Basic Business Statistics Concepts and Applications 5th Edition Berenson Solutions Manual by Ace Test Banks

Basic Business Statistics Concepts and Applications 5th Edition Berenson Solutions Manual

richard@qwconsultancy.com

1|Pa ge

Chapter 1: Defining and collecting data Learning objectives After studying this chapter you should be able to: 1. identify how statistics is used in business 2. recognise the sources of data used in business 3. identify the types of data used in business

4. 5.

distinguish between different survey sampling methods evaluate the quality of surveys

1.1

(a) (b)

1.2

Three sizes of coffee cups are classified into distinct categories — small, medium and large — in which order is implied.

1.3

(a) (b)

The type of beverage sold yields categorical or ‘qualitative’ responses. The type of beverage sold yields distinct categories in which no ordering is implied.

The time it takes to download a video from the Internet yields numerical or ‘quantitative’ responses. The download time is a ratio scaled variable because the true zero point in the measurement is zero units of time.

1.4

(a) (b) (c) (d)

Numerical, discrete, ratio scale Numerical, continuous, ratio scale Categorical, nominal scale Categorical, nominal scale

1.5

(a) (b) (c) (d)

Numerical, continuous, ratio scale Numerical, discrete, ratio scale Categorical, nominal scale Categorical, nominal scale

1.6

(a) (b) (c) (d)

Categorical, nominal scale Numerical, continuous, ratio scale Numerical, discrete, ratio scale Numerical, discrete, ratio scale

1.7

(a) Numerical, continuous, ratio scale* (b) Numerical, discrete, ratio scale (c) Numerical, continuous, ratio scale* (d) Categorical, nominal *Some researchers consider money as a discrete numerical variable because it can be ‘counted’.

1.8

(a) (b) (c)

1.9

(a) (b)

Income may be considered discrete if we ‘count’ our money. It may be considered continuous if we ‘measure’ our money; we are only limited by the way a country’s monetary system treats its currency. The first format is preferred because the responses represent data measured on a higher scale. The first format would bring a greater response because it has a large number of possible income levels, as opposed to a limited number of income groups in the second format. The population is ‘all Australian working women’. A systematic or random sample could be taken of Australian working women. The director might wish to collect both numerical and categorical data. Three categorical questions might be occupation, marital status and type of clothing purchased online. Numerical questions might be age, average monthly hours shopping online for clothing and income.

1.10

An experiment will be carried out in artificial conditions and will try to assess people’s attitudes to making charitable donations. These may differ in real-life situations. A survey of the same subjects will rely on self-reporting. Respondents may not always report accurately if they are trying to appear more generous.

1.11

The company is likely to have more data relating to its own sales. It can use invoices to retailers, if it is a wholesaler, or records of cash and credit card purchases, if it is a retailer. For competitors, it will have access to advertised prices from the media or online advertisements but may not know the sales volumes.

1.12

The answers to this question depend on which datafile is selected.

1.13

The answers to this question depend on which poll or news story is selected.

1.14

The answers to this question depend on which story is selected.

1.15

The data source will depend on what driving characteristics are being studied and the type of vehicle. It could be tachographs as used in heavy vehicles, dash cameras, GPS monitors and other equipment. However, surveys of drivers or log books might also be used.

1.16

The answer to this question will depend on the panel chosen.

1.17

(a) (b) (c)

1.18

Sample without replacement: read from left to right in 3-digit sequences and continue unfinished sequences from end of row to beginning of next row. Row 05: 338; 505; 855; 551; 438; 855; 077; 186; 579; 488; 767; 833; 170 Row 05-06: 897 Row 06: 340; 033; 648; 847; 204; 334; 639; 193; 639; 411; 095; 924 Row 06-07: 707 Row 07: 054; 329; 776; 100; 871; 007; 255; 980; 646; 886; 823; 920; 461 Row 08: 893; 829; 380; 900; 796; 959; 453; 410; 181; 277; 660; 908; 887 Row 08-09: 237 Row 09: 818; 721; 426; 714; 050; 785; 223; 801; 670; 353; 362; 449 Row 09-10: 406

001 040 902

Note: all sequences above 902 are discarded. 1.19

(a) (b)

Row 29: 12; 47; 83; 76; 22; 99; 65; 93; 10; 65; 83; 61; 36; 98; 89; 58; 86; 92; 71 Note: all sequences above 93 and all repeating sequences are discarded. Row 29: 12; 47; 83; 76; 22; 99; 65; 93; 10; 65; 83; 61; 36; 98; 89; 58; 86 Note: all sequences above 93 are discarded. Elements 65 and 83 are repeated.

1.20

A simple random sample would be less practical for personal interviews because of travel costs (unless interviewees are paid to attend a central interviewing location).

1.21

This is a probability sample because the selection is based on chance. It is not a simple random sample because A is more likely to be selected than B or C.

1.22

Here all members of the population are equally likely to be selected and the sample selection mechanism is based on chance. But not every sample of size 2 has the same chance of being selected. For example, the sample “’B and C’ is impossible.

1.23

(a)

(b) (c)

(d)

(e)

1.24

(a) (b)

As a complete list of voters exists, a simple random sample of 2,000 voters could be taken. If attitudes to urban consolidation randomly fluctuate across the alphabetical listing of voters, a systematic 1-in-20 sample could also be taken from the population frame. If attitudes may differ by gender and by ward of residence, a stratified sample using eight strata could be selected. If attitudes to urban consolidation are thought to fluctuate as much within clusters as between them, a cluster sample could be taken. A simple random sample is one of the simplest to select. The population frame is the council’s list of 40,000 voters’ names. A systematic sample is easier to select by hand from the council’s records than a simple random sample, since an initial person at random is selected and then every 20th person thereafter would be sampled. The systematic sample would have the additional benefit that the alphabetical distribution of sampled voters’ names would be more comparable to the alphabetic distribution of voters’ names in the population. However, if certain ethnic groups were more likely to be concentrated alphabetically and they have particular attitudes to urban consolidation due to their cultural background, then this method may introduce bias. If lists by gender and ward of residence are readily available, a stratified sample should be taken. Since attitudes to urban consolidation may indeed differ by gender and area of residence, the use of a stratified sampling design will ensure all strata are represented in the sample. It will also generate a more representative sample and produce estimates of the population parameter that have greater precision. People who already live in high-rise apartments are likely to have different views about urban consolidation from those living in low-density areas. Systematic sampling by street is likely to lead to selection bias if a high proportion of streets in one ward is chosen. A stratified sample should be taken so that each of the four strata will be proportionately represented. Since the stratum may differ in the invoice amount, it may be more important to sample a larger percentage of invoices in stratum 1 and stratum 2, and smaller percentages in stratum 3 and stratum 4. For example, 50/5000 = 1% so 1% of 500 = 5 invoices should be selected from stratum 1; similarly 10% = 50 should be selected from stratum 2, 20% = 100 from stratum 3, and 69% = 345 from stratum 4.

(c)

It is not simple random sampling because, unlike the simple random sampling, it ensures proportionate representation across the entire population.

1.25

Before accepting the results of a survey of university students, you might want to know, for example: Who funded the survey? Why was it conducted? What was the population from which the sample was selected? What sampling design was used? What mode of response was used: a personal interview, a telephone interview or a mail survey? Were interviewers trained? Were survey questions fieldtested? What questions were asked? Were they clear, accurate, unbiased, and valid? What operational definition of ‘vast majority’ was used? What was the response rate? What was the sample size?

1.26

(a) (b) (c) (d)

Possible coverage error: only employees in a specific division of the company were sampled. Possible non-response error: no attempt is made to contact non-respondents to urge them to complete the evaluation of job satisfaction. Possible sampling error: the sample statistics obtained from the sample will not be equal to the parameters of interest in the population. Possible measurement error: ambiguous wording in questions in the questionnaire.

1.27

Before accepting the results of the survey, you might want to know, for example: Who funded the study? Why was it conducted? What was the population from which the sample was selected? What sampling design was used? What mode of response was used: a computer based survey, a personal interview, a telephone interview or a mail survey? Were interviewers trained? Were survey questions field-tested? How were cybercrime and millennials defined? What other questions were asked? Were they clear, accurate, unbiased and valid? What was the response rate? What was the margin of error? What was the sample size? What was the frame being used?

1.28

Before accepting the results of the survey, the environmental economist might want to know, for example: Who funded the study? Was it a special interest group? Why was it conducted? What was the population from which the sample was selected? What was the sample size? What was the frame being used? What sampling design was used? What mode of response was used: in a less developed country telephones will not be commonly found in village homes, so was this a personal interview or a mail survey? In what languages were the questions written? What is the literacy rate in Kiribati? Were interviewers trained? Were survey questions field-tested? What other questions were asked? Were they clear, accurate, unbiased and valid? What was the response rate?

1.29

(a)

(b)

The SMS/phone/online voting system is a non-probability sampling method. It is open to bias as some viewers may choose not to vote, while supporters of a particular contestant can vote a large number of times and have the potential to influence the result. A random poll of viewers without replacement conducted by phone should produce a less-biased result, although it is possible some viewers without phones or with unlisted numbers might be excluded. A phone-in-voting system has cost benefits for the show’s owners as they can earn revenue from premium rate calls. However, SMS polling may incur a set up fee. Social media polling can show results such as the number of ‘likes’ quickly and can be useful for advertising purposes but has the disadvantage that later voters could be influenced by earlier votes. There may also be better ratings for the show if the audience feels more involved. Phoning viewers using research staff would be more costly but should give more reliable data.

1.30

The Internet survey described is a non-probability sampling method. It is open to bias as some diners may choose not to vote while those who favour the restaurant (e.g. friends, relatives) can participate each time they book through the online site and dine there. This has the potential to influence the result. Ratings obtained from such a survey could be biased. Ratings obtained from a random sample of all diners should produce a less biased result. It would include those who dined without making any booking and who possibly had no knowledge of prior reviews.

1.31

A population contains all the items of interest whereas a sample contains only a portion of the items in the population.

1.32

A statistic is a summary measure describing a sample whereas a parameter is a summary measure describing an entire population.

1.33

Descriptive statistical methods deal with the collection, presentation, summarisation and analysis of data whereas inferential statistical methods deal with decisions arising from the projection of sample information to the characteristics of a population.

1.34

Categorical random variables yield categorical responses, such as yes or no answers. Numerical random variables yield numerical responses, such as height in centimetres.

1.35

Discrete random variables produce numerical responses that arise from a counting process. Continuous random variables produce numerical responses that arise from a measuring process.

1.36

An operational definition is a universally accepted meaning that is clear to all associated with an analysis. Without an operational definition, confusion can occur.

1.37

The four types of measurement scales are (i) nominal scale, (ii) ordinal scale, (iii) interval scale and (iv) ratio scale.

1.38

Barrel draw methods of sampling do not allow for a thorough mixing of items.

1.39

Sampling with replacement means that once a person or item is selected, it is returned to the frame where it has the same probability of being selected again. Sampling without replacement means that a person or item once selected is not returned to the frame and therefore cannot be selected again.

1.40

In a simple random sample, each individual item is selected randomly. In a systematic sample, the N individuals or items in the population frame are partitioned into k groups by dividing the size of the population frame N by the desired sample size n. The first individual or item to be selected is chosen at random from the k individuals or items in the first partitioned group in the population frame, and the rest of the sample is obtained by selecting every kth individual or item thereafter from the entire population frame listing.

1.41

1.42

In a stratified sample, the N individuals or items in the population are first subdivided into separate subpopulations, or strata, according to some common characteristic. In a cluster sample, the N individuals or items in the population are divided into several clusters so that each cluster is representative of the entire population. A random sampling of clusters is then taken and all individuals or items in each selected cluster are then studied.

1.43 The answers to this question depend on which story is being selected. 1.44 The answers to this question vary from person to person according to their experiences. Answers for 1.45 to 1.47 provided below are just some of the many different possible answers. 1.45 (a)

(b) (c) (d)

1.46 (a) (b) (c) 1.47 (a) (b) (c)

This publication presents information about the Australian labour force, showing monthly data from 1978 onwards. It examines employment by gender and age group showing full time, part time, total employment and unemployment rates, which can be viewed by state. It is based on a survey sample of approximately 26,000 homes and non-private dwellings, such as hotels, with interviews carried out each month over an eight-month period. More recently, online electronic data collection methods have been used as well. Categorical variables: gender, employment status, industry, school or tertiary education attendance, state of residence etc. Numerical variables: number of hours worked, age Both hours worked and age are continuous numerical variables. Sometimes the way the data is collected results in integer values only, especially age in whole years. Categorical variables: gender, marital status, education, country of birth, religious affiliation, languages spoken other than English. Numerical variables: weekly income, age, weekly rent, monthly mortgage payment. Discrete numerical (weekly income, age in years, weekly rent, monthly mortgage payment). Numerical discrete, numerical continuous and categorical (nominal) data were used. Departing or arriving airline. Number of flights departed, arrived, cancelled.

1.48 (a) (b) (c)

1.49

(a) (b) (c) (d)

1.50

(a) (b) (c) (d)

The populations of interest were all school principals and P&C organisations in NSW and the ACT, and the general public of NSW and the ACT when the survey was conducted. Sample 1 consisted of respondents to the survey of principals and P&C organisations. Sample 2 consisted of respondents to the survey from the general public. If a greater proportion of responses were from the school community, there could be more concern about child safety and more observations made by pedestrians approaching the school. If a greater proportion of responses was from the general public, other issues such as irritation about poor signage for 40 km an hour school zones could predominate. The population of interest was all households in Tasmania. Categorical variables (where and what flavour of milk is purchased) and numerical variables (number of people living in the household and number of millilitres drunk). Answers could vary. Answers could vary. The population of interest was the residents living in a north-eastern region of Sydney when the study was conducted. Categorical (ticket type and main purpose of using the bus) and numerical (resident’s age and frequency of bus use). Numerical continuous questions (distance travelling) and numerical discrete (weekly income of passengers). Categorical questions (gender and social status such as students, full-time or part-time workers and pensioners).

1.51

Even though Internet polling is less expensive, faster and offers higher response rates than telephone surveys, it is a self-selection response method. Because respondents who choose to participate in the survey may not represent the view of the public, the data collected may not be appropriate for making inferences about the general population.

1.52

(a) (b)

Coverage error, non-response error, measurement error. All errors can be minimised by carefully conducting the survey and adopting a strict methodology. However, non-response error may be more difficult to manage than the other errors as it is totally dependent upon the respondents’ willingness to participate.

1.53

(a)

Given the low response rate, the researchers should be especially concerned with non-response bias. The low response rate will also increase the sampling error. Researchers can follow up on the non-responses by mail or email to encourage or remind those who have not returned the survey to complete it. A more extensive check of university records may have produced better address information; for example, alumni addresses held on a different database. It may have helped to include more information about the uses of the survey and the importance of completing it in order to encourage a better response rate.

(b) (c)

1.54

(a)

Before accepting the results of this survey, one would like to know: (i) how big the sample size is, (ii) what is the purpose of the survey, (iii) what sampling design was used, (iv) what sampling frame was being used, (v) what questions were asked, (vi) how were the questions being phrased, and (vii) what was the response rate?

(b) (c)

(d)

(e) (f)

1.55

(a)

(b)

The population is all people who are mobile phone users living in the geographical region. To reduce the chance of coverage error, an adequate and up-to-date frame should be used. If communication companies were willing to provide sensitive customer data, it would be possible to get a complete list of names, addresses and phone numbers. Alternatively, one could use the services of commercial data centres which provide databases for telemarketing firms. Non-sampling error can be minimised by following up on the non-responses by mail or telephone or by conducting personal interviews instead of mail or Internet surveys. The number of questions in the survey should be minimised to increase response rate. Sampling error can be reduced by increasing the response rate or by increasing the sample size. Measurement error can be minimised by asking unambiguous and neutral questions that use commonly accepted operational definitions. Interviewers should also be provided with adequate training to prevent the respondents from feeling obligated to please the interviewers. Before accepting the results of this survey, you would like to know: (i) what the purpose of the survey is, (ii) what sampling method was being used, (iii) what sampling frame was being used, (iv) what questions were being asked, (v) how were the questions being phrased, and (vi) what was the response rate? The population could be all workers in the region who were not self-employed or have no boss. The frame could be compiled by using a list of those residents of the region who file tax returns. A simple random sample could be appropriate.

Chapter 2: Presenting data in tables and charts Learning objectives After studying this chapter you should be able to: 1.

describe the distribution of a single categorical variable using tables and charts

describe the distribution of a single numerical variable using tables and graphs

describe the relationship between two categorical variables using contingency tables

describe the relationship between two numerical variables using scatter diagrams and time-series plots

correctly present data in graphs

2.1

(a)

(b)

Category

Frequency

Percentage

26%

56%

18%

Bar chart

Solutions to End-of-Section and Chapter Review Problems

2.2

(c)

Pie chart

(a)

Bar chart

737

Solutions to End-of-Section and Chapter Review Problems

2.3

(b)

Pie chart

(a)

ABC has approximately 15% to 18% of viewers (actual number is 16.5%).

(b)

Channel 7 has approximately 25% of viewers (actual number is 25.5%).

(c)

Channel 9 has approximately 25% of viewers (actual number is 24.5%).

(d)

Channel 10 has approximately 27% to 30% of viewers (actual number is 28.5%).

(e)

SBS has approximately 1% of viewers (actual number is 0.5%).

It is important that, whatever number is calculated, all percentages add to 100%.

(a)

Bar chart

Top Ten Websites Ranked by Number of Unique Visitors March 2017 Website

2.4

738

MSN Amazon 0

500 1000 1500 Monthly Number of Unique Visitors (Millions)

Pie chart

Solutions to End-of-Section and Chapter Review Problems

739

Top Ten Websites Ranked by Number of Unique Visitors March 2017 YouTube 17% Yahoo! 11% Wikipedia 7% Twitter MSN 4%4%

2.5

Amazon Bing 8%eBay 4% 4% Facebook 17%

Google 24%

(b)

Both graphical techniques highlight the relative market sizes effectively and the selection of one over the other is dependent upon the purpose of preparing the chart.

(c)

It is obvious that Google has the largest market share, followed by Facebook and YouTube with over a billion unique visitors per month. Bing, eBay, Twitter and MSN all have a similar market share with between 280 and 290 million unique visitors per month.

(a)

Pie chart

Solutions to End-of-Section and Chapter Review Problems

740

Bar chart

2.6

(b)

Over 25% of the customers prefer white and (as can be seen from the pie chart) in excess of 50% of the customers prefer the three top colours of white, red and blue. If Pat is not a niche supplier of cars, he should have mainly the top three preferred colours.

(a)

Bar chart Australian Labour Force Status (Aged 15 years & Over) January 2017

Unemployed Not Looking for Full-… Unemployed Looking for Full-time… Not in Labour Force

Employed Part-time Employed Full-time 0

2000 4000 6000 8000 Number Thousands

10000

Solutions to End-of-Section and Chapter Review Problems

741

Pie chart

Australian Labour Force Status (Aged 15 Not in & Over) Emplo January years Labou 2017 yed

Fullr Unemploy… Unemployed… time… Forc… Emplo yed Parttime…

2.7

(b)

A pie chart or a bar chart could be used; however, a pie chart gives percentages so may be preferable.

(c)

From the pie chart, approximately 36% of population is not in the labour force, so the participation rate is approximately 64%.

Pie chart

Solutions to End-of-Section and Chapter Review Problems

Bar chart

742

Solutions to End-of-Section and Chapter Review Problems

743

Country of Birth - Capital City

United Kingdom

Other

New Zealand

Australia

10%

20%

30%

40%

50%

Capital City

United Kingdom 9%

Australia 48% Other 37%

New Zealand 6%

2.8

Ordered array: 63 64 68 71 75 88 94

2.9

Stem-and-leaf display of finance exam scores – stem unit 10, leaf unit 1 5

38 n=7

2.10

Ordered array of information systems exam scores: 50 74 74 76 81 89 92

60%

Solutions to End-of-Section and Chapter Review Problems 2.11

(a)

744

Ordered array

Expense claims rounded to nearest $10

$110 $210 $290 $340 $450

$120 $230 $300 $350 $450

$140 $290 $310 $380 $460

$180 $290 $310 $390 $500

$190 $290 $320 $390 $510

$200 $290 $340 $410 $550

$200 $290 $340 $410 $560

(b) The stem-and-leaf display provides more information, as it shows the distribution of the data as well as the order. (c)

The most common expense claim is $290, which occurs 6 times. Most expense claims are in the range $200 to $390.

(d)

The expense claims are concentrated in the $200 and $300 stems, near the centre of distribution.

2.12

(a)

Ordered array

(b)

Stem-and-leaf display of late payment fee stem 10

unit: leaf unit 1

(c)

The stem-and-leaf display provides more information, as it shows the distribution of the data as well as the order.

(d)

The fees are concentrated between $35 and $45.

2.13

(a)

Ordered array

0.30

0.50

0.65

0.70

1.00

1.25

1.30

1.50

1.60

1.65

2.00

2.50

Solutions to End-of-Section and Chapter Review Problems (b)

Stem-and-leaf display of ATM fee stem $1

2.14

745

unit leaf unit 10 cents

(c)

The stem-and-leaf display provides more information, as it shows the distribution of the data as well as the order.

(d)

The ATM fees are concentrated between 30 and 70 cents.

(a)

Ordered arrays

Full cream milk 135

155

160

163

165

170

185

188

115

133

Low- or reduced-fat milk 110

113

115

118

120

125

128

133

140

108

No-fat or skim milk 85 (b)

Stem-and-leaf displays

Full Cream Milk - Calories per 250 ml cup Stem unit: 10 Calories Leaf unit: 1 Calorie 13 5 14 15 5 5 16 0 0 0 0 3 3 5 17 0 18 5 8 Low or Reduced Fat Milk - Calories per 250 ml cup Stem unit: 10 Leaf unit: 1 11 0 3 5 8 12 0 5 8 13 3 3 14 0

Solutions to End-of-Section and Chapter Review Problems

746

No-fat or skim milk - calories per 250 ml cup Stem unit: 10 Leaf unit: 1 8 588 9 000088 10 8 11 5 12 13 3 (c)

The stem-and-leaf display provides more information, as it shows the distribution of the data as well as the order.

(d)

Low-fat and no-fat milk have less calories than full cream milk but there is some overlap. 2.15 (a) The class boundaries of the 9 classes can be ‘10 to less than 20’, ‘20 to less than 30’, ‘30 to less than 40’, ‘40 to less than 50’, ‘50 to less than 60’, ‘60 to less than 70’, ‘70 to less than 80’, ‘80 to less than 90’ and ‘90 to less than 100’.

97.8 − 11.6 = 9.58  10 9

(b)

The class-interval width is =

(c)

The nine class mid-points are: 15, 25, 35, 45, 55, 65, 75, 85 and 95.

2.16

(a)

(b)

160 (0.8 x 200) customers spent at least $60.

(c)

The top 10% of customers spent between $140 and $200.

(d)

The bottom 10% of customers spent less than $40.

2.17

(a)

60% of customers spent less than $100.

Frequency distribution and percentage distribution

Electricity costs

Frequency

Percentage

Cumulative percentage

$80 to < $100

$100 to < $120

$120 to < $140

$140 to < $160

$160 to < $180

$180 to < $200

$200 to < $220

100

Solutions to End-of-Section and Chapter Review Problems (b)

Histogram and percentage polygon

747

Solutions to End-of-Section and Chapter Review Problems

748

(c)

Cumulative percentage polygon

(d)

Monthly electricity costs are concentrated between $140 and $160 a month, with more than 25% falling in that interval.

2.18

(a)

Frequency, percentage and cumulative distributions

Unleaded Petrol Frequency 115 to < 120 2 120 to < 125 3 125 to < 130 9 130 to < 135 15 135 to < 140 13 140 to < 145 3 Total 45

Percent 4.44% 6.67% 20.00% 33.33% 28.89% 6.67% 100.00%

Diesel Frequency 115 to < 120 1 120 to < 125 8 125 to < 130 24 130 to < 135 7 135 to < 140 4 140 to < 145 1 Total 45

Cumulative Cumulative Percent 2.22% 1 2.22% 17.78% 9 20.00% 53.33% 33 73.33% 15.56% 40 88.89% 8.89% 44 97.78% 2.22% 45 100.00% 100.00%

Cumulative Cumulative Frequency Percent 2 4.44% 5 11.11% 14 31.11% 29 64.44% 42 93.33% 45 100.00%

Solutions to End-of-Section and Chapter Review Problems (b)

Frequency histograms

Frequency

NSW Unleaded Petrol 16 14 12 10 8 6 4 2 0 117.5

122.5

127.5

132.5

137.5

142.5

Price per Litre (cents)

NSW Diesel

28 24

Frequency

20 16 12

8 4 0 117.5

122.5

127.5 132.5 Cents per Litre

137.5

142.5

749

Solutions to End-of-Section and Chapter Review Problems (c)

750

Percentage polygons

NSW Fuel Prices March 2017

26 24

22 20

Frequency

Unleaded Diesel

16 14 12

10 8 6 4 2 0 112.5

117.5

122.5

127.5

132.5

137.5

142.5

147.5

Cents per Litre

(d)

Cumulative percentage polygons

NSW Fuel Prices March 2017

100% 90%

80% 70% 60% 50%

Unleaded

40% 30%

20% 10% 0% 115

(e)

120

125

130

Cents per Litre

135

140

145

We can conclude that on this day unleaded petrol diesel prices varied approximately from 115 cents per litre to 145. We can also conclude that diesel prices were generally lower and less varied than unleaded petrol prices.

Solutions to End-of-Section and Chapter Review Problems 2.19

(a)

Frequency distribution and percentage distribution Manufacturer A Life in hours

Frequency

Percentage %

Cumulative %

5500 to < 6000

5.00%

6000 to < 6500

20.00%

6500 to < 7000

17.5

37.50%

7000 to < 7500

52.50%

7500 to < 8000

17.5

70.00%

8000 to < 8500

17.5

87.50%

8500 to< 9000

12.5

100.00%

9000 to < 9500

100.00%

9500 to < 10000

100.00%

Total

100

Life in hours

Frequency

Percentage

Cumulative %

5500 to < 6000

0.00%

6000 to < 6500

0.00%

6500 to < 7000

7.5

7.50%

7000 to < 7500

22.50%

7500 to < 8000

27.5

50.00%

8000 to < 8500

55.00%

8500 to< 9000

17.5

72.50%

9000 to < 9500

17.5

90.00%

9500 to < 10000

100.00%

Total

100

Manufacturer B

751

Solutions to End-of-Section and Chapter Review Problems (b)

Histogram and percentage polygon

Histogram: Manufacturer A - Light Bulbs

Frequency

8 7 6 5 4 3 2 1 0 5000

5500

6000

6500

7000

7500

8000

8500

9000

9500 10000

Lifetime (Hours)

Histogram: Manufacturer B - Light Bulbs

Frequency

8 6 4

2 0 5000

5500

6000

6500

7000

7500

8000

8500

9000

9500 10000

Lifetime (Hours)

Percentage Polygons: Light Bulbs Manufacturer A

30% 25% 20% 15% 10% 5%

0% 5250 5750 6250 6750 7250 7750 8250 8750 9250 9750 10250 Lifetime (Hours)

752

Solutions to End-of-Section and Chapter Review Problems (c)

753

Ogive Ogive: Light Bulbs

100% 90% 80% 70% 60% 50% 40% 30% 20% 10% 0% 5500

Manufacturer A Manufacturer B

6000

6500

7000

7500

8000

8500

9000

9500

10000

Lifetime (Hours)

2.20

(d)

Manufacturer B produces bulbs with longer lives than manufacturer A’s bulbs. The cumulative percentage for manufacturer B shows 50% of its bulbs lasted 8000 hours or less contrasted with 70% of manufacturer A’s bulbs. None of manufacturer A’s bulbs lasted more than 9000 hours, but 27.5% of manufacturer B’s bulbs did. At the same time, 20% of manufacturer A’s bulbs lasted less than 6500 hours, while all of manufacturer B’s bulbs lasted at least 6500 hours.

(a)

Table frequencies for all student responses Student major categories

(b)

Gender

Totals

Male

Female

Totals

Table percentages based on overall student responses Student major categories Gender

Totals

Male

35.0%

22.5%

5.0%

62.5%

Female

15.0%

7.5%

37.5%

Totals

50.0%

37.5%

12.5%

100.0%

Table percentages based on row percentages Student major categories Gender

Totals

Male

56.0%

36.0%

8.0%

100.0%

Female

40.0%

20.0%

100.0%

Totals

50.0%

37.5%

12.5%

100.0%

Solutions to End-of-Section and Chapter Review Problems Table percentages based on column percentages Student major categories

(c)

Gender

Totals

Male

70.0%

60.0%

40.0%

62.5%

Female

30.0%

40.0%

60.0%

37.5%

Totals

100.0%

Side-by-side bar chart

Major

Frequency Female

2.21

Male

Side-by-side bar chart

754

Solutions to End-of-Section and Chapter Review Problems

2.22

(a)

755

Cross-classification tables Table based on column percentages Study mode Employment Status

Studying Full-time

Studying Part-time

All Students

15.3

64.9

44.1

Employed full-time

43.2

21.0

30.3

Not employed

41.5

14.1

25.6

All students

100.0

Studying Full-time

Studying Part-time

All Students

14.5

85.5

100.0

Employed full-time

59.7

40.3

100.0

Not employed

68.0

32.0

100.0

All students

41.9

58.1

100.0

Employed time

part-

Table based on row percentages Study mode Employment Status Employed time

2.23

part-

(b)

Side-by-side bar charts

(c)

The majority of the full-time students are employed full-time while the majority of part-time students are employed part-time.

(a)

Side-by-side bar chart

Solutions to End-of-Section and Chapter Review Problems

756

Australian Road Fatalities 2012 to 2016 90 or more 80 to < 90 70 to < 80 60 to < 70 50 to < 60 40 to < 50 30 to < 40 20 to < 30 10 to < 20 < 10 0

100

200

300

400

500

600

Female

Male

700

900 1000

800

Number

(b)

In all age groups there are more male fatalities than female. While female fatalities in most age groups are approximately 200 male fatalities vary from under 100 to approximately 1000. Most male fatalities are in the 20 to 60 age group. Both male and female fatalities peak in the 20 to 30 age group. 2.24

(a)

Cross-classification table

Highest Level of Educational Attainment Below Year 10 Year 10 or equivalent Year 11 or equivalent Year 12 or equivalent Post Secondary Below Bachelor Degree Bachelor Degree or Higher Total

Male ('000) Female ('000) 8.7% 9.1% 11.9% 14.1% 3.7% 3.2% 17.9% 18.2% 30.6% 24.7% 27.3% 30.7% 100.0% 100.0%

Solutions to End-of-Section and Chapter Review Problems (b)

757

Side-by-side bar chart

Education Level and Gender Bachelor Degree or Higher Post Secondary Below Bachelor Degree

Year 12 or equivalent Year 11 or equivalent Year 10 or equivalent Below Year 10

Female ('000)

(c)

10%

15%

20%

25%

30%

Male ('000)

There is no discernible difference in the education level pattern for males and females. Approximately 55% of both genders have post-school qualifications. However, slightly more females have a bachelor’s degree or higher than males.

35%

Solutions to End-of-Section and Chapter Review Problems 2.25

(a)

758

Side-by-side bar chart

New Zealand Sales New Passenger Cars OTHER VOLVO VOLKSWAGEN TOYOTA TESLA SUZUKI SUBARU SSANGYONG SKODA RENAULT PORSCHE PEUGEOT NISSAN MITSUBISHI MINI MERCEDES-BENZ MAZDA MASERATI LEXUS LAND ROVER KIA JEEP JAGUAR HYUNDAI HONDA HOLDEN FORD DODGE CITROEN BMW AUDI 0

100 200 300 400 500 600 700 800 900 1000

February 2016

February 2017

(b) Sales were similar in February 2016 and 2017, with a slight increase in sales in February 2017 for some makes. Toyota has the largest market share for both periods.

Solutions to End-of-Section and Chapter Review Problems 2.26

(a)

759

Scatter diagram

Scatter Diagram 60

0 0

2.27

(b)

Yes, there appears to be a positive linear relationship between X and Y since both X and Y change in the same direction.

(a)

Time-series plot

Sales (in Millions of constant 2010$)

Time Series Plot - Sales 25 20

15 10 5 0 2007

2008

2009

2010

2011

2012

2013

2014

2015

2016

2017

Year (b)

Annual sales appear to be increasing before 2011 but start to decline after 2013.

Solutions to End-of-Section and Chapter Review Problems

2.28

(b)

(a)

760

Scatter diagram

There appears to be a weak linear relationship between overall rating and price. This relationship is positive, with restaurants with a higher overall rating generally having higher prices.

Solutions to End-of-Section and Chapter Review Problems 2.29

761

(a) Scatter diagrams

(b)

As expected, in general the older a car is or the more kilometres travelled, the cheaper the car is. There is a strong negative exponential relationship between age and price, while there is a weak, possibly linear, relationship between kilometres travelled and price.

Solutions to End-of-Section and Chapter Review Problems 2.30

(a)

762

Scatter diagram

New South Wales Fuel Prices Diesel Cents per Litre

136 134

132 130 128 126 124 122 120 110

115

120

125

130

135

140

Unleaded Petrol Cents per Litre (b)

There is a weak positive linear relationship between petrol and diesel prices; where petrol prices are high, diesel prices tend to be high. Diesel prices are generally higher but less varied than petrol prices. 2.31

(a)

Time-series plot

Seasonally Adjusted Australian Unemployment Rate 7 6 5

4 3 2

1 0 Mar 07 Apr 08 May 09 Jun 10

(b)

Jul 11 Sep 12 Oct 13 Nov 14 Dec 15 Feb 17

The (seasonally adjusted) unemployment rate increased from a minimum of 4% in 2007 and early 2008 to a maximum just below 6% in May 2009. The unemployment rate then decreased to approximately 5% during 2009 and 2010, then remained at just above 5% until the second half of 2012 when it steadily increased to just above 6% in 2014 and 2015. During 2016 and early 2017 the unemployment rate was slightly below 6%.

Solutions to End-of-Section and Chapter Review Problems 2.32

(a)

763

Time-series plots

Inflation Rate

6 5

3 2

1 0

Oct-16 Jul-16 Apr-16 Jan-16 Sep-15 Jun-15 Mar-15 Dec-14 Sep-14 Jun-14 Mar-14 Dec-13 Sep-13 Jun-13 Mar-13 Dec-12 Sep-12 Jun-12 Mar-12 Dec-11 Sep-11 Jun-11 Mar-11

Australia (b)

2.33(a)

New Zealand

The Australian and New Zealand inflation rates show similar variation for the six years, from a high of 3.5% and 5.3% respectively for the year ending June 2011 and a low 1.0% and 0.4% respectively for the year ending June 2016. From 2012 the inflation rate for New Zealand rate has been below 2% and less than the Australian rate which has been between 1 and 3%.

Excel output:

(b)

Solutions to End-of-Section and Chapter Review Problems

764

Bullet Graph Toyota Motor Sales Nissan Motors Corporation General Motors Ford Motor Company Chrysler LLC American Honda 0

100

200

300

400

500

600

Number of Complaints

(c)

The bullet graph is more effective at comparing the number of complaints for each automaker because it fosters the direct comparison of each measurement.

Solutions to End-of-Section and Chapter Review Problems 2.34(a)

765

Bullet graph for one-year return:

One-Year Return Long-term Long-term Long-term Long-term Long-term 1 Short-term Short-term Short-term Short-term Short-term 0.00

5.00

10.00

15.00

20.00

15.0

20.0

Bullet graph for three-year return:

Three-Year Return Long-term Long-term Long-term Long-term Long-term 1 Short-term Short-term Short-term Short-term Short-term 0.0

5.0

10.0

(b) (c)

Gauges take up too much visual space that would be needed for the 20 funds. The one-year return and the three-year return is much higher for the longterm funds than for the short-term funds.

Solutions to End-of-Section and Chapter Review Problems 2.35 (a)

Bullet graph of the P/B ratios:

P/B Ratio BAXTER INTERNATIONAL INC KINETIC CONCEPTS INC EMERGENT BIOSOLUTIONS INC SUPERGEN INC NEPTUNE TECH & BIORESSOURCES GEN-PROBE INC TRINITY BIOTECH PLC -ADR MERCK & CO NOVARTIS AG -ADR BRISTOL-MYERS SQUIBB CO MYLAN INC HOSPIRA INC MEDICIS PHARMACEUT CP -CL A PAR PHARMACEUTICAL COS INC QUESTCOR PHARMACEUTICALS INC RIGEL PHARMACEUTICALS INC LANNETT CO INC NATURAL ALTERNATIVES 0

B) Gauges take up too much visual space that would be needed for the 71 P/B ratios. C) The three groupings of P/B ratios are helpful in analyzing the data. For the purposes of information presentation, you would redefine or subdivide the current acceptable category if you feel that the 0 to 2, 2 to 5 and above 5 delineation does not service your purpose of subjective analysis.

766

Solutions to End-of-Section and Chapter Review Problems 2.36

741

(a)

Cost($) San Diego Philadelphia Miami Colorado Atlanta Texas Oakland Los Angeles Angels Cleveland Baltimore 0

(b) (c) (d)

2.37

100

150

200

250

300

350

The bullet graph shows that four teams have costs considered as expensive while the stem-and-leaf display shows that the costs are concentrated between $172 and $225. The bullet graph enables you to see the names of the individual teams and which teams are inexpensive, typical, and expensive whereas the stem-andleaf display only shows the distribution of the costs. The bullet graphs shows the costs for each team and which teams fall into the inexpensive, typical, and expensive categories.

(a)

Sparkline graph for the movie attendance between 2002 and 2012:

(b)

There was a slight decline in movie attendance between 2001 and 2012. During that time, movie attendance increased from 2001 to 2002 but then after 2004 began decreasing to levels below that in 2001.

(c)

You would use the sparkline graph if you want to summarize time-series data as small, compact graphs designed to appear as part of a table (or a written passage). Otherwise, you would use a time-series plot for the larger size makes it easier to discern the detail pattern. Yes, you might use both a sparkline graph and a time-series plot in the same analysis report for the reason described in (c).

(d)

2.38 (a) Stock Indexes

2006

2007

2008

2009

2010

2011

2012

DJIA

16.3

6.4

-33.8

18.8

11.0

5.5

7.3

S&P 500

13.6

3.5

-38.5

23.5

12.8

0.0

13.4

NASDAQ 9.5 9.8 -40.5 43.9 16.9 -1.8 15.9 (b) The rates of return of the three indices vary a great deal from year to year, but the pattern for the three indices are similar except for the NASDAQ in 2009 which had a much higher return in that year than the DJIA or the S&P500. (c) Unlike the three stock indices which had similar patterns between 2006–2012, Copyright © 2019 Pearson Australia (a division of Pearson Australia Group Pty Ltd) 9781488617249, Berenson, Basic Business Statistics 5e .

Solutions to End-of-Section and Chapter Review Problems the returns of the three metals differed greatly from year to year.

741

Solutions to End-of-Section and Chapter Review Problems

741

2.39 (a) Metal

2006

2007

2008

2009

2010

2011

2012

Platinum

15.9

36.9

-41.3

55.9

21.5

-21.1

8.7

Gold

23.2

31.9

4.3

23.9

29.8

10.2

0.1

Silver

(b) (c)

2.40

46.1 14.4 -26.9 49.3 83.7 -9.8 The rates of return of the three indices vary a great deal from year to year, and the pattern for the three indices are quite different. Unlike the three stock indices which had similar patterns between 2006– 2012. the returns of the three metals differed greatly from year to year.

7.1

(a) Year

Fast Food Chain

Burger King

178.45

171.33

178.20

162.22

173.37

160.52

Chick-Fil-A

219.39

198.81

194.80

167.21

150.57

146.38

McDonald's

177.59

167.02

169.88

170.85

162.72

156.92

Wendy's

171.30

150.29

141.73

134.67

127.21

116.22

Fast Food Chain

Year

Burger King

Chick-Fil-A

173.19

166.10

166.00

179.90

153.06

166.85

McDonald's

163.74

168.60

191.90

194.30

175.01

167.59

Wendy's

152.52

167.90

163.90

167.10

158.77

174.22

(b)

124.69 135.70 135.10 138.50 131.08 134.09 Wendy’s consistently has the fastest service time. The service time at McDonald’s and Burger King are similar over the years. The service time at Chick-Fil-A was slower in the earlier and later years than the other fast food chains.

Solutions to End-of-Section and Chapter Review Problems 2.41

(b)

(a)

General Motors has the highest sales at 234,071 while Subaru has the lowest sales at 35,994. The change in sales is the highest for Subaru at 42.9% and lowest for Volkswagen at 0.3%. The change in sales for General Motors is 16.3%. 2.42

(a)

742

Solutions to End-of-Section and Chapter Review Problems 2.42 (b) The values of the teams varied from $312 million for the Milwaukee Bucks to $1,100 million for the New York Knicks. The change in values was not consistent across the teams. The two most valuable teams, the Los Angeles Lakers, and the New York Knicks had very different increases in value (11% and 41% respectively.)

2.43

(a)

(b)

The “financial services” sector seems to have the best gains while the “automobiles & parts” and “software & computer services” sectors seem to have the worst gain.

743

Solutions to End-of-Section and Chapter Review Problems 2.43(c)

(d)

The treemap constructed in (c) allows easier recognition of the following: (i) the variation in market capitalization among companies in a particular country and (ii) in which countries are the larger market cap companies found; (iii) which country seems to have the most/more of the largest market cap companies. The treemap constructed in that in (a) allows easier recognition of the following: (i) the variation in market capitalization among companies in a particular country and a particular sector and (ii) in which countries and sectors are the larger market cap companies found; (iii) which countries and sectors seem to have the most/more of the largest market cap companies.

2.44 a)

744

Solutions to End-of-Section and Chapter Review Problems

(b)

(c)

Almost all the countries that had lower GDP had lower Internet use except for the Republic of Korea. The pattern of mobile cellular subscriptions does not seem to depend on the GDP of the country.

745

Solutions to End-of-Section and Chapter Review Problems 2.45 (a) Count of Market Cap Row Labels Growth

(b)

Mean Median Mode Minimum Maximum Range Variance Standard Deviation Coeff. of Variation Skewness Kurtosis Count Standard Error

Column Labels Average

Hig h 74

746

Lo Grand w Total

Large

10 14 3 3 90

227

Mid-Cap

Small Value

29 17

7 16 3 69

52 89

Large

Mid-Cap Small

4 9

15 9

19 20

Grand Total

13 21 2

316

103

There are 3 of such funds. PHStat output of the summary statistics of the variables: Assets Turno Beta S 1YrRet 3YrRet 5YrRetu 10YrRet Expense ver D urn% urn% rn% urn% Ratio Ratio 94.43333333 50.6666 2.2233333 36.346666 6667 33 67 5 #N/A 0 147 147 6966.33 33 83.4646

2.24 2.24 2.19 2.24 0.05 0.0008

36.63 #N/A 35.72 36.69 0.97 0.2954

33.72 22.173333 33 0.59666666 7 33.78 21.91 1.53 #N/A #N/A #N/A 33.4 21.89 -4.89 33.98 22.72 1.57 0.58 0.83 6.46 0.0868 0.2242 13.8249

0.0289

0.5435

0.2946

0.4735

3.7182

7.2750

0.1044

66.36% 164.73%

1.30%

1.50%

0.87%

2.14%

-623.16%

86.40%

5.64%

-1.7321 #DIV/0!

-1.7083 #DIV/0!

-0.8784 #DIV/0!

1.7286 #DIV/0!

-1.7318 #DIV/0!

-1.7313 #DIV/0!

-1.6608 #DIV/0!

3 0.0167

3 0.3138

3 0.1701

3 0.2734

3 2.1467

3 4.2002

3 0.0603

118.5 #N/A 23.3 141.5 118.2 3927.2133 62.6675

-1.4733 #DIV/0!

1.7251 #DIV/ 0! 3 3 36.1811 48.1883

8.42

1.85

12.55 #N/A 0.02 12.69 12.67 52.9249

1.9 #N/A 1.73 1.92 0.19 0.0109

Solutions to End-of-Section and Chapter Review Problems 2.46

747

(a) Count of Market Cap Column Labels Row Labels Five Four One Three Two Grand Total Growth 18 76 16 74 43 227

(b)

Large

103

Mid-Cap Small Value

7 2 5

28 17 22

4 7 7

20 17 36

13 9 19

72 52 89

Large

Mid-Cap Small

1 2

4 5

9 6

5 5

19 20

110

316

Grand Total There are 37 of such funds.

PHStat output of the summary statistics of the variables: Assets Turnover Ratio Mean Median Mode Minimum Maximum Range Variance Standard Deviation Coeff. of Variation Skewness Kurtosis Count Standard Error

649.048918 9 123.8 #N/A 0.1 6190 6189.9 1650071.13 35 1284.5509

Beta

S D

1YrRetur 3YrR 5YrR 10YrRet Expense n% eturn etur urn% Ratio % n%

57.6627027 1.0710810 16.97513 14.8743243 9.26324 0.8029 6.17864864 1.41837837 81 514 2 3243 72973 9 8 38 1.1 17.32 14.65 9.25 0.84 6.4 1.21 13 1.1 17.93 11.8 9.46 1.22 6.81 1.07 0 0.85 13.29 9.16 6.62 -3.27 0.9 0.58 413 1.25 19.9 20.85 12.59 3.36 9.69 6.97 413 0.4 6.61 11.69 5.97 6.63 8.79 6.39 5340.2103 0.0069 2.1347 7.3247 1.5361 1.7426 2.8816 1.0450 73.0767

0.0833

1.4610

2.7064 1.2394 1.3201

1.6975

1.0223

197.91%

126.73%

7.77%

8.61%

27.47%

72.07%

3.0908

3.4983

-1.0282

-0.7817

-1.1914

4.6897

10.4021 37 211.1789

15.5456 37 12.0137

1.1615 37 0.0137

0.5130 37 0.2402

18.20% 13.38% 164.40 % 0.1305 0.2500 0.8352 -0.3423 0.2923 1.8633 37 37 37 0.4449 0.2038 0.2170

2.6974 37 0.2791

25.3885 37 0.1681

Solutions to End-of-Section and Chapter Review Problems

2.47 (a) Count of Market Cap Column Labels Row Labels Five Four One Three Two Grand Total Large 11 44 10 58 30 153 Average

54 29

3 23 18

4 135 91

High Low Mid-Cap

10 8

42 32

1 6 4

Average

Low Small

6 4

25 22

1 9

16 23

4 14

52 72

Average

1 15

1 4

1 3

9 25

110

316

High Low (b)

Grand Total There is none of such fund.

748

Solutions to End-of-Section and Chapter Review Problems 2.48

749

(a) Count of Market Cap Column Labels Row Labels Five Four One Three Two Grand Total Growth 18 76 16 74 43 227

(b) 2.49

Average

High Low Value

15 5

1 60 22

5 5 7

1 45 36

3 18 19

10 143 89

Average

High Low

2 2

1 12

3 69

110

316

Grand Total There is only one such fund.

a) The fund with the highest five-year return is an average risk, large-cap growth fund with a five-star rating.

b) The five-year returns for the four funds that are small market cap funds that have a rating of five stars are 5.29, 6.97, 10.75, and 11.35.

Solutions to End-of-Section and Chapter Review Problems c) The fund with the lowest five-year return is an average risk, mid-cap growth fund with a one-star rating.

The highest five-year return of 12.33 is for a large cap growth fund.

2.50-2.53 Answers will vary

750

Solutions to End-of-Section and Chapter Review Problems

751

2.54

1600 1400 1200 1000 800 600 400 200 0 1969-70 1971-72 1973-74 1975-76 1977-78 1979-80 1981-82 1983-84 1985-86 1987-88 1989-90 1991-92 1993-94 1995-96 1997-98 1999-00 2001-02 2003-04 2005-06 2007-08 2009-10 2011-12 2013-14

Million Litres

Beverage Wine Production (million litres)

Financial Year

Litres per Capita (15+)

New Zealand Alcohol Consumption 10 9 8 7 6 5 4 3 2 1 0 2003

2004

2005

2006

2007

2008

2009

2010

2011

2012

2013

Year

2.55

Graph (a) probably represents the data best as the vertical scale starts at zero. However, as petrol prices usually differ by only a few cents, graph (b) may be useful in some applications if you wish to show variability of the prices. In this case, you would need to clearly indicate that the vertical axis does not start at zero. Graph (c) shows a very flat graph, as the vertical scale extends to 300; this is misleading, as it does not represent the variation in prices.

2.56

Answers will vary.

Solutions to End-of-Section and Chapter Review Problems 2.57

(a)

752

Doughnut chart, exploded pie chart, cone chart and pyramid chart

Doughnut Chart 10%

Comfortable environment

Competitive prices

20%

18%

Convenience Customer service

3% 13%

Variety/range of products 10%

Products well displayed Quality products

28%

Exploded Pie Chart

Variety/range of products

Comfortable environment 8%

Quality products 18%

Competitive prices 20%

Products well displayed 3% Customer service 13%

Convenience 28%

Solutions to End-of-Section and Chapter Review Problems

753

3D Pyramid Chart 28

18 20

Comfortable environment 10

Competitive prices Convenience

3 8

Customer service Products well displayed Quality products Variety/range of products

Percentage

Cone Chart

(b)

30 25 20 15 10 5 0

The bar chart and the pie chart should be preferred over the doughnut chart, the cone chart, the pyramid chart or the exploded pie chart, since the bar chart and pie chart are simpler and easier to interpret.

2.58 A histogram uses bars to represent each class while a polygon uses a single point. The histogram should be used for only one group, while several polygons can be plotted on a single graph. 2.59 A summary table allows one to determine the frequency or percentage of occurrences in each category.

Solutions to End-of-Section and Chapter Review Problems

754

2.60 A bar chart is useful for comparing categories. A pie chart is useful when examining the portion of the whole that is in each category. 2.61 The bar chart is plotted with the categories on the vertical axis and the frequencies or percentages on the horizontal axis. In addition, there is a separation between categories. The histogram is plotted with the class grouping on the horizontal axis and the frequencies or percentages on the vertical axis. This allows one to more easily determine the distribution of the data. There are no gaps between the classes in the histogram. 2.62

A time-series plot is a type of scatter diagram with the time on the x axis.

2.63

Percentage breakdowns according to the total percentage, the raw percentage and/or the column percentage allow the interpretation of data in a two-way contingency table from several different perspectives.

Solutions to End-of-Section and Chapter Review Problems 2.64

755

(a)

Website Types Visited Previous Week Weather User generated or up-load site TV Sport Social Network Shopping Search Engine Online music site News Gaming Email Dating Classifieds Banking Auction

100 200 300 400 500 600 700 800 Number visited out of 1000 Australians

Website Types Visited Previous Week Weather User generated or up-load site TV Sport Social Network Shopping Search Engine Online music site News Gaming Email Dating Classifieds Banking Auction 0% 10% 20% 30% 40% 50% 60% 70% 80%

(b)

Over 50% of Australian Internet users polled used the internet for searching, social networking and email, while more than 40% also used a user-generated or upload site such as YouTube in the previous week.

Solutions to End-of-Section and Chapter Review Problems 2.65

756

(a)

Time Spent Online Email & Communication 19% Social Networking 22% Searches 21%

Reading Content 20%

Multi-media Sites 13%

Online Shopping 5%

Time Spent Online Social Networking Searches Reading Content Online Shopping Multi-media Sites Email & Communication 0% (b)

10%

15%

20%

25%

Internet usage by time spent is evenly spread between social networking, Internet searches and email and communications. This compares with the data from problem 2.47, which shows that the majority of Australian Internet users polled had visited search, social networking and email sites during the previous week.

Solutions to End-of-Section and Chapter Review Problems 2.66

757

(a) Contingency table of year and crash type as percentage of year total

Crash Type Multiple vehicle Pedestrian Single vehicle Total

2012 44.1% 13.2% 42.8% 100.0%

2013 40.4% 13.3% 46.3% 100.0%

2014 43.7% 13.4% 42.9% 100.0%

2015 42.4% 13.4% 44.1% 100.0%

2016 42.8% 13.2% 44.1% 100.0%

The following are some of the appropriate side-by-side bar charts. Pie charts may be appropriate if reviewing one specific year or type of fatality.

Australian Road Fatalies - Crash Type 2016

2015

Single vehicle

2014

Pedestrian Multiple vehicle

2013 2012 0

100

200

300 400 Number

500

600

Australian Road Fatalies - Year

Crash Type

Single vehicle 2016 2015

Pedestrian

2014 2013

Multiple vehicle

2012 0

100

200

300 400 Number

500

600

Solutions to End-of-Section and Chapter Review Problems

758

Australian Road Fatalities - Crash Type 2016 2015

Single vehicle Pedestrian

2014

Multiple vehicle

2013 2012 0%

10%

15%

20%

25%

30%

35%

40%

45%

50%

Australian Road Fatalities - Year 2016 2015

Single vehicle

Crash Type

2014

2013

Pedestrian

2012

Multiple vehicle

0% 5% 10% 15% 20% 25% 30% 35% 40% 45% 50%

(b)

From the above graphs we can conclude that for all years between 40% and 45% of road fatalities involve multiple vehicles while approximately 13% are pedestrian fatalities and between 40 and 50% involve a single vehicle. We can also conclude that the number and percentage of fatalities in each classification are variable with no apparent increase or decrease.

Solutions to End-of-Section and Chapter Review Problems

2.67

(a)

759

Bar chart

Number of Drink-Driving Offences 500 450

400 350 300 250

200 150 100 50

0 Local: Seaside Town

Local: Not Seaside Town

Not Local: Intrastate

Not Local: Interstate

Not Local: International

Pie chart

NUMBER OF DRINK-DRIVING OFFENCES Not Local: International, 22, 2%

Local: Seaside Town, 151, 15%

Not Local: Interstate, 228, 23%

Not Local: Intrastate, 130, 13%

Local: Not Seaside Town, 462, 47%

(b)

The majority (62%) of drink driving offences are committed by residents of the local council area, either resident in the seaside town (15%) or elsewhere within the council area (47%).

(c)

The headline is correct, as only 38% of drink-driving offences are committed by residents from outside the council area.

Solutions to End-of-Section and Chapter Review Problems 2.68

760

(a)

Reason To save water To save on water costs Water restrictions on mains water Not connected to mains water Concerns about quality of mains water Water tank rebates Other

Brisbane

Rest of Queensland

65.6% 25.7% 25.5% 2.5% 2.5% 19.9% 22.4%

28.6% 20.2% 15.5% 35.7% 13.6% 3.8% 22.9%

Reason Why Rainwater Tank Installed Other

Water tank rebates Concerns about quality of mains water

Not connected to mains water

Rest of Queensland Brisbane

Water restrictions on mains water To save on water costs

To save water 0%

(b)

10%

20%

30%

40%

50%

60%

70%

There are differences between Brisbane and non-Brisbane households in the reasons for installing a rainwater tank. Brisbane households installed a rainwater tank mainly to save water, but also to save on water costs, to obtain a rebate and because of water restrictions. While non-Brisbane households also installed a rainwater tank for these reasons, additionally they were concerned about water quality, with the main reason being they were not connected to mains water.

Solutions to End-of-Section and Chapter Review Problems 2.69

(a)

Fresh Milk 25

Sugar, grams

20 15 10 5 0 0

Fat, grams

Fresh Milk 200

Calories

150 100 50 0

8 Fat, grams

Fresh Milk 200

Calories

150

100 50 0 0

Sugar, grams

761

Solutions to End-of-Section and Chapter Review Problems (b)

2.70

(a)

762

There appears to be a strong linear relationship between fat content and calories, with milks with higher fat content having more calories than those with low fat content. There may be a weak negative relationship between fat and sugar content, with a high level of fat tending to have lower sugar content and vice-versa. There is no relationship between sugar content and calories. Other graphs and tables may also be appropriate.

Queensland Unleaded Petrol Price Midpoint 115 to < 120 117.5 120 to < 125 122.5 125 to < 130 127.5 130 to < 135 132.5 A35 to < 140 137.5 140 to < 145 142.5 145 to < 150 147.5 150 to < 155 152.5 Total

Frequency 0 1 10 14 11 8 1 0 45

Percentage 0.00% 2.22% 22.22% 31.11% 24.44% 17.78% 2.22% 0.00% 100.00%

Cumulative Frequency 0 1 11 25 36 44 45 45

Cumulative Percentage 0.00% 2.22% 24.44% 55.56% 80.00% 97.78% 100.00% 100.00%

Cumulative Frequency 0 3 10 21 29 37 42 42 44 45 45

Cumulative Percentage 0.00% 6.67% 22.22% 46.67% 64.44% 82.22% 93.33% 93.33% 97.78% 100.00% 100.00%

Queensland Diesel Price 120 to < 122 122 to < 124 124 to < 126 126 to < 128 128 to < 130 130 to < 132 132 to < 134 134 to < 136 136 to < 138 138 to < 140 140 to < 142 Totals

Midpoint 121 123 125 127 129 131 133 135 137 139 141

Frequency 0 3 7 11 8 8 5 0 2 1 0

Percentage 0.00% 6.67% 15.56% 24.44% 17.78% 17.78% 11.11% 0.00% 4.44% 2.22% 0.00%

100.00%

Solutions to End-of-Section and Chapter Review Problems

Queensland Unleaded Petrol 16 14

Frequency

12 10 8 6 4 2

0 117.5

122.5

127.5

132.5

137.5

142.5

147.5

152.5

Price Cents per Litre

Queensland Diesel 12

Frequency

10 8 6 4 2 0 121

123

125

127

129

131

133

135

Price Cents per Litre

137

139

141

763

Solutions to End-of-Section and Chapter Review Problems

764

Queensland Fuel Prices 45% 40%

35% 30% 25%

ULP Diesel

20% 15%

10% 5%

118

122

126

130

134 Cents per Litre

138

142

146

150

The distribution of both diesel and unleaded petrol prices are mound shaped. We can conclude that on this day in March 2017 petrol prices in Queensland varied from 120 to 148 cents per litre while diesel prices varied from 122 to 140 cents per litre. Diesel prices concentrated between 124 and 136 cents per litre are generally less than petrol prices which are concentrated between 128 and144 cents per litre. (b)

Queensland Fuel Prices Diesel Cents per Litre

142 138 134 130 126 122 122

126

130

134

138

142

146

150

Unleaded Petrol Cents per Litre There is a very weak positive relationship between Queensland petrol and diesel prices, where petrol prices are high diesel prices tend to also be high. Diesel prices are generally lower than petrol prices

Solutions to End-of-Section and Chapter Review Problems (c)

765

Other graphs and tables may also be appropriate

New South Wales Unleaded Petrol Price 105 to < 110 110 to < 115 115 to < 120 120 to < 125 125 to < 130 130 to < 135 135 to < 140 140 to < 145 Totals

Midpoint 107.5 112.5 117.5 122.5 127.5 132.5 137.5 142.5

Frequency 0 13 8 4 3 11 6 0 45

Percentage 0.00% 28.89% 17.78% 8.89% 6.67% 24.44% 13.33% 0.00% 100.00%

Cumulative Frequency 0 13 21 25 28 39 45 45

Cumulative Percentage 0.00% 28.89% 46.67% 55.56% 62.22% 86.67% 100.00% 100.00%

Cumulative Frequency 0 6 7 9 20 31 40 44 45 45

Cumulative Percentage

New South Wales Diesel Price 118 to < 120 120 to < 122 122 to < 124 124 to < 126 126 to < 128 128 to < 130 130 to < 132 132 to < 134 134 to < 136 136 to < 138 Totals

Midpoint 119 121 123 125 127 129 131 133 135 137

Frequency

Percentage

0.00%

13.33%

2.22%

4.44%

24.44%

20.00%

8.89%

2.22%

0.00%

100.00%

0.00% 13.33% 15.56% 20.00% 44.44% 68.89% 88.89% 97.78% 100.00% 100.00%

Solutions to End-of-Section and Chapter Review Problems

New South Wales Unleaded Petrol Prices 14

Frequency

12 10 8 6

4 2 0

107.5

112.5

117.5

122.5

127.5

132.5

137.5

142.5

135

137

Cents per Litre

New South Wales Diesel Prices 12

Frequency

8 6 4 2 0 119

121

123

125

127

129

131

133

Price per Litre

New South Wales Fuel Prices

50%

45% 40% 35% 30% 25%

20%

ULP

Diesel

140

144

15% 10% 5% 0%

108

112

116

120

124 128 Cents per Litre

132

136

766

Solutions to End-of-Section and Chapter Review Problems

767

New South Wales Petrol Prices

60% 55% 50% 45% 40% 35% 30% 25% 20% 15% 10% 5% 0%

108

112

116

120

124

128

132

Regional

Capital

140

144

136

Cents per Litre

New South Wales Diesel Prices 60% 55% 50% 45% 40% 35% 30% 25% 20% 15% 10% 5% 0%

Regional

116

120

124

128

132

136

Capital

140

Cents per Litre

The distribution of both diesel and unleaded petrol prices are bimodal. The difference between capital city prices and regional prices which are more expensive. We can conclude that on this day in March 2017 petrol prices in New South Wales varied from 110 to 140 cents per litre while diesel prices varied from 120 to 136 cents per litre. Diesel prices concentrated between 126 and 134 cents per litre are generally more than petrol prices which are more varied.

Solutions to End-of-Section and Chapter Review Problems

768

(d) Unleaded Petrol 35% 30% 25% 20% 15%

NSW

QLD

10%

5% 0%

107.5

112.5

117.5

122.5

127.5

132.5

137.5

142.5

147.5

152.5

Cents per Litre

The distribution of unleaded petrol prices differs between New South Wales and Queensland. With Queensland prices higher than New South Wales prices (e) Diesel Prices 30% 25% 20%

15%

NSW

QLD

10% 5%

0% 119

121

123

125

127

129

131

133

135

137

139

141

Cents per Litre

Diesel prices in Queensland have a similar distribution to those in New South Wales. (f)

The answer to this part will vary.

Solutions to End-of-Section and Chapter Review Problems 2.71

(a)

769

Stem-and-leaf display Property Sales - Asking Price Stem unit:

$100,000 3 2233344456666678889999 4 0000011111112223333444555555666667788889999 5 00112222233344455678999 6 0012223688 7 8 44

Property prices range from $320,000 to $840,000. Asking prices are concentrated between $300,000 and $600,000. Only two asking prices are above $800,000. (b)

Frequency percentage and cumulative distributions

Asking Price, $ 300000 to < 350000 350000 to < 400000 400000 to < 450000 450000 to < 500000 500000 to < 550000 550000 to < 600000 600000 to < 650000 650000 to < 700000 700000 to < 750000 750000 to < 800000 800000 to < 850000 Total

Frequency 8 17 21 20 16 6 7 3 0 0 2 100

Percent 8.0% 17.0% 21.0% 20.0% 16.0% 6.0% 7.0% 3.0% 0.0% 0.0% 2.0% 100.0%

Cumulative Frequency 8 25 46 66 82 88 95 98 98 98 100

Cumulative Percent 8.0% 25.0% 46.0% 66.0% 82.0% 88.0% 95.0% 98.0% 98.0% 98.0% 100.0%

Solutions to End-of-Section and Chapter Review Problems (c)

Frequency histogram, percentage polygon and ogive

Frequency

Property Sales

25 20 15 10 5 0

Asking Price

Property Sales 100% 80% 60% 40% 20% 0%

Asking Price

770

Solutions to End-of-Section and Chapter Review Problems (d)

771

Can conclude that:

(e)

•

Property prices in the region range from approximately $300,00 to $850,000.

•

The asking prices are concentrated between $400,000 and $500,000.

•

Only two prices are above $700,000.

•

66% of asking prices are below $500,000.

•

74% of prices are between $350,000 and $550,000.

Scatter diagram

Selling Price

Asking and Selling Price

$900,000 $800,000 $700,000 $600,000 $500,000 $400,000 $300,000 $200,000 $100,000 $0 $300,000 $400,000 $500,000 $600,000 $700,000 $800,000 $900,000

Asking Price

There is a strong positive linear relationship between selling price and asking price.

Solutions to End-of-Section and Chapter Review Problems (f)

(g)

772

Cross-tabulation based on totals

Percent of Total: - Type House Unit Total

1 2.0% 4.0% 6.0%

2 7.0% 12.0% 19.0%

- Bedrooms 3 4 43.0% 24.0% 2.0% 0.0% 45.0% 24.0%

>4 6.0% 0.0% 6.0%

Total 82.0% 18.0% 100.0%

Percent of Rows: - Type House Unit Total

1 2.4% 22.2% 6.0%

2 8.5% 66.7% 19.0%

- Bedrooms 3 4 52.4% 29.3% 11.1% 0.0% 45.0% 24.0%

>4 7.3% 0.0% 7.3%

Total 100.0% 100.0% 100.0%

Percent of Columns: - Type House Unit Total

1 33.3% 66.7% 100.0%

2 36.8% 63.2% 100.0%

- Bedrooms 3 4 95.6% 100.0% 4.4% 0.0% 100.0% 100.0%

>4 100.0% 0.0% 100.0%

Total 82.0% 18.0% 100.0%

Side-by-side charts

Number of Bedrooms

Type and Number of Bedrooms >4 4 Unit House

3 2 1

(h)

10%

20%

30%

40%

50%

60%

70%

Units tend to have a maximum of three bedrooms while the number of bedrooms in a house varies from one to eight.

Solutions to End-of-Section and Chapter Review Problems

773

2.72

Feb-17

Jun-16

Jan-15

Sep-15

May-14

Jan-13

New Zealand

Apr-12

Aug-11

Apr-10

Dec-10

Aug-09

Nov-08

Mar-08

Jul-07

Feb-06

Nov-06

Jun-05

Oct-04

Feb-04

Jun-03

Jan-02

Sep-02

May-01

Jan-00

Australia

Sep-13

12 11 10 9 8 7 6 5 4 3 2 1 0

Sep-00

Mortgage Interest Rates

From approximately 2000 until October 2009, New Zealand interest rates have been higher than Australian rates and were slightly less until May 2014 when New Zealand rates increased and Australian decreased. Both rates follow a similar pattern with some variation. Since 2000 interest rates in both countries have decreased from a high of 9 to 11% to lows between 5 and 6%. 2.73

This graph distorts the data in Problem 2.55. This graph emphasises the decrease in Australian interest rates by not starting the vertical axis at zero. Furthermore, starting the horizontal axis at September 2011 after which rates began to fall is misleading as the graph does not show the larger variations and lower rates of previous years. In particular, the rates below 6% in 2009.

