIFETIME PORN PREMIUM ACCOUNT ","Wooody",9.11 EUR, Finland 13025," - LIFETIME PORN PREMIUM ACCOUNT ","Wooody",9.11 EUR, Finland 13026," - LIFETIME PORN PREMIUM ACCOUNT ","Wooody",9.11 EUR, Finland 13027," - LIFETIME PORN PREMIUM ACCOUNT ","Wooody",9.11 EUR, Finland 13028," - LIFETIME PORN PREMIUM ACCOUNT ","Wooody",9.

11 EUR, Finland 13030," - LIFETIME PORN PREMIUM ACCOUNT ","Wooody",9.11 EUR, Finland 13031," - LIFETIME PORN PREMIUM ACCOUNT ","Wooody",9.11 EUR, Finland 13032," - LIFETIME PORN PREMIUM ACCOUNT ","Wooody",9.11 EUR, Finland 13033," - LIFETIME PORN PREMIUM ACCOUNT ","Wooody",9.11 EUR, Finland 13034," - LIFETIME PORN PREMIUM ACCOUNT ","Wooody",9.

11 EUR, Finland 13035," - LIFETIME PORN PREMIUM ACCOUNT ","Wooody",9.11 EUR, Finland 13036,"Porn HD Collection - LIFETIME PORN PREMIUM ACCOUNTS ","Wooody",26.41 EUR, Finland 13037," - LIFETIME PORN PREMIUM ACCOUNT ","Wooody",9.11 EUR, Finland 13038," - LIFETIME PORN PREMIUM ACCOUNT ","Wooody",9.11 EUR, Finland 13039," - LIFETIME PORN PREMIUM ACCOUNT ","Wooody",9.

11 EUR, Finland 13040," - LIFETIME PORN PREMIUM ACCOUNT ","Wooody",9.11 EUR, Finland 13043," - LIFETIME PORN PREMIUM ACCOUNT ","Wooody",9.11 EUR, Finland 13044," - LIFETIME PORN PREMIUM ACCOUNT ","Wooody",9.11 EUR, Finland 13045," - LIFETIME PORN PREMIUM ACCOUNT ","Wooody",9.11 EUR, Finland 13046," - LIFETIME PORN PREMIUM ACCOUNT ","Wooody",9.

11 EUR, Finland 13047," - LIFETIME PORN PREMIUM ACCOUNT ","Wooody",9.11 EUR, Finland 13048," - LIFETIME PORN PREMIUM ACCOUNT ","Wooody",9.11 EUR, Finland 13050," - LIFETIME PORN PREMIUM ACCOUNT ","Wooody",9.11 EUR, Finland 13052," - LIFETIME PORN PREMIUM ACCOUNT ","Wooody",9.11 EUR, Finland 13053," - LIFETIME PORN PREMIUM ACCOUNT ","Wooody",9.

11 EUR, Finland 13054," - LIFETIME PORN PREMIUM ACCOUNT ","Wooody",9.11 EUR, Finland 13055," - LIFETIME PORN PREMIUM ACCOUNT ","Wooody",9.11 EUR, Finland 13056," - LIFETIME PORN PREMIUM ACCOUNT ","Wooody",9.11 EUR, Finland 13058," - LIFETIME PORN PREMIUM ACCOUNT ","Wooody",9.11 EUR, Finland 13060," - LIFETIME PORN PREMIUM ACCOUNT ","Wooody",9.

11 EUR, Finland 13061,"FULL PORN PACKAGE SPECIAL OFFER - LIFETIME PORN PREMIUM ACCOUNT ","Wooody",26.44 EUR, Finland 13062," - LIFETIME PORN PREMIUM ACCOUNT ","Wooody",9.11 EUR, Finland 13063," - LIFETIME PORN PREMIUM ACCOUNT ","Wooody",9.11 EUR, Finland 13064," - LIFETIME PORN PREMIUM ACCOUNT ","Wooody",9.

11 EUR, Finland 13065," - LIFETIME PORN PREMIUM ACCOUNT ","Wooody",9.

11 EUR, Finland 13108,"2gram of HQ Crystal Meth China ICE 99,8% Purity-EU Ship","Citra",114 EUR, Spain 13109,"100 Dutch LSD blotters 180ug-Free Shipp","Citra",346.05 EUR, Australia 13110,"10 gram Acrylfentanyl 99%-Ship From EU","Citra",716 EUR, Poland 13111,"Fentanyl HCL Blotter 1mg x100","Citra",382.9 EUR, Canada 13112,"Celebrity Nude, Naked, Topless photos, shots, leaks!","CardBoss",2.73 EUR, United States 13202,"1 oz OG Kush Trim **NEW STOCK!** Quality Scissor Trim","GreenLeafLabs",34.15 EUR, United States 13205,"1/2 LB OG Kush Trim **NEW STOCK!** Quality Scissor Trim","GreenLeafLabs",193.

52 EUR, United States 13206,"1 Rechargeable Vape Battery (510 thread) with USB Charger","GreenLeafLabs",21.42 EUR, United States 13211,"(6x)500mg CO2 Extracted Oil - (Topshelf) Sour OG Bud Run 85% THC!","GreenLeafLabs",176.41 EUR, United States 13230,"(2x) Kiva Chocolate Bar (180mg THC)","GreenLeafLabs",52.42 EUR, United States 13261," HGH EUROTROPIN 120IU 1 BOX ","stealthmeds",221.84 EUR, United Kingdom 13262," HYGETROPIN 100IU HGH","stealthmeds",170.

82 EUR, United Kingdom 13263," PROZAC / FLUOXETINE 20MG X 30 ","stealthmeds",27.73 EUR, United Kingdom 13264," VIAGRA (SILDAMAX) 10 X 100MG ","stealthmeds",14.42 EUR, United Kingdom 13265," VIAGRA (SILDAMAX) 20 X 100MG","stealthmeds",22.18 EUR, United Kingdom 13268," CODEINE PHOSPHATE 60MG X 28","stealthmeds",44.37 EUR, United Kingdom 13269," DIHYDOCODEINE 30MG X 28 ","stealthmeds",27.

79 EUR, United Kingdom 13272," DIAZEPAM (VALIUM) 10MG X 28 ","stealthmeds",26.62 EUR, United Kingdom 13273," DIAZEPAM (VALIUM) 56 X 10MG ","stealthmeds",46.59 EUR, United Kingdom 13274," HCG PREGNYL 5000IU ORGANON ","stealthmeds",38.9 EUR, United Kingdom 13275," ARIMIDEX / ANASTROZOLE 28 X 1MG ","stealthmeds",27.73 EUR, United Kingdom 13276," CLOMID 50MG X 24 ANFARM GREECE ","stealthmeds",27.

73 EUR, United Kingdom 13277," SERPAFAR CLOMIFENE CITRATE X20 TABLETS, 50MG ","stealthmeds",27.73 EUR, United Kingdom 13278," TAMOXIFEN 20MG X30 TABLETS ","stealthmeds",27.73 EUR, United Kingdom 13279," BAYER PROVIRON 25MG.79 EUR, United Kingdom 13283," - LIFETIME PORN PREMIUM ACCOUNT ","Wooody",9.

11 EUR, Finland 13284," T3 LIOTHYRONINE X30 25UP TABLETS ","stealthmeds",19.97 EUR, United Kingdom 13285," SERTRALINE 100MG X28 TABLETS ","stealthmeds",27.73 EUR, United Kingdom 13289," VALIUM (ROCHE) X100 10MG ","stealthmeds",84.3 EUR, United Kingdom 13290," ZOPICLONE 7.18 EUR, United Kingdom 13291," ZOPICLONE 56 X 7.04 EUR, United Kingdom 13292," CIALIS 20MG X 4 ","stealthmeds",28.28 EUR, United Kingdom 13293," TADAGRA / TADALAFIL 20MG x10","stealthmeds",22.18 EUR, United Kingdom 13294," TRAMADOL 50MG X 10 CAPSULES","stealthmeds",12.

2 EUR, United Kingdom 13295," ORGANON SUSTANON 250.TESTOSTRONE PROPIONATE X 10 AMPS PER BOx","stealthmeds",58.

79 EUR, United Kingdom 13407,"INTRO OFFER 1G COLOMBIAN FISHSCALE COCAINE 100% PURE LTD OFFER","drugkings",85.4 EUR, United Kingdom 13408,"2G COLOMBIAN FISHSCALE COCAINE LTD OFFER 100% PURE-UK SELLER","drugkings",157.5 EUR, United Kingdom 13409,"INTRO OFFER 1G PURE QUALITY FISHSCALE COCAINE,NXT DAY DEL-LTD OFF","drugkings",51.88 EUR, United Kingdom 13411,"2G PURE QUALITY FISHSCALE COCAINE-LTD OFFER-NEXT DAY DEL","drugkings",102.78 EUR, United Kingdom 13412,"3G PURE QUALITY FISHSCALE COCAINE-NXT DAY DELIVERY-LTD OFFER","drugkings",136.

41 EUR, United Kingdom 13413,"5G PURE QUALITY FISHSCALE COCAINE-UK SELLER NXT DAY DEL-LTD OFFER","drugkings",219.6 EUR, United Kingdom 13414,"7G PURE QUALITY FISHSCALE COCAINE-UK SELLER-NEXT DAY DELIVERY","drugkings",291.7 EUR, United Kingdom 13415,"14G PURE QUALITY COCAINE-UK SELLER-NEXT DAY DEL-LTD OFFER","drugkings",557.92 EUR, United Kingdom 13417,"25 tabs Diazepam (10mg tabs) (FE ONLY)","MedIndia",12.

42 EUR, United Kingdom 13418,"50 tabs Diazepam (10mg tabs) (FE ONLY)","MedIndia",22.

45 EUR, United Kingdom 13419,"100 tabs Diazepam (10mg tabs)(FE ONLY)","MedIndia",40.15 EUR, United Kingdom 13422,"25 tabs Zopiclone.3 EUR, United Kingdom 13423,"50 tabs Zopiclone.

01 EUR, United Kingdom 13424,"100 tabs Zopiclone.77 EUR, United Kingdom 13427,"kamagra oral jelly (viagra) x1 box","stealthmeds",22.18 EUR, United Kingdom 13432,"5 tabs Eszopiclone (4.99 EUR, United Kingdom 13435,"100 tabs DMAA (10 mg tabs) (FE ONLY)","MedIndia",12.42 EUR, United Kingdom 13436,"200 tabs DMAA (10 mg tabs) (FE ONLY)","MedIndia",21.

3 EUR, United Kingdom 13439,"50 tabs Dianabol (10mg tabs) (FE ONLY)","MedIndia",15.75 EUR, United Kingdom 13441,"100 tabs Dianabol (10mg tabs) (FE ONLY)","MedIndia",27.95 EUR, United Kingdom 13443,"200 tabs Dianabol (10mg tabs) (FE ONLY)","MedIndia",51.25 EUR, United Kingdom 13462,"Dibutylone 100 g","Jimmy chemicals",273.5 EUR, China 13463,"Dibutylone 500g","Jimmy chemicals",866.

08 EUR, China 13464,"Dibutylone 1kg","Jimmy chemicals",1640.5mg tabs (FE ONLY)","MedIndia",18 EUR, United Kingdom 13477,"Ostarine (MK-2866) 50 tabs.97 EUR, United Kingdom 13479,"FINALIZE EARLY 100.00g MDMA CLEAN & UNCUT","kacikz",1157.81 EUR, Lithuania 13510,"Denmark Drivers Licence","MikeDelacruz",185.98 EUR, China 13511,"Spain National ID Card","MikeDelacruz",277.14 EUR, China 13512,"French National ID Card","MikeDelacruz",185.

98 EUR, China 13513,"Belgium National ID Card","MikeDelacruz",277.14 EUR, China 13514,"Swiss Drivers Licence","MikeDelacruz",185.98 EUR, China 13515,"Ireland Drivers Licence","MikeDelacruz",185.98 EUR, China 13516,"Norway Drivers Licence","MikeDelacruz",185.98 EUR, China 13517,"United Kingdom Drivers Licence","MikeDelacruz",185.

98 EUR, China 13518,"South Africa Drivers Licence","MikeDelacruz",185.98 EUR, China 13519,"Italian Drivers Licence","MikeDelacruz",185.98 EUR, China 13520,"German National ID Card","MikeDelacruz",185.98 EUR, China 13522,"New Zealand Drivers Licence","MikeDelacruz",185.98 EUR, China 13524,"Sweden Drivers Licence","MikeDelacruz",185.

98 EUR, China 13526,"Poland Drivers Licence","MikeDelacruz",185.98 EUR, China 13529,"France Drivers Licence","MikeDelacruz",205.12 EUR, China 13530,"Singapore Drivers Licence","MikeDelacruz",185.98 EUR, China 13531,"Canada Permanent Resident Card","MikeDelacruz",185.98 EUR, China 13533,"Israel Drivers Licence","MikeDelacruz",185.

98 EUR, China 13534,"Czech Drivers Licence","MikeDelacruz",185.98 EUR, China 13537,"CUSTOM FAKE ID / DRIVERS LICENCE","MikeDelacruz",225 EUR, China 13538,"Singapore National ID Card","MikeDelacruz",185.98 EUR, China 13539,"German Drivers Licence","MikeDelacruz",185.98 EUR, China 13551,"1x 125ug Elephant LSD","Markovich",7.

29 EUR, Netherlands 13554,"3x 125ug Elephant LSD","Markovich",22.

77 EUR, Netherlands 13555,"5x 125ug Elephant LSD","Markovich",34.19 EUR, Netherlands 13557,"10x 125ug Elephant LSD","Markovich",63.82 EUR, Netherlands 13558,"25x 125ug Elephant LSD","Markovich",136.75 EUR, Netherlands 13559,"50x 125ug Elephant LSD","Markovich",227.92 EUR, Netherlands 13560,"100x 125ug Elephant LSD","Markovich",364.

66 EUR, Netherlands 13561,"1x 160ug Hanuman LSD","Markovich",9.12 EUR, Netherlands 13562,"3x 160ug Hanuman LSD","Markovich",25.5 EUR, Netherlands 13563,"5x 160ug Hanuman LSD","Markovich",40.98 EUR, Netherlands 13564,"10x 160ug Hanuman LSD","Markovich",72.93 EUR, Netherlands 13565,"25x 160ug Hanuman LSD","Markovich",159.

37 EUR, Netherlands 13566,"50x 160ug Hanuman LSD","Markovich",273.2 EUR, Netherlands 13567,"100x 160ug Hanuman LSD","Markovich",500.87 EUR, Netherlands 13568,"1x 300ug Flower Fish LSD","Markovich",12.75 EUR, Netherlands 13569,"3x 300ug Flower Fish LSD","Markovich",36.43 EUR, Netherlands 13570,"5x 300ug Flower Fish LSD","Markovich",59.

19 EUR, Netherlands 13571,"10x 300ug Flower Fish LSD","Markovich",109.28 EUR, Netherlands 13572,"25x 300ug Flower Fish LSD","Markovich",250.43 EUR, Netherlands 13573,"50x 300ug Flower Fish LSD","Markovich",455.83 EUR, Netherlands 13574,"100x 300ug Flower Fish LSD","Markovich",819.6 EUR, Netherlands 13577,"fluoxetine x1 box x30 tablets ","stealthmeds",27.

73 EUR, United Kingdom 13578,"Nitazepam - 1 box/28 tabs @ 5mg","stealthmeds",27.73 EUR, United Kingdom 13583,"levothyroxine activis 100mg x28 tablets ","stealthmeds",27.73 EUR, United Kingdom 13585,"Symbiosis - 1g MDMA, UK First Class","Saint Symbiosis",27.32 EUR, United Kingdom 13586,"Symbiosis - 5g MDMA, UK First Class","Saint Symbiosis",122.94 EUR, United Kingdom 13587,"Symbiosis - 7g MDMA, UK First Class","Saint Symbiosis",163.

92 EUR, United Kingdom 13588,"Symbiosis - 10g MDMA,UK First Class","Saint Symbiosis",227.67 EUR, United Kingdom 13589,"Symbiosis - 3.5g MDMA, UK First Class","Saint Symbiosis",91.07 EUR, United Kingdom 13590,"Symbiosis - 28g MDMA, UK First Class","Saint Symbiosis",628.36 EUR, United Kingdom 13591," Symbiosis - 14g MDMA, UK First Class","Saint Symbiosis",318.

73 EUR, United Kingdom 13592," Symbiosis - 0.5g Ketamine, UK First Class","Saint Symbiosis",31.87 EUR, United Kingdom 13593," Symbiosis- 1g Ketamine, UK First Class","Saint Symbiosis",50.09 EUR, United Kingdom 13594,"Symbiosis - 3.5g Ketamine, UK First Class","Saint Symbiosis",163.

92 EUR, United Kingdom 13595,"Symbiosis - 5g Ketamine, UK First Class","Saint Symbiosis",227.67 EUR, United Kingdom 13596,"Symbiosis - 7g Ketamine, UK First Class","Saint Symbiosis",300.52 EUR, United Kingdom 13598,"Symbiosis - 14g Ketamine, UK First Class","Saint Symbiosis",482.65 EUR, United Kingdom 13599,"Symbiosis - 28g Ketamine, UK First Class","Saint Symbiosis",902.54 EUR, United Kingdom 13600,"The Complete Collection of Carding Guides 100+","junkiepig666",9.

11 EUR, Germany 13603,"Amazon Carding Guide ","junkiepig666",4.55 EUR, Germany 13604,"Apple Carding Tutorial ","junkiepig666",4.55 EUR, Germany 13625,"Austria National ID Card","MikeDelacruz",185.98 EUR, China 13626,"Austria Drivers Licence","MikeDelacruz",185.98 EUR, China 13627,"Finland Drivers Licence","MikeDelacruz",185.

98 EUR, China 13628,"Spain Drivers Licence","MikeDelacruz",185.98 EUR, China 13629,"German Bahncard 100","MikeDelacruz",185.98 EUR, China 13630,"UK Durham University ID","MikeDelacruz",145 EUR, China 13631,"AUS Queensland University ID","MikeDelacruz",145 EUR, China 13632,"Amphetamine Syntheses Overview & Reference Guide for Professional","junkiepig666",4.55 EUR, Germany 13635,"Amphetamine Synthesis - Very Easy With Photos","junkiepig666",4.