2.74

(a)

A pie chart of grade distribution. Please note that a frequency or percentage bar chart would also be appropriate.

Grade Distribution

CREDIT

20% 44% 13%

DIST FAIL

16%

PASS

Solutions to End-of-Section and Chapter Review Problems

774

16% of students failed this statistics unit while 84% passed. Further 20% obtained either a credit or above. (b)

A histogram of total marks. Please note that a percentage histogram, polygon or ogive would also be appropriate.

Histogram Total Marks 18 16

Frequency

14 12 10 8 6 4 2 0 0

100

Total Mark out of 100

The majority of students obtained a pass mark of 50%; the most common mark is 50 to 60. A few students obtained a total mark of 10 to 50 and hence failed the unit. (c)

A scatter diagram of semester marks and final marks

Exam Mark out of 50

Scatter Diagram 50 45 40 35 30 25 20 15 10 5 0 0

Semester Marks out of 50

There is a positive liner relationship between a student’s semester mark and their exam mark. In general, students with high semester marks obtain higher exam marks.

Solutions to End-of-Section and Chapter Review Problems

775

2.75 and 2.76 Answers will vary. Case Study - Tasman University Student Survey Below is a sample of graphs and charts appropriate to this data. Other graphs and tables are also appropriate (a)

BBus student survey. Gender Tasman University Bachelor of BusinessPie Chart

Male 47%

Female 53%

Age Tasman University Bachelor of Business 25

Frequency

20 15 10 5 0 18

Age

Major Tasman Univeristy Bachelor of Business Undecided Retailing/Marketing Other Management

Major

International Business Information Systems Economics/Finance Accounting 0

8 Frequency

Solutions to End-of-Section and Chapter Review Problems Weighted Average Mark

Tasman University Bachelor of Business 20

Frequency

10 5 0 5

Weighted Average Mark (WAM) Employment Status Tasman University Bachelor of Business Employment Statius Full-Time 16%

Not Employed 15%

Part-Time 69%

Expected Salary

Bachelor of Business - Expected Salary 20

Frequency

15 10 5

0 35

Thousand Dollars

105

776

Solutions to End-of-Section and Chapter Review Problems Social Networking

Stem-and-Leaf Display Bachelor of Business Social Networking Stem unit: 1 0 000 1 000000000000000000000000000000000 2 0000000000000000000 3 00000 4 00 Satisfaction with On-campus Food Services

Stem-and-Leaf Display Bachelor of Business:Satisfaction on campus food services Seven point scale Stem unit: 1 1 00000 2 00 3 000000000000000 4 00000000000000000000000000 5 0000000000 6 0000 Textbook Cost

Bachelor of Business Textbook Cost 16 14

Frequency

12 10 8 6 4

2 0 50

150 250 350 450 550 650 750 850 950 1050 1150 1250 1350 1450 $

777

Solutions to End-of-Section and Chapter Review Problems Computer Bachelor of Business - Computer Tablet 3%

Desktop 8%

Laptop 89%

Number of Text Messages

Stem-and-Leaf Display Bachelor of Business - Number of Text Messages Stem unit: 100 0 01344555555567 1 0000000011455558 2 00000055 3 000000000055 4 00 5 000 6 00 7 005 8 0 9 0 Bachelor of Business 18 16

Frequency

14 12 10 8 6

4 2

0 50

150

250

350

450

550

650

750

850

Number of Text Messages

950

1050

778

Solutions to End-of-Section and Chapter Review Problems (b)

Gender

Tasman University MBA

Female 43% Male 57%

Age

Tasman University - MBA 25

Frequency

20 15 10 5 0 17.5

22.5

27.5

32.5

37.5

42.5

47.5

52.5

Age (years)

Major Tasman Universoty MBA Retailing/Marketing

MBA Major

Other Management International Business Information Systems Economics/Finance Accounting 0

8 Frequency

779

Solutions to End-of-Section and Chapter Review Problems Weighted Average Mark

Tasman University - MBA

Frequency

20 15

10 5 0 45

Weighted Average Mark

Employment Status

Number of Full-time Jobs

Stem-and-Leaf Display MBA - Number of Full-time Jobs Stem unit: 1 0 0000 1 000000000000000 2 00000000000000000 3 00000 4 000

780

Solutions to End-of-Section and Chapter Review Problems Textbook Cost

16 14 12 10 8 6 4 2 0

50 150 250 350 450 550 650 750 850 950 1050 1150 1250 1350 1450 1550 1650 1750 1850 1950 2050 2150 2250 2350

Frequency

Tasman University - MBA

Textbook Cost $

Advisory Services

Stem-and-Leaf Display Tasman University MBA - Advisory Rating Seven point scale Stem unit: 1 2 0 3 000 4 00000000000000000 5 0000000000000000 6 000000 7 0 Computer

781

Solutions to End-of-Section and Chapter Review Problems Number of Text Messages

Tasman University MBA 25

Frequency

15 10 5 0

150

250

350

450

550

650

750

850

950

1050

1150

1250

1350

Number of Text Message

(c)

Age and Gender Bachelor of Business Student Survey

Bachelor of Business Gender 18 19 Female 0 2 Male 1 3 Total 1 5

Age 20 9 5 14

21 13 9 22

22 5 6 11

23 3 2 5

24 0 3 3

26 1 0 1

Total 33 29 62

Tasman University Bachelor of Business Side-By-Side Chart 26

Male

Age Years

Female

22 21 20

19 18 0

Frequency

782

Solutions to End-of-Section and Chapter Review Problems

783

Age and Gender MBA Student Survey

Tasman University - MBA 60%

50% 40% Male

30%

Female

20%

10% 0% 17.5

22.5

27.5

32.5

37.5

42.5

47.5

52.5

Age (years)

Weighted Average Mark and Gender Bachelor of Business Student Survey Tasman University Bachelor of Business 40%

35% 30% 25%

20%

Female Male

15% 10% 5% 0% 5

45 55 65 Weighted Average Mark

Weighted Average Mark and Gender MBA Student Survey Tasman University MBA 35% 30% 25% 20%

Male Female

15% 10% 5% 0% 52.5

57.5

62.5

67.5

72.5

77.5

82.5

87.5

Weighted Average Mark

92.5

97.5

Solutions to End-of-Section and Chapter Review Problems

784

Weighted Average Mark and Employment Status Bachelor of Business Student Survey Tasman University Bachelor of Business 60%

Full-Time

50%

Part-Time Not Employed

40% 30% 20% 10% 0% 5

45 55 65 Weighted Average Mark (WAM)

105

Weighted Average Mark – BBus and MBA Tasman University 40%

BBus WAM MBA WAM

35% 30% 25% 20% 15% 10% 5% 0% 15

Weighted Average Mark (WAM)

Undergraduate and MBA WAM – MBA Student Survey

MBA Weighted Average Mark

Tasman University MBA 95 90

85 80 75

70 65 60

55 50 60

Undergraduate Weighted Average Mark

100

Solutions to End-of-Section and Chapter Review Problems

785

Textbook Cost – BBus and MBA Tasman University

35% 30% 25% 20%

BBus

MBA

15% 10% 5% 0%

50 150 250 350 450 550 650 750 850 950 1050 1150 1250 1350 1450 1550 1650 1750 1850 1950 2050 2150 2250 2350

Textbook Cost $

Number of Text Messages – BBus and MBA

Tasman University 50% 45% 40% 35% 30%

BBus

MBA

25% 20% 15% 10% 5%

0% 50

150

250

350

450

550

650

750

850

950

1050

1150

1250

1350

Number of Text Messages

MBA and Undergraduate Major - MBA

Tasman University MBA Major Accounting Economics/Finance Information Systems International Business Management Other Retailing/Marketing Total

Undergraduate Major Biological Sciences 0 1 0 0 0 1 1 3

Business 3 10 0 1 5 3 3 25

Engineering 0 1 0 0 2 0 0 3

Other 2 2 2 1 2 1 3 13

Total 5 14 2 2 9 5 7 44

Solutions to End-of-Section and Chapter Review Problems

786

Case Study - Safe-As Houses Real Estate Below is a sample of graphs and charts appropriate to this data. Other graphs and tables are also appropriate (a)(i) Regional City 1 State A Price

Residential Property Prices Regional City 1 State A 60

Frequency

40 30 20 10 0 50

150

250

350

450

550

650

750

Price $000 Internal Area

Frequency

Residental Properties: Regional City 1 State A 50 45 40 35 30 25 20 15 10 5 0

125

175

225

275

325

Internal Area Square Metres

375

Solutions to End-of-Section and Chapter Review Problems

787

Bedrooms 6 1%

Regional City 1 State A - Number of Bedrooms

9 1%

5 5%

1 1%

2 30%

4 23%

3 39%

Bathrooms Residental Properties Regional City 1 State A Number of Bathrooms

Frequency

Garages

Stem-and-Leaf Display Regional City 1 State A: Number of Garages Stem unit: 1 0 000000 1 00000000000000000000000000000000000000000000000000 2 00000000000000000000000000000000000000000000000000000000 3 00000000000 4 0 5 6 0

Solutions to End-of-Section and Chapter Review Problems Type Regional City 1 State A

Unit 33%

House 67%

(a)(ii) Regional City 1 State A Residental Properties Regional City 1 State A 45% 40% 35%

30% 25%

House Unit

20% 15% 10% 5% 0%

100

150

200

250

300

350

400

450

500

550

600

650

Price $000

Internal Area Residental Properties: Regional City 1 State A 80% 70% 60% 50% 40%

House Unit

30% 20% 10% 0% 37.5

62.5

87.5

112.5

137.5

162.5

187.5

212.5

237.5

262.5

287.5

312.5

Internal Area Square Metres

337.5

362.5

788

Solutions to End-of-Section and Chapter Review Problems Price and Internal Area

Residental Properties: Regional City 1 State A 700

Price $000

600 500 400 300

200 100

100

150

200

250

300

350

Internal Area Square Metres

Bedrooms and Bathrooms

Regional City 1 Bathrooms 1 2 3 Total

1 1 0 0 1

2 34 3 0 37

3 38 10 1 49

Bedrooms 4 8 20 1 29

5 1 5 1 7

6 0 1 0 1

9 0 0 1 1

Total 82 39 4 125

2 7 30 37

Bedrooms 3 4 39 29 10 0 49 29

5 7 0 7

6 1 0 1

9 1 0 1

Total 84 41 125

Bedrooms and Type

Regional City 1 Type House Unit Total

1 0 1 1

789

Solutions to End-of-Section and Chapter Review Problems (b)(i) Coastal City 1 State A Price

Residential Property Prices Coastal City 1 State A

Frequency

45 40 35 30 25 20 15 10 5 0 50 150 250 350 450 550 650 750 850 950 1050 1150 1250 1350 Price $000

Internal Area

Residental Properties: Coastal City 1 State A 50

Frequency

40 30 20 10 0 25

125

175

225

275

325

Internal Area Square Metres

Bedrooms Coastal City 1 State A Number of Bedrooms

7 6 5 4

3 2

1 0

Frequency

790

Solutions to End-of-Section and Chapter Review Problems

791

Bathrooms

Stem-and-Leaf Display Coastal City 1: Number of Bathrooms Stem unit: 1 1 00000000000000000000000000000000000000000000000000000 2 0000000000000000000000000000000000000000000000000000000 3 0000000000000 4 0000 Garages

Coastal City 1 Number of Garages 2 45%

3 7%

4 2% 8 1% 0 2%

1 43%

Type Residental Properties: Coastal City 1 State A

Unit 42% House 58%

Solutions to End-of-Section and Chapter Review Problems (b)(ii) Coastal City 1 State A Price Residental Property Prices - Coastal City 1 State A

40% 35% 30% 25% 20%

House

Unit

1150

1250

15% 10% 5% 0% 50

150

250

350

450

550

650

750

850

950

1050

1350

Price $000

Internal Area Residental Properties: Coastal City 1 State A 50% 45% 40% 35% 30% 25% 20% 15% 10% 5% 0%

House Unit

37.5

62.5

87.5

112.5

137.5

162.5

187.5

212.5

237.5

262.5

287.5

Internal Area Square Metres

Price and Internal Area

Residental Properties: Coastal City 1 State A 1400

Price $000s

1200

1000 800 600

400 200 0 50

100

150

200

Internal Area Square Metres

250

300

792

Solutions to End-of-Section and Chapter Review Problems Bedrooms and Bathrooms

Coastal City 1 Bathrooms 1 2 3 4 Total

1 4 0 0 0 4

2 26 2 0 0 28

Bedrooms 3 4 5 22 1 0 20 25 7 1 9 2 0 0 1 43 35 10

6 0 1 1 2 4

7 0 0 0 1 1

Total 53 55 13 4 125

2 6 22 28

Bedrooms 3 4 5 20 31 10 23 4 0 43 35 10

6 4 0 4

7 1 0 1

Total 72 53 125

Bedrooms and Type

Coastal City 1 Type House Unit Total

(c)

1 0 4 4

Regional City 1 and Coastal City 1 State A Price Residental Property Prices 45% 40% 35%

30% 25%

Regional City 1 Coastal City 1

20% 15%

10% 5%

0% 50

150

250

350

450

550

650 750 Price $000

850

950

1050

1150

1250

1350

793

Solutions to End-of-Section and Chapter Review Problems Internal Area Residental Properties: Internal Area 40% 35% 30% 25% 20%

Regional City 1 Coastal City 1

15% 10% 5% 0%

125

175

225

275

Internal Area Square metres

325

375

794

Chapter 3: Numerical descriptive measures Learning objectives After studying this chapter you should be able to: 1.

calculate and interpret numerical descriptive measures of central tendency, variation and shape for numerical data

calculate and interpret descriptive summary measures for a population

construct and interpret a box-and-whisker plot

calculate and interpret the covariance and the coefficient of correlation for bivariate data

3.1

Excel output: X Mean

Median

Mode

#N/A

Standard Deviation

2.915476

Sample Variance

8.5

Range

Minimum

Maximum

Sum

Count

First Quartile

Third Quartile

8.5

Interquartile Range

5.5

Coefficient of Variation

48.5913%

(a)

Mean = 6

Median = 7

(b)

Range = 7

Variance = 8.5

Standard deviation = 2.9

There is no mode. Coefficient of variation = (2.915/6)•100% = 48.6%

3.2

Excel output: X Mean

Median

Mode

Standard Deviation

3.286335

Sample Variance

10.8

Range

Minimum

Maximum

Sum

Count

First Quartile

Third Quartile

Interquartile Range

Coefficient of Variation

46.9476 %

(a)

Mean = 7

Median = 7

Mode = 7

(b)

Range = 9

Variance = 10.8

Standard deviation = 3.286 Coefficient of variation = (3.286/7)•100% = 46.948% (c)

Z scores: 0, –0.913, 0.609, 0, –1.217, 1.522 None of the Z scores is larger than 3.0 or smaller than –3.0. There is no outlier.

3.3

Excel output: X Mean

Median

Mode

Standard Deviation

Sample Variance

Kurtosis

-0.34688

Minimum

Maximum

Sum

Count

First Quartile

Third Quartile

Interquartile Range

Coefficient of Variation

66.6667 %

(a)

Mean = 6

Median = 7

(b)

Range = 12

Variance = 16

Mode = 7

Standard deviation = 4 Coefficient of variation = (4/6)•100% = 66.67%

3.4

Excel output: X Mean

Median

Mode

Standard Deviation

7.874007874

Sample Variance

Range

Minimum

-8

Maximum

Sum

Count

First Quartile

-6.5

Third Quartile

Interquartile Range

14.5

Coefficient of Variation

393.7004%

(a)

Mean = 2

Median = 7

Mode = 7

(b)

Range = 17

Variance = 62

Standard deviation = 7.874 Coefficient of variation = (7.874/2)•100% = 393.7%

3.5

RG = (1 + 0.1)(1 + 0.3) 

− 1 = 19.58%

3.6

(a)

Grade X

Grade Y

Mean

575

575.4

Median

575

Standard deviation

6.4

2.1

1/ 2

(b)

If quality is measured by central tendency, Grade X tyres provide slightly better quality because X’s mean and median are both equal to the expected value, 575 mm. If, however, quality is measured by consistency, Grade Y provides better quality because, even though Y’s mean is only slightly larger than the mean for Grade X, Y’s standard deviation is much smaller. The range in values for Grade Y is 5 mm compared to the range in values for Grade X, which is 16 mm.

(c)

Excel output: Grade X

Grade Y

Mean

575

Mean

577.4

Median

575

Median

575

Mode

#N/A

Mode

#N/A

Standard Deviation

6.40312 4

Standard Deviation

6.10737 3

Sample Variance

37.3

Range

Minimum

568

Minimum

573

Maximum

584

Maximum

588

Sum

2875

Sum

2887

Count

Grade X

Grade Y, Altered

Mean

575

577.4

Median

575

Standard deviation

6.4

6.1

When the fifth tyre measures 588 mm rather than 578 mm, Y’s mean inner diameter becomes 577.4 mm, which is larger than Xs mean inner diameter, and Y’s standard deviation increases from 2.07 mm to 6.11 mm. In this case, Xs tyres are providing better quality in terms of the mean inner diameter, with only slightly more variation among the tyres than Ys. 3.7

(a) and (b)

Full cream Mean 163 Standard Deviation 13.260 Sample Variance 175.833 Coefficient of variation 8.14% Minimum 135 First Quartile 157.5 Median 160.0 Third Quartile 167.5 Maximum 188 Interquartile Range 10.0 Range 53

Non-fat or Low or reduced skim milk fat milk 97.75 123.5 14.265 9.902 203.477 98.056 14.59% 8.02% 85 110 88.0 115.0 90.0 122.5 108.0 133.0 133 140 20.0 18.0 48 30

(c)

As expected low- and no-fat milk have fewer calories than full cream milk as indicated by the mean, median and quartiles. The standard deviation and interquartile range indicate that there is more variation in the calories in non-fat/skim milk than in either low-fat or full cream milk.

3.8 Sales per day

3.9

Mean

2200.00

Standard Error

176.03

Median

2360.00

Mode

2390.00

Standard Deviation

746.81

Sample Variance

557726.47

Kurtosis

-0.68

Skewness

0.63

Range

2230.00

Minimum

1350.00

Maximum

3580.00

Sum

39600.00

Count

18.00

(a)

Mean = $2,200; Median = $2,360; (The data has two modes) Mode = $2,390 and $2,400; Q1 = $1,525; Q3 = $2,545.

(b)

Variance = 557,726.47; Standard deviation = $746.81; Range = $2,230; IQR = $1,020; Coefficient of variation = 33.95%.

(c)

Descriptive statistics given above reveal that this store is able to maintain average daily sales of $2,200. However, daily sales can vary significantly as indicated by a significant SD of $746.81. Moreover, the store has a minimum daily sales of $1,350 and a maximum sales of $3,580.

(a)

Mean 2.45 minutes Median 2.5 minutes Mode 1.4 minutes First quartile 1.4 minutes Third quartile 3.1 minutes

(b)

Variance 2.271 Standard deviation 1.507 minutes

Range 5.5 minutes Interquartile range 1.7 minutes Coefficient of variation 61.55%

Time 0.6 0.9 1.4 1.4 1.5 2.4 2.6 2.7 2.8 3.1 3.9 6.1

Z score -1.227 -1.028 -0.697 -0.697 -0.630 -0.033 0.100 0.166 0.232 0.431 0.962 2.422

(c)

There are no outliers. There is no clear indication of skewness, as the mean and median are close and positive and negative Z scores are evenly spread.

(d)

The average time to serve a customer is approximately 2.5 minutes, but this varies from 0.6 minutes to 6.1 minutes, probably indicating that some customers need only a ‘yes’ or ‘no’ answer or a brochure, but that others need detailed information.

3.10

(a) and (b)

Summary statistics A

Mean

7377.33

8260.90

Standard Error

135.38

143.86

Median

7316.50

8140.50

Mode

8416.00

#N/A

6667.00

7569.00

8091.00

9036.00

IQR

1424.00

1467.00

Standard Deviation

856.23

909.83

Sample Variance

733128.43

827788.91

Range

3187.00

3043.00

11.61%

11.01%

Minimum

5544.00

6701.00

Maximum

8731.00

9744.00

Sum

295093.00

330436.00

Count

Coefficient Variation

(c)

The average life for CFL bulbs of manufacturer A is 7,377 hours while the mean life for manufacturer B is 8,260 hours. In terms of the average life and the median life, manufacturer B has a considerably higher lifetime. However, based on standard deviation, manufacturer B has a fairly high dispersion.

3.11 Price($) Mean

347.86

Standard error

18.73

Median

340.00

Mode

340.00

Standard Deviation

70.07

Sample Variance

4910.44

Kurtosis

-0.74

Skewness

-0.07

Range

230.00

Minimum

220.00

Maximum

450.00

Sum

4870.00

Count

14.00

Price($)

Price Z Score

340

-0.1122

450

1.4577

450

1.4577

280

-0.9685

220

-1.8247

340

-0.1122

290

-0.8257

370

0.3160

400

0.7441

310

-0.5403

340

-0.1122

430

1.1723

270

-1.1112

380

0.4587

(a)

Mean = $347.86; Median = $340; Q1 = $290; Q3 = $400

(b)

Variance = 4,910.44; Standard deviation = $70.07; Range = $230; IQR = $110; CV = 20.14%. None of the Z scores is less than –3 or greater than 3. There is no outlier in the price data.

(c)

The price of the digital cameras is rather symmetrical.

(d)

3.12

The mean price is $347.86 while the middle-ranked price is $340. The average scatter of price around mean is $70. 50% of the price is scattered over $110 while the difference between the highest and the lowest price is $230.

(a) and (b) Distance km

(c)

Mean

25.46

Standard Error

1.53

Median

26.00

Mode

27.00

Standard Deviation

10.84

Sample Variance

117.60

Kurtosis

-0.42

Skewness

-0.03

Range

43.00

Minimum

4.00

Maximum

47.00

Sum

1273.00

Count

50.00

The mean and median km driven by people are similar. On average, these people drive approximately 26 km to work. 50% of the people drive less or equal to 26 km and the rest drive more than 26 km.

3.13

(a) Hours: Original

Hours: Altered

Mean

473.46

Mean

550.38

Standard Error

58.46

Standard Error

87.46

Median

451.00

Median

492.00

Mode

#N/A

Mode

#N/A

Standard Deviation

210.77

Standard Deviation

315.33

Sample Variance

44422.44

Sample Variance

99435.26

Kurtosis

4.18

Kurtosis

2.70

Skewness

1.75

Skewness

1.70

Range

785.00

Range

1078.00

Minimum

264.00

Minimum

264.00

Maximum

1049.00

Maximum

1342.00

Sum

6155.00

Sum

7155.00

Count

13.00

Count

13.00

Based on results (hours/original), the median seems to be a better descriptive measure of the data since the outlier of 1,049 affects the mean. (b)

As shown in Hours: original table above, Range = 785; Variance = 44422.44; Standard deviation = 210.77.

(c)

From the the manufacturer’s viewpoint, the worst measure would be to compute the percentage of batteries that last over 400 hours (8/13 = .61). The median (451) and the mean (473.5) are both over 400, and would be a better measure for the manufacturer to use in advertisements.

(d)

Descriptive statistics for the altered hours are given in the table above. The median seems to be better a descriptive measure of the data as the two outliers affect the mean. From the manufacturer’s viewpoint, the worst measure remains the percentage of batteries that last over 400 hours (9/13 = .69). The median (492) and the mean (550.38) are both well over 400, and would be a better measure for the manufacturer to use in advertisements. The average life of batteries is higher in the second sample but the second sample appears to have a considerably higher deviation from the mean. The shape of the distribution of both is likely to be right-skewed as the mean is higher than the median.

3.14

Excel output: Waiting time

Mean

4.286667

Standard Error

0.422926

Median

4.5

Mode

#N/A

Standard Deviation

1.637985

Sample Variance

2.682995

Kurtosis

0.832925

Skewness

-0.83295

Range

6.08

Minimum

0.38

Maximum

6.46

Sum

64.3

Count

First Quartile

3.2

Third Quartile

5.55

Interquartile Range

2.35

Coefficient of Variation

38.2112%

(a)

Mean = 4.287

Median = 4.5

(b)

Variance = 2.683

Standard deviation = 1.638 Range = 6.08

Coefficient of variation = 38.21%

Waiting Time 4.21 5.55 3.02 5.13 4.77 2.34 3.54 3.20 4.50 6.10 0.38 5.12 6.46 6.19 3.79

Z-Score -0.05 0.77 -0.77 0.51 0.30 -1.19 -0.46 -0.66 0.13 1.11 -2.39 0.51 1.33 1.16 -0.30

There are no outliers. (c)

Since the mean is less than the median, the distribution may be left-skewed.

(d)

The mean and median are both under 5 minutes and the distribution is left-skewed, meaning that there are more unusually low observations than there are high observations. But six of the 15 bank customers sampled (or 40%) had wait times in excess of 5 minutes. So, although the customer is more likely to be served in less than 5 minutes, the manager may have been overconfident in responding that the customer would ‘almost certainly’ not wait longer than 5 minutes for service.

3.15

Excel output: Waiting Time

(a)

Mean

7.114667

Standard Error

0.537619

Median

6.68

Mode

#N/A

Standard Deviation

2.082189

Sample Variance

4.335512

Kurtosis

-1.05627

Skewness

0.072493

Range

6.67

Minimum

3.82

Maximum

10.49

Sum

106.72

Count

First Quartile

5.64

Third Quartile

8.73

Interquartile Range

3.09

Coefficient of Variation

29.2662 %

Mean = 7.114

Median = 6.68

(b)

Variance = 4.336

Standard deviation = 2.082 Range = 6.67

Coefficient of variation = 29.27% Waiting Time

Z Score

9.66

1.222431

5.90

-0.58336

8.02

0.434799

5.79

-0.63619

8.73

0.775786

3.82

-1.58231

8.01

0.429996

8.35

0.593286

10.49

1.62105

6.68

-0.20875

5.64

-0.70823

4.08

-1.45744

6.17

-0.45369

9.91

1.342497

5.47

-0.78987

There is no outlier since none of the observations is greater than 3 standard deviations away from the mean. (c)

Because the mean is greater than the median, the distribution is likely to be rightskewed.

(d)

The mean and median are both greater than 5 minutes. The distribution is likely to be right-skewed, meaning that there are some unusually high values. Further, 13 of the 15 bank customers sampled (or 86.7%) had waiting times greater than 5 minutes. So the customer is likely to experience a waiting time in excess of 5 minutes. The manager overstated the bank’s service record in responding that the customer would ‘almost certainly’ not wait longer than 5 minutes for service.

3.16

Summary statistics

Asking Price Mean

472440

Median

457000

First Quartile

397000

Third Quartile

529000

Standard Deviation

102394.989

Sample Variance

1.0485E+10

Kurtosis

1.92131411

Range Interquartile Range

3.17

524000 132000

(a)

The average asking price of properties sold is $472,440. 50% of the properties sold had an asking price equal to or less than $457,000 and 50% of the properties had an asking price over $457,000. The average price is less than the median price.

(b)

The lowest 25% of the properties had an asking price equal to or below $397,000 while the highest 25% of the properties had an asking price equal to or above $529,000.

(c)

Properties sold had an asking price range of $524,000. The range of the middle 50% of the properties was $132,000.

(d)

The properties sold had an average deviation of $102,394 from the mean.

(a)

Year 2012 2013 2014 2015 2016 Geometric Rate of Return

(b)

3.18

Annual Return Hang ASX Seng 200 22.9% 14.6% 2.9% 15.1% 1.3% 1.1% -7.2% -2.1% 0.4% 7.0% 3.6% 6.9%

The average rate of return for the Hang Seng is lower; however, it also is more variable.

(a)

Fund Conservative Balanced Balanced High Growth Sustainable Balanced

3.19

3.20

3.21

Historical crediting rate for year ending 30 2017 2016 2015 2014 2013

Average returns 30 June 2017 % 3 year 5 year

5.3

7.5

10.2

11.6

11.7

7.65%

9.23%

9.2 16.6 12.4

6.1 0.0 0.0

11.0 13.9 15.0

13.9 18.9 15.7

15.9 20.7 15.9

8.75% 9.92% 8.93%

11.17% 13.77% 11.63%

(b)

The conservative balanced fund gave lower returns over both three and five years, the balanced and sustainable balanced give similar returns over three and five years. The high growth fund gave higher returns than the other funds. All funds had higher returns over five years than over three years.

(a)

Population mean = 6

(b)

Population standard deviation = 3.23

(a)

Population mean = 6

(b)

Population standard deviation = 1.76

(a)

(b)

Population Data

Female

Male

Mean

28.083

79.833

Variance (pop)

16.910

51.306

Standard Deviation (pop)

4.112

7.163

Male m  m = 79.833  7.163 = ( 72.670,86.996 ) Nine out of 12 or 75% of months are in this range.

m  2m = 79.833  2  7.163 = ( 65.508,94.159 ) 100% of months are in this range. Female

 f   f = 28.083  4.112 = (23.971,32.195 ) Eight out of 12 or 66.7% of months are in this range.

 f  2 f = 28.083  2  4.112 = (19.859,36.307 ) Eleven out of 12 or 91.7% of months are in this range.

 f  3 f = 28.083  3  4.112 = (15.747,40.419 ) 100% of months are in this range. (c)

The proportion within one, two and three standard deviations of the mean for both distributions is approximately what would be expected from the empirical rule. Therefore, these distributions are probably approximately mound-shaped.

3.22

As have weekly sales for all 52 weeks in the year this is population data.

Population Data Mean Variance (Pop) Standard Deviaiton (Pop)

Sunday Morning Wednesday Evening 581.692 508.019 37295.213 58800.980 193.120 242.489

Weekly sales in the previous year were higher and less varied for the Sunday morning market than for the Wednesday evening market. However, as the population standard deviation is large for both markets, we can conclude that the weekly sales for both markets were highly variable

Sunday Morning Within 1 standard deviation Within 2 standard deviations Within 3 standard deviations

Lower Value 388.57 195.45 2.33

Upper Value 774.81 967.93 1161.05

Number Percentage 35 67.31% 51 98.08% 52 100.00%

Wednesday Evening Within 1 standard deviation Within 2 standard deviations Within 3 standard deviations

Lower Value 265.53 23.04 -219.45

Upper Value 750.51 993.00 1235.49

Number Percentage 39 75.00% 50 96.15% 52 100.00%

On average 582 and 508 bars where sold per week at the Sunday morning and Wednesday evening markets respectively. Furthermore, in a typical week sales at the Sunday morning market were between 389 and 775 bars and between 266 and 751 bars at the Wednesday evening markets. The percentage of weeks with sales within one, two, and three standard deviations of the mean approximately follows the empirical rule for the Sunday morning market. Therefore, the weekly sales distribution for this market may be mound-shaped with most weeks’ sales being close to the mean of 582 and a few weeks having very low or high sales. However, weekly sales for the Wednesday evening market have a larger percentage of weeks within one standard deviation than that given by the empirical rule so may not be mound-shaped. As all values are within three standard deviations of the mean there are no outliers. However, there is one possible outlier for the Sunday morning market 1039 bars in Week 1 (New Year) and two for the Wednesday evening markets, sales of 1032 and 1057 bars in weeks 49 and 51 (leading up to Christmas). As you would expect sales of this product to increase over the Christmas period, these quantities are probably not outliers. 3.23

(a) and (b)

As data is all employee ages, this is population data.

Age

Z-Score -0.427 -0.427 2.935 -0.297 -0.168 -0.168 -0.556 -0.297 0.091 -0.685 22.3 59.81 7.7337

19 19 45 20 21 21 18 20 23 17 Mean Variance Standard Devistion (c)

The mean age of all employees is 22.3 years with a standard deviation of 7.75 years. However, only 20% of employees (2 out of 10) have ages above the mean, so the mean is not a good measure of the typical age of an employee. Furthermore, 90% (9 out of 10) of employees have ages within one standard deviation of the mean. There is one extreme age of 45 years, 2.9 standard deviations above the mean, which has unduly affected the mean.

3.24 Mean Median Variance Standard Deviation

35.000 37.500 182.167 13.497

(a)

mean = 35. On average, the 30 employees worked 35 hours last week.

(b)

variance = 182.17, standard deviation = 13.50. The average squared distance between the 30 employees’ working hours and the mean is 182. If the distribution is approximately symmetrical, about 68% of the work hours will be within 13.5 hours of the mean value of 35.

(c)

Since the median = 37.5 it is highly likely that the distribution is not symmetrical. According to the Chebyshev rule at least 75% of the values will be found within 2 standard deviations of the mean. 93% (28/30) of the values are within 2 standard deviations of the mean and 100% of the values are within thre standard deviations of the mean.

(d)

The preceding suggests there are no outliers within the sample set.

3.25 mj

mjfj

mj fj2 250

300

4,500

1,000

25,000

700

24,500

450

20,250

100

2,500

Sum

(a)

Mean = 2,500/100 = 25

(b)

74,500

74500 − 100  252 = 121.212... = 11.009... 99

3.26 mj

mjfj

mj fj2

200

1,000

375

5,625

375

9,375

525

18,375

225

10,125

100

1,700

44,500

Sum

(a)

Mean = 1,700/100 = 17

(b)

44500 − 100  172 = 157.575... = 12.552... 99

3.27

March

(a)

Midpoint

Frequency

fjmj

fjmj2

1,000 3,000 5,000 7,000 9,000 11,000 Totals April

6 13 17 10 4 50

6,000 39,000 85,000 70,000 36,000 236,000

6,000,000 117,000,000 425,000,000 490,000,000 324,000,000 1,362,000,000

Midpoint

Frequency

fjmj

fjmj2

1,000 3,000 5,000 7,000 9,000 11,000 Totals

10 14 13 10 3 50

10,000 42,000 65,000 70,000 33,000 220,000

10,000,000 126,000,000 325,000,000 490,000,000 363,000,000 1,314,000,000

March: Mean = 236,000/50 = 4,720 April:

Mean = 220,000/50 = 4,400

March: Standard deviation =

April: Standard deviation =

3.28

1362000000 − 50  47202 = 2250.079... 49

1314000000 − 50  44002 = 2657.296... 49

(b)

The arithmetic mean has declined by $320 while the standard deviation has increased by $407.21.

(a)

Five-number summary: 2; 3; 7; 8.5; 9

(b)

Box-and-whisker plot

The distribution is left skewed. There is a longer distance from Q 1 to Q2 than Q2 to Q3, confirming our conclusion that the data are left-skewed.

3.29

(a)

Five-number summary: 3; 4; 7; 9; 12

(b)

Box-and-whisker plot

The distribution is fairly symmetrical. The data set is almost symmetrical since the median line almost divides the box in half but the whiskers show right-skewness. 3.30

(a)

Five-number summary: 0; 3; 7; 9; 12

(b)

Box-and-whisker plot

The distribution is left-skewed. The box-and-whisker plot shows a longer left box from Q1 to Q2 than from Q2 to Q3 visually confirming our conclusion that the data are left-skewed. 3.31

(a)

Five-number summary: -8; -6.5; 7; 8; 9

(b)

Box-and-whisker plot

The distribution is left-skewed. The box-and-whisker plot shows a longer left box from Q1 to Q2 than from Q2 to Q3 visually confirming our conclusion that the data are left-skewed.

3.32

(a) Manufacturer A Five-number Summary

(b)

Minimum

5544

First Quartile

6667

Median

7316.5

Third Quartile

8091

Maximum

8731

Box-and-whisker plot

The distribution is left-skewed. The left-side whisker is longer than the right side, confirming that the life of CFL bulbs of manufacturer A is left-skewed. Manufacturer B Five-number Summary Minimum

6701

First Quartile

7569

Median

8140.5

Third Quartile

9036

Maximum

9744

The distribution is rightly skewed. The right-hand side of the box is longer than the left-hand side though the whiskers are fairly similar in distance. In summary the life of the bulbs shows a marginal skewness towards the right. 3.33

(a) (b)

Five-number summary 1350; 1525; 2360; 2545; 3580 Box-and-whisker plot

The results are inconsistent. The right-hand whisker is far longer than the left-hand and the distance from the median to X largest is greater than the distance from the median to Xsmallest indicating right-skewness. However, the left-hand side of the box is far longer than the right-hand side indicating left-skewness. 3.34

(a)

Five-Number Summary Rolls and lean burgers Minimum 3.70 First Quartile 5.40 Median 8.15 Third Quartile 19.60 Maximum 34.60

Salads 3.90 5.30 15.80 24.60 27.90

Traditional Burgers and Rolls 19.80 20.80 23.05 39.30 65.10

(b)

Box-and-whisker plot Fat Content per Serve, grams

Traditional Burgers and Rolls

Salads

Rolls and lean burgers

3.35

(c)

The traditional items have the highest average fat content, followed by salads with rolls and lean burgers the last. However, the fat content of the traditional items and rolls and lean burgers vary more than that of the salads. Therefore some rolls and lean burgers have more fat than any salads and some traditional items have less fat than some of the healthier options. Rolls and lean burgers and traditional items are both skewed to the right while the fat content of salads seems fairly symmetric.

(a)

Five-number summary of waiting times

Five-Number Summary Commerical Residential District Area Minimum 0.38 3.82 First Quartile 3.20 5.64 Median 4.50 6.68 Third Quartile 5.55 8.73 Maximum 6.46 10.49

(b)

Box-and-whisker plot Boxplot - Waiting Time

Residential Area

Commercial District

Minutes

Residential area: the distribution is skewed slightly to the right. Commercial district: the distribution is skewed to the left.

3.36

(c)

The central tendency of the waiting times for the bank branch located in the commercial district of a city is lower than that of the branch located in the residential area. There are a few longer-than-normal waiting times for the branch located in the residential area, whereas there are a few exceptionally short waiting times for the branch located in the commercial area.

(a)

cov(X,Y) = 65.2909

(b)

S X2 = 21.7636, SY2 = 195.8727 r=

3.37

cov ( X , Y ) S X2

SY2

65.2909 = +1.0 21.7636 195.8727

(c)

There is a perfect positive linear relationship between X and Y; all the points lie exactly on a straight line with a positive slope.

(a)

cov(X,Y) = 49.88 and r = 0.816, calculation of r is as follows:

𝑆𝑆𝑋 = 5,202 − (10 × 222 ) = 182 𝑆𝑆𝑌 = 22,822 − (10 × 462 ) = 1,682

𝑟=

(b)

𝑆𝑆𝑋𝑌 √𝑆𝑆𝑋√𝑆𝑆𝑌

449 √182√1,662

= 0.816

Based on covariance, it can be concluded that there is a positive linear relationship between the number of people in a sales team and the sales generated. Based on correlation coefficient, it is concluded that there is a fairly strong positive linear relationship between number of people in a sales team and the sales generated.

3.38

(a)

New South Wales cov(ULP, Diesel) = 18.0007… r = 0.6224…

(b)

Queensland cov(ULP, Diesel) = 5.5199…… r = 0.6318…

(c)

From the values of the correlation coefficients can conclude that there is a moderate positive linear relationship between petrol and diesel prices in New South Wales and in Queensland.

Diesel Cents per Litre

This is relationship is illustrated by the scatter plots below. 146 144 142 140 138 136 134 132 130 128 126 124 122 120

New South Wales

120

122

124

126

128 130 132 134 136 138 Unleaded Petrol Cents per Litre

140

142

144

146

Diesel Cents per Litre

Queensland 134 132 130 128 126 124 122 120 118 116 116

118

120

122

124 126 128 130 Unleaded Petrol Cents per Litre

132

134

136

138

3.39

(a)

Scatter diagram

Annual water usage kilolitres

Scatter diagram of water usage 1000 900 800 700 600 500 400 300 200 100 0 0

Number of seats The scatter diagram shows a weak relationship between size of local restaurants (as measured in number of seats) and annual water usage.

3.40

(b)

cov(seats, water) = 1546, r = 0.7941. A positive relationship was expected.

(c)

Based on correlation coefficient, there is a fairly strong positive linear relationship between the size of local restaurants (as measured in number of seats) and annual water usage.

(a)

cov(Calories,Fat) = 67.813K

(b)

r = 0.862K

(c)

The correlation coefficient is more valuable for expressing the relationship between calories and fat as it does not depend on the measurement units.

(d)

We can conclude that there is a relatively strong positive linear relationship between calories and fat content.

3.41

Central tendency or location refers to the fact that most sets of data show a distinct tendency to group or cluster about a certain central point.

3.42

The arithmetic mean is a simple average of all the values, but is subject to the effect of extreme values. The median is the middle-ranked value, but varies more from sample to sample than the arithmetic mean, although it is less susceptible to extreme values. The mode is the most common value, but is extremely variable from sample to sample.

3.43

The first quartile is the value below which ¼ of the total ranked observations will fall, the median is the value that divides the total ranked observations into two equal halves and the third quartile is the observation above which ¼ of the total ranked observations will fall.

3.44

Variation is the amount of dispersion, or ‘spread’, in the data.

3.45

The Z score measures how many standard deviations an observation in a data set is away from the mean.

3.46

The range is a simple measure, but only measures the difference between the extremes. The interquartile range measures the range of the centre 50% of the data. The standard deviation measures variation around the mean while the variance measures the squared variation around the mean, and these are the only measures that take into account each observation. The coefficient of variation measures the variation around the mean relative to the mean. The range, standard deviation, variance and coefficient of variation are all sensitive to outliers, while the interquartile range is not.

3.47

The empirical rule relates the mean and standard deviation to the percentage of values that will fall within a certain number of standard deviations of the mean. The Chebyshev rule applies to any type of distribution while the empirical rule applies only to data sets that are approximately bell-shaped. The empirical rule is more accurate than the Chebyshev rule in approximating the concentration of data around the mean.

3.48

Excel output: Tea-bags

Mean

5.5014

Standard Error

0.014967

Median

5.515

Mode

5.53

Standard Deviation

0.10583

Sample Variance

0.0112

Kurtosis

0.127022

Skewness

-0.15249

Range

0.52

Minimum

5.25

Maximum

5.77

Sum

275.07

Count

First Quartile

5.44

Third Quartile

5.57

Interquartile Range

0.13

1.9237%

(a)

Mean = 5.5014, median = 5.515, first quartile = 5.44, third quartile = 5.57

(b)

Range = 0.52, interquartile range = 0.13, variance = 0.0112, standard deviation = 0.10583, coefficient of variation = 1.924%

(c)

The mean weight of the tea-bags in the sample is 5.5014 grams while the middleranked weight is 5.515. The company should be concerned about the central tendency because that is where the majority of the weight will cluster around. The average of the squared differences between the weights in the sample and the sample mean is 0.0112, whereas the square root of it is 0.106 grams. The difference between the lightest and the heaviest tea-bags in the sample is 0.52. 50% of the tea-bags in the sample weigh between 5.44 and 5.57 grams. According to the empirical rule, about 68% of the tea-bags produced will have weight that falls within 0.106 grams around 5.5014 grams. The company producing the tea-bags should be concerned about the variation because tea-bags will not weigh exactly the same due to various factors in the production process (e.g. temperature and humidity inside the factory, differences in the density of the tea, etc). Having some idea about the amount of variation will enable the company to adjust the production process accordingly.

(d)

Box-and-whisker plot Box-and-whisker Plot

Teabags

5.2

5.4

5.6

5.8

The data are slightly left skewed. (e)

3.49

On average, the weight of the tea-bags is quite close to the target of 5.5 grams. Even though the mean weight is close to the target weight of 5.5 grams, the standard deviation of 0.106 indicates that approximately 95% of the tea-bags will fall within 0.212 grams around the target weight of 5.5 grams. The interquartile range of 0.13 also indicates that half of the tea-bags in the sample fall in an interval 0.13 grams around the median weight of 5.515 grams. If possible, adjust the process to reduce the variation of the weight around the target mean.

(a) and (b)

NSW Unleaded

NSW Diesel

Queensland Queensland Unleaded Diesel

Mean

123.2222

128.0533

134.9133

128.9133

Standard Deviation

9.4173

3.6089

5.3967

3.7931

Sample Variance

88.6863

13.0244

29.1244

14.3875

Minimum

111.7

120.8

123.9

122.2

First Quartile

113.7

126.05

130.05

126.15

Median

120.5

128.8

133.9

128.9

Third Quartile

131.05

131.25

139.2

131.05

Maximum

139.9

134.6

146.2

139.9

Range

28.2

13.8

22.3

17.7

17.35 7.643%

5.2 2.818%

9.15 4.000%

4.9 2.942%

Interquartile Range Coefficient of Variation

We can conclude that on this day, that in New South Wales diesel prices were on average higher but less variable than petrol prices. However, in Queensland, unleaded petrol prices were higher and more variable than diesel prices.

(c)

Box-and-whisker plot

Fuel Prices March 2017

Queensland Diesel

Queensland Unleaded

NSW Diesel

NSW Unleaded

110

115

120

125

130 Cents Per Litre

135

140

145

150

Petrol prices are generally more variable than diesel prices. Unleaded petrol prices are positively skewed. Diesel prices are similar in New South Wales and Queensland while unleaded petrol prices are higher in Queensland. (d)

New South Wales cov(ULP, Diesel) = 20.2588…

r = 0.5960…

As expected from the scatter plot in Problem 2.30 (a) there is a weak positive linear relationship between petrol and diesel prices. Queensland cov(ULP, Diesel) = 7.8407……

r = 0.3830…

As expected from the scatter plot in Problem 2.53 (b) there is a very weak positive relationship between Queensland petrol and diesel prices.

3.50

(a) and (b)

Excel output

Total Mark Mean

60.782

Standard Error

2.440

Median

63.000

Mode

57.000

Standard Deviation

18.099

Sample Variance

327.581

Kurtosis

0.675

Skewness

-0.788

Range

80.000

Minimum

14.000

Maximum

94.000

Sum

3343.000

Count

55.000

First Quartile

56.000

Third Quartile

73.000

IQR

17.000

0.298

Sample mean 60.782 is slightly less than the median 63, which indicates that the data may be skewed to the left with a few small values (marks). From the standard deviation of 18 we can conclude that the majority of marks, or a typical student’s mark, will be in the range of approximately 61 ± 18; that is, 43 to 79. 50% of the students have marks in the range of 56 to 73 with 25% less than 56 and 25% more than 73. (c)

Box-and-whisker plot

Data seems skewed to the left as the distance from median to the lowest mark is longer than the distance from the median to the highest mark, and the left-hand whisker is longer than the right-hand whisker. However, the right-hand box is longer than the left-hand side.

3.51

(d)

cov(semester mark, exam mark) = 39.7714, r = 0.7939

(e)

There is a strong positive linear relationship between a student’s semester mark and their exam mark.

(a) To 30 June

2013

2016

Males

Female

Males

Females

Mean

37.138…

38.653…

37.513…

39.010…

Standard deviation

22.463…

23.142…

22.720…

23.294…

Note: these are population parameters and are approximations as calculated from the frequency distribution with the mid-point of the last class (100 and over) estimated to be 102. (b)

3.52

These statistics show that the average age for females is higher than for males and slightly more varied. Furthermore, the Australian population is growing older, with the average age of both males and females increasing from 2013 to 2016.

(a) and (b) WIP Plant A

Plant B

Mean

9.3820

11.3535

Standard Error

0.8939

1.1462

Median

8.5150

11.9600

7.2900

6.2500

11.4200

14.2500

IQR

4.1300

8.0000

Standard Deviation

3.9977

5.1262

Sample Variance

15.9812

26.2775

Range

17.2000

23.4200

Minimum

4.4200

2.3300

Maximum

21.6200

25.7500

Coefficient of variation

42.61%

45.15%

Count

(c)

Box-and-whisker plot

The distribution of processing times for plant A is right skewed. (d)

Processing times for plants A and B are quite different. Plant B has a greater range of processing times, much more dispersion among data values, a higher median, a higher value for the third quartile and a greater extreme value than plant A.

3.53 (a) & (c)

Daily Water Usage 2.5

Kilolitres (Kl)

2.0 1.5

1.0 0.5 0.0 0

100

120

140

Day

Daily water usage varies from approximately 0.5 Kl to just above 2Kl. However, on most days usage is between 1.0 and 1.5 Kl.

Daily Water Usage

Daily Water Usage Kl (kilolitres)

0.4

0.6

0.8

1.0

1.2

1.4

1.6

1.8

2.0

2.2

1.9

2.1

2.4

Histogram of Daily Water Usage 35

Frequency

30 25 20 15 10 5 0 0.3

0.5

0.7

0.9

1.1

1.3

1.5

1.7

2.3

Kilolitres The distribution of daily water usage is approximately mound shaped, with a usage of between 0.8 and 1.8 Kilolitres on the majority of days. (b) & (c)

Daily Water Usage Five-Number Summary Minimum 0.5536 First Quartile 1.0362 Median 1.2670 Third Quartile 1.5020 Maximum 2.1830 Range 1.6294 Interquartile Range 0.4658 On 50% of days water usage is between 1.03 and 1.50 Kilolitres, with usage below 1.03 Kilolitres on 25% of days and above 1.50 Kilolitres on the remaining 25% of days.

Daily Water Usage Kl (kilolitres) Mean 1.270353 Standard Error 0.028812 Median 1.26700 Mode 1.289872 Standard Deviation 0.332281 Sample Variance 0.110411 Kurtosis -0.08876 Skewness 0.360474 Range 1.6294 Sum 168.957 Coefficient of Varition 26.16% Count 133 On average the flats use 1.27 Kl of water per day. 3.54

3.55

(a)

As petrol prices between petrol stations vary by a few cents, $0.03 is a reasonable value for the standard deviation.

(b)

As starting salaries will vary by a few thousand dollars, $5,000 is a reasonable value for the standard deviation.

(c)

As weights between individuals vary by kilograms, 10 kg is a reasonable value for the standard deviation.

(a)

Geometric mean annual change in CPI for Australia for five years December 2012 to December 2016 is 1.96% Geometric mean annual change in CPI for New Zealand for five years December 2012 to December 2016 is 0.94%.

(b)

3.56

For the given period New Zealand's annual inflation rate was on average approximately half of Australia's.

Sunday morning markets, r = -0.7321…. There is a moderate negative linear relationship between price and quantity sold at the Sunday morning markets. That is, quantity sold increases when the price is reduced. Wednesday evening market, r = -0.0549…. There is an extremely weak negative or no relationship between price and quantity sold at the Wednesday evening markets. That is, factors other than price are influencing the quantity sold.

3.57 to 3.58 Answers will vary.

Case Study - Tasman University Student Survey (a) & (c)

Bachelor of Business

Mean Standard Error Median Mode Standard Deviation Sample Variance Kurtosis Skewness Coefficient of Variation Sum Count

Expected Social Textbook Text Age WAM Salary Networking Satisfaction Cost Messages Wealth 21.1290 63.9403 63.5484 1.5161 3.7419 482.0161 246.2097 7.1512 0.1818 1.6317 1.5343 0.1072 0.1542 28.1882 27.2372 2.4903 21 64.6 65 1 4 500 200 1 21 72.5 55 1 4 500 300 1 1.43131 12.84768 12.08091 0.84430 1.21379 221.954 214.46595 19.60850 2.05 165.06 145.95 0.71 1.47 49263.49 45995.64 384.49 1.40 0.36 0.42 1.10 0.47 4.56 1.14 16.98 0.74 -0.11 0.53 0.96 -0.51 1.59 1.30 4.16 6.77% 20.09% 19.01% 55.69% 32.44% 46.05% 87.11% 274.20% 1310 3964.3 3940 94 232 29885 15265 443.375 62 62 62 62 62 62 62 62

Five-Number Summary

Minimum First Quartile Median Third Quartile Maximum Range Interquartile Range

Age 18 20 21 22 26 8 2

WAM 27.9 56.3 64.7 71.15 93.9 66 14.85

Expected Social Textbook Text Salary Networking Satisfaction Cost Messages Wealth 40 0 1 100 0 0.1 55 1 3 300 100 0.725 65 1 4 500 200 1 70 2 4 600 300 5 95 4 6 1400 900 100 55 4 5 1300 900 99.9 15 1 1 300 200 4.275

In this sample the median and mean age are similar at 21 and 21.1 respectively. Therefore, can conclude that the average age of students enrolled in the Bachelor of Business (BBus) is approximately 21 years. The youngest student is 18 years and the oldest 26 with 50% of students aged between 20 and 22 years. Half of BBus students in the sample have a Weighted Average Mark (WAM) of at most 64.7 and half at least 64.7. The average WAM is 63.9. The middle 50% of BBus students in the sample have a WAM of between 56.3 and 71.2, with 25% of students having a WAM above 71.1. The lowest WAM in the sample is 27.9 and the highest is 93.9. 50% of these students expect their starting salary to be at least $65,000 while the mean expected starting salary is $63,550. The average scatter of expected starting salary about the mean is $12,000. Half of students are registered for at least one social networking site while the mean number is 1.51. The lowest number of social networking sites is zero and the highest is 4. Less than 25% of students are satisfied with the on-campus food services, rating of 5 or above, with at least 25% dissatisfied, rating of 3 or below. Half of students spent at least $500 on textbooks last semester, while mean amount students spent was $482. The average scatter about the mean is $221.95. The middle 50% of

students in the sample spent between $300 and $600 on textbooks last semester. The lowest amount spent was $100 and the highest was $1400. The median number of text messages in a typical week is 200 while the mean is 246.2. The average scatter of the number of text messages about the mean is 214.47. The middle 50% of students send between 100 and 300 text messages during a typical week. The lowest number of text message sent in a typical week is 0 and the highest is 900. Half the students say that they would need to accumulate at least one million dollars to consider themselves rich. While the mean amount of wealth to accumulate before considering themselves rich is 7.2 million dollars. The average scatter about the mean is 19.61 million dollars. The middle 50% of students in the sample say that to consider themselves rich they would need to accumulate between $725,000 and five million dollars.

Bachelor of Businnes

Age

22 Age Years

Age is slightly left skewed

Bachelor of Business

WAM

55 60 65 70 Weighted Average Mark (WAM)

100

Weighted average mark has a slight left skew.

Bachelor of Business

Expected Salary

100

105

Expected Starting Salary ($000)

No clear skewness Bachelor of Business

Text Messages

100 150 200 250 300 350 400 450 500 550 600 650 700 750 800 850 900 950 1000 Number of Text Messages Sent In A Typical Week

Number of text messages sent per week is right skewed. Bachelor of Business

Wealth

30 40 50 60 70 Accumulated Wealth to be Rich (Millions Dollars)

100

Wealth is extremely right skewed.

(b) & (c)

Master of Business Administration

MBA UG Full-time Expected Textbook Advisory Text Age WAM WAM Jobs Salary Cost Rating Messages Wealth Mean 26.455 73.805 80.816 1.727273 102.9773 363.8636 4.590909 232.4545 10.6409 Standard Error 0.8053 1.2993 1.3446 0.153754 6.701128 54.20989 0.146554 44.75937 3.36994 Median 25.5 72.75 80.9 2 90 300 5 100 2.75 Mode 26 75.3 80.2 2 90 300 4 100 1 Standard Deviation 5.3416 8.6185 8.9188 1.019887 44.45025 359.5877 0.972127 296.9001 22.3537 Sample Variance 28.533 74.279 79.545 1.040169 1975.825 129303.3 0.945032 88149.65 499.686 Kurtosis 6.2328 -0.389 -0.935 0.067159 12.9347 17.69406 0.495591 3.230509 10.9721 Skewness 2.3611 0.2315 0.1085 0.447546 3.131929 3.968429 -0.02959 1.833591 3.33381 Coefficient of Variation 26 34.5 32.1 4 260 2130 5 1250 99.9 Sum 1164 3247.4 3555.9 76 4531 16010 202 10228 468.2 Count 44 44 44 44 44 44 44 44 44

Five-Number Summary

Age Minimum 21 First Quartile 23 Median 25.5 Third Quartile 27 Maximum 47 Range 26 Interquartile Range 4

MBA UG Full-time Expected Textbook Advisory Text WAM WAM Jobs Salary Cost Rating Messages Wealth 57.6 65.5 0 60 70 2 0 0.1 67.7 72.7 1 80 200 4 20 1 72.75 80.9 2 90 300 5 100 2.75 80.4 88.4 2 120 400 5 400 10 92.1 97.6 4 320 2200 7 1250 100 34.5 32.1 4 260 2130 5 1250 99.9 12.7 15.7 1 40 200 1 380 9

Correlation (MBA and Undergraduate WAM) = 0.5744 In this sample the median and mean age are similar at 25.5 and 26.45 respectively. Therefore, can conclude that the average age of students enrolled in the Master of Business Administration is approximately 26 years. The youngest student is 21 years and the oldest 57 with 50% of students aged between 23 and 27 years. Half of MBA students in the sample have a Weighted Average Mark (WAM) of at most 72.75 which is lower than the students’ undergraduate median WAM of 80.9. The mean MBA WAM is 73.81 again lower than the undergraduate mean WAM for these students of 80.82. The middle 50% of MBA students in the sample have a MBA WAM of between 67.7 and 80.4, with 25% of students having a MBA WAM above 80.4. The lowest WAM in the sample is 57.6 and the highest is 92.1. The correlation coefficient of 0.5744 between MBA and Undergraduate WAM indicates a moderate, positive linear relationship between undergraduate and MBA WAM. That is, a higher Undergraduate WAM indicate that a student’s MBA WAM is also likely to be higher, and vice versa. Copyright © 2019 Pearson Australia (a division of Pearson Australia Group Pty Ltd) 9781488617249, Berenson, Basic Business Statistics 5e

Half of the MBA students in the sample have held at least two full-time jobs in the past 10 years, with a mean of 1.73 jobs. The average scatter of the number of full-time jobs held in the past ten years is 1.02. The middle 50% of MBA students in the sample have held either 1 or 2 full-time jobs in the past ten years. 50% of these students expect their starting salary after graduating from the MBA to be at least $90,000 while the mean expected starting salary is $103,000. The average scatter of expected starting salary about the mean is $44,500. Half of students spent at least $300 on textbooks last semester, while mean amount students spent was $363.86. The average scatter about the mean is $359.59. The middle 50% of students in the sample spent between $200 and $400 on textbooks last semester. The lowest amount spent was $70 and the highest was $2200. At least 50% of students are satisfied with the MBA Program advisory services, rating of 5 or above. The median number of text messages in a typical week is 100 while the mean is 232.45. The average scatter of the number of text messages about the mean is 296.9. The middle 50% of students send between 20 and 400 text messages during a typical week. The lowest number of text message sent in a typical week is 0 and the highest is 1250. Half the students say that they would need to accumulate at least 2.75 million dollars to consider themselves rich. While the mean amount of wealth to accumulate before considering themselves rich is 10.6 million dollars. The average scatter about the mean is 22.35 million dollars. The middle 50% of students in the sample say that to consider themselves rich they would need to accumulate between one and ten million dollars.

Tasman University BBus Age

MBA Age

Age in Years

The age of MBA students is skewed to the right. MBA students are generally older with more varied ages than Bachelor of Business students.

Tasman University

BBus WAM

UG WAM (MBA Students) MBA WAM

50 60 70 Weighted Average Mark (WAM)

100

MBA students Weighted Average Mark (WAM) for the MBA is lower than their undergraduate WAM. However, as expected MBA students undergraduate WAM is higher and less varied than BBus students WAM. Tasman University BBus Expected Salary

MBA Expected Salary

100

120

140 160 180 200 220 240 Expected Salary on Graduation ($000s)

260

280

300

320

340

MBA students in general expect a higher salary on graduation than BBus students. Expected salary for MBA students is extremely right skewed, and more varied than expected salary for BBus students. Tasman University BBus Textbook Cost

MBA Textbook Cost

250

500

750

1000 1250 1500 Cost of Textbooks Last Semester, $

1750

2000

2250

Textbook cost for the MBA is generally less but more varied than textbook cost for the BBus.

Tasman University

BBus Text Messages

MBA Text Messages

100

200

300

400 500 600 700 800 900 Number of Text Messages Sent in a Typical Week

1000

1100

1200

1300

Number of text messages sent in a typical week is skewed to the right. Tasman University

BBus Wealth

MBA Wealth

30 40 50 60 70 Accoumulated Wealth to be Rich (Million Dollars)

100

Wealth is extremely right skewed. Case Study - Safe-As Houses Real Estate (a) & (c) Regional City 1 State A

Internal Regional City 1 State A Price $000 Area m^2 Bedrooms Bathrooms Garages Mean 320.8144 140.4104 3.104 1.376 1.64 Standard Error 9.89216514 3.88945022 0.09559592 0.04907401 0.07569037 Median 298 141.5 3 1 2 Mode 269 90.4 3 1 2 Standard Deviation 110.597769 43.4853754 1.06879492 0.54866407 0.84624407 Sample Variance 12231.8664 1890.97787 1.14232258 0.30103226 0.71612903 Kurtosis -0.0777445 2.94267655 6.41272014 0.23020394 4.89373279 Skewness 0.77490082 1.00718542 1.60209846 1.10402999 1.17299372 Coeffient of Variation 34.47% 30.97% 34.43% 39.87% 51.60% Sum 40101.8 17551.3 388 172 205 Count 125 125 125 125 125

Five-Number Summary Regional City 1 State A Price $000 Minimum 165 First Quartile 231.5 Median 298 Third Quartile 383.5 Maximum 645 Range 480 Interquartile Range 152

Internal Area m^2 66.5 93.45 141.5 171.6 349.6 283.1 78.15

Bedrooms Bathrooms 1 1 2 1 3 1 4 2 9 3 8 2 2 1

Garages 0 1 2 2 6 6 1

Internal Correlation Coefficient, r Price $000 Area m^2 Bedrooms Bathrooms Price $000 1.00000 0.77726 0.67079 0.67454 Internal Area m^2 0.77726 1.00000 0.94052 0.65530 Bedrooms 0.67079 0.94052 1.00000 0.59289 Bathrooms 0.67454 0.65530 0.59289 1.00000 Garages 0.52692 0.41752 0.33597 0.34599

Garages 0.52692 0.41752 0.33597 0.34599 1.00000

In this sample the cheapest property was $165,000 and most expensive $645,000. The median, the middle value, indicates that 50% of properties in the sample had prices less than $298,000 and 50% more than. The mean, is slightly more than the median at $320,814. As the mean is affected by the extreme value of $645,000 and the median is not, the median is the better measure of the average price to use. The standard deviation indicates that on average prices vary from the mean price by approximately $110,598. The five number summary above divides the sample into quarters, with 25% of prices being below the first quartile, $231,500, and 25% of prices above the third quartile, $383,500. Therefore, the middle 50% of prices vary by $152,000. That is, between $231,500 and $383,500. The mean internal area for this sample of residential sales in Regional City 1 is 140.4 square metres, with the average scatter about this mean value of 43.5 square metres. This is, similar to the median internal area of 141.5. 50% of residential properties sales in the sample had at most three bedrooms, 1 bathroom and 2 garages. The values of the correlation coefficients indicate that there is a strong positive linear relationship between the number of bedrooms and internal area and a weaker positive relationship between internal area and price.

Regional City 1 State A

Garages

Bathrooms

Bedrooms

Number of garages, bathrooms and bedrooms are all right skewed.

Regional City 1, State A

Unit

House

150

200

250

300

350

400 Price $000

450

500

550

600

650

Both house and unit prices are skewed to the right, with house prices generally higher and more varied than unit prices.

Regional City 1, State A

Unit

House

100

150

200 Internal Area, Square Metres

250

300

350

The internal area for both houses and units in Regional City 1, State A are skewed to the right, with the internal area of houses generally higher and more varied than the internal area of units. (b) & (c) Coastal City 1 State A

Coastal City 1 State A

Internal Price $000 Area m^2 Bedrooms Bathrooms Garages Mean 458.6384 148.8096 3.28 1.744 1.672 Standard Error 19.0468138 4.20198976 0.10198039 0.06900584 0.08252155 Median 399 146.5 3 2 2 Mode 429 92.4 3 2 2 Standard Deviation 212.949852 46.9796738 1.14017543 0.77150879 0.92261899 Sample Variance 45347.6393 2207.08975 1.3 0.59522581 0.85122581 Kurtosis 2.09470154 -0.3420367 0.38636578 0.56796717 17.0807461 Skewness 1.28109336 0.38922437 0.49126502 0.90629126 2.83089913 Coefficient of Variation 46.43% 31.57% 34.76% 44.24% 55.18% Sum 57329.8 18601.2 410 218 209 Count 125 125 125 125 125

Five-Number Summary

Coastal City 1 State Internal A Price $000 Area m^2 Bedrooms Bathrooms Minimum 122 66.5 1 1 First Quartile 325 115.5 2 1 Median 399 146.5 3 2 Third Quartile 549.5 183.4 4 2 Maximum 1260 289.1 7 4 Range 1138 222.6 6 3 Interquartile Range 224.5 67.9 2 1 Correlation Coefficient, r Price $000 Internal Area m^2 Bedrooms Bathrooms Garages

Price $000 1.00000 0.65894 0.62097 0.63667 0.48465

Internal Area m^2 0.65894 1.00000 0.94092 0.77334 0.61095

Bedrooms 0.62097 0.94092 1.00000 0.75139 0.63231

Bathrooms 0.63667 0.77334 0.75139 1.00000 0.64017

Garages 0 1 2 2 8 8 1 Garages 0.48465 0.61095 0.63231 0.64017 1.00000

In this sample the cheapest property sold for $122,000 and most expensive for $1,260,000. The median, the middle value, indicates that 50% of properties in the sample sold for at most $399,000 and 50% more than. The mean, is more than the median at $458,638. As the mean is affected by the extreme value of $1,260,000 and the median is not, the median is the better measure of the average price to use. The standard deviation indicates that on average prices vary from the mean price by approximately $212,950. The five number summary above divides the sample into quarters, with 25% of prices being below the first quartile, $325,000, and 25% of prices above the third quartile, $549,500. Therefore, the middle 50% of prices vary by $224,500. That is, between $325,000 and $549,500. The mean internal area for this sample of residential sales in Coastal City 1 is 148.8 square metres, with the average scatter about this mean value of 47.9 square metres. This is, similar to the median internal area of 146.5. 50% of residential properties sales in the sample had at most three bedrooms, 2 bathroom and 2 garages. The values of the correlation coefficients indicate that there is a strong positive linear relationship between the number of bedrooms and internal area and a weaker positive relationship between internal area and number of bathrooms.