55 EUR, Germany 13638,"Amazon Master Guide With Video","junkiepig666",4.

55 EUR, Germany 13639,"Canada Social Security Card","MikeDelacruz",195 EUR, China 13641,"Malaysian Drivers Licence","MikeDelacruz",195 EUR, China 13643,"A Complete MDMA Synthesis for The First Time Chemist","junkiepig666",4.55 EUR, Germany 13646,"PRESSE ID France","MikeDelacruz",145 EUR, China 13647,"PRESSE ID Denmark","MikeDelacruz",145 EUR, China 13648,"(U.98 EUR, China 13649,"Udemy - Weight Loss By Eating What You Want - Kettlebell","junkiepig666",18.22 EUR, Germany 13650,"UK Manchester University iD","MikeDelacruz",145 EUR, China 13651,"CAN Waterloo University ID","MikeDelacruz",145 EUR, China 13652,"UK London University ID","MikeDelacruz",145 EUR, China 13655,"NZ Auckland University ID","MikeDelacruz",144 EUR, China 13656," Symbiosis - 0.5g Uncut Cocaine, UK First Class","Saint Symbiosis",68.3 EUR, United Kingdom 13657," Symbiosis - 1g Uncut Cocaine, UK First Class","Saint Symbiosis",132.05 EUR, United Kingdom 13658," Symbiosis - 3.

5g Uncut Cocaine, UK First Class","Saint Symbiosis",409.8 EUR, United Kingdom 13659," Symbiosis - 5g Uncut Cocaine, UK First Class","Saint Symbiosis",546.4 EUR, United Kingdom 13660," Symbiosis - 7g Uncut Cocaine, UK First Class","Saint Symbiosis",737.64 EUR, United Kingdom 13661," Symbiosis - 10g Uncut Cocaine, UK First Class","Saint Symbiosis",1047.26 EUR, United Kingdom 13662,"Symbiosis, 14g Uncut Cocaine, UK First Class","Saint Symbiosis",1411.

53 EUR, United Kingdom 13663,"Symbiosis, 20g Cocaine uncut flake, UK First Class","Saint Symbiosis",2003.46 EUR, United Kingdom 13668,"Symbiosis - 50 x LSD tabs, 250ug, UK First Class","Saint Symbiosis",291.41 EUR, United Kingdom 13669,"Symbiosis - 25 x LSD tabs, 250ug, UK First Class","Saint Symbiosis",200.35 EUR, United Kingdom 13670,"Symbiosis - 5 x LSD tabs, 250ug, UK First Class","Saint Symbiosis",50.09 EUR, United Kingdom 13671,"one doze 2cb 18-20mg Perfect Pure","LoveIsTheKey",8.

2 EUR, United Kingdom 13673," Symbiosis - 5 x 2cb, 18mg, UK First Class","Saint Symbiosis",34.61 EUR, United Kingdom 13675,"1000mg DMT Pure Crystal MHRB Hand Extraction","LoveIsTheKey",90.16 EUR, United Kingdom 13677,"500mg DMT Pure Crystal MHRB Hand Extraction","LoveIsTheKey",48.27 EUR, United Kingdom 13682,"250mg DMT Pure Crystal MHRB Hand Extraction","LoveIsTheKey",36.47 EUR, United Kingdom 13685,"100mg DMT Pure Crystal MHRB Hand Extraction","LoveIsTheKey",16.

39 EUR, United Kingdom 13690,"100x 100uq LSD White Crystal Swiss Lab","LoveIsTheKey",384.3 EUR, United Kingdom 13691,"10x 100uq LSD White Crystal Swiss Lab","LoveIsTheKey",45.53 EUR, United Kingdom 13692,"1x 100uq LSD White Crystal Swiss Lab","LoveIsTheKey",7.29 EUR, United Kingdom 13693,"10g MDMA 84% Pure Crystal Best World Quality Free Shipping ","LoveIsTheKey",162.1 EUR, United Kingdom 13694,"5g MDMA 84% Pure Crystal Best World Quality Free shipping","LoveIsTheKey",89.

25 EUR, United Kingdom 13695,"1g MDMA 84% Pure Crystal Best World Quality","LoveIsTheKey",20.95 EUR, United Kingdom 13696,"3g MDMA 84% Pure Crystal Best World Quality Free Shipping","LoveIsTheKey",55.55 EUR, United Kingdom 13710,"100mg Perfect Pure 2 CB HBr","LoveIsTheKey",20.95 EUR, United Kingdom 13712,"250mg Perfect Pure 2 CB HBr","LoveIsTheKey",36.43 EUR, United Kingdom 13713,"500mg Perfect Pure 2 CB HBr","LoveIsTheKey",43.

71 EUR, United Kingdom 13714,"1000mg Perfect Pure 2 CB HBr","LoveIsTheKey",79.31 EUR, United Kingdom 13717,"500mg Changa 50% DMT Pure Crystal MHRB Extraction","LoveIsTheKey",35.52 EUR, United Kingdom 13718,"1000mg Changa 50% DMT Pure Crystal MHRB Extraction","LoveIsTheKey",61.01 EUR, United Kingdom 13719,"5000mg Changa 50% DMT Pure Crystal MHRB Extraction","LoveIsTheKey",272.59 EUR, United Kingdom 13720,"1X Liquid Drop of Truth LSD 150uq Candy,Jelly Best World Quality","LoveIsTheKey",10.

02 EUR, United Kingdom 13721,"10x Truth LSD 150uq Gold Crystal Pure Experience","LoveIsTheKey",90 EUR, United Kingdom 13722,"5x Truth LSD 150uq Gold Crystal Pure experience","LoveIsTheKey",47 EUR, United Kingdom 13723,"1x Truth LSD 150uq Gold Crystal Pure Experience","LoveIsTheKey",11 EUR, United Kingdom 13724,"6x 25i Nbome 1200uq","LoveIsTheKey",18.23 EUR, United Kingdom 13725,"10x 25i Nbome 1200uq","LoveIsTheKey",25.5 EUR, United Kingdom 13726,"30x 25i Nbome 1200uq","LoveIsTheKey",64.66 EUR, United Kingdom 13727,"50x 25i Nbome 1200uq","LoveIsTheKey",97.44 EUR, United Kingdom 13728,"100x 2 CB HBr Capsule 18-20mg each Perfect Pure","LoveIsTheKey",322.

38 EUR, United Kingdom 13729,"50x 2 CB HBr Capsule 18-20mg Each Perfect Pure","LoveIsTheKey",213.1 EUR, United Kingdom 13735,"Etizolam 99% 10 gram's | FE","Empereor",455.83 EUR, China 13912,"Etizolam 99% purity 100 Gram's | FE","Empereor",1823.32 EUR, China 13918,"Furanyl fentanyl / Fu-F 99% pure 100 Gram`s ","Empereor",911.

66 EUR, China 13919,"Etizolam 99% purity 500 Gram`s | FE","Empereor",4558.

3 EUR, China 13921,"Etizolam 99% purity 50 Gram`s | FE","Empereor",1276.32 EUR, China 13929,"U-47700 10 Gram`s ","Empereor",364.66 EUR, China 13962,"Audemars Piguet - Box (AAA Grade Replica)","RepAAA",113.96 EUR, China 13963,"Audemars Piguet - Box "50% DISCOUNTED" (AAA Grade Replica)","RepAAA",68.37 EUR, China 13964,"Audemars Piguet - Royal Oak 15400 RG Black Dial AAA+ ","RepAAA",404.

78 EUR, China 13965,"Audemars Piguet - Royal Oak 15400 SS Black Dial AAA+ ","RepAAA",386.54 EUR, China 13966,"Audemars Piguet - Royal Oak 15400 SS Blue Dial AAA+ ","RepAAA",386.54 EUR, China 13967,"CARDING - CC to BTC - Donation Method","junkiepig666",4.55 EUR, Germany 13968,"Audemars Piguet - Royal Oak 15400 SS Silver Dial AAA+ ","RepAAA",386.54 EUR, China 13970,"CARDING - NEW CASHOUT METHOD CC OR PP TO BTC","junkiepig666",4.

55 EUR, Germany 13971,"CARDING - CC Cashout Guide - Rabbit Method","junkiepig666",4.55 EUR, Germany 14003,"Audemars Piguet - Royal Oak Lady SS Black Dial AAA+ ","RepAAA",295.38 EUR, China 14008,"Audemars Piguet - Royal Oak Offshore Arnold Schwarzenegger AAA+ ","RepAAA",623.58 EUR, China 14010,"Audemars Piguet - Royal Oak Offshore Chronograph Blue Dial AAA+ ","RepAAA",596.23 EUR, China 14011,"Audemars Piguet - Royal Oak Offshore Chronograph Forged Carbon A","RepAAA",723.

86 EUR, China 14013,"Audemars Piguet - Royal Oak Offshore Diver Carbon Ultimate AAA+ ","RepAAA",386.54 EUR, China 14014,"Audemars Piguet - Royal Oak Offshore Diver Ceramic AAA+ ","RepAAA",386.54 EUR, China 14015,"Audemars Piguet - Royal Oak Offshore Diver SS Ultimate AAA+ ","RepAAA",359.19 EUR, China 14016,"Audemars Piguet - Royal Oak Offshore Don Ramon de la Cruz Ultima","RepAAA",596.23 EUR, China 14017,"Audemars Piguet - Royal Oak Offshore Grand Prix Rose Gold AAA+ ","RepAAA",815.

02 EUR, China 14018,"Audemars Piguet - Royal Oak Offshore Rubberclad SS Ultimate AAA+","RepAAA",577.99 EUR, China 14019,"Audemars Piguet - Royal Oak Offshore Rubens Baricello II Titanium","RepAAA",641.81 EUR, China 14020,"Audemars Piguet - Royal Oak Offshore Sebastien Buemi AAA+ ","RepAAA",641.81 EUR, China 14048,"Breitling - Bentley Flying B Chronograph Black/Leather AAA+ ","RepAAA",568.88 EUR, China 14050,"Breitling - Box Large (AAA Grade Replica)","RepAAA",103.

93 EUR, China 14051,"Breitling - Box Small (AAA Grade Replica)","RepAAA",85.7 EUR, China 14052,"Breitling - Chronomat Evolution Black Dial / Metal AAA+ ","RepAAA",386.54 EUR, China 14053,"Breitling - Navitimer Black Dial Metal Band AAA+ ","RepAAA",432.13 EUR, China 14054,"Breitling - Navitimer Blue Dial Metal Band AAA+ ","RepAAA",432.13 EUR, China 14055,"Breitling - Navitimer Leather Strap AAA+ ","RepAAA",432.

13 EUR, China 14056,"Breitling - Super Avenger Black N-Dial / Metal AAA+ ","RepAAA",432.13 EUR, China 14058,"Breitling - Super Avenger White N-Dial / Leather AAA+ ","RepAAA",432.13 EUR, China 14059,"Breitling - Super Avenger White N-Dial / Metal AAA+ ","RepAAA",432.13 EUR, China 14060,"Breitling - Transocean Chronograph GMT LE Leather AAA+ ","RepAAA",432.13 EUR, China 14061,"Breitling - Transocean Chronograph GMT LE Metal AAA+ ","RepAAA",432.

13 EUR, China 14062,"Breitling - Transocean Chronograph RG AAA+ ","RepAAA",432.13 EUR, China 14063,"Breitling - Transocean Chronograph SS AAA+ ","RepAAA",386.54 EUR, China 14088,"MDMA 83% LAB TESTED 25g","the-company",133 EUR, Netherlands 14110,"WHITE INSTAGRAMS 200MG 50pcs","the-company",73 EUR, Netherlands 14117,"Cartier - Ballon Bleu Rose Gold Diamonds AAA+ ","RepAAA",240.68 EUR, China 14120,"Cartier - Calibre de Cartier Rose Gold Leather AAA+ ","RepAAA",322.73 EUR, China 14123,"Chanel - Box for J12 (AAA Grade Replica)","RepAAA",85.

7 EUR, China 14125,"Chanel - J12 Black Ceramic 38mm Ultimate AAA+ ","RepAAA",523.29 EUR, China 14128,"Chanel - J12 White Ceramic 38mm Ultimate AAA+ ","RepAAA",523.29 EUR, China 14130,"Chanel - Jewelry Watch J4281 Replica ","RepAAA",204.21 EUR, China 14134,"Chanel - Jewelry Watch J4282 RG Replica ","RepAAA",222.45 EUR, China 14194,"U-47700 100 Gram`s ","Empereor",1367.

49 EUR, China 14196,"U-47700 500 Gram`s ","Empereor",4102.47 EUR, China 14256,"Lyrica 150mg 7kpl *EI POSTIKULUJA* SILKKITIEN HALVIN*","Nappi Kauppa",31 EUR, Finland 14257,"Lyrica 150mg 14kpl *EI POSTIKULUJA* SILKKITIEN HALVIN*","Nappi Kauppa",50 EUR, Finland 14258,"Lyrica 150mg 2kpl *EI POSTIKULUJA* SILKKITIEN HALVIN*","Nappi Kauppa",12 EUR, Finland 14262,"0.3G 88% PURE FISHSCALE COCAINE","CocaineCowboy",33.5G OF SOCIABLE COCAINE","CocaineCowboy",155.

29 EUR, United Kingdom 14264,"14G OF SOCIABLE COCAINE","CocaineCowboy",527.92 EUR, United Kingdom 14270,"Dutch MDMA crystals 84% 1 gram ","Depanage",13 EUR, Netherlands 14299,"19 capsules filled with cannabis and coconut oil - Ship to USA","laWnmoWermAn",44.62 EUR, United States 14346,"Hublot - Big Bang All Black Carbon Fiber GR Ultimate AAA+ ","RepAAA",714.74 EUR, China 14347,"Hublot - Big Bang All Black Carbon Fiber Rubber Ultimate AAA+ ","RepAAA",714.74 EUR, China 14348,"Hublot - Big Bang All Black Ceramic Ultimate AAA+ ","RepAAA",568.

88 EUR, China 14349,"Hublot - Big Bang Black Magic Ceramic Ultimate AAA+ ","RepAAA",568.88 EUR, China 14352,"Hublot - Big Bang Ladies 38mm SS white Replica ","RepAAA",176.86 EUR, China 14353,"Hublot - Big Bang Red Magic Vendome Carbon Fiber Ultimate AAA+ ","RepAAA",714.74 EUR, China 14355,"Hublot - Big Bang Rose Gold Ultimate AAA+ ","RepAAA",568.88 EUR, China 14356,"1g Fresh Maroccan Double Zero","peaceandpot",11.

2 EUR, Spain 14357,"Hublot - Big Bang Rose Gold Ceramic Ultimate AAA+ ","RepAAA",587.11 EUR, China 14359,"5g Fresh Maroccan Double Zero - FREE EU SHIPPING","peaceandpot",42 EUR, Spain 14360,"10g Fresh Maroccan Double Zero - FREE EU SHIPPING ","peaceandpot",81 EUR, Spain 14363,"Hublot - Big Bang Stainless Steel Ceramic Ultimate AAA+ ","RepAAA",568.88 EUR, China 14365,"Hublot - Box (AAA Grade Replica)","RepAAA",131.28 EUR, China 14367,"Hublot - Box Window (AAA Grade Replica)","RepAAA",145.87 EUR, China 14371,"Hublot - Box Window "50% DISCOUNTED" (AAA Grade Replica)","RepAAA",86.

61 EUR, China 14377,"Hublot - Classic Fusion SS Black Dial ETA AAA+ ","RepAAA",432.13 EUR, China 14378,"Hublot - King Power Foudroyante Rose Gold AAA+ ","RepAAA",386.54 EUR, China 14381,"IWC - Portofino Automatic Black Dial AAA+ ","RepAAA",249.79 EUR, China 14382,"IWC - Portuguese Chronograph Black AAA+ ","RepAAA",368.31 EUR, China 14383,"IWC - Portuguese Chronograph Blue AAA+ ","RepAAA",368.

31 EUR, China 14384,"IWC - Portuguese Chronograph RG Silver Dial AAA+ ","RepAAA",386.54 EUR, China 14385,"IWC - Top Gun Pilots Chronograph ETA AAA+ ","RepAAA",432.13 EUR, China 14386,"IWC - Watch Box (AAA Grade Replica)","RepAAA",113.96 EUR, China 14390,"Jaeger-LeCoultre - Master Control RG ETA AAA+ ","RepAAA",249.79 EUR, China 14391,"Jaeger-LeCoultre - Master Control SS ETA AAA+ ","RepAAA",249.

79 EUR, China 14437,"BESTBUY CARDING TUTORIAL","junkiepig666",4.55 EUR, Germany 14462,"prestige pharma deca-300.15 EUR, United Kingdom 14463,"prestige pharma tren- enan 200 10ml vial ","stealthmeds",33.28 EUR, United Kingdom 14464,"prestige pharma tri-tren 225 10ml vial ","stealthmeds",33.

28 EUR, United Kingdom 14465,"prestige pharma test- enan 300 10ml vial ","stealthmeds",33.34 EUR, United Kingdom 14466,"prestige pharma test-prop 100 10ml vial ","stealthmeds",33.28 EUR, United Kingdom 14467,"prestige pharma mast-prop 10ml vial ","stealthmeds",33.34 EUR, United Kingdom 14468,"prestige pharma oxy 50.9 EUR, United Kingdom 14694,"Furanyl Fentanyl nasal spray","HappyDrugs",55 EUR, Czech Republic 14695,"0.3g China White Synthetic Heroin","HappyDrugs",35 EUR, Czech Republic 14696,"Opioid Liquid - O-Liquid 5 ML","HappyDrugs",85 EUR, Czech Republic 14707,"100mg Furanyl-Fentanyl - powerful cousin of fentanyl and heroin","HappyDrugs",59.26 EUR, Czech Republic 14708,"50 mg Furanyl-Fentanyl - powerful cousin of fentanyl and heroin","HappyDrugs",40 EUR, Czech Republic 14709,"250 mg Furanyl-Fentanyl - powerful cousin of fentanyl and heroin","HappyDrugs",109.28 EUR, Czech Republic 14710,"500 mg Furanyl-Fentanyl - powerful cousin of fentanyl and heroin","HappyDrugs",175 EUR, Czech Republic 14711,"1g Furanyl-Fentanyl - powerful cousin of fentanyl and heroin","HappyDrugs",290 EUR, Czech Republic 14712,"5g Furanyl-Fentanyl - powerful cousin of fentanyl and heroin","HappyDrugs",1140 EUR, Czech Republic 14718,"0.