Coastal City 1, State A

Garages Bathrooms Bedrooms

Number of garages, bathrooms and bedrooms are all right skewed. Coastal City 1, State A

House

Unit

100

200

300

400

500

600

700 800 Price ($000)

900

1000

1100

1200

1300

1400

Both house and unit prices are skewed to the right, with house prices generally higher and more varied than unit prices.

Coastal City 1, State A

House

Unit

100

150 200 Internal Area Square Metres

250

300

The internal area of houses in Coastal City 1 is generally higher and more varied than the internal area of units. State A

Coastal City 1

Regional City 1

100

200

300

400

500

600

700 Price ($000)

800

900

1000

1100

1200

1300

Prices are skewed to the right in both locations. Prices in Coastal City 1 are generally higher and more varied than in Regional City 1. State A

Coastal City 1

Regional City 1

100

150

200 Internal Area (Square Metres)

250

300

350

Generally, as measured by internal area, properties sold in Coastal City 1 are generally slightly larger than those in Regional City 1. However, house size is more varied in Regional City 1.

Chapter 4: Basic probability Learning objectives After studying this chapter you should be able to: 1.

recognise basic probability concepts

calculate probabilities of simple, marginal and joint events

calculate conditional probabilities and determine whether events are independent or not

revise probabilities using Bayes’ theorem

use counting rules to calculate the number of possible outcomes

4.1

(a)

Simple events include tossing a head or tossing a tail.

(b)

Joint events include tossing two heads (HH), a head followed by a tail (HT), a tail followed by a head (TH), and two tails (TT).

(c)

Tossing a tail on the first toss.

(a)

Simple events include selecting a red ball.

(b)

Selecting a white ball.

(a)

30/90 = 1/3 = 0.33

(b)

60/90 = 2/3 = 0.67

(c)

10/90 = 1/9 = 0.11

(d)

30 30 10 50 5 + − = = = 0.556 90 90 90 90 9

(a)

60/100 = 3/5 = 0.6

(b)

10/100 = 1/10 = 0.1

(c)

35/100 = 7/20 = 0.35

(d)

60 65 35 90 9 + – = = = 0.9 100 100 100 100 10

4.2

4.3

4.4

4.5

4.6

4.7

4.8

(a)

A priori

(b)

Subjective

(c)

A priori

(d)

Empirical classical

(a)

Mutually exclusive, not collectively exhaustive. ‘An exit poll in an Australian federal election asking voters if they had voterd for Labour, the Coalition or another candidate’ will be mutually exclusive and collectively exhaustive.

(b)

Mutually exclusive, not collectively exhaustive. ‘If respondents were classified by country of manufacture of car owned and used for the majority of their driving, into the categories Australian, American, European, Japanese, other country or none’ will be mutually exclusive and collectively exhaustive. People can own more than one car.

(c)

Mutually exclusive, not collectively exhaustive. ‘People being asked, “Do you currently live in (i) an apartment, (ii) a house or (iii) none of the above?”’ will be mutually exclusive and collectively exhaustive.

(d)

Mutually exclusive, collectively exhaustive.

(a)

The joint probability of mutually exclusive events (as a day is not simultaneously Christmas and Easter, mutually exclusive) is zero.

(b)

The joint probability of mutually exclusive events (being defective and not defective) is zero.

(c)

The joint probability of mutually exclusive events (being a Ford and a Toyota) is zero.

(a)

P(ABC) =

(b)

P(ABC or SBS) =

(c)

P(Not(ABC or SBS)) = P(Seven or Nine or Ten or Other)

1290 = 0.129 10000 1290 + 430 1720 = = 0.172 10000 10000

2850 + 2060 + 1695 + 1675 10000 8280 = 10000 = 0.828

(d)

P(Seven or Nine or Ten) =

2850 + 2060 + 1695 6605 = = 0.6605 10000 10000

P(Not(Seven or Nine or Ten)) = P(ABC or SBS or Other) 1290 + 430 + 1675 = 10000 3395 = = 0.3395 10000

(e)

4.9

4.10

Enjoy clothes shopping

Male

Female

Yes

136

224

360

104

140

Total

240

260

500

Total

(a)

P(enjoys clothes shopping) = 360/500 = 18/25 = 0.72

(b)

P(female and enjoys clothes shopping) = 224/500 = 56/125 = 0.448

(c)

P(female or enjoys clothes shopping) = 396/500 = 99/125 = 0.792

(d)

P(male or female) = 500/500 = 1.00

(a)(i) P(A | B) = 10/30 = 1/3 = 0.33 (ii) P(A | B’) = 20/60 = 1/3 = 0.33 (iii) P(A’ | B’) = 40/60 = 2/3 = 0.67 (b)

4.11

As P(A | B) = P(A) = 1/3, events A and B are statistically independent.

(a)(i) P(A | B) = 10/35 = 2/7 = 0.2857 (ii) P(A’ | B’) = 35/65 = 7/13 = 0.5385 (iii) P(A | B’) = 30/65 = 6/13 = 0.4615 (b)

As P(A | B) = 0.2857 and P(A) = 0.40, events A and B are not statistically independent.

P(A and B) 0.4 1 = = = 0.5 P(B) 0.8 2

4.12

P(A | B) =

4.13

P(A and B) = P(A) P(B) = 0.7 • 0.6 = 0.42

4.14

As P(A and B) = .20 and P(A) P(B) = 0.12, events A and B are not statistically independent.

4.15

10075.4 = 0.5081... 19829.2

(a)

P(female) =

(b)

P(not employed | male) =

(c)

P(female | employed full-time) =

(d)

From parts (a) and (c) P(female)  P(female | employed full-time) so the

277.6 + 90.3 + 2859.4 3227.3 = = 0.3308... 9753.8 9753.8 3001.2 = 0.3617... 8297.2

two events ‘employed full-time’ and ‘female’ are not statistically independent. (e)

P(male and employed full-time) =

(f)

Unemployment rate

5296.0 = 0.2670... 19829.2

P(unemployed | labour force) 483.0 + 220.4 = 8297.2 + 3908.7 + 483.0 + 220.4 703.4 = 12909.3 = 0.05448...  5.45% Female unemployment rate P(unemployed and female | female labour force) 335.5 6014.9 = 0.05577...  5.58%

Male unemployment rate P(unemployed and male | male labour force) 367.9 6894.4 = 0.05336...  5.34%

(g)

Participation rate

P(labour force | population) =

12909.3 = 0.6510...  65.1% 19829.2

Female participation rate P(female labour force | female population) 6014.9 10075.4 = 0.5969...  59.7%

Male participation rate

P(male labour force | male population) 6894.4 9753.8 = 0.7068...  70.7%

4.16

(a)(i) P(subscribe) = 5,600/16,000 = 0.35 (ii) P(high income) = 4,000/16,000 = 0.25 (iii) P(subscribe and high income) 3,200 / 16,000 = 0.2 (iv) P(subscribe | high income) = 3,200 /4,000 = 0.8 (v) P(high income | subscribe) = 3,200 /5,600= 0.5714 (b)

4.17

As P(subscribe | high income) = 0.8, P(subscribe) = 0.35, they are not statistically independent.

Let M = student male and B = student in business faculty Given P(B) = 0.25, P(M | B) = 0.66 and P(M) = 0.52

4.18

4.19

(a)

P(M and B) = P(M | B) P(B) = 0.66 x 0.25 = 0.165

(b)

P(M or B) = P(M) + P(B) – P(M and B) = 0.52 + 0.25 – 0.165= 0.605

(c)

P(B | M) = P(M and B)/ P(M) = 0.165 / 0.52 = 0.3173, approximately 31.7% of male students are in the business faculty.

(a)

P(does not enjoy clothes shopping | female) = 36/260 = 9/65 = 0.1385

(b)

P(male | enjoys clothes shopping) = 136/360 = 17/45 = 0.378

(c)

As P(male | enjoys clothes shopping) = 0.378 and P(male) = 240/500 or 0.48, the two events are not statistically independent.

(a)(i)

P(A) =

750 = 0.75 1000

(ii)

P( B) =

675 = 0.675 1000

(iii)

P( A and B) =

504 = 0.504 1000

(iv) P( A or B ) = P ( A) + P ( B ) -

P( A and B)

750 675 504 + 1000 1000 1000 921 = 1000 = 0.921

(v) P (C ¢) =

4.20

(b)

P( B | A) =

P( A and B) 0.504 = = 0.672 P( A) 0.75

(c)

P( A | B) =

P( A and B) 0.504 = = 0.7466 K P( B) 0.675

(d)

Not statistically independent, P( A | B) ¹ P( A)

(a)

P(S&P finished higher Year) = 55/88 = 0.625

(b)

P(S&P finished higher for the year | higher first 5 days) = 41/59 = 0.6949

(c)

As P(S&P finished higher) ≠ P(S&P finished higher | higher first 5 days), the events ‘first-week performance’ and ‘annual performance’ are not statistically independent.

(d) and (e)

4.21

1000 - 631 369 = = 0.369 1000 1000

(a) (b) (c) (d)

answers vary.

4 3 12 1  = = = 0.0045 52 51 2,652 221 4 8 32 8 P(10 followed by 5 or 6) =  = = = 0.012 52 51 2,652 663 4 4 16 1 P(both queens) =  = = = 0.0059 52 52 2,704 169 16 4 4 16 128 32 P(blackjack) =  +  = = = 0.0483 52 51 52 51 2,652 663 P(both queens) =

4.22

P( B | A) =

P( A | B ) P ( B ) 0.8  0.05 0.04 = = = 0.095 P( A | B) P( B) + P( A | B ') P( B ') ( 0.8  0.05) + ( 0.4  0.95) 0.42

4.23

P( B | A) =

P( A | B ) P( B ) 0.6  0.3 0.18 = = = 0.340 P( A | B) P( B) + P( A | B ') P( B ') ( 0.6  0.3) + ( 0.5  0.7 ) 0.53

4.24

Let D = has disease; T = test positive; D' = does not have disease; T' = test negative Given P(D) = 0.03, P(T | D) = 0.90, P(T | D') = 0.01, P(D') = 0.97, P(T' | D) = 0.10, P(T' | D') = 0.99

(a)

P(D | T) =

P(T | D)P(D) P(T | D)P(D) + P(T | D ')P(D ')

0.9  0.03 ( 0.9  0.03) + (0.01  0.97 )

0.027 0.0367 = 0.7356...

(b)

P(D' | T) = 1 – 0.7356… = 0.2643…

(c)

P(D ' | T ') =

P(T ' | D ')P(D ') P(T ' | D ')P(D ') + P(T ' | D)P(D)

0.99  0.97 ( 0.99  0.97 ) + (0.10  0.03 )

0.9603 0.9633 = 0.9968...

4.25

(d)

P(D | T') = 1 – 0.9968… = 0.0031…

(a)

H = husband watching

W = wife watching

Given P(H) = 0.60 P(W | H) = 0.40 and P(W | H') = 0.30

P(H |W ) = =

(

P(W | H )P(H ) P(W | H )P(H ) + P(W | H ')P(H ') 0.4 ´ 0.6 0.4 ´ 0.6 + 0.3´ 0.4

) (

)

0.24 0.36 = 0.666...

4.26

(b)

P(W) = 0.24 + 0.12 = 0.36

(a)

P(huge success | favourable review) = 0.099/0.459 = 0.2157 P(moderate success | favourable review) = 0.14/0.459 = 0.3050 P(break even | favourable review) = 0.16/0.459 = 0.3486 P(loser | favourable review) = 0.06/0.459 = 0.1307

(b)

P(favourable review) = 0.99(0.1) + 0.7(0.2) + 0.4(0.4) + 0.2(0.3) = 0.459

4.27

Let U = borrower unemployed; D = borrower defaults; U' = borrower not unemployed; D' = borrower does not default Given P(D) = 0.1; P(U | D) = 0.32; P(U' | D) = 0.68; P(D') =0.9; P(U | D') = 0.02; P(U' | D') = 0.98 (a)

P(D U) =

P(U D)P(D) P(U D)P(D) + P(U D)P(D)

(0.32  0.1) (0.32  0.1) + (0.02  0.9) 0.032 = 0.032 + 0.018 0.032 = 0.05 = 0.64

64% of unemployed borrowers default. (b)

5% of borrowers are unemployed, so 95% are not unemployed.

(c)

P(D U) =

P(U D)P(D) (0.98  0.9) = = 0.9284 K P(U) 0.95

Approximately 92.84% of borrowers who are not unemployed do not default. 4.28

4.29

(a)

310 = 59,049

(b)

P(all correct) =

(a)

(30)(30)(30) = 27,000

(b)

P(able to open the bank vault) = (1/30)(1/30)(1/30) = .000037

(c)

In ‘dial combination’, the order of the combination is important whereas order is irrelevant in the mathematical combination expressed by equation (4.14).

1 = 1.69...´10-5 » 0 59049

4.30

(7)(3)(3) = 63

4.31

(8)(4)(3)(3) = 228

4.32

(a)

7! = 5,040 = (7*6*5*4*3*2*1)

(b)

5! = 120

(c)

6! = 720

(d)

4! = 24

4.33

26 ´ 26 ´ 10 ´ 10 ´ 26 ´ 26 = 264 ´ 102 = 45,697,600

4.34

2 letter initials + 3 letter initials = (262 × 102) + (263 × 102) = 1,825,200

4.35

(a)

12 P3 =

(b)

P(win trifecta) =

12! 12! = = 1320 (12 - 3)! 9!

4.36

9C4 =

4.37

(a)

100 C2 =

(b)

P(win ) =

4.38

20C3 =

4!(9−4)!

1 = 7.575K ´ 10- 4 1320

= 126

20! 3!(20−3)!

100! 100! 100 ´ 99 = = = 4950 2!(100 - 2)! 2!98! 2

1 = 2.0202K ´ 10- 4 4950

20! 3!17!

(20)(19)(18) (3)(2)(1)

= 1,140

4.39 to 4.40 Answers will vary. 4.41

With a priori probability, the probability of success is based on prior knowledge of the process involved. With empirical probability, outcomes are based on observed data. Subjective probability refers to the chance of occurrence assigned to an event by a particular individual.

4.42

A simple event can be described by a single characteristic. Joint probability refers to phenomena containing two or more events.

4.43

The general addition rule is used by adding the probability of A and the probability of B and then subtracting the joint probability of A and B.

4.44

Events are mutually exclusive if both cannot occur at the same time. Events are collectively exhaustive if one of the events must occur.

4.45

If events A and B are statistically independent, the conditional probability of event A given B is equal to the probability of A.

4.46

When events A and B are independent, the probability of A and B is the product of the probability of event A and the probability of event B. When events A and B are not independent, the probability of A and B is the product of the conditional probability of event A given event B and the probability of event B.

4.47

Bayes’ theorem uses conditional probabilities to revise the probability of an event in the light of new information.

4.48

In a permutation, the order of the outcomes is important whereas the order is irrelevant in combination.

4.49

(a)

P(Local) =

(b)

P(Another State) =

(c)

P(Another State|Not Local) =

(d)

P(SeasideTown| Local) =

(e)

P(Outside State) =

4.50

151 + 462 613 = = 0.6173... 993 993 228 = 0.2296... 993 228 = 0.6745... 380

151 = 0.2463... 613

250 = 0.2517... 993

Let V = student vaccinated; D = gets disease; V' = student not vaccinated; D' = does not get disease Given P(V) = 0.95; P(V') = 0.05; P(D) = 20/200 = 0.1; P(D') = 0.9 P(D and V) = 11/200 = 0.55; P(D and V') = 9/200 = 0.045

(a)(i)

P(V | D ) =

P(V and D ) 0.055 = = 0.55 P( D ) 0.1

(ii) P( D | V ) =

P(V and D ) 0.055 = = 0.0578K P(V ) 0.95

(iii) P ( D | V ¢) =

P(V ¢ and D ) 0.045 = = 0.9 0.05 P(V ¢)

(b)

Parent is confusing P(V | D) and P(D | V). Vaccination is not 100% effective so during an outbreak some vaccinated children will get the disease, and as 95% of children have been vaccinated this proportion is likely to be more than 50%. However, only 6% of vaccinated children get

the disease compared with 90% of unvaccinated children. So vaccination is effective. 4.51

4.52

(a)

P(win prize) = 3/24 = 0.125

(b)

24P3 =

(c)

3! – right order = 6 – 1 = 5

(d)

24C3 =

24! (24−3)!

24! 3!(24−3)!

24!

= 12,144

21!

24! 3!21!

= 2,044, P(prize) = 1/2024 = 4.940 × 10-4

(a) Plumbed dwelling

into

Not plumbed into dwelling

Total

Rainwater tank

14.2308

12.5692

26.8000

No tank

0.0000

73.2000

73.20000

14.2308

85.7692

100.0000

rainwater

Total

(b)(i) P(rainwater tank and not plumbed into dwelling) = 0.125692 (ii) P(not plumbed into dwelling | rainwater tank) =

12.5692 = 0.469 = 46.9% 26.8

Alternatively, since in 53.1% of dwellings with a rainwater tank it was plumbed into the dwelling, in 46.9% (100 – 53.1%) it was not. (iii) P(no rainwater tank) = 0.732 (c)

P(rainwater tank and plumbed into dwelling) × number of dwellings = 0.142308×2268800 = 322,868

4.53

Female (a)

P(alive at 30) =

99288 = 0.99288 100000

(b)

P(alive at 45) =

97934 = 0.97934 100000

(c)

P(alive at 30 | alive at 20) =

98949 = 0.99658K 99288

(d)

P(alive at 40 | alive at 20) =

98427 = 0.99132 K 99288

(e)

P(dead at 35 | alive at 30) =

(f)

Male P(alive at 30) =

99031 = 0.99031 100000

P(alive at 45) =

4.54

98949 − 98726 223 = = 0.00225K 98949 98949

96649 = 0.96649 100000

P(alive at 30 | alive at 20) =

98312 = 0.99273K 99031

P(alive at 40 | alive at 20) =

97381 = 0.98333K 99031

P(dead at 35 | alive at 30) =

98312 − 97899 413 = = 0.00420K 98312 98312

Bayes’ theorem Let

V = student vaccinated

D = gets diease

V = student not vaccinated

D = does not get diease

Given P(D) = 0.001

P(D) = 0.999

P(V | D) = 0.3

P(V ' | D) = 0.7

P(V | D ') = 0.8 P(V ' | D') = 0.2 (a)

Approximately 0.04% of vaccinated children caught the disease since:

P(V | D)P(D) P(V | D)P(D) + P(V | D ')P(D ') 0.3  0.001 = ( 0.3  0.001) + ( 0.8  0.999 )

P(D | V) =

0.0003 0.7995 = 0.0003752K  0.04%

(b)

Approximately 0.4% of unvaccinated children caught the disease since:

P(V ' | D)P(D) P(V ' | D)P(D) + P(V ' | D ')P(D ') 0.7  0.001 = ( 0.7  0.001) + ( 0.2  0.999 )

P(D | V ') =

0.0007 0.2005 = 0.00349K  0.4% =

4.55

4.56

(c)

Approximately 80% are vaccinated since P(V) = 0.7995

(a)

100C10 =

(b)

100P10 =

(a)

Contingency table

100! 10!(100−10)!

100! (100−10)!

100! 10!90!

100! 90!

= 1.731 ∗ 1013

= 6.3 ∗ 1019

Marital status Education

Single

Not Single

University degree

340

400

No university degree

160

200

100

500

600

P(single or degree) =

(c)

P(degree | single) =

(d)

Gender and educational background not statistically independent since:

P(degree) =

4.57

100 + 400 − 60 440 = = 0.7333K 600 600

(b)

60 = 0.6 = 60% 100

400 = 0.666K  P(degree | single) 600

(a)

P(TV One) = 3,160/10,000 = 0.316

(b)

P(TV2 or TV3) = 4,130/10,000 = 0.413

(c)

P(Prime) = 860/10,000 = 0.086

(d)

P(Not TV One, TV2, TV3) = 2,710/10,000 = 0.271

4.58(a)(i)

P(Male) =

7466 = 0.4888... 15275

(ii)

P(Female AND  65) =

(iii)

P(Child) =

703 + 492 + 327 1522 = = 0.09963... 15275 15275

779 + 1813 2592 = = 0.16968... 15275 15275 410 + 952 1362 = = 0.5254... 779 + 1813 2592

(b)

P(Male | Child) =

(c)

Not statistically independent since P(Male) ≠ P(Male | Child)

(d)

P( 65 | Female) =

4.59 (a)(i)

779 + 1813 2592 = = 0.16968... 15275 15275

P(Unpaid Domestic Work) 6206 + 5396 + 1509 + 694 + 4246 + 6459 + 3524 + 2840 19194 + 20998 30874 = 40192 = 0.7681...

(ii)

P(No Unpaid Domestic Work AND Female) 3929 19194 + 20998 3929 = 40192 = 0.09775...

(iii)

P(Unpaid Domestic Work AND Male) 6206 + 5396 + 1509 + 694 19194 + 20998 13805 = 40192 = 0.34347...

(iv)

P(Unpaid Domestic Work  15 hours Work AND Male) 1509 + 694 19194 + 20998 2203 = 40192 = 0.05481...

(v)

P(No Unpaid Domestic Work AND Male) 5389 = 40192 = 0.13408...

(b)

P(Unpaid Domestic Work | Male) 6206 + 5396 + 1509 + 694 19194 13805 = 19194 = 0.71923... =

(c)

P(Unpaid Domestic Work | Female) 4246 + 6459 + 3524 + 2840 20998 17069 = 20998 = 0.81288...

(d) (e)(i)

P(Unpaid Domestic Work | Male) ≠ P(Unpaid Domestic Work) so not statistically independent.

P(Unpaid Domestic Work  15 hours| Male) 1509 + 694 19194 2203 = 19194 = 0.114775...

(ii)

P(Unpaid Domestic Work < 5 hours| Male) 6206 + 5389 19194 11595 = 19194 = 0.60409...

(f)(i) P(Unpaid Domestic Work  15 hours| Female)

3524 + 2840 20998 6464 = 20998 = 0.303076...

(ii)

P(Unpaid Domestic Work < 5 hours| Female) 4246 + 3929 20998 8175 = 20998 = 0.389322...

(g)

From parts (e) and (f) 60% of mean do less than 5 hours of unpaid domestic work peer work compared with 39% of women. While 30% of women do more than 15 hours of unpaid domestic work compared with 11% of men. From this we can conclude that men do less unpaid domestic work than women. (h)

P(No Unpaid Domestic Work | Male  65 years) 780 + 496 + 273 2469 + 1266 + 504 1549 = 4239 = 0.36541...

(i)

P(No Unpaid Domestic Work | Female  65 years) 639 + 537 + 483 2602 + 1450 + 825 1659 = 4877 = 0.34016...

(j)

P(Unpaid Domestic Work | Male < 35 years) 785 + 802 + 2132 1387 + 1326 + 2844 3719 = 5557 = 0.66924...

(k)

P(Unpaid Domestic Work | Female < 35 years) 830 + 1023 + 2770 1310 + 1313 + 3112 4623 = 5735 = 0.8061...

(l)

While in both age groups women did more unpaid domestic work than men, from parts (i) to (j) the difference between the proportion of of men and women doing unpaid domestic work has decreased from from 13.7% for the under 35’s to 2.5% for 65 years and over. The proportion of those 65 years and over doing unpaid domestic work is less than for those under 35 years. The decrease is greatest for women from 81% to 66%, while for men the decrease is 67% to 64%.

4.60

(a) P

P’

Total

C’

Total

100

P(P’ AND C’) =

34 = 0.34 = 34% 100

(b) (i) (ii)

P(P’ | C ) =

21 = 0.6 35

Alternatively, told 40% of all households with children have pets so 60% = 0.6 of household with children do not have pets. (iii)

P(C | P) =

14 = 0.3111... 45

Approximately 31.1% of households with pets have children.

Case Study - Tasman University Student Survey (a)

Bachelor of Business – other probabilities and contingency tables can be calculated Age and Gender

Age 18 19 20 21 22 23 24 26 Total

Female 0 2 9 13 5 3 0 1 33

Male 1 3 5 9 6 2 3 0 29

Total 1 5 14 22 11 5 3 1 62

Age 18 19 20 21 22 23 24 26 Marginal Probabilities

Joint Probabilities Female Male 0.0000 0.0161 0.0323 0.0484 0.1452 0.0806 0.2097 0.1452 0.0806 0.0968 0.0484 0.0323 0.0000 0.0484 0.0161 0.0000 0.5323

Marginal Probabilities 0.0161 0.0806 0.2258 0.3548 0.1774 0.0806 0.0484 0.0161

0.4677

1.0000

Conditional Probability (Age | Gender) Age Female Male 18 0.0000 0.0345 19 0.0606 0.1034 20 0.2727 0.1724 21 0.3939 0.3103 22 0.1515 0.2069 23 0.0909 0.0690 24 0.0000 0.1034 26 0.0303 0.0000 Conditional Probability (Gender | Age) Age Female Male 18 0.0000 1.0000 19 0.4000 0.6000 20 0.6429 0.3571 21 0.5909 0.4091 22 0.4545 0.5455 23 0.6000 0.4000 24 0.0000 1.0000 26 1.0000 0.0000 Age and Gender are not statistically independent Gender and Major

Major Accounting Economics/Finance Information Systems International Business Management Other Retailing/Marketing Undecided Total

Female 3 7 3 4 4 3 9 0 33

Major Accounting Economics/Finance Information Systems International Business Management Other Retailing/Marketing Undecided Marginal Probabilities

Joint Probabilities Marginal Female Male Probabilities 0.0484 0.0645 0.1129 0.1129 0.0645 0.1774 0.0484 0.0161 0.0645 0.0645 0.0323 0.0968 0.0645 0.0968 0.1613 0.0484 0.0645 0.1129 0.1452 0.0806 0.2258 0.0000 0.0484 0.0484 0.5323 0.4677 1.0000

Male 4 4 1 2 6 4 5 3 29

Total 7 11 4 6 10 7 14 3 62

Conditional Probability ( Gender | Major) Major Female Male Accounting 0.4286 0.5714 Economics/Finance 0.6364 0.3636 Information Systems 0.7500 0.2500 International Business 0.6667 0.3333 Management 0.4000 0.6000 Other 0.4286 0.5714 Retailing/Marketing 0.6429 0.3571 Undecided 0.0000 1.0000 Conditional Probability (Major | Gender) Major Female Male Accounting 0.0909 0.1379 Economics/Finance 0.2121 0.1379 Information Systems 0.0909 0.0345 International Business 0.1212 0.0690 Management 0.1212 0.2069 Other 0.0909 0.1379 Retailing/Marketing 0.2727 0.1724 Undecided 0.0000 0.1034 Gender and major are not statistically independent

Gender and Employment Status

Employment Full-Time Not Employed Part-Time Total

Female 3 6 24 33

Employment Full-Time Not Employed Part-Time Marginal Probabilities

Male 7 3 19 29

Total 10 9 43 62

Joint Probabilities Female Male 0.0484 0.1129 0.0968 0.0484 0.3871 0.3065 0.5323 0.4677

Marginal Probabilities 0.1613 0.1452 0.6935 1.0000

Conditional Probability( Gender | Employment) Employment Female Male Full-Time 0.3000 0.7000 Not Employed 0.6667 0.3333 Part-Time 0.5581 0.4419 Conditional Probability( Employment| Gender) Employment Female Male Full-Time 0.0909 0.2414 Not Employed 0.1818 0.1034 Part-Time 0.7273 0.6552 Gender and Employment Status are not statistically independent Enrolment Status and Postgraduate Study

Status Completed Continuing New Total

Postgraduate Study No Undecided Yes 3 6 5 6 11 15 3 5 8 12 22 28

Status Completed Continuing New Marginal Probabilities

Total 14 32 16 62

Postgraduate Study Joint Probabilities No Undecided Yes 0.0484 0.0968 0.0806 0.0968 0.1774 0.2419 0.0484 0.0806 0.1290 0.1935 0.3548 0.4516

Marginal Probabilities 0.2258 0.5161 0.2581 1.0000

Postgraduate Study Conditional Probability( Status| Postgraduate Study) Status No Undecided Yes Completed 0.2500 0.2727 0.1786 Continuing 0.5000 0.5000 0.5357 New 0.2500 0.2273 0.2857

Postgraduate Study Conditional Probability( Postgraduate Study | Status) Status No Undecided Yes Completed 0.2143 0.4286 0.3571 Continuing 0.1875 0.3438 0.4688 New 0.1875 0.3125 0.5000 Enrolment Status and intention to enroll in Postgraduate Study are not statistically independent Major and Postgraduate Study

Major Accounting Economics/Finance Information Systems International Business Management Other Retailing/Marketing Undecided Total

Postgraduate Study No Undecided Yes 0 2 5 1 4 6 2 1 1 2 2 2 2 3 5 0 3 4 4 6 4 1 1 1 12 22 28

Major Accounting Economics/Finance Information Systems International Business Management Other Retailing/Marketing Undecided Marginal Probabilities

Postgraduate Study Joint Probabilities No Undecided Yes 0.0000 0.0323 0.0806 0.0161 0.0645 0.0968 0.0323 0.0161 0.0161 0.0323 0.0323 0.0323 0.0323 0.0484 0.0806 0.0000 0.0484 0.0645 0.0645 0.0968 0.0645 0.0161 0.0161 0.0161 0.1935 0.3548 0.4516

Total 7 11 4 6 10 7 14 3 62

Marginal Probabilities 0.1129 0.1774 0.0645 0.0968 0.1613 0.1129 0.2258 0.0484 1.0000

Postgraduate Study Conditional Probability( Major| Postgraduate Study) Major No Undecided Yes Accounting 0.0000 0.0909 0.1786 Economics/Finance 0.0833 0.1818 0.2143 Information Systems 0.1667 0.0455 0.0357 International Business 0.1667 0.0909 0.0714 Management 0.1667 0.1364 0.1786 Other 0.0000 0.1364 0.1429 Retailing/Marketing 0.3333 0.2727 0.1429 Undecided 0.0833 0.0455 0.0357 Postgraduate Study Conditional Probability(Postgraduate Study| Major) Major No Undecided Yes Accounting 0.0000 0.2857 0.7143 Economics/Finance 0.0909 0.3636 0.5455 Information Systems 0.5000 0.2500 0.2500 International Business 0.3333 0.3333 0.3333 Management 0.2000 0.3000 0.5000 Other 0.0000 0.4286 0.5714 Retailing/Marketing 0.2857 0.4286 0.2857 Undecided 0.3333 0.3333 0.3333 Major and intention to enroll in Postgraduate Study are not statistically independent Gender and Postgraduate Study

Gender and Computer Preference

Gender and computer preference not statistically independent.

(b)

Master of Business Administration – other probabilities and contingency tables can be calculated Graduate and Undergraduate Majors

Undergraduate Major Biological MBA Major Sciences Business Accounting 0 3 Economics/Finance 1 10 Information Systems 0 0 International Business 0 1 Management 0 5 Other 1 3 Retailing/Marketing 1 3 Total 3 25

Engineering 0 1 0 0 2 0 0 3

Other 2 2 2 1 2 1 3 13

Total 5 14 2 2 9 5 7 44

Undergraduate Major Joint Probabilities Biological MBA Major Sciences Business Accounting 0.0000 0.0682 Economics/Finance 0.0227 0.2273 Information Systems 0.0000 0.0000 International Business 0.0000 0.0227 Management 0.0000 0.1136 Other 0.0227 0.0682 Retailing/Marketing 0.0227 0.0682 Marginal Probabilities 0.0682 0.5682

Engineering 0.0000 0.0227 0.0000 0.0000 0.0455 0.0000 0.0000 0.0682

Other 0.0455 0.0455 0.0455 0.0227 0.0455 0.0227 0.0682 0.2955

Undergraduate Major Conditional Probability(MBA Major | Undergraduate Major) Biological MBA Major Sciences Business Engineering Accounting 0.0000 0.1200 0.0000 Economics/Finance 0.3333 0.4000 0.3333 Information Systems 0.0000 0.0000 0.0000 International Business 0.0000 0.0400 0.0000 Management 0.0000 0.2000 0.6667 Other 0.3333 0.1200 0.0000 Retailing/Marketing 0.3333 0.1200 0.0000

Other 0.1538 0.1538 0.1538 0.0769 0.1538 0.0769 0.2308

Undergraduate Major Conditional Probability (Undergraduate Major | MBA Major ) Biological MBA Major Sciences Business Engineering Accounting 0.0000 0.6000 0.0000 Economics/Finance 0.0714 0.7143 0.0714 Information Systems 0.0000 0.0000 0.0000 International Business 0.0000 0.5000 0.0000 Management 0.0000 0.5556 0.2222 Other 0.2000 0.6000 0.0000 Retailing/Marketing 0.1429 0.4286 0.0000

Other 0.4000 0.1429 1.0000 0.5000 0.2222 0.2000 0.4286

Marginal Probabilities 0.1136 0.3182 0.0455 0.0455 0.2045 0.1136 0.1591 1.0000

MBA and undergraduate majors are not statistically independent.

MBA Major and Gender

MBA Major Accounting Economics/Finance Information Systems International Business Management Other Retailing/Marketing Total

Female 3 6 0 1 4 1 4 19

Major Accounting Economics/Finance Information Systems International Business Management Other Retailing/Marketing Marginal Probabilities

Joint Probabilities Marginal Female Male Probabilities 0.0682 0.0455 0.1136 0.1364 0.1818 0.3182 0.0000 0.0455 0.0455 0.0227 0.0227 0.0455 0.0909 0.1136 0.2045 0.0227 0.0909 0.1136 0.0909 0.0682 0.1591 0.4318 0.5682 1.0000

Male 2 8 2 1 5 4 3 25

Grand Total 5 14 2 2 9 5 7 44

Conditional Probability( Gender | Major) Major Female Male Accounting 0.6000 0.4000 Economics/Finance 0.4286 0.5714 Information Systems 0.0000 1.0000 International Business 0.5000 0.5000 Management 0.4444 0.5556 Other 0.2000 0.8000 Retailing/Marketing 0.5714 0.4286 Conditional Probability(Major | Gender) Major Female Male Accounting 0.1579 0.0800 Economics/Finance 0.3158 0.3200 Information Systems 0.0000 0.0800 International Business 0.0526 0.0400 Management 0.2105 0.2000 Other 0.0526 0.1600 Retailing/Marketing 0.2105 0.1200 Gender and MBA major are not statistically independent.

Employment Status and Gender

Employment Full-Time Not Employed Part-Time Grand Total Employment Full-Time Not Employed Part-Time Marginal Probabilities

Female 12 5 2 19

Male 16 4 5 25

Joint Probabilities Female Male 0.2727 0.3636 0.1136 0.0909 0.0455 0.1136 0.4318 0.5682

Grand Total 28 9 7 44 Marginal Probabilities 0.6364 0.2045 0.1591 1.0000

Conditional Probability( Gender | Employment) Employment Female Male Full-Time 0.4286 0.5714 Not Employed 0.5556 0.4444 Part-Time 0.2857 0.7143 Conditional Probability (Employment| Gender) Employment Female Male Full-Time 0.6316 0.6400 Not Employed 0.2632 0.1600 Part-Time 0.1053 0.2000 Gender and employment status are not statistically independent. MBA Major and Employment Status

MBA Major Accounting Economics/Finance Information Systems International Business Management Other Retailing/Marketing Total

Employment Status Full-Time Not Employed Part-Time 3 2 0 8 3 3 1 0 1 1 0 1 6 1 2 4 1 0 5 2 0 28 9 7

Total 5 14 2 2 9 5 7 44

MBA Major Accounting Economics/Finance Information Systems International Business Management Other Retailing/Marketing Marginal Probabilities

Employment Status Joint Probabilities Full-Time Not Employed Part-Time 0.0682 0.0455 0.0000 0.1818 0.0682 0.0682 0.0227 0.0000 0.0227 0.0227 0.0000 0.0227 0.1364 0.0227 0.0455 0.0909 0.0227 0.0000 0.1136 0.0455 0.0000 0.6364 0.2045 0.1591

Marginal Probabilities 0.1136 0.3182 0.0455 0.0455 0.2045 0.1136 0.1591 1.0000

Employment Status Conditional Probability( MBA Major| Employment Status) MBA Major Full-Time Not Employed Part-Time Accounting 0.1071 0.2222 0.0000 Economics/Finance 0.2857 0.3333 0.4286 Information Systems 0.0357 0.0000 0.1429 International Business 0.0357 0.0000 0.1429 Management 0.2143 0.1111 0.2857 Other 0.1429 0.1111 0.0000 Retailing/Marketing 0.1786 0.2222 0.0000 Employment Status Conditional Probability(Employment Status| MBA Major) MBA Major Full-Time Not Employed Part-Time Accounting 0.6000 0.4000 0.0000 Economics/Finance 0.5714 0.2143 0.2143 Information Systems 0.5000 0.0000 0.5000 International Business 0.5000 0.0000 0.5000 Management 0.6667 0.1111 0.2222 Other 0.8000 0.2000 0.0000 Retailing/Marketing 0.7143 0.2857 0.0000 Employment status and MBA major are not statistically independent Gender and Undergraduate Major

Gender and undergraduate major are not statistically independent. Gender and Computer Preference

Gender and computer preference are not statistically independent Case Study – Safe-As-Houses Real Estate (a)

Regional City 1 State A – other probabilities and contingency tables can be calculated Number of Bedrooms and Type

Regional City 1 Type House Unit Total

1 0 1 1

2 7 30 37

Bedrooms 3 4 39 29 10 0 49 29

5 7 0 7

6 1 0 1

9 1 0 1

Total 84 41 125

Regional City 1

Joint Probabilities Bedrooms Marginal Type 1 2 3 4 5 6 9 Probabilities House 0.0000 0.0560 0.3120 0.2320 0.0560 0.0080 0.0080 0.6720 Unit 0.0080 0.2400 0.0800 0.0000 0.0000 0.0000 0.0000 0.3280 Marginal Probabilities 0.0080 0.2960 0.3920 0.2320 0.0560 0.0080 0.0080 1.0000

Regional City 1 Type House Unit

Conditional Probability (Bedrooms | Type) Bedrooms 1 2 3 4 5 6 9 0.0000 0.0833 0.4643 0.3452 0.0833 0.0119 0.0119 0.0244 0.7317 0.2439 0.0000 0.0000 0.0000 0.0000

Conditional Probability (Type | Bedrooms) Bedrooms 1 2 3 4 5 6 9 0.0000 0.1892 0.7959 1.0000 1.0000 1.0000 1.0000 1.0000 0.8108 0.2041 0.0000 0.0000 0.0000 0.0000

Number of Bedrooms and Bathrooms

Regional City 1 Bathrooms 1 2 3 Total

1 1 0 0 1

2 34 3 0 37

Bedrooms 3 4 38 8 10 20 1 1 49 29

5 1 5 1 7

6 0 1 0 1

9 0 0 1 1

Total 82 39 4 125

Regional City 1

Joint Probabilities Bedrooms Marginal Bathrooms 1 2 3 4 5 6 9 Probabilities 1 0.0080 0.2720 0.3040 0.0640 0.0080 0.0000 0.0000 0.6560 2 0.0000 0.0240 0.0800 0.1600 0.0400 0.0080 0.0000 0.3120 3 0.0000 0.0000 0.0080 0.0080 0.0080 0.0000 0.0080 0.0320 Marginal Probabilities 0.0080 0.2960 0.3920 0.2320 0.0560 0.0080 0.0080 1.0000

Regional City 1 Bathrooms 1 2 3

Regional City 1 Bathrooms 1 2 3 (b)

Conditional Probability (Bedrooms | Bathrooms) Bedrooms 1 2 3 4 5 6 9 0.0122 0.4146 0.4634 0.0976 0.0122 0.0000 0.0000 0.0000 0.0769 0.2564 0.5128 0.1282 0.0256 0.0000 0.0000 0.0000 0.2500 0.2500 0.2500 0.0000 0.2500

Conditional Probability (Bathrooms | Bedrooms) Bedrooms 1 2 3 4 5 6 9 1.0000 0.9189 0.7755 0.2759 0.1429 0.0000 0.0000 0.0000 0.0811 0.2041 0.6897 0.7143 1.0000 0.0000 0.0000 0.0000 0.0204 0.0345 0.1429 0.0000 1.0000

Coastal City 1 State A – other probabilities and contingency tables can be calculated Number of Bedrooms and Type

Coastal City 1 Type House Unit Total

1 0 4 4

2 6 22 28

3 20 23 43

Bedrooms 4 31 4 35

5 10 0 10

6 4 0 4

7 1 0 1

Total 72 53 125

Coastal City 1

Joint Probabilities Bedrooms Marginal Type 1 2 3 4 5 6 7 Probabilities House 0.0000 0.0480 0.1600 0.2480 0.0800 0.0320 0.0080 0.5760 Unit 0.0320 0.1760 0.1840 0.0320 0.0000 0.0000 0.0000 0.4240 Marginal Probabilities 0.0320 0.2240 0.3440 0.2800 0.0800 0.0320 0.0080 1.0000

Coastal City 1 Type House Unit

Conditional Probability (Bedrooms | Type) Bedrooms 1 2 3 4 5 6 7 0.0000 0.0833 0.2778 0.4306 0.1389 0.0556 0.0139 0.0755 0.4151 0.4340 0.0755 0.0000 0.0000 0.0000

Conditional Probability (Type | Bedrooms) Bedrooms 1 2 3 4 5 6 7 0.0000 0.2143 0.4651 0.8857 1.0000 1.0000 1.0000 1.0000 0.7857 0.5349 0.1143 0.0000 0.0000 0.0000

Number of Bedrooms and Bathrooms

Coastal City 1 Bathrooms 1 2 3 4 Total

1 4 0 0 0 4

2 26 2 0 0 28

3 22 20 1 0 43

Bedrooms 4 5 1 0 25 7 9 2 0 1 35 10

6 0 1 1 2 4

7 0 0 0 1 1

Total 53 55 13 4 125

Coastal City 1

Joint Probabilities Bedrooms Marginal Bathrooms 1 2 3 4 5 6 7 Probabilities 1 0.0320 0.2080 0.1760 0.0080 0.0000 0.0000 0.0000 0.4240 2 0.0000 0.0160 0.1600 0.2000 0.0560 0.0080 0.0000 0.4400 3 0.0000 0.0000 0.0080 0.0720 0.0160 0.0080 0.0000 0.1040 4 0.0000 0.0000 0.0000 0.0000 0.0080 0.0160 0.0080 0.0320 Marginal Probabilities 0.0320 0.2240 0.3440 0.2800 0.0800 0.0320 0.0080 1.0000

Coastal City 1 Bathrooms 1 2 3 4 Coastal City 1 Bathrooms 1 2 3 4

Conditional Probability (Bedrooms | Bathrooms) Bedrooms 1 2 3 4 5 6 7 0.0755 0.4906 0.4151 0.0189 0.0000 0.0000 0.0000 0.0000 0.0364 0.3636 0.4545 0.1273 0.0182 0.0000 0.0000 0.0000 0.0769 0.6923 0.1538 0.0769 0.0000 0.0000 0.0000 0.0000 0.0000 0.2500 0.5000 0.2500 Conditional Probability (Bathrooms | Bedrooms) Bedrooms 1 2 3 4 5 6 7 1.0000 0.9286 0.5116 0.0286 0.0000 0.0000 0.0000 0.0000 0.0714 0.4651 0.7143 0.7000 0.2500 0.0000 0.0000 0.0000 0.0233 0.2571 0.2000 0.2500 0.0000 0.0000 0.0000 0.0000 0.0000 0.1000 0.5000 1.0000

Chapter 5: Some important discrete probability distributions Learning outcomes After studying this chapter you should be able to: 1.

recognise and use the properties of a probability distribution

calculate the expected value and variance of a probability distribution

calculate average return and measure risk associated with various investment proposals

identify situations that can be modelled by a binomial distribution and calculate binomial probabilities

identify situations that can be modelled by a Poisson distribution and calculate Poisson probabilities

identify situations that can be modelled by a hypergeometric distribution and calculate hypergeometric probabilities

5.1

PHStat output for Distribution A: Probabilities & Outcomes:

X 0.5 0.2 0.15 0.1 0.05

Y 0 1 2 3 4

Statistics E(X) E(Y) Variance(X) Standard Deviation(X) Variance(Y) Standard Deviation(Y) Covariance(XY) Variance(X+Y) Standard Deviation(X+Y)

1 0 1.5 1.224745 0 0 0 1.5 1.224745

PHStat output for Distribution B:

Probabilities & Outcomes:

0.05

0.1

0.15

0.2

0.5

Statistics E(X)

E(Y)

Variance(X)

1.5

Standard Deviation(X)

1.224745

Variance(Y)

Standard Deviation(Y)

Covariance(XY)

Variance(X+Y)

1.5

Standard d\Deviation(X+Y)

(a)

Distribution A

1.224745

Distribution B

P(X)

XP(X)

P(X)

XP(X)

0.50

0.00

0.05

0.00

0.20

0.10

0.15

0.30

0.15

0.30

0.10

0.30

0.20

0.60

0.05

0.20

0.50

2.00

1.00

3.00

 = 1.00

 = 3.00

(b)

Distribution A X

(X–  )2

P(X)

(X–  )2P(X)

(–1)2

0.50

(0)

0.20

0.00

(1)2

0.15

(2)

0.10

0.40

(3)2

0.05

0.45

 2=

1.50

 = (X – )2 P(X) = 1.2247… Distribution B X

(X–  )2

P(X)

(X–  )2P(X)

(–3)2

0.05

0.45

(–2)2

0.10

0.40

(–1)2

0.15

(0)2

0.20

0.00

(1)2

0.50

 =

1.50

 = (X – )2 P(X) = 1.2247… (c)

Distribution A: the distribution is right skewed. Distribution B: the distribution is left skewed. The means are different but the variances are the same.

5.2

(a)

Not valid probability distribution. P(2) = –0.1 is not a valid probability.

(b)

Not valid probability distribution. Sum of probabilities not equal to one since: P(X) = 0.1 + 0.2 + 0.3 + 0.3 = 0.9

(c)

Valid probability distribution. Sum of probabilities is one since: P(X) = 0.50 + 0.25 + 0.25 = 1.0

(b)

Not valid probability distribution. Sum of probabilities not equal to one since: P(X) = 0.2 + 0.1 + 0.4 + 0.5 = 1.1

5.3

(a)

Probability distribution is given below:

P(X)

XP(X)

X2P(X)

0 1 2 3 4 5 6 7 8 9 10 11

40 100 142 66 36 30 26 20 16 14 8 2

0.080 0.200 0.284 0.132 0.072 0.060 0.052 0.040 0.032 0.028 0.016 0.004 Sum

0.000 0.200 0.568 0.396 0.288 0.300 0.312 0.280 0.256 0.252 0.160 0.044 3.056

0.000 0.200 1.136 1.188 1.152 1.500 1.872 1.960 2.048 2.268 1.600 0.484 15.408

(b)

Mean:  =  XP(X) = 3.056

(c)

Variance 2 =  X2P(X) − 2 = 15.408 − 3.0562 = 6.068864 Standard deviation  =

6.068864 = 2.4635...

5.4

P(X)

XP(X)

X2P(X)

0 1 2 3 4 5 6

0.32 0.35 0.18 0.08 0.04 0.02 0.01 Sum

0.00 0.35 0.36 0.24 0.16 0.10 0.06 1.27

0.00 0.35 0.72 0.72 0.64 0.50 0.36 3.29

(a)

Mean:  =  XP(X) = 1.27

(b)

Variance 2 =  X2P(X) −2 = 3.29 − 1.272 = 1.6771 Standard deviation  =

1.6771 = 1.2950...

5.5

Let HHi = heads on ith spin

TTi = tails on ith spin

HTi = odds on ith spin Spinning coins independent of each other and each spin so:

P(HHi ) =

1 4

P(TT) = i

1 4

P(HT) = i

1 2

1 P(HT1 and HT2 and K andHT) = i 2  

i−1

1 1 P(HHi |HT1 and HT2 and K andHTi−1) =    4 2 (a) and (b)

5.6

Win

Probability

2 3 4 1  1 1   1  1    1  1    1  1   31 +    +     +     +     = 4  4 2   4  2    4  2    4  2   64

- $1

2 3 4 5 1  1 1   1  1    1  1    1  1    1  33 +    +     +     +     +   = 4  4 2   4  2    4  2    4  2    2  64

(c)

33  2  31   E(X) = 1  +  −1  = − = −0.03125 64  64  64  

(a)

E(X) = (0.4)($100) + (0.6)($200) = $160 E(Y) = (0.4)($200) + (0.6)($100) = $140

(b)

(

) (

)

2 X = 1002  0.4 + 2002  0.6 − 1602 = 28000 − 25600 = 2400

X = 2400 = 48.9897...

(

) (

)

2 Y = 2002  0.4 + 1002  0.6 − 1402 = 22000 − 19600 = 2400

X = 2400 = 48.9897... (c) 

= (100  200  0.4 ) + (200  100  0.6 ) − 160  140 = 20000 − 22400 = −2400

(d)

E(X + Y) = E(X) + E(Y) = $160 + $140 = $300

5.7

(a)

E(X) = (0.2)(-100)+(0.4)(50)+(0.3)(200)+(0.1)(300) = $90 E(Y) = (0.2)(50)+(0.4)(30)+(0.3)(20)+(0.1)(20) = $30

(b)

 X = (0.2)(−100 − 90) 2 + (0.4)(50 − 90) 2 + (0.3)(200 − 90) 2 + (0.1)(300 − 90)2 = 15900 = 126.10

 Y = (0.2)(50 – 30) 2 + (0.4)(30 – 30)2 + (0.3)(20 – 30)2 + (0.1)(20 – 30)2 = 120 = 10.95 (c)

 XY

= (0.2)( –100 – 90)(50 – 30) + (0.4)(50 – 90)(30 – 30) + (0.3)(200 – 90)(20 – 30) + (0.1)(300 – 90)(20 – 30) = –1300

(d)

E(X + Y) = E(X) + E(Y) = $90 + $30 = $120 PHStat output: Probabilities & Outcomes:

Weight Assigned to X

0.2

-100

0.4

0.3

200

0.1

300

0.5

Statistics E(X)

E(Y)

Variance(X) Standard Deviation(X)

15900 126.0952

Variance(Y) Standard Deviation(Y)

120 10.95445

Covariance(XY)

-1300

Variance(X+Y)

13420

Standard Deviation(X+Y)

115.8447

Portfolio Management Weight Assigned to X

0.5

Weight Assigned to Y

0.5

Portfolio Expected Return

Portfolio Risk

5.8

(a)

57.92236

E(P) = (0.4)($50) + (0.6)($100) = $80

5.9

5.10

(b)

 P = (.4) 2 (9000) + (.6) 2 (15000) + 2(.4)(.6)(7500) = 102.18

(a)

E(total time) = E(time waiting) + E(time served) = 4 + 5.5 = 9.5 minutes

(b)

 (total time) =

1.22 + 1.52 = 1.9209 minutes

(a)(i) E(P) = 0.3(105) + 0.7(35) = $56 𝜎𝑝 = √(0.3)2 (14725) + (0.7)2 (11025) + 2(0.3)(0.7)(−12675) = $37.47 (ii) E(P) = 0.7(105) + 0.3(35) = $ 84 𝜎𝑝 = √(0.7)2 (14,725) + (0.3)2 (11,025) + 2(0.7)(0.3)(−12675) = $53.70 (b)

5.11

Investing 50% in each fund will yield the lowest risk. This will be the investment recommendation if you are a risk-averse investor.

PHStat output for (a)–(c): Covariance Analysis Probabilities & Outcomes:

0.1

-100

0.3

150

0.3

-20

0.3

150

-100

Statistics E(X)

E(Y)

Variance(X)

6189

Standard Deviation(X) Variance(Y)

78.6702 9924

Standard Deviation(Y)

99.61928

Covariance(XY)

-6306

Variance(X+Y)

3501

Standard Deviation(X+Y) (a)(i) E(X) =

 X P ( X ) = 59 i

i =1

E(Y) =

59.16925

Y P (Y ) = 14 i =1

(ii)

X =

  X − E ( X ) P ( X ) = 78.6702 2

i =1

Y =

 Y − E (Y ) P (Y ) = 99.62 2

i =1

(iii)

 XY =   X i − E ( X ) Yi − E (Y ) P ( X iYi ) =

6,306

i =1

(b)

5.12

Share X gives the investor a lower standard deviation while yielding a higher expected return, so the investor should select share X.

(a)(i) PHStat2 output: Probabilities & Outcomes:

Weight Assigned to X

0.1

–100

0.3

150

0.3

-20

0.3

150

–100

0.3

Statistics E(X)

E(Y)

Variance(X)

6189

Standard Deviation(X)

78.6702

Variance(Y)

9924

Standard Deviation(Y)

99.61928

Covariance(XY)

–6306

Variance(X+Y)

3501

Standard Deviation(X+Y)

59.16925

Portfolio management Weight Assigned to X

0.3

Weight Assigned to Y

0.7

Portfolio Expected Return Portfolio Risk E(P) = $27.5

 P = 52.64

27.5 52.64266

CV =

P 52.64 = (100%) = 191.42% E ( P ) 27.5

(ii)

PHStat output: Probabilities & Outcomes:

Weight Assigned to X

0.1

–100

0.3

150

0.3

-20

0.3

150

-100

0.5

Statistics E(X)

E(Y)

Variance(X)

6189

Standard Deviation(X)

78.6702

Variance(Y)

9924

Standard Deviation(Y)

99.61928

Covariance(XY)

-6306

Variance(X+Y)

3501

Standard Deviation(X+Y)

59.16925

Portfolio Management Weight Assigned to X

0.5

Weight Assigned to Y

0.5

Portfolio Expected Return Portfolio Risk

E(P) = $36.5

 P = 29.59

36.5 29.58462

CV =

P 29.59 = (100%) = 81.07% E ( P ) 36.5

(iii)

PHStat output:

Probabilities & Outcomes:

P 0.1 0.3 0.3 0.3 0.7

Weight Assigned to X Statistics E(X) E(Y) Variance(X)

59 14 6189

Standard Deviation(X)

78.6702

Variance(Y)

99.61928

Covariance(XY) Variance(X+Y) Standard Deviation(X+Y) Portfolio Management Weight Assigned to X Weight Assigned to Y

-6306 3501 59.16925

Portfolio Expected Return

45.5

Portfolio Risk

(b)

5.13

 P = 35.74

-100 0 80 150

50 150 -20 -100

9924

Standard Deviation(Y)

E(P) = $45.5

0.7 0.3

35.73863

CV =

P

E ( P)

35.74 (100%) = 78.55% 45.5

Based on the results of (i)–(iii), you should recommend a portfolio with 70% of stock X and 30% of stock Y because it has the lowest risk per unit average return.

Let X = corporate bond fund; Y = common share fund (a)(i)

E(X) = $77; E(Y) = $97

(ii) σX = 39.76; σY = 108.95 (iii) σXY = 4,161 (b)

Common share fund gives the investor a higher expected return than corporate bond fund, but also has a higher standard deviation (about 2.5 times higher) which reflects the higher risk. An investor should carefully weigh the increased risk.

5.14

(a)(i)

E(P) = $91; σP = 87.79 CV = (σP/E(P))*100 = (87.79/91)*100 = 96.47%

(ii) E(P) = $87; σP = 73.78 CV = (σP/E(P))*100 = (73.78/87)*100 = 84.80% (iii) E(P) = $83; σP = 59.92 CV = (σP/E(P))*100 = (59.92/83)*100 = 72.19% (b)

5.15

5.16

Based on the results of (a)–(c), you should recommend a portfolio with 70% of corporate bonds and 30% of common shares because it has the lowest risk per unit average return.

(a)

0.5997

(b)

0.0016

(c)

0.0439

(d)

0.4018

PHStat output: Binomial Probabilities Data Sample Size

Probability of an Event of Interest

0.4

Statistics Mean

Variance Standard Deviation

1.2 1.09544

Binomial Probabilities Table X

P(X)

P(<=X)

0 0.07776 0.07776

P(X = 4) = 0.0768

(b)

P(X  3) = 0.9130

(c)

P(X < 2) = 0.3370

P(>X)

P(>=X)

0 0.92224

0.2592 0.33696 0.07776 0.66304 0.92224

0.3456 0.68256 0.33696 0.31744 0.66304

0.2304 0.91296 0.68256 0.08704 0.31744

0.0768 0.98976 0.91296 0.01024 0.08704

5 0.01024 (a)

P(<X)

1 0.98976

0 0.01024

(d)

P(X > 1) = 0.6630

5.17

PHStat output for (a): Data Sample Size

Probability of an Event of Interest

0.1

Statistics Mean

0.4

Variance

0.36

Standard Deviation

0.6

PHStat output for (b): Data Sample Size

Probability of an Event of Interest

0.4

Statistics Mean

1.6

Variance

0.96

Standard Deviation

0.979796

PHStat output for (c): Data Sample Size

Probability of an Event of Interest

0.8

Statistics Mean Variance Standard Deviation

4 0.8 0.894427

PHStat output for (d): Data Sample Size

Probability of an Event of Interest

0.5

Statistics Mean

1.5

Variance Standard Deviation

0.75 0.866025

Mean Standard Deviation (a)

0.40

0.60

(b)

1.60

0.980

(c)

4.00

0.894

(d)

1.50

0.866

5.18

Given p = 0.5 and n = 5, P(X = 5) = 0.0312.

5.19

Given p = 0.6 and n = 5, (a)(i) P(X = 5) = 0.0778 (ii) P(X ≥ 3) = 0.6826 (iii) P(X < 2) = 0.0870 (b)(i) P(X = 5) = 0.3277 (ii) P(X ≥ 3) = 0.9421 (iii) P(X < 2) = 0.0067

5.20

PHStat output: Data

Sample Size

Probability of an Event of Interest

0.25

Statistics Mean

1.25

Variance

0.9375

Standard Deviation

0.968246

Binomial Probabilities Table X

P(X)

P(<=X)

P(<X)

P(>X)

P(>=X)

0.237305

0.762695

0.395508

0.632813

0.237305

0.367188

0.762695

0.263672

0.896484

0.632813

0.103516

0.367188

0.087891

0.984375

0.896484

0.015625

0.103516

0.014648

0.999023

0.984375

0.000977

0.015625

0.000977

0.999023

0.000977

If p = 0.25 and n = 5, (a)

P(X = 5) = 0.0010

(b)

P(X  4) = P(X = 4) + P(X = 5) = 0.0146 + 0.0010 = 0.0156

(c)

P(X = 0) = 0.2373

(d)

P(X  2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.2373 + 0.3955 + 0.2637 = 0.8965

5.21

Given p = 0.90 and n = 3, (a)(i) 𝑃(𝑋 = 3) = (ii) 𝑃(𝑋 = 0) =

𝑛! 𝑋!(𝑛−𝑋)!

𝑝 𝑥 (1 − 𝑝)𝑛−𝑥 =

𝑛! 𝑋!(𝑛−𝑋)!

𝑝 𝑥 (1 − 𝑝)𝑛−𝑥 =

3! 3!0!

(0.9)3 (0.1)0 = 0.729

3! 0!3!

(0.9)0 (0.1)3 = 0.001

(iii) 𝑃(𝑋 ≥ 2) = 𝑃(𝑋 = 2) + 𝑃(𝑋 = 3) = (b)

3! 3! (0.9)2 (0.1)1 + (0.9)3 (0.1)0 = 0.972 2! 1! 3! 0!

E(X) = np = 3(0.9) = 2.7 𝜎𝑥 = √𝑛𝑝(1 − 𝑝) = √3(0.9)(0.1) = 0.5196

5.22

Let X = number of major prizes won in week . Then X is binomial with n = 5 and

1 = 0.05 . 20

(a)(i)

P( X = 0) =

5! (0.05)0 (1 - 0.05)5- 0 = 0.77378K 0!(5 - 5)!

(ii)

P( X = 1) =

5! (0.05)1 (1 - 0.05)5- 1 = 0.20362 K 1!(5 - 1)!

(iii)

P( X = 2) =

5! (0.05)2 (1 - 0.05)5- 2 = 0.02143K 2!(5 - 2)!

P( X £ 2) = P( X = 0) + P( X = 1) + P( X = 2) = 0.77378K + 0.20362K + 0.02143K = 0.9988K (iv) (b)

P( X ³ 3) = 1 - P( X £ 2) = 1 - 0.9988K = 0.0011K E ( X ) = 5´ 0.05 = 0.25 s =

(c)

5´ 0.05´ 0.95 = 0.48733K

100,000 ´ E ( X ) = 100,000 ´ 0.25 = $25,000 per week.

5.23

Partial PHStat output: Binomial Probabilities Data Sample Size

Probability of an Event of Interest

0.05

Statistics Mean

Variance

0.95

Standard Deviation

0.974679

Binomial Probabilities Table X

P(X)

P(<=X)

0 0.358486 0.358486 1 0.377354

5.24

Mean = 1

(b)

P(X = 0) = 0.3585

(c)

P(X = 1) = 0.3774

(d)

P(X  2) = 0.2642

P(>X)

P(>=X)

0 0.641514

0.73584 0.358486

2 0.188677 0.924516 (a)

P(<X)

0.26416

0.641514

0.73584 0.075484

0.26416

Standard deviation = 0.9747

Let X = number of patients drug effective. Then X is binomial with n = 20 and p = 0.9. (a)

The 90% figure is most likely obtained by examining patient records or drug company statistics from drug trials, hence is an empirical classical probability.

(b)(i) P(X < 5) ≈ 0 (ii) P(X ≥ 10) ≈ 1 (iii) P(X = 20) ≈ 0.1216

5.25

(a)

Partial PHStat output: Poisson Probabilities

Data Average/Expected Number of Successes:

2.5

Poisson Probabilities Table X

P(X)

P(<=X)

P(<X)

P(>X)

P(>=X)

0.256516

0.543813

0.287297

0.456187

0.712703

Using the equation, if  = 2.5, P ( X = 2) =

(b)

e −2.5  (2.5) 2 = 0.2565 2!

Partial PHStat output:

Poisson Probabilities Data Average/Expected Number of Successes:

Poisson probabilities table X

P(X)

P(<=X)

P(<X)

P(>X)

P(>=X)

0.139587

0.592547

0.452961

0.407453

0.547039

If  = 8.0, P(X = 8) = 0.1396 (c)

Partial PHStat output:

Poisson Probabilities Data Average/Expected Number of Successes:

0.5

Poisson Probabilities Table X

P(X)

P(<=X)

P(<X)

P(>X)

P(>=X)

0.606531

0.000000

0.393469

1.000000

0.303265

0.909796

0.606531

0.090204

0.393469

If  = 0.5, P(X = 1) = 0.3033

(d)

Partial PHStat output:

Poisson Probabilities Data Average/Expected Number of Successes:

3.7

Poisson Probabilities Table X 0

P(X)

P(<=X)

P(<X)

P(>X)

P(>=X)

0.024724 0.024724 0.000000 0.975276 1.000000

If  = 3.7, P(X = 0) = 0.0247 5.26

If  = 2.0, P(X  2) = 1 – [P(X = 0) + P(X = 1)] = 1 – [0.1353 + 0.2707]

(a)

= 0.5940 (b)

Partial PHStat output: Poisson Probabilities Data

Average/Expected Number of Successes:

Poisson Probabilities Table X

P(X)

P(<=X)

P(<X)

P(>X)

P(>=X)

0.000335

0.000000

0.999665

1.000000

0.002684

0.003019

0.000335

0.996981

0.999665

0.010735

0.013754

0.003019

0.986246

0.996981

0.028626

0.042380

0.013754

0.957620

0.986246

If  = 8.0, P(X  3) = 1 – [P(X = 0) + P(X = 1) + P(X = 2)] = 1 – [0.0003 + 0.0027 + 0.0107] = 1 – 0.0137 = 0.9863 (c)

Partial PHStat output:

Poisson Probabilities Data Average/Expected Number of Successes:

0.5

Poisson Probabilities Table X

P(X)

P(<=X)

P(<X)

P(>X)

P(>=X)

0.606531

0.000000

0.393469

1.000000

0.303265

0.909796

0.606531

0.090204

0.393469

If  = 0.5, P(X  1) = P(X = 0) + P(X = 1) = 0.6065 + 0.3033 = 0.9098 Copyright © 2019 Pearson Australia (a division of Pearson Australia Group Pty Ltd) 9781488617249, Berenson, Basic Business Statistics 5e

(d)

Partial PHStat output: Poisson Probabilities Data

Average/Expected Number of Successes:

Poisson Probabilities Table X

P(X)

P(<=X)

P(<X)

P(>X)

P(>=X)

0.018316

0.000000

0.981684

1.000000

0.073263

0.091578

0.018316

0.908422

0.981684

If  = 4.0, P(X  1) = 1 – P(X = 0) = 1 – 0.0183 = 0.9817 (e)

Partial PHStat output:

Poisson Probabilities Data Average/Expected Number of Successes:

Poisson Probabilities Table X

P(X)

P(<=X)

P(<X)

P(>X)

P(>=X)

0.006738

0.000000

0.993262

1.000000

0.033690

0.040428

0.006738

0.959572

0.993262

0.084224

0.124652

0.040428

0.875348

0.959572

0.140374

0.265026

0.124652

0.734974

0.875348

If  = 5.0, P(X  3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0067 + 0.0337 + 0.0842 + 0.1404 = 0.2650

5.27

PHStat output for (a)–(d): Poisson Probabilities Table X

P(X)

P(<=X)

P(<X)

P(>X)

P(>=X)

0.006738

0.000000

0.993262

1.000000

0.033690

0.040428

0.006738

0.959572

0.993262

0.084224

0.124652

0.040428

0.875348

0.959572

0.140374

0.265026

0.124652

0.734974

0.875348

0.175467

0.440493

0.265026

0.559507

0.734974

0.175467

0.615961

0.440493

0.384039

0.559507

0.146223

0.762183

0.615961

0.237817

0.384039

0.104445

0.866628

0.762183

0.133372

0.237817

0.065278

0.931906

0.866628

0.068094

0.133372

0.036266

0.968172

0.931906

0.031828

0.068094

0.018133

0.986305

0.968172

0.013695

0.031828

0.008242

0.994547

0.986305

0.005453

0.013695

0.003434

0.997981

0.994547

0.002019

0.005453

0.001321

0.999302

0.997981

0.000698

0.002019

0.000472

0.999774

0.999302

0.000226

0.000698

0.000157

0.999931

0.999774

0.000069

0.000226

0.000049

0.999980

0.999931

0.000020

0.000069

0.000014

0.999995

0.999980

0.000005

0.000020

0.000004

0.999999

0.999995

0.000001

0.000005

0.000001

1.000000

0.999999

0.000000

0.000001

0.000000

1.000000

0.000000

Given  =5.0, (a)

P(X = 1) = 0.0337

(b)

P(X < 1) = 0.0067

(c)

P(X > 1) = 0.9596

(d)

P(X  1) = 0.0404

5.28

Portion of PHStat output: Data Average/Expected Number of Successes:

Poisson Probabilities Table

(a)

P(X)

P(<=X)

P(<X)

P(>X)

P(>=X)

0.002479

0.000000

0.997521

1.000000

0.014873

0.017351

0.002479

0.982649

0.997521

0.044618

0.061969

0.017351

0.938031

0.982649

0.089235

0.151204

0.061969

0.848796

0.938031

0.133853

0.285057

0.151204

0.714943

0.848796

(b) 0.160623

0.445680 (a) 0.285057

0.554320 (c) 0.714943

0.160623

0.606303

0.445680

0.393697

0.554320

0.137677

0.743980

0.606303

0.256020

0.393697

0.103258

0.847237

0.743980

0.152763

0.256020

0.068838

0.916076

0.847237

0.083924

0.152763

0.041303

0.957379

0.916076

0.042621

0.083924

0.022529

0.979908

0.957379

0.020092

0.042621

0.011264

0.991173

0.979908

0.008827

0.020092

0.005199

0.996372

0.991173

0.003628

0.008827

0.002228

0.998600

0.996372

0.001400

0.003628

0.000891

0.999491

0.998600

0.000509

0.001400

0.000334

0.999825

0.999491

0.000175

0.000509

0.000118

0.999943

0.999825

0.000057

0.000175

P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) =

e −6 ( 6 )

e−6 ( 6 )

e −6 ( 6 ) 2!

e −6 ( 6 )

= 0.002479 + 0.014873 + 0.044618 + 0.089235 + 0.133853 = 0.2851

e −6 ( 6 )

(b)

P(X = 5) =

(c)

P(X  5) = 1 – P(X < 5) = 1 – 0.2851 = 0.7149

(d)

= 0.1606

P(X = 4 or X = 5) = P(X = 4) + P(X = 5) =

e −6 ( 6 ) 4!

e −6 ( 6 ) 5!