5g China White Synthetic Heroin","HappyDrugs",35 EUR, Czech Republic 14719,"1g China White Synthetic Heroin","HappyDrugs",45 EUR, Czech Republic 14720,"3g China White Synthetic Heroin","HappyDrugs",70 EUR, Czech Republic 14721,"5g China White Synthetic Heroin","HappyDrugs",100 EUR, Czech Republic 14722,"10g China White Synthetic Heroin","HappyDrugs",170 EUR, Czech Republic 14723,"25g China White Synthetic Heroin","HappyDrugs",310 EUR, Czech Republic 14727,"50g China White Synthetic Heroin","HappyDrugs",525 EUR, Czech Republic 14728,"100g China White Synthetic Heroin","HappyDrugs",910 EUR, Czech Republic 14729,"250g China White Synthetic Heroin","HappyDrugs",2060 EUR, Czech Republic 14730,"500g China White Synthetic Heroin","HappyDrugs",3910 EUR, Czech Republic 14789,"1g - Salvia Divinorum - 10x Extract","b1g1mpact",22.95 EUR, Poland 14790,"1g - Salvia Divinorum - 20x Extract","b1g1mpact",34.5g - Salvia Divinorum - 10x Extract","b1g1mpact",15.5g - Salvia Divinorum - 20x Extract","b1g1mpact",22.95 EUR, Poland 14793,"28g/1oz - Kratom - "White Vein Sumatra"","b1g1mpact",29.63 EUR, Poland 14794,"28g/1oz - Kratom - "Green Malay"","b1g1mpact",29.63 EUR, Poland 14795,"28g/1oz - Kratom - "Red Vein Borneo"","b1g1mpact",29.

63 EUR, Poland 14796,"28g/1oz - Kratom - "Red Thai"","b1g1mpact",29.

6 EUR, Poland 14797,"28g/1oz - Kratom - "Maeng Da" Pimps Kratom","b1g1mpact",34.42 EUR, Poland 14799," 10 x 500mg capsules - Yohimbine 8% - Prescription","b1g1mpact",36.1 EUR, Poland 14800,"90g - Ephedra (Organic) - 3% Ephedrine, 3% Pseudoephedrine","b1g1mpact",61.47 EUR, Poland 14801,"1kg - Ephedra (Organic) - 3% Ephedrine, 3% Pseudoephedrine","b1g1mpact",312.61 EUR, Poland 14802,"100g / ~4oz - Mimosa Hostilis Rootbark - High Yield, Mexico","b1g1mpact",62.

36 EUR, Poland 14803,"28g / ~1oz - Mimosa Hostilis Rootbark - High Yield, Mexico","b1g1mpact",34.46 EUR, Poland 14804,"1kg - Mimosa Hostilis Rootbark - High Yield, Mexico","b1g1mpact",345.43 EUR, Poland 14805,"28g - Dioscorea Dregeana (African Wild Yam) - PHYSICAL SEDATIVE","b1g1mpact",34.46 EUR, Poland 14841,"Omega - Seamaster Aqua Terra 15'000 Gauss AAA+ ","RepAAA",295.38 EUR, China 14842,"Omega - Seamaster Aqua Terra White Dial AAA+ ","RepAAA",295.

38 EUR, China 14843,"Omega - Box (AAA Grade Replica)","RepAAA",113.96 EUR, China 14844,"Omega - Constellation Brushed RG Replica ","RepAAA",222.45 EUR, China 14845,"Omega - DeVille Ladymatic 2-tone brown Replica ","RepAAA",140.4 EUR, China 14846,"Omega - DeVille Ladymatic 2-tone MoP AAA+ ","RepAAA",249.79 EUR, China 14847,"Omega - DeVille Ladymatic 2-tone white Replica ","RepAAA",140.

4 EUR, China 14849,"Omega - Seamaster Planet Ocean 600m Orange ETA AAA+ ","RepAAA",432.13 EUR, China 14860,"Omega - Speedmaster '57 Chronograph ETA AAA+ ","RepAAA",505.06 EUR, China 14874,"SQL INJECTION PACKAGE + DORKS + GUIDE","Citra",4 EUR, United Kingdom 14875,"10GB FAKE ID + CC FORGERY + MONEY COUNTERFEIT MEGA","Citra",49 EUR, Australia 14876,"Lithuanian Passport+DL Premium Bank quality","Citra",1691.13 EUR, Lithuania 14877,"XTC Coca Cola 200-220mg 5 pcs","Depanage",20 EUR, Netherlands 14878,"Lithuanian Passport Premium Bank quality","Citra",1326.47 EUR, Lithuania 14887,"U-47700 1 Kg ","Empereor",5469.

96 EUR, China 14890,"Furanyl fentanyl / Fu-F 99% pure 500 Gram`s ","Empereor",3190.81 EUR, China 14892,"XTC Coca Cola 200-220mg 10 pcs","Depanage",29 EUR, Netherlands 14893,"Quality cocaine 0.5 gram 80%","Depanage",32 EUR, Netherlands 14894,"Quality cocaine 1 gram 80%","Depanage",50 EUR, Netherlands 14908,"DUTCH SPEED 74% PURE 5GR","hapygarden",15.48 EUR, Netherlands 14913,"DUTCH SPEED 74% PURE 1GR","hapygarden",7.29 EUR, Netherlands 14914,"DUTCH SPEED 74% PURE 10GR","hapygarden",25.

5 EUR, Netherlands 14915,"DUTCH SPEED 74% PURE 25GR","hapygarden",52.82 EUR, Netherlands 14916,"DUTCH SPEED 74% PURE 50GR","hapygarden",89.25 EUR, Netherlands 14918,"dutch mdma cristals 84% pure 5gr","hapygarden",49.18 EUR, Netherlands 14919,"dutch mdma cristals 84% pure 1gr","hapygarden",13.66 EUR, Netherlands 14922,"dutch mdma cristals 84% pure 10gr","hapygarden",87.

42 EUR, Netherlands 14923,"dutch mdma cristals 84% pure 25gr","hapygarden",192.15 EUR, Netherlands 14940,"Omega - Speedmaster Dark Side Of The Moon ETA AAA","RepAAA",523.29 EUR, China 14941,"Omega - Speedmaster Lady Chrono Diamond Replica ","RepAAA",149.51 EUR, China 14942,"Omega - Speedmaster Racing Black SS Band ETA AAA+ ","RepAAA",386.54 EUR, China 14943,"Omega - Speedmaster Racing Red ETA AAA+ ","RepAAA",350.

08 EUR, China 14944,"Panerai - Box (AAA Grade Replica)","RepAAA",113.96 EUR, China 14945,"Panerai - Luminor 1950 3 Days GMT PAM441 Replica ","RepAAA",277.14 EUR, China 14959,"Panerai - Luminor 1950 3 Days GMT PAM441 Ultimate AAA+ ","RepAAA",432.13 EUR, China 14992,"Alprazolam 99% pure 10 Gram's ","Empereor",455.83 EUR, China 14993,"Alprazolam 99% pure 50 Gram`s ","Empereor",1367.

49 EUR, China 14994,"Alprazolam 99% pure 100 Gram`s ","Empereor",2005.65 EUR, China 14996,"Alprazolam 99% pure 250 Gram`s ","Empereor",3641.74 EUR, China 15002,"S-isomer Ketamine 1 gram","Depanage",23 EUR, Netherlands 15056,"Panerai - Luminor 1950 Regatta Rattrapante PAM332 AAA+ ","RepAAA",523.29 EUR, China 15057,"Panerai - Luminor Marina 1950 PAM312 AAA+ ","RepAAA",441.24 EUR, China 15058,"Panerai - Luminor Marina 1950 PAM565 Ultimate AAA+ ","RepAAA",386.

54 EUR, China 15059,"Panerai - Luminor Marina 1950 Titanium PAM351 Ultimate AAA+ ","RepAAA",432.13 EUR, China 15060,"Panerai - Luminor Marina PAM111 ETA Ultimate AAA+ ","RepAAA",368.31 EUR, China 15061,"Panerai - Luminor Power Reserve Leather PAM090 Replica ","RepAAA",277.14 EUR, China 15062,"Panerai - Luminor Submersible 1950 Amagnetic PAM389 Ultimate AAA","RepAAA",450.

36 EUR, China 15063,"Panerai - Luminor Submersible 1950 Bronzo PAM382 Ultimate AAA+ ","RepAAA",523.

29 EUR, China 15064,"Panerai - Luminor Submersible 1950 DLC PAM508 AAA+ ","RepAAA",432.13 EUR, China 15065,"Panerai - Luminor Submersible Left-Hand PAM569 Ultimate AAA+ ","RepAAA",432.13 EUR, China 15066,"Panerai - Luminor Submersible PAM305 Replica ","RepAAA",295.38 EUR, China 15068,"Panerai - Radiomir Base PAM231 Rose Gold Ultimate AAA+ ","RepAAA",477.71 EUR, China 15069,"Panerai - Radiomir California Special Edition PAM448 AAA+ ","RepAAA",340.

96 EUR, China 15072,"Patek Philippe - Nautilus RG Black ETA AAA+ ","RepAAA",368.31 EUR, China 15073,"Patek Philippe - Nautilus SS Blue ETA AAA+ ","RepAAA",432.13 EUR, China 15074,"Patek Philippe - Watch Box (AAA Grade Replica)","RepAAA",140.4 EUR, China 15075,"Richard Mille - RM011 Felipe Massa Chronograph Black AAA+ ","RepAAA",523.29 EUR, China 15076,"Richard Mille - RM011 Felipe Massa Chronograph White AAA+ ","RepAAA",523.

29 EUR, China 15077,"Richard Mille - Tourbillon RM 052 Skull AAA+ ","RepAAA",523.29 EUR, China 15078,"Rolex - Cellini Rose Gold white AAA+ ","RepAAA",295.38 EUR, China 15079,"Rolex - Cellini Stainless Steel white AAA+ ","RepAAA",295.38 EUR, China 15086,"Rolex - Datejust II 2-tone Gold Dial ETA AAA+ ","RepAAA",322.73 EUR, China 15087,"Rolex - Datejust II SS Black Dial ETA AAA+ ","RepAAA",295.

38 EUR, China 15089,"Rolex - Datejust II SS Black Roman Dial ETA AAA+ ","RepAAA",295.38 EUR, China 15091,"Rolex - DateJust Lady 26mm MoP Jubilee ETA AAA+ ","RepAAA",295.38 EUR, China 15164,"2 gram quality fishscale cocaine","tennin",131 EUR, Netherlands 15165,"1 gram quality fishscale cocaine","tennin",68 EUR, Netherlands 15166,"1.5 gram quality fishscale cocaine","tennin",100 EUR, Netherlands 15167,"0.63 EUR, China 15216,"100g "Mapacho" - Nicotiana Rustica 9x stronger than Tobacco","b1g1mpact",85.33 EUR, Poland 15217,"28g "Mapacho" - Nicotiana Rustica 9x stronger than Tobacco","b1g1mpact",28.69 EUR, Poland 15218,"9g "Yopo" / "Cebil" Seeds (Bufotenine & 5-Meo-DMT)","b1g1mpact",24.59 EUR, Poland 15219,"Ayahuasca - 100% Authentic pack for 1 or 2","b1g1mpact",85.24 EUR, Poland 15220,"Pharmahuasca - 100% Authentic pack for 1","b1g1mpact",41.

53 EUR, Poland 15221,"10g High Quality 99.9% pure 5-HTP - 5-HydroxyTryptophan","b1g1mpact",32.82 EUR, Poland 15222,"1g - Iboga TA extraction","b1g1mpact",150.5g - Iboga TA (Total Alkaloid) extraction","b1g1mpact",85.

33 EUR, Poland 15224,"3g - Iboga TA (Total Alkaloid) Extract","b1g1mpact",429.94 EUR, Poland 15225,"20g - Iboga Rootbark - High Quality, Cameroon","b1g1mpact",101.74 EUR, Poland 15226,"108g - Iboga Rootbark - High Quality, Cameroon","b1g1mpact",477.73 EUR, Poland 15227,"54g - Iboga Rootbark - High Quality, Cameroon","b1g1mpact",269.12 EUR, Poland 15228,"10g Syrian Rue Seeds (Peganum Harmala) - Iran","b1g1mpact",21.

52 EUR, Poland 15229,"100g Syrian Rue Seeds (Peganum Harmala) - Iran","b1g1mpact",88.61 EUR, Poland 15297,"Flavored 30ml E-Liquid 8mg Nicotine BUY 1 GET 1 FREE ","ESLight",15.48 EUR, United States 15315,"Starbuzz Electronic Hookah Cartridges Pack of 4 ","ESLight",10.94 EUR, United States 15318,"Rechargeable Liquid Oil Vaporizer Single 510 thread FREE ELIQUID ","ESLight",23.68 EUR, United States 15352,"5g Swazi Gold Kief","Swazibudbud888",18.

23 EUR, South Africa 15354,"1oz Swazi Gold Kief (30g)","Swazibudbud888",81.14 EUR, South Africa 15355,"1oz(30g) Royal Swazi (Sativa)","Swazibudbud888",79.77 EUR, South Africa 15356,"2oz(60g) Royal Swazi (Sativa)","Swazibudbud888",136.75 EUR, South Africa 15357,"4oz(120g) Royal Swazi (Sativa)","Swazibudbud888",250.71 EUR, South Africa 15359,"7g Royal Swazi (Sativa)","Swazibudbud888",36.

47 EUR, South Africa 15361,"2g Royal Swazi (Sativa) ","Swazibudbud888",9.11 EUR, South Africa 15362,"1oz Swazi Rooibaard","Swazibudbud888",63.82 EUR, South Africa 15364,"2oz Swazi Rooibaard","Swazibudbud888",113.96 EUR, South Africa 15365,"4oz Swazi Rooibaard (120g)","Swazibudbud888",209.

45 EUR, South Africa 15373,"BK-ETHYL-K CRYSTAL / Ephylone / BK-EBDP 1 Kg ","Empereor",2734.

98 EUR, China 15374,"BK-ETHYL-K CRYSTAL / Ephylone / BK-EBDP 500 Gram`s ","Empereor",1640.99 EUR, China 15383,"1 x Hanumen 160ug","Petch",13.09 EUR, United Kingdom 15385,"2 x Hoffman Acid 100ug","Petch",7.54 EUR, United Kingdom 15391,"2 x Hanumen 160ug","Petch",19.74 EUR, United Kingdom 15392,"5 x Hanumen 160ug","Petch",38.

6 EUR, United Kingdom 15393,"10 x Hanumen 160ug","Petch",59.12 EUR, United Kingdom 15394,"15 x Hanumen 160ug","Petch",83.52 EUR, United Kingdom 15395,"25 x Hanumen 160ug","Petch",123.45 EUR, United Kingdom 15412,"Afghan "Mazari-i-Sharif" - 20g","DutchMagic",59 EUR, Netherlands 15413,"Afghan "Mazari-i-Sharif" - 50g","DutchMagic",144 EUR, Netherlands 15414,"Afghan "Mazari-i-Sharif" - 100g","DutchMagic",270 EUR, Netherlands 15416,"Afghan "Mazari-i-Sharif" - 200g","DutchMagic",526 EUR, Netherlands 15418,"Afghan "Mazari-i-Sharif" - 1 KILOGRAM","DutchMagic",2310 EUR, Netherlands 15420,"White Widow - 5g","DutchMagic",46 EUR, Netherlands 15421,"White Widow - 10g","DutchMagic",88 EUR, Netherlands 15422,"White Widow - 20g","DutchMagic",168 EUR, Netherlands 15423,"White Widow - 50g","DutchMagic",399 EUR, Netherlands 15424,"White Widow - 100g","DutchMagic",776 EUR, Netherlands 15425,"White Widow - 250g","DutchMagic",1805 EUR, Netherlands 15427,"White Widow - 1 KILOGRAM","DutchMagic",5990 EUR, Netherlands 15430,"Tracked Shipping","DutchMagic",20 EUR, Netherlands 15431,"Amnesia Haze - 5g","DutchMagic",51 EUR, Netherlands 15471,"Rechargeable EVOD Blister Kit Liquid Oil Vaporizer 510 thread ","ESLight",17.32 EUR, United States 15477,"5 x Hydromorphone 8mg tabs (Dilaudid) 100% Genuine","fasterndsaffer",132.

19 EUR, United States 15478,"5 roxy30 mg only $32each !!!!! A215/K9/M30/a51","fasterndsaffer",145.71 EUR, United States 15480,"10 pills methadone 10mg PRICE ONLY $125 US stock ","fasterndsaffer",113.96 EUR, United States 15482,"10 pc Adderall 30 mg real deal","fasterndsaffer",182.13 EUR, United States 15483,"adderalL 20 mg real deal US Pharm 10 pills","fasterndsaffer",136.6 EUR, United States 15494,"FINALIZE EARLY 200.

00g MDMA CLEAN & UNCUT","kacikz",2023.89 EUR, Lithuania 15519,"Rolex - Datejust Silver & Diamond Jubilee ETA AAA+ ","RepAAA",249.79 EUR, China 15520,"Rolex - Datejust Silver & Diamond Oyster AAA+ ","RepAAA",249.79 EUR, China 15521,"Rolex - Datejust SS Black Dial Roman ETA AAA+ ","RepAAA",249.79 EUR, China 15522,"Rolex - Daydate II Ice Blue Dial Replica ","RepAAA",158.

63 EUR, China 15524,"Rolex - Daydate President Full Yellow Gold AAA+ ","RepAAA",295.38 EUR, China 15527,"Suboxone FILM 8mg sa! 2kpl.","rehtitukkku2",83 EUR, Finland 15531,"Rolex - Daydate President Full Yellow Gold Iced-Out AAA+ ","RepAAA",477.71 EUR, China 15532,"Rolex - Daydate President SS black AAA+ ","RepAAA",295.38 EUR, China 15533,"Rolex - Daydate President Stainless Steel Iced-Out AAA+ ","RepAAA",459.

48 EUR, China 15534,"Rolex - Daytona "Project X Design" DS1 LE AAA+ ","RepAAA",340.96 EUR, China 15535,"Rolex - Daytona 2-Tone Black Dial ETA AAA+ ","RepAAA",322.73 EUR, China 15537,"Rolex - Daytona 2-Tone Gold Dial ETA AAA+ ","RepAAA",322.73 EUR, China 15539,"Rolex - Daytona 2-Tone White Dial ETA AAA+ ","RepAAA",322.73 EUR, China 15548,"Rolex - Daytona 50th Anniversary ETA AAA+ ","RepAAA",350.

08 EUR, China 15549,"Rolex - Daytona RG Ceramic Bezel AAA+ ","RepAAA",340.96 EUR, China 15550,"Rolex - Daytona RG Chocolate Dial Ceramic Bezel AAA+ ","RepAAA",340.96 EUR, China 15552,"Rolex - Daytona Stainless Steel Black Dial ETA AAA+ ","RepAAA",322.73 EUR, China 15553,"Rolex - Daytona Stainless Steel White Dial ETA AAA+ ","RepAAA",322.73 EUR, China 15554,"Rolex - Daytona Yellow Gold ETA AAA+ ","RepAAA",322.

73 EUR, China 15555,"Rolex - Daytona Yellow Gold Rainbow ETA AAA+ ","RepAAA",432.13 EUR, China 15623,"4-CL-PVP Crystal 500 Gram`s ","Empereor",1640.99 EUR, China 15624,"2-NMC Crystal 500 Gram`s ","Empereor",1549.82 EUR, China 15625,"2-NMC Crystal 1 Kg ","Empereor",2552.65 EUR, China 15650,"Ambien (zolpidem) Generic 10 MG 30 Pills","DeepMeds",82.

87 EUR, United States 15658,"Restoril (temazepam) 30 MG 100 Caps","DeepMeds",173.94 EUR, United States 15660,"Restoril (temazepam) 30 MG 500 Caps","DeepMeds",629.96 EUR, United States 15661,"Restoril (temazepam) 30 MG 1000 Caps","DeepMeds",1135.93 EUR, United States 15685,"Ativan (Lorazepam) 2 MG 500 Pills","DeepMeds",424.83 EUR, United States 15686,"Ativan (Lorazepam) 2 MG 1000 Pills","DeepMeds",762.

15 EUR, United States 15690,"Xanax (Alprazolam) 1 MG 200 Pills","DeepMeds",210.59 EUR, United States 15695,"Rolex - Explorer ETA AAA+ ","RepAAA",249.79 EUR, China 15696,"Rolex - Explorer II Black Dial Replica ","RepAAA",158.63 EUR, China 15697,"Rolex - Explorer II White Dial Replica ","RepAAA",158.

63 EUR, China 15698,"Rolex - GMT Master II 2-tone AAA+ ","RepAAA",268.

03 EUR, China 15699,"Rolex - GMT Master II Black AAA+ ","RepAAA",295.38 EUR, China 15700,"Rolex - GMT Master II Black Replica ","RepAAA",158.63 EUR, China 15701,"Rolex - GMT Master II Black Ultimate AAA+ ","RepAAA",477.71 EUR, China 15702,"Rolex - GMT Master II Black/Blue 2013 Replica ","RepAAA",158.63 EUR, China 15703,"Rolex - GMT Master II Pro-Hunter ETA AAA+ ","RepAAA",322.

73 EUR, China 15704,"Rolex - GMT Master II Vintage Pepsi Bezel ETA AAA+ ","RepAAA",295.38 EUR, China 15705,"Rolex - GMT Master II Yellow Gold AAA+ ","RepAAA",322.73 EUR, China 15712,"S-isomer Ketamine 3 gram","Depanage",69 EUR, Netherlands 15729,"Valium (Diazepam) 10 MG By Roche 500 Pills ","DeepMeds",424.83 EUR, United States 15750,"HIGH LIMIT SWISS VISA CARD RELOADABLE LEGIT TUT","bank",9.12 EUR, United States 15751,"18 Extensions For Turning Firefox Into a Penetration Testing Tool","bank",4.

56 EUR, United States 15752,"PAYPAL CASHOUT 2015 TUTORIALS + BONUS","bank",13.67 EUR, United States 15753,"5 ATM HACKING TUTORIALS","bank",8.2 EUR, United States 15754,"Amazon Stealth Complete Guide to a New Amazon Identity","bank",4.56 EUR, United States 15755,"ANAL SEX TUTORIAL : HOW TO MAKE HER LOVE ANAL SEX","bank",3.65 EUR, United States 15756,"ANDROID SECURITY PRIVACY AND ENCRYPTION PACK","bank",4.

56 EUR, United States 15758,"ANY PRODUCT FROM CHINA , FREE CONSULTATION","bank",0 EUR, United States 15760,"ATM Skimmer Cashing / Installing Safety","bank",4.56 EUR, United States 15762,"how to AVOID CAMERA DETECTION","bank",5.47 EUR, United States 15763,"BANK and CC TRANSFERS FOR BEGINNERS","bank",4.56 EUR, United States 15764,"HOW TO BE A PERPETUAL TRAVELLER","bank",4.56 EUR, United States 15766,"BITCOIN MINING GUIDE","bank",4.

56 EUR, United States 15767,"BYE BYE BIG BROTHER BBBB - 500 usd orginal value","bank",22.79 EUR, United States 15768,"Bypass BIOS Passwords","bank",3.65 EUR, United States 15770,"Carding Online ","bank",4.56 EUR, United States 15771,"Carding tutorial for Noobs","bank",4.56 EUR, United States 15774,"carding with cvv2","bank",4.

56 EUR, United States 15775,"BASIC CARD PRINTING TUTORIAL","bank",4.56 EUR, United States 15777,"CASHOUT METOD ","bank",4.56 EUR, United States 15778,"Cell phone tracking TUTORIAL ","bank",4.56 EUR, United States 15805,"INVITE TO DREAM MARKET + SILKKITIE + HANSA MARKETS","bank",0 EUR, United States 15806,"█ █ ★ HOW TO DISAPPEAR - BIG PACK ★█ █","bank",13.67 EUR, United States 15807,"█ █★ CREATE HIGH CREDIT SCORE GHOSTS IN BOREAUS SYSTEMS ★ █ █","bank",13.

67 EUR, United States 15808,"★ ★ ★ 200$ DAILY WITH AUTOPILOT - 100% WORKING OR I REFUND ★ ★ ★","bank",9.12 EUR, United States 15809,"█ █ ★ CHANGE YOUR IDENTITY 2015 ★ █ █","bank",18.23 EUR, United States 15810,"█ █★ E- BOOK OF FORBIDDEN KNOWLEDGE ★ █ █","bank",8.2 EUR, United States 15811,"CREDIT SECRETS PACK - BOOST CREDIT SCORE - GET A NEW CREDIT PROFI","bank",13.67 EUR, United States 15813,"█ █ CC TO BTC 2015 - 100% WORKING OR I REFUND █ █","bank",13.

67 EUR, United States 15815,"10,000 or higher credit card even if your credit sucks","bank",9.12 EUR, United States 15817,"RANDOM POLISH ID + UB SCANS","bank",13.67 EUR, United States 15818,"█ █ Document Fraud and Other Crimes Of Decerption █ █","bank",8.2 EUR, United States 15827,"█ Starbucks Cashout Method █","bank",9.12 EUR, United States 15828,"ANONYMOUS VBA TO SEND AND RECEIVE BANK TRANSFERS","bank",6.

38 EUR, United States 15829,"█ HIGH LIMITS BITCOIN DEBIT CARD █ PLASTIC MASTERCARD █","bank",13.67 EUR, United States 15830,"★ GUIDE TO RESOLVING LIMITED PAYPAL ACCOUNTS★","bank",3.65 EUR, United States 15831,"★ PAYPAL VERIFIED PREMIER OR BUSINESS ACCOUNT ★ ★ ★","bank",50.14 EUR, United States 15832,"★ ★ ★ INCOME TAX RETURNS FRAUD U.67 EUR, United States 15833,"★ HOW TO START YOUR OWN INTERNATIONAL BANK ★","bank",4.56 EUR, United States 15834,"SECURITY MEGA TUTORIAL","bank",13.67 EUR, United States 15835,"BIGGEST COLLECTION OF .

56 EUR, United States 15836,"★ Credit building credit cards tips and tricks ★","bank",3.65 EUR, United States 15837,"█ █★ 100% LEGAL TACTICAL PEN - SELF DEFENCE ★ █ █","bank",19.14 EUR, China 15838,"★ ESSENTIAL UNDERGROUND HANDBOOK OF SECRET INFORMATIONS★","bank",5.47 EUR, United States 15857,"LIQUID MUSHROOMS Pure Psilocybin No Nausea, Faster Trip, Cleaner","TripWithScience",18.19 EUR, United States 15870,"HEROIN # 3 ECELLENT QUALITY 130$ OR 90£ UK Vendor","pure12",120.

21 EUR, United Kingdom 15871,"HEROIN #3 AFGHANI HEROIN 80$ = 0.67 EUR, United Kingdom 15888,"10 Endocet - the ORIGINALS of Percs , Yellow Legend","fasterndsaffer",150.26 EUR, United States 15914,"30x (Zolpidem 10 mg) (ambien)","fasterndsaffer",109.28 EUR, United States 15917,"POWERPLANT 3 GRAM freeshipping","DutchDreams",36 EUR, Netherlands 15919,"POWERPLANT 5 GRAM freeshipping","DutchDreams",56 EUR, Netherlands 15921,"POWERPLANT 25 GRAM freeshipping","DutchDreams",236 EUR, Netherlands 15922,"POWERPLANT 50 GRAM freeshipping","DutchDreams",446 EUR, Netherlands 15923,"POWERPLANT 100 GRAM freeshipping","DutchDreams",834 EUR, Netherlands 15930,"AMNESIA HAZE 5 GRAM freeshipping","DutchDreams",58 EUR, Netherlands 15931,"AMNESIA HAZE 3 GRAM freeshipping","DutchDreams",40 EUR, Netherlands 15932,"AMNESIA HAZE 10 GRAM freeshipping","DutchDreams",110 EUR, Netherlands 15933,"AMNESIA HAZE 25 GRAM freeshipping","DutchDreams",265 EUR, Netherlands 15934,"AMNESIA HAZE 50 GRAM freeshipping","DutchDreams",485 EUR, Netherlands 15935,"AMNESIA HAZE 100 GRAM freeshipping","DutchDreams",917 EUR, Netherlands 15960,"POLM 3 GRAM freeshipping","DutchDreams",36 EUR, Netherlands 15961,"POLM 5 GRAM freeshipping","DutchDreams",56 EUR, Netherlands 15962,"POLM 10 GRAM freeshipping","DutchDreams",101 EUR, Netherlands 15963,"POLM 25 GRAM freeshipping","DutchDreams",236 EUR, Netherlands 15964,"POLM 50 GRAM freeshipping","DutchDreams",446 EUR, Netherlands 15966,"POLM 100 GRAM freeshipping","DutchDreams",834 EUR, Netherlands 16028,"█ █ ★ HOW TO DISAPPEAR - BIG PACK ★█ █","bank",13.

67 EUR, United States 16032,"█ █★ CREATE HIGH CREDIT SCORE GHOSTS IN BOREAUS SYSTEMS ★ █ █","bank",13.67 EUR, United States 16034,"★ ★ ★ 200$ DAILY WITH AUTOPILOT - 100% WORKING OR I REFUND ★ ★ ★","bank",9.12 EUR, United States 16037,"█ █★ E- BOOK OF FORBIDDEN KNOWLEDGE ★ █ █","bank",8.2 EUR, United States 16038,"█ █ ★ CHANGE YOUR IDENTITY 2016 ★ █ █","bank",18.23 EUR, United States 16040,"CREDIT SECRETS PACK - BOOST CREDIT SCORE - GET A NEW CREDIT PROFI","bank",13.

67 EUR, United States 16041,"★ INSIDE THE RUSSIAN UNDERGROUND HACKERS COMMUNITY ★","bank",5.47 EUR, United States 16042,"█ █ CC TO BTC 2015 - 100% WORKING OR I REFUND █ █","bank",13.67 EUR, United States 16043,"11 FBI FORENSIC TOOLS","bank",4.56 EUR, United States 16044,"RANDOM POLISH ID + UB SCANS","bank",13.67 EUR, United States 16045,"10,000 or higher credit card even if your credit sucks","bank",8.

2 EUR, United States 16046,"█ █ Document Fraud and Other Crimes Of Decerption █ █","bank",8.2 EUR, United States 16047,"CVV to Moneygram cashout now! IF NEVER WORK I REFUND","bank",4.56 EUR, United States 16048,"★ ★ ★ ESSENTIAL UNDERGROUND HANDBOOK OF SECRET INFORMATIONS ★ ★ ★","bank",3.65 EUR, United States 16049,"HOW TO CASHOUT STOLEN PAYPAL ACCOUNTS - 100% WORKING OR I REFUND!","bank",13.67 EUR, United States 16050,"█ Starbucks Cashout Method █","bank",9.

12 EUR, United States 16052,"ANONYMOUS VBA TO SEND AND RECEIVE BANK TRANSFERS","bank",6.38 EUR, United States 16054,"█ HIGH LIMITS BITCOIN DEBIT CARD █ PLASTIC MASTERCARD █","bank",13.67 EUR, United States 16057,"★ ★ ★ INCOME TAX RETURNS FRAUD U.67 EUR, United States 16058,"★ HOW TO START YOUR OWN INTERNATIONAL BANK ★","bank",4.56 EUR, United States 16059,"SECURITY MEGA TUTORIAL","bank",13.67 EUR, United States 16061,"★ Credit building credit cards tips and tricks ★","bank",3.65 EUR, United States 16062,"█ █★ 100% LEGAL TACTICAL PEN - SELF DEFENCE ★ █ █","bank",10.03 EUR, United States 16063,"★ ESSENTIAL UNDERGROUND HANDBOOK OF SECRET INFORMATIONS★","bank",5.

47 EUR, United States 16064,"★ Burner Phone Tips ★","bank",4.56 EUR, United States 16065,"█ █ Chaining a Socks5 With Tor TUTORIAL █ █","bank",4.56 EUR, United States 16066,"★ How to Create Transaction History for Your Paypal Account ★","bank",4.56 EUR, United States 16067,"★ HOW TO PASS COUNTERFEIT MONEY ★","bank",9.12 EUR, United States 16068,"ULTIMATE EBAY GUIDE FOR DROP SHIPPING","bank",8.

2 EUR, United States 16069,"★ ★ ★ LOCKPICKING 007 STEALTH CARD ★ ★ ★","bank",31.91 EUR, China 16070,"█ █★ Enter the Dark Net ★ █ █","bank",4.56 EUR, United States 16073,"★ HOW TO FIND RDP SERVERS AND ACCESS IT ★","bank",4.56 EUR, United States 16074,"Norge: 3 Ferdigrullede Sativa Jointer","hogstandard",33.58 EUR, Norway 16075,"★ Make $4000 a month with only 8 hours of work per week on Fiverr","bank",4.

56 EUR, United States 16076,"Norge: 1 Ferdigrullet Sativa Joint","hogstandard",15.96 EUR, Norway 16078,"ID TEMPLATES: CANADA","bank",18.23 EUR, United States 16079,"WHERE TO GET CVV","bank",13.67 EUR, United States 16080,"★ WHERE TO BUY DUMPS CVVs BINS ★","bank",9.

12 EUR, United States 16081,"★ FIND OUT IF BIN IS BLOCKED IN YOUR REGION ★","bank",6.

38 EUR, United States 16082,"★ BANK and CC TRANSFERS FOR BEGINNERS ★","bank",9.12 EUR, United States 16083,"★ WENSTERN UNION FLAW FRAUD ★","bank",4.56 EUR, United States 16084,"How to Remove Find my iPhone app with iCloud id Activations","bank",9.12 EUR, United States 16085,"★ WHERE TO BUY STOLEN BANK , PAYPAL ACCOUNTS ★","bank",13.67 EUR, United States 16087,"█ █ CVV ssn dob + Background Check & Credit Report █ █","bank",13.

67 EUR, United States 16088,"█ █ ★ VENMO CASHOUT GUIDE ★ █ █","bank",22.79 EUR, United States 16089,"★ ★ BASIC CARD PRINTING TUTORIAL ★ ★","bank",6.38 EUR, United States 16091,"★ 2 TRAVEL HACKING GUIDEs ★","bank",9.12 EUR, United States 16096,"★ COVER YOUR TRACKS - for hackers ★","bank",3.65 EUR, United States 16097,"★ ★ ★ PENIS ENLARGEMENT + CUM INCREASE ★ ★ ★","bank",6.

38 EUR, United States 16098,"★ UK ACCOUNT - LOAN FRAUD GUIDE ★","bank",13.67 EUR, United States 16100,"23 various editable utility bills + cards templates UK","bank",4.56 EUR, United States 16101,"★ SPANISH ID + UB HIGH QUALITY PICS ESPANA ★","bank",4.56 EUR, United States 16102,"★ GET FULL ACCOUNT INFO FROM WELLS FARGO ★","bank",9.12 EUR, United States 16103,"★ HIGH LIMITS ANON BTC DEBIT CARD EUR USD GBP WITH IBAN ★","bank",0.

91 EUR, United States 16104,"★ How to Find Unclaimed Money and Property ★","bank",4.56 EUR, United States 16105,"★ FACEBOOK MANIAC GOLDEN PACK ★","bank",22.79 EUR, United States 16109,"★ no limits black visa debit card ★","bank",9.12 EUR, United States 16110,"ID TEMPLATES: France","bank",18.23 EUR, United States 16111,"★ FAKE ID OR DRIVER LICENCE - WHERE TO BUY IT ★","bank",4.

56 EUR, United States 16112,"█ █★ HOW TO SHIP DRUGS STEALTHY - TUTORIALS AND VIDEO ★ █ █","bank",13.67 EUR, United States 16113,"★ DUMPS generating algorithms ★","bank",4.56 EUR, United States 16114,"★ HIGH LIMIT SWISS VISA CARD RELOADABLE LEGIT TUT★","bank",13.67 EUR, United States 16116,"★ ★ ★ KNUCKLE BLASTER STUN GUN 950,000 volt ★ ★ ★","bank",58.35 EUR, China 16117,"18 Things Every 18 Year Old Should Know","bank",2.

73 EUR, United States 16119,"DUMPS generating algorithms","bank",4.56 EUR, United States 16120,"█ █ ★ BASIC CARD PRINTING TUTORIAL ★ █ █","bank",9.12 EUR, United States 16121,"█ █★ HOW TO SHIP DRUGS STEALTHY - TUTORIALS AND VIDEO ★ █ █","bank",13.67 EUR, United States 16124,"★ TRAVEL GHOST - travel without ID ★","bank",8.2 EUR, United States 16125,"Amazon Stealth Complete Guide to a New Amazon Identity","bank",4.