= 0.2945

5.29

Partial PHStat2 output: Poisson Probabilities Data Average/Expected Number of Successes:

Poisson Probabilities Table X

P(X)

P(<=X)

P(<X)

P(>X)

P(>=X)

0.089235

0.151204

0.061969

0.848796

0.938031

If  = 6.0, P(X  3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0025 + 0.0149 + 0.0446 + 0.0892 = 0.1512 n × P(X  3) = 100×(0.1512) = 15.12, so 15 or 16 cookies will probably be discarded.

5.30

Let X = number of floods in one year . Then X is Poisson with l =

(a)

(i) P ( X = 1) =

(ii) (b)

3 = 0.3 10

e- 0.3 0.31 = 0.2222 K 1! e- 0.3 0.30 = 1 - 0.7408K = 0.2591K 0!

P( X ³ 1) = 1 - P( X = 0) = 1 -

Let Y = number of floods in three years . Then Y is Poisson with

l = 0.3´ 3 = 0.9 (i) P(Y = 1) =

e- 0.9 0.91 = 0.3659K 1!

(ii) P(Y ³ 1) = 1 - P(Y = 0) = 1 -

5.31

(a)

λ = 0.2 0.0176

e- 0.9 0.90 = 1 - 0.4065K = 0.5934 K 0!

P(X ≥ 2) = 1 – P(X = 0) + P(X = 1) = 1 – (0.8187 + 0.1637) =

Partial PHStat output: Poisson Probabilities Data Average/Expected Number of Successes:

0.2

Poisson Probabilities Table

(b)

P(X)

P(<=X)

P(<X)

P(>X)

P(>=X)

0.016375

0.998852

0.982477

0.001148

0.017523

If there are 0.2 flaws per metre on average, then there are on average λ = 10 × 0.2 = 2 flaws in a 10-metre roll. P(X ≥ 1) = 1 – P(X = 0) = 1 – 0.1353 = 0.8647

(c)

λ = 50 × 0.2 = 10 flaws on average in a 15-metre roll. P(5 ≤ X ≤ 15) = 0.9220

Partial PHStat output: Poisson Probabilities Data Average/Expected Number of Successes:

Poisson Probabilities Table

5.32

(a)

P(X)

P(<=X)

P(<X)

P(>X)

P(>=X)

0.037833

0.067086

0.029253

0.932914

0.970747

0.063055

0.130141

0.067086

0.869859

0.932914

0.090079

0.220221

0.130141

0.779779

0.869859

0.112599

0.332820

0.220221

0.667180

0.779779

0.125110

0.457930

0.332820

0.542070

0.667180

0.125110

0.583040

0.457930

0.416960

0.542070

0.113736

0.696776

0.583040

0.303224

0.416960

0.094780

0.791556

0.696776

0.208444

0.303224

0.072908

0.864464

0.791556

0.135536

0.208444

0.052077

0.916542

0.864464

0.083458

0.135536

0.034718

0.951260

0.916542

0.048740

0.083458

For the number of phone calls received in a 1-minute period to be distributed as a Poisson random variable, we need to assume that (i) the

probability that a phone call is received in a given 1-minute period is the same for all the other 1-minute periods, (ii) the number of phone calls received in a given 1-minute period is independent of the number of phone calls received in any other 1-minute period, and (iii) the probability that two or more phone calls received in a time period approaches zero as the length of the time period becomes smaller. (b)(i) λ = 0.4, P(X = 0) = 0.6703 (ii) λ = 0.4, P(X  3) = 0.0079

5.33

(c)

λ = 0.4, P(X  4) = 0.999939. A maximum of four phone calls will be received in a 1-minute period 99.99% of the time.

(a)

PHStat output: Hypergeometric Probabilities Data Sample Size

No. of Successes in Population

Population Size

Hypergeometric Probabilities Table X

P(X)

0.238095

 5 10 − 5  5  4  3! 5  4!    3 4 − 3  3! 2 1  4!1! 5 P ( X = 3) =   = = = 0.2381 10  9  8  7  6! 3  7 10    6! 4  3  2 1 4  (b)

PHStat output: Hypergeometric Probabilities Data Sample Size

No. of Successes in Population

Population Size

Hypergeometric Probabilities Table X

P(X)

0.2

0.6

0.2

 3 6 – 3 3 2! 3!       1  4 –1 1 P(X = 1) = = 2!1 3!0! = = 0.2 6 5  4!  6 5   4!21  4 (c)

Partial PHStat output: Hypergeometric Probabilities Data Sample Size

No. of Successes in Population

Population Size

Hypergeometric Probabilities Table X

P(X)

0.159091

 3  12 – 3 3! 9 8 7 6  5!       0  5 – 0  3!0! 5!4  3 2 1 7 P(X = 0) = = = = 0.1591 12 1110  9 8 7! 12   44   7!5  4 3 2 1 5 (d)

Partial PHStat output: Hypergeometric Probabilities Data Sample Size

No. of Successes in Population

Population Size

Hypergeometric Probabilities Table X

P(X)

0.008333

 3  10 – 3  3! 7!       3 3 – 3     3 !  0 ! 7!0! = 1 = 0.0083 P( X = 3) = = 10  9  8  7! 120 10    7!3  2  1 3

5.34

(a)

=

nA 4  5 = =2 N 10

=

nA ( N − A) N − n = 0.8165 N2 N −1

5.35

nA ( N − A) N − n = 0.6325 N2 N −1

(b)

=

nA 4  3 = =2 6 N

(c)

=

nA 5  3 = = 1.25 12 N

=

nA ( N − A) N − n = 0.7724 N2 N −1

(d)

=

nA 3  3 = = 0.9 N 10

=

nA ( N − A) N − n = 0.7 N2 N −1

=

(a)(i) Partial PHStat2 output: Hypergeometric Probabilities Data Sample Size No. of Successes in Population Population Size

6 25 100

Hypergeometric Probabilities Table X

P(X)

0.168918

0.361968

0.305888

0.130286

0.029448

0.003343

0.000149

If n = 6, A = 25, and N = 100, P(X  2) = 1 – [P(X = 0) + P(X = 1)]

 25 100 − 25   25 100 − 25        0  6 − 0  1  6 − 1  ] =1–[ + 100  100      6  6  = 1 – [0.1689 + 0.3620] = 0.4691

(ii)

Partial PHStat output: Hypergeometric Probabilities Data Sample Size No. of Successes in Population Population Size

6 30 100

Hypergeometric Probabilities Table X

P(X)

0.109992

0.304593

0.33459

0.186438

0.05552

0.008368

0.000498

If n = 6, A = 30, and N = 100, P(X  2) = 1 – [P(X = 0) + P(X = 1)]

 30 100 − 30   30 100 − 30        0  6 − 0  1  6 − 1  ] =1–[ + 100  100      6  6  = 1 – [0.1100 + 0.3046] = 0.5854

(iii)

Partial PHStat output: Hypergeometric Probabilities Data Sample Size

No. of Successes in Population

Population Size

100

Hypergeometric Probabilities Table X

P(X)

0.729085

0.243028

0.026706

0.001161

1.87E-05

7.97E-08

If n = 6, A = 5, and N = 100, P(X  2) = 1 – [P(X = 0) + P(X = 1)]

 5 100 − 5   5 100 − 5         0  6 − 0  + 1  6 − 1  ] =1–[ 100  100      6  6  = 1 – [0.7291 + 0.2430] = 0.0279

(iv)

Partial PHStat output: Hypergeometric Probabilities Data Sample Size

No. of Successes in Population Population Size

10 100

Hypergeometric Probabilities Table X

P(X)

0.522305

0.368686

0.096458

0.011826

0.000706

1.9E-05

1.76E-07

If n = 6, A = 10, and N = 100, P(X  2) = 1 – [P(X = 0) + P(X = 1)]

10 100 − 10  10 100 − 10        0  6 − 0  1  6 − 1   =1–[ + ] 100  100      6  6  = 1 – [0.5223 + 0.3687] = 0.1090 (b)

The probability that the entire group will be audited is very sensitive to the true number of returns with errors in the population. If the true number is very low (A = 5), the probability is very low (0.0279). When the true number is increased by a factor of six (A = 30), the probability the group will be audited increases by a factor of almost 21 (0.5854).

5.36

(a)

Partial PHStat output: Hypergeometric Probabilities Data Sample Size

No. of Successes in Population

Population Size

Hypergeometric Probabilities Table

(i)

P(X)

0.306039

0.437198

0.211061

0.042212

0.003404

8.51E-05

If n = 5, A = 8, and N = 40, P(X = 0) = 0.3060

(ii) If n = 5, A = 8, and N = 40, P(X  1) = 1 – 0.3060 = 0.6940 (iii) If n = 5, A = 8, and N = 40, P(X  1) = P(X = 0) + P(X = 1) = 0.3060 + 0.4372 = 0.7432 (b)

Partial PHStat output: Hypergeometric Probabilities Data Sample Size

No. of Successes in Population

Population Size

Hypergeometric Probabilities Table X

P(X)

0.180537

If n = 7, A = 8, and N = 40, P(X = 0) = 0.1805

5.37

N = 45, A = 7, S = 2, n = 7 (a)

P(Division 1) = P( X = 7) æ 7 öæ 38 ö ç ÷ç ÷ è 7 øè 0 ø = æ 45 ö ç ÷ è 7 ø 1 = 45,379,620 » 2.2 ´ 10-8

(b)

P(Division 2) = P( X = 6,Y = 1) æ 7 öæ 2 öæ 36 ö ç ÷ç ÷ ÷ç è 6 øè 1 øè 0 ø = æ 45 ö ç ÷ è 7 ø 7´2 = 45,379,620 » 3.08 ´10-7

(c)

P(Division 3) = P( X = 6,Y = 0) æ 7 öæ 2 öæ 36 ö ç ÷ç ÷ç ÷ è 6 øè 0 øè 1 ø = æ 45 ö ç ÷ è 7 ø 7 ´ 36 = 45,379,620 251 = 45,379,620 » 5.53 ´10-6

(d)

5 winning numbers plus 1 or 2 supplementary numbers

P(Division 4) = P( X = 5,Y = 1) + P( X = 5,Y = 2) æ 7 öæ 2 öæ 36 ö æ 7 öæ 2 öæ 36 ö ç ÷ç ÷ç ÷ ÷ç ÷ ç ÷ç è 5 ø è 1 øè 1 ø è 5 ø è 2 øè 0 ø = + æ 45 ö æ 45 ö ç ÷ ç ÷ è 7 ø è 7 ø 21´ 2 ´ 36 21 = + 45,379,620 45,379,620 1533 = 45,379,620 » 3.38 ´10-5 (e)

P(Division 5) = P( X = 5,Y = 0) æ 7 öæ 2 öæ 36 ö ç ÷ç ÷ç ÷ è 5 øè 0 øè 2 ø = æ 45 ö ç ÷ è 7 ø 21´ 630 = 45,379,620 13230 = 45,379,620 » 2.92 ´10-4

(f)

Four winning numbers, and three fails which includes supplementary numbers.

P(Division 6) = P( X = 4) æ 7 öæ 38 ö ç ÷ç ÷ è 4 øè 3 ø = æ 45 ö ç ÷ è 7 ø 35 ´ 8436 = 45,379,620 295,260 = 45,379,620 » 0.0065

(g)

Three winning numbers plus one or two supplementary numbers.

P(Division 7) = P( X = 3,Y = 1) + P( X = 3,Y = 2) (h) æ 7 öæ 2 öæ 36 ö æ 7 öæ 2 öæ 36 ö ç ÷ç ÷ ç ÷ç ÷ ÷ç ÷ç è 3 øè 1 øè 3 ø è 3 øè 2 øè 2 ø = + æ 45 ö æ 45 ö ç ÷ ç ÷ è 7 ø è 7 ø 35 ´ 2 ´ 7140 35 ´ 1´ 630 = + 45,379,620 45,379,620 521,850 = 45,379,620 » 0.0115 P(No winning or supplementary numbers) = P( X = 0,Y = 0) æ 7 öæ 2 öæ 36 ö ç ÷ç ÷ ÷ç è 0 øè 0 ø è 7 ø = æ 45 ö ç ÷ è 7 ø 8,347,680 = 45,379,620 » 0.1840 5.38

N = 45, A = 6, S = 2, n = 6 (a)

P(Division 1) = P( X = 6) æ 6 öæ 39 ö ç ÷ç ÷ è 6 øè 0 ø = æ 45 ö ç ÷ è 6 ø 1 = 8,145,060 » 1.2 ´ 10-7

(b)

P(Division 2) = P( X = 5,Y = 1) æ 6 öæ 2 öæ 37 ö ç ÷ç ÷ ÷ç è 5 øè 1 øè 0 ø = æ 45 ö ç ÷ è 6 ø 6´2 = 8,145,060 » 1.47 ´ 10-6

(c)

P(Division 3) = P( X = 5,Y = 0) æ 6 öæ 2 öæ 37 ö ç ÷ç ÷ç ÷ è 5 øè 0 øè 1 ø = æ 45 ö ç ÷ è 6 ø 6 ´ 37 = 45,379,620 222 = 8,145,060 » 2.73 ´10-5

(d)

Four winning numbers and two fails which includes supplementary numbers.

P(Division 4) = P( X = 4) æ 6 öæ 39 ö ç ÷ç ÷ è 4 øè 2 ø = æ 45 ö ç ÷ è 6 ø 15 ´ 741 = 8,145,060 11115 = 8,145,060 » 0.0014 (e)

3 winning numbers plus 1 or 2 supplementary numbers

P(Division 5) = P( X = 3,Y = 1) + P( X = 3,Y = 2) æ 6 öæ 2 öæ 37 ö æ 6 öæ 2 öæ 37 ö ç ÷ç ÷ ç ÷ç ÷ ÷ç ÷ç è 3 øè 1 øè 2 ø è 3 øè 2 ø è 1 ø = + æ 45 ö æ 45 ö ç ÷ ç ÷ è 6 ø è 6 ø 20 ´ 2 ´ 666 20 ´ 37 = + 8,145,060 8,145,060 27380 = 8,145,060 » 0.0034

(f)

P(No winning or supplementary numbers) = P( X = 0,Y = 0) æ 6 öæ 2 öæ 37 ö ç ÷ç ÷ ÷ç è 0 øè 0 ø è 6 ø = æ 45 ö ç ÷ è 6 ø 2,324,784 = 8,145,060 » 0.2854

5.39

(a)

PHStat output: Hypergeometric Probabilities Data Sample Size

No. of Events of Interest in Population

Population Size

Hypergeometric Probabilities Table

(i)

P(X)

0.15385

0.43956

0.32967

0.07326

0.00366

0.43956

(ii) 0.84615 (iii) 0.92308 (b)

µ = nA/N = (4×5)/15 = 1.33

5.40

The expected value is the average of a probability distribution. It is the value that can be expected to occur on the average, in the long run.

5.41

The four properties of a situation that must be present in order to use the binomial distribution are (i) the sample consists of a fixed number of observations, n, (ii) each observation can be classified into one of two mutually exclusive and collectively exhaustive categories, usually called ‘an event of interest’ and ‘not an event of interest’, (iii) the probability of an observation being classified as ‘an event of interest’, , is constant from observation to observation and (iv) the outcome (i.e. ‘an event of interest’ or ‘not an event of interest’) of any observation is independent of the outcome of any other observation.

5.42 The four properties of a situation that must be present in order to use the Poisson distribution are (i) you are interested in counting the number of times a particular event occurs in a given area of opportunity (defined by time, length, surface area, and so forth), (ii) the probability that an event occurs in a given area of opportunity is the same for all of the areas of opportunity, (iii) the number of events that occur in one area of opportunity is independent of the number of events that occur in other areas of opportunity and (iv) the probability that two or more events will occur in an area of opportunity approaches zero as the area of opportunity becomes smaller. 5.43

The hypergeometric distribution should be used when the probability of an event of interest of a sample containing n observations is not constant from trial to trial due to sampling without replacement from a finite population.

5.44

(a)

0.590

(b)

0.5902 = 0.3481

X = number of months open higher than previous month, n = 5, p = 0.590 (c)

P(X = 4) =

5! 0.5904 (1 − 0.590)1 = 0.24840 K 4!(5 − 4)!

(d)

P(X = 0) =

5! 0.5900 (1 − 0.590)5 = 0.001158K 0!(5 − 0)!

(e)

Share prices tend to rise when economy is expanding and fall during a recession. Therefore, the probability that the All Ordinaries Index will open higher than the previous month may not be independent and/or constant from month to month

5.45 Let X = number who voted for the Greens candidate. Then X is binomial with n = 10 and p = 0.12 (a) (b) (c) 5.46

(a)

P(X = 4) = 0.02022 P(X ≤ 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.2785 + 0.3797 + 0.2330 + 0.0847 + 0.2022 = 0.9963 P(X > 5) = 1 – P(X ≤ 4) + P(X = 5) = 1 – (0.9963 + 0.0033) = 0.0004 Let X = Payout, $, 35 year old female then

P(Alive at 40th birthday|alive at 35th birthday) =

98427 = 0.9969714... 98726

So the probability distribution for X is X

$1000000

P(X)

0.996971…

0.0030285…

E(X) = ( 0  0.996971...) + (1000000  0.0030285...) = 3028.584... (b)

To make a profit the insurance company will charge more than $3028.58 for this policy.

(c)

Let Y = Payout, $, 40 year old female and

P(Alive at 45th birthday | alive at 40th birthday) =

97740  0.994556 98275

then the probability distribution for Y is Y

$1000000

P(Y)

0.994556

0.005444

and the insurance company’s expected payout is $5444 since

E(Y) = (0  0.0.994556) + (1000000  0.005444) = 5444 (d)

Let X = Payout, $, 35 year old male then

P(Alive at 40th birthday | alive at 35th birthday) =

97381 = 0.9947088... 97899

So the probability distribution for X is X

$1000000

P(X)

0.9947088…

0.00529116…

and the insurance company's expected payout is $5291.17 since

E(X) = ( 0  0.9947088.. ) + (1000000  0.00529116... ) = 5291.167... To make a profit the insurance company will charge more than $5291.17 for this policy. Let Y = Payout, $, 40 year old male then

P(Alive at 45th birthday | alive at 40th birthday) =

96649 = 0.99248313... 97381

The probability distribution for Y is Y

$1000000

P(Y)

0.9924831…

0.00751686…

and the insurance company's expected payout is $7516.87 since

E(Y) = ( 0  0.99248313...) + (1000000  0.00751686...) = 7516.866... 5.47 (a)

Let X = number of times facility used in a week.

Then X is Poisson with  = 120/60 = 2

5.48

(b)

P(X = 0) = 0.1353

(c)

P(X ≥ 2) = 1 – (P(X = 0) + P(X = 1)) = 1 – (0.1353 + 0.2707) = 0.5939

(d)

Let Y = number of times facility used in two weeks. Then Y is Poisson with  = 2 and P(X ≥ 1) = 1 – P(X = 0) = 1– 0.0183 = 0.9816

Binomial probabilities Data Sample Size

Probability of an Event of Interest

0.58

Statistics Mean

11.6

Variance

4.872

Standard Deviation

2.207261

Binomial Probabilities Table X

P(X) 0

2.92E-08

8.06E-07

1.06E-05

8.76E-05

0.000514

0.002272

0.007842

0.02166

0.048605

0.089495

0.135948

0.17067

0.176765

0.150218

0.103722

0.057294

0.024725

0.008034

(a)(i)

P(X = 0) = 2.92E-08

(ii)

P(X ≤ 5) = 0.002885

0.001849

0.000269

1.86E-05

(iii) P(X > 10) = 0.693566

5.49

(b)

Assumption: independence of card holders.

(a)

Let X = number married during 2016 Since unmarried women were randomly selected X is binomial with n = 60 and p =

10.95 = 0.01095 1000

(i) P(X  3) = 1 − P(X  3)

= 1 − (P(0) + P(1) + P(2))

(

) (

)

 C  0.010950  0.9890560 + C  0.010951  0.9890559  60 1  60 0  =1−  2 48 + C  0.01095  0.98905   60 2   = 1 − (0.51652... + 0.34311... + 0.11206...) = 1 − 0.97170... = 0.02829 K

(

)

(ii) P(X  2) = P(0) + P(1) + P(2)

= 0.51652... + 0.34311... + 0.11206... = 0.97170... (iii) Mean Standard deviation

 = np = 60×0.01095 = 0.657

 = np(1 − p) = 60  0.01095  0.98905 = 0.64980... = 0.89783...

Sample size Probability of an event of interest

60 0.01095

Statistics Mean Variance Standard deviation

0.657 0.6498 0.8061

Binomial Probabilities Table X 0 1 2 3 4 5 (b)

P(X) 0.5165 0.3431 0.1121 0.0240 0.0038 0.0005

P(<=X) 0.5165 0.8596 0.9717 0.9957 0.9995 0.9999

P(<X) 0.0000 0.5165 0.8596 0.9717 0.9957 0.9995

P(>X) 0.4835 0.1404 0.0283 0.0043 0.0005 0.0001

P(>=X) 1.0000 0.4835 0.1404 0.0283 0.0043 0.0005

Let Y = number divorced during 2016 Since married couples were randomly selected Y is binomial with n = 60 and p =

8.7 = 0.0087 1000

(i)

P(Y = 0) =60 C0  0.00870  0.991360 = 0.59197...

(ii)

P(Y  2) = P(0) + P(1) + P(2) = 0.59197... + 60 C1  0.00871  0.991359 + 60 C2  0.00872  0.991358 = 0.59197... + 0.31172... + 0.08070... = 0.98441...

(iii) Mean  = np = 60×0.0087 = 0.522 Standard deviation

 = np(1 − p) = 60  0.0087  0.9913 = 0.51745... = 0.71934...

Sample size Probability of an event of interest

60 0.0087

Statistics Mean Variance Standard deviation

0.522 0.5175 0.7193

Binomial Probabilities Table X 0 1 2 3 5.50

P(X) 0.5920 0.3117 0.0807 0.0137

P(<=X) 0.5920 0.9037 0.9844 0.9981

P(<X) 0.0000 0.5920 0.9037 0.9844

P(>X) 0.4080 0.0963 0.0156 0.0019

P(>=X) 1.0000 0.4080 0.0963 0.0156

(a)

You can use the Poisson distribution to model the number of calls received in one minute, since we are interested in counting the number of times a particular event occurs in a given area of opportunity (defined by time, length, surface area, and so forth).

(b)

Let X = number of calls received in a minute. Then X is Poisson with = 180/60 = 3, so: (i) P(X = 2) = 0.224042 (ii) P(X  2) = 1 − P(X  2)

= 1 −  P(X = 0) + P(X + 1) + P(X = 2)  = 1 −  0.04978K + 0.1493K + 0.2240 K  = 1 − 0.4231K = 0.5768K Let Y = number of calls received in 5 minutes. Then Y is Poisson with  = 3×5 = 15 (iii)

P(Y  20) = 0.124781

(iv)

P(X < 10) = 0.069854

(c)

5.51

Based on (b)(iii) and (iv), in approximately 7% of five-minute intervals less than 10 calls are received, with increased staff idle time. While in approximately 12.5% of five-minute intervals, 20 or more calls are received, hence increasing the likelihood that a call is not answered within 20 seconds.

Let X = number of students who log on in 1 minute. Then X is Poisson with = 4.45/5 = 0.89, so (a) (b)

P(X = 6) = 0.000283 Let Y = number of students who log on in 2 minutes. Then Y is Poisson with  = 0.89×2 = 1.78, so

P(X < 6) = P(X = 0) + P(X = 1) +………. P(X = 5) = 0.1686 + 0.3002 + 0.2671 + 0.1585 + 0.0705 + 0.0251 = 0.9901 5.52

5.53

(a)

λ = 0.4, P(X = 0) = 0.6703

(b)

λ = 0.4, P(X ≥ 5) = 0.00006

(c)

λ = 0.16, P(X = 0) = 0.8521

(d)

λ = 0.16, P(X ≥ 10) = 0.0000

Let X = number of questions in second test also in first test Hypergeometric, with N = 100 , n = A = 10 (a)

10  90     0 10 P(X = 0) =    = 0.3304K 100     10  (b)

P(X  1) = 1 − P(X = 0) = 1 − 0.3304K = 0.6695K

(c)

10  90     5 5 P(X = 5) =    = 0.000639 K 100     10  10  90     10 0 P(X = 10) =     0 100     10 

(d)

5.54

(a)

P(passing grade) = 0.85

(b)

P(credit or above) = 0.45

(c)(i) n = 15; p = 0.15; P(X = 3) = 0.21843 (ii) n = 15; p = 0.15; P(X > 5) = 0.01681 (iii) n = 15; p = 0.15; P(X = 0) = 0.087354 or n = 15; p = 0.85; P(X = 15) = 0.087354 (iv) n = 15; p = 0.45; P(X = 0) = 0.000127 (v) n = 15; p = 0.45; P(X = 5) = 0.14036 (vi) n = 15; p = 0.20; P(X ≥ 1) = 1 – (P(X = 0) = 1- 0.035184 = 0.96482

(d)(i) The expected number E(X), variance and standard deviation of number of fail grades are given below: Binomial probabilities Data Sample size

Probability of an event of interest

0.15

Statistics Mean: E(X)

2.25

Variance

1.9125

Standard deviation

1.382932

(ii) The expected number E(X), variance and standard deviation of number of grades pass or above are given below: Binomial probabilities Data Sample size

Probability of an event of interest

0.85

Statistics Mean

12.75

Variance

1.9125

Standard deviation

1.382932

(e)

In a sample of 15 we can expect on average 2.25 fail grades and 12.75 grades pass or above. In both cases the variability from the mean is the same.

(f)

Let G = grade point

G = ( 0  0.15) + ( 4  0.40) + ( 5  0.25) + ( 6  0.15) + ( 7  0.05) = 4.1

(

) (

)

2 G = 02  0.15 + 42  0.40 + 52  0.25 + 62  0.15 + 72  0.05 − 4.12

= 3.69 G = 3.69 = 1.9209K

5.55

Let

A = return on $1000 for share A B = return on $1000 for share B

(a)(i) Expected return share A:

E(A) =  aP(a) = (240  0.25) + (150  0.5) + ( −100  0.25) = 110 Expected return share B:

E(B) =  bP(b) = ( −100  0.25) + (150  0.5) + (240  0.25) = 110 (ii)

Variance share A:

2A =  a 2 P(a) −  2 = (2402  0.25) + (1502  0.5) + ((−100) 2  0.25) − 1102 = 14400 + 11250 + 2500 − 12100 = 16050 Variance share B same as variance share A, so:

2B = 16050 Standard deviation share A and B:

A = B = 16050 = 126.688K (iii) Covariance of share A and share B:

AB =  abP(ab) −  A  B

= ( 240  (−100)  0.25 ) + (150  150  0.5 ) + ( −100  240  0.25 ) − (110  110 ) = −6000 + 11250 − 6000 − 12100 = −12850

(b)

Since E(A) = E(B) Portfolio expected return E(P) = 110 for all proportions of share A.

Portfolio risk: (i)

w = 0.4 = 0.6 P =

( 0.42 16050) + ( 0.62 16050) + ( 2  0.4  0.6  (−12850) )

= 2178 (ii) (iii)

= 46.669 K w = 0.5 P = 40

w = 0.6

P = 46.669...

5.56

Hypergeometric with N = 993, n = 20 (a)

A = 613 local

 613  380     20  0   P(X = 20) = = 5.725K 10−5  0  993     20  (b)

A = 613 local

 613  380     15  5   P(X = 15) = = 0.0910K  993     20  (c)

A = 228 (Interstate)

 228  765     5  15   P(X = 5) = = 0.1997 K  993     20  (d)

A = 380 (Interstate)

P(X  10) = 0.1943K Non-local Offenders Data Sample size No. of successes in population Population size

20 380 993

Hypergeometric Probabilities Table X 10 11 12 13 14 15 16 17 18 19 20 Sum

P(X) 0.099952 0.055662 0.025462 0.009515 0.002876 0.000693 0.00013 1.82E-05 1.8E-06 1.12E-07 3.31E-09 0.19431

5.57

Let X = number HDs in a tutorial group Since students are randomly allocated to a tutorial group X is binomial with n = 30 and p = 0.06 (a)(i) (ii)

P(X = 0) =30 C0  0.060  0.9430 = 0.15625... P(X  2) = P(0) + P(1) + P(2) = 0.15625... + 30 C1  0.061  0.94 29 + 30 C2  0.062  0.94 28 = 0.15625... + 0.29921... + 0.27693... = 0.73239...

P(X  4) = 1 − P(X  4)

(iii)

= 1 − (P( 2) + P(3) + P(4)) = 1 − ( 0.73239... + 30 C3  0.063  0.94 27 + 30 C 4  0.06 4  0.94 26 ) = 1 − (0.51652... + 0.16498... + 0.07108...) = 1 − 0.96846... = 0.03153 (b)

Mean

 = np = 30×0.06 = 1.8

Standard deviation

 = np(1 − p) = 30  0.06  0.94 = 1.692 = 1.30076...

Sample size Probability of an event of interest

30 0.06

Statistics Mean Variance Standard deviation

1.8 1.6920 1.3008

Binomial Probabilities Table X 0 1 2 3 4 5

P(X) 0.1563 0.2992 0.2769 0.1650 0.0711 0.0236

P(<=X) 0.1563 0.4555 0.7324 0.8974 0.9685 0.9921

P(<X) 0.0000 0.1563 0.4555 0.7324 0.8974 0.9685

P(>X) 0.8437 0.5445 0.2676 0.1026 0.0315 0.0079

P(>=X) 1.0000 0.8437 0.5445 0.2676 0.1026 0.0315

5.58

Let X = number of traffic accidents in time interval, with average  = 2.6 an hour 7:00am to 7:00pm Assume X is Poisson. (a)

9:00 am to 9:30 am = 30 minutes so  = 1.3

P(X = 4) = (b)

e−1.3 1.34 = 0.0324... 4!

2:00pm to 4:00pm = 2 hours so  = 5.2

e−5.2 5.25 P(X = 5) = = 0.17478... 5! (c)

2:00 pm to 3:00 pm = 1 hour so  = 2.6

P(3 or 4) = P(3) + P(4) e−2.6 2.63 e−2.6 2.64 + 3! 4! = 0.21757... + 0.13142... =

= 0.35899... (d)

4:00 pm to 5:30 pm = 1.5 hour so  = 3.9

P(X  1) = 1 − P(0) e−3.9 3.90 0! = 1 − 0.02020...

=1−

= 0.97975... 5.59

Hypergeometric with N = 52, n = 5 (a) (b) (c) Let X = number of red cards so A = 26 red cards

(a)

 26  26     5 0 P(X = 5) =    = 0.02531...  52    5

(b)

 26  26     2 3 P(X = 2) =    = 0.32513...  52    5

(c)

P(X  1) = 1 − P(X = 0)  26   26     0 5 = 1−     52    5 = 1 − 0.02531... = 0.97468... Hypergeometric Probabilities Data Sample size No. of events of interest in population Population size Hypergeometric Probabilities Table

5 26 52 X 0 1 2 3 4 5

P(X) 0.0253 0.1496 0.3251 0.3251 0.1496 0.0253

(d) & (e) Let K = number of Kings so A = 4 kings

(d)

 4  48     4 1 P(K = 4) =    = 0.00001846...  52    5

(e)

P(K  1) = 1 − P(K = 0)  4   48     0 5 = 1−     52    5 = 1 − 0.65884... = 0.34115...

Hypergeometric Probabilities Data Sample size No. of events of interest in population Population size Hypergeometric Probabilities Table

5 4 52 X 0 1 2 3 4

P(X) 0.6588420 0.2994736 0.0399298 0.0017361 0.0000185

(f) Let H = number of hearts so A = 13 hearts

13  39     5 0 P(H = 5) =    = 0.000495K  52    5 (g)

P(all same suit) = P(all hearts) + P(all diamonds) + P(all clubs) + P(all spades) = 4  0.000495... = 0.0019807...

5.60

Let X = number of cars sold in day, with average  = 3.6 a day so  = 3.6

e−3.6 3.65 = 0.13768... 5!

(a)

P(X = 5) =

(b)

P(X  2) = P(0) + P(1) + P(2) e−3.6 3.60 e−3.6 3.61 e−3.6 3.62 + + 0! 1! 2! = 0.02732... + 0.09836... + 0.17705...

= 0.30274...

POISSON.DIST Probabilities Data Mean/Expected number of events of interest:

3.6

POISSON.DIST Probabilities Table X P(X) P(<=X) 0 0.0273 0.0273 1 0.0984 0.1257 2 0.1771 0.3027 3 0.2125 0.5152 4 0.1912 0.7064 5 0.1377 0.8441

P(<X) 0.0000 0.0273 0.1257 0.3027 0.5152 0.7064

(c)

Mean

 =  = 10×3.6 = 36

Standard deviation

5.61

(a)

P(>X) P(>=X) 0.9727 1.0000 0.8743 0.9727 0.6973 0.8743 0.4848 0.6973 0.2936 0.4848 0.1559 0.2936

 =  = 36 = 6

Assume binomial assumptions are satisfied. Let X = Number of potential customers who subscribe to 3-For-All X binomial with n = 50, p = 0.02

(i)

P(X  3) = P(0) + P(1) + P(2) 50! 50! 50! 0.98500.020 + 0.98490.021 + 0.98480.022 0!50! 1!49! 2!48! = 0.36416... + 0.37160... + 0.18580...

= 0.92157... (ii) P(X  1) = P(0) + P(1) = 0.36416... + 0.37160... = 0.73577...

(iii)

P(X  4) = 1 − P(X  4) = 1 − ((P(0) + P(1) + P(2) + P(3) + P(4)) 50! 0.98470.023 3!47! 50! + 0.98460.024 ) 4!46! = 1 − (0.36416... + 0.37160... + 0.18580... + 0.06066... + 0.01454...)

= 1 − (0.36416... + 0.37160... + 0.18580... +

= 1 − 0.99679... = 0.00320... (iv) P(X  5) = P(X  4) = 0.00320.. The probability of getting five or more subscribers from a sample of 50 if the probability of subscribing is 0.02 is small at 0.00320. Therefore, the probability of getting a new subscriber when no free premium channels are offered is likely to be higher than 0.02. Binomial Probabilities Data Sample size Probability of an event of interest Statistics Mean Variance Standard deviation

50 0.02

1 0.9800 0.9899

Binomial Probabilities Table X 0 1 2 3 4 5

(b)

P(X) 0.3642 0.3716 0.1858 0.0607 0.0145 0.0027

P(<=X) 0.3642 0.7358 0.9216 0.9822 0.9968 0.9995

P(<X) 0.0000 0.3642 0.7358 0.9216 0.9822 0.9968

P(>X) 0.6358 0.2642 0.0784 0.0178 0.0032 0.0005

X binomial with n = 50, p = 0.06

(i)

P(X  3) = P(0) + P(1) + P(2) 50! 50! 50! 0.94500.060 + 0.94490.061 + 0.94480.062 0!50! 1!49! 2!48! = 0.045330... + 0.144672... + 0.226243...

= 0.416246... (ii) P(X  1) = P(0) + P(1) = 0.045330... + 0.144672... = 0.190003...

P(>=X) 1.0000 0.6358 0.2642 0.0784 0.0178 0.0032

(iii)

P(X  4) = 1 − P(X  4) = 1 − ((P(0) + P(1) + P(2) + P(3) + P(4)) 50! 0.9447 0.063 3!47! 50! + 0.94460.064 ) 4!46! = 1 − (0.045330... + 0.144672... + 0.226243... + 0.231056... + 0.173292...)

= 1 − (0.045330... + 0.144672... + 0.226243... +

= 1 − 0.820596... = 0.179403... Binomial Probabilities Data Sample size Probability of an event of interest Statistics Mean Variance Standard deviation

50 0.06

3 2.8200 1.6793

Binomial Probabilities Table X 0 1 2 3 4 5 6

P(X) 0.0453 0.1447 0.2262 0.2311 0.1733 0.1018 0.0487

P(<=X) 0.0453 0.1900 0.4162 0.6473 0.8206 0.9224 0.9711

P(<X) 0.0000 0.0453 0.1900 0.4162 0.6473 0.8206 0.9224

P(>X) 0.9547 0.8100 0.5838 0.3527 0.1794 0.0776 0.0289

P(>=X) 1.0000 0.9547 0.8100 0.5838 0.3527 0.1794 0.0776

(c) The probability of getting more than 4 subscribers in a sample of 50 if the probability of a subscription is 0.06 is 0.1794. This is higher than when no free premium channels were offered. (d)

P(X  5 | p = 0.06) = P(X  4 | p = 0.06) = 0.179403... The probability of getting five or more subscribers from a sample of 50 if the probability of subscribing is 0.06 is 0.1794. Therefore, there is no evidence that the probability of getting a new subscriber when two free premium channels are offered is not 0.06.

(e)

Offering free premium channels increases the probability of subscriptions.

Chapter 6: The normal distribution and other continuous distributions Learning objectives After studying this chapter you should be able to: 1.

calculate probabilities from the normal distribution

determine whether a set of data is approximately normally distributed

calculate probabilities from the uniform distribution

calculate probabilities from the exponential distribution

use the normal distribution to approximate probabilities from the binomial distribution

6.1

PHStat output: Normal Probabilities Common Data Mean

Standard Deviation

1 Probability for a Range

Probability for X <=

From X Value

1.57

X Value

1.57

To X Value

1.84

Z Value

1.57

Z Value for 1.57

1.57

0.94179 24

Z Value for 1.84

1.84

P(X<=1.57)

Probability for X > X Value

1.84

Z Value

1.84

P(X>1.84)

0.0329

P(X<=1.57)

0.9418

P(X<=1.84)

0.9671

P(1.57<=X<=1.84)

0.0253

Find X and Z Given Cum. Pctage Cumulative Percentage

95.00%

Probability for X<1.57 or X>1.84

Z Value

1.644854

P(X<1.57 or X>1.84)

X Value

1.644854

0.9747

6.2

(a)

P(Z < 1.57) = 0.9418

(b)

P(Z > 1.84) = 1 – 0.9671 = 0.0329

(c)

P(1.57 < Z < 1.84) = 0.9671 – 0.9418 = 0.0253

(d)

P(Z < 1.57) + P(Z > 1.84) = 0.9418 + (1 – 0.9671) = 0.974

PHStat output: Normal Probabilities Common Data Mean

Standard Deviation

1 Probability for a Range

Probability for X <=

From X Value

1.57

X Value

–1.57

To X Value

1.84

Z Value

–1.57

Z Value for 1.57

1.57

0.0582076

Z Value for 1.84

1.84

P(X<=–1.57)

Probability for X > X Value

1.84

Z Value

1.84

P(X>1.84)

0.0329

P(X<=1.57)

0.9418

P(X<=1.84)

0.9671

P(1.57<=X<=1.84)

0.0253

Find X and Z Given Cum. Pctage Cumulative Percentage

84.13%

Probability for X<–1.57 or X >1.84

Z Value

0.999815

P(X<–1.57 or X>1.84)

X Value

0.999815

0.0911

(a)(i) P(– 1.57 < Z < 1.84) = 0.9671 – 0.0582 = 0.9089 (ii) P(Z < – 1.57) + P(Z > 1.84) = 0.0582 + 0.0329 = 0.0911 (b)

If P(Z > A) = 0.025, P(Z < A) = 0.975. A = + 1.96.

(c)

If P(–A < Z < A) = 0.6826, P(Z < A) = 0.8413. So 68.26% of the area is captured between –A = – 1.00 and A = + 1.00.

6.3

(a)

Partial PHStat output:

Normal Probabilities Common Data Mean

Standard Deviation

1 Probability for a Range

Probability for X <=

From X Value

1.57

X Value

1.08

To X Value

1.84

Z Value

1.08

Z Value for 1.57

1.57

0.8599289

Z Value for 1.84

1.84

P(X<=1.08)

Probability for X > X Value

-0.21

Z Value

-0.21

P(X>–0.21)

0.5832

P(X<=1.57)

0.9418

P(X<=1.84)

0.9671

P(1.57<=X<=1.84)

0.0253

Find X and Z Given Cum. Pctage Cumulative Percentage

Probability for X<1.08 or X>–0.21 P(X<1.08 or X>–0.21)

1.4431

84.13%

Z Value

0.999815

X Value

0.999815

P(Z < 1.08) = 0.8599 (b)

P(Z > – 0.21) = 1.0 – 0.4168 = 0.5832

(c)

Partial PHStat output: Probability for X<–0.21 or X>0 P(X<–0.21 or X>0)

0.9168

P(Z < – 0.21) + P(Z > 0) = 0.4168 + 0.5 = 0.9168

(d)

Partial PHStat output: Probability for X<–0.21 or X>1.08 P(X<–0.21 or X >1.08)

0.5569

P(Z < – 0.21) + P(Z > 1.08) = 0.4168 + (1 – 0.8599) = 0.5569 6.4

PHStat output: Normal Probabilities Common Data Mean

Standard Deviation

1 Probability for a Range

Probability for X <=

From X Value

–1.96

X Value

–0.21

To X Value

–0.21

Z Value

–0.21

Z Value for –1.96

–1.96

0.4168338

Z Value for –0.21

–0.21

P(X<–0.21)

Probability for X > X Value

1.08

Z Value

1.08

P(X>1.08)

P(X<=–1.96)

0.0250

P(X<=–0.21)

0.4168

P(–1.96<=X<=–0.21)

0.3918

0.1401

Find X and Z Given Cum. Pctage Cumulative Percentage

84.13%

Probability for X<–0.21 or X>1.08

Z Value

0.999815

P(X<–0.21 or X>1.08)

X Value

0.999815

0.5569

(a)(i) P(Z > 1.08) = 1 – 0.8599 = 0.1401 (ii) P(Z < – 0.21) = 0.4168 (iii) P(– 1.96 < Z < – 0.21) = 0.4168 – 0.0250 = 0.3918 (b)

P(Z > A) = 0.1587, P(Z < A) = 0.8413.

A = + 1.00

6.5

(a)

Partial PHStat output: Common Data Mean

100

Standard Deviation

Probability for X <= X Value

Z Value

–3

P(X<=70)

0.0013499

Probability for X > X Value

Z Value

–2.5

P(X>75)

0.9938

X–



75 –100 = – 2.50 10

P(X > 75) = P(Z > – 2.50) = 1 – P(Z< – 2.50) = 1 – 0.0062 = 0.9938

(b)

X–



70 – 100 = – 3.00 10

P(X < 70) = P(Z < – 3.00) = 0.00135 (c)

Partial PHStat output: Common Data Mean

100

Standard Deviation

Probability for X <= X Value

Z Value

–2

P(X<=80)

0.0227501

Probability for X > X Value

110

Z Value

P(X>110)

0.1587

Probability for X<80 or X >110 P(X<80 or X >110)

X–



80 – 100 = –2.00 10

0.1814

X–



110 – 100 =1.00 10

P(X < 80) = P(Z < – 2.00) = 0.0228 P(X > 110) = P(Z > 1.00) = 1 – P(Z < 1.00) = 1.0 – 0.8413 = 0.1587 P(X < 80) + P(X > 110) = 0.0228 + 0.1587 = 0.1815 (d)

P(Xlower < X < Xupper) = 0.80 P(– 1.28 < Z) = 0.10 and P(Z < 1.28) = 0.90

Z = –1.28 =

Xlower –100 10

Z = +1.28 =

Xupper – 100 10

Xlower = 100 – 1.28(10) = 87.20 and Xupper = 100 + 1.28(10) = 112.80 6.6

(a)

Partial PHStat output: Common Data

Mean

Standard Deviation

4 Probability for a Range

Probability for X <=

From X Value

X Value

To X Value

Z Value

–2

Z Value for 42

–2

0.0227501

Z Value for 43

–1.75

P(X<=42)

Probability for X > X Value

Z Value

–1.75

P(X>43)

0.9599

P(X<=42)

0.0228

P(X<=43)

0.0401

P(42<=X<=43)

0.0173

Find X and Z Given Cum. Pctage Cumulative Percentage

Probability for X<42 or X>43 P(X<42 or X>43)

0.9827

5.00%

Z Value

–1.644854

X Value

43.42059

P(X > 43) = P(Z > – 1.75) = 1 – 0.0401 = 0.959 (b)

P(X < 42) = P(Z < – 2.00) = 0.0228

(c)

P(X < A) = 0.05,

Z = −1.645 =

A − 50 4

A = 50 – 1.645(4) = 43.42

(d)

Partial PHStat output: Find X and Z Given Cum. Pctage Cumulative Percentage

80.00%

Z Value

0.841621

X Value

53.36648

P(Xlower < X < Xupper) = 0.60 P(Z < – 0.84) = 0.20 and P(Z < 0.84) = 0.80

Z = –0.84 =

Xlower – 50 4

Z = +0.84 =

Xupper – 50 4

Xlower = 50 – 0.84(4) = 46.64 and Xupper = 50 + 0.84(4) = 53.36 6.7

6.8

Let X = Credit card balance, given X normal with µ = 3,325 and σ = 1,500. 2,500−3,325

(a)

P(X < 2500) = 𝑃 (𝑍 <

(b)

P(X > 5,000) = 𝑃(𝑍 > 1.12 ) = 1 − 0.8686 = 0.1314

(c)

P(3,000 < X < 4,000) = P(-0.22 < Z < 0.45) = 0.6736 – 0.4129 = 0.2607

(d)

P(X  A) = 0.99 so Z  2.33 and X =  + Z = 3325 + 2.33 1500 = 6820

1,500

) ≈ 𝑃(𝑍 < −0.55) = 0.2912

Let X = annual kilometres, 000's kms Given X normal with m= 100 and s = 20 (a)

68.26% since:

æ 80 - 100 120 - 80 ö P(80 < X < 120) = P ç <Z< ÷ 20 ø è 20 = P(-1 < Z < 1) = P(Z < 1) - P(Z £ -1) = 0.8413 - 0.1587 = 0.6826 (b)

4.56% since:

P( X < 60) + P( X > 140) = P( Z < - 2) + P( Z > 2) = P( Z < - 2) + (1 - P( Z £ 2)) = 0.0228 + (1 - 0.9772) = 0.0456 (c)

P( X ³ A) = 0.80 so P( X < A) = 0.20 so Z » - 0.84 and X = m+ Z s = 100 + (- 0.84) ´ 20 = 100 - 16.8 = 83.2 80% of trucks will travel at least 83,200 kilometres per year.

(d)

6.9

Given X normal with m= 100 and s = 10 (a)

P(80 < X < 120) = P(- 2 < Z < 2) = 0.9772 - 0.0228 = 0.9544

(b)

P( X < 60) + P( X > 140) = P( Z < - 4) + P( Z > 4) = 0.000063342

(c)

X = m+ Z s = 100 + (- 0.84) ´ 10 = 100 - 8.4 = 91.6

Let X = breaking strength, kPa, given X normal with µ = 35 and σ = 10. (a)(i) P(X < 20) = 𝑃 (𝑍 <

20−35 10

) ≈ 𝑃(𝑍 < −1.5) = 0.0668

(ii) P(X ≥30) = P(Z ≥ –0.5) = 1- 0.3085 = 0.6915 (iii) P(25 < X < 45) = P(–1 < Z < 1) = 0.8413 - 0.1587 = 0.6826 (b)

P(XL < X < XU) = P(–1.96 < Z < 1.96) = 0.95 XL = 35 + (–1.96) × 10 = 15.4 XU = 35 + 1.96 × 10 = 54.6

6.10

(a)

P(X < 91) = P(Z < 2.25) = 0.9878

(b)

P(65 < X < 89) = P(– 1.00 < Z < 2.00) = 0.9772 – 0.1587 = 0.8185

(c)

P(X ≥ A) = 0.15, so P(X < A) = 0.85, so Z ≈ 1.04 and X = µ + Zσ = X = 73 + 1.04 × 8 = 81.32, a mark of at or more than 81 is required for a distinction.

(d)

Exam 1: P(X > 80) = P(Z > 0.88) = 1 – 0.8186 = 0.1894; not in top 15% so will probably receive a Credit grade. Exam 2: P(X > 68) = P(Z > 2) = 1 – 0.9792 = 0.0228; in top 5% so receive a High Distinction. Better.

6.11

PHStat output: Common Data Mean

240

Standard Deviation

40 Probability for a Range

Probability for X <=

From X Value

180

X Value

180

To X Value

300

Z Value

–1.5

Z Value for 180

–1.5

0.0668072

Z Value for 300

1.5

P(X<=180)

Probability for X > X Value

Z Value

–3.975

P(X>81)

1.0000

P(X<=180)

0.0668

P(X<=300)

0.9332

P(180<=X<=300)

0.8664

Find X and Z Given Cum. Pctage Cumulative Percentage

95.00%

Probability for X<180 or X>81

Z Value

1.644854

P(X<180 or X>81)

X Value

305.7941

1.0668

(a)

P(X < 180) = P(Z < – 1.50) = 0.0668

(b)

P(180 < X < 300) = P(– 1.50 < Z < 1.50) = 0.9332 – 0.0668 = 0.8664

(c)

PHStat output: Common Data Mean

240

Standard Deviation

40 Probability for a Range

Probability for X <=

From X Value

110 180

X Value

180

To X Value

Z Value

–1.5

Z Value for 110

–3.25

0.0668072

Z Value for 180

–1.5

P(X<=180)

Probability for X > X Value

Z Value

–3.975

P(X>81)

1.0000

P(X<=110)

0.0006

P(X<=180)

0.0668

P(110<=X<=180)

0.0662

Find X and Z Given Cum. Pctage Cumulative Percentage

Probability for X<180 or X>81 P(X<180 or X>81)

1.0668

1.00%

Z Value

–2.326348

X Value

146.9461

P(110 < X < 180) = P(– 3.25 < Z < – 1.50) = 0.0668 – 0.00058 = 0.06622 (d)

P(X < A) = 0.01 P(Z < – 2.33) = 0.01 A = 240 – 2.33(40) = 146.80 seconds

6.12

Let X = volume of trade 2008, million shares Assume X normal with  = 992 and  = 252 (a)(i)

500 − 992   P(X  500) = P  Z    P(Z  −1.95) = 0.0256 252  

(ii) P(750  X  1000)  P( −0.96  Z  0.03) = 0.5120 − 0.1685 = 0.3435 (iii) P(X  1500) = P(Z  2.02) = 0.9783 (iv) P(X  1200) = P(Z  0.83) = 1 − 0.7976 = 0.2024 (b)

P(X  2125) = P(Z  4.50)  0 As this probability is approximately zero, this indicates that our assumption that the number of shares traded daily is normal is incorrect. Alternatively, our assumption may be approximately correct and this may be an extremely rare event, due to the GFC.

6.13

(a)(i) P(21.9 < X < 22.00) = P(– 20.4 < Z < – 0.4) = 0.3446 (ii) P(21.9 < X < 22.01) = P(– 20.4 < Z < 1.6) = 0.9452 (iii) Partial PHStat output: Find X and Z Given Cum. Pctage Cumulative Percentage

98.00%

Z Value

2.05375

X Value

22.0123

P(X > A) = 0.02

Z = 2.05

A = 22.0123

(b)(i) P(21.9 < X < 22.00) = P(– 25.5 < Z < – 0.5) = 0.3085 (ii) P(21.9 < X < 22.01) = P(– 25.5 < Z < 2) = 0.9772 (iii) P(X > A) = 0.02

Z = 2.05

A = 22.0102

6.14

With 39 values, the smallest of the standard normal quantile values covers an area under the normal curve of 0.025. The corresponding Z value is –1.96. The middle (20th) value has a cumulative area of 0.50 and a corresponding Z value of 0.0. The largest of the standard normal quantile values covers an area under the normal curve of 0.975, and its corresponding Z value is +1.96.

6.15

Area under normal curve covered: 0.1429 0.2857 0.4286 0.5714 0.7143 0.8571 Standardised normal quantile value: –1.07 –0.57 –0.18 +0.18 +0.57 +1.07

6.16

(a) Full Rate A$ Mean Median

364.32 312

Mode

#N/A

Standard Deviation

135.19

Range

462

Minimum

200

Maximum

662

Count

Largest(5)

441

Smallest(5)

279

Mean of $364.32 is higher than the median of $312. The box-and-whisker plot below is right skewed. Interquartile range = Q3 – Q1 = 441 – 279 = 162, is 162/135.19 = 1.198 standard deviations. The range of 462 is equal to 441/135.19 = 3.262 standard deviations. Copyright © 2019 Pearson Australia (a division of Pearson Australia Group Pty Ltd) 9781488617249, Berenson, Basic Business Statistics 5e

It can be concluded that the daily rate is not normal.

Boxplot Full Rate A$

190

290

390

490

590

690

(b)

Normal Probability Plot 700 600

Full Rate A$

500 400 300 200 100 0 -2

-1.5

-1

-0.5

0.5

1.5

Z Value The normal probability plot suggests that the data is skewed to the right.

6.17

Office 1 Time

(a)

Mean

2.214

Median

1.540

Mode

1.480

Standard Deviation

1.718

Skewness

1.127

Range

5.800

Minimum

0.52

Maximum

6.32

3.93

0.93

The mean (2.214) is higher than the median (1.540). The interquartile range (3/1.718 = 1.74) is slightly larger than 1.33 times the standard deviation. The range of 5.8 (5.8/1.718 = 3.37) is much smaller than six times the standard deviation. From the boxplot the distribution is right skewed. The data do not satisfy the theoretical properties of a normal distribution and appear to be right skewed. Office 2 Time Mean

2.012

Median

1.505

Mode

3.750

Standard Deviation

1.892

Sample Variance

3.579

Skewness

1.466

Range

7.470

Minimum

0.080

Maximum

7.550

3.75

0.6

The mean (2.012) is higher than the median (1.505). The interquartile range (3.15/1.718 = 1.83) is slightly larger than 1.33 times the standard deviation. The range of 7.47 (7.47/1.718 = 4.35) is smaller than six times the standard deviation. From the boxplot the distribution is right skewed. The data do not

satisfy the theoretical properties of a normal distribution and appear to be right skewed. Phone

Time Seconds

(b)

Office 1

Normal Probability Plot 7 6

Time

5 4 3

2 1 0 -2

-1.5

-1

-0.5

0.5

1.5

Z Value

Office 2

Normal Probability Plot 8 7 6

Time

5 4 3 2 1 0 -2

-1.5

-1

-0.5

0.5

1.5

Z Value According to the normal probability plot, the data appear to be right skewed. 6.18

(a) Plant A Mean

9.3820

Median

8.5150

7.2900

11.4200

IQR

4.1300

Standard Deviation

3.9977

Range

17.2000

Minimum

4.4200

Maximum

21.6200

The mean (9.382) is higher than the median (8.5150). The interquartile range (4.13/3.9977 = 1.033) is smaller than 1.33 times the standard deviation. The range of 17.2 (17.2/3.977 = 4.32) is much smaller than six times the standard deviation. From the boxplot the distribution is right skewed. The data do not satisfy the theoretical properties of a normal distribution and appear to be right skewed.

Plant B Mean

11.3535

Median

11.9600

6.2500

14.2500

IQR

8.0000

Standard Deviation

5.1262

Range

23.4200

Minimum

2.3300

Maximum

25.7500

The mean (11.354) is slightly smaller than the median (11.9600). The interquartile range (8/5.1262 = 1.56) is larger than 1.33 times the standard deviation. The range of 23.42 (23.42/5.1262 = 4.568) is smaller than six times the standard deviation. The data do not satisfy the theoretical properties of a normal distribution. From the boxplot below the data is not symmetric. WIP

Plant B

Plant A

Time Days

(b)

Plant A 25

Processing time

20 15 10 5 0 -2

-1.5

-1

-0.5

0.5

1.5

Z Value The normal probability plot suggests that the data is not normal. Normal Probability Plot Plant B 30

Processing Time, plant B

0 -2

-1.5

-1

-0.5

0.5

Z Value

The normal probability plot suggests that the data is not normal.

6.19

(a) Total Mark Mean

60.7818

Median

63.0000

Mode

57.0000

Standard Deviation

18.0992

Skewness

-0.7880

Range

Minimum

Maximum

Box-and-Whisker Plot

100

Total Mark

The mean (60.78) is slightly less than the median (63). The interquartile range of 17 (17/18.09 = 0.94) is smaller than 1.33 times the standard deviation. The range of 80 (80/18.09 = 4.44) is much smaller than six times the standard deviation. The box-and-whisker plot does not indicate a symmetric distribution. The data do not satisfy the theoretical properties of a normal distribution.

(b)

Normal Probability Plot

Total Mark

100 90 80 70 60 50 40 30 20 10 0

-3

-2

-1

Z Value . The normal probability plot suggests that the data is not normal. 6.20

(a)

Excel output: Exam Mark Mean

30.1000

Median

28.5000

Mode

25.0000

Standard Deviation

6.6922

Skewness

0.0890

Range

Minimum

Maximum

Box-and-Whisker Plot

25Exam Mark 30

The mean (30.1) is higher than the median (28.5). The range is smaller than six times the standard deviation. The interquartile range of 9 (9/6.692 = 1.344) is marginally over 1.33 times the standard deviation. The range of 30 (30/6.692 = 4.48) is smaller than six times the standard deviation. The boxand-whisker plot indicates a slightly right-skewed distribution. It can be concluded that while exam marks are slightly right skewed, they are approximately normal. (b)

Normal Probability Plot 50 45

40 Exam Mark

35 30 25 20 15 10 5 0 -3

-2

-1

Z Value As the majority of points lie on or close to a straight line, normal probability plot suggests that the data is approximately normal. 6.21

(a)(i) P(5 < X < 7) = (7 – 5)/10 = 0.2 (ii) P(2 < X < 3) = (3 – 2)/10 = 0.1 (b)

=

0 + 10 =5 2

(c)

6.22

(10 − 0 )

 =

= 2.8868

(a)(i) P(0 < X < 20) = (20 – 0) /120 = 0.1667 (ii) P(10 < X < 30) = (30 – 10)/120 = 0.1667 (iii) P(35 < X < 120) = (120 – 35)/120 = 0.7083

6.23

(120 − 0 ) = 34.6410 2

(b)

0 + 120 = = 60 2

(a)

P(5:55 am < X < 7:38 pm) = P(355 < X < 1178)

=

= (1178 – 355)/(1440) = 0.5715 (b)

P(10 pm < X < 5 am) = P(1320 < X < 1440) + P(0 < X < 300) = (1440 – 1320)/(1440) + (300)/(1440) = 0.2917

(c)

Let X be duration between the occurrence of a failure and its detection, a = 0, b = 60. P(0 < X < 10) = (10) /60 = 0.1667

(d)

6.24

P(40 < X < 60) = (60 – 40) /60 = 0.3333

Let X = waiting time, minutes, is uniform with, a = 0, b = 3. Height is 1

(a)(i) P(X ≤ 1) = (1 − 0) × =

(ii) P(1 < X < 2) = (2 − 1) × = 1

(iii) P(X > 2) = (3 − 2) × = (b)

𝜇=

𝑎+𝑏 2

(0+3) 2

= 1.5

𝜎=√

(𝑏−𝑎)2 12

(3−0)2

=√

9 12

1 𝑏−𝑎

1 3

= 0.8660

6.25

(a) Exponential Probability Data l X Value

10 0.1

Results P(<=X) P(>X)

0.6321 0.3679

P(X  0.1) = 1 – e (b)

– lx

= 1– e

–(10)( 0.1)

= 0.6321

P(X > 0.1) = 1 – P(X  0.1) = 1 – 0.6321 = 0.3679

10 0.2

Results P(<=X) P(>X)

0.8647 0.1353

P(0.1 < X < 0.2) = P(X < 0.2) – P(X ≤ 0.1) = 0.8647 – 0.6321 = 0.2326 (d) 6.26

P(X < 0.1) + P(X > 0.2) = 0.6321 + 0.1353 = 0.7674

(a) Exponential Probability Data l X Value

30 0.1

Results P(<=X) P(>X)

0.9502 0.0498

P(X  0.1) = 1 − e (b)

− lx

= 1 − e −(30 )(0.1) = 0.9502

P(X > 0.1) = 1 – P(X  0.1) = 1 – 0.9502 = 0.0498

30 0.2

Results P(<=X) P(>X)

0.9975 0.0025

P(0.1 < X < 0.2) = P(X < 0.2) – P(X < 0.1) = 0.9975 – 0.9502 = 0.0473

6.27

(d)

P(X < 0.1) + P(X > 0.2) = 0.9502 + 0.0025 = 0.9527

(a)

P(X  4) = 1 − e−lx = 1 − e−(204) ≈ 1

(b)

P(X > 0.4) = 1 – P(X  0.4) = 1 – 0.9997 = 0.0003

(c)

P(0.4 < X < 0.5) = P(X < 0.5) – P(X ≤ 0.4) = 0.99996 – 0.99967 = 0.00029

(d) 6.28

P(X < 0.4) + P(X > 0.5) = 0.999665 + 0.000045 = 0.99971

Let X = time between arrivals, minutes, is exponential with λ= 240/60 = 4 per minute. (a)

P(X ≤ 1) = 1 − 𝑒 −(4)(1) = 1 − 0.01835 = 0.98168

(b)

P(X > 0.5) = 𝑒 −(4)(0.5) = 0.13533

(c)

𝜇 = = = 0.25 minutes or 15 seconds

(d)

λ= 300/60 = 5 per minute

P(X ≤ 1) = 1 − 𝑒 −(5)(1) = 1 − 0.00673 = 0.99326 P(X > 0.5) = 𝑒 −(5)(0.5) = 0.0820 1

𝜇 = = = 0.2 minutes or 12 seconds (e)

λ= 210/60 = 3.5 per minute P(X ≤ 1) = 1 − 𝑒 −(3.5)(1) = 1 − 0.03019 = 0.96980 P(X > 0.5) = 𝑒 −(3.5)(0.5) = 0.17377 1

3.5

𝜇= =

= 0.285 minutes or 17.142 seconds

6.29

(a) Exponential Probabilities Data l

X Value

1 Results

P(<=X)

0.8647

P(arrival time  1) = 0.8647

(b) Exponential Probabilities Data l

X Value

5 Results

P(<=X)

0.999955

P(arrival time  5) = 0.999955

(c)

Exponential Probabilities Data l

X Value

1 Results

P(<=X)

0.6321 Data

X Value

5 Results

P(<=X) If l = 1,

0.993262 P(arrival time  1) = 0.6321 P(arrival time  5) = 0.9933

6.30 (a) Exponential Probabilities Data l

0.05

X Value

14 Results

P(<=X)

0.5034

P(X  14) = = 1– e

–(1/ 20)(14)

= 0.5034

(b) Exponential Probabilities Data l

0.05

X Value

21 Results

P(<=X)

0.6501

P(X > 21) = 𝑒

1 −( )×21 20

= 0.3499

0.05

X Value

7 Results

P(<=X) P(X  7) = = 1– e

0.2953 –(1/ 20)(7)

= 0.2953

6.31

(a)(i) PHStat output: Exponential Probabilities Data l

X Value

0.25 Results

P(<=X)

0.8647

P(arrival time  0.25) = 0.8647

(ii)

PHStat output: Exponential Probabilities Data l

X Value

0.05 Results

P(<=X)

0.3297

P(arrival time  0.05) = 0.3297 (b)

PHStat output: Exponential Probabilities Data l

X Value

0.25 Results

P(<=X)

0.9765

Exponential Probabilities Data l

X Value

0.05 Results

P(<=X)

0.5276

If l = 15, P(arrival time  0.25) = 0.9765, P(arrival time  0.05) = 0.5276

6.32

X = time between floods, years exponential with l = (a)(i) (ii)

3 = 0.3 per year 10

P( X £ 1) = 1- e- 0.3´ 1 = 1- 0.74081K = 0.25918K

P( X > 2) = e- 0.3´ 2 = 0.54881K æ è

(iii) P ç çX £

ö 1÷ = 1 - e- 0.3´ 1/12 = 1 - 0.9753K = 0.02469 K ÷ ÷ 12 ø

(iv) P( X £ 0.5) = 1- e- 0.3´ 0.5 = 1- 0.86070K = 0.13929K

(b)

6.33

1 1 = = 3.333K years or 3 years 4 months l 0.3

n =100, p = 0.20, np = 20 ≥ 5 and n(1 – p) = 80 ≥ 5 µ = np = 20; 𝜎 = √𝑛𝑝(1 − 𝑝) = 4

6.34

(a)

𝑃(𝑋 = 25) ≅ 𝑃(24.5 ≤ 𝑋 ≤ 25.5) = 𝑃(1.125 ≤ 𝑍 ≤ 1.375) = 0.0457

(b)

𝑃(𝑋 > 25) = 𝑃(𝑋 ≥ 26) ≅ 𝑃(𝑋 ≥ 25.5) = 𝑃(𝑍 ≥ 1.375) = 0.0846

(c)

𝑃(𝑋 ≤ 25) ≅ 𝑃(𝑋 ≤ 25.5) = 𝑃(𝑍 ≤ 1.375) = 0.9154

(d)

𝑃(𝑋 < 25) = 𝑃(𝑋 ≤ 24) ≅ 𝑃(𝑋 ≤ 24.5) = 𝑃(𝑍 ≤ 1.125) = 0.8697

n =100, p = 0.40, np = 40 ≥ 5 and n(1 – p) = 60 ≥ 5 µ = np = 40; 𝜎 = √𝑛𝑝(1 − 𝑝) = 4.899

6.35

(a)

𝑃(𝑋 = 40) ≅ 𝑃(39.5 ≤ 𝑋 ≤ 40.5) = 𝑃(−0.1021 ≤ 𝑍 ≤ 0.1021) = 0.0813

(b)

𝑃(𝑋 > 40) = 𝑃(𝑋 ≥ 41) ≅ 𝑃(𝑋 ≥ 40.5) = 𝑃(𝑍 ≥ 0.1021) = 0.4594

(c)

𝑃(𝑋 ≤ 40) ≅ 𝑃(𝑋 ≤ 40.5) = 𝑃(𝑍 ≤ 0.1021) = 0.5406

(d)

𝑃(𝑋 < 40) = 𝑃(𝑋 ≤ 39) ≅ 𝑃(𝑋 ≤ 39.5) = 𝑃(𝑍 ≤ −0.1021) = 0.4594

n =10, p = 0.50, np = 5 ≥ 5 and n(1 – p) = 5 ≥ 5 µ = np = 5; 𝜎 = √𝑛𝑝(1 − 𝑝) = 1.5811 PHStat output:

(a)(i) 𝑃(𝑋 = 4) = 0.2051 (ii) 𝑃(𝑋 ≥ 4) = 0.8281 (iii) 𝑃(4 ≤ 𝑋 ≤ 7) = 0.9453 − 0.1719 = 0.7734 (b)

𝑃(𝑋 = 4) ≅ 𝑃(3.5 ≤ 𝑋 ≤ 4.5) = 𝑃(−0.9487 ≤ 𝑍 ≤ −0.3162) = 0.2045

𝑃(𝑋 ≥ 4) ≅ 𝑃(𝑋 ≥ 3.5) = 𝑃(𝑍 ≥ −0.9487) = 0.8286 𝑃(4 ≤ 𝑋 ≤ 7) ≅ 𝑃(3.5 ≤ 𝑋 ≤ 7.5) = 𝑃(−0.9487 ≤ 𝑍 ≤ 1.5812) = 0.7717 6.36

n =90, p = 0.3333, np = 30 ≥ 5 and n(1 – p) = 60 ≥ 5 µ = np = 30; 𝜎 = √𝑛𝑝(1 − 𝑝) = 4.4721 (a)

𝑃(𝑋 ≥ 20) ≅ 𝑃(𝑋 ≥ 19.5) = 𝑃(𝑍 ≥ −2.3479) = 0.9906

(b)

𝑃(𝑋 = 20) ≅ 𝑃(19.5 ≤ 𝑋 ≥ 20.5) = 𝑃(−2.3479 ≤ 𝑍 ≥ −2.1243) = 0.0074

(c)

𝑃(𝑋 < 20) = 𝑃(𝑋 ≤ 19) ≅ 𝑃(𝑋 ≤ 19.5) = 𝑃(𝑍 ≤ −2.3479) = 0.0094

6.37

Using Table E.2, first find the cumulative area up to the larger value, and then subtract the cumulative area up to the smaller value.

6.38

Find the Z value corresponding to the given percentile and then use the equation X =  + z .

6.39

The normal distribution is bell-shaped; its measures of central tendency are all equal; its middle 50% is within 1.33 standard deviations; and 99.7% of its values are contained within three standard deviations of its mean.

6.40

If the distribution is normal, the plot of the Z values on the horizontal axis and the original values on the vertical axis will be a straight line.

6.41

A binomial distribution is a discrete probability distribution with only discrete (integer) values. A continuity correction will make the normal approximation more accurate.

6.42

You can use the normal distribution to approximate the binomial distribution when both np and n(1 - p) are at least 5.

6.43

Let X = distance between flaws, metres exponential with λ= 1/5 = 0.2 per metre. (a)(i) P(X > 10) = 𝑒 −0.2×10 = 0.13533 (ii) P(X ≤ 1) = 1 − 𝑒 −0.2×1 = 0.18126 (iii) P(X ≤ 5) = 1 − 𝑒 −0.2×5 = 0.63212 (b)

6.44

0.2

𝜇= =

= 5 metres

Let X = inter-arrival time, minutes,

(a) (i)

Let X be exponential with λ = 30/60 = 0.5 per minute.

P(X  2) = e−0.52 = e −1 = 0.36789...

(ii)

P(X  0.5) = 1 − e−0.50.5 = 1 − e−0.25 = 0.22119...

(iii)

=

(b)

Let X be uniform with a = 0 and b = 4.

1 ( 4 − 2 ) = 0.5 4

(i)

P(X  2) =

(ii)

P(X  0.5) =

(iii)

=

(c)

6.45

1 1 = = 2 minutes l 0.5

1 ( 0.5 − 0 ) = 0.125 4

0+4 = 2 minutes 2

Let X be normal with

2 and

= 0.6.

(i)

P(X  2) = P ( Z  0 ) = 0.5

(ii)

0.5 − 2   P(X  0.5) = P  Z   = P(Z  −2.5) = 0.0062 0.6  

Let X = juice per orange, ml, given X normal with π = 135, σ = 12 (a)

P(135 < X < 140) = P(0 < Z < 0.42) = 0.6628 – 0.5000 = 0.1628

(b)

P(140 < X < 155) = P(0.42 < Z < 1.67) = 0.9525 – 0.6628 = 0.2897

(c)

P(X ≥ A) = 0.77 so P(X < A) = 0.23 so Z ≈ -0.74 and X = µ + Zσ = X = 135 + (-0.74) × 12 = 126.12 ml

(d)

P(XL < X < XU) = P(–1.28 < Z < 1.28) = 0.80 XL = 135 + (–1.28) × 12 = 119.64 ml XU = 135 + 1.28 × 12 = 150.36

6.46

(a) Hot Deals Rate A$ (Sunday evening) Mean

159.421

Median

152

Mode

189

Standard Deviation

51.848

Skewness

0.857

Range

208

189

122

The mean of $159.42 is higher than the median of 152. The box-and-whisker plot below is slightly right skewed. The interquartile range of 189 (189/51.8484 = 1.2922) is smaller than 1.33 times the standard deviation. Copyright © 2019 Pearson Australia (a division of Pearson Australia Group Pty Ltd) 9781488617249, Berenson, Basic Business Statistics 5e

The range of 208 (208/51.8484 = 4.011) is much smaller than six times the standard deviation. It can be concluded that the daily rate is right skewed, so is not normal.

Boxplot Hot Deals Rate A$ (a Sunday evening)

120

170

220

270

(b)

Normal Probability Plot Hot Deals Rate A$ (Sunday evening)

300 250 200

150 100 50 0 -2

-1

Z Value The normal probability plot shows a right skew, which suggests that the data is not normal. 6.47

Let X = time from 2020 to next major earthquake, years Assume X is exponential with l = (a)

1 = 0.004 per year 250

P(X  10) = e−0.00410 = e−0.04 = 0.96078...

6.48

(b)

P(X  50) = 1 − e−0.00450 = 1 − e−0.2 = 0.18126...

(c)

P(X  180) = e−0.004180 = e−0.72 = 0.48675...

X = number of questions correct, is binomial with n = 40 and p = 0.25.

m= np = 40 ´ 0.25 = 10 and s 2 = np(1-

p) = 40´ 0.25´ 0.75 = 7.5

Use normal approximation to the binomial. Therefore,

P(at least 50% correct) » P( X a ³ 19.5) æ 19.5 - 10 ö = Pç Z ³ ÷ è 7.5 ø = P(Z ³ 3.47) = 1- 0.99974 = 0.00026 »0 6.49

(a)(i) and (b)

All Ords - 2016-17 Financial Year 90 80

Frequency

70 60 50 40 30 20 10 0

-125

-125-100-100-75 -75 -50 -50 -25 -25 0 0 25 25 50 50 75 75 100 100125 125150 150175 175200

Change in Closing Price

All Ordinaries 2016-17 Financial year

Change in Closing price

-125

-100

-75

-50

-25

100

125

150

175

Both the Histogram and the Box-and-Whisker plot show that the change in closing price is mound shaped and appears to be approximately normal. However, there appear to an outlier representing a change in closing price of above 150. (a)(ii) & (b)

Change in Closing price Mean 1.792885 Median 4.6 Mode 16.6 Standard Deviation 37.1297 Skewness 0.009321 Range 292

Five-Number Summary Minimum First Quartile Median Third Quartile Maximum Interquartile Range

-121.4 -20.4 4.6 26.15 170.6 46.55

•

The mean of 1.79 is slightly less than the median of 4.6, indicating a possible slight left skew

•

The interquartile range of 46.55 is approximately 1.253 standard deviations.

•

The range of 292 is equal to 7.86 standard deviations, due to the outlier.

•

69% of values are within one standard deviation of the mean, 96% within two and 100% within three. These properties, also suggest that the change in change in closing prices is approximately normal.

(a)(iii) & (b) All Ordinaries 2016-17 Financial Year 200

Change in Closing price

150 100 50

Change in Closing price

-50 -100

-150 -3

-2

-1

0 Z Value

The normal probability plot also suggests that the change in closing prices is approximately normal, except for the outlier representing a change in closing price of above 150. 6.50

Let X = the age of houses, years, is uniform with, a = 20, b = 40. Height is

6.51

1 𝑏−𝑎

1 40−20

1 20

. 1

(a)

P(X > 30) = (40 − 30) ×

(b)

P(25 < X < 35) = (35 − 25) ×

(c)

P(X < 35) = (35 − 20) ×

1 20

20 15

1 2

Let X = time on hold, minutes normal with  = 20 ,  = 10 . (a)

P(X  40) = P(Z  2) = 0.0228

(b)

P(X  30) = P(Z  1) = 0.8413

(c)

P(X  10) = P(Z  −1) = 0.1587 = 15.87% 16% of calls are answered within 10 minutes.

6.52

6.53

(a)

P(X ≤ 20) = 1 − 𝑒 −(15)×20 = 1 − 0.2636 = 0.7364

(b)

P(X < 20) = 1 − 𝑒 −(15)×10 = 1 − 0.5134 = 0.4866

(c)

P(X > 20) = 𝑒 −(15)×40 = 0.0694

Let X = weekly household expenditure, $ assumed normal with  = 1300 and  = 350 (a)

P(X  500) = P(Z  −2.29) = 0.0110

(b)

P(X  1750) = P(Z  1.29) = 0.0985

(c)

P(1250  X  1500) = P(−0.14  Z  0.57) = 0.2714

(d)

Z = 2.33. X = 1300 + 2.33×350 = $2116

(e)

Z = –1.645, X = 1300 – 1.645×350 = $724

Normal Probabilities Common Data Mean 1300 Standard Deviation 350 Probability for X <= X Value 500 Z Value -2.285714 P(X<=500) 0.0111 Probability for X > X Value 1750 Z Value 1.2857143 P(X>1750) 0.0993 Find X and Z Given Cum. Pctage. Cumulative Percentage 5.00% Z Value -1.6449 X Value 724.3012

Probability for a Range From X Value 1250 To X Value 1500 Z Value for 1250 -0.142857 Z Value for 1500 0.5714286 P(X<=1250) 0.4432 P(X<=1500) 0.7161 P(1250<=X<=1500) 0.2729 Find X and Z Given Cum. Pctage. Cumulative Percentage 99.00% Z Value 2.3263 X Value 2114.2218

6.54

(a)(i)

Daily Water Usage

Daily Water Usage Kl (kilolitres)

0.4

0.6

0.8

1.0

1.2

1.4

1.6

1.8

2.0

2.2

2.4

Histogram of Daily Water Usage 35

Frequency

30 25 20 15 10

5 0 0.3

0.5

0.7

0.9

1.1

1.3

1.5

1.7

1.9

2.1

2.3

Kilolitres Both the Histogram and the Box-and-Whisker plot show that daily water usage is mound shaped and appears to be approximately normal.

Daily Water Usage Five-Number Summary Minimum First Quartile Median Third Quartile Maximum Range Interquartile Range

0.5536 1.0362 1.2670 1.5020 2.1830 1.6294 0.4658

The mean and median, 1.27, are the same to two decimal places

•

The interquartile range is approximately 1.4 standard deviations.

•

The range is approximately 4.9 standard deviations.

•

65% of values are within one standard deviation of the mean, 95% within two and 100% within three.

These properties, also suggest that daily water usage is approximately normal. (a)(ii) Normal Probability Plot

Daily Water Usage Kl (kilolitres)

2.5 2.0 1.5

Daily Water Usage Kl (kilolitres)

1.0 0.5 0.0 -3

-2

-1

0 Z Value

The normal probability plot also suggests that daily water usage is approximately normal. (b)

Let

X = Daily water usage, kl

Given X is normally distributed with mean  = 1.27 kl and standard deviation  = 0.33 kl

(i)

1.0 − 1.27   P(X  1) = P  Z   0.33   = P(Z  −0.82) = 0.2061 = 20.61%

1.27

− 0.82

On approximately 21% of days water usage is less than 1.0 kl.

(ii)

1.4 − 1.27   0.8 − 1.27 P(0.8  X  1.4) = P  Z  0.33   0.33 = P(−1.42  Z  0.39) = P(Z  0.39) − P(Z  −1.42) = 0.6517 − 0.0778 = 0.5739

0.8

1.27

1.4

−1.42

0.39

The proportion of days with water usage between 0.8 and 1.4 kl is approximately 0.57.

(iii)

2.0 − 1.27   P ( X  2) = P  Z   0.33   = P(Z  2.12) = 1 − P(Z  2.12) = 1 − 0.9830 = 0.0170

1.27

2.12

Probability that water usage tomorrow is over 3.0Kl is 0.017

Normal Probabilities Common Data Mean 1.27 Standard Deviation 0.33 Probability for X <= X Value 1 Z Value -0.818182 P(X<=1) 0.2066 Probability for X > X Value 2 Z Value 2.2121212 P(X>2) 0.0135 Probability for X<1 or X >2 P(X<1 or X >2) 0.2201 Probability for a Range From X Value 0.8 To X Value 1.4 Z Value for 0.8 -1.424242 Z Value for 1.4 0.3939394 P(X<=0.8) 0.0772 P(X<=1.4) 0.6532 P(0.8<=X<=1.4) 0.5760

6.55

Let X = Time to wait for bus, minutes X is uniform with a = 0 and b = 20.

6.56

1 ( 5 − 0 ) = 0.25 20

(a)

P(X  5) =

(b)

P(10  X  15) =

(c)

P(X  12) =

1 (15 − 10 ) = 0.25 20

1 (20 − 12 ) = 0.4 20

Let X = Lifetime, years X is normal with mean  years and standard deviation 0.75 years (a)

7−5 ) 0.75 = P(Z  2.67) = 1 − P(Z  2.67) = 1 − 0.9962

P(X  7) = P(Z 

= 0.0038

3−5 ) = P(Z  −2.67) = 0.0038 0.75

(b)

P(X  3) = P(Z 

(c)

Want x such that P(X < x) = 0.01

P (X < x) = P(Z < z) = 0.01

Find z such that P(Z < z) = 0.01, from tables z = −2.33 gives closest area of 0.0099 Therefore, warranty period should be at most 39 months since X =  + z = 5 + (−2.33 ×0.75) = 5 − 1.7474 = 3.2525 years

Common Data Mean 5 Standard Deviation 0.75 Probability for X <= X Value 3 Z Value -2.666667 P(X<=3) 0.0038 Probability for X > X Value 7 Z Value 2.6666667 P(X>7) 0.0038 Find X and Z Given Cum. Pctage. Cumulative Percentage 1.00% Z Value -2.3263 X Value 3.2552 Case Study - Tasman University Student Survey (a)(i)(ii) & (c) Bachelor of Business

Mean Standard Error Median Mode Standard Deviation Sample Variance Kurtosis Skewness Coefficient of Variation Sum Count

Five-Number Summary

Age 18 20 21 22 26 8 2

Minimum First Quartile Median Third Quartile Maximum Range Interquartile Range

WAM 27.9 56.3 64.7 71.15 93.9 66 14.85

Age

Tasman University Bachelor of Business 25

Frequency

20 15 10 5 0 18

Age

Bachelor of Businnes

Age

22 Age Years

Both the Histogram and the Box-and-Whisker plot show that age is mound shaped and appears to be approximately normal. • • •

The mean and median to 0 decimal places are the same at 21 years. The interquartile range, 2, is approximately 1.40 standard deviations The range, 8, is approximately 5.6 standard deviations.

These properties, also suggest that age is approximately normal. As does the normal probability plot below.

Normal Probability Plot 30 25

Age

20 15 Age

10 5 0 -2.5

-2

-1.5

-1

-0.5

0 Z Value

0.5

1.5

2.5

Weighted Average Mark (WAM)

Tasman University Bachelor of Business 20

Frequency

10 5 0 5

Weighted Average Mark (WAM) Bachelor of Business

WAM

55 60 65 70 Weighted Average Mark (WAM)

100

WAM

Normal Probability Plot

100 90 80 70 60 50 40 30 20 10 0

WAM

-2.5

-2

-1.5

-1

-0.5

0 Z Value

0.5

1.5

2.5

Both the Histogram and the Box-and-Whisker plot show that WAM is mound shaped and appears to be approximately normal. • The mean and median are similar at 63.9 and 64.6. • The interquartile range, 14.85, is approximately 1.20 standard deviations • The range, 66, is approximately 5.1 standard deviations. These properties, also suggest that WAM is approximately normal. As does the normal probability plot above. Expected Salary

Bachelor of Business - Expected Salary 20

Frequency

15 10 5

0 35

105

Thousand Dollars

Bachelor of Business

Expected Salary

100

105

Expected Starting Salary ($000)

Expected Salary

Normal Probability Plot 100 90 80 70 60 50 40 30 20 10 0

Expected Salary

-2.5

-2

-1.5

-1

-0.5

0 Z Value

0.5

1.5

2.5

Both the Histogram and the Box-and-Whisker plot show that expected salary is mound shaped and approximately normal. • • •

The mean at 63.5 is slightly less than the median 65. The interquartile range is approximately 1.24 standard deviations The range, 66, is approximately 4.55 standard deviations.

These properties, also suggest that expected salary is approximately normal. As does the normal probability plot above. Social Networking

Stem-and-Leaf Display Bachelor of Business Social Networking Stem unit: 1 0 000 1 000000000000000000000000000000000 2 0000000000000000000 3 00000 4 00

Normal Probability Plot 4.5

Social Networking

4.0 3.5 3.0 2.5 2.0

Social Networking

1.5 1.0 0.5 0.0 -2.5

-1

-0.5

The stem-and leaf plot approximately normal.

show

• •

-2

-1.5

0 Z Value

that

0.5

social

1.5

networking

2.5

mound

shaped

and

The mean at 1.5 is slightly more than the median 1.0. The interquartile range is approximately 1.18 standard deviations slightly less than 1.33. The range, 66, is approximately 4.73 standard deviations slightly less than 6.

•

These properties, also suggest that social networking is approximately normal. As does the normal probability plot above Satisfaction with On-campus Food Services

7 6

Satisfaction

5 4 3

Satisfaction

2 1 0 -2.5

-2

-1.5

-1

-0.5

0 Z Value

0.5

1.5

2.5

The stem-and leaf plot show that satisfaction is bi-modal so not normal. • •

The mean at 3.7 is similar median 4.0. The interquartile range is approximately 0.82 standard deviations significantly less than 1.33.

•

The range, 5 is approximately 4.1 standard deviations less than 6.

These properties, also suggest that satisfaction is not normal. As does the normal probability plot above Textbook Cost

Bachelor of Business Textbook Cost 16 14

Frequency

12 10 8 6 4

2 0 50

150 250 350 450 550 650 750 850 950 1050 1150 1250 1350 1450 $

Bachelor of Business

Textbook Cost

100

200

300

400

500

600 700 800 900 1000 Cost of Textbooks Last Semester ($)

1100

1200

1300

1400

Normal Probability Plot 1600 1400 Textbook Cost

1200 1000 800 Textbook Cost

600 400 200 0 -2.5

-2

-1.5

-1

-0.5

0 Z Value

0.5

1.5

2.5

1500

Both the Histogram and the Box-and-Whisker plot show that textbook cost is mound shaped but very right skewed. The normal probability plot also shows this right skew. • • •

The mean at 482 is less than the median 500. The interquartile range is approximately 1.33 standard deviations The range is approximately 6 standard deviations.

Text Messages

Bachelor of Business 18 16

Frequency

14 12 10 8 6

4 2

0 50

150

250

350

450

550

650

750

850

950

1050

Number of Text Messages

Bachelor of Business

Text Messages

100 150 200 250 300 350 400 450 500 550 600 650 700 750 800 850 900 950 1000 Number of Text Messages Sent In A Typical Week

Normal Probability Plot

Text Messages

1000 900 800 700 600 500 400 300 200 100 0

Text Messages

-2.5

-2

-1.5

-1

-0.5

0 Z Value

0.5

1.5

2.5

Both the Histogram and the Box-and-Whisker plot show that textbook cost is right skewed. The normal probability plot also shows this right skew. • •

The mean at 246 is larger than the median 200. The interquartile range is approximately 0.932 standard deviations, significantly less than 1.33 The range is approximately 4.2 standard deviations.

•

These properties, also suggest that the number of text messages is right skewed. Wealth Bachelor of Business

Wealth

30 40 50 60 70 Accumulated Wealth to be Rich (Millions Dollars)

100

Normal Probability Plot

120 100

Wealth

80 60 Wealth

40 20 0 -2.5

-2

-1.5

-1

-0.5

0 Z Value

0.5

1.5

2.5

The Box-and-Whisker plot shows that wealth is right skewed. The normal probability plot also shows this right skew. • • •

The mean at 7.2 is larger than the median 1.0. The interquartile range is approximately 0.22 standard deviations, significantly less than 1.33 The range is approximately 5.1 standard deviations.

These properties, also suggest that wealth is right skewed.

Tasman University Student Survey (b)(i)(ii) & (c)

Master of Business Administration

Five-Number Summary

Age Minimum 21 First Quartile 23 Median 25.5 Third Quartile 27 Maximum 47 Range 26 Interquartile Range 4

Age

Tasman University - MBA 25

Frequency

20 15 10 5 0 17.5

22.5

27.5

32.5

37.5

42.5

47.5

52.5

Age (years)

Tasman University

MBA Age

Age in Years

Age

Normal Probability Plot 50 45 40 35 30 25 20 15 10 5 0

Age

-2.5

-2.0

-1.5

-1.0

-0.5

0.0 Z Value

0.5

1.0

1.5

2.0

2.5

Both the Histogram and the Box-and-Whisker plot show that MBA age is right skewed. The normal probability plot also shows this right skew. • • •

The mean at 26.5 is slightly larger than the median 25.5. The interquartile range is approximately 0.75 standard deviations, significantly less than 1.33 The range is approximately 4.9 standard deviations.

These properties, also suggest that MBA age is not normal. Weighted Average Mark

Tasman University - MBA

Frequency

20 15

10 5 0 45

Weighted Average Mark

Tasman University

UG WAM (MBA Students)

MBA WAM

75 80 Weighted Average Mark (WAM)

100

Normal Probability Plot

80 MBA WAM

70 60 50 40

MBA WAM

30 20 10 0

Undergraduate WAM

-2.5

-2.0

-1.5

-1.0

-0.5 Z Value

0.0

0.5

1.0

1.5

2.0

2.5

Normal Probability Plot

100 90 80 70 60 50 40 30 20 10 0

Undergraduate WAM

-2.5

-2.0

-1.5

-1.0

-0.5

0.0 Z Value

0.5

1.0

1.5

2.0

2.5

The Histogram and the Box-and-Whisker plots show that MBA and Undergraduate WAM are mound shaped and appear approximately normal. The normal probability plots also shows that WAM for both MBA and undergraduate is approximately normal. • • •

The MBA and UG WAM’s means and medians are similar. The interquartile ranges are approximately 1.5 and 1.8 standard deviations. The ranges are approximately 4.0 and 3.6 standard deviations.

These properties, also suggest that MBA and UG WAM may be approximately normal.

Expected Salary Tasman University

MBA Expected Salary

100

120

140 160 180 200 220 240 Expected Salary on Graduation ($000s)

260

280

300

320

340

Normal Probability Plot 350

Expected Salary

300 250 200 150

Expected Salary

100 50 0 -2.5

-2.0

-1.5

-1.0

-0.5

0.0 Z Value

0.5

1.0

1.5

2.0

2.5

The Box-and-Whisker plot shows that MBA expected salary is right skewed. The normal probability plot also shows this right skew. • • •

The mean at 103 is larger than the median 90. The interquartile range is approximately 0.9 standard deviations, significantly less than 1.33 The range is approximately 5.8 standard deviations.

These properties, also suggest that MBA expected salary is not normal. Number of Full-time Jobs

Stem-and-Leaf Display MBA - Number of Full-time Jobs Stem unit: 1 0 0000 1 000000000000000 2 00000000000000000 3 00000 4 000

Normal Probability Plot

4.5 4.0

Full-time Jobs

3.5 3.0 2.5 2.0

Full-time Jobs

1.5 1.0 0.5 0.0 -2.5

-2.0

-1.5

-1.0

-0.5

0.0 Z Value

0.5

1.0

1.5

2.0

2.5

The stem-and-leaf plot shows that number of full-time jobs is mound shaped but may not be normal. • • •

The mean 1.7 is slightly less than the median 2. The interquartile range is approximately 0.99 standard deviations, significantly less than 1.33 The range is approximately 3.9 standard deviations.

These properties with the normal probability plot suggest that number of full-time jobs is not normal. Textbook Cost

16 14 12 10 8 6 4 2 0

50 150 250 350 450 550 650 750 850 950 1050 1150 1250 1350 1450 1550 1650 1750 1850 1950 2050 2150 2250 2350

Frequency

Tasman University - MBA

Textbook Cost $ Tasman University

MBA Textbook Cost

250

500

750

1000 1250 1500 Cost of Textbooks Last Semester, $

1750

2000

2250

Normal Probability Plot 2500

Textbook Cost

2000 1500 Textbook Cost

1000 500

0 -2.5

-2.0

-1.5

-1.0

-0.5

0.0 Z Value

0.5

1.0

1.5

2.0

2.5

Both the Histogram and the Box-and-Whisker plot show that MBA textbook cost is right skewed. The normal probability plot also shows this right skew. • •

The mean at 364 is larger than the median 300. The interquartile range is approximately 0.56 standard deviations, significantly less than 1.33 The range is approximately 5.9 standard deviations.

•

These properties, also suggest that MBA textbook cost is not normal. Advisory Services

Stem-and-Leaf Display Tasman University MBA - Advisory Rating Seven point scale Stem unit: 1 2 0 3 000 4 00000000000000000 5 0000000000000000 6 000000 7 0 Normal Probability Plot 8

Advisory Rating

7 6 5 4 Advisory Rating

3 2 1 0

-2.5

-2.0

-1.5

-1.0

-0.5

0.0 Z Value

0.5

1.0

1.5

2.0

2.5

The stem-and-leaf plot shows that number of full-time jobs is mound shaped and may be approximately normal. • •

The mean 4.6 is slightly less than the median 5. The interquartile range is approximately 1.03 standard deviations

•

The range is approximately 5.14 standard deviations.

These properties with the normal probability plot suggest that advisory rating may be approximately normal. Number of Text Messages

Tasman University MBA 25

Frequency

15 10 5 0

150

250

350

450

550

650

750

850

950

1050

1150

400 500 600 700 800 900 Number of Text Messages Sent in a Typical Week

1000

1250

1350

Number of Text Message

Tasman University

MBA Text Messages

100

200

300

1400

1100

1200

1300

Normal Probability Plot

Text Messages

1200 1000

800 600

Text Messages

400 200 0 -2.5

-2.0

-1.5

-1.0

-0.5

0.0 Z Value

0.5

1.0

1.5

2.0

2.5

Both the Histogram and the Box-and-Whisker plot show that MBA number of text messages is right skewed. The normal probability plot also shows this right skew. • • •

The mean 232.5 is significantly larger than the median 100. The interquartile range is approximately 1.28 standard deviations The range is approximately 4.21 standard deviations.

These properties, also suggest that MBA number of text messages is not normal. Copyright © 2019 Pearson Australia (a division of Pearson Australia Group Pty Ltd) 9781488617249, Berenson, Basic Business Statistics 5e

Wealth Tasman University

MBA Wealth

30 40 50 60 70 Accoumulated Wealth to be Rich (Million Dollars)

100

Normal Probability Plot

120 100

Wealth

80 60 Wealth 40 20 0 -2.5

-2.0

-1.5

-1.0

-0.5

0.0 Z Value

0.5

1.0

1.5

2.0

2.5

The Box-and-Whisker plot shows that MBA wealth is right skewed. The normal probability plot also shows this right skew. • • •

The mean 10.64 is significantly larger than the median 2.75. The interquartile range is approximately 0.4 standard deviations, significantly smaller than 1.33 standard deviations. The range is approximately 4.4 standard deviations.

These properties, also suggest that MBA wealth is not normal. (d)

Let X = BBus WAM X is assumed normal with mean 63.9 and standard deviation 12.8 (i)

65 − 63.9   P(X  65) = P  Z   12.8   = P(Z  0.09) = 1 − P(Z  0.09) = 1 − 0.5359 = 0.4641

(ii)

P(X  75) = P(Z  0.87) = 0.1922

(iii)

P(X  85) = P(Z  1.65) = 0.0495

(iv)

P(X  50) = P(Z  −1.09) = 0.1379

(v)

P(50  X  70) = P(−1.09  Z  0.48) = P(Z  0.48) − P(Z  −1.09) = 0.6844 − 0.1379 = 0.5465

(vi)

Want x such that P(X < x) = 0.1

P(X < x) = P(Z < z) = 0.1

63.9

Find z such that P(z < Z) = 0.1, from tables z = −1.28 gives closest area of 0.1003 Therefore, lowest 10% of BBus students have WAM scores below 47.5 since X =  + z = 63.9 + (−1.28×12.8) = 63.9 − 16.384 ≈ 47.5 (vii)

Want x such that P(X < x) = 0.95 Find z such that P(z < Z) = 0.95, from tables z = 1.64/5 gives closest area of 0.9495/0.9505 Therefore, highest 95% of BBus students have WAM scores above 84.9/85.0 since X =  + z = 63.9 + (1.64/5×12.8) = 84.9/85.0

(e)

Let X = MBA WAM X is assumed normal with mean 73.8 and standard deviation 8.6 (i)

65 − 73.8   P(X  65) = P  Z   8.6   = P(Z  −1.02) = 1 − P(Z  −1.02) = 1 − 0.1539 = 0.8461

(ii)

P(X  75) = P(Z  0.14) = 0.4443

(iii)

P(X  85) = P(Z  1.30) = 0.0968

(iv)

P(X  50) = P(Z  −2.77) = 0.0028

(v)

P(50  X  70) = P(−2.77  Z  −0.44) = P(Z  −0.44) − P(Z  −2.77) = 0.3300 − 0.0028 = 0.3272

(vi)

Want x such that P(X < x) = 0.1

P(X < x) = P(Z < z) = 0.1

73.8

Find z such that P(z < Z) = 0.1, from tables z = −1.28 gives closest area of 0.1003 Therefore, lowest 10% of MBA students have WAM scores below 62.8 since X =  + z = 73.8 + (−1.28×8.6) = 73.8 − 11.008 ≈ 62.8 (vii)

Want x such that P(X < x) = 0.95 Find z such that P(z < Z) = 0.95, from tables z = 1.64/5 gives closest area of 0.9495/0.9505 Therefore, highest 95% of MBA students have WAM scores above 87.9/88.0 since X =  + z = 73.8 + (1.64/5×8.6) = 87.9/88.0

Safe-As Houses Real Estate (a)(i) Regional City 1 State A

Five-Number Summary Regional City 1 State A Price $000 Minimum 165 First Quartile 231.5 Median 298 Third Quartile 383.5 Maximum 645 Range 480 Interquartile Range 152

Internal Area m^2 66.5 93.45 141.5 171.6 349.6 283.1 78.15

Bedrooms Bathrooms 1 1 2 1 3 1 4 2 9 3 8 2 2 1

Garages 0 1 2 2 6 6 1

Price

Residential Property Prices Regional City 1 State A 60

Frequency

40 30 20 10 0 50

150

250

350

450

550

650

750

Price $000

Normal Probability Plot 700 600

Price $000

500 400 300

Price $000

200 100

0 -3

-2

-1

0 Z Value

The histogram and normal probability plot show that price is skewed to the right. • • •

The mean 321 is larger than the median 298. The interquartile range is approximately 1.37 standard deviations The range is approximately 4.3 standard deviations.

These properties, also suggest that price is skewed to the right. Internal Area

Frequency

Residental Properties: Regional City 1 State A 50 45 40 35 30 25 20 15 10 5 0

125

175

225

275

325

375

Internal Area Square Metres Normal Probability Plot 400

Internal Area m^2

350 300

250 200 Internal Area m^2

150 100 50 0 -3

-2

-1

0 Z Value

The histogram and normal probability plot show that internal area is skewed to the right, with an extreme outlier. However, internal area may be approximately normal. • • •

The mean and median are similar. The interquartile range is approximately 1.8 standard deviations. The range is approximately 6.5 standard deviations.

These properties, also suggest that internal area may be approximately normal. Bedrooms, Bathrooms and Garages Regional City 1 State A

Garages

Bathrooms

Bedrooms

Normal Probability Plot 10 9 8 7 6 5 4 3 2 1 0

Bedrooms

-3

-2

-1

0 Z Value

Normal Probability Plot 3.5 3.0

Bathrooms

2.5 2.0

Bathrooms

1.5 1.0 0.5 0.0

-3

-2

-1

0 Z Value

Normal Probability Plot 7 6

Garages

5 4 3

Garages

2 1

0 -3

-2

-1

0 Z Value

The Box-and-Whisker plots and the normal probability plots show that the number of bedrooms, bathrooms and garages are all skewed to the right. Bedrooms • • •

The mean is slightly larger than the median. The interquartile range is approximately 1.9 standard deviations The range is approximately 7.5 standard deviations.

Bathrooms • • •

The mean is slightly larger than the median. The interquartile range is approximately 1.7 standard deviations The range is approximately 3.64 standard deviations.

Garages • • •

The mean is slightly less than the median. The interquartile range is approximately 1.2 standard deviations The range is approximately 7.09 standard deviations.

These properties, also suggest that the number of bedrooms, bathrooms and garages may not be normal. (b)(i) Coastal City 1 State A

Coastal City 1 State A

Five-Number Summary

Garages 0 1 2 2 8 8 1

Price

Frequency

Residential Property Prices Coastal City 1 State A 45 40 35 30 25 20 15 10 5 0 50 150 250 350 450 550 650 750 850 950 1050 1150 1250 1350 Price $000

Normal Probability Plot 1400 1200

Price $000

1000 800 Price $000

600 400 200 0 -3

-2

-1

0 Z Value

The histogram and normal probability plot show that price is skewed to the right. • • •

The mean 459 is larger than the median 399. The interquartile range is approximately 1.1 standard deviations The range is approximately 5.3 standard deviations.

These properties, also suggest that price is skewed to the right.

Internal Area

Residental Properties: Coastal City 1 State A 50

Frequency

40 30 20 10 0 25

125

175

225

275

325

Internal Area Square Metres

Normal Probability Plot 350

Internal Area m^2

300 250 200 150

Internal Area m^2

100 50

0 -3

-2

-1

0 Z Value

The histogram and normal probability plot show that internal area is skewed to the right. However, internal area may be approximately normal. • • •

The mean and median are similar. The interquartile range is approximately 1.4 standard deviations. The range is approximately 4.7 standard deviations.

These properties, also suggest that internal area may be approximately normal.

Bedrooms, Bathrooms and Garages Coastal City 1, State A

Garages Bathrooms Bedrooms

Normal Probability Plot 8 7

Bedrooms

6 5

4 Bedrooms

3 2 1 0 -3

-2

-1

0 Z Value

Normal Probability Plot 4.5 4.0

Bathrooms

3.5 3.0 2.5

2.0

Bathrooms

1.5 1.0 0.5 0.0 -3

-2

-1

0 Z Value

Normal Probability Plot 9 8 7

Garages

6 5 4

Garages

3 2

1 0 -3

-2

-1

0 Z Value

The Box-and-Whisker plots and the normal probability plots show that the number of bedrooms, bathrooms and garages are all skewed to the right. Bedrooms • • •

The mean is slightly larger than the median. The interquartile range is approximately 1.8 standard deviations The range is approximately 5.2 standard deviations.

Bathrooms • • •

The mean is slightly less than the median. The interquartile range is approximately 1.3 standard deviations The range is approximately 3.9 standard deviations.

Garages • • •

The mean is slightly less than the median. The interquartile range is approximately 1.1 standard deviations The range is approximately 8.7 standard deviations.

These properties, also suggest that the number of bedrooms, bathrooms and garages may not be normal.

Chapter 7: Sampling distributions Learning objectives After studying this chapter you should be able to: 1. Interpret the concept of the sampling distribution 2. calculate probabilities related to the sample mean 3. recognise the importance of the Central Limit Theorem 4. calculate probabilities related to the sample proportion Please note that the answers to some problems in this chapter have been calculated using PHStat2. There may be slightly different results when Table E2 is used. 7.1

PHStat2 output: Common Data Mean Standard Deviation

100 2

Probability for X <= X Value 95 Z Value –2.5 P(X<=95) 0.0062097 Probability for X > X Value Z Value P(X>102.2) Probability >102.2

for

102.2 1.1 0.1357 X<95

P(X<95 or X>102.2) 0.1419

Probability for a Range From X Value 95 To X Value 97.5 Z Value for 95 –2.5 Z Value for 97.5 –1.25 P(X<=95) 0.0062 P(X<=97.5) 0.1056 P(95<=X<=97.5) 0.0994 Find X and Z Given Cum. Pctage Cumulative Percentage 35.00% – Z Value 0.38532 99.2293 X Value 6

(a)(i) P( X < 95) = P(Z < – 2.50) = 0.0062 (ii) P(95 < X < 97.5) = P(– 2.50 < Z < – 1.25) = 0.1056 – 0.0062 = 0.0994 (iii) P( X > 102.2) = P(Z > 1.10) = 1.0 – 0.8643 = 0.1357 (b)

P( X > A) = P(Z > – 0.39) = 0.65

X = 100 – 0.39(

10 ) = 99.22 25

7.2

PHStat2 output: Common Data Mean 50 Standard Deviation 0.5 Probability for X <= X Value 47 Z Value –6 P(X <=47) 9.866E-10 Probability for X > X Value Z Value P(X >51.5)

51.5 3 0.0013

Probability for X <47 or X>51.5 P(X<47 or X>51.5) 0.0013 Probability X> X Value Z Value P(X>51.1)

Probability for a Range From X Value 47 To X Value 49.5 Z Value for 47 –6 Z Value for 49.5 –1 P(X<=47) 0.0000 P(X<=49.5) 0.1587 P(47<=X<=49.5) 0.1587 Find X and Z Given Cum. Pctage Cumulative Percentage 65.00% Z Value 0.38532 X Value 50.19266

for 51.1 2.2 0.0139

(a)(i) P( X < 47) = P(Z < – 6.00) = virtually zero (ii) P(47 < X < 49.5) = P(– 6.00 < Z < – 1.00) = 0.1587 – 0.00 = 0.1587 (iii) P( X > 51.1) = P(Z > 2.20) = 1.0 – 0.9861 = 0.0139 X = 50 + 0.39(0.5) = 50.195 (b) P( X > A) = P(Z > 0.39) = 0.35 7.3

(a) (b)

(c)

For samples of 25 travel expense vouchers for a university in an academic year, the sampling distribution of sample means is the distribution of means from all possible samples of 25 vouchers that could occur. For samples of 25 absentee records in 2009 for employees of a large manufacturing company, the sampling distribution of sample means is the distribution of means from all possible samples of 25 records that could occur. For samples of 25 sales of E10 at service stations located in a particular state, the sampling distribution of sample means is the distribution of means from all possible samples of 25 sales that could occur.

7.4

(a)

Sampling Distribution of the Mean for n = 2 (without replacement)

Sample Number

(a)

Outcomes

Sample Means X i

1, 3

X1 = 2

1, 6

X 2 = 3.5

1, 7

X3 = 4

1, 9

X4 = 5

1, 10

X 5 = 5.5

3, 6

X 6 = 4.5

3, 7

X7 = 5

3, 9

X8 = 6

3, 10

X 9 = 6.5

6, 7

X10 = 6.5

6, 9

X 11 = 7.5

6, 10

X 12 = 8

7, 9

X13 = 8

7, 10

X 14 = 8.5

9, 10

X15 = 9.5

Mean of All Possible Sample Means:

90 X = =6 15

Mean of All Population Elements:

=

1 + 3 + 6 + 7 + 9 + 10 =6 6

Both means are equal to 6. This property is called unbiasedness.

7.4

(b)

Sampling Distribution of the Mean for n = 3 (without replacement)

Sample Number

Outcomes

Sample Means X i

1, 3, 6

X 1 = 3 1/3

1, 3, 7

X 2 = 3 2/3

1, 3, 9

X 3 = 4 1/3

1, 3, 10

X 4 = 4 2/3

1, 6, 7

X 5 = 4 2/3

1, 6, 9

X 6 = 5 1/3

1, 6, 10

X 7 = 5 2/3

3, 6, 7

X 8 = 5 1/3

3, 6, 9

X9 = 6

3, 6, 10

X10 = 6 1/3

6, 7, 9

X 11 = 7 1/3

6, 7, 10

X 12 = 7 2/3

6, 9, 10

X13 = 8 1/3

7, 9, 10

X 14 = 8 2/3

1, 7, 9

X15 = 5 2/3

1, 7, 10

X16 = 6

1, 9, 10

X17 = 6 2/3

3, 7, 9

X18 = 6 1/3

3, 7, 10

X19 = 6 2/3

3, 9, 10

X 20 = 7 1/3

X =

120 =6 20

This is equal to  , the population mean.

(c)

The distribution for n = 3 has less variability. The larger sample size has resulted in sample means being closer to  .

(d)

Sampling Distribution of the Mean for n = 2 (with replacement)

Sample Number

Outcomes

Sample Means X i

1, 1

X1 = 1

1, 3

X2 = 2

1, 6

X 3 = 3.5

1, 7

X4 = 4

1, 9

X5 = 5

1, 10

X 6 = 5.5

3, 1

X7 = 2

3, 3

X8 = 3

3, 6

X 9 = 4.5

3, 7

X10 = 5

3, 9

X 11 = 6

3, 10

X 12 = 6.5

6, 1

X13 = 3.5

6, 3

X 14 = 4.5

6, 6

X15 = 6

6, 7

X16 = 6.5

6, 9

X17 = 7.5

6, 10

X18 = 8

7, 1

X19 = 4

7, 3

X 20 = 5

7, 6

X 21 = 6.5

7, 7

X 22 = 7

7, 9

X 23 = 8

7, 10

X 24 = 8.5

9, 1

X 25 = 5

9, 3

X 26 = 6

9, 6

X 27 = 7.5

9, 7

X 28 = 8

9, 9

X 29 = 9

9, 10

X 30 = 9.5

10, 1

X 31 = 5.5

10, 3

X 32 = 6.5

10, 6

X 33 = 8

10, 7

X 34 = 8.5

10, 9

X 35 = 9.5

10, 10

X 36 = 10

(d)

(a)

Mean of All Possible Sample Means:

X = (b)

(c)

7.5

(a)

(b) (c) (d)

7.6

(a)

(b) (c) 7.7

(a)

Mean of All Population Elements:

216 =6 36

=

1+ 3 +6 + 7 + 7 +12 =6 6

Both means are equal to 6. This property is called unbiasedness. Repeat the same process for the sampling distribution of the mean 3 for n = 3 (with replacement). There will be 6 = 216 different samples. X = 6 This is equal to  , the population mean. The distribution for n = 3 has less variability. The larger sample size has resulted in more sample means being close to  .

For samples of 16 days at the South Asian airport, the sampling distribution of sample means is the distribution of means from all possible samples of 16 days that could occur. 98,000−110,000 P( X < 98,000) = 𝑃(𝑍 < = −2.38 = 0.0087 5050

P(102,000 < X < 104,500) = P(–1.58 < Z < –1.09) = 0.1379 – 0.0571 = 0.0808 P(A < X < B) = P(–0.84 < Z < 0.84) = 0.60 Lower bound: X = 110,000 – 0.84(5050) = 105,758 Upper bound: X = 110,000 + 0.84(5050) = 114,242 For samples of size 2, the sampling distribution of X will resemble the right-skewed population distribution. Because the mean is larger than the median, the distribution of the sales price of new houses is skewed to the right, and so is the sampling distribution of X although it will be less skewed than the population. When the sample size is 100, the sampling distribution of X will be very close to a normal distribution as a result of the central limit theorem. 1,235,000−1,236,450 P( X < 1,235,000) = 𝑃(𝑍 < = 𝑃(𝑍 − 0.0967) = 0.4602

X =



15,000

2 = = 0.4 n 25

P(6.9 < X < 8.2) = P(– 2.75 < Z < 0.50) = 0.6915 – 0.0028 = 0.6887 (b)

PHStat2 output:

Probability for a Range From X Value 7.5 To X Value 8 Z Value for 7.5 –1.25 Z Value for 8 0 P(X<=7.5) 0.1056 P(X<=8) 0.5000 P(7.5<=X<=8) 0.3944 P(7.5 < X < 8.0) = P(– 1.25 < Z < 0) = 0.5 – 0.1056 = 0.3944

(c)

X = (d)

7.8

(a) (b) (c)

(d) 7.9

(a) (b) (c)

7.11

7.12

7.13

= 0 .2 100 P(6.9 < X < 8.2) = P(– 6.50 < Z < 1.00) = 0.8413 – 0.0000 = 0.8413 n

With the sample size increasing from n = 25 to n = 100, more sample means will be closer to the distribution mean. The standard error of the sampling distribution of size 100 is much smaller than that of size 25, so the likelihood that the sample mean will fall within 0.2 minutes of the mean is much higher for samples of size 100 (probability = 0.8413) than for samples of size 25 (probability = 0. 3830). P( X < 38) = P(Z < –2.8284) = 0.0023 P(39 < X < 41) = P(–1.4142 < Z < 1.4142) = 0.8427 P(A < X < B) = P(–1.2816 < Z < 1.2816) = 0.80 A = 40 – 1.2816 (0.7071) = 39.0938; B = 40 + 1.2816 (0.7071) = 40.9062 P( X < A) = P(Z < 1.2816) = 0.90; A = 40 + 1.2816 (0.7071) = 40.9062

(a)

P( X < 3.75) = P(Z < 1.5336) = 0.9370 P(3.7 < X < 3.9) = P(0.9859 < Z < 3.1768) = 0.1604 P(A < X < B) = P(–1.2816 < Z < 1.2816) = 0.80 A = 3.61 – 1.2816 (0.0913) = 3.4930; B = 3.61 + 1.2816 (0.09131) = 3.7270 P( X < A) = P(Z < 1.2816) = 0.90; A = 3.61 + 1.2816 (0.09130) = 3.7270 p = 48/64 = 0.75

(b)

p =

(a)

p = 16/50 = 0.32

(b)

p =

(a)

p = 14/40 = 0.35

(b)

p =

(a)(i)

 p =  = 0.501 ,  p =

(d) 7.10



0.70(0.30) = 0.0573 64

0.4  0.6 = 0.0693 50

0.30(0.70) = 0.0725 40

 (1 −  ) n

0.501(1 − 0.501) = 0.05 100

Partial PHStat2 output: Probability for X > X Value 0.55 Z Value 0.98 P(X>0.55) 0.1635 P(p > 0.55) = P (Z > 0.98) = 1 – 0.8365 = 0.1635 (ii)

 p =  = 0.60 ,  p =

 (1 −  ) n

0.6 (1 − 0.6 ) = 0.04899 100

Partial PHStat2 output: Probability for X > X Value 0.55 Z Value –1.020621 P(X>0.55) 0.8463 P(p > 0.55) = P (Z > – 1.021) = 1 – 0.1539 = 0.8461 (iii)

 p =  = 0.49 ,  p =

 (1 −  ) n

0.49 (1 − 0.49 ) = 0.05 100

Partial PHStat2 output: Probability for X > X Value 0.55 Z Value 1.2002401 P(X>0.55) 0.1150 P(p > 0.55) = P (Z > 1.20) = 1 – 0.8849 = 0.1151 (b)

Increasing the sample size by a factor of 4 decreases the standard error by a factor of 2. (i)

Partial PHStat2 output: Probability for X > X Value 0.55 Z Value 1.9600039 P(X>0.55) 0.0250 P(p > 0.55) = P (Z > 1.96) = 1 – 0.9750 = 0.0250

(ii)

Partial PHStat2 output: Probability for X > X Value 0.55 Z Value –2.041241 P(X>0.55) 0.9794 P(p > 0.55) = P (Z > – 2.04) = 1 – 0.0207 = 0.9793

(iii)

Partial PHStat2 output: Probability for X > X Value 0.55 Z Value 2.4004801 P(X>0.55) 0.0082 P(p > 0.55) = P (Z > 2.40) = 1 – 0.9918 = 0.0082 If the sample size is increased to 400, the probability in (i), (ii) and (iii) is smaller, larger and smaller, respectively because the standard error of the sampling distribution of the sample proportion becomes smaller and, hence, the sampling distribution is more concentrated around the true population proportion.

7.14

(a) Partial PHStat2 output: Probability for a Range From X Value 0.5 To X Value 0.6

Z Value for 0.5 0 Z Value for 0.6 2.828427 P(X<=0.5) 0.5000 P(X<=0.6) 0.9977 P(0.5<=X<=0.6) 0.4977 P(0.50 < p < 0.60) = P(0 < Z < 2.83) = 0.4977 (b) Partial PHStat2 output: Find X and Z Given Cum. Pctage Cumulative Percentage 95.00% Z Value 1.644854 X Value 0.558154 P(– 1.645 < Z < 1.645) = 0.90 p = .50 – 1.645(0.0354) = 0.4418 p = .50 + 1.645(0.0354) = 0.5582 (c) Partial PHStat2 output: Probability for X > X Value 0.65 Z Value 4.2426407 P(X>0.65) 0.0000 P(p > 0.65) = P (Z > 4.24) = virtually zero (d)

Partial PHStat2 output:

Probability for X > X Value 0.6 Z Value 2.8284271 P(X>0.6) 0.0023 If n = 200, P(p > 0.60) = P (Z > 2.83) = 1.0 – 0.9977 = 0.0023 Probability for X > X Value 0.55 Z Value 3.1622777 P(X>0.55) 0.00078 If n = 1000, P(p > 0.55) = P (Z > 3.16) = 1.0 – 0.99921 = 0.00079 More than 60% correct in a sample of 200 is more likely than more than 55% correct in a sample of 1000. 7.15

(a) (b)

P(p < 0.24) = P(Z < –0.2066) = 0.4168 P(0.242 < p < 0.25) = P(–0.1378 < Z < 0.1378) = 0.5557-0.4443 = 0.1114 (c) P(0.245 < p < 0.247) = P(–0.0344 < Z < 0.0344) = 0.5120 – 0.4880 = 0.0240 (d)(i) P(p < 0.24) = P(Z < –0.1393) = 0.4443 (ii) P(0.242 < p < 0.25) = P(–0.0929 < Z < 0.0929) = 0.5359 – 0.4641= 0.0718 (iii) P(0.245 < p < 0.247) = P(–.0232 < Z < 0.0230) = 0.5080 – 0.4920 = 0.0160

7.17

(b) (c)(i) (ii) (a) (b) (c)

7.18

(a) (b) (c)

(d)

P(p < 0.028) = P(Z < 0.1690) = 0.5675 P(p > 0.091) = P(Z > 0.1432) = 0.4443 P(p < 0.028) = P(Z <0.4297) = 0.6664 P(0.075 < p < 0.082) = P(–0.1843 < Z < 0.0737) = 0.5279 – 0.4286 = 0.0993 P(A < p < B) = P(–1.6449 < Z < 1.6449) = 0.90 A = 0.08 – 1.6449(0.0271) = 0.0354 B = 0.08 + 1.6449(0.0271) = 0.1246 P(A 0.502) = P(Z > 0.48) = 1 – 0.6844 = 0.3156. If the true population proportion is 49%, the proportion of samples with 50.2% or more is about 32% and the previous population proportion is a fairly good indicator. P(p > 0.502) = P(Z > 0.24) = 1 – 0.5948 = 0.4052. If the true population proportion is 49%, the proportion of samples with 50.2% or more is about 41%. Hence, the previous population proportion is likely to be good indicator of the true value. When the sample size is smaller in (c) compared with (b), the standard error of the sampling distribution of the sample proportion is larger and hence a higher portion of samples will have 50.2% or more part-time workers.

7.19

(a) (b) (c)

P(0.24 0.30) = P(Z > 1.1547) = 0.1241

7.20

(a) (b)

P(p > 0.18) = P(Z > -1.1656) = 0.8790 P(p < 0.335) = P(Z < –0369) = 0.4840

7.21

Because the average of all the possible sample means of size n is equal to the population mean.

7.22

The variation of the sample means becomes smaller as larger sample sizes are taken. This is due to the fact that an extreme observation will have a smaller effect on the mean in a larger sample than in a small sample. Thus, the sample means will tend to be closer to the population mean as the sample size increases.

7.23

As larger sample sizes are taken, the effect of extreme values on the sample mean becomes smaller and smaller. With large enough samples, even though the population is not normally distributed, the sampling distribution of the mean will be approximately normally distributed.

7.24

The probability distribution is the distribution of a particular variable of interest, while the sampling distribution represents the distribution of a statistic.

7.25

When the items of successes and the number of failures are at least 5, the normal distribution can be used to approximate the binomial distribution.

7.26

µ = 0.503; σ = 0.004; n = 25 (a)(i) P(0.5 < X < 0.503) = P(–3.75 < Z < 0) = 0.4999 (ii) P(0.49 < X < 0.5) = P(–16.25 < Z < –3.75) = 0.0009 (iii) P( X > 0.51) = P(Z > 8.75) = Virtually zero (iv) P( X < 0.49) = P(Z < –16.25) = Virtually zero (b) P( X > A) = P(Z > 1.50) = 0.9332 A = 0.503 – 1.5*0.0008 = 0.5018

7.27

µ = 2.0; σ = 0.05; n = 25 (a)(i) P(1.99 < X < 2.0) = P(–1.00 < Z < 0) = 0.5 – 0.1587 = 0.3413 (ii) P( X < 1.98) = P(Z < –2.00) = 0.0228 (iii) P( X > 2.01) = P(Z > 1.00) = 1 – 0.8413 = 0.1587 (iv) P( X > A) = P(Z > –2.33) = 0.99, A = 2.00 – 2.33*0.01 =1.9767 (b) P(A < X < B) = P(–2.58 < Z < 2.58) = 0.99 A = 2.00 – 2.58*0.01 = 1.9742 B = 2.00 + 2.58*0.01 = 2.0258

7.28

(a) (b) (c)

P(p > 0.262) = P(Z > -0.2071) = 0.5832 P(0.27 0.7927) = 0.2533 P(p > A) = P(Z > –0.7389) = 0.77 A = 0.2678 – 0.7389(0.0280) = 0.2471

7.29

(a) (b) (c) (d)

P(0.18 –0.94) = 0.1736 – 0.0025 = 0.1711 P(0.16 0) = 0.5 – 0.00009 = 0.4999 P(0.14 0.94) = 0.8264 – 0.0000 = 0.8264 P(0.12 1.87) = 0.9693 – 0.0000 = 0.9693

7.30

(a) (b) (c)

P( X < 270) = P(Z <1.05) = 0.8531 P(265 < X < 275) = P(0.5270< Z <1.5811) = 0.9429 – 0.7019 = 0.2410 P( X > 282) = P(Z > 2.3190) = 1 – 0.9898 = 0.0102

7.31

(a)(i) P(X < 8.2) = P(Z < 1.0) = 0.8413 (ii) P(6.9 < X < 7.8) = P(–0.8571 < Z < 0.4286) = 0.6664 – 0.1949 = 0.4715 (iii) P(X > 7.9) = P(Z > 0.5714) = 1 – 0.7157 = 0.2843 (b) (i) P( X < 8.2) = P(Z < 3.1623) =0.99921 (ii) P(6.9 < X < 7.8) = P(–2.7105 < Z < 1.3553) = 0.9131 – 0.0034 = 0.9097 (iii) P( X > 7.9) = P(Z > 1.8070) = 1 – 0.9649 = 0.0351

7.32

(a) (b) (c) (d) (e) (f) (g)

P(X < 0)=P(Z < -0.154)=0.4404 P(-20 < X < -10) = P(-2.154 < Z < -1.154) = 0.1251 – 0.0158 = 0.1093 P(X > -5) = P (Z > -0.654) =1- 0.7422 = 0.2578 P( X < 0)=P(Z <-0.308) = 0.3783 P(-20 < X <-10) = P(-4.308 < Z < -2.308)= 0.0104 P( X > -5)= P(Z > -1.308)=1-0.0951 = 0.9049 The results for the sampling distribution in parts (d)-(f) show that the sample mean values are clustered closer to the mean than individual values in the population.

7.33 to 7.36 Answers will vary.

Chapter 8: Confidence interval estimation Learning objectives After studying this chapter you should be able to: 1. construct and interpret confidence interval estimates for the mean and the proportion 2. determine the sample size necessary to develop a confidence interval for the mean or proportion 3. recognise how to use confidence interval estimates in auditing

= 85 ± 1.96 ×

8 =83.04 £ m £ 86.96 64

= 125 ± 2.58 ×

24 =114.68 £ m £ 135.32 36

8.1

X ± Z×

8.2

X ± Z×

8.3

If all possible samples of the same size n are taken, 95% of them include the true population average monthly sales of the product within the interval developed. Thus you are 95% confident that this sample is one that does correctly estimate the true average amount.

8.4

Since the results of only one sample are used to indicate whether salmon are large enough to go to market, the farmer can never know with 100% certainty that the specific interval obtained from the sample includes the true population mean. In order to have 100% confidence, the entire population (sample size N) would have to be selected.

8.5

To the extent that the sampling distribution of sample means is approximately normal, it is true that approximately 95% of all possible sample means taken from a sample of that same size will fall within 1.96 times the standard error away from the true population mean. But the population mean is not known with certainty. Since the farmer estimated the mean would fall between 3501.67 g and 3712.33 g based on a single sample, it is not necessarily true that 95% of all sample means will fall within those same bounds.

8.6

Approximately 5% of the intervals will not include the true population. Since the true population mean is not known, we do not know for certain whether it is contained in the interval (between 3501.67 g and 3712.33 g) that we have developed.

s n

8.7

(a) (b)

8.8

√

(c)

Since the value of 4.0 is included in the interval, there is no reason to believe that the mean is different from 4 litres. No. Since the population standard deviation is known and n = 50, from the Central Limit Theorem, we may assume that the sampling distribution of 𝑋̅ is approximately normal.

(a)

𝜎 3000 𝑋̅ ± 𝑍. 𝑛 = 34,000 ± 1.96. 64 = 33,265 ≤ 𝜇 ≤ 34,735

(b) (c)

(d)

8.9

𝜎 0.08 𝑋̅ ± 𝑍. 𝑛 = 3.98 ± 2.58. 50 = 3.9508 ≤ 𝜇 ≤ 4.0092

(a) (b)

√

No. The manufacturer does not have the right to state that the light globes last an average of 35,000 hours with 95% confidence. No. Since the population standard deviation is known and n = 64, from Central Limit Theorem, we may assume that the sampling distribution of 𝑋̅ is approximately normal. The confidence interval is wider based on a standard deviation of 6,000 hours rather than the original assumption of 3,000 hours. (a)

𝜎 6000 𝑋̅ ± 𝑍. = 34,000 ± 1.96. = 32,530 ≤ 𝜇 ≤ 35,470

(b)

Yes as 35,000 now lies within the interval the claim can be made by the manufacturer with 95% confidence.

√𝑛

√64

𝜎 0.05 𝑋̅ ± 𝑍. 𝑛 = 1.99 ± 1.96. 100 = 1.9802 ≤ 𝜇 ≤ 1.9998 √

√

No. Since the population standard deviation is known, and n = 1000, from the Central Limit Theorem, we may assume that the sampling distribution of 𝑋̅ is approximately normal. An individual value of 2.02 is only 0.60 standard deviations above the sample mean of 1.99. The confidence interval represents bound on the estimate of the average of a sample of 100, not an individual value. A shift of 0.02 units in the sample average shifts the confidence interval by the same distance without affecting the width of the resulting interval. (a)

𝜎 0.05 𝑋̅ ± 𝑍. 𝑛 = 1.97 ± 1.96. 100 = 1.9602 ≤ 𝜇 ≤ 1.9798 √

√

8.10

(a) (b) (c) (d) (e)

t9 = 2.2622 t9 = 3.2498 t31 = 2.0395 t64 = 1.9977 t15 = 1.7531

8.11

𝑆 24 𝑋̅ ± 𝑡. = 75 ± 2.0301. = 66.8796 ≤ 𝜇 ≤ 83.1204

8.12

̅ ± t. S = 50 ± 2.9467. 15 = 38.9499 ≤ μ ≤ 61.0501 X

8.13

Set 1: 𝑋̅ ± 𝑡.

√𝑛

√n

√36

√16

𝑆 3.7417 = 4.5 ± 2.3060. 8 = 1.4494 ≤ 𝜇 ≤ 7.5506 √𝑛 √ 𝑆 2.4495 Set 2: 𝑋̅ ± 𝑡. = 4.5 ± 2.3060. = 2.5029 ≤ 𝜇 ≤ 6.4971 √𝑛 √8

The data sets have different confidence interval widths because they have different values for the standard deviation.

8.14

𝑆 6.4660 = 5.8571 ± 2.4469. = −0.1229 ≤ 𝜇 ≤ 11.8371 √7 √𝑛 𝑆 2.1602 𝑋̅ ± 𝑡. 𝑛 = 4.00 ± 2.4469. 7 = 2.0022 ≤ 𝜇 ≤ 5.9978 √ √

Original data: 𝑋̅ ± 𝑡. Altered data:

The presence of an outlier in the original data increases the value of the sample mean and greatly inflates the sample standard deviation. 8.15

(a) (b)

8.16

8.18

√

𝑆 25 𝑋̅ ± 𝑡. = 380 ± 2.0096. = 372.895 ≤ 𝜇 ≤ 387.105

(b)

Sydney water can be 95% confident that the average household usage is somewhere between 372.895 litres and 387.105 litres. Water usage seems to have changed in the second summer by a significant amount (at least at a 5% significance level).

√𝑛

√50

(a)

𝑆 30 𝑋̅ ± 𝑡. = 367 ± 2.1098. = 352.08 ≤ 𝜇 ≤ 381.92

(b) (c)

No, the advertised usage of 355 is within the interval. It is not unusual. An energy usage of 350 for a particular refrigerator is only 0.57 standard deviations below the sample mean of 355.

(a)

𝑆 4.6024 𝑋̅ ± 𝑡. 𝑛 = 23 ± 2.0739. 23 = $21.01 ≤ 𝜇 ≤ $24.99

(b) (c)

8.19

√

The store owner can be 95% confident that the population mean retail value of greeting cards that the store has in its inventory is somewhere between $4.56 and $5.34. The store owner could multiply the ends of the confidence interval by the number of cards to estimate the total value of his inventory.

(a)

(c)

8.17

𝑆 0.82 𝑋̅ ± 𝑡. 𝑛 = 4.95 ± 2.0930. 20 = 4.566 ≤ 𝜇 ≤ 5.334

(a)

√𝑛

√

The population distribution needs to be normally distributed. You can be 95% confident that the mean bounced-cheque fee for the population is somewhere between $21.01 and $24.99.

X  t.

s n

(b) (c)

√18

= 43.04  2.0096

27.97133

= 35.09    50.99

The population distribution needs to be normally distributed. Both the normal probability plot and the box-and-whisker plot suggest that the distribution is skewed to the right.

8.20

(d)

Even though the population distribution is not normally distributed, with a sample mean of 50, the t distribution can still be used due to the Central Limit Theorem.