56 EUR, United States 16126,"CONNECTING TOR > VPN FOR WINDOWS USERS","bank",4.56 EUR, United States 16127,"where to buy fake ids","bank",4.56 EUR, United States 16128,"█ █★ BYE BYE BIG BROTHER BBBB - 500 usd orginal value ★ █ █","bank",13.67 EUR, United States 16129,"GET PERFECT MIDDLEMAN PAYPAL ACCOUNT","bank",17.32 EUR, United States 16130,"★ MEGA LIST OF .

ONION WEB SITES , DEEP WEB LINKS ★","bank",6.38 EUR, United States 16131,"HOW TO MAKE FAKE USD BILLS","bank",13.67 EUR, United States 16132,"★ CC TO PAYPAL - IF NEVER WORK I REFUND YOU ★","bank",9.12 EUR, United States 16133,"Create perfect PayPal accounts for beginners - VIDEOGUIDE","bank",13.67 EUR, United States 16134,"HOW TO PAY DIRECT INTO EU PAYPAL PREMIER USING CCs AND FRAUDFOX","bank",9.

12 EUR, United States 16135,"★ The Ultimate Guide to Texting Girls ★","bank",5.47 EUR, United States 16136,"★ CHANGE ID BIBLE ★","bank",6.38 EUR, United States 16137,"★ EBAY FRAUDS BIG PACK , buy anything for free ★","bank",13.67 EUR, United States 16138,"★ TRAVEL INDUSTRY SECRETS ★","bank",2.73 EUR, United States 16139,"█ █ ★ MEGAPACK PROJECT - extreme collection ★ █ █","bank",147.

18 EUR, United States 16140,"★ USA PASSPORTS .32 EUR, United States 16141,"AUSTRALIAN BANK DROPS TUTORIAL","bank",13.67 EUR, United States 16142,"★ how to AVOID CAMERA DETECTION ★","bank",4.

56 EUR, United States 16143,"★ THE TOR USER SECURITY BIBLE -133 PAGES ★","bank",4.

56 EUR, United States 16144,"★ BANK AND ARMORED CAR ROBBERY ★","bank",4.56 EUR, United States 16145,"E-WHORING MEGAPACK","bank",4.56 EUR, United States 16146,"29+ selling on eBay tricks","bank",4.56 EUR, United States 16147,"REVENGE - MAKE 'EM PAY ! -2 EBOOKS","bank",8.2 EUR, United States 16148,"★█ █ ★ MEGA PACKAGE OF TUTORIALS ★ █ █★","bank",8.

2 EUR, United States 16149,"★ CARDING TUTORIALS MEGA COLLECTION ★","bank",13.67 EUR, United States 16150,"█ █ ★ COCAINE FULL CULTIVATION AND SYNTESIS guide ★ █ █","bank",8.2 EUR, United States 16151,"500 usd bank cashout for noobs- tutorial","bank",4.56 EUR, United States 16152,"Free Lawn Mower and Cash Out Guide ($400) Value","bank",4.56 EUR, United States 16153,"ID TEMPLATES: GERMANY","bank",9.

12 EUR, United States 16155,"★ THE ULTIMATE PARANOIA SECURITY GUIDE - 67 PAGES★","bank",5.47 EUR, United States 16156,"★ AMAZON BIG PACK v2 ★","bank",22.79 EUR, United States 16161,"★ SUPER DOLLARS COUNTERFEITING ★","bank",9.12 EUR, United States 16162,"★ How To Be An International Fugitive ★","bank",4.56 EUR, United States 16163,"BIG PACK OF CC TEMPLATES","bank",18.

23 EUR, United States 16164,"★ THE SATANIC BIBLE ★","bank",4.67 EUR, United States 16168,"★ ★ CREDIT SECRETs BIBLE ★ ★","bank",8.

2 EUR, United States 16169,"★ BOTNETS AND CYBERCRIME ★","bank",2.73 EUR, United States 16170,"50 eur TEMPLATES, VERY DETAILED 350 MB","bank",13.67 EUR, United States 16172,"★ ★ THE LARGEST CARDING TUTORIAL - 330 PAGES ★ ★","bank",22.79 EUR, United States 16173,"ANDROID SECURITY PRIVACY AND ENCRYPTION PACK","bank",4.56 EUR, United States 16175,"★ CREATING ANONYMOUS WALLETT ★","bank",4.

56 EUR, United States 16177,"201 Dump cashout method","bank",6.38 EUR, United States 16182,"★ EXTREME Hardening Windows 7 64 Bit Guide ★","bank",9.12 EUR, United States 16183,"★ BECOME AN ALPHA MALE - 7 EBOOKS ★","bank",9.12 EUR, United States 16185,"BLACKMAIL GUIDE","bank",8.2 EUR, United States 16186,"A Guide to Kernel Exploitation: Attacking the Core","bank",3.

000 USD/EUR/CHF SWISS VISA RELOADABLE WITH IBAN","bank",17.32 EUR, United States 16201,"DEFINITIVE HERPES CURE","bank",3.65 EUR, United States 16202,"★ ATM MANUALS BIG PACK - HACK INTO ATM ADMIN PANEL","bank",13.67 EUR, United States 16203,"█ █★ MAIL DROP ,THE ULTIMATE GUIDE 2015 ★█ █","bank",17.

32 EUR, United States 16204,"█ █ BTC TO ANY CC + HIGH LIMIT ANONYMOUS DEBIT CARD - HOW TO █ █ ","bank",13.67 EUR, United States 16207,"★ THE MOST IMPORTANT BOOK YOU WILL EVER READ ★","bank",4.56 EUR, United States 16208,"★ 200$ DAILY WITH AUTOPILOT - 100% WORKING OR I REFUND ★","bank",13.67 EUR, United States 16209,"★ ANTI EAVESDROPPING devices ★","bank",68.37 EUR, United States 16210,"★ THE ULTIMATE MDMA MAKING GUIDE TOTAL SYNTHESIS II ★","bank",8.

2 EUR, United States 16211,"★Complete Drop Shipping Course★","bank",4.56 EUR, United States 16214,"DUMPS CARDING - BIG TUTORIAL","bank",8.2 EUR, United States 16215,"★★★ FIVERR MEGA PACK - 18 tutorials ★★★","bank",13.67 EUR, United States 16217,"★ Never Get Busted Again Vol 1 & 2 + Bonus! ★","bank",8.2 EUR, United States 16218,"AndroRAT - Android Remote Administration Tool Setup Tutorial","bank",4.

56 EUR, United States 16219,"50 MONEY MAKING METHODS - BULK","bank",9.12 EUR, United States 16220,"CPA Hotel Monopoly","bank",8.2 EUR, United States 16221,"ATM Skimmer Cashing / Installing Safety","bank",4.56 EUR, United States 16222,"understanding Anti-fraud systems","bank",4.

56 EUR, United States 16223,"★ HOW TO BECOME INVISIBLE - 262 PAGES ★","bank",5.

47 EUR, United States 16224,"★ HOW TO DISAPPEAR IN THE USA ★","bank",4.56 EUR, United States 16225,"HOW TO CHANGE YOUR SSN - 100% LEGAL METHOD!","bank",9.12 EUR, United States 16226,"WHERE TO GET HACKED RDP SERVERS","bank",8.2 EUR, United States 16229,"50 Ways To Succeed - Make Money","bank",4.56 EUR, United States 16230,"★ Drop Shipping Amazon To Ebay method ★","bank",4.

56 EUR, United States 16231,"To Be or Not to Be Intimidated?","bank",4.56 EUR, United States 16232,"★ DATA EXFILTRATION ★","bank",5.47 EUR, United States 16233,"US passport ID template","bank",18.23 EUR, United States 16235,"★ BANK TRANSFER CASHOUT ROBBERY GOLDEN PACK ★","bank",31.91 EUR, United States 16236,"█ █ CIA Flaps And Seals Manual █ █","bank",0.

91 EUR, United States 16237,"30 Arduino Projects for the Evil Genius","bank",1.82 EUR, United States 16238,"1,787 Kindle eBooks MEGA COLLECTION","bank",8.2 EUR, United States 16239,"POKER Tracker 4 - Improve Your Online Poker Game","bank",9.12 EUR, United States 16240,"★ Guide to Opiate Detoxification ★","bank",8.2 EUR, United States 16241,"★ ★ HOW TO CHEAT AT EVERYTHING ★ ★","bank",4.

56 EUR, United States 16242,"█ █ CIA Flaps And Seals Manual █ █","bank",0.91 EUR, United States 16243,"★ How to Find Out Anything on Google - 2 ebooks ★","bank",4.56 EUR, United States 16244,"★ BOTNET - The Killer Web Applications ★","bank",4.56 EUR, United States 16245,"★ Cruise Tips and Tricks ★","bank",4.56 EUR, United States 16246,"★ The Dark Art Of Death ★","bank",2.

73 EUR, United States 16247,"█ █★ ESOTERIC GUNS ★█ █","bank",4.56 EUR, United States 16248,"★ THE MODERN IDENTITY CHANGER ★","bank",8.2 EUR, United States 16249,"█ █ 20 eur templates and counterfitting manual █ █","bank",9.12 EUR, United States 16250,"█ █ ★ MY LITTLE DIRTY BITCOIN SECRETS ★ █ █","bank",4.56 EUR, United States 16251,"★ Hack Your School Grading System ★","bank",13.

67 EUR, United States 16252,"★ How ToAchieve Maximum Anonymity + Legitimacy ★ ★","bank",4.56 EUR, United States 16253,"★ You Are Being Lied To ★","bank",2.73 EUR, United States 16254,"HOW TO CHANGE YOUR SSN + HOW TO GET A BANK ACOCUNT WITHOUT SSN","bank",13.67 EUR, United States 16255,"★ THE CARDING TUTORIAL ★","bank",6.38 EUR, United States 16256,"★ ★ ★ MAIL DROP INTERNATIONAL USA EU UK ★ ★ ★","bank",13.

67 EUR, United States 16257,"★Turn an Old Mac Into a VPN with OS X Server★","bank",2.73 EUR, United States 16260,"★ The Web Application Hacker's Handbook 2 ★","bank",4.56 EUR, United States 16261,"★ LITTLE BROTHER ★","bank",1.82 EUR, United States 16263,"█ █ PAYPAL CASHOUT EXTREME PACK █ █ 100% CUSTOMER SATISFACTION WA","bank",13.67 EUR, United States 16265,"★ HOW TO USE TWO FACTOR AUTHENTICATION SAFELY ★","bank",0.

91 EUR, United States 16266,"★ 48 Hour Client 2015 - bank up to $30,000/Month ★","bank",9.12 EUR, United States 16267,"★ THE NARCOTICS COOKBOOK ★","bank",4.56 EUR, United States 16268,"500 Ways To Make Money Online","bank",4.56 EUR, United States 16269,"★ The Prolong Ejaculation Guide ★","bank",4.56 EUR, United States 16270,"█ █ ★ MY LITTLE DIRTY BITCOIN SECRETS ★ █ █","bank",4.

56 EUR, United States 16271,"★ HOW TO BE A PERPETUAL TRAVELLER ★","bank",6.38 EUR, United States 16272,"★ 4 DATING EBOOKS ★","bank",7.29 EUR, United States 16273,"13 Ways to Know if the Government is Reading Your Email","bank",4.56 EUR, United States 16274,"EQUIFAX + EXPERIAN CREDIT REPORT TEMPLATES","bank",13.67 EUR, United States 16275,"★ THE MODERN IDENTITY CHANGER ★","bank",8.

2 EUR, United States 16276,"BUSINESS VERIFIED FRESH PAYPAL ACCOUNT U.75 EUR, United States 16277,"★ TICKETS CARDING ★","bank",4.

56 EUR, United States 16278,"█ █ Are You Talking to an Agent Provocateur? █ █","bank",4.

56 EUR, United States 16279,"To Be or Not to Be Intimidated?","bank",4.56 EUR, United States 16280,"IF AN AGENT KNOCKS - protect yourself from police investigations","bank",4.56 EUR, United States 16281,"█ █ MONEY LAUDERING SCIENCE █ █","bank",9.12 EUR, United States 16282,"UK ID PASSPORT TEMPLATES","bank",18.23 EUR, United States 16283,"██ PayMe Instant Profit Spree ██","bank",8.

2 EUR, United States 16284,"QUIT OPIOIDS WITHOUT WITHDRAWAL SYMPTOMS","bank",7.29 EUR, United States 16285,"TUTORIAL: Photoshop CS6 Unlocked","bank",4.56 EUR, United States 16286,"CRIMINAL PSYCHOLOGY and FORENSIC TECHNOLOGY","bank",4.56 EUR, United States 16287,"10 Ways to Find People on Facebook","bank",6.38 EUR, United States 16288,"1G Ketamine AAA+++","HiddenHands",27.

73 EUR, United Kingdom 16289,"BITCOIN TRADING GUIDE","bank",4.56 EUR, United States 16290,"2G Ketamine AAA+++","HiddenHands",55.46 EUR, United Kingdom 16291,"ATM MANUALS BIG PACK - HACK INTO ATM","bank",13.74 EUR, United Kingdom 16293,"UK PASSPORT TEMPLATE","bank",18.23 EUR, United States 16294,"7G Ketamine AAA+++","HiddenHands",166.38 EUR, United Kingdom 16295,"14G Ketamine AAA+++","HiddenHands",311.19 EUR, United Kingdom 16296,"28G Ketamine AAA+++","HiddenHands",621.5G Colombian Flake Cocaine 82-85% AAA+++","HiddenHands",50.01 EUR, United Kingdom 16299,"1G Colombian Flake Cocaine 82-85% AAA+++","HiddenHands",77.5 Colombian Flake Cocaine 82-85% AAA+++","HiddenHands",255.12 EUR, United Kingdom 16301,"2G Colombian Flake Cocaine 82-85% AAA+++","HiddenHands",155.

29 EUR, United Kingdom 16302,"XOOM CASHOUT METHOD","bank",17.32 EUR, United States 16303,"CANADA SIN CARD TEMPLATE","bank",4.56 EUR, United States 16304,"HOW TO ENCODE DUMP TO PLASTIC","bank",5.47 EUR, United States 16305,"100K Bucks a Year","bank",8.2 EUR, United States 16306,"★ Where to Get Anybodies SSN MMN ★","bank",7.

29 EUR, United States 16307,"50 THINGS YOU’RE NOT SUPPOSED TO KNOW","bank",2.73 EUR, United States 16308,"5 ATM HACKING TUTORIALs","bank",17.32 EUR, United States 16309,"★ RAPID FAT LOSS + INCREDIBLE SEX ★","bank",4.56 EUR, United States 16310,"★ 7 reasons a credit card get blocked ★","bank",1.82 EUR, United States 16311,"█ █ ★ CARDING TRICKS FROM AN INSIDER ★ █ █","bank",4.

56 EUR, United States 16312,"★ 19gb FREE OFFSHORE CLOUD STORAGE FOREVER★","bank",8.2 EUR, United States 16313,"█ █ CIA Flaps And Seals Manual █ █","bank",0.91 EUR, United States 16314,"ANATOMY OF A DATABASE ATTACK","bank",2.73 EUR, United States 16315,"★ BEST BITCOIN BLENDER ★","bank",0 EUR, United States 16317,"Never Get a "Real" Job: How to Dump Your Boss, Build a Business a","bank",8.2 EUR, United States 16318,"★ WHERE TO BUY HACKED SERVERS ★","bank",4.

56 EUR, United States 16319,"Creating an Anonymous Wallet for Covert Practices","bank",3.65 EUR, United States 16320,"How 2 intercept or "Hack" a phone call","bank",6.38 EUR, United States 16321,"★ HOW TO MAKE OPIUM ★","bank",4.56 EUR, United States 16322,"★ Anonymous RAT setup for dummies ★","bank",8.2 EUR, United States 16324,"★ ENCYCLOPEDIA OF ANCIENT AND FORBIDDEN SECRETS ★","bank",4.

56 EUR, United States 16325,"Creating a WordPress Blog in Whonix","bank",4.56 EUR, United States 16326,"DEFINITIVE MDMA SYNTHESYS MEGAGUIDE","bank",22.79 EUR, United States 16327,"█ █ PENETRATION TESTING PACK █ █","bank",4.65 EUR, United States 16329,"Anonymity and Privacy for Advanced Linux Users","bank",9.12 EUR, United States 16330,"Videopusher - 50000 FREE views a day for your video !!","bank",44.67 EUR, United States 16332,"★ Guide to Opiate Detoxification ★","bank",8.2 EUR, United States 16333,"How to legally accept any drug package","bank",9.12 EUR, United States 16337,"Go BIG or Go HOME","bank",3.

65 EUR, United States 16338,"★ BANK TRANSFER CASHOUT ROBBERY GOLDEN PACK ★","bank",31.91 EUR, United States 16346,"★ AMAZON SCAMS : What to do if [email protected] emails you★","bank",4.56 EUR, United States 16347,"█ █ How to Delete Everything Google Knows About You █ █","bank",2.73 EUR, United States 16349,"ANAL SEX TUTORIAL","bank",3.65 EUR, United States 16350,"Amex Gold Editable Scan","bank",0.

91 EUR, United States 16351,"How to Find WELLS FARGO Full Accounts Info","bank",18.23 EUR, United States 16352,"█ █ HOW TO MAKE HER LOVE ANAL SEX █ █","bank",4.56 EUR, United States 16354,"★ HOW TO USE TWO FACTOR AUTHENTICATION SAFELY ★","bank",0.91 EUR, United States 16359,"★ 2 FACEBOOK PICKUP TUTORIALS ★","bank",5.47 EUR, United States 16361,"█ █ 68,000,000 PASSWORD LIST █ █","bank",17.

32 EUR, United States 16365,"How to Own Your House and Car Anonymously","bank",7.29 EUR, United States 16369,"█ █ ★ USA PASSPORTS .79 EUR, United States 16370,"40 Ways to Sabotage Your School","bank",4.56 EUR, United States 16371,"★ ENCYCLOPEDIA OF DRUGS ★","bank",2.

73 EUR, United States 16375,"Rolex - Milgauss Blue Dial Green Crystal 2014 Ultimate AAA+ ","RepAAA",432.13 EUR, China 16376,"Rolex - Milgauss Green Crystal SS AAA+ ","RepAAA",295.38 EUR, China 16377,"Rolex - Milgauss Pro-Hunter Orange Replica ","RepAAA",185.98 EUR, China 16378,"Rolex - Milgauss Stainless Steel AAA+ ","RepAAA",295.38 EUR, China 16379,"Rolex - Sea-Dweller 4000 ETA Ultimate AAA+ ","RepAAA",459.

48 EUR, China 16380,"Rolex - Sea-Dweller Deepsea 3135 Clone Ultimate AAA+ ","RepAAA",468.59 EUR, China 16381,"Rolex - Sea-Dweller Deepsea ETA Ultimate AAA+ ","RepAAA",395.66 EUR, China 16382,"Rolex - Submariner 2Tone YG/SS Black Replica ","RepAAA",158.63 EUR, China 16383,"Rolex - Submariner 2Tone YG/SS Black ETA AAA+ ","RepAAA",295.38 EUR, China 16384,"Rolex - Submariner 2Tone YG/SS Blue Replica ","RepAAA",158.

63 EUR, China 16385,"Rolex - Submariner 2Tone YG/SS Blue ETA AAA+ ","RepAAA",295.38 EUR, China 16387,"Rolex - Submariner 50th Anniversary ETA AAA+ ","RepAAA",295.38 EUR, China 16396,"Rolex - Submariner Blue Dial ETA Ultimate AAA+ ","RepAAA",477.71 EUR, China 16398,"Rolex - Submariner Blue/Blue Replica ","RepAAA",158.63 EUR, China 16399,"Rolex - Submariner Classic Black Replica ","RepAAA",158.

63 EUR, China 16400,"Rolex - Submariner Classic Black ETA AAA+ ","RepAAA",273.5 EUR, China 16401,"Rolex - Submariner Classic Black ETA Ultimate AAA+ ","RepAAA",413.89 EUR, China 16403,"Rolex - Submariner Green/Green Replica ","RepAAA",158.63 EUR, China 16405,"Rolex - Submariner Green/Green ETA AAA+ ","RepAAA",273.5 EUR, China 16406,"Rolex - Submariner Pro-Hunter Replica ","RepAAA",185.

98 EUR, China 16407,"Rolex - Submariner Pro-Hunter ETA AAA+ ","RepAAA",322.73 EUR, China 16408,"Rolex - Submariner Yellow Gold AAA+ ","RepAAA",322.73 EUR, China 16409,"Rolex - Submariner Yellow Gold Replica ","RepAAA",158.63 EUR, China 16411,"Rolex - Watch Box "50% DISCOUNTED" (AAA Grade Replica)","RepAAA",68.37 EUR, China 16412,"Rolex - Watch Box (AAA Grade Replica)","RepAAA",113.

96 EUR, China 16415,"Rolex - Yacht-Master 116655 Rose Gold AAA+ ","RepAAA",313.61 EUR, China 16435,"1g ORGANIC Ketama Hashish","danhash",9.56 EUR, Spain 16436,"3g ORGANIC Ketama Hashish","danhash",24.15 EUR, Spain 16437,"5g ORGANIC Ketama Hashish ","danhash",43.

29 EUR, Spain 16438,"10g ORGANIC Ketama Hashish","danhash",75.

2 EUR, Spain 16440,"Premium Bayer Opium (1g)","danhash",48.31 EUR, Spain 16441,"Premium Bayer Opium (3g) ","danhash",136.74 EUR, Spain 16442,"Premium Bayer Opium (5g)","danhash",209.67 EUR, Spain 16626,"Hublot - Big Bang Stainless Steel 38mm AAA+ ","RepAAA",204.21 EUR, China 16627,"Hublot - Big Bang Ladies Chronograph Iced-Out Case Replica ","RepAAA",249.

79 EUR, China 16628,"Hublot - Big Bang Ladies Full Iced-Out Swiss Quartz Replica ","RepAAA",295.38 EUR, China 16629,"Omega - Seamaster Aqua Terra Blue Dial AAA+ ","RepAAA",295.38 EUR, China 16630,"Panerai - Luminor 1950 3 Days GMT PAM438 Tuttonero Ultimate AAA+","RepAAA",432.13 EUR, China 16638,"Clenbuterol - 100 tb/40 mcg - Body Pharm","SteroidWarehouse",53 EUR, Netherlands 16639,"Rolex - Datejust 2-tone Black Dial ETA AAA+ ","RepAAA",249.79 EUR, China 16644,"TAG Heuer - Carrera Calibre 16 Day Date AAA+ ","RepAAA",432.

13 EUR, China 16646,"TAG Heuer - Carrera Calibre 1887 Black Dial Ultimate AAA+ ","RepAAA",523.29 EUR, China 16647,"TAG Heuer - Carrera Calibre 1887 Grey Dial Ultimate AAA+ ","RepAAA",523.29 EUR, China 16652,"TAG Heuer - Grand Carrera Calibre 17 RS2 Titanium/Leather AAA+ ","RepAAA",550.64 EUR, China 16655,"TAG Heuer - Grand Carrera Calibre 36 RS AAA+ ","RepAAA",432.13 EUR, China 16656,"TAG Heuer - Grand Carrera Calibre 36 RS Titanium/Rubber AAA+ ","RepAAA",550.

64 EUR, China 16657,"TAG Heuer - Watch Box (AAA Grade Replica)","RepAAA",113.96 EUR, China 16658,"Tudor - Heritage Black Bay 2012 Ultimate AAA+ ","RepAAA",432.13 EUR, China 16659,"Tudor - Heritage Black Bay Leather Strap Ultimate AAA+ ","RepAAA",432.13 EUR, China 16660,"Tudor - Heritage Ranger Stainless Steel Bracelet Ultimate AAA+ ","RepAAA",386.54 EUR, China 16661,"Tudor - Pelagos Titanium Ultimate AAA+ ","RepAAA",477.

71 EUR, China 16662,"Vacheron Constantin - Malte Tourbillion SS AAA+ ","RepAAA",842.37 EUR, China 16680,"Bell & Ross - BR03-92 Aviation Golden Heritage AAA+ ","RepAAA",340.96 EUR, China 16685,"Cartier - Santos 100 XL Chronograph Stainless Steel AAA+ ","RepAAA",386.54 EUR, China 16686,"TAG Heuer - Carrera Calibre 1887 Automatic Chronograph AAA+ ","RepAAA",340.96 EUR, China 16687,"Tudor - Fastrider Black Shield Chronograph Alcantara AAA+ ","RepAAA",386.

54 EUR, China 16705,"Rivotril (Clonazepam) 2 MG 60 Pills","DeepMeds",96.64 EUR, United States 16715,"Rivotril (Clonazepam) 2 MG 160 Pills","DeepMeds",187.8 EUR, United States 16719,"Rivotril (Clonazepam) 2 MG 500 Pills","DeepMeds",433.95 EUR, United States 16720,"Rivotril (Clonazepam) 2 MG 1000 Pills","DeepMeds",771.26 EUR, United States 16752,"Valium (Diazepam) 10 MG By Roche 80 Pills ","DeepMeds",105.

75 EUR, United States 16754,"Valium (Diazepam) 10 MG By Roche 90 Pills ","DeepMeds",114.87 EUR, United States 16755,"Valium (Diazepam) 10 MG By Roche 120 Pills ","DeepMeds",142.22 EUR, United States 16767,"1 gram TSC's World Famous Culero - Ketama Hashish ","The Scurvy Crew",10.01 EUR, Denmark 16773,"Xanax (Alprazolam) 1 MG 40 Pills","DeepMeds",69.29 EUR, United States 16869,"Tudor - Fastrider Black Shield Chronograph Red AAA+ ","RepAAA",432.

13 EUR, China 16875,"custom ordr (2) for SafeB","MikeDelacruz",227.92 EUR, China 16877,"Masteron-100 (Drostanolone Propionate) - Malay Tiger 10x1ml","arnold",95 EUR, Finland 16906,"Amnesia Haze - 10g","DutchMagic",95 EUR, Netherlands 16907,"Amnesia Haze - 20g","DutchMagic",184 EUR, Netherlands 16908,"Amnesia Haze - 50g","DutchMagic",457 EUR, Netherlands 16909,"Amnesia Haze - 100g","DutchMagic",848 EUR, Netherlands 16910,"Amnesia Haze - 250g","DutchMagic",1975 EUR, Netherlands 16911,"Amnesia Haze - 500g","DutchMagic",3700 EUR, Netherlands 16912,"Amnesia Haze - 1 KILOGRAM","DutchMagic",6690 EUR, Netherlands 16914,"Hexagram Hash - 50g","DutchMagic",125 EUR, Netherlands 16915,"Hexagram Hash - 100g","DutchMagic",235 EUR, Netherlands 16916,"Hexagram Hash - 200g","DutchMagic",440 EUR, Netherlands 16917,"Hexagram Hash - 500g","DutchMagic",1070 EUR, Netherlands 16918,"Hexagram Hash - 1 KILOGRAM","DutchMagic",1995 EUR, Netherlands 16919,"Magic Mix - 10g","DutchMagic",34.5 EUR, Netherlands 16926,"Restoril (temazepam) 30 MG 60 Caps","DeepMeds",119.43 EUR, United States 16927,"Restoril (temazepam) 30 MG 70 Caps","DeepMeds",133.1 EUR, United States 16967,"50 euro counterfeit notes - 10 pcs ","charliebrown1",67 EUR, Sweden 17153,"IWC - Pilot Chronograph IW377701 Ultimate AAA+ ","RepAAA",432.

13 EUR, China 17154,"IWC - Pilot Chronograph IW377704 Ultimate AAA+ ","RepAAA",432.13 EUR, China 17162,"Cypabol (Testosterone Cypionate) 1x10lm","arnold",55 EUR, Finland 17179,"Windows 7 Activator","junkiepig666",4.55 EUR, Germany 17180," - LIFETIME PORN PREMIUM ACCOUNT ","Wooody",9.11 EUR, Finland 17183,"THC berlingot (candy) STRONG VERSION","projeccao",39.9 EUR, France 17217,"Ayahuasca One Person Kit","blackhand",43.

76 EUR, United States 17218,"Ayahuasca Two Person Kit","blackhand",68.37 EUR, United States 17219,"Ayahuasca Three Person Kit","blackhand",99.37 EUR, United States 17220,"Modafinil 200mg x 1","blackhand",9.57 EUR, United States 17221,"Stoplok Pegboard/DVD Magnetic Security Key","blackhand",51.

96 EUR, United States 17222,"Clothing Tag Remover - Sensormatic Hook","blackhand",33.

73 EUR, United States 17223,"600-700 Morning Glory Seeds - LSA","blackhand",17.32 EUR, United States 17224,"Kratom - Maeng Da Red 28g","blackhand",23.7 EUR, United States 17225,"Kratom - Green Indo 28g","blackhand",23.7 EUR, United States 17226,"Platinum Hide IP + Patch","junkiepig666",4.55 EUR, Germany 17227,"Alexander Shulgin Book Collection","blackhand",1.

37 EUR, United States 17228,"Altered states of consciousness","blackhand",1.37 EUR, United States 17229,"antipodes of the mind - charting the phenomenology of the ayahua","blackhand",1.37 EUR, United States 17230,"Arrest-Proof Yourself","blackhand",1.37 EUR, United States 17231,"Body Language: How To Read Others' Thoughts By Their Gestures","blackhand",1.37 EUR, United States 17232,"Brainwave Generator + Crack","blackhand",4.

56 EUR, United States 17233,"Breaking Open the Head: A Psychedelic Journey","blackhand",1.37 EUR, United States 17234,"Cannabis Alchemy: Art of Modern Hashmaking","blackhand",1.37 EUR, United States 17239,"IWC - Portuguese Automatic IW500702 RG AAA+ ","RepAAA",477.71 EUR, China 17240,"IWC - Portuguese Automatic IW500703 SS AAA+ ","RepAAA",459.48 EUR, China 17241,"Audemars Piguet - Royal Oak Offshore Lebron James AAA+ ","RepAAA",778.

56 EUR, China 17271,"EXCEL PHARMA nandroject 250 ","stealthmeds",33.28 EUR, United Kingdom 17273,"EXCEL PHARMA TNT 450 10mil vial ","stealthmeds",33.28 EUR, United Kingdom 17274,"EXCEL PHARMA super rip 200 ","stealthmeds",33.28 EUR, United Kingdom 17295,"Treasure Island Media - Fuck Holes 3 ","junkiepig666",4.55 EUR, Germany 17314,"!!!!SPECIALT!!!! 1G PREMIER Organic Moroccan Hashish","danhash",15.

03 EUR, Denmark 17315,"Cheaper, Better, Faster: Over 2,000 Tips and Tricks to Save You","blackhand",1.37 EUR, United States 17316,"Cleansing the Doors of Perception: The Religious Significance of ","blackhand",1.37 EUR, United States 17317,"Cocaine Handbook: An Essential Reference","blackhand",1.37 EUR, United States 17318,"Cocaine Nation: How the White Trade Took Over the World","blackhand",1.37 EUR, United States 17319,"Cocaine: the Straight Facts","blackhand",1.

37 EUR, United States 17320,"Collection of Alex Grey Works","blackhand",1.37 EUR, United States 17321,"Complete Atari 2600 Games collection + Emulator","blackhand",1.37 EUR, United States 17322,"Complete Commander Keen Collection","blackhand",1.37 EUR, United States 17323,"Designer Drugs Directory","blackhand",1.37 EUR, United States 17324,"DMT & Ayahuasca Book Collection","blackhand",1.

37 EUR, United States 17325,"Do It Yourself Advertising and Promotion","blackhand",1.37 EUR, United States 17326,"Dr Atomic's Marijuana Multiplier (1st ed.37 EUR, United States 17327,"Ecstasy : The Complete Guide : A Comprehensive Look at the Risks ","blackhand",1.37 EUR, United States 17328,"Essential Psychedelic Guide","blackhand",1.

37 EUR, United States 17329,"Essential Underground Handbook 2003 PML Publishing","blackhand",1.37 EUR, United States 17330,"expert resumes for career changers","blackhand",1.37 EUR, United States 17331,"Fooling The Bladder Cops: The Complete Drug Test Guide","blackhand",1.37 EUR, United States 17332,"Gray Hat Hacking The Ethical Hackers Handbook, 3rd Edition","blackhand",1.37 EUR, United States 17333,"Hacking Exposed Computer Forensics, Second Edition: Computer Fore","blackhand",1.

37 EUR, United States 17334,"HACKING EXPOSED WEB APPLICATIONS, 3rd Edition","blackhand",1.37 EUR, United States 17335,"Hacking GMail","blackhand",1.37 EUR, United States 17336,"Hacking Matter: Levitating Chairs, Quantum Mirages, and the Infin","blackhand",1.37 EUR, United States 17337,"Hacking Windows XP (ExtremeTech)","blackhand",1.37 EUR, United States 17338,"Hallucinogenic Plants (A Golden Guide)","blackhand",1.

37 EUR, United States 17339,"Hallucinogens and Shamanism","blackhand",1.37 EUR, United States 17340,"How to Develop a Perfect Memory","blackhand",1.37 EUR, United States 17341,"How to Get Ideas","blackhand",1.37 EUR, United States 17342,"How to Talk to Anyone, Anytime, Anywhere: The Secrets of Good Com","blackhand",1.

37 EUR, United States 17343,"How to Win Friends & Influence People","blackhand",1.

37 EUR, United States 17344,"I-Doser - Complete Doses","blackhand",1.37 EUR, United States 17345,"It's None of Your Business: A Complete Guide to Protecting Your P","blackhand",1.37 EUR, United States 17346,"Jitbit Macro Recorder","blackhand",1.37 EUR, United States 17347,"Ketamine: Dreams and Realities","blackhand",1.37 EUR, United States 17348,"Ketamine: Journeys into the bright world","blackhand",1.

37 EUR, United States 17349,"Lockpicking book collection","blackhand",1.37 EUR, United States 17350,"LSD: A Documentary Report","blackhand",1.37 EUR, United States 17351,"LSD: The Consciousness-Expanding Drug","blackhand",1.37 EUR, United States 17353,"Marijuana Chemistry: Genetics, Processing, Potency","blackhand",1.37 EUR, United States 17354,"Mastering Astral Projection: 90-day Guide to Out-of-Body Experien","blackhand",1.

37 EUR, United States 17355,"Men's Fitness Workout Manual - Your Guide To Building Muscle And ","blackhand",1.37 EUR, United States 17356,"Ninja Hacking: Unconventional Penetration Testing Tactics and Tec","blackhand",1.37 EUR, United States 17357,"No Tech Hacking: A Guide to Social Engineering, Dumpster Diving, ","blackhand",1.37 EUR, United States 17358,"Occult Chemistry: Investigations by Clairvoyant Magnification int","blackhand",1.37 EUR, United States 17359,"Organic Chemistry 8th Ed by John McMurry","blackhand",1.

37 EUR, United States 17360,"Organic Chemistry Demystified 2/E","blackhand",1.37 EUR, United States 17361,"Otto Snow Complete Collection","blackhand",1.37 EUR, United States 17362,"Past-Life Regression - Ultra-Depth Hypnosis","blackhand",1.37 EUR, United States 17363,"Persephone's Quest: Entheogens and the Origins of Religion","blackhand",1.37 EUR, United States 17364,"Pharmacotheon: Entheogenic Drugs, Their Plant Sources and History","blackhand",1.

37 EUR, United States 17365,"Cocaine Senior, 1 gram","projeccao",82 EUR, France 17366,"Phenomenology and Mysticism: The Verticality of Religious Experie","blackhand",1.37 EUR, United States 17367,"Piracy: The Intellectual Property Wars from Gutenberg to Gates","blackhand",1.37 EUR, United States 17368,"Plants of the Gods: Their Sacred, Healing, and Hallucinogenic Pow","blackhand",1.37 EUR, United States 17369,"Psilocybin: Magic Mushroom Grower's Guide: A Handbook for Psilocy","blackhand",1.37 EUR, United States 17370,"Psilocybin Mushroom Handbook: Easy Indoor and Outdoor Cultivation","blackhand",1.

37 EUR, United States 17371,"Psychedelic Chemistry","blackhand",1.37 EUR, United States 17372,"Psychedelic Drugs Reconsidered","blackhand",1.37 EUR, United States 17373,"Psychosis and Spirituality: Consolidating the New Paradigm","blackhand",1.37 EUR, United States 17374,"Rational Mysticism: Spirituality Meets Science in the Search for ","blackhand",1.37 EUR, United States 17375,"Ready Player One - Ernest Cline Audiobook","blackhand",1.