(a)

𝑆 25.2835 𝑋̅ ± 𝑡. 𝑛 = 43.8889 ± 2.0555. 27 = 33.89 ≤ 𝜇 ≤ 53.89

(b) (c)

√

The population distribution needs to be normally distributed. Both the normal probability plot and the box-and-whisker plot show that the population distribution is not normally distributed and is skewed to the right.

8.21

(a)

𝑆 17.39 𝑋̅ ± 𝑡. = 134.35 ± 2.0930. = 126.21 ≤ 𝜇 ≤ 142.49

(b)

𝑆 17.39 𝑋̅ ± 𝑡. 𝑛 = 134.35 ± 2.8609. 20 = 123.23 ≤ 𝜇 ≤ 145.47

√𝑛

√20

√

The population distribution needs to be normally distributed. The population distribution for motel room cost is not normally distributed and is skewed to the right. 𝑋

𝑝(1−𝑝) 0.25(0.75) = 0.25 ± 1.96√ 200 = 0.19 ≤ 𝜋 𝑛

8.22

𝑝 = 𝑛 = 200 = 0.25

8.23

𝑝=

8.24

(a)

𝑝=

(b)

The manager in charge of promotional programs for mobile customers can infer that the proportion of customers that would purchase the phone plan at the reduced cost is between 0.3240 and 0.4360 with a 99% level of confidence.

(a)

𝑝=

(b)

You can be 95% confident that the population proportion of highly educated women who have postponed their return to work due to difficulty in making suitable childcare arrangements is between 0.62 and 0.70.

(a)

𝑝 = 0.45

𝑝 ± 𝑍. √

(b)

𝑝 = 0.45

𝑝 ± 𝑍. √

(c)

The 95% confidence interval is wider. The loss in precision reflected as a wider confidence interval is the price you have to pay to achieve a higher level of confidence.

(a)

𝑝 = 0.27

(b)

At a 95% level of confidence, the proportion of older Australians who feel confused when making financial decisions is between 25% and 28.99%.

(a)

𝑝 = 0.36

(b)

The proportion found by the Ombudsman, 0.341, lies within this interval so it is possible that the population proportion is still 0.341. If the proportion of complaints about faults has not changed in the period since 2016 it would not be unusual to find a proportion of 0.36 in a survey of only 1000 responses.

(a)

p = 0.69 p  Z

8.25

8.26

8.27

8.28

8.29

(b)

𝑋 𝑛

25 400

= 0.0625 𝑋 𝑛

𝑋 𝑛

190 500

330 500

𝑝 ± 𝑍. √

𝑝(1−𝑝)

𝑝 ± 𝑍. √

= 0.0625 ± 2.58√

𝑝(1−𝑝)

𝑝 ± 𝑍. √

= 0.38

𝑛

𝑝(1−𝑝)

𝑝 ± 𝑍. √

= 0.66

𝑝(1−𝑝) 𝑛

𝑝(1−𝑝)

𝑝 ± 𝑍. √

𝑛

𝑝(1−𝑝)

𝑝 ± 𝑍. √

p = 0.69 p  Z

𝑛

0.0625(0.9375) 400

= 0.38 ± 2.58√

= 0.66 ± 1.96√

0.38(1−0.38) 500

0.66(1−0.66) 500

0.45(1−0.45)

= 0.45 ± 1.96√

045(1−0.45) 293

0.27 (1−0.27)

= 0.27 ± 1.96√

1900

0.36(0.64)

= 0.36 ± 1.96√

p (1 − p )

1000

= 0.69  1.96

n p (1 − p )

= 0.69  2.58

= 0.0313 ≤ 𝜋 ≤ 0.0937 = 0.3240 ≤ 𝜋 ≤ 0.4360

= 0.62 ≤ 𝜋 ≤ 0.70

= 0.3930 ≤ 𝜋 ≤ 0.5070

293

= 0.45 ± 1.645√

≤ 0.31

= 0.4022 ≤ 𝜋 ≤ 0.4978

= 0.2500 ≤ 𝜋 ≤ 0.2899

= 0.3302 ≤ 𝜋 ≤ 0.3898

0.69 ( 0.31) 518 0.69 ( 0.31) 518

= 0.6512    0.7298 = 0.6376    0.7424

The

99% interval is wider. 8.30

(a)

𝑝=

𝑋 𝑛

126 600

= 0.21

𝑝(1−𝑝)

𝑝 ± 𝑍. √

𝑛

= 0.21 ± 1.96√

0.21(1−0.21) 600

= 0.18 ≤ 𝜋 ≤ 0.24

𝑋

126

𝑝(1−𝑝)

0.21(1−0.21)

(b)

𝑝=

(c)

You are 95% confident that the population proportion of employers who have used a recruitment service within the past two months to find new staff is between 0.17 and 0.24. You are 99% confident that the population proportion of employers who have used a recruitment service within the past two months to find new staff is between 0.17 and 0.25. When the level of confidence is increased, the confidence interval becomes wider. The loss in precision reflected as a wider confidence interval is the price you have to pay to achieve a higher level of confidence.

(d)

𝑛

600

= 0.21

𝑝 ± 𝑍. √

𝑛

= 0.21 ± 2.5758√

8.31

𝑛=

𝑍2 ×𝜎 2 1.962 ×152 = = 34.57 2 𝑒 52

8.32

𝑛=

𝑍2 ×𝜎 2 2.582 ×1002 = = 166.41 𝑒2 202

8.33

𝑛=

𝑍2 ×𝜋(1−𝜋) 2.582 ×(0.5)(0.5) = = 1,040.06 𝑒2 (0.04)2

𝑈𝑠𝑒 𝑛 = 1,041

8.34

𝑛=

𝑍2 ×𝜋(1−𝜋) 1.962 ×(0.4)(0.6) = = 2,304.96 2 𝑒 (0.02)2

𝑈𝑠𝑒 𝑛 = 2,305

8.35

(a) (b)

Use n = 97

𝑍2 ×𝜎 2 1.962 ×0.052 = 0.012 = 96.04 𝑒2

𝑈𝑠𝑒 𝑛 = 97

8.38

𝑛=

8.39

(a)

8.41 8.42

𝑈𝑠𝑒 𝑛 = 984

Z2 ×σ2 1.962 ×50002 = = 96.04 e2 10002

8.37

8.40

𝑈𝑠𝑒 𝑛 = 246

𝑈𝑠𝑒 𝑛 = 97

𝑛=

(b)

𝑈𝑠𝑒 𝑛 = 167

𝑍2 ×𝜎 2 1.962 ×0.0752 = = 96.04 2 𝑒 0.0152

8.36

= 0.17 ≤ 𝜋 ≤ 0.25

𝑈𝑠𝑒 𝑛 = 35

𝑍2 ×𝜎 2 1.962 ×4002 = = 245.86 2 𝑒 502 2 2 2 2 𝑍 ×𝜎 1.96 ×400 𝑛 = 𝑒2 = = 983.41 252

𝑛=

600

𝑍2 ×𝜎 2 2.582 ×602 = 152 = 106.5024 𝑈𝑠𝑒 𝑛 = 107 𝑒2 𝑍2 ×𝜎 2 1.962 ×602 𝑛 = 𝑒 2 = 152 = 61.4656 𝑈𝑠𝑒 𝑛 = 62

𝑛=

𝑍2 ×𝜎 2 1.6452 ×452 = = 219.19 𝑈𝑠𝑒 𝑛 = 220 2 𝑒 52 2 2 2 2 𝑍 ×𝜎 2.58 ×45 (b) 𝑛 = 𝑒2 = = 539.17 𝑈𝑠𝑒 𝑛 = 540 52 𝑍2 ×𝜎 2 1.962 ×202 𝑛= 2 = = 61.47 𝑈𝑠𝑒 𝑛 = 62 𝑒 52

(a)

𝑛=

(a)

𝑛=

(b) (c)

𝑍2 ×𝜋(1−𝜋) 1.962 ×(0.17)(0.83) = = 150.57 𝑈𝑠𝑒 𝑛 = 151 𝑒2 (0.06)2 2 2 𝑍 ×𝜋(1−𝜋) 1.96 ×(0.17)(0.83) 𝑛= = = 3338.78 𝑈𝑠𝑒 𝑛 = 339 𝑒2 (0.04)2 𝑍2 ×𝜋(1−𝜋) 1.962 ×(0.17)(0.83) 𝑛= = = 1,355.12 𝑈𝑠𝑒 𝑛 = 1,356 𝑒2 (0.02)2

8.43

(a) (b) (c)

8.44

(d)

The sample sizes differ because the estimated sample proportions are different. As using both questions concern buying Asian vegetables, it is appropriate and less costly to use one sample and ask the respondents both questions. However, drawing two separate samples is also appropriate for the setup.

(a)

𝑝=

(b)

You are 95% confident that the population proportion of persons in the SSO audience who are visitors is between 0.101 and 0.173.

(c)

𝑛=

(d) 8.45

8.46

𝑍2 ×𝜋(1−𝜋) 1.962 ×(0.345)(1−0.345) = = 2,170.26 𝑈𝑠𝑒 𝑛 = 2,171 2 𝑒 (0.02)2 2 2 𝑍 ×𝜋(1−𝜋) 1.96 ×(0.47)(1−0.47) 𝑛= = = 2,392.36 𝑈𝑠𝑒 𝑛 = 2,393 𝑒2 (0.02)2

𝑛=

𝑋 𝑛

48 350

= 0.137

𝑝(1−𝑝)

𝑝 ± 𝑍. √

𝑛

0.137(1−0.137)

= 0.137 ± 1.96√

𝑍 2 ×𝜋(1−𝜋) 1.962 ×(0.137)(1−0.137) = = 504.662 𝑒2 (0.03)2 2 2 𝑍 ×𝜋(1−𝜋) 2.58 ×(0.137)(1−0.137) 𝑛= = = 874.436 𝑒2 (0.03)2 𝑝(1−𝑝)

0.36(0.64)

350

= 0.101 ≤ 𝜋 ≤ 0.173

𝑈𝑠𝑒 𝑛 = 505 𝑈𝑠𝑒 𝑛 = 875

(a)

𝑝 = 0.36

(b)

You are 95% confident that the proportion of Australian adults who held shares in late 2014 was between 0.3451 and 0.3749.

(c)

𝑛=

𝑁. 𝑋̅ ± 𝑁. 𝑡

𝑆 √𝑛

𝑝 ± 𝑍. √

𝑛

= 0.36 ± 1.96√

4009

𝑍2 ×𝜋(1−𝜋) 1.962 ×(0.36)(0.64) = = 8,851.05 𝑒2 (0.01)2

√

𝑁−𝑛 𝑁−1

= 500 × 25.7 ± 500 × 2.7969 ×

7.8 √25

= 0.3451 ≤ 𝜋 ≤ 0.3749

𝑈𝑠𝑒 𝑛 = 8,852

500−25

×√

500−1

= $10,721.53 ≤ 𝑃𝑜𝑝𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑇𝑜𝑡𝑎𝑙 ≤ $14,978.47

8.47

𝑁. 𝑋̅ ± 𝑁. 𝑡

𝑆 √𝑛

√

𝑁−𝑛 𝑁−1

= 10000 × 1.00315 ± 10000 × 1.972 ×

4.9985 √200

×√

10000−200 10000−1

= 3,131.37 ≤ 𝑃𝑜𝑝𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑇𝑜𝑡𝑎𝑙 ≤ 16,931.63

8.48

8.49

𝑝(1−𝑝) 0.04(1−0.04) = 0.04 + 1.2816 . √ 𝑛 300

(a)

𝑝 + 𝑍. √

(b)

𝑝 + 𝑍. √

(c)

𝑝 + 𝑍. √

𝑝(1−𝑝) 0.04(1−0.04) = 0.04 + 1.645 . √ 𝑛 300

𝑁. 𝑋̅ ± 𝑁. 𝑡

𝜋 < 0.0586

𝑝(1−𝑝) 0.04(1−0.04) = 0.04 + 2.3263 . √ 𝑛 300

𝑆 √𝑛

√

𝑁−𝑛 𝑁−1

= 300 × 5.45 ± 300 × 2.0930 ×

0.82 √20

𝜋 < 0.0545

𝜋 < 0.0663

300−20

×√

300−1

= 1,523.59 ≤ 𝑃𝑜𝑝𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑇𝑜𝑡𝑎𝑙 ≤ 1,746.41

8.50

𝑁. 𝑋̅ ± 𝑁. 𝑡

𝑆 √𝑛

√

𝑁−𝑛 𝑁−1

= 3000 × 1349 ± 3000 × 1.8331 ×

946.5035 √10

3000−10

×√

3000−1

= $2,403,472.42 ≤ 𝑃𝑜𝑝𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑇𝑜𝑡𝑎𝑙 ≤ $5,690,527.59

8.51

𝑁. 𝑋̅ ± 𝑁. 𝑡

𝑆 √𝑛

√

𝑁−𝑛 𝑁−1

= 1546 × 252.28 ± 1546 × 2.0096 ×

93.67 √50

×√

1546−50 1546−1

= $349,526.64 ≤ 𝑃𝑜𝑝𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑇𝑜𝑡𝑎𝑙 ≤ $430,523.12

8.52

𝑁. 𝑋̅ ± 𝑁. 𝑡

𝑆 √𝑛

√

𝑁−𝑛 𝑁−1

= 4000 × 7.45907 ± 4000 × 2.6092 ×

29.5523 √150

4000−150

×√

4000−1

= $5,125.99 ≤ 𝑃𝑜𝑝𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑇𝑜𝑡𝑎𝑙 ≤ $54,546.57

8.53

Historical cost ̅ ± 𝑁. 𝑡 𝑁. 𝐷

𝑆𝐷 √𝑛

√

𝑁−𝑛 𝑁−1

= 1200 × 26.625 ± 1200 × 1.9801 ×

75.97 √120

×√

1200−120 1200−1

= $16,310.53 ≤ 𝑇𝑜𝑡𝑎𝑙 𝐷𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑃𝑜𝑝𝑢𝑙𝑎𝑡𝑖𝑜𝑛 ≤ $47,589.47

Audited value ̅ ± 𝑁. 𝑡 𝑁. 𝐷

𝑆𝐷 √𝑛

√

𝑁−𝑛 𝑁−1

= 1200 × 27.583 ± 1200 × 1.9801 ×

75.85 √120

1200−120

×√

1200−1

= $17,484.83 ≤ 𝑇𝑜𝑡𝑎𝑙 𝐷𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑃𝑜𝑝𝑢𝑙𝑎𝑡𝑖𝑜𝑛 ≤ $48,714.37

8.54

𝑝(1−𝑝) 𝑁−𝑛 0.0367(1−0.0367) 10000−300 √ √ = 0.0367 + 1.645 . √ 𝑛 𝑁−1 300 10000−1

(a)

𝑝 + 𝑍. √

(b)

Since the upper bound is higher than the tolerable exception rate of 0.04, the auditor should request a larger sample.

𝜋 < 0.0542

8.55

The only way to have 100% confidence is to obtain the parameter of interest, rather than a sample statistic. From another perspective, the range of the normal and t distribution is infinite, so a Z or t value that contains 100% of the area cannot be obtained.

8.56

The t distribution is used for obtaining a confidence interval for the mean when the population standard deviation is unknown.

8.57

If the confidence level is increased, a greater area under the normal or t distribution needs to be included. This leads to an increased value of Z or t, and thus a wider interval.

8.58

When estimating the rate of non-compliance, it is commonplace to use a one-sided confidence interval instead of a two-sided confidence interval since only the upper bound on the rate of non-compliance is of interest.

8.59

In some applications such as auditing, interest is primarily on the total amount of a variable rather than the mean amount.

8.60

Difference estimation involves determining the difference between two amounts, rather than a single amount.

8.61

(a) The population from which this sample was drawn was the collection of all hospital workers who visited the union’s website. (b) The sample is not a random sample from this population. The sample consisted of only those who visited the union’s website and chose to fill out the survey. (c) This is not a statistically valid study. There was a selection bias since only those who visited the union’s website and chose to answer the survey were represented. There was possibly non-response bias as well. Visitors to the

(d)

website who chose to fill out the survey might not answer all questions and there was no way for the union to get back to them to follow up on the nonresponses if this was an anonymous survey. To avoid potential pitfalls, the union could draw a random sample from the list of all hospital workers and offer them the option of filling out the survey over the Internet or on the survey form that is mailed to the workers. The union should also keep track of the workers who are invited to fill out the survey and follow up on the non-responses after a specified period of time by mail, email or telephone to encourage them to participate in the survey. The sample size needed is

𝑛= 8.62

8.63

8.64

𝑍2 ×𝜋(1−𝜋) 1.962 ×(0.6195)(1−0.6195) = = 2,264 2 𝑒 (0.02)2 𝑝(1−𝑝)

(a)

𝑝 = 0.492

𝑝 ± 𝑍. √

(b)

𝑝 = 0.60

𝑝 ± 𝑍. √

(c)

𝑝 = 060

𝑝 ± 𝑍. √

(a)

𝑝 = 0.25

𝑝 ± 𝑍. √

(b)

𝑝 = 0.31

𝑝 ± 𝑍. √

𝑛

𝑝(1−𝑝) 𝑛

𝑝(1−𝑝)

Understood well 𝑝 = 0.21

𝑛

𝑝(1−𝑝) 𝑛

0.492(0.508)

= 0.492 ± 1.96√

= 0.60 ± 1.96√ = 0.60 ± 2.58√

𝑝 ± 𝑍. √

𝑝(1−𝑝) 𝑛

𝑝 = 0.33

𝑝 ± 𝑍. √

A little

𝑝 = 0.34

𝑝 ± 𝑍. √

Not understand 𝑝 = 0.12

𝑝 ± 𝑍. √

477

0.25(0.75) 210

0.31(0.69) 240

𝑝(1−𝑝) 𝑛

𝑝(1−𝑝)

= 0.4714 ≤ 𝜋 ≤ 0.5126

= 0.5560 ≤ 𝜋 ≤ 06440 = 0.5421 ≤ 𝜋 ≤ 06579

= 0.1914 ≤ 𝜋 ≤ 0.3086 = 0.2515 ≤ 𝜋 ≤ 0.3685

0.21(079) 1,402

= 0.33 ± 1.96√

𝑛

477

= 0.21 ± 1.96√

𝑝(1−𝑝)

Adequately

0.60(0.40)

= 0.25 ± 1.96√ = 0.31 ± 1.96√

2,256

= 0.1887 ≤ 𝜋 ≤ 0.2313

0.33(0.67) 1,402

= 0.34 ± 1.96√

0.34(0.66) 1,402

0.12(0.88)

= 0.12 ± 1.96√

1,402

= 0.3054 ≤ 𝜋 ≤ 0.3546 = 0.3152 ≤ 𝜋 ≤ 0.3648 = 0.1030 ≤ 𝜋 ≤ 0.1370

The results indicate that only around half of clients understand adequately or well what their legal costs are likely to be. 8.65

(a)

𝑆 2.8 𝑋̅ ± 𝑡. 𝑛 = 12.9 ± 1.96. 2,000 = 12.7773 ≤ 𝜇 ≤ 13.0227 √

√

Note: since n is large the value obtained from the normal distribution may be used for t. (b)

𝑝 = 0.433

(c)

𝑛=

𝑝(1−𝑝)

𝑝 ± 𝑍. √

𝑛

= 0.433 ± 1.96√

𝑍2 ×𝜎 2 1.962 ×32 = 22 = 8.6436 𝑒2

0..433(0.567) 500

= 0.3896 ≤ 𝜋 ≤ 0.4764

𝑈𝑠𝑒 𝑛 = 9, however in this case the sample

size is very small and unless the population has a normal distribution a sample of 9 may not be appropriate. If the t distribution is used a slightly larger sample size is necessary. Copyright © 2019 Pearson Australia (a division of Pearson Australia Group Pty Ltd) 9781488617249, Berenson, Basic Business Statistics 5e

8.66

8.67

𝑛=

(a)

𝑆 70 𝑋̅ ± 𝑡. 𝑛 = 350 ± 2.6490. 70 = 327.84 ≤ 𝜇 ≤ 372.16 √

𝑝(1−𝑝)

0.30(0.70)

𝑝 ± 𝑍. √

(a)

𝑆 4.0 𝑋̅ ± 𝑡. = 9.7 ± 2.0639. = 8.049 ≤ 𝜇 ≤ 11.351

= 0.30 ± 1.96√

𝑛

√𝑛

(b)

𝑝 ± 𝑍. √

(c)

𝑛=

= 0.19 ≤ 𝜋 ≤ 0.41

√25

𝑝(1−𝑝) 𝑛

= 0.48 ± 1.96√

0.48(0.52)

= 0.284 ≤ 𝜋 ≤ 0.676

𝑍2 ×𝜎 2 1.962 ×4.52 = 1.52 = 34.57 𝑈𝑠𝑒 𝑛 = 35 𝑒2 𝑍2 ×𝜋(1−𝜋) 1.6452 ×(0.5)(0.5) 𝑛= = = 120.268 𝑈𝑠𝑒 𝑛 = 121 𝑒2 (0.075)2

(e)

If a single sample were to be selected for both purposes, the larger of the two sample sizes (n = 121) should be used.

(a)

𝑛=

(b)

8.69

√

𝑈𝑠𝑒 𝑛 = 784

(b)

(d)

8.68

𝑍2 ×𝜋(1−𝜋) 1.962 ×(0.5)(0.5) = = 784 2 𝑒 (0.035)2

(d)

𝑍2 ×𝜎 2 2.582 ×64.72 = = 1114.57 𝑈𝑠𝑒 𝑛 = 1115 2 𝑒 52 𝑍2 ×𝜋(1−𝜋) 1.6452 ×(0.5)(0.5) 𝑛= = = 334.07 𝑈𝑠𝑒 𝑛 = 335 𝑒2 (0.045)2

(c)

If a single sample were to be selected for both purposes, the larger of the two sample sizes (n = 1115) should be used.

(a)

𝑆 113.90 𝑋̅ ± 𝑡. 𝑛 = 528.9 ± 1.96. 270 = 515.3138 ≤ 𝜇 ≤ 542.4862 √

√

Note: since n is large the value obtained from the normal distribution may be used for t. (b) (c) (d)

8.70

𝑛=

𝑝(1−𝑝) 𝑛

𝑍2 ×𝜎 2

= 0.40 ± 1.645√ 1.962 ×102

0.40(0.60) 70

= 266.77

= 0.3037 ≤ 𝜋 ≤ 0.4963

𝑈𝑠𝑒 𝑛 = 267

𝑒2 1.22 𝑍2 ×𝜋(1−𝜋) 1.6452 ×(0.5)(0.5) 𝑛= = = 422.82 𝑒2 (0.04)2

𝑈𝑠𝑒 𝑛 = 423

(e)

If a single sample were to be selected for both purposes, the larger of the two sample sizes (n = 423) should be used.

(a)

𝑆 9.22 𝑋̅ ± 𝑡. 𝑛 = 21.34 ± 1.9949. 70 = 19.14 ≤ 𝜇 ≤ 23.54

(b)

8.71

𝑝 ± 𝑍. √

√

𝑝 ± 𝑍. √

√

𝑝(1−𝑝) 𝑛

𝑍2 ×𝜎 2

= 0.3714 ± 1.645√ 1.962 ×102

0.3714(0.6286) 70

= 0.2764 ≤ 𝜋 ≤ 0.4664

(c)

𝑛=

(d)

𝑛=

(e)

If a single sample were to be selected for both purposes, the larger of the two sample sizes (n = 335) should be used.

(a)

𝑆 8.60 𝑋̅ ± 𝑡. = 47.20 ± 2.0010. = 44.98 ≤ 𝜇 ≤ 49.42

(b) (c)

𝑒2 𝑍2 ×𝜋(1−𝜋) 𝑒2

√𝑛

𝑝 ± 𝑍. √

𝑛=

𝑈𝑠𝑒 𝑛 = 171 𝑈𝑠𝑒 𝑛 = 335

√60

𝑝(1−𝑝) 𝑛

𝑍2 ×𝜎 2 𝑒2

= 170.74

1.52 1.6452 ×(0.5)(0.5) = = 334.08 (0.045)2

= 0.37 ± 1.645√ 1.962 ×92 1.52

0.30(0.70)

= 138.30

= 0.2027 ≤ 𝜋 ≤ 0.3973

𝑈𝑠𝑒 𝑛 = 139

8.72

𝑍2 ×𝜋(1−𝜋) 1.6452 ×(0.5)(0.5) = = 422.82 2 𝑒 (0.04)2

(d)

𝑛=

(e)

If a single sample were to be selected for both purposes, the larger of the two sample sizes (n = 423) should be used.

(a)

𝑛=

𝑍2 ×𝜋(1−𝜋) 1.962 ×(0.5)(0.5) = = 384.16 𝑒2 (0.05)2

𝑈𝑠𝑒 𝑛 = 423

𝑈𝑠𝑒 𝑛 = 385

If we assume that the population proportion is only 0.50, then a sample of 385 would be required. If the population proportion is 0.90, the sample size required is cut to 103.

8.73

8.74

𝑝(1−𝑝)

0.84(0.16)

(b)

𝑝 ± 𝑍. √

(c)

The representative can be 95% confident that the actual proportion of pavers that have required breaking strength is between 74.5% and 93.5%. He/she can accordingly advise the public of the paver performance.

(a)

𝑝=

(b)

Since the upper bound is higher than the tolerable exception rate of 0.15, the auditor should request a larger sample.

(a)

𝑛0 =

(b)

7 50

= 0.84 ± 1.96√

𝑛

𝑝(1−𝑝)

= 0.14 𝑝 + 𝑍. √

𝑛

𝑁−𝑛

√

𝑁−1

= 0.7384 ≤ 𝜋 ≤ 0.9416

0.14(1−0.14)

= 0.14 + 1.2816 . √

𝑍 2 ×𝜎 2 1.962 ×602 = = 138.298 2 𝑒 102 𝑛0 𝑁 138.298×25056 𝑛 = 𝑛 +(𝑁−1) = 138.298+(25056−1) = 137.544 0

𝑝=

12 138

𝑝(1−𝑝)

= 0.087 𝑝 ± 𝑍. √

𝑛

𝑁−𝑛

√

𝑁−1

1000−50

√

1000−1

𝜋 < 0.2013

𝑈𝑠𝑒 𝑛 = 138

= 0.087 ± 1.645 . √

0.087(1−0.087) 138

√

25056−138 25056−1

0.0476 ≤ 𝜋 ≤ 0.1263

(c)

𝑆 𝑁−𝑛 44.55 25056−138 𝑋̅ ± 𝑡. . √ = 98.70 ± 1.9774 . .√ = $91.22 ≤ 𝜇 ≤ $106.18

(d)

𝑁. 𝑋̅ ± 𝑁. 𝑡

√𝑛

𝑁−1

𝑆 √𝑛

√

𝑁−𝑛 𝑁−1

√138

25056−1

= 25056 × 98.70 ± 25056 × 1.9774 ×

44.55 √138

×√

25056−138 25056−1

= $2,285,857.25 ≤ 𝑃𝑜𝑝𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑇𝑜𝑡𝑎𝑙 ≤ $2,660,197.15

(e)

̅ = Σ𝐷 = 241 = 1.7463768 𝐷 𝑛

138 𝑆𝐷

̅ ± 𝑁. 𝑡 𝑁. 𝐷

√𝑛

√

𝑁−𝑛 𝑁−1

= 25056 × 1.7463768 ± 25056 × 1.9774 ×

6.85199 √138

×√

25056−138 25056−1

= $14,937.30 ≤ 𝑇𝑜𝑡𝑎𝑙 𝐷𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑃𝑜𝑝𝑢𝑙𝑎𝑡𝑖𝑜𝑛 ≤ $72,577.14

8.75

8.76

𝑛=

(b)

𝑁. 𝑋̅ ± 𝑁. 𝑡

(a) (b)

8.77

𝑍2 ×𝜎 2 2.582 ×452 = = 25.48 2 𝑒 232

(a)

𝑆 √𝑛

√

𝑁−𝑛 𝑁−1

𝑈𝑠𝑒 𝑛 = 26

= 258 × 575 ± 258 × 2.7874 ×

72.20 √26

×√

258−26 258−1

= $138,675.09 ≤ 𝑃𝑜𝑝𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑇𝑜𝑡𝑎𝑙 ≤ $158,024.91 𝑆 0.1058 𝑋̅ ± 𝑡. 𝑛 = 5.5014 ± 2.6800 . 50 = 5.46 ≤ 𝜇 ≤ 5.54 √

√

Since 5.5 g is within the 99% confidence interval, the company can claim that the mean weight of tea in a bag is 5.5 g with a 99% level of confidence. 𝑆 0.4611 𝑋̅ ± 𝑡. = 204.21 ± 2.0106 . = 204.08 ≤ 𝜇 ≤ 204.34 √𝑛

√49

8.78

(b)

With 95% confidence, the population mean width of trough is somewhere between 204.08 mm and 204.34 mm.

(a)

𝑆 0.0347 𝑋̅ ± 𝑡. 𝑛 = 4.0242 ± 1.9665 . 368 = 4.0206 ≤ 𝜇 ≤ 4.0278

(b)

√

𝑆 0.0467 𝑋̅ ± 𝑡. 𝑛 = 4.0040 ± 1.9672. 330 = 3.9989 ≤ 𝜇 ≤ 4.009

(c)

(d)

8.79

(a) (b) (c)

The volume for the cypress pine mulch is slightly skewed to the right while the volume for cedar mulch appears to be slightly skewed to the left. Since the two confidence intervals do not overlap, the mean volume of cypress pine mulch is greater than the mean volume of cedar mulch. 𝑆 0.1424 𝑋̅ ± 𝑡. 𝑛 = 0.2641 ± 1.9741 . 170 = 0.2425 ≤ 𝜇 ≤ 0.2856 √

√

𝑆 0.1227 𝑋̅ ± 𝑡. 𝑛 = 0.218 ± 1.9772 . 140 = 0. .1975 ≤ 𝜇 ≤ 0.2385 √

√

(d) 8.80

The amount of particle loss for both shingles is skewed to the right. Since the two confidence intervals do not overlap, we can conclude that the mean particle loss of Bondi shingles is higher than that of Vincentia shingles.

(a) FOOD North Island South Island PRESENTATION North Island South Island SERVICE North Island South Island TOILETS North Island South Island (b)

8.81

LCL

UCL

6.66 6.70

8.13 7.84

6.39 6.54

7.92 7.66

5.92 6.05

7.90 7.48

5.88 6.11

7.27 7.16

With 95% confidence you can conclude that the mean score in North Island is more varied than the mean score in South Island. With 95% confidence, you can conclude that there are no significant differences in the ratings for food, presentation, service and toilets between the two islands. If anything, South Island scores are more consistent, which can be seen as a good thing.

Answers will vary.

Chapter 9: Fundamentals of hypothesis testing: One-sample tests Learning objectives After studying this chapter you should be able to: 1. identify the basic principles of hypothesis testing 2. use hypothesis testing to test a mean or proportion 3. explain the assumptions of each hypothesis-testing procedure, how to evaluate them and the consequences if they are seriously violated 4. recognise the pitfalls involved in hypothesis testing 5. identify the ethical issues involved in hypothesis testing 9.1

H0 is used to denote the null hypothesis.

9.2

H1 is used to denote the alternative hypothesis.

9.3

a is used to denote the significance level, or the chance of committing a Type I error.

9.4

b is used to denote the risk or the chance of committing a Type II error.

9.5

1- b represents the power of a statistical test—that is, the probability of correctly rejecting the null hypothesis when in reality the null hypothesis is false and should be rejected.

9.6

a is the probability of making a Type I error—that is, the probability of incorrectly rejecting the null hypothesis when in reality the null hypothesis is true and should not be rejected.

9.7

b is the probability of a Type II error—that is, the probability of incorrectly failing to reject the null hypothesis when it is false.

9.8

The power of a test is the complement, which is (1- b ), of the probability of making a Type II error.

9.9

It is possible to incorrectly reject a true null hypothesis because the mean of a single sample can fall in the rejection region even though the hypothesised population mean is true.

9.10

It is possible not to reject a false null hypothesis because the mean of a single sample can fall in the non-rejection region even though the hypothesised population mean is false.

9.11

b will increase.

9.13

a is the probability of incorrectly convicting the defendant when he is innocent (this is a Type I error). b is the probability of incorrectly failing to convict the defendant when he is guilty (this is a Type II error).

9.14

Under the French judicial system, the null hypothesis assumes the defendant is guilty, the alternative assumes the defendant is innocent. A Type I error would be not convicting a guilty person and a Type II error would be convicting an innocent person.

9.15

(a) (b) (c)

9.16

9.17

9.18

9.19

Type I error is the mistake of approving an unsafe drug. A Type II error is not approving a safe drug. Both errors are serious but approving an unsafe drug may have more immediate consequences resulting in deaths so should be avoided. To lower both Type I and Type II errors, select a smaller value for a and increase the sample size.

HO : m = 10 minutes

10 minutes is adequate travel time between classes.

H1 : m ¹ 10 minutes

10 minutes is not adequate travel time between classes.

HO : m = 40 kg

The cloth has a mean breaking strength of 40 kg.

H1 : m ¹ 40 kg

The cloth does not have a mean breaking strength of 40 kg.

HO : m = 4 L

The mean amount of paint per can is 4 litres.

H1 : m ¹ 4 L

The mean amount of paint per can differs from 4 litres.

HO : m = 10 hours hours. H1 : m ¹ 10 hours hours.

The mean life of the manufacturer’s batteries is equal to 10 The mean life of the manufacturer’s batteries differs from 10

9.20

Decision rule: Reject H0 if Z < -1.96 or Z > +1.96. Decision: Since Zcalc = +2.21 > +1.96, reject HO .

9.21

Decision rule: Reject H0 if Z  −1.645 or Z  +1.645.

9.22

Decision rule: Reject H0 if Z  −2.58 or Z  +2.58.

9.23

Decision: Since, Zcalc = -2.61 < -2.58, reject HO .

9.24

p-value = 2(1–0.9772) = 0.0456

9.25

Since the p-value of 0.0456 is less than the 0.10 level of significance, the statistical decision is to reject the null hypothesis.

9.26

p-value =2(0.0838)= 0.1676

9.27

Since the p-value of 0.1676 is greater than the 0.01 level of significance, the statistical decision is to not reject the null hypothesis.

9.28

(a)

HO : m = 30 kg H1 : m ¹ 30 kg Decision rule: Reject H0 if Z < -1.96 or Z > +1.96 . Test statistic: Z =

X −

/ n

29.3 − 30 3.5 / 49

= −1.40

Decision: Since Zcalc = -1.40 is between the critical bounds of ± 1.96, do not reject H0 . There is not enough evidence to conclude that the cloth has a mean breaking strength that differs from 30 kg at the 5% level of significance. (b)

p-value = 2(0.0808) = 0.1616 Interpretation: The probability of getting a sample of 49 pieces that yield a mean strength that is farther away from the hypothesised population mean than this sample is 0.1616 or 16.16%.

(c)

Decision rule: Reject H0 if Z < -1.96 or Z > +1.96. Test statistic: Z =

X −  29.3 − 30 = = −2.72  / n 1.8 / 49

Decision: Since Zcalc = -2.72 is lower than the critical bound of –1.96, reject

H0 . There is enough evidence to conclude that the cloth has a mean breaking strength that differs from 30 kg at the 5% level of significance. (d)

Decision rule: Reject H0 if Z < -1.96 or Z > +1.96. Test statistic: Z =

X −

/ n

29 − 30 1.7 / 49

= − 4.12

Decision: Since Zcalc = - 4.12 is less than the critical bound of –1.96, reject Ho. There is enough evidence to conclude that the cloth has a mean breaking strength that differs from 30 kg at the 5% level of significance. 9.29

(a)

HO :  = 90 MB/s H1 :   90 MB/s Decision rule: Reject H0 if Z < -1.96 or Z > +1.96. Test statistic: Z =

X − 87.5 − 90 = = − 2.50 / n 8 / 64

Decision: Since Zcalc = − 2.50 is less than the critical bound of –1.96, reject H0 . There is enough evidence to conclude that the mean data transfer rate differs from 90 MB/s at the 5% level of significance.

9.30

(b)

p-value = 2(0.0228) = 0.0456. Interpretation: The probability of getting a sample of 64 cards that will yield a mean transfer rate that is farther away from the hypothesised population mean than this sample is 0.0456.

(c)

X ±Z i

(d)

The results are the same. The confidence interval formed does not include the hypothesised value of 10 MB/s.

(a)

H0: m = 2.00L

s n

= 9.8 ±1.96 i

0.8

9.604    9.996

H1: m ¹ 2.00L Decision rule: Reject H0 if Z < -1.96 or Z > +1.96. Test statistic: Z =

X-m

s/ n

1.99- 2.00 0.05/ 100

= -2.00

Decision: Since Zcalc = –2.00 is less than the critical bound of –1.96, reject H0. There is enough evidence to conclude that the amount of soft drink placed in 2-litre bottles at the local bottling plant differs from 2 litres at the 5% level of significance.

9.31

(b)

p-value = 2(0.0228) = 0.0456 Interpretation: The probability of getting a sample of 100 bottles that will yield a mean amount that is farther away from the hypothesised population mean than this sample is 0.0456.

(c)

X ±Z i

(d)

The results here are the same. The confidence interval formed does not include 2.00.

(a)

H0: m = 330mL

s n

= 1.99±1.96 i

0.05 100

1.9802    1.9998

H1: m ¹ 330mL Decision rule: Reject H0 if Z < -1.96 or Z > +1.96. Test statistic: Z =

X-m

s/ n

329.7- 330 4 / 50

= -0.53.

Decision: Since Zcalc = − 0.53 is in between the critical bounds of ±1.96, do not reject H0. There is insufficient evidence to conclude that the amount of salad dressing placed in a bottle is significantly different from 330 mL at the 5% level of significance. (b)

p-value = 2(0.2981) = 0.5962 Interpretation: The probability of observing a Z statistic more extreme than –0.53 is 0.5962 if the population mean is 330 mL.

(c)

Decision rule: Reject H0 if Z  −1.96 or Z  +1.96 . Test statistic: Z =

X −

/ n

329.7 − 330 0.9/ 50

= −2.36

Decision: Since Zcalc = − 2.36 is lower than the critical bound of –1.96, reject H0. There is enough evidence to conclude that the machine is filling bottles improperly, at the 5% level of significance. (d)

Decision rule: Reject H0 if Z < -1.96 or Z > +1.96. Test statistic: Z =

X −

/ n

328.9 − 330 3/ 50

= −2.59

Decision: Since Zcalc = − 2.59 is lower than the critical bound of –1.96, reject H0. There is enough evidence to conclude that the machine is filling bottles improperly, at the 5% level of significance. 9.32

(a)

H 0 :  = 80 H 1 :   80 Decision rule: Reject H0 if Z  −1.96 or Z  +1.96 . Test statistic: Z = X - m = 84.2 - 80 = 2.31 s / n 10.90 / 36 Decision: Since Z calc = 2.31 is greater than the critical bound of +1.96, reject H0. There is enough evidence to conclude that the average withdrawal amount is not $80 at the 5% level of significance.

(b)

p-value =2(0.0082) = 0.0208. Interpretation: The probability of observing a Z test statistic more extreme than 2.31 is 0.0208 if the population mean is indeed $80. Since this is lower than the 5% significance level, we reject the null hypothesis.

(c)

Using a 0.01 level of significance, we would not reject the null, given that the p-value is greater than this.

(d)

Test statistic: Z =

X -m

s/ n

84.2 - 80 9 / 36

= 2.8

p-value = 2(0.0026) = 0.0052 Interpretation: The probability of observing a Z test statistic more extreme than 2.8 is 0.0052 if the population mean is indeed $80. Since this is lower than the 1% and 5% significance levels, we reject the null hypothesis in both circumstances. 9.33

Z = 2.33

9.34

Since Zcalc = 2.39 is greater than Zcrit = 2.33, reject H0.

9.35

Z = –2.33

9.36

Since Zcalc = –1.15 is greater than Zcrit = –2.33, do not reject H0.

9.37

p-value = 1 – 0.9772 = 0.0228

9.38

Since the p-value = 0.0228 is less than a = 0.05, reject H0.

9.39

p-value = 0.0838

9.40

Since the p-value = 0.0838 is greater than a = 0.01, do not reject H0.

9.41

p-value = P(Z<1.38) = 0.9162

9.42

Since the p-value = 0.9162 > 0.01, do not reject the null hypothesis.

9.43

(a)

H0: m ³ 855 mm H1: m < 855 mm Decision rule: Reject H0 if Z  −1.645 . Test statistic: Z = X - m = 832 - 855 = 1.77 s / n 65 / 25 Decision: Since Zcalc = -1.77 is less than the critical bound of –1.645, reject H0. There is enough evidence to conclude the production equipment needs adjustment, at the 5% level of significance.

9.44

(b)

p-value = P(Z < –1.77) = 0.0384. Decision rule: If p-value < 0.05, reject H0. Decision: Since p-value < 0.05, reject H0. There is enough evidence to conclude that the production equipment needs adjustment, at the 5% significance level.

(c)

The probability of obtaining a sample whose mean is 832 mm or less when the null hypothesis is true is 0.0384.

(d)

The conclusions are the same.

(a)

H0: m ³ 25 minutes

H1: m < 25 minutes Decision rule: Reject H0 if Z < -1.645. Test statistic: Z = X - m = 22.4 - 25 = -2.6 s / n 6 / 36 Decision: Since Zcalc = -2.6 is less than the critical bound of -1.645, reject Ho. There is enough evidence to conclude the population mean delivery time has been reduced below the previous value of 25 minutes, at the 5% level of significance. (b)

(a)

(b)

(d)

9.47

Decision: Since p-value = 0.0047 is less than a = 0.05, reject H0. There is enough evidence to conclude the population mean delivery time has been reduced below the previous value of 25 minutes. The probability of obtaining a sample whose mean is 22.4 minutes or less when the null hypothesis is true is 0.0047. The conclusions are the same.

H0 :   10 A teenager will post photographs on social networking sites 10 times a week or less. H1 :   10 A teenager will post photographs on social networking sites more than 10 times a week. A Type I error occurs when you conclude the mean number of times that teenagers post photos on social networking sites is more than 10 when in fact the mean number is not more than 10. A Type II error occurs when you conclude the mean number of times that teenagers post photos on social networking sites is not more than 10 when in fact the mean number is more than 10. Decision rule: Reject H0 if Z > + 2.33. X - m 10.87 -10 Test statistic: Z = = = 5.4375 s / n 1.6 / 100

(c)

9.46

Decision rule: If p-value < 0.05, reject H0. p-value = 0.0047

Decision: Since Zcalc = 5.4375 is greater than the critical bound of +2.33, reject H0. There is enough evidence to conclude the population mean number of times teenagers post photos on social networking sites is more than 10 times per week, at the 5% level of significance. When the null hypothesis is true, the probability of obtaining a sample whose mean is 10.87 times or more is virtually zero (p-value = P(Z > 5.4375) = 0.000). X -m

S/ n

56 - 50 12 / 16

= 2.00

df = n – 1 = 16 – 1 = 15

9.48

(a) (b)

For a two-tailed test with a 0.05 level of significance, tcrit = ± 2.1315. For an upper-tailed test with a 0.05 level of significance, tcrit = ± 1.7531.

9.49

(a) (b)

Since tcalc = 2.00 is between the critical bounds, do not reject H0. Since tcalc = 2.00 is above the critical bound, reject H0.

9.50

No, you should not use the t test to test the null hypothesis that m = 60 on a population that is left-skewed because the sample size (n = 16) is less than 30. The t test assumes that, if the underlying population is not normally distributed, the sample size is sufficiently large to enable the test statistic t to be influenced by the Central Limit Theorem. If sample sizes are small (n < 30), the t test should not be used because the sampling distribution does not meet the requirements of the Central Limit Theorem.

9.51

Yes, you may use the t test to test the null hypothesis that m = 60 even though the population is left skewed because the sample size is sufficiently large (n = 160). The t test assumes that, if the underlying population is not normally distributed, the sample size is sufficiently large to enable the test statistic t to be influenced by the Central Limit Theorem.

9.52

(a)

H0 :   $665 H1 :   $665 Decision rule: df = 99. If t > 1.2902, reject H0. X −  675.60 − 665 Test statistic: t = = = 2.345 S/ n

45.20 / 100

Decision: Since tcalc = 2.345 is greater than the critical bound of +1.2902, reject H0. There is enough evidence to conclude that the mean cost of textbooks per semester at a large university is more than $665 at the 1% significance level. (b)

Decision rule: df = 99. If t >1.6604, reject H0. Test statistic: t =

X −  675.60 − 665 = = 1.413 S/ n 75 / 100

Decision: Since tcalc = 1.413 is less than the critical bound of +1.6604, do not reject H0. There is not enough evidence to conclude that the mean cost of textbooks per semester at a large university is more than $665 at the 5% significance level. (c)

Decision rule: df = 99. If >1.2902, reject H0. Test statistic: t =

X −  669.60 − 665 = = 1.018 S / n 45.20 / 100

Decision: Since tcalc = 1.018 is less than the critical bound of +1.2902, do not reject H0. There is insufficient evidence to conclude that the mean cost of

textbooks per semester at a large university is more than $665 at the 1% significance level. 9.53

H0 : m = 90 seconds H1 : m ¹ 90 seconds Decision rule: df = 49. If t  1.2991 , reject H0. Test statistic: t =

X -m S/ n

101- 90 26.5 / 50

= 2.935

Decision: Since tcalc = 2.935 is greater than the critical bound of  1.2991 , reject H0. There is enough evidence to conclude that the processing time is different from the target at the 10% significance level. 9.54

H0 : m ³ 3.7 minutes

H1 : m < 3.7 minutes Decision rule: df = 63. If t < –1.6694, reject H0. Test statistic: t =

X -m S/ n

3.57 - 3.7 0.8 / 64

= 1.3

p-value = 0.0992 Decision: Since tcalc = −1.3 is greater than the critical bound of –1.6994, do not reject H0. There is not enough evidence to conclude that the mean waiting time is less than 3.7 minutes. 9.55

(a)

H0 : m = 250.4 grams H1 : m ¹ 250.4 grams Decision rule: df = 49. If Test statistic: t = Decision: Since

9.56

X -m S/ n

t > 2.0096 , reject H0.

250.35 - 250.4 0.2 / 50

= 1.7678

tcalc = 1.7678 is less than the critical bound of 2.0096, do not

(b)

reject H0. There is not enough evidence to conclude that the mean amount is different from 250.4 g at the 5% significance level. The p-value is 0.0833. If the population mean is indeed 250.4 g, the probability of obtaining a sample mean that is more than 0.05 g away from 250.4 g is 0.0833.

(a)

H0 : m = 45 days

H1 : m ¹ 45 days Decision rule: df = 26. If t  2.0555 , reject H0. (a)

Test statistic: t =

X -m S/ n

43.8889 - 45 25.2835 / 27

= -0.2284

Decision: Since tcalc = 0.2284 is less than the critical bound of 2.0555, do not reject H0. There is not enough evidence to conclude that the mean processing time has changed from 45 days at the 5% significance level. (b)

The population distribution needs to be normal.

(c)

The normal probability plot indicates that the distribution is not normal and is skewed to the right. The assumption in (b) has been violated. 9.57

(a)

H0 : m = 2L

H1 : m ¹ 2L Decision rule: df = 49. If t > 2.0096 , reject H0. Test statistic: t =

X -m S/ n

2.0007 - 2 0.0446 / 50

= 0.1143

Decision: Since tcalc = 0.1143 is less than the critical bound of 2.0096, do not reject H0. There is not enough evidence to conclude that the mean amount of soft drink filled is different from 2.0 litres at the 5% significance level.

(b)

p-value = 0.9095. If the population mean amount of soft drink filled is indeed 2.0 litres, the probability of observing a sample of 50 soft drinks that will result in a sample mean amount of fill more different from 2.0 litres is 0.9095.

(c)

The normal probability plot suggests that the data are rather normally distributed. Hence, the results in (a) are valid in terms of the normality assumption. (d)

The time series plot of the data reveals that there is a downward trend in the amount of soft drink filled. This violates the assumption that the data are drawn independently from a normal population distribution because the amount of fill in consecutive bottles appears to be closely related. As a result, the t test in (a) becomes invalid. 9.58

(a)

H0 : m ³ 42 days

H1 : m < 42 days

Decision Rule: df = 49. If, t < –1.6766, reject H0. Mean is 40.18, thus, test statistic: X -m 40.81- 42 t =

S/ n

27.97 / 50

= - 0.46009

Thus decision: Since tcalc = –0.46009 is greater than the critical bound of – 1.6766, do not reject H0. There is not enough evidence to conclude that the mean number of days between receipt of the order and delivery of the sofa is 42 days or less, at the 5% significance level. (b)

The population distribution needs to be normal.

(c)

The normal probability plot indicates that the distribution is skewed to the right.

9.59

(d)

Even though the population distribution is probably normally distributed, the result obtained in (a) should be valid due to the Central Limit Theorem as a result of the relatively large sample size of 50.

(a)

H0 : m ³ 15.5 H1 : m ¹ 15.5 Decision rule: df = 119. If t > 1.6577 , reject H0. Test statistic: t =

X -m S/ n

14.9775 -15.5 1.007 / 120

= - 5.684

Decision: Since tcalc = 5.684 is greater than the critical bound of 1.6577, reject H0. There is enough evidence to conclude that the mean viscosity has changed from 15.5, at the 5% significance level. (b)

The population distribution needs to be normal.

(c)

The normal probability plot indicates that the distribution is slightly skewed to the right. Even though the population distribution is probably not normally distributed, the result obtained in (a) should still be valid due to the Central Limit Theorem as a result of a relatively large sample size of 120. 9.60

(a)

H0 : m = 0

H1 : m ¹ 0 Decision rule: df = 99. If | t > 1.9842 , reject H0. Test statistic: t =

X -m S/ n

-0.0058 0.0424 / 100

= -1.3563

Decision: Since tcalc < 1.9842 , do not reject H0. There is not enough evidence to conclude that the mean difference is different from 0.0 mm, at the 5% significance level. (b)

p-value = 0.1781 Interpretation: If the population mean difference is indeed 0.0 mm, the probability of observing a sample of 100 steel parts that will result in a sample mean difference further away from the hypothesised value than this sample is 0.1781.

(c)

In order for the t test to be valid, the data are assumed to be independently drawn from a population that is normally distributed. Since the sample size is 100, which is considered quite large, the t distribution will provide a good approximation to the sampling distribution of the mean as long as the population distribution is not very skewed.

(d)

The box-and-whisker plot suggests that the data has a distribution that is skewed slightly to the right. Given the relatively large sample size of 100 observations, the t distribution should still provide a good approximation to the sampling distribution of the mean. 9.61

(a)

H0 : m = 0 H1 : m ¹ 0 Decision rule: df = 29. If | t > 2.0452 , reject H0. Test statistic: t =

X -m S/ n

-0.25 - 0 0.2823 / 30

= - 4.9290

Decision: Since tcalc > 2.0452 , and the p-value is virtually zero, reject H0. There is enough evidence to conclude that the population mean fund value changed, at the 5% significance level. (b) (c)

Since the sample size is 30, we do not need to assume that the population fund value change is to be normally distributed. The p-value is virtually zero. If there is indeed no change in the population mean fund value, the probability of obtaining a sample mean fund change that is more than 0.25 dollars is virtually zero.

9.62

p =

X 88 = = 0.22 n 400

9.63

Z =

p-p 0.22 - 0.20 = = 1.00 p (1- p ) 0.20(0.80) n

9.64

400

H0 : p = 0.20

H1 : p ¹ 0.20 Decision rule: If Z < –1.96 or Z > 1.96, reject H0.

Test statistic: Z =

p-p

0.22 - 0.20

= 1.00 0.20(0.80) n 400 Decision: Since Zcalc = 1.00 is between the critical bounds of Z = ±1.96, do not reject H0.

9.65

(a)

p (1- p )

H0 : p £ 0.5

H1 : p > 0.5 Decision rule: If Z > 1.645, reject H0. Test statistic: Z =

p-p 0.5702 - 0.5 = = 4.5273 p (1- p ) 0.5(0.5)

n 1040 Decision: Since Zcalc = 4.5723 is above the critical bound of Z = +1.96, reject H0. There is enough evidence to show that more than half of all Sydney residents would rather have reduced tolls.

(b)

9.66

p-value = virtually zero. If the population proportion is indeed no more than 0.5, the probability is virtually zero of observing 593 or more out of 1040 Sydney residents who would rather have reduced tolls.

H 0 :  = 0.306 H1 :   0.306

Decision rule: Z < –1.96 or Z > 1.96, reject H0. Test statistic: Z =

p − (315 / 975) − 0.306 = = − 1.157  (1 −  ) 0.306(0.694) n 975

p-value = 0.2460 Decision: Since Zcalc = –1.157 is between the critical bounds of Z = ± 1.96 and the p-value is greater than 0.05, do not reject H0. There is not sufficient evidence to show that the proportion of unemployed people looking only for part-time work has changed. 9.67

(a)

H 0 :  = 0.64 H1 :   0.64 Decision rule: If Z > 1.645, reject H0. Test statistic: Z =

p − 0.6833 − 0.64 = = 1.6122  (1 −  ) 0.6833(0.3167) n

300

(b)

Decision: Since Zcalc = 1.6122 is not above the critical bound of Z = 1.645, do not reject H0. There is not enough evidence to show that the proportion of shoppers supporting the ban and using the alternative bags is higher than 0.64. p-value = 0.0535 Interpretation: The probability of obtaining a test statistic equal to or more extreme than the one from this sample, given that the null hypothesis is true, is approximately 5.35% so we would not reject the null hypothesis at the 5% level.

(c)

(a)

Test statistic:

p − =  (1 −  ) n

0.6833 − 0.64 = 2.9435 0.6833(0.3167) 1000

Decision: Since Zcalc = 2.9435 is above the critical bound of Z = 1.645, reject H0. There is evidence to show that the proportion of shoppers supporting the ban and using the alternative bags is higher than previously. (b)

9.68

p-value = 0.0016 Interpretation: The probability of obtaining a test statistic equal to or more extreme than the one from this sample, given that the null hypothesis is true, is approximately 0.16%.

(d)

A larger sample size implies that there is more information about the population, and reduces the standard error (variation) of the sample proportion.

(a)

H 0 :   0.382 H1 :   0.382 Decision rule: If Z > 1.645, reject H0. Z=

Test statistic:

p − =  (1 −  ) n

0.3991 − 0.382 = 2.2248 0.382(1 − 0.382) 3,996

Decision: Since Zcalc = 2.2248 is above the critical bound of Z = +1.645, reject H0. There is enough evidence to show that the proportion of businesses with a social media presence has increased from 38.2%.

9.69

(b)

p-value = 0.0130 Decision: Since p-value is smaller than 0.05, reject H0.

(a)

H 0 :   0.709 H1 :   0.709 Decision rule: If Z < -2.33, reject H0.

Test statistic: Z =

p − 0.681 − 0.709 = = −1.9493  (1 −  ) 0.709(0.291) n 1000

Decision: Since Zcalc = -1.9493 is not below the critical bound of Z = -2.33, do not reject H0.

9.70

(b)

There is not enough evidence to show that fewer than 70.9% of new graduates are finding work with four months of completing their degrees.

(c)

The sample used in (a) and (b) needs to be random. As the sample size is large, the conditions that np > 5 and n(1–p) > 5 should have been met.

H0 : m ³ 200

H1 : m < 200 a = 0.05, n = 16, s = 6

Lower critical value: ZL = –1.6449

æ s ö æ 6 ö X = m + ZL ç = 200 -1.6449 ç = 197.5327 ÷ è nø è 16 ÷ø (a)

X L - m1

s/ n

197.5327 - 197 = 0.3551 6 16

power = 1- b = P( X < X L ) = P(Z < 0.3551) = 0.6388

b = 1- 0.6388 = 0.3612

(b)

X L - m1

s/ n

197.5327 - 194 = 2.3551 6 16

power = 1- b = P( X < X L ) = P(Z < 2.3551) = 0.9907

b = 1- 0.9907 = 0.0093 9.71

H0 : m ³ 200

H1 : m < 200 a = 0.01, n = 16, s = 6

Lower critical value: ZL = –2.3263

æ s ö æ 6 ö X = m + ZL ç = 200 - 2.3263 ç = 196.5106 ÷ è nø è 16 ÷ø (a)

X L - m1

s/ n

196.5106 - 197 = -0.3263 6 16

power = 1- b = P( X < X L ) = P(Z < 0.4263) = 0.3721

b = 1- 0.3721 = 0.6279 (b)

X L - m1

s/ n

196.5106 - 194 = 1.6737 6

16 power = 1- b = P( X < X L ) = P(Z < 1.6737) = 0.9529

b = 1- 0.9529 = 0.0471 (c)

9.72

Holding everything else constant, the greater the distance between the true mean and the hypothesised mean, the higher the power of the test will be and the lower the probability of committing a Type II error will be. Holding everything else constant, the smaller the level of significance, the lower the power of the test will be and the higher the probability of committing a Type II error will be.

H0 : m ³ 200

H1 : m < 200 a = 0.05, n = 25, s = 6

Lower critical value: ZL = –1.6449

æ s ö æ 6 ö X = m + ZL ç = 200 -1.6449 çè ÷ = 198.0261 è n ÷ø 25 ø (a)

X L - m1

s/ n

198.0261 - 197 = 0.8551 6 25

power = 1- b = P( X < X L ) = P(Z < 0.8551) = 0.8038

b = 1- 0.8038 = 0.1962

(b)

X L - m1

s/ n

198.0261 -194 = 3.3551 6 25

power = 1- b = P( X < X L ) = P(Z < 3.3551) = 0.9996

b = 1- 0.9996 = 0.0004 (c)

9.73

Holding everything else constant, the larger the sample size, the higher the power of the test will be and the lower the probability of committing a Type II error will be.

H0 : m ³ 40,000

H1 : m < 40,000

a = 0.05, n = 100, s = 5,600

Lower critical value: ZL = –1.6449

æ s ö æ 5600 ö X = m + ZL ç = 40,000 -1.6449 ç = 39078.86 ÷ è nø è 100 ÷ø Copyright © 2019 Pearson Australia (a division of Pearson Australia Group Pty Ltd) 9781488617249, Berenson, Basic Business Statistics 5e

(a)

X L - m1

s/ n

39078.86 - 38400 = 1.2123 5600 100

power = 1- b = P( X < X L ) = P(Z < 1.2123) = 0.8873

b = 1- 0.8038 = 0.1127 (b)

X L - m1

s/ n

39078.86 - 39840 = -1.3592 5600 100

power = 1- b = P( X < X L ) = P(Z < -1.3592) = 0.0870

b = 1- 0.0870 = 0.9130 9.74

H1 : m < 40,000

H0 : m ³ 40,000

a = 0.01, n = 100, s = 5,600

Lower critical value: ZL = –2.3263

æ s ö æ 5600 ö X = m + ZL ç = 40,000 2.3263 çè ÷ = 38,697.27 è n ÷ø 100 ø (a) Z=

X L - m1

s/ n

38,697.27 - 38,400 = 0.5308 5600 100

power = 1- b = P( X < X L ) = P(Z < 0.5308) = 0.7019

b = 1- 0.7019 = 0.298

(b)

X L - m1

s/ n

38697.27 - 39840 = -2.0406 5600 100

power = 1- b = P( X < X L ) = P(Z < -2.0406) = 0.0206

b = 1- 0.0206 = 0.9794

(c)

9.75

Holding everything else constant, the greater the distance between the true mean and the hypothesised mean, the higher the power of the test will be and the lower the probability of committing a Type II will be. Holding everything else constant, the smaller the level of significance, the lower the power of the test will be and the higher the probability of committing a Type II error will be.

H0 : m ³ 40,000

H1 : m < 40,000

a = 0.05, n = 25, s = 5,600

Lower critical value: ZL = –1.6449

æ s ö æ 5600 ö X = m + ZL ç = 40,000 -1.6449 ç = 38,157.71 ÷ è nø è 25 ÷ø (a)

X L - m1

s/ n

38157.71 - 38400 = -2.2163 5600 25

power = 1- b = P( X < X L ) = P(Z < -0.2163) = 0.4144

b = 1- 0.4144 = 0.5856 (b)

X L - m1

s/ n

38157.71 - 39840 = -1.5020 5600 25

power = 1- b = P( X < X L ) = P(Z < -1.5020) = 0.0665

b = 1- 0.0665 = 0.9335 (c)

9.76

Holding everything else constant, the larger the sample size, the higher the power of the test will be and the lower the probability of committing a Type II error will be.

H0 : m = 40,000

H1 : m ¹ 40,000

a = 0.05, n = 100, s = 5,600

Critical values: ZL = –1.960, ZU = 1.960

æ s ö æ 5600 ö X = m + ZL ç = 40,000 -1.960 ç = 38,902.40 ÷ è nø è 100 ÷ø

æ s ö æ 5600 ö XU = m + Z L ç = 40,000 +1.960 ç = 41,097.60 ÷ è nø è 100 ÷ø (a)

power = 1- b = P( X L < X < XU ) = P(0.8971 < Z < 4.8171) = 0.1848

b = 1- 0.1848 = 0.8152

(b)

power = 1- b = P( X L < X < XU ) = P(-1.6743 < Z < 2.2457) = 0.9406

b = 1- 0.9406 = 0.0594

(c)

A one-tail test is more powerful than a two-tail test, holding everything else constant.

9.77

(a) (b)

Answers will vary. Students should be informed of the purpose for which data is being collected and give their consent, They should be told whether it will be treated as confidential or not.

9.78

(a)

Answers will vary. Avoidance of coverage errors will be necessary.

(b)

Participants should be informed of confidentiality arrangements and give consent to providing information. Reports should disclose the sampling method and sample size.

9.79

(a) (b)

Answers will vary. The supermarket should not report results as applying to all customers if they simply use data from those in their loyalty program because these data are easier to access.

9.80

The null hypothesis represents the status quo or the hypothesis that is to be disproved. The null hypothesis includes an equal sign in its definition of a parameter of interest. The alternative hypothesis is the opposite of the null hypothesis and usually represents taking an action. The alternative hypothesis includes either a less than sign, a not equal to sign or a greater than sign in its definition of a parameter of interest.

9.81

A Type I error represents rejecting a true null hypothesis, while a Type II error represents not rejecting a false null hypothesis.

9.82

The power of the test is the probability that the null hypothesis will be rejected when the null hypothesis is false.

9.83

In a one-tailed test for a mean or proportion, the entire rejection region is contained in one tail of the distribution. In a two-tailed test, the rejection region is split into two equal parts, one in the lower tail of the distribution, and the other in the upper tail.

9.84

The p-value is the probability of obtaining a test statistic equal to or more extreme than the result obtained from the sample.

9.85

Assuming a two-tailed test is used, if the hypothesised value for the parameter does not fall into the confidence interval, the null hypothesis can be rejected.

9.86

The following steps would be used in all hypothesis tests: State the null hypothesis. State the alternative hypothesis. Choose the level of significance. Choose the sample size. Determine the appropriate statistical technique and corresponding test statistic to use. Set up the critical values that divide the rejection and non-rejection regions. Collect the data and compute the sample value of the appropriate test statistic. Determine whether the test statistic has fallen into the rejection or the non-rejection region. The computed value of the test statistic is compared with the critical values for the appropriate sampling distribution to determine whether it falls into the rejection or non-rejection region. Make the statistical decision. If the test statistic falls into the non-rejection region, the null hypothesis H0 cannot be rejected. If the test statistic falls into the rejection region, the null hypothesis is rejected. Express the statistical decision in terms of a particular situation.

9.87

Some of the ethical issues that arise when dealing with hypothesis testing methodology include the data collection method, informed consent from human subjects being ‘treated’, the type of test—two-tailed or one-tailed, the choice of level of significance, data snooping, the cleansing and discarding of data, and reporting findings.

9.88

Among the questions to be raised are: What is the goal of the experiment or research? Can it be translated into a null and alternative hypothesis? Is the hypothesis test going to be two-tailed or one-tailed? Can a random sample be drawn from the underlying population of interest? What kinds of measurements will be obtained from the sample? Are the sampled outcomes of the random variable going to be numerical or categorical? At what significance level, or risk of committing a Type I error, should the hypothesis test be conducted? Is the intended sample size large enough to achieve the desired power of the test for the level of significance chosen? What statistical test procedure is to be used on the sampled data and why? What kind of conclusions and interpretations can be drawn from the results of the hypothesis test?

9.89

(a)

H0 : p = 0.6 H1 : p > 0.6

(b)

A Type I error is to reject the null hypothesis when it is true. In this context that means believing that the new drug is more effective than the current one when this is not so. A Type II error is not rejecting the null hypothesis when it is false. This means not deciding that the new drug is more effective, although it is. Both errors can have serious consequences for patients. A Type I error can result in a new drug coming onto the market that is not as effective as claimed. A Type II error could result in a more effective drug not being approved so patients do not get its benefit.

(c)

9.90

(a) (b)

9.91

(a)

TasmanTrading commits a Type I error when it purchases a site that is not profitable. Type II error occurs when TasmanTrading fails to purchase a profitable site. The cost to the company when a Type II error is committed is the loss on the amount of profit the site could have generated had the company had decided to purchase the site. The executives at TasmanTrading are trying to avoid a Type I error by adopting a very stringent decision criterion. Only sites that are classified as capable of generating high profit will be purchased. If the executives adopt a less-stringent rejection criterion by buying sites that the econometric model predicts will produce moderate or large profit, the probability of committing a Type I error will increase. On the other hand, the less-stringent rejection criterion will lower the probability of committing a Type II error as now more potentially profitable sites will be purchased.

H 0 :  = 0.787 H1 :   0.787 Decision rule: Reject H0 if Z < –2.58 or Z > 2.58 Test statistic: Z =

p − (412 / 500) − 0.787 = = 2.0207  (1 −  ) 0.787(0.213) n 500

Decision: Since Zcalc = 2.0207 falls within the upper and lower critical bounds, do not reject H0. There is not enough evidence to show that the proportion of medical practitioners who will bulk bill has changed. (b)

p-value = 0.0433. If the proportion of medical practitioners who bulk bill is 78.7% the probability of obtaining a percentage more extreme than 82.4% in a sample of 500 doctors is 0.0433.