37 EUR, United States 17376,"Recreational Drugs by Professor Buzz","blackhand",1.37 EUR, United States 17377,"Rule by Secrecy: The Hidden History That Connects the Trilateral ","blackhand",1.37 EUR, United States 17378,"Satanic Bible Underground Edition","blackhand",1.37 EUR, United States 17379,"Social Engineering: The Art of Human Hacking","blackhand",1.37 EUR, United States 17380,"Sources by Strike","blackhand",1.

37 EUR, United States 17381,"Steal This Book","blackhand",1.37 EUR, United States 17382,"Storming Heaven: LSD and the American Dream","blackhand",1.37 EUR, United States 17383,"Techniques of the Professional Pickpocket","blackhand",1.37 EUR, United States 17384,"The Art of Deception: Controlling the Human Element of Security","blackhand",1.37 EUR, United States 17385,"The Basics of Hacking and Penetration Testing: Ethical Hacking an","blackhand",1.

37 EUR, United States 17386,"The Best Damn IT Security Management Book Period","blackhand",1.37 EUR, United States 17387,"The Beyond Within: The L.37 EUR, United States 17388,"The Book: On the Taboo Against Knowing Who You Are","blackhand",1.37 EUR, United States 17389,"The Construction of Secret Hiding Places","blackhand",1.37 EUR, United States 17390,"The Construction & Operation Of Clandestine Drug Laboratories","blackhand",1.37 EUR, United States 17392,"The Dream Drugstore: Chemically Altered States of Consciousness","blackhand",1.

37 EUR, United States 17393,"the good in bed guide to orally pleasuring a woman","blackhand",1.

37 EUR, United States 17394,"The Joyous Cosmology: Adventures in the Chemistry of Consciousnes","blackhand",1.37 EUR, United States 17395,"The Organic Chem Lab Survival Manual 3rd edition","blackhand",1.37 EUR, United States 17396,"The Popular Life - Make People Like, Respect, and Befriend You","blackhand",1.37 EUR, United States 17398,"The Private Sea: LSD & the Search for God","blackhand",1.37 EUR, United States 17399,"The Psychology of Religion, Fourth Edition: An Empirical Approach","blackhand",1.

37 EUR, United States 17400,"The Road to Eleusis: Unveiling the Secret of the Mysteries","blackhand",1.37 EUR, United States 17401,"The Six-Figure Second Income: How To Start and Grow A Successful ","blackhand",1.37 EUR, United States 17403,"The Varieties of Psychedelic Experience: The Classic Guide to the","blackhand",1.37 EUR, United States 17404,"Total Synthesis II - Ecstasy & Amphetamines Synthesis","blackhand",1.37 EUR, United States 17405,"Transforming Your Self: Becoming Who You Want to Be","blackhand",1.

37 EUR, United States 17406,"Ultimate Guide to Using Tor privately and anonymously","blackhand",1.37 EUR, United States 17407,"Vin DiCarlo- The Red Pill Guide to Picking up Women","blackhand",1.37 EUR, United States 17409,"When a Man Makes Love to a Woman","blackhand",1.37 EUR, United States 17411,"Where God and Science Meet - Volums 1-3","blackhand",1.37 EUR, United States 17476,"1g GOLD Shatter/Dab 95%THC","hashishmaster",72.

85 EUR, Canada 17477,"Cash Out Via Tax Return-tutorial","Citra",4.47 EUR, United States 17478,"EMAIL SPAM-PRO ESSENTIAL TOOLS PACKAGE","Citra",10.94 EUR, Sweden 17480,"Stolen Bank Drop to BITCOIN","Citra",9.12 EUR, Lithuania 17481,"How to hack wireless networks tutorial","Citra",1.82 EUR, Lithuania 17482,"Mac Address Change softwares(2)Windows","Citra",1.

82 EUR, Denmark 17483,"Make Cocaine at home-99,9% Purity-Full Guide","Citra",4.55 EUR, Denmark 17484,"Make Cocaine at home-99,9% Purity-Full Guide","Citra",4.56 EUR, Norway 17485,"VPN LIFETIME FOR Windows,Linux,Android,Mac,Iphone,","Citra",2.64 EUR, Finland 17486,"12GB FAKE ID + CC FORGERY + MONEY COUNTERFEIT MEGA","Citra",44.62 EUR, United Kingdom 17487,"AlphaBay market invitation link","Citra",0.

09 EUR, Finland 17488,"SUPER HACK PACKAGE +50 HACKING TOOLS 2016","Citra",44.67 EUR, France 17489,"ANTI DETECT PATCH 5 FFTOOLS CRACKED","Citra",3.65 EUR, Norway 17490,"Anarchist Cookbook","Citra",4.26 EUR, Norway 17491,"Carding with dumps-2015-TUTORIAL","Citra",1.82 EUR, Iceland 17492,"PC Remote Administrator Tool","Citra",1.

82 EUR, Iceland 17493,"SQL INJECTION PACKAGE + DORKS + GUIDE","Citra",3.64 EUR, Sweden 17494,"New Stolen HONDA CBR 1000RR","Citra",3063.18 EUR, Sweden 17495,"How to create CYPRUS Anonymous Bank Account with CC and Online-tu","Citra",4.55 EUR, Denmark 17496,"How to create Latvia Anonymous Bank Account with CC and Online-Tu","Citra",3.65 EUR, Latvia 17498,"Getting Cash from a CC using Western Union tutorial ","Citra",2.

20-2016 Professional/Bussines Edition","Citra",2.73 EUR, Denmark 17500,"2015 Modern hackers desk reference","Citra",2.73 EUR, Denmark 17501,"DL real and valid Australia scan copy","Citra",22.

79 EUR, Australia 17502,"Carding Tutorial 2015","Citra",2.28 EUR, Sweden 17536,"3 Grams PREMIER Organic Moroccan Hashish","danhash",40.56 EUR, Denmark 17537,"1 Grams PREMIER Organic Moroccan Hashish","danhash",15.03 EUR, Denmark 17538,"Dream market registration referreal","Citra",0.09 EUR, Finland 17550,"50 Dutch LSD blotters 180ug-Free Shipp","Citra",113.

96 EUR, United States 17552,"225 "Best Kaif" blotters 180ug-Free Shipp","Citra",545.49 EUR, United Kingdom 17556,"20 Dutch LSD blotters 180ug-Free Shipp","Citra",92.99 EUR, Australia 17562,"300mg Freebase DMT “Desert Dream” ","Noumena",23.22 EUR, Germany 17565,"150mg Freebase DMT “Desert Dream” ","Noumena",15.

48 EUR, Germany 17567,"600mg Freebase DMT “Desert Dream” ","Noumena",39.

15g Freebase DMT “Desert Dream” ","Noumena",69.67 EUR, Germany 17569,"150mg Freebase DMT “Dream Sand” ","Noumena",16.39 EUR, Germany 17574,"300mg Freebase DMT “Dream Sand” ","Noumena",27.32 EUR, Germany 17599,"UK 1g Amphetamine Sulphate DRY POWDER > FREE SHIPPING","AllHoursOpen",13.

31 EUR, United Kingdom 17600,"UK 1g Amphetamine Sulphate DRY POWDER > FREE SHIPPING","AllHoursOpen",13.5g Amphetamine Sulphate DRY POWDER > FREE SHIPPING","AllHoursOpen",7.5g Amphetamine Sulphate DRY POWDER > FREE SHIPPING","AllHoursOpen",7.

76 EUR, United Kingdom 17603,"2g Amphetamine Sulphate DRY POWDER > FREE SHIPPING","AllHoursOpen",24.45 EUR, United Kingdom 17604,"2g Amphetamine Sulphate DRY POWDER > FREE SHIPPING","AllHoursOpen",24.4 EUR, United Kingdom 17608,"10 x Frontin Alprazolam 0.5mg (XANAX) FREE SHIPPING","AllHoursOpen",8.87 EUR, United Kingdom 17609,"10 x Frontin Alprazolam 0.

5mg (XANAX) FREE SHIPPING","AllHoursOpen",8.87 EUR, United Kingdom 17636,"Modafinil Modalert 200mg 10tabs Free UK NDD","NextGeneration",22.77 EUR, United Kingdom 17661,"Hacking Wireless Networks For Dummies","color",0.91 EUR, United States 17768,"The Hacking Bible The Dark secrets of the hacking world","color",0.91 EUR, United States 17770,"The Hacker's Manual (2015)","color",0.

91 EUR, United States 17771,"The Basics of Web Hacking","color",0.91 EUR, United States 17772,"Python Hacking Essentials (2015)","color",0.91 EUR, United States 17773,"Penetration Testing A Hands-On Introduction to Hacking 2014","color",0.91 EUR, United States 17774,"Michal Zalewski - Il rumore dell hacking","color",0.91 EUR, United States 17775,"Linux (Hacking Exposed) by Brian Hatch","color",0.

91 EUR, United States 17776,"IPhone 5 All-in-One For Dummies, 2nd edition ","color",0.91 EUR, United States 17777,"Hacking Web Intelligence","color",0.91 EUR, United States 17779,"Hacking Secrets To Becoming A Genius Hacker GLODLS ","color",0.91 EUR, United States 17780,"Hacking Secrets Exposed - A","color",0.91 EUR, United States 17781,"Hacking Secret Ciphers with Python (2015)","color",0.

91 EUR, United States 17782,"Hacking Raspberry Pi 2013","color",0.91 EUR, United States 17783,"Hacking Magazine (IT Security Magazine) VOL 5.91 EUR, United States 17784,"Hacking For Dummies, 4th Edition","color",0.91 EUR, United States 17786,"Hacking Exposed Wireless, Second Edition by Johnny Cache","color",0.

91 EUR, United States 17787,"Hacking Exposed 7 Network Security Secrets & Solutions","color",0.91 EUR, United States 17788,"Hacking Electronics - An Illustrated DIY Guide","color",0.91 EUR, United States 17789,"Copy of Hacking and Penetration Testing with Low Power Devices","color",0.91 EUR, United States 17790,"Hacking - Firewalls And Networks How To Hack Into Remote Com","color",0.91 EUR, United States 17791,"Hacking - Basic Security,","color",0.

91 EUR, United States 17793,"Hacking - 101 Hacking Guide - 2nd Edition (2015)","color",0.91 EUR, United States 17794,"Growth Hacking Techniques, Disruptive Technology","color",0.91 EUR, United States 17795,"Gray Hat Hacking - The Ethical Hacker's Handbook - 4th Edition","color",0.91 EUR, United States 17825,"Rocco's Perfect Slaves 7","junkiepig666",4.55 EUR, Germany 17826,"Windows 7 AIO - 11 in 1 Editions - Activated","junkiepig666",4.

55 EUR, Germany 17827,"Windows 10 Spying Removal Tool","junkiepig666",4.55 EUR, Germany 17852,"BIO JACK HERER 5 GRAM freeshipping","DutchDreams",56 EUR, Netherlands 17855,"BIO JACK HERER 3 GRAM freeshipping","DutchDreams",36 EUR, Netherlands 17857,"BIO JACK HERER 10 GRAM freeshipping","DutchDreams",101 EUR, Netherlands 17858,"BIO JACK HERER 25 GRAM freeshipping","DutchDreams",236 EUR, Netherlands 17859,"BIO JACK HERER 50 GRAM freeshipping","DutchDreams",446 EUR, Netherlands 17860,"BIO JACK HERER 100 GRAM freeshipping","DutchDreams",836 EUR, Netherlands 17882,"Windows Hacking crack all cool stuff of windows GLODLS ","color",0.91 EUR, United States 17901,"Ephedrin 50tabs/50mg-kpl","arnold",85 EUR, Finland 17910,"ALDIWEED 10 GRAM freeshipping","DutchDreams",56 EUR, Netherlands 18049,"1/4oz (7+grams) AGENT ORANGE - 100% ORGANIC - FREE SHIPPING","joeybagadonuts",90.98 EUR, United States 18059,"$5 TIP JAR - these markets closings are killing us !!!","joeybagadonuts",4.

56 EUR, United States 18062,"6-PK POTENT CHOCOLATE CHIP POT COOKIES w/FREE PRIORITY SHIPPING!","joeybagadonuts",54.

55 EUR, United States 18092,"BULK WHOLESALE GLASS - bulk orders starting at $99.07 EUR, United States 18094,"BULK WHOLESALE GLASS - bulk orders starting at $99.98 EUR, United States 18123,"Purple Candy 7g- 1/4 oz *Indoor AAA Quality","BCBUDKING",45.

56 EUR, Canada 18139,"Snoop Dogg Dog G Pen Dry Herb Pen Newest Version Herbal Vaporizer","ESLight",40.11 EUR, United States 18140,"Snoop Dogg Dog G Pen Dry Herb Pen Newest Version Herbal Vaporizer","ESLight",40.07 EUR, United States 18151,"Penetration Testing A Hands-On Introduction to Hacking 2014","color",0.91 EUR, United States 18153,"Internet Security Online Protection From Computer Hacking","color",0.91 EUR, United States 18154,"ETHICAL HACKING UNDERSTANDING ETHICAL HACKING","color",0.

91 EUR, United States 18155,"Hacking with Kali Practical Penetration Testing Techniques","color",0.91 EUR, United States 18156,"Corporate Hacking and Technology-Driven","color",0.91 EUR, United States 18157,"Gray Hat Hacking The Ethical Hackers Handbook - THE LATEST STRATE","color",0.91 EUR, United States 18158,"The Hacker’s Underground Handbook","color",0.91 EUR, United States 18159,"Ethical Hacking and Penetration Testing Guide","color",0.

91 EUR, United States 18160,"HACKING SECRET CIPHERS WITH GOONER","color",0.91 EUR, United States 18161,"THE BASICS OF WEB HACKING TOOLS AND TECHNIQUES TO ATTACK THE WEB","color",0.91 EUR, United States 18162,"HACKING WIRELESS NETWORKS FOR DUMMIES EBOOK","color",0.91 EUR, United States 18163,"Wiley - Hacking Wireless Networks for Dummies 2005 ","color",0.91 EUR, United States 18164,"HACKING WIRELESS NETWORKS FOR DUMMIES 2014","color",0.

91 EUR, United States 18165,"LOW TECH HACKING STREET SMARTS FOR SECURITY PROFESSIONALS","color",0.91 EUR, United States 18166,"Hacking Firefox - More Than 150 Hacks, Mods, and Customizations","color",0.91 EUR, United States 18167,"HACKING EDUCATION - 10 QUICK FIXES FOR EVERY SCHOOL","color",0.91 EUR, United States 18168,"Electronics from the Ground Up Learn by Hacking, Designing, and I","color",0.91 EUR, United States 18169,"HACKING HAPPINESS","color",0.

91 EUR, United States 18170,"THE HACKERS MANUAL 2016","color",0.91 EUR, United States 18171,"GAME CONSOLE HACKING - HAVE FUN WHILE VOIDING YOUR WARRANTY","color",0.91 EUR, United States 18172,"Hacking For Dummies, 4th Edition","color",0.91 EUR, United States 18173,"GOOGLE HACKING TRICKS","color",0.91 EUR, United States 18174,"THE HACKING BIBLE","color",0.

91 EUR, United States 18175,"EBOOK HACKERS SECRETS REVEALED","color",0.91 EUR, United States 18176,"HACKING GPS TRICKS TO IMPROVE ","color",0.91 EUR, United States 18177,"TCP IP ILLUSTRATED, VOLUME 1 THE PROTOCOLS 2ND EDITION ONLYGILL","color",0.91 EUR, United States 18178,"CODING FREEDOM THE ETHICS AND AESTHETICS OF HACKING GLODLS ","color",0.91 EUR, United States 18179,"THE ART OF HUMAN HACKING","color",0.

91 EUR, United States 18181,"Hacking For Dummies ","color",0.91 EUR, United States 18183,"The Basics of Web Hacking ","color",0.91 EUR, United States 18185,"LOW TECH HACKING","color",0.91 EUR, United States 18186,"Asterisk Hacking - Stay Safe by Learning the secrets the bad guys","color",0.91 EUR, United States 18187,"Android Magazine UK - Hacking Master Class -","color",1.

37 EUR, United States 18188,"HACKING FOR DUMMIES, THIRD EDITION","color",0.91 EUR, United States 18189,"HACK ATTACKS REVEALED A COMPLETE REFERENCE WITH CUSTOM SECURIT","color",0.91 EUR, United States 18190,"Hacking Exposed - Web Applications Security Secrets and Solutions","color",0.91 EUR, United States 18191,"Hacking Wireless Networks For Dummies - Find And Fix Network ","color",0.91 EUR, United States 18192,"Android Magazine UK - Hottest Phones 2013 + Amazing Hacking ","color",0.

91 EUR, United States 18193,"The Basics of Web Hacking","color",0.91 EUR, United States 18194,"Hacking Exposed 7 Network Security Secrets & Solutions","color",0.91 EUR, United States 18212,"Treasure Island Media - Fuck Off Homo","junkiepig666",13.67 EUR, Germany 18267,"7 reasons your credit card gets blocked - Interesting Article","HappyEyes",0.

91 EUR, United States 18276,"4-CL-PVP Crystal 250 Gram`s ","Empereor",910.

66 EUR, China 18285,"2-NMC Crystal 250 Gram`s ","Empereor",911.66 EUR, China 18288,"2-NMC Crystal 100 Gram`s ","Empereor",547 EUR, China 18292,"BK-ETHYL-K CRYSTAL / Ephylone / BK-EBDP 250 Gram`s ","Empereor",911.66 EUR, China 18294,"BK-ETHYL-K CRYSTAL / Ephylone / BK-EBDP 100 Gram`s ","Empereor",546.4 EUR, China 18299,"Penetration Testing,A Hands-on Introduction to Hacking","color",0.91 EUR, United States 18300,"Micro Mart - Hacking can it be a Good thing (13 August 2015)","color",0.

91 EUR, United States 18301,"HACKING EXPOSED WEB APPLICATIONS","color",0.91 EUR, United States 18302,"HACKING FOR DUMMIES, 4TH EDITION 2","color",0.91 EUR, United States 18303,"HACKING GUIDE V3.91 EUR, United States 18304,"HACKING EXPOSED 6TH EDITION - NETWORK SECURITY SECRETS AND SOL","color",0.