9.92

(a)

H0 : m = 38 litres H1 : m ¹ 38 litres Decision rule: df = 59. If t < –2.0010 or t > 2.0010, reject H0. Test statistic:

(b) (c)

X-m S/ n

42.8-38 11.7/ 60

= 3.1778

Decision: Since tcalc = 3.1778 is greater than the upper critical value of t = 2.0010, reject H0. There is enough evidence to conclude that the mean petrol purchase differs from 38 litres. p-value = 0.0024 Note: The p-value was found using Excel.

H0 : p ³ 0.20 H1 : p < 0.20

Decision rule: If Z < –1.645, reject H0. Test statistic: Z =

(d)

9.93

p-p

0.1833- 0.20

= -0.32 0.20(1- 0.20) n 60 Decision: Since Zcalc = –0.32 is greater than the bound of Z = –1.645, do not reject H0. There is not sufficient evidence to conclude that less than 20% of the motorists purchase premium unleaded petrol. X -m 39-38 Test statistic: t = = = -0.6620 S / n 11.7 / 60 Decision: Since the test statistic tcalc = 0.6620 is between the critical bounds of ± 2.0010, do not reject H0. There is not enough evidence to conclude that the mean petrol purchase differs from 38 litres.

p (1- p )

p-p

(e)

Test statistic: Z =

(a)

H0 : m ³ $100 H1 : p < $100

0.1167- 0.20

= -1.61 0.20(1- 0.20) n 60 Decision: Since Zcalc = –1.61 is greater than the critical bound of Z = –1.645, do not reject H0. There is not sufficient evidence to conclude that less than 20% of the motorists purchase premium unleaded petrol.

p (1- p )

Decision rule: df = 84. If t < –1.6632, reject H0. Test statistic: t =

(b)

X − S/ n

98.95 − 100 34.55 / 85

= −2.6818

Decision: Since tcalc = –2.6818 is below the critical bound of t = –1.6632, reject H0. There is enough evidence to conclude that the mean reimbursement for consultations with specialists paid by Medicare is less than $100. H0 :   0.10 H1 :   0.10 Decision rule: If Z > 1.645, reject H0.

Test statistic: Z =

p −

 (1 −  )

(c)

(d)

0.1176 − 0.10 0.10(1 − 0.10) 85

= 0.5409

Decision: Since Zcalc = 0.5409 is lower than the critical bound of Z = 1.645, do not reject H0. There is not sufficient evidence to conclude that more than 10% of all reimbursements for consultations with specialists paid by Medicare are incorrect. To perform the t test on the population mean, you must assume that the observed sequence in which the data were collected is random and that the data are approximately normally distributed.

H0 : m ³ $100 H1 : m < $100

Decision rule: df = 84. If t < –1.6632, reject H0. Test statistic: t =

X − S/ n

90 − 100 34.55 / 85

= −2.6685

Decision: Since tcalc = –2.6685 is less than the critical bound of t = –1.6632, reject H0. There is enough evidence to conclude that the mean reimbursement for consultations with specialists paid by Medicare is less than $100. (e)

H0 : p £ 0.10 H1 : p > 0.10 Decision rule: If Z > 1.645, reject H0. Test statistic: Z =

p −

 (1 −  )

0.1765 − 0.10 0.10(1 − 0.10) 85

= 2.3510

Decision: Since Zcalc = 2.3510 is greater than the critical bound of Z = 1.645, reject H0. There is sufficient evidence to conclude that more than 10% of all reimbursements for consultations with specialists paid by Medicare are incorrect. 9.94

(a)

H0 : m ³ 5 minutes H1 : m < 5 minutes Decision rule: df = 14. If t < –1.7613, reject H0. X -m 4.2866-5.0 Test statistic: t = = = -1.6867 S / n 1.637985/ 15

(b) (c)

Decision: Since tcalc = –1.6867 is greater than the critical bound of t = – 1.7613, do not reject H0. There is not enough evidence to conclude that the mean waiting time at a bank branch in a commercial district of the city is less than 5 minutes during the 12.00 pm to 1.00 pm lunch period. To perform the t test on the population mean, you must assume that the observed sequence in which the data were collected is random and that the data are approximately normally distributed. Normal probability plot:

With the exception of one extreme point, the data are approximately normally distributed.

9.95

(d)

Based on the results of (a), the manager does not have enough evidence to make that statement.

(a)

H 0 :  = 1700 H1 :   1700 Decision rule: df = 29. If |t |> 2.0452, reject H0. Test statistic: t =

(b)

X -m S/ n

1723.4 -1700 89.55/ 30

= 1.4312

Decision: Since tcalc = 1.4312 is less than the critical bound of t = 2.0452, do not reject H0. There is not enough evidence to conclude that the mean height is different from 1,700 mm. In order for the t test to be valid, the data are assumed to be independently drawn from a population that is normally distributed.

(c)

The box-and-whisker plot suggests that the distribution of the data is skewed only slightly to the left. With a sample size of 30, the t distribution will provide a good approximation to the sampling distribution of the sample mean. 9.96

(a)

H0 : m = 0.5

H1 : m ¹ 0.5

Decision rule: df = 169. If |t| > 1.9741, reject H0. Test statistic: t =

(b)

(c)

X -m S/ n

0.2641-0.5 0.1424 / 170

= 21.6059

Decision: Since tcalc < –1.9741, reject H0. There is enough evidence to conclude that the mean particle loss is different from 0.5 grams. p-value is virtually zero. If the population of observing a sample of 170 shingles that will yield a test statistic more extreme than –21.6059 is virtually zero.

H0 : m = 0.5

H1 : m ¹ 0.5

Decision rule: df = 139. If |t| > 1.977, reject H0. Test statistic: t =

(d) (e)

9.97

X -m S/ n

0.218-0.5 0.1227/ 140

= -27.1940

Decision: Since tcalc < –1.977, reject H0. There is enough evidence to conclude that the mean particle loss is different from 0.5 grams. p-value is virtually zero. The probability of observing a sample of 140 shingles that will yield a test statistic more extreme than –27.1940 is virtually zero if the population mean particle loss is in fact 0.5 grams. In order for the t test to be valid, the data are assumed to be independently drawn from a population that is normally distributed. Since the sample sizes are 170 and 140, respectively, which are considered large enough, the t distribution will provide a good approximation to the sampling distribution of the mean even if the population is not normally distributed.

Answers will vary.

Chapter 10: Hypothesis testing: Two-sample tests After studying this chapter you should be able to: 1. conduct hypothesis tests for the means of two independent populations 2. conduct hypothesis tests for the means of two related populations 3. conduct hypothesis tests for the variances of two independent populations 4. conduct hypothesis tests for two population proportions

(X1 - X2 )-( m1 - m2 ) (72-66)-0 = = 1.8229 s 12 s 22 102 152 + + 30 30 n1 n2

10.1

10.2

Z = 1.8229 Decision rule: If Zcalc > 1.96 or Z < –1.96, reject H0. Decision: Since Zcalc = 1.8229 is greater than the critical bound of 1.96, reject H0. There is enough evidence to conclude that the first population mean is different to the second population mean.

10.3

p-value = 2(1.0–0.9656) = 0.0688, where 0.9656 is a cumulative probability for Z = 1.82.

10.4

(a)

S p2 =

t =

(n1 − 1) S12 + (n2 − 1) S22 (20) 42 + (17) 52 = 21.29 = (n1 − 1) + (n2 − 1) 19 + 16

( X1 − X 2 ) − ( 1 − 2 ) 1 1 S  +   n1 n2  2 p

10.5

10.6

(53 − 48) − 0 1   1 21.29 +   20 17 

= 3.29

(both formula updated)

(b)

df = (n1 – 1) + (n2 – 1) = 19 + 16 = 35

(c)

Decision rule: df = 35. If tcalc > 2.4377, reject H0.

(d)

Decision: Since t = 3.29 is greater than the critical bound of XXXX, reject H0. There is enough evidence to conclude that the first population mean is larger than the second population mean.

Assume that you are sampling from two independent normal distributions with equal variances.

1 1 1   1 ( X1 − X2 )  t S p2  +  = (53 − 48)  2.031 21.29 +  formula changed  20 17   n1 n2  1.81  1 − 2  8.09

changed

10.7 question needs to say assume unequal variances (a) H 0 : 1 −  2  0 The mean estimated amount of calories in the cheeseburger is not

lower for the people who thought about the cheesecake first than for the people who thought about the organic fruit salad first H 0 : 1 − 2  0 The mean estimated amount of calories in the cheeseburger is lower for the people who thought about the cheesecake first than for the people who thought about the organic fruit salad first (b) PHStat output: Separate-Variances t Test for the Difference Between Two Means (assumes unequal population variances) Data Hypothesized Difference 0 Level of Significance 0.01 Population 1 Sample Sample Size 20 Sample Mean 780 Sample Standard Deviation 128 Population 2 Sample Sample Size 20 Sample Mean 1041 Sample Standard Deviation 140 Intermediate Calculations Numerator of Degrees of Freedom Denominator of Degrees of Freedom Total Degrees of Freedom Degrees of Freedom Separate Variance Denominator Difference in Sample Means Separate-Variance t Test Statistic Lower-Tail Test Lower Critical Value p-Value Reject the null hypothesis

3237120.6400 85867.8232 37.6989 37 42.4170 -261 -6.1532

-2.4314 0.0000

Decision rule: if tcalc < -2.4314, reject H0 t=

( X 1 − X 2 ) − ( 1 − 2 ) S12 S 22 + n1 n2

= −6.1532

Decision: Since -6.1532 < 2.4314, reject H0. There is evidence that the mean estimated

amount of calories in the cheeseburger is lower for the people who thought about the cheesecake first than for the people who thought about the organic fruit salad first 10.8

(a) H0: μ1 – μ2 ≤ 0 H1: μ1 – μ2 > 0 where population 1 = private school students

population 2 = public school students. df = (n1 – 1) + (n2 – 1) = 53 + 31 = 84 Decision rule: If tcalc > 2.3733, reject H0. S p2 =

(n1 − 1) S12 + (n2 − 1) S22 (53)122 + (31)102 = = 127.76 (n1 − 1) + (n2 − 1) 53 + 31

Test statistic: t =

( X1 − X 2 ) − ( 1 − 2 ) 1 1 S p2  +   n1 n2 

76.5 − 74.7  1 1  127.76  +   54 32 

= 0.7138

Decision: Since 0.7138 < 2.3733, do not reject H0. There is not enough evidence that private school students outperform public school students. 10.9

Assuming that the variance e of the weight loss of the high-protein diet and high-carbohydrate diet are the same, the appropriate test to perform is the pooled-variance test. (a)

H0 : 1 − 2 = 0 H1 : 1 − 2  0

(b)

A Type I error is committed when one concludes that there is a difference in mean weight loss between the two diets when there is no significant difference. A Type II error is committed when one concludes that there is no significant difference in mean weight loss between the two diets when there is indeed significant difference.

(c)

(d)

Sp2 =

(n1 -1) S12 +(n2 -1) S22 (n1 -1)+(n2 -1)

(X1 - X 2 )-( m1 - m2 ) æ 1 1ö S ç + ÷ è n1 n2 ø 2 p

(99)3.22 +(99)3.92 = 12.725 (199) (7.6- 6.7)- 0

æ 1 1 ö 12.725ç + è 100 100 ÷ø

= 1.7840

Critical values = ±1.9720 Decision: Since tcalc = 1.7840 is between the critical bounds of ±1.9720, do not reject H0. There is no evidence of a difference between the mean weight loss of obese patients in the high-protein and high-carbohydrate diets. 10.10

H0 : 1 − 2  0 H1 : 1 − 2  0 where Campbelltown = population 1 and rest of Sydney = population 2 df = (n1 – 1) + (n2 – 1) = 222 + 222 = 444 Decision rule: If tcalc > 1.645, reject H0.

S p2 =

(n1 − 1) S12 + (n2 − 1) S22 (222)12.12 + (222)12.72 = = 153.85 (n1 − 1) + (n2 − 1) 222 + 222

t calc =

( X1 − X 2 ) − ( 1 − 2 ) 1 1 S  +   n1 n2 

2 p

140.1 − 131.4 1   1 153.85 +   222 222 

= 7.3898

Decision: Since tcalc = 7.3898 > 1.645, reject H0. There is evidence to suggest that the average petrol price in Campbelltown is greater than the rest of Sydney. 10.11

(a)

Populations 1 = female, 2 = male

H0 : 1 − 2 = 0 H1 : 1 − 2  0 Decision rule: df = 58. If t > 2.0017 or < -2.0017, reject H0. S p2 =

t =

(n1 − 1) S12 + (n2 − 1) S22 (29)57.782 + (29)205.722 = = 22830.24 new (n1 − 1) + (n2 − 1) 29 + 29

( X1 − X 2 ) − ( 1 − 2 ) 1 1 S  +   n1 n2  2 p

(79 − 157.17) − 0  1 1  22830.24 +   30 30 

= −2.0036 new

Decision: Since tcalc = -2.0036 is less than the lower critical bound of 2.0017, reject H0. There is sufficient evidence to conclude that there is

enough evidence of a difference in the mean time spent on Facebook per day between males and females (assumes equal population variances) Data Hypothesized Difference Level of Significance Population1Sample Sample Size Sample Mean Sample StandardDeviati o Population2Sample Sample Size Sample Mean Sample StandardDeviati o

0 0.05

Confidence Interval Estimate forthe Difference BetweenTwoMeans Data

30 79 57.78079862 30 157.1666667 205.7227936

Intermediate Calculations Population 1Sample Deg 29 Population 2Sample Deg 29 Total Degrees of Freedom 58 Pooled Variance 22830.24425 Standard Error 39.0130 Difference in Sample Me -78.16666667 t Test Statistic -2.0036 Two-Tail Test Lower Critical Value Upper Critical Value p-Value Reject the null hypothesis

Confidence Level

95%

Intermediate Calculations Degrees of Freedom 58 t Value 2.0017 Interval Half Width 78.0931 Confidence Interval Interval Lower Limit -156.2597 Interval Upper Limit -0.0736

-2.0017 2.0017 0.0498

(b) You must assume that each of the two independent populations is normally

10.12

H0 : 1 − 2 = 0 where populations 1 = line A, 2 = line B H1 : 1 − 2  0

(a)

Decision rule: df = 25. If t calc  2.0595, reject H0. S p2 =

(n1 − 1) S12 + (n2 − 1) S22 (n1 − 1) + (n2 − 1)

S p2 =

(10) 0.615 + (15) 0.706 = 0.0751 10 + 15

t =

( X1 − X 2 ) − ( 1 − 2 ) 1 1 S p2  +   n1 n2 

(410.25 − 409.85) − 0

= 3.7276 1   1 0.0751  +   11 16  Since t = 3.7276 > 2.0595 or p-value = 0.0010 < 0.05, reject Ho. There is sufficient evidence of a difference in the mean weight of cans filled on the two lines.

H0 : 1 − 2 = 0 where populations 1= line A, 2 = line B H1 : 1 − 2  0

(b)

Decision rule: df = 12. If t calc  2.1788, reject H0. Test statistic:

10.13

(X1 - X 2 )-( m1 - m2 ) S12 S22 + n1 n2 (410.25- 409.89)- 0

= 1.5625 0.6152 0.7062 + 11 16 Since t = 1.5625 < 2.1788 or p-value = 0.1441 > 0.05, do not reject H0. There is not sufficient evidence of a difference in the mean weight of cans filled on the two lines.

(c)

The results from (a) and (b) are different. The results obtained from (b) may be more reliable because the sample variances from both samples suggest that the two population variances are not likely to be equal.

(a)

H0 : 1 − 2  0 H1 : 1 − 2  0 S p2 =

(n1 − 1) S12 + (n2 − 1) S22 (33)(6.25) + (44)(25) = = 16.96 (n1 − 1) + (n2 − 1) 77

( X1 − X2 ) − ( 1 − 2 ) 1 1 S  +   n1 n2 

2 P

(8 − 9.5) − 0  1 1  16.96  +   34 45 

= 1.603

df = 45 + 34 – 2 = 77 Decision rule: Reject H0 if tcalc < –1.6649. Since tcalc = –1.603 > –1.6649, do not reject H0. There is not enough evidence that the mean waiting time at the Bank of Singapore is lower than that at the competitor’s bank.

( X - X ) ±t S çè n1 + n1 ÷ø = (8- 9.5)±1.99 16.96çè 341 + 451 ÷ø

(b)

2 P

–3.362 ≤ µ1 – µ2 ≤ 0.362 We are 95% confident that the difference in mean time waiting time between Bank of Singapore and the competitor’s bank is between –3.362 and 0.362. 10.14

H0 : 1 − 2  0 H0 : 1 − 2  0 Degrees of freedom = 68

t calc =

t =

( X - X )-(m - m ) 1

æS S ö çn +n ÷ è 1 2ø 2 1

( 9.5 − 8 ) − 0 6.25 25 + 34 25

2 2

= − 1.744

Decision rule: Reject H0 if tcalc < –1.667. Since tcalc = –1.744 < –1.667, do not reject H0. There is sufficient evidence that the mean waiting time at the Bank of Singapore is lower than that at the competitor’s bank. The outcomes of the tests in 10.13(a) and 10.14 are different. The value of pooled tcalc in 10.13(a) is slightly lower and should be more reliable as variances at the two banks appear to be different. 10.15

(a)

H0 : 1 − 2 = 0 M Mean times to clear problems at Office I and Office II are

the same. H1 : 1 − 2  0 Mean times to clear problems at Office I and Office II are

different. S

2 P

t =

( n1 − 1) S12 + ( n2 − 1) S22 = = 3.265 new ( n1 − 1) + ( n2 − 1) ( X1 − X2 ) − ( 1 − 1 ) 1 1 S  +  n2   n1

= 0.354 new

2 P

Reject H0 if tcalc > 2.024 or tcalc < –2.024. Since tcalc = 0.354, that is < 2.024, do not reject. (b) t Test for Differences in Two Means Data Hypothesized Difference Level of Significance Population 1 Sample Sample Size Sample Mean Sample Standard Deviation Population 2 Sample Sample Size Sample Mean Sample Standard Deviation Intermediate Calculations Population 1 Sample Degrees of Freedom Population 2 Sample Degrees of Freedom Total Degrees of Freedom Pooled Variance Difference in Sample Means t-Test Statistic Two-Tailed Test Lower Critical Value Upper Critical Value p-Value Do not reject the null hypothesis

0 0.05 20 2.214 1.718039 20 2.0115 1.891706 19 19 38 3.265105 0.2025 0.354386 -2.02439 2.024394 0.725009

p-value = 0.725. The probability of obtaining a sample that will yield a t test statistic more extreme than 0.354 is 0.725 if, in fact, the mean times for Office I and II are the same. (c)We need to assume that the two populations are normally distributed.

1 1  1 1  (d) ( X1 − X2 )  t S P2  +  = (2.214 − 2.0115)  2.024 3.265 +  new  20 20   n1 n2  –0.9543 ≤ µ1 – µ2 ≤ 1.3593

Since the Confidence Interval contains 0, we cannot claim that there’s a difference between the two means 10.16 H0 : I − II = 0 Mean times to answer queries by Team I and Team II are the same.

H1 : I − II  0 Mean times to answer queries by Team I and Team II are different. Degrees of freedom =38

t =

( X − X ) − ( −  ) 1

2 1

2 2

S S + n1 n2

t =

(2.719 − 2.6615) − 0 2.482 2.813 + 20 20

= 0.112

Since tcalc < 2.024 (same as in 10.15) do not reject the null. There is not enough evidence to conclude that the time to answer queries by the two groups is different. The conclusions from both the pooled-variance t test and the separate variance t test are exactly the same.

H0 : 1 − 2  0 H1 : 1 − 2  0 Where females = population 1 and males = population 2 10.17

(a)

Excel output: t-Test: Two-Sample Variances

Assuming

Equal

Female

male

Mean

49925.94444

77478.18182

Variance

489142253.2

673473852.5

Observations

Pooled Variance

591009716

Hypothesized Mean Difference

df t Stat

38 3.565965675

P(T<=t) one-tail

0.00049959

t Critical one-tail

1.68595446

P(T<=t) two-tail

0.000999181

t Critical two-tail

2.024394164

Decision rule: Reject H0 if tcalc < –1.686 Decision: Since tcalc = –3.566 is less than –1.686, reject H0. There is evidence that male graduate salaries exceed those of females. (b)

p-value = 0.0005. The probability of obtaining two samples with a mean difference of -3.566 or less is 0.0005 if the mean female salaries are equal to those of males.

(c)

Since both sample sizes are smaller than 30, you need to assume that the population of male and female graduate salaries is normally distributed.

(d)





1 1  +  ( X − X )  t S  n1 + n1  = (49925.94 − 77478.18)  1.686 591009716  18 22 1

2 P

 1







–40578.7 ≤ µ1 – µ2 ≤ –14525.8 10.18

H0 : 1 − 2  0 H1 : 1 − 2  0

Excel output: t-Test: Two-Sample Assuming Unequal Variances female

male

Mean

49925.94444

77478.18182

Variance

489142253.2

673473852.5

Observations

Hypothesized Mean Difference

t Stat

–3.624446816

P(T<=t) one-tail

0.000422668

t Critical one-tail

1.68595446

P(T<=t) two-tail

0.000845335

t Critical two-tail

2.024394164

Decision rule: reject H0 if tcalc< –1.686 Decision: Since tcalc = -3.624 is less than the-1.686, reject H0. The value of pooled-variance t test statistic and the separate-variance t test statistic are almost identical. 10.19 (a)

Population 1 = computer-assisted individual-based, 2 = team-based. H0 : 1 − 2 = 0

H1 : 1 − 2  0 S P2 =

S P2 =

t =

( n1 − 1) S12 + ( n2 − 1) S22 ( n1 − 1) + ( n2 − 1)

(20) 1.93332 + (20) 4.57672 20 + 20

= 12.3419

( X − X ) − ( −  ) 1

1 1 S P2  +  n2   n1

(17.5571 − 198905) − 0 1   1 12.3419  + 21   21

= −2.1522

Decision rule: df = 40. If tcalc < –2.0211 or > 2.0211, reject H0. Decision: Since tcalc = –2.1522 is below the lower critical bound of – 2.0211, reject H0. There is enough evidence to conclude that the mean assembly times in seconds are different between employees trained in a computer-assisted, individual-based program and those in a team-based program. (b)

You must assume that each of the two independent populations is normally distributed.

H0 : 1 − 2 = 0 H1 : 1 − 2  0

(c)

t Test: Two-Sample Unequal Variances

Assuming

Computerassisted program

Mean

17.55714286

Variance

3.737571429

Observations Hypothesized Mean Difference Df t Stat P(T<=t) one-tail t Critical one-tail P(T<=t) two-tail t Critical two-tail

21 0 27 –2.152203195 0.020240852 1.703288035 0.040481703 2.051829142

t =

(17.5571 − 19.8905) − 0 1.93332 4.57672 + 21 21

Teambased program 19.8904761 9 20.9459047 6 21

= −2.1522

Decision rule: df = 27. If tcalc < –2.052 or > 2.052, reject H0. Decision: Since tcalc = –2.1522 is below the lower critical bound of –2.052, reject H0. There is enough evidence to conclude that the mean assembly times in seconds are different between employees trained in a computerassisted, individual-based program and those in a team-based program. (d) The results in (a) and (c) are the same. (e)

( X - X ) ±t S çè n1 + n1 ÷ø = (17.557-19.89) ±2.021 12.3419çè 211 + 211 ÷ø 1

2 P

–4.524 ≤ µ1 – µ2 ≤ –0.142 10.20

df = n – 1 = 12 – 1 = 11

10.21

df = n – 1 = 11 – 1 = 10

10.22 (a)

Population 1 = June 2011 daily room rates, 2 = March 2015 daily rates.

H0 : mD = 0 H1 : mD ¹ 0 n

åD

D = i=1 n

= 27.39042

(

SD2 = å Di - D i=1

) n 2

= 750.2353

D − D SD

7.333 − 0 27.39

= 1.1359

df = ( n − 1) = 17 t 0.05 = 2.1098 2015

2011

Diffs

120

173

–53

139

133

105

156

167

–11

170

139

141

–2

Mean

7.333333

201

180

27.39042

247

223

122

116

125

167

–42

1.135897

156

142

316

273

148

143

177

124

165

135

191

176

163

159

235

262

–27

Decision: Since tcalc = 1.1359 is less than the upper critical value of 2.1098, do not reject H0. There is insufficient evidence to conclude that there is a difference in the mean daily hotel rates in 2011 and 2015.

10.23 (a)

(b)

You must assume that the distribution of the differences between the hotel daily room rate in 2011 and 2015 is approximately normal.

(c)

p-value is 0.27. The probability of obtaining a mean difference in daily hotel rates that gives rise to a test statistic that deviates from 0 by 1.1359 or more in either direction is 0.27 if there is no difference in the mean daily hotel rate in 2011 and 2015.

(d)

-6.2876 £ mD £ 20.9536 n 18 You are 95% confident that the mean difference in hotel rate between 2011 and 2015 is somewhere between –$6.29 and $20.95. D ±t

= 7.333 ± 2.1098

27.39

H0 : mD = 0 vs. H1 : mD ¹ 0 Excel Output:

HypothesizedMeanDifference Level of significance

0 0.05

Intermediate Calculations Sample Size 13 DBar -8.9231 Degrees of Freedom 12 SD Standard Error t Test Statistic

3.0403 0.8432 -10.5820

Two-Tail Test Lower Critical Value Upper Critical Value p-Value Reject the null hypothesis

-2.1788 2.1788 0.0000

Test statistic: t =

D − D = -10.582 SD

n Decision: Since tcalc = –10.582 falls below the lower critical values -2.1788, reject H 0 . There is enough evidence of a difference in the mean service

rating between TV and phone. (b)

You must assume that the distribution of the differences between the mean measurements is approximately normal.

(c)

Both the boxplot and normal probability plot suggest that the distribution does not deviate too far from normal.

(d)

D t

= − 8.9231  2.1788

n −10.76  D  −7.09

3.0403 13

new

You are 95% confident that the difference in the mean service rating between TV and phone is between -10.76 and -7.09 10.24

Excel output: Cola A Adindex Cola B(Test Cola) Adindex Mean 18.55263158 21.31578947 StandardError 0.978937044 0.822086011 Median 18 21 Mode 24 21 StandardDeviation 6.034573222 5.067678519

Sample Variance 36.41607397 Kurtosis -0.640865482 Skewness -0.077015645 Range 24 Minimum 6 Maximum 30 Sum 705 Count 38 First Quartile 15 Third Quartile 24 Interquartile Range 9 1.33*StdDev 8.025982385 5*StdDev 30.17286611

25.68136558 -0.294923931 -0.173917096 21 9 30 810 38 18 24 6 6.740012431 25.3383926

From the descriptive statistics provided in the Microsoft Excel output there does not seem to be any violation of the assumption of normality. The mean and median are similar and the skewness value is near 0. Without observing other graphical devices such as a stem-and-leaf display, boxplot, or normal probability plot, the fact that the sample size (n = 38) is not very small enables us to assume that the paired t test is appropriate here. PHStat output: Pairedt Test Data HypothesizedMeanDifference Level of significance

0 0.05

Intermediate Calculations Sample Size 38 DBar -2.7632 Degrees of Freedom 37 SD Standard Error t Test Statistic

6.6309 1.0757 -2.5688

Lower-Tail Test Lower Critical Value p-Value Reject the null hypothesis

-1.6871 0.0072

The PHStat output for the paired t test indicates the p-value is 0.0072 < 0.05, and, hence, reject H0 that the mean Cola A Adindex is no less than Cola B (Test Cola) Adindex. There is enough evidence that the cola video ad is significant in raising the Adindex of the test Cola.

10.25

(a) Population 1 = Fruit Shop, 2 = Supermarket H0: μD ≥ 0 H1: μD < 0

Excel output

t-Test: Paired Two Sample for Means Thursday

Friday

Mean

130.9

148

Variance

94.54444

177.3333

Observations

Pearson Correlation Hypothesized Difference

0.496847 Mean 0

t Stat

-4.51864

P(T<=t) one-tail

0.000725

t Critical one-tail

2.821438

P(T<=t) two-tail

0.00145

t Critical two-tail

3.249836

(b)

10.26

(a)

df = 10-1 = 9 Decision rule: Reject H0 is tcalc < –2.821 Decision: Since t = –4.519 is less than the lower critical value of – 2.821, reject H0. There is sufficient evidence at α = 0.01 to conclude that petrol prices increase on public holidays. The p-value of 0.0007 indicates that there is a 0.0007 probability of observing a calculated value of –4.519 or less if petrol prices on public and non-public holidays are equal. Population 1 = Before Lumosity, 2 = After Lumosity

H0: μD ≥ 0 H1: μD < 0 Excel output t-Test: Paired Two Sample for Means Before

After

Mean

108.2857

111.4286

Variance

299.5714

300.2857

Observations

Pearson Correlation Hypothesized Difference

0.955863 Mean 0

t Stat

-1.61602

P(T<=t) one-tail

0.078609

t Critical one-tail

3.142668

P(T<=t) two-tail

0.157218

t Critical two-tail

3.707428 Decision rule: Reject H0 if tcalc < –3.143 Decision: Since t = –1.616 is not less than the lower critical value of –3.143, do not reject H0. There is insufficient evidence at α = 0.01 to conclude that the mean IQ has increased after using Lumosity.

(b) (c)

The differences between the IQ before and after using Lumosity is approximately normally distributed. The p-value for this test is 0.079; it needs to be less than 0.01 in order to reject H0.

10.27 (a) change to 0.05  = 0.05, n1 =8, n2 =7, FU = 4.21, FL = 0.2375 i.e. df1= 7, df2= 6 (b)

a =0.05, n1 =9, n2 =6, FU = 4.82, FL = 0.2075

(c)

a =0.025, n1 =7, n2 =5, FU = 9.20, FL = 0.1087

(d)

a = 0.01, n1 =9, n2 =9, FU = 6.03, FL = 0.1658

(a)

a =0.05, n1 =8, n2 =7, F0.05 = 4.21, i.e. df1= 7, df2= 6

(b)

a =0.025, n1 =9, n2 =6, F0.025 = 6.76

(c)

a =0.01, n1 =7, n2 =5, F0.01 = 15.21

(d)

a =0.005, n1 =9, n2 =9, F0.005 = 7.50

(a)

a =0.05, n1 =16, n2 =21, F0.95 = 0.4296, i.e. df1= 15, df2= 20

(b)

a =0.025, n1 =16, n2 =21, F0.975 = 0.3629

(c)

a =0.01, n1 =16, n2 =21, F0.99 = 0.2966

(d)

a =0.005, n1 =16, n2 =21, F0.995 = 0.2576

10.28

10.29

10.30

Fcalc =

S22

= 2

S 2 133.7 161.9 = 1.2109 (or alternatively, Fcalc = 22 = = 0.8258 ) 133.7 S1 161.9

10.31 There are  2 = 15 and 1 = 10 degrees of freedom respectively in numerator and denominator of F =

S22

S12

S22

(or alternatively, 10 and 15 in F = 2

)

10.32 FU = 3.52 and FL = 0.327 (or alternatively FU =3.06 = 1/0.327 and FL = 0.826 = 1/3.52) 10.33 Since Fcalc = 1.2109 < FU = 3.52 and Fcalc = 1.2109 > FL = 0.327, then Fcalc is not in the rejection region. Therefore, do not reject H0. 10.34 No, since the F test is very sensitive to the normality assumption, it cannot be validly used when that assumption is clearly violated by the statement that the data, unlike normally distributed data, is very skewed. 10.35 (a)

Fcalc = S12 S22 =47.3 36.4 =1.299 suggesting that s

2 1

which

considerably

greater

than

s >1, that is, s > s . 2 2

2 1

2 2

(

)

(

)

FU = F a , u1 , u2 = F 0.05, 15, 12 = 2.62 where υ1 = df1–1 = 16–1 and

(b)

υ2 = df2–1 = 13–1. Thus, since Fcalc = 1.299 < FU = 2.62, then do not reject H0: s 12 = s 22 in favour of H1: s 12 > s 22 . There is little evidence to support the claim that s 12 > s 22 .

(

)

(

)

FL = F 1- a , u1 , u2 = F 0.95, 15, 12 = 0.4040 and since Fcalc = 1.299 is not even

(c)

less than 1, it is certainly not in the lower rejection region, i.e. do not reject H0: s 12 = s 22 in favour of H1: s 12 < s 22 . What little evidence there is that s 12 ¹ s 22 is that s 12 > s 22 . 10.36

H0 : s 12 - s 22 = 0

(a)

H1 : s 12 - s 22 ¹ 0 PHStat2 output F Test for Differences in Two Variances Data Level of Significance

0.05

Larger-Variance Sample Sample Size

Sample Variance

144

Smaller-Variance Sample Sample Size

Sample Variance

100

Intermediate Calculations F Test Statistic Population 1 Sample Degrees of Freedom Population 2 Sample Degrees of Freedom

1.4400 53 31

Two-Tail Test Upper Critical Value p-Value Do not reject hypothesis

1.9409 0.2780 the

null

Since the p-value = 0.278 > 0.05, then do reject H0, i.e. underlying variances are equal. (b)

The test assumes that the two populations are both normally distributed.

(c)

The pooled variance t test is appropriate.

10.37 (a)

H0: s 12 = s 22 The population variances are the same. H1: s 12 ¹ s 22 The population variances are different. Decision rule: If Fcalc > 2.9786, reject H0. S2 2.08222 Test statistic: F = 12 = = 1.6159 S2 1.63802 Decision: Since Fcalc = 1.6159 is below the upper critical bound of Fa /2 = 2.9786, do not reject H0. There is not enough evidence to conclude that

(b) (c)

the two population variances are different. p-value = 0.715. The probability of obtaining a sample that yields a test statistic more extreme than 1.6159 is 0.715 if the null hypothesis that there is no difference in the two population variances is true. The test assumes that the two populations are both normally distributed. Normal Probability Plot of Waiting Time (Bank 1) 7

Waiting Time (Bank 1)

0 -2

-1.5

-1

-0.5

0.5

1.5

Z Value

Normal Probability Plot of Waiting Time (Bank 2) 12

Waiting Time (Bank2)

0 -2

-1.5

-1

-0.5

0.5

1.5

Z Value

Waiting Time (Bank2)

Waiting Time (Bank 1)

(d)

Based on the results of (b), it is not appropriate to use the pooled-variance t-test to compare the means of the two branches. That is, the F-ratio test for testing equality of variances is not justified and since we are not able to assume that the two population variances are equal, we cannot pool the sample variances.

10.38 (a)

H0 :s 12 - s 22 = 0 H1 :s 12 - s 22 ¹ 0

PHStat2 output: F Test for Differences in Two Variances Data Level of Significance

0.05

Larger-Variance Sample Sample Size

Sample Variance

51.09090909

Smaller-Variance Sample Sample Size

Sample Variance

20.47272727

Intermediate Calculations F Test Statistic Population 1 Sample Degrees of Freedom Population 2 Sample Degrees of Freedom

2.4956 10 10

Two-Tail Test Upper Critical Value

3.7168

p-Value

0.1653

not

reject

the

null

hypothesis

Decision: Since the p-value is 0.1653 > 0.05, do not reject H0. There is not enough evidence to conclude that the two population variances are different. The p-value = 0.1653. If the population variances of both groups were equal, the probability of a sample F ratio falling in the lower or upper rejection regions is 0.1653. The test assumes that the two populations are both normally distributed. The pooled-variance t test can be validly carried out.

(a)

H0 :  12 - 22 = 0 The population variances in petrol prices are the same for

(b)

10.39

Campbelltown and the rest of Sydney. H1 :  12 -  22  0 The population variances in petrol prices are not the same for Campbelltown and the rest of Sydney.. Decision rule: if F < 0.74 or F > 1.35, reject the null hypothesis (using n=120 as a proxy). S2 146.41 Test statistics F = 12 = = 0.91 . 161.29 S2

(b)

10.40 (a)

Decision: Since F = 0.74 < 0.91 < 1.35 we do not reject the null hypothesis. There is not enough evidence to conclude the population variances in petrol prices are different. Assuming the underlying normality in the two populations is met, based on the results obtained in part (a), it is more appropriate to use the pooledvariance t test to compare petrol prices for Campbelltown and the rest of Sydney . H0 :  1 −  2 = 0 H1 :  1 −  2  0

Z =

new

( p1 − p2 ) − ( 1 −  2 ) 1 1 p(1 − p ) +   n1 n2  ( p1 − p2 ) − ( 1 −  2 ) = 1 1 p(1 − p ) +   n1 n2  =

where p =

X1 + X 2 55 + 30 85 = = = 0.459 new n1 + n2 120 + 65 185

 55 30   120 − 65  − 0   new 1   1 0.459(1 − 0.459) +   120 65 

= −0.04 Decision rule: if Z > 1.96 or , -1.96reject H0 Since -0.04 > -1.96, do not reject H0.. Thus there is no evidence that the two group population proportions are not equal. (b)

( p1 − p2 )  Z

p1 (1 − p1 ) p2 (1 − p2 ) + n1 n2

55  55  30  30  1−   1 −  120  120  65  65   55 30  new = −   1.96 + 120 65  120 65  = −0.003  0.1505 10.41 (a)

H 0 : 1 −  2 = 0 H1 : 1 −  2  0

( p1 − p2 ) − (1 −  2 ) 1 1 p (1 − p )  +   n1 n2 

new

where p =

x1 − x2 45 + 25 70 = = = 0.467 n1 + n2 100 + 50 150

( p1 − p2 ) − (1 −  2 ) 1 1 p (1 − p )  +   n1 n2 

 45 25  − −0  100 50   = 1   1 0.467 (1 − 0.467 )  +   100 50  = −0.5787 Z 0.005 = 2.576 Thus, fail to reject H0 as the calculated z is in the non-rejection region. Thus there is not enough evidence that the two group population proportions are unequal.

(b)

( p1 − p2 )  Z

p1 (1 − p1 ) p2 (1 − p2 ) + n1 n2

45  45  25  25  1−   1 −  100  100  50  50   45 25  = −   2.576 + 100 50  100 50  = −0.05  0.0132 =  −0.27272, 0.172716

10.42 (a)

H 0 : 1 −  2  0 H1 : 1 −  2  0

Where population 1 = Christchurch and population 2 = Brisbane

p1 =

12 32 12 + 32 = 0.0329, p2 = = 0.0360, p = = 0.0348 365 890 365 + 890

Using the 0.01 level of significance Decision rule: Reject null

( p1 − p2 ) − (1 −  2 ) = −0.2703

hypothesis

Zcalc

–2.33.

1 1 p (1 − p )  +   n1 n2 

Since –0.2703 > –2.33, we do not reject the null hypothesis at 1% level and conclude that there is not enough evidence that there is a significant improvement in the rate of resignations of Christchurch vs Brisbane factories. (b)

−0.003078  −2.33

0.0329(1 − 0.0329) 0.0360(1 − 0.0360) + = −0.003078  0.011228 365 890

= [–0.014306, 0.00815] 10.43 (a)

H 0 : 1 −  2 = 0

new

H1 : 1 −  2  0 p1 = 0.38, p2 = 0.33, p = 0.34 Using the 0.05 level of significance Decision, reject null hypothesis if Zcalc < –1.96 or > 1.96.

Z =

( p1 − p2 ) − (1 −  2 ) = 2.6765 new 1 1 p (1 − p )  +   n1 n2 

Since 2.6765 > 1.96, we reject the null hypothesis. There

is sufficient evidence of a significant difference in the proportion of technology crowd-funding projects and games crowd-funding projects that were successful.

ZTest for Differences inTwoProportions Data HypothesizedDifference Level of Significance Group1 Number of Items of Interest Sample Size Group2 Number of Items of Interest Sample Size

0 0.05 316 831 923 2796

Intermediate Calculations Group 1Proportion 0.380264741 Group 2Proportion 0.330114449 Difference in Two Proportions 0.050150292 Average Proportion 0.3416 Z Test Statistic 2.6765 Two-Tail Test Lower Critical Value -1.9600 Upper Critical Value 1.9600 p-Value 0.0074 Reject the null hypothesis (b)

p-value = 0.0074. The probability of obtaining a difference in proportions that

gives rise to a test statistic that deviates from 0 in either direction by 2.6765 or more in either direction is virtually 0 if there is no difference in the proportion of technology crowd- funding projects and games crowd-funding projects that were successful 10.44 (a)

H 0 : 1 −  2 = 0 H1 : 1 −  2  0

new

where Populations: 1 = males, 2 = females Decision rule: If Zcalc < – 2.58 or > 2.58, reject H0. Zcalc = -3.5080

Decision: Since Zcalc = -3.5080 is less than the lower critical bound, reject H0. There is sufficient evidence to conclude that a significant difference exists in the proportion of males and females who enjoy shopping clothing for themselves Z Test for Differences in Two Proportions Data Hypothesized Difference Level of Significance Group 1 Number of Items of Interest

0 0.01 218

Sample Size

542

Group 2 Number of Items of Interest

276

Sample Size

543

Intermediate Calculations Group 1 Proportion Group 2 Proportion Difference in Two Proportions Average Proportion Z Test Statistic

0.402214022 0.508287293 -0.106073271 0.455299539 -3.50802898

Two-Tail Test Lower Critical Value -2.575829304 Upper Critical Value 2.575829304 p-Value 0.00045144 Reject the null hypothesis

(b)

p-value = 0.0005. The probability of obtaining a difference in two sample proportions of 0.1061 or more in either direction when the null hypothesis is true is 0.0005. (c)

Change question to 99%-0.1835  1 − 2  -0.0286

You are 99% confident that the difference in the proportions of males and females who enjoy shopping clothing for themselves is between -0.1835 and - 0.0286. Confidence Interval Estimate of the Difference Between Two Proportions Data Confidence Level

99%

Intermediate Calculations Z Value

-2.575829304

Std. Error of the Diff. between two Proportions Interval Half Width

0.030064781 0.077441743

Confidence Interval Interval Lower Limit Interval Upper Limit

-0.183515014 -0.028631527

10.45 H 0 :  1 −  2  0 The proportion of car drivers in Malaysia that have converted to LPG fuel is no more than the proportion of car drivers in Singapore.

H1 :  1 −  2  0 The proportion of car drivers in Malaysia that have converted to LPG fuel is greater than the proportion of car drivers in Singapore.

p1 = 0.4, p2 = 0.4, p = 0.4

Using the 0.01 level of significance, reject H0, if Zcalc > 2.326.

Z =

( p1 − p2 ) − (1 −  2 ) = 0.00 1 1 p (1 − p )  +   n1 n2 

Since 0.00 < 2.326, we do not reject the null hypothesis, there is not enough evidence to show that the proportion of car drivers in Malaysia that have converted to LPG fuel is different as the proportion of car drivers in Singapore. 10.46 H 0 :  M −  F  0 The proportion of males who prefer margarine is at least the proportion of females who prefer margerine H1 :  M −  F  0 The proportion of males that prefer margarine is less than the proportion of females that prefer margarine

pM = 0.4024, pF = 0.4887, p = 0.4558

Using the 0.05 level of significance Decision, reject null hypothesis if Zcalc < –1.645.

Z =

( pM − pF ) − ( M −  F ) = −1.235  1 1  p (1 − p )  +   nM nF 

Since –1.235 > –1.645, we do not reject the null hypothesis at 5% level, there is insufficient evident to conclude that the proportion of males who prefer margarine is less than the proportion of females who prefer margarine. 10.47 Among the criteria to be used in selecting a particular hypothesis test are the type of data, whether the samples are independent or paired, whether the test involves central tendency or variation, whether the assumption of normality is valid and whether the variances in the two populations are equal. 10.48 The separate variance t test is used when the variances of independent populations are unequal. 10.49 The F test can be used to examine the differences in two variances when each of the two populations is assumed to be normally distributed. 10.50 With independent populations, the outcomes in one population do not depend on the outcomes in the second population. With two related populations, either repeated measurements are obtained on the same set of items or individuals, or items or individuals are paired or matched according to some characteristic. 10.51 Repeated measurements represent two measurements on the same items or individuals, while paired measurements involve matching items according to a characteristic of interest. 10.52 When you have obtained data from either repeated measurements or paired data. 10.53 They are two different ways of investigating the concern of whether there is a significant difference between the means of two independent populations. If the hypothesised value of 0 for the difference in two population means is not in the confidence interval, then, assuming a two-tailed test is used, the null hypothesis of no difference in the two population means can be rejected.

10.54 When parametric assumptions can be met that the data is normally distributed and measured at least at the interval scale.

10.55 One year return

H 0 : s 12 - s 22 = 0 H1 : s 12 - s 22 ¹ 0

F Test for Differences in Two Variances Data Level of Significance 0.05 Larger-Variance Sample Sample Size 10 Sample Variance 8.925161111 Smaller-Variance Sample Sample Size 10 Sample Variance 3.29496 Intermediate Calculations F Test Statistic Population 1 Sample Degrees of Freedom Population 2 Sample Degrees of Freedom

2.7087 9 9

Two-Tail Test Upper Critical Value p -Value Do not reject the null hypothesis

4.0260 0.1538

Since p-value > 0.05, do not reject H0. There is not enough evidence to conclude that the two population variances are different. Hence, the appropriate test for the difference in two means is the pooled-variance t test

H 0 : 1 − 2 = 0 H1 : 1 − 2  0 Populations: 1 = short-term, 2 = long-term Pooled-Variancet TestfortheDifferenceBetweenTwo (assumes equal populationvariances) Data HypothesizedDifference 0 Level of Significance 0.05 Population1Sample Sample Size 10 Sample Mean 5.046 Sample StandardDeviation 1.815202468 Population2Sample Sample Size 10 Sample Mean 11.365 Sample StandardDeviation 2.987500814

Intermediate Calculations Population1Sample Degrees of Freedom 9 Population2Sample Degrees of Freedom 9 Total Degrees of Freedom 18 Pooled Variance 6.1101 Standard Error 1.1054 Difference in Sample Means -6.3190 t Test Statistic -5.7162 Two-Tail Test Lower Critical Value -2.1009 Upper Critical Value 2.1009 p-Value 0.0000 Reject the null hypothesis

Since the p-value = 0.0000 is less than 0.05, reject H0. There is sufficient evidence to conclude that the mean 1-year return is different between the long-term and short-term bond funds Three year return H 0 : s 12 - s 22 = 0

H1 : s 12 - s 22 ¹ 0 F Test for Differences inTwoVariances Data Level of Significance 0.05 Larger-Variance Sample Sample Size 10 Sample Variance 4.965444444 Smaller-Variance Sample Sample Size 10 Sample Variance 2.279555556 Intermediate Calculations F Test Statistic 2.1783 Population1Sample Degrees of Freedom 9 Population2Sample Degrees of Freedom 9 Two-Tail Test UpperCritical Value 4.0260 p-Value 0.2617 Donot reject the null hypothesis

H1 : 1 − 2  0 Populations: 1 = short-term, 2 = long-term

Pooled-Variancet TestfortheDifferenceBetweenTwo (assumes equal populationvariances) Data HypothesizedDifference 0 Level of Significance 0.05 Population1Sample Sample Size 10 Sample Mean 4.82 Sample StandardDeviation 1.50981971 Population2Sample Sample Size 10 Sample Mean 11.59 Sample StandardDeviation 2.228327724 Intermediate Calculations Population1Sample Degrees of Freedom 9 Population2Sample Degrees of Freedom 9 Total Degrees of Freedom 18 Pooled Variance 3.6225 Standard Error 0.8512 Difference in Sample Means -6.7700 t Test Statistic -7.9537 Two-Tail Test Lower Critical Value -2.1009 Upper Critical Value 2.1009 p-Value 0.0000 Reject the null hypothesis

Since the p-value < 0.05, reject H0. There is sufficient evidence to conclude that the mean 3year return is different between the long-term and short-term funds

10.56

(a)

H 0 : s 12 - s 22 = 0 H1 : s 12 - s 22 ¹ 0 F Test for Differences in Two Variances Data Level of Significance Larger-Variance Sample Sample Size Sample Standard Deviation Smaller-Variance Sample Sample Size Sample Standard Deviation

0.05 20 5.714421 38 5.406387

Intermediate Calculations

F Test Statistic Population 1 Sample Degrees of Freedom Population 2 Sample Degrees of Freedom

1.117198 19 37

Two-Tail Test Upper Critical Value p-Value Do not reject the null hypothesis

2.11685 0.749246

Since the p value > 0.05 there is not enough evidence of any difference in the variance of the study time for male students and female students (b)

Since there is not enough evidence of any difference in the variance of the study time for male students and female students, a pooled-variance t test should be used

H1 : 1 − 2  0 Pooled-Variance t Test for the Difference Between Two Means (assumes equal population variances) Data Hypothesized Difference 0 Level of Significance 0.05 Population 1 Sample Sample Size 20 Sample Mean 16.625 Sample Standard Deviation 5.714421 Population 2 Sample Sample Size 38 Sample Mean 11.02632 Sample Standard Deviation 5.406387 Intermediate Calculations Population 1 Sample Degrees of Freedom Population 2 Sample Degrees of Freedom Total Degrees of Freedom Pooled Variance Difference in Sample Means t Test Statistic

19 37 56 30.39127 5.598684 3.676244

Two-Tail Test Lower Critical Value Upper Critical Value p-Value Reject the null hypothesis

(d)

-2.00324 2.003241 0.000532

Since the p value < 0.05 there is enough evidence of a difference in the mean study time for male and female students

10.57 (a) (b) (c) (d)

(e)

10.58

(a)

At the 5% significance level, there is not enough evidence that the means of responses to ‘I have been sunburnt at least once during the summer’ are not equal. At the 5% significance level, there is enough evidence to conclude that the mean responses to ‘I would willing to pay more for a sunscreen that I know will be more effective while I am swimming’ are not equal. At the 5% significance level, there is enough evidence that the means of responses for males and females to ‘skin cancer due to sun exposure is something I want to prevent’ are not equal. At the 5% significance level, there is not enough evidence to conclude that the means of responses for males and females to ‘I was not aware that sunscreen needs to be applied at least twenty minutes before exposure to the sun’ are not equal. The means of the responses for males and females to two out of the four questions being asked in the survey are significantly different, at the 5% significance level.

H 0 :  A −  S = 0 Mean petrol is the same in Adelaide and rural South

Australia

H1 :  A −  S  0 Mean petrol is different in Adelaide and rural South

Australia Assuming that the samples are from underlying normal populations with equal variances, we can use pooled-variance t test. The t test statistics follow a t distribution with 44 degrees of freedom. Using a level of significance of 0.01, the critical values are –2.692 and 2.692. Reject H0 if tcalc < –2.692 or > 2.692 Test statistics:

( X − X ) − ( −  ) 1

1 1 S p2  +   n1 n2 

Where:

2 2 n1 − 1) S12 + ( n2 − 1) S 22 24 ( 0.23 ) + 20 ( 0.74 ) ( S = = = 0.278 24 + 20 ( n1 − 1) + ( n2 − 1) (1.23 − 1.43) − 0 = −1.281 t= 2 p

 1 1  0.278  +   25 21 

Since, –1.281 > –2.692, we do not reject the null hypothesis and at 1% significance level, we conclude that there is not enough evidence that mean price petrol in Adelaide is higher than that of rural South Australia.

(b)

H0 :  A2 −  S2 = 0 The population variances for petrol in Adelaide and rural South Australia are the same

H1 :  A2 −  S2  0 The population variances for petrol in Adelaide and rural South Australia are different Decision rule: Reject null if F > 3.22, or F < 0.33

S A2 0.232 F= 2= = 0.097 S S 0.74 2

Test statistics:

Decision: Since F = 0.097 is less than 0.33, we reject H0. There is enough evidence to conclude that Adelaide and rural South Australia have different population variances for mean petrol prices.

(c)

(X + X )t 1

n1+ n 2 − 2

1 1 S p2  +   n1 n2 

(1.23 -1.43) ± 2.692 ( 0.278) æçè 251 + 211 ö÷ø -0.62 £ m1 - m2 £ 0.22 We are 95% confident that the difference in mean petrol prices between Adelaide and rural South Australia is between –0.62 to 0.22. From a hypothesis-testing perspective, since the interval includes zero, we do not reject the null hypothesis of no difference between the means of the two populations.

10.59

(a)

H0 : m R £ 6 : The mean processing time in the research department is not greater than 6 seconds H 0 :  R  6 : The mean processing time in the research department is greater than 6 seconds Decision rule: df = 5. If tcalc > 2.015, reject null hypothesis. Test statistics: t =

X - m 85 - 6 = = 1.9463 S 3.1464 n

Decision: Since t = 1.9643 < 2.015, do not reject the null hypothesis. There is not enough evidence to conclude that the mean processing time in the research department is greater than 6 seconds. (b)

H0 :  A2 −  R2 = 0 : The population variances for processing times are the same for the accounting department and research department H1 :  A2 −  R2  0 : The population variances for processing times are different for the accounting department and the research department Decision rule: if F < 0.107 or F > 7.39, reject the null hypothesis.

S A2 3.27112 Test statistics F = 2 = = 1.08 . S R 3.14642 Decision: Since F = 1.08 is between critical bounds of 0.107 and 7.39, do not reject the null hypothesis. There is not enough evidence to conclude

that the population variances for processing times are different for the accounting department and the research department. (c)

H 0 :  A −  R = 0 : The two departments have the same mean processing time

H1 :  A −  R  0 : The two departments have different mean processing time

Assuming that the samples are from underlying normal populations with equal variances, we can use pooled-variance t test. The t test statistics follow a t distribution with 9 degrees of freedom. Using a level of significance of 0.05, the critical values are –2.2622 and 2.2622. Reject H0 if tcalc < –2.2622 or > 2.2622 Test statistics:

( X − X ) − ( −  ) 1

1 1 S p2  +   n1 n2 

Where:

2 2 n1 − 1) S12 + ( n2 − 1) S 22 4 ( 3.2711 ) + 5 ( 3.1464 ) ( S = = = 10.2556 4+5 ( n1 − 1) + ( n2 − 1) ( 7.8 − 8.5) − 0 = −0.3610 t= 2 p

1 1 10.2556  +  5 6

Since, –0.3610 > –2.2622, we do not reject the null hypothesis and at 5% significance level, we conclude that there is not enough evidence that the two departments have different mean processing times. (d)

(a) p–value = 0.0545. Since the p-value > 0.05, the probability of obtaining a t test statistic value that is 1.9643 or greater is 5.46% if the mean processing time in the research department is no more than 6 seconds. (b) Given F = 1.08, numerator df = 4 and denominator df = 5 for a twotailed hypothesis test:  1 1  P   = 0.4551 F P F5,4 > 1.08 = 0.4551  5,4 1.08 

(

)

 1 1  p − value = P    + P ( F5,4  1.08) = 0.9102 F  5,4 1.08  The probability of obtaining the F statistic value that is smaller than 1.108 or larger than 1.08 is 91.02% if the population variances for processing times are the same for the accounting department and the research department. (c) Given t = -0.3610, df = 9 for a two-tailed hypothesis test, the pvalue = 0.7264 using excel. (e)

( X - X ) ±t A

n1+n2-2

æ 1 1ö Sp2 ç + ÷ è n1 n2 ø

æ 1 1ö -0.7 ±2.2622 10.2556ç + ÷ = -5.0867 £ m A - mR £ 3.6867 è 5 6ø You are 95% confident that the mean difference in the mean processing times between the accounting and research departments is between – 5.0867 and 3.6867 seconds. 10.60 (a)

H0 : Z −  A  0 New Zealand watches no more television on average than Australia H1 : Z −  A  0 New Zealand has the higher average than Australia Assuming that the samples are from underlying normal populations with equal variances, we can use pooled-variance t test. The t test statistics follow a t distribution with 100 degrees of freedom. Use a level of significance of 0.05. The critical value is 2.692. Reject H0 if tcalc > 2.692. X - X A - mZ - m A Test statistics: t = Z æ 1 1ö Sp2 ç + ÷ è nZ nA ø

(

) (

)

(nZ − 1) S Z2 + (nA − 1) S A2 40(342 ) + 60(232 ) = = 779.8 100 (nZ − 1) + (nA − 1) (245 − 220) − ( Z −  A ) t= = 4.43

Where: S p2 =

1   1 779.8  +   41 61  Since 4.43 > 2.692, we reject the null hypothesis and we conclude that at 5% significance level, there is enough evidence that New Zealand has a higher mean of watching television compared to Australia.

(b)

H0 : Z −  A  0 New Zealand watches no more television on average than Australia H1 : Z −  A  0 New Zealand has the higher average than Australia Test statistics: df = 64 X Z - X A - mZ - m A 25- 0 t= = = 4.117 6.072 S Z2 S A2 + nZ nA

(

) (

)

Decision: reject H0 if tcalc > 1.669. Since 4.117 > 1.669, we reject the null hypothesis. We conclude that at 5% significance level, there is enough evidence that New Zealand has a higher mean of watching television compared to Australia. (c)

H0 :  Z2 −  A2 = 0 : The population variances for mean time to watch TV are the

same for New Zealand and Australia H1 :  Z2 −  A2  0 The population variances for mean time to watch TV are different for New Zealand and Australia Decision rule: if Fcalc < 0.556 or > 1.74, reject the null hypothesis. S 2 342 Test statistics: F = Z2 = 2 = 2.185 . S A 23 Decision: Since F = 2.185>1.74 we reject the null hypothesis. There is enough evidence to conclude that the population variances for mean time to watch TV are different for New Zealand and Australia.

(d)

Test in part (b) is appropriate since the variances of Australia and New Zealand are different.

(e)

As illustrated in part (d) in which there is enough evidence that the population variances are different for the mean time of watching television for Australia and New Zealand, the t test with unequal variances is appropriate in this case. The p-value is virtually zero. The probability of observing a sample tcalc that is greater than 4.117 is 0% which is an unlikely event. The test in (a) is not appropriate since based on the result of hypothesis testing in (d), we know that the variances are unequal. The p-value in this case is also virtually zero. In this case, two tests give us the same result.

10.61 (a)

H 0 : s 12 - s 22 = 0 H1 : s 12 - s 22 ¹ 0

Population 1 = Pinterest, 2 = Facebook F Test for Differences inTwoVariances Data Level of Significance 0.05 Larger-Variance Sample Sample Size 500 Sample Variance 22500 Smaller-Variance Sample Sample Size 500 Sample Variance 6400 Intermediate Calculations F Test Statistic 3.5156 Population1Sample Degrees of Freedom 499 Population2Sample Degrees of Freedom 499 Two-Tail Test Upper Critical Value 1.1921 p-Value 0.0000 Reject the null hypothesis

Since the p-value < 0.05, reject H0 . There is enough evidence of a difference in the variances of the order values between Pinterest shoppers and Facebook shoppers. Hence, a separate-variance t test is appropriate H 0 : 1 − 2 = 0 (b)

H1 : 1 − 2  0

Separate-Variancest TestfortheDifferenceBetweenT (assumes unequal populationvariances) Data HypothesizedDifference 0 Level of Significance 0.05 Population1Sample Sample Size 500 Sample Mean 153 Sample StandardDeviation 150.0000 Population2Sample Sample Size 500 Sample Mean 85 Sample StandardDeviation 80.0000 Intermediate Calculations Numerator of Degrees of Freedom 3340.8400 Denominator of Degrees of Freedom 4.3865 Total Degrees of Freedom 761.6268 Degrees of Freedom 761 Standard Error 7.6026 Difference in Sample Means 68.0000 Separate-Variance t Test Statistic 8.9443 Two-Tail Test Lower Critical Value -1.9631 Upper Critical Value 1.9631 p-Value 0.0000 Reject the null hypothesis

Since the p-value is virtually zero, reject H0 . There is enough evidence of a difference in the mean order value between Pinterest shoppers and Facebook shoppers (c) 53.0754  1 − 2  82.92462003

10.62

(a)

H 0 : 1 −  2 = 0

H1 : 1 −  2  0 where Populations: 1 = Males, 2 = Females PHStat2 output: ZTest for Differences inTwoProportions

Data HypothesizedDifference Level of Significance Group1 Number of Items of Interest Sample Size Group2 Number of Items of Interest Sample Size

0 0.05 50 300 96 330

Intermediate Calculations Group 1Proportion 0.166666667 Group 2Proportion 0.290909091 Difference in Two Proportions -0.12424242 Average Proportion 0.2317 Z Test Statistic -3.6911 Two-Tail Test Lower Critical Value -1.9600 Upper Critical Value 1.9600 p-Value 0.0002 Reject the null hypothesis

Since the p-value is smaller than 0.05, reject H0. There is enough evidence of a difference between males and females in the proportion who order dessert

(b) (c)

Decision rule: Reject H0 if Zcalc > 1.645 Decision: Since Zcalc = 1.07 < 1.645, we do not reject H0. There is insufficient evidence to suggest a greater proportion of younger voters vote for the Greens compared to older voters. From the PHStat2 output, the p-value is 0.1422.

F Test Two-Sample for Variances

Mean Variance Observations df F P(F<=f) one-tail F Critical one-tail

Male 88683.23684 3834187011 114 113 1.709298042

Female 74575.17544 2243135437 114 113

0.002354401 1.553351444

H0 :  M2 −  F2 = 0 : The population variances are the same H1 :  M2 −  F2  0 : The population variances are different Decision rule: if Fcalc < 0.614 or > 1.629, reject null hypothesis. Test statistics: F =

S M2 = 1.709 S F2

Decision: Since F = 1.709 > 1.629, reject null hypothesis. There is enough evidence to conclude that the two population variances are different.

10.63

The two normal probability plots do not suggest any departure from the normality assumption. You can perform F test on the different variances. H0 :  C2 −  S2 = 0 , H1 :  C2 −  S2  0 F Test Two-Sample for Variances

Mean Variance Observations df F P(F<=f) onetail

Variable 1 1313.566667 8088.529885 30 29 0.541408325

Variable 2 1435 14939.79592 50 49

0.039847629

F Critical onetail

0.562623603

Since the p-value = 0.07965 > 0.05, we do not reject the null hypothesis. There is not sufficient evidence to conclude that two variances are different. We can perform pooled variance t test for differences in means.

H 0 : C −  S = 0

H1 : C −  S  0 t Test: Two-Sample Assuming Equal Variances

Mean Variance Observations Pooled Variance Hypothesized Difference df t Stat P(T<=t) one-tail t Critical one-tail P(T<=t) two-tail t Critical two-tail

Variable 1 1313.566667 8088.529885 30 12392.53034

Variable 2 1435 14939.79592 50

Mean 0 78 -4.72344168 5.03164E-06 1.664624645 1.00633E-05 1.990847069

Since the p-values are essentially zero, reject the null hypothesis. There is sufficient evidence to conclude that means of transport costs are different for NSW and Sydney trainees.

Both normal probability plots suggest that both distributions are not normal. It is inappropriate to perform an F test on the difference in variances. Values on sample variances, SC2 = 617974.01 and S S2 = 1306292.396 , suggest that a separate variance t test is more appropriate.

H 0 : C −  S = 0

H1 : C −  S  0 t Test: Two-Sample Assuming Unequal Variances

Mean Variance Observations Hypothesized Difference Df t Stat P(T<=t) one-tail t Critical one-tail P(T<=t) two-tail t Critical two-tail

Variable 1 5986.7 617974.0103 30

Variable 2 7917.18 1306292 50

Mean 0 76 –8.93080768 9.05559E-14 1.665151353 1.81112E-13 1.99167261

Since the p-value is essentially zero, reject the null hypothesis. There is sufficient evidence to conclude that the means of the rent cost are different for NSW country and Sydney.

Both normal probability plots suggest that both distributions are not normal. It is inappropriate to perform an F test on the difference in variances. Values on sample variances, SC2 = 9410128.179 and S S2 = 15180521.27 , suggest that a separate variance t test is more appropriate.

H 0 : C −  S = 0

H1 : C −  S  0 t Test: Two-Sample Assuming Unequal Variances

Mean Variance Observations Hypothesized

Variable 1 27478.4 9410128 30 0

Variable 2 29124.44 15180521 50

Mean Difference df t Stat P(T<=t) one-tail t Critical one-tail P(T<=t) two-tail t Critical two-tail

72 -2.09507 0.019841 1.666294 0.039683 1.993464

Since the p-values for a two-tail test is 0.0397 < 0.05, reject the null hypothesis. There is sufficient evidence to conclude that the means of the annual wages are different for NSW country and Sydney trainees. 10.64 A

Mean Standard Error Median Mode Standard Deviation Sample Variance Kurtosis Skewness Range Minimum Maximum Sum Count

7377.325 135.3817224 7316.5 8416 856.2291925 733128.4301 –1.010573077 -0.17096025 3187 5544 8731 295093 40

Mean Standard Error Median Mode Standard Deviation Sample Variance Kurtosis Skewness Range Minimum Maximum Sum Count

8260.9 143.8566051 8140.5 #N/A 909.8290569 827788.9128 –1.311672475 0.047944167 3043 6701 9744 330436 40

From the descriptive statistics above, we know that both data seem to have come from rather symmetrical distributions that are quite normally distributed since the value of skewness is close to zero. F Test Two-Sample for Variances

Mean Variance Observations df F P(F<=f) one-tail F Critical one-tail

7377.325 733128.4301 40 39 0.885646593

8260.9 827788.9 40 39

0.35321677 0.586694336

The following F test p-value = 0.7064. Do not reject the null hypothesis. There is insufficient evidence that the two population variances are significantly different at 5% significance level.