91 EUR, United States 18305,"HACKING EDUCATION 10 QUICK FIXES FOR EVERY SCHOOL","color",0.91 EUR, United States 18306,"Hacking the Cable Modem ","color",0.91 EUR, United States 18308,"Android Magazine UK - Hacking Masterclass - ","color",0.91 EUR, United States 18309,"Hacking and Penetration Testing with Low Power Devices","color",0.91 EUR, United States 18310,"GRAY HAT HACKING, 3RD EDITION [email protected]~~ WBRG ","color",0.

91 EUR, United States 18311,"Hacking VoIP - Protocols, Attacks And Countermeasures","color",0.91 EUR, United States 18313,"BIT WARS Cyber Crime, Hacking & Information Warfare Volume 2","color",0.91 EUR, United States 18314,"ETHICAL HACKING STUDENT GUIDE","color",0.91 EUR, United States 18315,"HACKING EXPOSED WINDOWS MICROSOFT WINDOWS SECURITY SECRETS","color",0.91 EUR, United States 18316,"WEB HACKING ATTACKS AND DEFENSE","color",0.

91 EUR, United States 18317,"BIT WARS Cyber Crime, Hacking & Information Warfare Volume 2","color",0.91 EUR, United States 18318,"ETHICAL HACKING STUDENT GUIDE","color",0.91 EUR, United States 18319,"HACKING EXPOSED WINDOWS MICROSOFT WINDOWS SECURITY SECRETS","color",0.91 EUR, United States 18320,"HACKING FOR DUMMIES","color",0.91 EUR, United States 18321,"HACKING EXPOSED WEB APPLICATIONS 3RD EDITION","color",0.

91 EUR, United States 18322,"APRESS HACKING THE KINECT 2012 EBOOK-REPACKB00K","color",0.91 EUR, United States 18323,"Wired UK - The web Is Under Threat -","color",0.91 EUR, United States 18324,"NINJA HACKING UNCONVENTIONAL PENETRATION TESTING TACTICS","color",0.91 EUR, United States 18325,"APRESS HACKING THE KINECT 2012 EBOOK-REPACKB00K","color",0.91 EUR, United States 18326,"HACKING FOR DUMMIES, 3RD EDITION BY KEVIN BEAVER","color",0.

91 EUR, United States 18327,"HACKING FOR DUMMIES-3RD EDITION TEAM NANBAN TPB ","color",0.91 EUR, United States 18328,"BLUETOOTH HACKING AND HACK THE NET","color",0.91 EUR, United States 18329,"HARDWARE HACKING PROJECTS FOR GEEKS","color",0.91 EUR, United States 18330,"Hacking Happiness Why Your Personal Data Counts ","color",0.91 EUR, United States 18331,"Website Magazine - E- Commerce Growith Hacking (August 2014","color",0.

91 EUR, United States 18332,"Hacking Happiness Why Your Personal Data Counts ","color",0.91 EUR, United States 18333,"Hacking Windows VistaExtremeTech Take Control of The Taskbar ","color",0.91 EUR, United States 18334,"Hacking For Dummies, 3 edition","color",0.91 EUR, United States 18335,"AI +KAOS 10 ANNI DI HACKING E MEDIATTIVISMO 2012 PDF TXT","color",0.91 EUR, United States 18336,"HACKING THE CODE - WEB APPLICATION SECURITY","color",0.

91 EUR, United States 18337,"Android Magazine UK - Issue 15, 2012","color",0.91 EUR, United States 18338,"Ninja Hacking Unconventional Penetration Testing Tactics and Tech","color",0.91 EUR, United States 18339,"Android Apps Security - Create apps that are safe from Hacking, A","color",0.91 EUR, United States 18340,"HACKING EXPOSED COMPUTER FORENSICS","color",0.91 EUR, United States 18341,"Gray Hat Hacking - The Ethical Hackers Handbook (2 Volume Collect","color",0.

91 EUR, United States 18342,"Hacking Exposed - Network Security Secrets and Solutions (6th Edi","color",0.91 EUR, United States 18343,"Gray Hat Hacking, Third Edition","color",0.91 EUR, United States 18345,"Hacking Exposed - Network Security Secrets and Solutions (6th Edi","color",0.91 EUR, United States 18346,"Hacking Exposed Windows - Microsoft Windows Security Secrets and ","color",0.

91 EUR, United States 18347,"CODING FREEDOM THE ETHICS AND AESTHETICS OF HACKING","color",0.

91 EUR, United States 18348,"Hacking - The Next Generation","color",0.91 EUR, United States 18349,"HACKING THE PSP NEUROSIS ","color",0.91 EUR, United States 18350,"Reality Kings+Naughty America Porn Account Lifetime+Freebies ","cyberzen",10 EUR, Finland 18351,"HACKING THE CABLE MODEM","color",0.91 EUR, United States 18353,"Low Tech Hacking - Street Smarts for Security Professionals","color",0.91 EUR, United States 18354,"HACKING INTRANET WEBSITES","color",0.

91 EUR, United States 18355,"Brazzers+Mofos+Bangbros+Tiny4k Porn Account Lifetime+Freebies ","cyberzen",10 EUR, Finland 18356,"THE BASICS OF HACKING AND PENETRATION TESTING","color",0.91 EUR, United States 18357,"HACKING DIGITAL CAMERAS","color",0.91 EUR, United States 18358,"HACKING FOR DUMMIES - ACCESS TO OTHER PEOPLES SYSTEMS MADE SIM","color",0.91 EUR, United States 18359,"WIRELESS HACKING","color",0.91 EUR, United States 18360,"HACKING EXPOSEDMOBILE SECURITY","color",0.

91 EUR, United States 18361," ALMOST FREE! BRAZZERS or MOFOS PREMIUM PORN ACCOUNT","cyberzen",2 EUR, Finland 18363," Premium Porn Account LIFETIME + FREEBIES ","cyberzen",10 EUR, Finland 18364," Premium Porn Account LIFETIME + FREEBIES ","cyberzen",10 EUR, Finland 18365," Premium Porn Account LIFETIME + FREEBIES ","cyberzen",10 EUR, Finland 18366," Premium Porn Account LIFETIME + FREEBIES ","cyberzen",10 EUR, Finland 18368,"REALITY KINGS Premium Porn Account LIFETIME + FREEBIES ","cyberzen",10 EUR, Finland 18369," Premium Porn Account LIFETIME + FREEBIES ","cyberzen",10 EUR, Finland 18370,"Naughty America Premium Porn Account LIFETIME + FREEBIES ","cyberzen",10 EUR, Finland 18371," Premium Porn Account LIFETIME + FREEBIES ","cyberzen",10 EUR, Finland 18372,"Twistys Premium Porn Account LIFETIME + FREEBIES ","cyberzen",10 EUR, Finland 18373," Premium Porn Account LIFETIME + FREEBIES ","cyberzen",10 EUR, Finland 18374,"Digital Playground Premium Porn Account LIFETIME + FREEBIES ","cyberzen",10 EUR, Finland 18375," Premium Porn Account LIFETIME + FREEBIES ","cyberzen",10 EUR, Finland 18376,"3DXSTAR Premium Porn Account LIFETIME + FREEBIES ","cyberzen",10 EUR, Finland 18377," Premium Porn Account LIFETIME + FREEBIES ","cyberzen",10 EUR, Finland 18378,"HD PORN PASS Premium Porn Account LIFETIME + FREEBIES ","cyberzen",10 EUR, Finland 18380," Premium Porn Account LIFETIME + FREEBIES ","cyberzen",10 EUR, Finland 18381,"TEEN PORNO PASS Premium Porn Account LIFETIME + FREEBIES ","cyberzen",10 EUR, Finland 18383," Premium Porn Account LIFETIME + FREEBIES ","cyberzen",10 EUR, Finland 18384," Premium Porn Account LIFETIME + FREEBIES ","cyberzen",10 EUR, Finland 18385,"TINY4K Premium Porn Account LIFETIME + FREEBIES ","cyberzen",10 EUR, Finland 18386,"FANTASY-HD Premium Porn Account LIFETIME + FREEBIES ","cyberzen",10 EUR, Finland 18387," Premium Porn Account LIFETIME + FREEBIES ","cyberzen",10 EUR, Finland 18388,"PUREMATURE Premium Porn Account LIFETIME + FREEBIES ","cyberzen",10 EUR, Finland 18389," Premium Porn Account LIFETIME + FREEBIES ","cyberzen",10 EUR, Finland 18390,"VIDEOBOX Premium Porn Account LIFETIME + FREEBIES ","cyberzen",10 EUR, Finland 18391," Premium Porn Account LIFETIME + FREEBIES ","cyberzen",10 EUR, Finland 18392,"COLLEGERULES Premium Porn Account LIFETIME + FREEBIES ","cyberzen",10 EUR, Finland 18393," Premium Porn Account LIFETIME + FREEBIES ","cyberzen",10 EUR, Finland 18396,"PLAYBOY Premium Porn Account LIFETIME + FREEBIES ","cyberzen",10 EUR, Finland 18397,"POVD Premium Porn Account LIFETIME + FREEBIES ","cyberzen",10 EUR, Finland 18398," Premium Porn Account LIFETIME + FREEBIES ","cyberzen",10 EUR, Finland 18399,"The Real Me Premium Porn Account LIFETIME + FREEBIES ","cyberzen",10 EUR, Finland 18400,"Porn Pros+Twistys+Babes+Playboy Porn Account Lifetime+Freebies ","cyberzen",10 EUR, Finland 18401,"Reality Kings+Naughty America Porn Account Lifetime+Freebies ","cyberzen",10 EUR, Finland 18402,"Brazzers+Mofos+Bangbros+Tiny4k Porn Account Lifetime+Freebies ","cyberzen",10 EUR, Finland 18405,"Super Skunk 3g - laadukas -Extrabonus! Tarjous 50e per 3g","Suomen Kukkaparatiisi",55 EUR, Finland 18406,"Super Skunk 5g - laadukas - Extrabonus! Tarjous 85e per 5g","Suomen Kukkaparatiisi",90 EUR, Finland 18420,"7g DJ Shorts Blueberry (Medical Guaranteed) ","BudsNRoses",65.57 EUR, United States 18421,"14g DJ Shorts Blueberry (Medical Guaranteed) ","BudsNRoses",111.1 EUR, United States 18422,"1oz DJ Shorts Blueberry (Medical Guaranteed) ","BudsNRoses",196.7 EUR, United States 18423,"1/2oz Premium Shake","BudsNRoses",60.1 EUR, United States 18425,"1oz Premium Shake","BudsNRoses",96.

53 EUR, United States 18435,"100 Dutch LSD blotters 180ug-Free Shipp","Citra",255.26 EUR, United Kingdom 18449," Exploit Kit Bleeding Life (2.0)","OnePiece",1 EUR, Philippines 18450,"Account Creator Extreme 4.2","OnePiece",1 EUR, Philippines 18451,"Alien Spy Rat 5.0 with plugin (No Crack)","OnePiece",1 EUR, Philippines 18452,"Antidetect FF Browser Ver 5.

50 Best For Carders","OnePiece",1 EUR, Philippines 18453,"Apple Scam Phishing Page Get your own Fullz","OnePiece",1 EUR, Philippines 18454,"4 gram quality fishscale cocaine","tennin",251 EUR, Netherlands 18455,"5 gram quality fishscale cocaine","tennin",308 EUR, Netherlands 18456,"AVG Internet Security 2015 Serial Keys","OnePiece",1 EUR, Philippines 18458,"BlackShades RAT 5.1 + User Guide","OnePiece",1 EUR, Philippines 18459,"CCleaner Pro and Business Edition Serial Keys","OnePiece",1 EUR, Philippines 18460,"Citadel Rain Version Enter the Fort’s Dungeons","OnePiece",1 EUR, Philippines 18461,"DarkComet RAT","OnePiece",1 EUR, Philippines 18462,"DROID JACK Andriod RAT","OnePiece",0.

55 EUR, Philippines 18463,"Free VPS Windows Linux for a year+ Set Up guide","OnePiece",1 EUR, Philippines 18464,"Hotspot Shield Elite VPN","OnePiece",1 EUR, Philippines 18465,"Hydra 8.1 The Hacker's Choice Network Login cracker","OnePiece",1 EUR, Philippines 18466,"Lizard Stresser DB","OnePiece",1 EUR, Philippines 18467,"Pony Loader 2.0","OnePiece",1 EUR, Philippines 18468,"RAT4A v.0 (Spy On Cam","OnePiece",1 EUR, Philippines 18469,"Rise Up Mail Invitation Code","OnePiece",1.

82 EUR, Philippines 18470,"SmsBot Android Botnet","OnePiece",1 EUR, Philippines 18471,"The Jerm DUMP Writer + User Guide","OnePiece",1 EUR, Philippines 18472,"VM WARE Virtual Machine 12 Serial Keys","OnePiece",1 EUR, Philippines 18475,"MDMB CHMICA 100 Gram`s ","Empereor",638.16 EUR, China 18476,"MDMB CHMICA 250 Gram`s ","Empereor",1093.99 EUR, China 18477,"MDMB CHMICA 500 Gram`s ","Empereor",1640.99 EUR, China 18478,"MDMB CHMICA 1 Kg ","Empereor",2734.98 EUR, China 18496,"A-PVP 500g","Jimmy chemicals",3281.

98 EUR, China 18499,"1g High Grade Crystal Methamphetamine USA>USA","junkiejesus",36.42 EUR, United States 18500,"2g High Grade Crystal Methamphetamine USAtoUSA","junkiejesus",71.93 EUR, United States 18507,"3g High Grade Crystal Methamphetamine USAtoUSA","junkiejesus",108.36 EUR, United States 18508,"8g High Grade Crystal Meth USAtoUSA","junkiejesus",264.08 EUR, United States 18510,"NO-FE 1g High Grade Crystal Meth USAtoUSA","junkiejesus",54.

63 EUR, United States 18514,"10gr Great Quality Cocaine 70%-75%","amsterdam2015",401.13 EUR, Netherlands 18516,"5gr Great Quality Cocaine 70%-75%","amsterdam2015",203.3 EUR, Netherlands 18517,"2gr Great Quality Cocaine 70%-75%","amsterdam2015",86.61 EUR, Netherlands 18518,"1gr Great Quality Cocaine 70%-75%","amsterdam2015",48.32 EUR, Netherlands 18519," DK 1g AB-CHMINACA-jauhe 100 €/g","Hamppukauppa",105 EUR, Finland 18520,"0.

5gr Great Quality Cocaine 70%-75% ","amsterdam2015",25.53 EUR, Netherlands 18521,"10gr Pure Uncut Cocaine 90%-92% ((( get 1 for freee))))))","amsterdam2015",542.44 EUR, Netherlands 18522,"5gr Pure Uncut Cocaine 90%-92%((( get 0.06 EUR, Netherlands 18523,"2gr Pure Uncut Cocaine 90%-92%((( get 0.

52 EUR, Netherlands 18525,"1gr Pure Uncut Cocaine 90%-92% ","amsterdam2015",57.5gr Pure Uncut Cocaine 90%-92%","amsterdam2015",30.08 EUR, Netherlands 18528,"10gr Pure Heroin HEROIN #3(((( get 1 for freee))))","amsterdam2015",350.

99 EUR, Netherlands 18530,"5gr Pure Heroin HEROIN #3(((( get 0.33 EUR, Netherlands 18532,"2gr Pure Heroin HEROIN #3((( get 0.84 EUR, Netherlands 18533,"1gr Pure Heroin HEROIN #3","amsterdam2015",38.

5gr Pure Heroin HEROIN #3","amsterdam2015",21.88 EUR, Netherlands 18545,"bk-ebdp 1000g","eleven88",1366 EUR, China 18548,"Dibutylone Crystal 1000g","eleven88",1366 EUR, China 18550,"NM2201 1000g","eleven88",1730.26 EUR, China 18551,"MAM2201 1000G","eleven88",1732.15 EUR, China 18553,"fentanyl hcl 2g 99%","eleven88",364.

27 EUR, China 18561,"Reindeer meat jerky 60g","faceless",14.5 EUR, Finland 18562,"Gerodorm (diazepam) 100 kpl/40mg - TARJOUS!","arnold",194 EUR, Finland 18566,"METHADONE HCL - 40 mg BOX of 7 CAPS | THE STRONGEST !","appylove",100 EUR, France 18567,"METHADONE HCL - 40 mg - 1 CAP (strong)","appylove",15 EUR, France 18569,"20 x Frontin Alprazolam 0.5mg (XANAX) FREE SHIPPING","AllHoursOpen",16.67 EUR, United Kingdom 18570,"20 x Frontin Alprazolam 0.5mg (XANAX) FREE SHIPPING","AllHoursOpen",16.

64 EUR, United Kingdom 18571,"40 x Frontin Alprazolam 0.5mg (XANAX) FREE SHIPPING","AllHoursOpen",30.01 EUR, United Kingdom 18572,"40 x Frontin Alprazolam 0.5mg (XANAX) FREE SHIPPING","AllHoursOpen",30.01 EUR, United Kingdom 18625,"TH-PVP POWDER 100 Gram`s ","Empereor",547 EUR, China 18626,"TH-PVP POWDER 250 Gram`s ","Empereor",911.

66 EUR, China 18627,"TH-PVP POWDER 500 Gram`s ","Empereor",1595.41 EUR, China 18628,"TH-PVP POWDER 1 Kg ","Empereor",2734.98 EUR, China 18650,"2g Marocco Rifman Hash A+","Tr4derJo3",20.95 EUR, Germany 18667," 12 x C-99 F2 Female Seeds *welcome special*","Tr4derJo3",16.95 EUR, Germany 18668,"12 x Iced Grapefruit F2 Female Seeds *welcome special*","Tr4derJo3",16.

95 EUR, Germany 18718,"Cialis Pharma Grade Lilly brand 20 x 20mg tablets","cerberus",68.91 EUR, United Kingdom 18730,"100 Gram HQ DUTCH Pink Speed, Pep, Paste 200€ instead of 250€","Meyer Lansky",200 EUR, Netherlands 18748,"3.52 EUR, Australia 18750,"14g ( oz) Hydro Weed","Medibank",97.9 EUR, Australia 18752,"56g (2 oz) Hydro Weed","Medibank",340.

27 EUR, Australia 18755,"1g Uncut MDMA","Medibank",82.

87 EUR, Australia 18756,"2g Uncut MDMA","Medibank",158 EUR, Australia 18757,"2.34 EUR, Australia 18758,"3g Uncut MDMA","Medibank",233.5