Since both data are drawn from independent populations, the most appropriate test for any difference in the life of the bulbs between two manufacturers is the pooled variance t test. t Test: Two-Sample Assuming Equal Variances

Mean Variance Observations Pooled Variance Hypothesized Difference df t Stat P(T<=t) one-tail t Critical one-tail P(T<=t) two-tail t Critical two-tail

A 7377.325 733128.4301 40 780458.6715

B 8260.9 827788.9 40

Mean 0 78 –4.472841019 1.29478E-05 1.664624645 2.58957E-05 1.990847069

Since the p-value is virtually zero, at the 5% significance level, there is sufficient evidence to reject the null hypothesis of no difference in the mean life of the bulbs between two manufacturers. Based on the above analysis we can conclude that there is significant difference in the life of the bulbs between two manufacturers. 10.65 Female Current points

Age Mean Standard Error Median Mode Standard Deviation Sample Variance Kurtosis Skewness Range Minimum Maximum Sum Count

24.48387 0.483871 24 23 2.69408 7.258065 0.050729 0.013505 12 18 30 759 31

Mean Standard Error Median Mode Standard Deviation Sample Variance Kurtosis Skewness Range Minimum Maximum Sum Count

Top speed 184.9032 2.599981 183 184 14.47608 209.557 2.586411 1.489346 64 163 227 5732 31

Mean Standard Error Median Mode Standard Deviation Sample Variance Kurtosis Skewness Range Minimum Maximum Sum Count

Season points 71.70968 0.464187 72 72 2.584486 6.67957 -0.51209 0.157465 10 67 77 2223 31

Mean Standard Error Median Mode Standard Deviation Sample Variance Kurtosis Skewness Range Minimum Maximum Sum Count

3460.323 105.6994 3505 3590 588.5093 346343.2 -1.2441 -0.17929 1830 2485 4315 107270 31

Male Current points

Age Mean Standard Error Median Mode Standard Deviation Sample Variance Kurtosis Skewness Range Minimum Maximum Sum Count

22.15556 0.456309 21 21 4.328923 18.73958 8.172932 2.47886 24 17 41 1994 90

Mean Standard Error Median Mode Standard Deviation Sample Variance Kurtosis Skewness Range Minimum Maximum Sum Count

Top speed 187.9778 1.33931 189 201 12.70581 161.4377 –0.29096 –0.2567 60 155 215 16918 90

Mean Standard Error Median Mode Standard Deviation Sample Variance Kurtosis Skewness Range Minimum Maximum Sum Count

Season points 71 0.331079 71 68 3.140887 9.865169 –0.0965 0.556294 14 65 79 6390 90

Mean Standard Error Median Mode Standard Deviation Sample Variance Kurtosis Skewness Range Minimum Maximum Sum Count

3391.722 50.79488 3427.5 3990 481.8825 232210.8 -0.50866 -0.29968 2165 2150 4315 305255 90

From the descriptive statistics above, we can see that the value of skewness and kurtosis are quite different from normal distribution. The F test for the difference in variances, which is sensitive to departure from normal probability distribution assumption, will not be appropriate. The variances of all the variables between female and male cyclists are also quite different. Hence, you perform separate variance t tests on the difference on means.

H0 : F −  M = 0 H1 :  F −  M  0 t Test: Variances

Two-Sample

Mean Variance Observations Hypothesized Difference Df t Stat P(T<=t) one-tail t Critical one-tail P(T<=t) two-tail t Critical two-tail

Assuming

Unequal

Variable 1 24.48387 7.258065 31

Variable 2 22.15556 18.73958 90

Mean 0 85 3.500737 0.000371 1.662978 0.000742 1.988268

Since the p-value is 0.0007 < 0.05, we reject the null hypothesis. There is sufficient evidence to conclude that the mean ages are different between males and females.

Current points:

H0 : F −  M = 0 H1 :  F −  M  0

t Test: Variances

Two-Sample

Mean Variance Observations Hypothesized Difference df t Stat P(T<=t) one-tail t Critical one-tail P(T<=t) two-tail t Critical two-tail

Assuming

Unequal

Variable 1 184.9032 209.557 31

Variable 2 187.9778 161.4377 90

Mean 0 47 –1.05125 0.14926 1.677927 0.298519 2.011741

Since, the p-values = 0.298 > 0.05, we do not reject the null hypothesis. There is not enough evidence to conclude that the mean current points are different between males and females.

H0 : F −  M = 0

H1 :  F −  M  0 t Test: Variances

Two-Sample

Mean Variance Observations Hypothesized Difference df t Stat P(T<=t) one-tail t Critical one-tail P(T<=t) two-tail t Critical two-tail

Assuming

Unequal

Variable 1 71.70968 6.67957 31

Variable 2 71 9.865169 90

Mean 0 63 1.244698 0.108927 1.669402 0.217853 1.998341

Since p-value = 0.2179 > 0.05, we do not reject the null hypothesis. There is not enough evidence to conclude that the mean top speed is different between males and females. Season points H0 : F − M = 0

t Test: Variances

Two-Sample

Mean Variance Observations Hypothesized Difference Df t Stat P(T<=t) one-tail t Critical one-tail P(T<=t) two-tail t Critical two-tail

Assuming

Unequal

Variable 1 3460.323 346343.2 31

Variable 2 3391.722 232210.8 90

Mean 0 45 0.584973 0.280744 1.679427 0.561488 2.014103

Since the p-values = 0.5615 > 0.05, we do not reject the null hypothesis. There is not sufficient evidence that the mean season points are different between males and females. 10.66

H0 : D = 0 H1 : D  0 Choosing the level of significance = 0.1 and assuming the differences are normally distributed, use paired t test. Degrees of freedom = 7 Reject null hypothesis if tcalc >1.895 or <–1.895.

D - mD SD

-35.125 = -0.781 44.95

n Since –0.781 > –1.895, we do not reject the null hypothesis and there is not enough evidence that the husband and wife of a couple have different spending patterns at 10% significance level.

10.67

H 0 : s 12 - s 22 = 0 H1 : s 12 - s 22 ¹ 0 F TestforDifferencesinTwoVariances Data Level of Significance 0.01 Larger-Variance Sample Sample Size 100 Sample Variance 15625 Smaller-Variance Sample Sample Size 100 Sample Variance 10000 Intermediate Calculations F Test Statistic 1.5625

Population1SampleDegreesofFreedo Population2SampleDegreesofFreedo

99 99

Two-Tail Test Upper Critical Value 1.6854 p-Value 0.0274 Donot reject the null hypothesis Since p > 0.01 there is not enough evidence of a difference in the variances of the amount of

time spent talking between women and men (b)It is more appropriate to use a pooled-variance t test H 0 : 1 − 2 = 0

H1 : 1 − 2  0

Pooled-Variancet Testforthe DifferenceBetwee (assumes equal populationvariances) Data HypothesizedDifference 0 Level of Significance 0.01 Population1Sample Sample Size 100 Sample Mean 818 Sample StandardDeviation 125 Population2Sample Sample Size 100 Sample Mean 716 Sample StandardDeviation 100 Intermediate Calculations Population1SampleDegreesof Freedo 99 Population2SampleDegreesof Freedo 99 Total Degrees of Freedom 198 Pooled Variance 12812.5 Standard Error 16.0078 Difference in Sample Means 102 t Test Statistic 6.3719 Two-Tail Test Lower Critical Value -2.6009 Upper Critical Value 2.6009 p-Value 0.0000 Reject the null hypothesis Copyright © 2019 Pearson Australia (a division of Pearson Australia Group Pty Ltd) 9781488617249, Berenson, Basic Business Statistics 5e

Since p value is virtually zero there is enough evidence of a difference in the mean amount of

(c)

time spent talking between women and men. H 0 : s - s 22 = 0 2 1

H1 : s 12 - s 22 ¹ 0 F TestforDifferencesinTwoVariances Data Level of Significance 0.01 Larger-Variance Sample Sample Size 100 Sample Variance 22500 Smaller-Variance Sample Sample Size 100 Sample Variance 15625 Intermediate Calculations F Test Statistic 1.4400 Population1SampleDegreesof Freedo 99 Population2SampleDegreesof Freedo 99 Two-Tail Test Upper Critical Value 1.6854 p-Value 0.0711 Donot reject the null hypothesis Since p value > 0.01 there is not enough evidence of a difference in the variances of

the number of text messages sent per month by women and men. (d) It is more appropriate to use a pooled-variance t test.

H 0 : 1 − 2 = 0 H1 : 1 − 2  0

Pooled-Variancet TestfortheDifferenceBetwee

(assumes equal populationvariances) Data HypothesizedDifference Level of Significance Population1Sample Sample Size Sample Mean Sample StandardDeviation Population2Sample Sample Size Sample Mean Sample StandardDeviation

0 0.01 100 716 150 100 555 125

Intermediate Calculations Population1SampleDegreesof Freedo 99 Population2SampleDegreesof Freedo 99 Total Degrees of Freedom 198 Pooled Variance 19062.5 Standard Error 19.5256 Difference in Sample Means 161 t Test Statistic 8.2456 Two-Tail Test Lower Critical Value -2.6009 Upper Critical Value 2.6009 p-Value 0.0000 Reject the null hypothesis Since p value is virtually zero there is enough evidence of a difference in the mean number of

text messages sent per month by women and men

10.68 H0 :  L2 −  S2  0 , H1 :  L2 −  S2  0

Decision rule: If Fcalc > 4.54, reject null hypothesis. S2 Test statistics: F = l2 = 1.345 SS Decision: Since F = 1.345 < 4.54, we do not reject the null hypothesis. There is not enough evidence to conclude that two population variances are different at 1% significance level. 10.69 (a)

As the data prices for the same items at two different stores, a paired t test is appropriate. H0 : C − W = 0 The mean price of stationery at Coles and Woolworths are the same in the week H1 : C − W  0 The mean price of stationery at Coles and Woolworths are different in the week t Test: Paired Two-Sample for Means

Mean Variance Observations Pearson Correlation Hypothesized Mean Difference Df t Stat P(T<=t) one-tail t Critical one-tail P(T<=t) two-tail t Critical two-tail

Coles 0.719333 0.37575 15 0.960864

Woolworths 0.702667 0.371807 15

0 14 0.377318 0.355798 2.624494 0.711595 2.976843

Since t = 0.377 < 2.9768, we do not reject the null hypothesis and conclude that there is insufficient evidence that the mean price of stationery was different at Coles and Woolworths. (b) The p-value for two-tail test is 0.7116 > 0.01, so we do not reject the null hypothesis. The p-value represents the probability of obtaining samples that will yield a test statistic more extreme than 0.3773 if the means are equal. 10.70

H0 :  p −  r  0 H1 :  p −  r  0

pp = 0.74, pr = 0.4635, p = 0.5976 Using the 0.01 level of significance Decision, reject null hypothesis if Zcalc > 2.326 Z=

( p − p ) − ( −  ) = 11.521 p

1 1 p (1 − p )  +  n n  r   p

Since 11.521 > 2.326, we reject the null hypothesis and conclude that there is enough evidence that the proportion of city dwellers who have access ADSL is higher than the proportion of regional dwellers. 10.71

H 0 : 1 − 2 = 0 H1 : 1 − 2  0

df = 20 + 20 -2 = 38 Decision rule: Reject H0 if tcalc < –2.0244 or > 2.0244. Since tcalc = 5.1615 > 2.0244, reject H0.

There is enough evidence of a difference in the mean delivery time in the two wings of the hotel.

10.72 To construct the 95% confidence interval estimate in question 10.70

0.2487 + = éë0.2318,0.32125ùû (0.74 - 0.4635) ±1.96 0.1824 800 850 You have 95% confidence that the difference between the population proportion of city dwellers and regional dwellers who have access to ADSL is between 0.2315 and 0.3213. 10.73

H0 :  p −  F  0 H1 :  p −  F  0

p1 = 0.76, p2 = 0.78, p = 0.773 Using the 0.1 level of significance Decision, reject null hypothesis if Zcalc > 2.326 Z=

( p − p ) − ( −  ) = −0.323 p

1 1  p (1 − p )  +  n n  f   p Since –0.323 < 1.282, we reject the null hypothesis at 10% level and conclude that there is not enough evidence that the proportion of young males is higher than the proportion of males speeding on a regular basis.

Chapter 11: Analysis of variance Learning objectives After studying this chapter you should be able to: 1. explain basic concepts of experimental design 2. apply the one-way analysis of variance to test for differences between the means of several groups 3. construct and apply a randomised block design 4. conduct a two-way analysis of variance and interpret the interaction 11.1

(a) (b) (c)

df B = c – 1 = 4 – 1 = 3 df W = n – c = 20 – 4 = 16 df T = n – 1 = 20 – 1 = 19

11.2

(a)

SSW = SST – SSB = 120 – 60 = 60

(b)

MSB =

SSB 60 = = 20 new c −1 4 −1 SSW 60 MSW = = = 3.75 n − c 20 − 4 MSB 20 Fcalc = = = 5.33 new MSW 3.75

(a) Source Among groups Within groups Total

11.4

SS 3 16 19

60 60 120

F 20 3.75

5.33

(b) (c) (d)

F3, 16 = 3.24 Decision rule: If Fcalc > 3.24, reject H0. Decision: Since Fcalc = 5.33 is greater than the critical bound 3.24, reject H0.

(a) (b) (c)

df B = c – 1 = 6 – 1 = 5 df W = n – c = 42 – 6 =36 df T = n – 1 = 42– 1 = 41

11.5 Source Among groups Within groups Total 11.6

(a) (b)

Df 6–1=5 30 – 6 = 24 30 – 1 = 29

SS (80) 400 560

(5)

MS 80 560/24 23.33

F 80/23.33 3.43

400 + 560 = 960

Decision rule: If Fcalc > 2.62, reject H0. Since Fcalc = 3.43 is greater than the critical bound of F5,24 > 2.62, reject H0.

(c)

There are 5 degrees of freedom in the numerator and n – c = 30 – 6 = 24 degrees of freedom in the denominator. The critical value, Q = 4.17.

(d)

perform

Q

MSW  1 1  23.33  1 1   +  = 4.17  +  = 9.01 2  n j n j  2 5 5

the

Tukey–Kramer

procedure,

the

critical

range

11.7 Source of variation

Degrees of freedom

Sum of squares

Mean square

F calc

Between groups

c–1=4

400

100

8.036

Within groups

n – c = 45

560

12.444

Total

n – 1 = 49

960

112.444

11.8

(a)

Source of variation

All of these tests should start with stating the hypotheses (see 11.9 as an example) Mean Degrees of freedom Sum of squares square F calc

Between groups

c–1=3

3,799,436

1,266.479

Within groups

n – c = 16

2,917.665

182.354

Total

n – 1 = 19

6,717,101

1448.833

(b) (c)

11.9

(a)

0.144

𝐹3,16 = 3.24. Since 𝐹𝑐𝑎𝑙𝑐 = 0.144 < 3.24 do not reject the null. Conclude there is no significant difference in the average grades. Since there is not enough evidence of a difference in means it is inappropriate to perform the Tukey–Kramer procedure.

H 0 : 1 = 2 = 3 H1: Not all the means are equal

ANOVA: Factor

Single

SUMMARY Groups urban regional rural

Count 6 6 6

Sum 25 39 58

Average 4.166666667 6.5 9.666666667

Variance 2.9667 5.1000 2.2667

ANOVA Source of Variation Between Groups Within Groups

SS 91.4444 51.6667

df 2 15

MS 45.7222 3.4444

F 13.2742

Total

143.1111

17 Level of significance

Pvalue 0.0005

F crit 3.6823

0.05

Decision rule: if Fcalc > 3.68 reject H0 Since Fcalc = 13.27 > 3.68 reject the null. Conclude there is enough evidence of a difference in the mean unemployment rates Copyright © 2019 Pearson Australia (a division of Pearson Australia Group Pty Ltd) 9781488617249, Berenson, Basic Business Statistics 5e

(b)

The Tukey–Kramer procedure is used to establish which of the means are significantly different from one another. 𝑄𝑈(𝑐,𝑛−𝑐) = Q(3,15) = 3.67

Critical range=𝑄𝑈(𝑐,𝑛−𝑐) √

Pairwise comparisons urban-regional urban-rural regional-rural

𝑀𝑆𝑊 2

(

1 𝑛𝑗

1 𝑛𝑗′

) = 3.67

3.4444  1 1   +  = 2.8707 2 6 6

2.33 5.5 3.17

The average return is different between urban and rural, and regional and rural.

H0 : 12 =  22 =  32

(c)

H1: not all variances are equal Using absolute differences urban regional Rural 1

ANOVA: Factor

Single

SUMMARY Groups

Count

Sum

Average

Variance

urban

1.5

0.3000

regional

1.833333333

1.3667

rural

1.2000

Pvalue

F crit

Between Groups

2.1111

1.0556

1.1047

0.3568

3.6823

Within Groups

14.3333

0.9556

Total

16.4444

ANOVA Source Variation

Decision rule: If Fcalc > 3.68 reject H0 Since Fcalc = 1.1047 < 3.68 do not reject the null. Conclude there is no significant difference between three variances.

(a) H 0 : 1 = 2 = 3 = 4 where 1 = Main, 2 = Satellite1, 3 = Satellite2, 4 = Satellite3

11.10

H1 : Not all j are equal Excel output: Source of Variation Between Groups Within Groups

Total

P-value

F crit

6.371691

0.000859

2.769431

6312.444

2104.148

18493.09

330.2338

24805.54

Decision Rule: If p-value < 0.05, reject H0. Since p-value = 0.0009 < 0.05, reject the null hypothesis. There is enough evidence to conclude that there is a significant difference in the mean waiting time in the four locations (b) PHStat output of the Tukey-Kramer procedure: Sample Mean

Sample Size

69.8427

41.254

56.5613

51.972

Group

Other Data

Level of significance Numerator d.f.

0.05

Denominator d.f. MSW

Comparison

Absolute Difference

Std. Error of Difference

Critical Range

Group 1 to Group 2 Group 1 to Group 3

28.588667

4.692077

17.548

13.281333

4.692077

17.548

Group 1 to Group 4 Group 2 to Group 3

17.870667

4.692077

17.548

15.307333

4.692077

17.548

Group 2 to Group 4

10.718

4.692077

17.548

Group 3 to Group 4

4.5893333

4.692077

17.548

Results Means are different Means are not different Means are different Means are not different Means are not different Means are not different

330.2338

Q Statistic

3.74

From the Tukey-Kramer procedure, there is a difference in mean waiting time between the main campus and Satellite1, and the main campus and Satellite3 (c) H0: 

=  2 =  2 =  2

H1: At least one variance is different Source of Variation Between Groups Within Groups

P-value

F crit

310.979 7078.435

3 56

103.6597 126.4006

0.8201

0.4883

2.7694

Total

7389.414

Since the p-value = 0.4883 > 0.05, do not reject H0. There is not enough evidence to conclude there is a significant difference in the variation in waiting time among the four locations.

Since 𝐹𝑐𝑎𝑙𝑐 = 22.375 > 2.45 reject the null. Conclude there is enough evidence of a difference in the variance of price of petrol between the days of the week. 11.11 (a)

H 0 : m1 = m2 = m3 H1: Not all the means are equal

ANOVA: Factor

Single

SUMMARY Groups young mature old

Count 10 10 10

Sum 120 75 163

Average 12 7.5 16.3

Variance 42.8889 33.8333 69.3444

ANOVA Source of Variation Between Groups Within Groups

SS 387.2667 1314.6000

df 2 27

MS 193.6333 48.6889

F 3.9770

Total

1701.8667

Pvalue 0.0306

F crit 3.3541

Decision rule: If Fcalc > 3.35 reject H0 Since Fcalc = 3.977 > 3.65 reject H0. There is evidence that the mean number of sick days differs across the three age groups. . (b)

The Tukey–Kramer procedure is used to establish which means are significantly different from one another QU(c,n-c) = QU(3,27) = 3.49

Critical range = QU ( c , n −c )

MSW  1 1  48.68889  1 1   +  = 3.49  +  = 7.7009 2  nj nj'  2  10 10 

Pairwise differences Young–mature

4.5

Young–old

4.3

Mature–old

8.8 The means of mature and old are different.

(c)

H0 : 12 =  22 =  32 H1: not all variances are equal

ANOVA: Factor

Single

SUMMARY Groups young mature old

Count 10 10 10

Sum 52 51 65

Average 5.2 5.1 6.5

Variance 13.9556 4.9333 22.4444

ANOVA Source of Variation Between Groups Within Groups

SS 12.2000 372.0000

df 2 27

MS 6.1000 13.7778

F 0.4427

Total

384.2000

Pvalue 0.6469

F crit 3.3541

Decision rule: If Fcalc > 3.35 reject H0 Since Fcalc = 4427 < 3.35 do not reject H0. Conclude there is no significant difference in the variances across the 3 age groups. (d)

The results in (a) and (b) are valid as the assumption of equal variances is reasonable based on results in (c).

11.12 (a)

(b)

skill

H0: 1 = 2 = 3 H1: At least one mean is different.

Since FStat = 8.7558 > 3.00, reject H0. There is evidence of a difference in the mean softscore of the different groups

Group Nocourseworkinlea Certificate inleaders Degree inleadership

Sample Sample Mean Size 3.29 109 3.362 90 3.471 102

Absolute Std. Error Critical Comparison Difference of Difference Range Results Group1toGroup2 0.072 0.032989588 0.1092 Means are notdifferent Group1toGroup3 0.181 0.031908968 0.1056 Means are different Group2toGroup3 0.109 0.033497634 0.1109 Means are notdifferent

Other Data Level of significance 0.05 Numerator d.f. 3 Denominator d.f. 297 MSW 0.1073 QStatistic 3.31

There is evidence of a difference in the mean soft-skill score between those who had no coursework in leadership and those who had a degree in leadership.

Since 𝐹𝑐𝑎𝑙𝑐 = 1.0927 < 2.76 do not reject the null. Conclude there is no significant difference in the variance of rating of the five advertisements.

(d) A

15 18 17 19 19 20 18.000

16 17 21 16 19 17 17.667

8 7 10 15 14 14 11.333

5 6 13 11 9 10 9.000

12 19 18 12 17 14 15.333

The average rating of the advertisements C and D are not different and both have the lowest ratings. The advertisements C and D should not be used. Then, the advertisement E is not different compared with A, B and C. So, it should not be used. 11.13 (a) SUMMARY Groups Front Middle Rear

Count 6 6 6

Sum 36.400 12.400 22.400

Average 6.067 2.067 3.733

Variance 2.715 0.427 2.011

ANOVA Source of Variation Between Groups Within Groups

SS 48.444 25.760

Df 2 15

MS 24.222 1.717

F 14.105

Total

74.204

Pvalue 0.000

F crit 3.682

Since 𝐹𝑐𝑎𝑙𝑐 = 14.11 > 3.68 reject the null. Conclude there is enough evidence of a difference in the mean sales volumes across three store aisle locations. (b)

The Tukey–Kramer procedure is used to establish which of the means are significantly different from one another. 𝑄𝑈(𝑐,𝑛−𝑐) = 𝑄𝑈(3,15) = 3.67 Critical range = 𝑄𝑈(𝑐,𝑛−𝑐) √

Front-Middle Front-Rear Middle-Rear

𝑀𝑆𝑊 2

(

1 𝑛𝑗

1 𝑛𝑗′

) = 3.67√

1.717 1 2

( + ) = 1.963 6

4.000 2.333 -1.667 Because 4.000 > 1.963 and 2.333 > 1.963, conclude that there is a significant difference between the mean sales volumes between front location and middle location as well as between front location and rear location.

Count 6 6 6

Sum 7.6 2.8 6.8

Average 1.267 0.467 1.133

Variance 0.875 0.199 0.603

ANOVA Source of Variation Between Groups Within Groups

SS 2.204 8.380

Df 2 15

MS 1.102 0.559

F 1.973

Total

10.584

Pvalue 0.174

F crit 3.682

Since 𝐹𝑐𝑎𝑙𝑐 = 1.973 < 3.682 do not reject the null. Conclude there is no significant difference in the variance of sales volumes across different locations. (d)

The front aisle appears to be the best location for the sale of fluffy toys. The manager should consider switching the location of fluffy toys to the front aisle if it yields marginal profit that is greater than that of the product currently displayed at the front aisle.

11.14 (a) SUMMARY Groups Design1 Design2 Design3 Design4

Count 10 10 10 10

Sum 2365.810 2366.210 2437.200 2434.180

Average 236.581 236.621 243.720 243.418

Variance 110.407 139.665 43.163 60.300

ANOVA Source of Variation Between Groups Within Groups

SS 485.994 3181.820

df 3 36

MS 161.998 88.384

F 1.833

Total

3667.814

Pvalue 0.159

F crit 2.866

Since 𝐹𝑐𝑎𝑙𝑐 = 1.83 < 2.87 do not reject the null. Conclude there is no significant difference in the mean distance travelled by the golf balls with different designs. (b)

Since there is not enough evidence of a difference in means it is inappropriate to perform the Tukey–Kramer procedure.

(c)

Three assumptions needed in (a) are (i) samples are drawn randomly and independently, (ii) populations have normal distributions, and (iii) populations have equal variances.

(d) SUMMARY Groups abs_dif_des1 abs_dif_des2 abs_dif_des3 abs_dif_des4

ANOVA Source of Variation Between Groups Within Groups Total

Count 10 10 10 10

Sum 96.710 103.030 52.200 64.660

Average 9.671 10.303 5.220 6.466

Variance 7.679 24.026 13.563 13.859

Pvalue

F crit

181.487 532.142

3 36

60.496 14.782

4.093

0.013

2.866

713.629

Since 𝐹𝑐𝑎𝑙𝑐 = 4.09 > 2.87 reject the null. Conclude there is enough evidence of a difference in the mean distance travelled by the golf balls with different designs. 11.15 (a) (b) (c) (d)

df B = c – 1 = 4 – 1 = 3 df BL = r – 1 = 6 – 1 = 5 df E = (r – 1)(c – 1) = (6 – 1)( 4 – 1) = 15 df T = rc – 1 = 6 x 4 – 1 = 23

11.16 (a)

SSE = SST – SSB – SSBL = 210 – 60 – 75 = 75

(b)

SSA 60 = = 20 c −1 3 SSBL 75 MSBL = = = 15 r −1 5 SSE 75 MSE = = =5 (r − 1)  (c − 1) 5  3 MSB =

(c)

𝐹𝑐𝑎𝑙𝑐 =

(d)

𝐹𝑐𝑎𝑙𝑐 =

𝑀𝑆𝐴

𝑀𝑆𝐸 𝑀𝑆𝐵𝐿 𝑀𝑆𝐸

5 15 5

11.17 (a) Source Variation treatment block Error Total (b)

of SS 60 75 75

Df 3 5 15

210

MS 20 15 5

F 4 3

Using the 0.05 level of significance to test for the difference between four population means, reject the null of equal means if calculated F is greater than 3.29, the upper-tail critical value from F distribution with 3 and 15 degrees of freedom. Since 𝐹𝑐𝑎𝑙𝑐 = 4 > 𝐹𝑈 = 3.29, reject the null and conclude that there is evidence of a difference in the four means.

(c)

11.18

Using the 0.05 level of significance to test for the difference between the six blocks, reject the null of equal means if calculated F is greater than 2.90, the upper-tail critical value from F distribution with 5 and 15 degrees of freedom. Since 𝐹𝑐𝑎𝑙𝑐 = 3 > 𝐹𝑈 = 2.90, reject the null and conclude that there is evidence of a difference in the means between the blocks.

(a) There are 4 degrees of freedom in the numerator and 15 degrees of freedom in the denominator. (b)

Q = 4.08

(c)

Critical range = QU

MSE 5 = 4.08 = 3.72 r 6

11.19 (a) (b) (c) (d)

df B = c – 1 = 3 – 1 = 2 df BL = r – 1 = 7 – 1 = 6 df E = (r – 1)(c – 1) = (7 – 1)(3 – 1) = 12 df T = rc – 1 = 7 x 3 – 1 = 20

11.20 (a)

𝑀𝑆𝐵 =

𝑆𝑆𝐵 𝑐−1 𝑀𝑆𝐵

36 3 12

= 12

𝑀𝑆𝐸 = = =2 𝐹 6 𝑆𝑆𝐸 = 𝑀𝑆𝐸 × (𝑟 − 1)(𝑐 − 1) = 2 × 30 = 60 (b)

𝑆𝑆𝐵𝐿 = 𝑀𝑆𝐵𝐿 × (𝑟 − 1) and 𝑀𝑆𝐵𝐿 = 𝐹 × 𝑀𝑆𝐸 = 4 × 2 = 8 𝑆𝑆𝐵𝐿 = 8 × 10 = 80

(c)

SST = SSB + SSBL + SSE = 36 + 80 + 60 = 176

(d)

Since Fcalc = 6.0 > F0.01,3,30 = 4.51, reject the null hypothesis of no treatment effect. There is adequate evidence to conclude there is a treatment effect. Since Fcalc = 4.0 > F0.01,10,30 = 2.98, reject the null hypothesis of no block effect. There is adequate evidence to conclude there is a block effect.

11.21 Source Among groups Among blocks Error Total

4 – 1 =3

3 x 80 = 240

8 – 1 = 7 3 x 7 = 21 32 – 1 = 31

540 15.43 x 21 = 324.03 240 + 540 + 324 = 1104

MS 80 540  7 = 77.14 77.14  5 = 15.43

F 80  15.43 = 5.19 5.00

11.22 (a)

Reject the null of equal means if calculated F is greater than 2.92, the uppertail critical value from F distribution with 3 and 30 degrees of freedom. Decision: Since Fcalc = 5.19 is greater than the critical bound 2.92, reject H0. There is enough evidence to conclude that the treatment means are not all equal.

(b)

Reject the null of equal means if calculated F is greater than 2.17, the uppertail critical value from F distribution with 10 and 30 degrees of freedom. Decision: Since Fcalc = 5.00 is greater than the critical bound 2.17, reject H0. There is enough evidence to conclude that the block means are not all equal.

11.23 ANOVA Source Variation

of SS

Pvalue

F crit

Rows

153.222

19.153

19.065

0.000

2.355

Columns

79.639

26.546

26.424

0.000

3.009

Error

24.111

1.005

Total

256.972

H0: mA = mB = mC = mD H1: At least one mean differs. Decision rule: If Fcalc > 3.009, reject H0. Test statistic: Fcalc = 26.42 Decision: Since Fcalc = 26.42 is greater than the critical bound 3.009, reject H0. There is adequate evidence to conclude that there is a difference in the mean summed ratings of the four brands of Colombian coffee. From Table E.10, QU = 3.9 Critical range = Q =

MSE 1.005 = 3.9 = 1.303 r 9

Pairs of means that differ at the 0.05 level are marked with * below.

X A − X B = 1.56*

X A − X C = 0.89

X A − X D = 2.56*

X B − X C = 2.45*

X B − X D = 4.12*

X C − X D = 1. 67*

Brand B is rated highest with a sample mean rating of 25.56. 11.24 (a)

H0: .1 = .2 = .3 where 1 = TV, 2 = Phone, 3 = Internet

H1: Not all . j are equal where j = 1, 2, 3 Excel Output: ANOVA Sourceof Variation SS Rows 1028.256 Columns 540.1538 Error 70.51282 Total

1638.923

MS F P-value F crit 12 85.68803 29.16509 2.23E-11 2.18338 2 270.0769 91.92436 5.62E-12 3.402826 24 2.938034 38

Fcalc = 91.9244. Since the p-value is virtually 0 < 0.05, reject H0. There is evidence of a difference in the mean rating between TV, phone, and Internet. (b) Excel output for the Tukey procedure Tukey-Kramer Multiple Comparisons

Group 1: TV 2: Phone

Sample Sample Mean Size 64.84615 13 73.76923 13

Absolute Std. Error Critical Comparison Difference of Difference Range Results Group1toGroup2 8.9230769 0.475397339 1.6782 Means are different Group1toGroup3 2.8461538 0.475397339 1.6782 Means are different

3: Internet

67.69231

Group2toGroup3

6.0769231

0.475397339 1.6782 Means are different

Other Data Level of significance 0.05 Numerator d.f. 3 Denominator d.f. 24 MSW 2.938034 QStatistic 3.53

The mean rating for the three services are significantly different from each other with TV at the lowest, followed by Internet, and finally Phone

11.25 (a) ANOVA Source Variation

of SS

Pvalue

F crit

Rows

2013.655

287.665

107.415

0.000

2.764

Columns

11.060

5.530

2.065

0.164

3.739

Error

37.493

2.678

Total

2062.207

Reject H0 if Fcalc > 2.76. Since Fcalc = 107.4 is greater than the critical bound 2.76, reject H0. There is enough evidence to conclude that the treatment means are not all equal. (b)

One must assume: - samples obtained randomly and independently - drawn from normally distributed populations - population variances approximately equal - no interacting effects between the treatments and the blocks.

(c)

From Table E.10, QU = 3.58 Critical range = Q

Shop 1- Shop 2

1.445

Shop 1- Shop 3

1.435

Shop 2- Shop 3

0.010

MSE 2.678 = 3.58 = 2.071 r 8

None of the shops have different means. (d)

Test statistic: Fcalc = 2.065 Decision: Since Fcalc = 2.065 is less than the critical bound 3.739, do not reject H0. There is not enough evidence to conclude that the block means are not all equal.

11.26 (a) ANOVA Source Variation

of SS

Pvalue

F crit

Rows

21.930

5.483

0.227

0.919

3.007

Columns

20.377

5.094

0.211

0.928

3.007

Error

385.838

24.115

Total

428.145

Test statistic: Fcalc = 0.227 Decision: Since Fcalc = 0.227 is less than the critical bound 3.007, do not reject H0. There is not enough evidence to conclude that there is a difference in the mean prices for books at the five locations. (b)

(c)

As we have not rejected the null, the Tukey–Kramer procedure is not appropriate here.

(d)

Decision rule: If Fcalc > 3.007, reject H0. Decision: Since Fcalc = 0.211 is less than the critical bound 3.007, do not reject H0. There is not enough evidence to conclude that the block means are not all equal.

11.27

H0: .1 = .2 = .3 = .4 where 1 = Money Market, 2 = One Year CD, 3 = Two Year CD, 4 = Five Year CD H1: Not all . j are equal where j = 1, 2, 3, 4 Excel output: ANOVA SourceofVariation SS Rows 4.854975 Columns 7.931888 Error 1.192113 Total

13.97898

MS F P-value F crit 15 0.323665 12.21774 3.07E-11 1.894875 3 2.643963 99.8046 6.6E-20 2.811544 45 0.026491 63

Fcalc = 99.8046. Since the p-value is virtually 0, reject H0. There is evidence of a difference in the mean rates for these investments. (b)

The assumptions needed are: (i) samples are randomly and independently drawn, (ii) populations are normally distributed, (iii) populations have equal variances and (iv) no interaction effect between treatments and blocks (c) Excel output of the Tukey procedure: Tukey-Kramer Multiple Comparisons Sample Sample Group Mean Size 1: MoneyMarket 0.31875 16 2: One Year CD 0.515625 16

Absolute Std. Error Critical Comparison Difference of Difference Range Results Group1toGroup2 0.196875 0.040690439 0.1542 Means are different Group1toGroup3 0.5275 0.040690439 0.1542 Means are different

3: TwoYear CD 4: Five Year CD

0.84625 1.246875

16 16

Other Data Level of significance 0.05 Numerator d.f. 4 Denominator d.f. 45 MSW 0.026491 QStatistic 3.79

Group1toGroup4 0.928125 Group2toGroup3 0.330625 Group2toGroup4 0.73125 Group3toGroup4 0.400625

0.040690439 0.1542 Means are different 0.040690439 0.1542 Means are different 0.040690439 0.1542 Means are different 0.040690439 0.1542 Means are different

Using Q= 3.79 for numerator d.f. = 4 and denominator d.f. = 40, the mean rates of these investments are all different with Money Market being the lowest, followed by One Year CD, Two Year CD and finally Five Year CD.

(d) H0: 1. = 2. =· = 16.

H1: Not all i. are equal where i = 1, 2,·,16 Fcalc = 12.2177. Since the p-value is virtually 0, reject H0. There is enough evidence of a significant block effect in this experiment. The blocking has been advantageous in reducing the experimental error.

11.28 (a) ANOVA Source SS of Variation Rows 21.170 Columns 50.628 Error 7.361 Total 79.159

39 2 78 119

0.543 25.314 0.094

5.752 0.000 268.256 0.000

H0: 1 =  2 = 3

P-value

F crit

1.553 3.114

where 1 = 2 days, 2 = 7 days, 3 = 28 days

H1: At least one mean differs. Decision rule: If F > 3.114, reject H0. Test statistic: F = 268.256 Decision: Since Fcalc = 268.256 is greater than the critical bound F = 3.114, reject H0. There is enough evidence to conclude that there is a difference in the mean compressive strength after 2, 7 and 28 days. (b)

From Table E.10, QU(3, 78)  QU(3, 60) = 3.4. Critical range = QU ( c ,( r −1)( c −1))

X1 − X 2 = 0.5531*

MSE = 3.4 r

X 1 − X 3 = 1.5685*

0.094 = 0.1651 40 X 2 − X 3 = 1.0154*

At the 0.05 level of significance, all of the comparisons are significant. This is consistent with the results of the F test indicating that there is significant difference in the mean compressive strength after 2, 7 and 28 days. (c)

RE =

(r − 1) MSBL = r (c − 1) MSE 39 0.5428 + 40 2 0 0.0943 = = 2.558 (rc − 1) MSE 119  0.0943

(d) Box-Whisker Plot Comparison

28 days

Seven days

Two days

(e)

The compressive strength of the concrete increases over the three time periods.

11.29 (a)

(a)

df A = r – 1 = 3 – 1 = 2 df B = c – 1 = 3 – 1 = 2 (b) df AB = (r – 1)(c – 1) = (3 – 1)( 3 – 1) = 4 (c) df E = rc(n’ – 1) = 3 x 3 x (4 – 1) = 27 d) df T = n – 1 = 35

11.30

11.30 df 2 2 4 27 35

Source Factor A Factor B Interaction, AB Error E Total T

SS 120 110 (a) 540–120–110–270=40 270 540

MS (b) 120 2 = 60 (b) 110 2 = 55 (c) 40  4 = 10 270=

F 60 10 = 6 55 10 = 5.5 10 10 = 1

11.31 Source Factor A Factor B

df 2 2

SS 120 110

Interaction, AB

Error, E Total, T

27 35

540 – 120 – 110 – 270 = 40 270 540

11.32 F(2, 27) = 3.35

MS 120  2 = 60 110  2 = 55 40  4 = 10

F (b) 60  10 = 6 (c) 55  10 = 5.5 (a) 10  10 = 1

270  27 = 10

F(4, 27) = 2.73

(a) (b) (c)

Decision: Since Fcalc = 6.00 is greater than the critical bound of 3.35, reject H0. There is evidence of a difference among factor A means. Decision: Since Fcalc = 5.50 is greater than the critical bound of 3.35, reject H0. There is evidence of a difference among factor B means. Decision: Since Fcalc = 1.00 is less than the critical bound of 2.73, do not reject H0. There is insufficient evidence to conclude there is an interaction effect.

11.33 (a) Source Factor A Factor B Interaction, AB

Df 2–1 = 1 5–1 = 4 1•4 = 4

Error, E

2•5•3= 30 39

Total, T

SS 18 64 150 – 18 – 64 – 60 =8 60

MS 18  1 = 18 64  4 = 16 8 4 = 2

F 18  2 = 9 16  2 = 8 2 2 = 1

60  30 = 2

150

F(1, 30) = 4.1709

F(4, 30) = 2.6896

(b)(i) Decision: Since Fcalc = 9 is greater than the critical bound of 4.1709, reject H0. There is evidence of a difference among factor A means. (ii) Decision: Since Fcalc = 8 is greater than the critical bound of 2.6896, reject H0. There is evidence of a difference among factor B means. (iii) Decision: Since Fcalc = 1 is less than the critical bound of 2.6896, do not reject H0. There is insufficient evidence to conclude there is an interaction effect. 11.34 Source Factor A Factor B Interaction, AB Error, E Total, T

Df 2 82 4 8 30 44

SS 2 x 80 = 160 220 8 x 10 = 80 30 x 5 = 150 160+220+80+150 610

11.35 F(2, 30) = 3.32 (a) (b) (c)

MS 80 220  4 = 55 10 55  11 = 5

F 80  5 = 16 11 10  5 = 2

F(4, 30) = 2.69

F(8, 30) = 2.27

Decision: Since Fcalc = 16 is greater than the critical bound of 3.32, reject H0. There is evidence of a difference among factor A means. Decision: Since Fcalc = 11 is greater than the critical bound of 2.69, reject H0. There is evidence of a difference among factor B means. Decision: Since Fcalc = 2 is less than the critical bound of 2.27, do not reject H0. There is insufficient evidence to conclude there is an interaction effect.

11.36 ANOVA Source Variation

of SS

Pvalue

F crit

Sample

52.563

23.579

0.000

4.747

Columns

1.563

0.701

0.419

4.747

Interaction

3.063

1.374

0.264

4.747

Within

26.750

Total

83.938

2.229

Excel Two-way ANOVA output: (a) H0: There is no interaction between development time and developer strength. H1: There is an interaction between development time and developer strength. Decision rule: If Fcalc > 4.747, reject H0. Test statistic: Fcalc = 1.374. Decision: Since Fcalc = 1.374 is less than the critical bound of 4.747, do not reject H0. There is insufficient evidence to conclude that there is any interaction between development time and developer strength. (b) H0: 1 =  2 H1: 1   2

(c)

Decision rule: If Fcalc > 4.747, reject H0. Test statistic: Fcalc = 23.58. Decision: Since Fcalc = 23.579 is greater than the critical bound of 4.747, reject H0. There is sufficient evidence to conclude that developer strength affects the density of the photographic plate film. H0: 10 = 14 H1: 10  14 Decision rule: If Fcalc > 4.747, reject H0. Test statistic: Fcalc = 0.701. Decision: Since Fcalc = 0.701 is less than the critical bound of 4.747, do not reject H0. There is inadequate evidence to conclude that development time affects the density of the photographic plate film.

(d) 8

Plot of density vs strength

Density

5 4

3 2

10 minutes

14 minutes

0 1

2 Developer strength

(e)

At 5% level of significance, developer strength has a positive effect on the density of the photographic plate film while the developer time does not have any impact on the density. There is no significant interaction between developer time and developer strength on the density.

11.37 ANOVA Source Variation

of SS

Pvalue

F crit

Sample

528.125

24.143

0.008

7.709

Columns

5253.125

240.143

0.000

7.709

Interaction

28.125

1.286

0.320

7.709

Within

87.500

Total

5896.875

(a)

(b)

21.875

H0: There is no interaction between the type of pasta and cooking time. H1: There is an interaction between the type of pasta and cooking time. Decision rule: If Fcalc > 7.709, reject H0. Test statistic: Fcalc = 1.286 Decision: Since Fcalc = 1.286 is less than the critical bound of 7.709, do not reject H0. There is insufficient evidence to conclude that there is any interaction between the type of pasta and cooking time. H0:  Australian =  Italian H1:  Australian   Italian

(c)

Decision rule: If Fcalc > 7.709, reject H0. Test statistic: Fcalc = 24.143 Decision: Since Fcalc = 24.143 is greater than the critical bound of 7.709, reject H0. There is sufficient evidence to conclude that the type of pasta affects the weight of the pasta. H0:  4 = 8 H1:  4  8 Decision rule: If Fcalc > 7.709, reject H0. Test statistic: Fcalc = 240.143 Decision: Since Fcalc = 240.143 is greater than the critical bound of 7.709, reject H0. There is adequate evidence to conclude that cooking time affects the weight of the pasta.

(d)

Average weight vs type of pasta 340 320 300 280

260 240

4 minutes

220

8 minutes

200 Australian (e)

Italian

At 5% level of significance, cooking time has a positive effect on the weight of the pasta while the type of pasta also affects the weight of the pasta. There is no significant interaction between cooking time and the type of pasta on the weight of the pasta.

11.38 ANOVA Source Variation

of SS

Pvalue

F crit

Sample

1998.375

15.361

0.001

4.414

Columns

6825.083

3412.542

26.231

0.000

3.555

Interaction

270.750

135.375

Within

2341.750

130.097

Total

11435.958

(a)

1.041

0.374

3.555

H0: There is no interaction between marketing technique and location of the market. H1: There is an interaction between marketing technique and location of the market.

H0.05,2,18 = 3.555

(b)

Since Fcalc = 1.041 <3.555, do not reject H0. There is no significant evidence of an interaction between marketing technique and location of the market. H0:  A =  B = C

(c)

H1: the means are different Decision rule: If Fcalc > 3.555, reject H0. Test statistic: Fcalc = 26.231 Decision: Since Fcalc = 26.231 is greater than the critical bound of 3.555, reject H0. There is adequate evidence to conclude that marketing technique affects the sales of a new product. H0: Newcastle = Wollongong H1: Newcastle  Wollongong Decision rule: If Fcalc > 4.414, reject H0. Test statistic: Fcalc = 15.361 Decision: Since Fcalc = 15.361 is greater than the critical bound of 4.414, reject H0. There is sufficient evidence to conclude that the location of the market affects the sales of a new product.

(d)

Average sales vs marketing technique

100 90

Average sales

70 60 50

40 30

Newcastle

Wollongong

10 0 A

Marketing technique (e)

The difference in the average sales of a new product depends both on the location of the market and on the marketing technique. The plot of the average sales versus marketing technique illustrates that on average sales are higher in Newcastle compared with Wollongong. The marketing technique A is the most successful. The marketing technique C provides the smallest difference between average sales in Newcastle and Wollongong.

11.39 The between-groups variance MSB represents variation among the means of the different groups. The within-groups variance MSW measures variation within each group. 11.40 In a completely randomised design, individual items in different samples are randomly and independently drawn. In a randomised block design, individual items in different samples are matched using common characteristics, or repeated measurements are taken to reduce within-group variation. 11.41 The completely randomised design evaluates one factor of interest, in which sample observations are randomly and independently drawn. The randomised block design also evaluates one factor of interest, but sample observations are divided into blocks according to common characteristics to reduce within-group variation. The two-factor factorial design evaluates two factors of interest and the interaction between these two factors. 11.42 The major assumptions of ANOVA are randomness and independence, normality and homogeneity of variance. 11.43 If the populations are approximately normally distributed and the variances of the groups are approximately equal, you should select the one-way ANOVA F test to examine possible differences among the means of c independent populations. 11.44 When the ANOVA has indicated that at least one of the groups has a different population mean than the others, you should use multiple comparison procedures for evaluating pairwise combinations of the group means. In such cases, the Tukey–Kramer procedure should be used to compare all pairs of means. 11.45 In the randomised block design, individual items in different samples are matched using common characteristics, or repeated measurements are taken to reduce within-group variation. In the two-factor factorial design, more than one observation can be obtained for each treatment combination to measure the interaction of two factors. 11.46 The ANOVA tests for differences across two or more population means whereas the Levene test is for the differences in population variances. 11.47 You should use the two-way ANOVA F test to examine possible differences among the means of each factor in a factorial design when there are two factors of interest that are to be studied and more than one observation can be obtained for each treatment combination (to measure the interaction of the two factors). 11.48 Interaction measures the difference in the effect of one variable for the different levels of the second factor. If there is no interaction, any difference between levels of one factor will be the same at each level of the second factor. 11.49 You can obtain the interaction effect and carry out an F test for its significance. In addition, you can develop a plot of the response for each level of one factor at each level of a second factor. 11.50 ANOVA: Two-Factor With Replication SUMMARY

Sydney

Singapore

Wellington

London

Total

pre-crisis Count

Sum

145

Average

7.4

8.6

6.6

6.4

7.25

Variance

2.3

1.3

4.3

2.7236842

Count

Sum

118

Average

6.4

8.6

5.2

3.4

5.9

Variance

1.3

0.3

0.7

1.3

4.5157894

Count

Sum

Average

6.9

8.6

5.9

4.9

Variance

1.877777778

0.711111111

1.433333333

4.988888889

Source of Variation

P-value

Sample

18.225

11.390625

0.0019489

Columns

74.675

24.89166667

15.55729167

2.03661E-

Interaction

11.675

3.891666667

2.432291667

0.0830940

Within

51.2

1.6

Total

155.775

post-crisis

Total

ANOVA

(a)

(b)

H0: There is no interaction between location and the GFC. H1: There is an interaction between location and the GFC. Decision rule: If Fcalc > 2.901, reject H0. Since Fcalc = 2.43 < 2.901 do not reject H0. There is not enough evidence to conclude that there is an interaction between location and GFC on consumer confidence.

H 0 : 1 = 2 = 3 = 4

H1: at least one mean differs Decision rule: If Fcalc > 2.901, reject H0. Since Fcalc = 15.557 > 2.901, reject H0. There is sufficient evidence to conclude mean consumer confidence differs by location. (c)

H 0 : 1 =  2

H1: at least one mean differs Decision rule: If Fcalc > 4.149, reject H0. Since Fcalc = 11.39 > 4.149, reject H0. There is adequate evidence to conclude that mean consumer confidence differs pre vs post_GFC.

(d) 10 9 8 7 6 5

pre-crisis

post-crisis

3 2 1 0 Sydney (e) (f)

Singapore

Wellington

London

The Tukey procedure for pairwise comparisons is not used. The significant interaction makes the study of the main effects difficult. Both location and the GFC had individual (main) effects on mean consumer confidence separately. However, as there was no significant interaction effect there was not a combination of location and affected consumer confidence. 10 9 8 7 6 5

pre-crisis

post-crisis

3 2 1 0 Sydney 11.51 (a)

Singapore

Wellington

London

MS 2 0.035 15 0.468667

F 0.07468

H0: 1 =  2 =  3 2

H1: Not all  j are the same. 2

ANOVA Source of Variation Between Groups Within Groups Total

SS 0.07 7.03 7.1

P-value 0.928383

F crit 3.682317

Since the p-value = 0.928 > 0.05, do not reject H0. There is not enough evidence of a significant difference in the variances of the breaking strengths for the three air-jet pressures.

(b)

H0: 1 =  2 = 3 H1: At least one of the means differs. Decision rule: If Fcalc > 3.89, reject H0.

ANOVA: Single Factor

ANOVA Source Variation

of SS

Pvalue

F crit

Between Groups

8.074

4.037

4.09

0.038

3.8853

Within Groups

14.815

0.988

Total

22.889

Test statistic: Fcalc = 4.09 Decision: Since Fcalc = 5.89 is greater than the critical bound of 3.68, reject H0.

There is enough evidence to conclude that the mean breaking strengths differ for the three air-jet pressures. (c)

Breaking strength scores under 30 psi are significantly higher than those under 50 psi. (d)

Other things being equal, use 30 psi.

11.52 (a) Source of Variation Sample Columns Interaction Within

P-value

F crit

112.5603 46.01025 0.70225 133.105

1 1 1 36

112.5603 46.01025 0.70225 3.697361

30.4434 12.4441 0.1899

3.07E-06 0.001165 0.665575

4.113165 4.113165 4.113165

Total

292.3778

(a)

(a) H0: There is no interaction between type of

breakfast and desired time. H1: There is an interaction between type of breakfast and desired time. Decision rule: If FSTAT > 4.1132, reject H0. Copyright © 2019 Pearson Australia (a division of Pearson Australia Group Pty Ltd) 9781488617249, Berenson, Basic Business Statistics 5e

Test statistic: FSTAT = 0.1899. Decision: Since FSTAT = 0.1899 is less than the critical bound of 4.1132, do not reject H0. There is insufficient evidence to conclude that there is any interaction between type of breakfast and desired time. (b)

H0: 1 − 2=0 H1: 1 -2 Population 1 = Continental, 2 = American Decision rule: If Fcalc > 4.1132, reject H0. Test statistic: Fcalc = 30.4434. Decision: Since Fcalc = 30.4434 is greater than the critical bound of 4.1132, reject H0. There is sufficient evidence to conclude that there is an effect that is due to type of breakfast.

. ©H0: 1 − 2=0 H1: 1 -2 Population 1 = Early, 2 = Late

Decision rule: If Fcalc > 4.1132, reject H0. Test statistic: Fcalc =12.4441. Decision: Since Fcalc =12.4441 is greater than the critical bound of 4.1132, reject H0. There is sufficient evidence to conclude that there is an effect that is due to desired time (d)

Time (minutes)

c .

. (e) At the 5% level of significance, both the type of breakfast ordered and the desired time have an effect on delivery time difference. There is no interaction between the type of breakfast ordered and the desired time.

11.53 (a)

ANOVA Source Variation

H0: There is no interaction between department and store. H1: There is an interaction between department and store. Decision rule: If Fcalc > 3.885, reject H0.

of SS

Pvalue

F crit

Sample

1000.536

22.531

0.000

4.747

Columns

685.738

342.869

7.721

0.007

3.885

Interaction

80.458

40.229

0.906

0.430

3.885

Within

532.887

44.407

Total

2299.618

Test statistic: Fcalc = 0.906 Decision: Since Fcalc = 0.906 is less than the critical bound of 3.885, do not reject H0. There is insufficient evidence to conclude there is an interaction between department and store. (b)

H0: mSports = mPerfume H1: mSports ¹ mPerfume

(c)

Decision rule: If Fcalc > 4.747, reject H0. Test statistic: Fcalc = 22.531 Decision: Since Fcalc = 22.531 is greater than the critical bound of 4.747, reject H0. There is sufficient evidence to conclude that mean customer service rating does differ between departments. H0: 1 =  2 = 3 H1: At least one of the means differ. Decision rule: If Fcalc > 3.885, reject H0. Test statistic: Fcalc = 7.721 Decision: Since Fcalc = 7.721 is greater than the critical bound of 3.885, reject H0. There is enough evidence to conclude that mean customer service rating does differ between three stores.

(d)

Average customer service rating

85 80

Average customer service rating vs store

75 70

65 60 55

Sports

Perfume

45 40 Store 1

(e) (f) (g)

11.54 (a)

Store 2

Store 3

The Tukey procedure should not be used since the null hypothesis in part (a) above was not rejected. The mean customer service rating is higher for the sports department and it is the highest for store 3. In 11.51 part (d) the conclusion was that the level of customer service across three stores is about the same. However, in this question, the two-factor experiment gave a more complete, refined set of results than the one-factor experiment. To test the homogeneity of variance, you perform a Levene test:

H0 : 12 =  22 =  32

H1: Not all  j are the same. 2

ANOVA: Single Factor SUMMARY Groups

Count

Sum

Average

Variance

Boost juice

7.8

1.56

0.908

Naked juice

3.4

0.68

0.232

Funky juice

3.9

0.78

0.837

ANOVA Source of Variation

P-value

F crit

Between Groups

2.321333

1.160667

1.761254

0.21346

3.885294

Within Groups

7.908

0.659

Total

10.22933

Decision rule: if Fcalc > 3.885 reject H0 Fcalc = 1.76 < 3.885, do not reject H0. There is not enough evidence to suggest a significant difference in variances for the three outlets.

(b)

H 0 : 1 = 2 = 3

H1: At least one of the means differ.

ANOVA: Single Factor SUMMARY Groups

Count

Sum

Average

Variance

Boost juice

35.6

7.12

3.852

Naked juice

32.5

6.5

0.76

Funky juice

38.5

7.7

1.285

P-value

F crit

Between Groups

3.601333

1.800667

0.916059

0.426339

3.885294

Within Groups

23.588

1.965667

Total

27.18933

ANOVA Source Variation

Decision rule: If Fcalc > 3.885, reject H0. Fcalc = 0.916 < 3.885, do not reject H0. There is insufficient evidence to conclude that the mean price differs across juice outlet. The Tukey procedure should not be used since the null hypothesis in part (b) above was not rejected. There is inadequate evidence on which to form a conclusion about differences in mean prices across the three juice outlets.

11.55 ANOVA: Two-Factor With Replication Boost juice

Naked juice

Funky juice

Total

Count

Sum

30.9

36.4

39.3

106.6

Average

7.725

9.1

9.825

8.883333

Variance

0.669167

0.913333

0.255833

1.328788

SUMMARY University

Shopping centre Count

Sum

33.2

39.5

34.6

107.3

Average

8.3

9.875

8.65

8.941667

Variance

4.92

1.395833

1.71

2.686288

Count

Sum

22.1

24.5

31.3

77.9

Average

5.525

6.125

7.825

6.491667

Variance

2.369167

1.449167

2.449167

2.74447

Count

Sum

86.2

100.4

105.2

Average

7.183333

8.366667

8.766667

Variance

3.730606

3.875152

1.938788

P-value

F crit

Sample

46.90389

23.45194

13.08405

0.000106

3.354131

Columns

16.26889

8.134444

4.538279

0.019993

3.354131

Interaction

9.691111

2.422778

1.351689

0.276712

2.727765

Within

48.395

1.792407

Total

121.2589

Street

Total

ANOVA Source Variation

(a)

H0: There is no interaction between outlet name and location. H1: There is an interaction between. Decision rule: If Fcalc > 2.7278, reject H0. Since Fcalc = 1.35 < 2.7278, do not reject H0. There is insufficient evidence to conclude there is an interaction between outlet name and location (b) H 0 : 1 = 2 = 3 H1: At least one of the means differ. Decision rule: If Fcalc > 3.35, reject H0. Fcalc = 4.538 > 3.35, reject H0. (c)

H 0 : 1 = 2 = 3

H1: At least one of the means differ. Decision rule: If Fcalc > 3.35, reject H0. Fcalc = 13.08 > 3.35, reject H0.

(d) 12.00

10.00 8.00 University 6.00

Shopping centre Street

4.00 2.00 0.00 Boost juice

Funky juice

As there was no interaction effect we may interpret the main effect in (b) and (c) Mean price differs by outlet and location but there is no interaction between the two. In the completely randomised design, there is not enough evidence of a difference between mean price for the three outlets. In the two-factor factorial design, there is evidence for differences in mean prices by outlet as well as location. Allowing for the difference in location was necessary to better isolate the difference in price due to outlet.

(e) (f)

11.56

Naked juice

(a)

Let 1 = more expensive French white, 2 = Italian white, 3 = Italian red, 4 = less expensive French burgundy (red), 5 = more expensive French burgundy (red), 6 = California beaujolais (red), 7 = less expensive French white, and 8 = California white. H0: 1 = 2 = 3 = 4 = 5 = 6 = 7 = 8 H1: At least one of the means differs. Randomised block design output:

ANOVA Source Variation Rows Columns Error Total

P-value

F crit

521.5 440.333 630.167

11 7 77

47.409 62.905 8.184

5.793 7.686

0.000 0.000

1.915 2.131

1592

Decision rule: If Fcalc > 2.131, reject H0. Test statistic: Fcalc = 7.686 Decision: Since Fcalc = 7.686 is well greater than the critical bound of 2.131, reject H0. There is enough evidence of a difference in the mean rating scores among the wines.

(b)

(i)

You will need to assume (i) items are randomly selected from the eight groups or are randomly assigned to the eight groups, (ii) the sample values in each group are from a normally distributed population, (iii) the variances of the eight groups are equal, and (iv) there is no interacting effect between the groups and the blocks. The first assumption is satisfied by the design of the experiment.

French white

Normal Probability Plot 18 16 14 12 10 8 6 4 2 0

-2

-1

Z Value

Italian white

Normal Probability Plot 20 18 16 14 12 10 8 6 4 2 0

-2

-1

Z Value

Normal Probability Plot 16

14 Italian red

10 8

6 4 2

0 -2

-1

0 Z Value

Normal Probability Plot French burgundy (red)

12 10 8

6 4 2

0 -2

-1

Z Value

French burgundy (red)

Normal Probability Plot 20 18 16 14 12 10 8 6 4 2 0

-2

-1

0 Z Value

California Beaujolais (red)

Normal Probability Plot 18 16 14 12 10 8 6 4 2 0

-2

-1

0 Z Value

French white

Normal Probability Plot 18 16 14 12 10 8 6 4 2 0

-2

-1

0 Z Value

California white

Normal Probability Plot 18 16 14 12 10 8 6 4 2 0

-2

-1

Z Value

(b)

With the exception of California beaujolais (red) and California white, the distribution of the other brands shows departure from the normal distribution. However, the randomised block design is quite robust to departure from the normality assumption. Or, alternatively, a Friedman rank sum test can be performed. ANOVA output for Levene test for homogeneity of variance: H0: 1 =  2 =  3 =  4 =  5 =  6 =  7 =  8 2

H1: At least one of the variances differs. ANOVA of SS

Source Variation Between Groups Within Groups Total

32.167 515.333

7 88

4.595 5.856

0.785

547.5

Pvalue 0.602

F crit 2.116

Since Fcalc = 0.785<2.116, do not reject the null hypothesis. There is not enough evidence of a difference in the variances. To test for the blocking effect: H0: 1 =  2 = = 12

(c)

H1: At least one of the means differs. Decision rule: If Fcalc > 1.915, reject H0. Test statistic: Fcalc = 5.793 Decision: Since Fcalc = 5.793 is greater than the critical bound of 1.915, reject H0. There is enough evidence to conclude that the mean rating scores differ across the 12 club members. Hence, the blocking has been advantageous in reducing the random error. To determine which of the wines’ mean rating is significantly different from one another, use the Tukey multiple comparisons procedure for randomised block designs. With r = 8 and c = 12, the numerical degrees of freedom is 8 and the denominator degrees of freedom is (r − 1)(c − 1) = 7(11) = 77. But you will use 8 numerator and 60 denominator degrees of freedom to establish the critical range from Table E.10: Q = 4.44 Partial output from the Tukey multiple comparisons procedure for randomised block designs showing only the pairs of groups that are significantly different at the 5% level of significance:

Tukey Multiple Comparisons

Group

Sampl e Mean

Sampl e Size

10.417

14.667

11.167

9.333

14.417

8.417

11.833

9.750

Other Data Level of significance Numerator d.f. Denominator d.f. MSE Q Statistic

Absolute Comparison

Difference

Group 1 to Group 2 Group 1 to Group 5 Group 2 to Group 4 Group 2 to Group 6 Group 2 to Group 8 Group 4 to Group 5 Group 5 to Group 6 Group 5 to Group 8

4.250

Std. Error of Differenc e 0.826

Critical

4.000

0.826

3.667

5.333

0.826

3.667

6.250

0.826

3.667

4.917

0.826

3.667

5.083

0.826

3.667

6.000

0.826

3.667

4.667

0.826

3.667

Range

Results

3.667

Means are different Means are different Means are different Means are different Means are different Means are different Means are different Means are different

0.05 8 60 8.18398 4.44

(d)

(e)

There are two clusters of wines that have different summated rating. The first group of Italian white, the more expensive French burgundy (red), the less expensive French white and Italian red does not have means that are different from each other significantly within this group while the second group of the more expensive French white, California white, the less expensive French burgundy (red) and California beaujolais (red) does not have means that are different from each other significantly within this group. However, Italian white and the more expensive French burgundy (red) in the first group are significantly rated higher than all the wines in the second group. From the results in (c), it does not appear that country of origin, the type of wine, or the price has had an effect on the ratings.

RE =

( r − 1) MSBL + r (c − 1) MSE (11)(47.4091) + 12(7)(8.1840) = 1.5550 = (95)(8.1840) (rc − 1) MSE

If blocking is not effective, it should be avoided. In this study, there is evidence of a blocking effect, so the completely randomised design model is inferior to the randomised complete block design model. Notice that the relative efficiency measure indicates about 55.5% more observations would be needed in the one-way ANOVA design to obtain the same precision for comparison of

treatment group means as would be needed for the randomised complete block design. 11.57 (a) ANOVA Source of Variation Between Groups Within Groups Total

(b)

(c)

P-value

F crit

440.333

62.905

4.807

0.000

2.115

1151.66 7

13.087

1592

Since Fcalc = 4.807 > 2.115, reject the null hypothesis. There is enough evidence of a difference in the mean rating scores among the wines. SSBL = 521.5, SSE = 630.167, SSW = 1151.667. The sum of SSBL and SSE in the randomised block design equals the SSW in the completely randomised design. This has to be true because the SST, which measures the total variation of the observations around the grand mean, is the same for both designs and the SSA, which measures the difference from group to group, should also be the same for both designs. The SSW in the completely randomised design is subdivided into SSBL and SSE in the randomised block design. The p-value of the randomised block design ( 5.0725 (10) −7 ) is much smaller than that in the completely randomised design ( 0.000129 ). Hence, the amount of evidence against the null hypothesis of no treatment effect is much stronger using the randomised block effect, which controls for the variation across the raters.

11.58 Answers may vary.

Chapter 12: Simple linear regression

Commented [AC1]: I have left as is but needs some work with the formatting.

Learning objectives After studying this chapter you should be able to: 1. conduct a simple regression and the meaning of the regression coefficients b0 and b1 2. use regression analysis to predict the value of a dependent variable based on an independent variable 3. assess the adequacy of your estimated model 4. evaluate the assumptions of regression analysis 5. make inferences about the slope and correlation coefficient 6. estimate mean values and predict individual values 7. comprehend the pitfalls in regression and ethical issues 12.1

(a) (b)

When X = 0, the estimated expected value of Y is 15. For each increase in the value X by 1 unit, you can expect an increase by an estimated 25 units in the value of Y.

(c)

Yˆ = 15 + 25(12) = 315

12.2

(a) (b) (c) (d)

yes no no yes

12.3

(a) (b)

When X = 0, the estimated expected value of Y is 24. For each increase in the value X by 1 unit, you can expect a decrease in an estimated 0.2 units in the value of Y.

(c)

Yˆ = 24 − 0.2(10) = 22

12.4

(a)

12.5

12.6

(b) (c)

b0 = –63667.9 and b1 = 1442.216 For each terawatt hour of increase in energy generation, there is an expected increase in CO2 emission of an estimated 1442.216 (‘000) metric tonnes.

(d)

Ŷ = -63667.9 +1442.216 X = -63667.9 +1442.216(1000) Ŷ = 1378548('000s) metric tonnes

(a)

(b)

Yˆ = 0.7500 + 0.5000 X

(c)

For each increase of one additional plate gap, the estimated mean tear rating will increase by 0.5. (d) Yˆ= 0.7500 + 0.5000 X = 0.7500 + 0.5000 (0) = 0.7500

(a)

12.7

(b) (c)

b0 = 20.936 and b1 = 0.668 For each additional unit of increase in income, spending is expected to increase by $0.667.

(d)

Ŷ = 20.936 + 0.668 X Ŷ = 20.936 + 0.668(225) = $171.16

(a)

(b) (c)

b0 = 91.192 and b1 = –1.057 For each additional unit of increase in the tutorial class size, average mark is expected to decline by 1.057

(d)

Yˆ = 91 .192 − 1.057 X

Ŷ = 91.192 -1.057(20) = 70.06 12.8

(a)

Scatter Plot

g ni at R r a e T

-4.00

-3.00

-2.00

5.00 4.50 4.00 3.50 3.00 2.50 2.00 1.50 1.00 0.50 0.00 -1.00 0.00

1.00

2.00

3.00

4.00

PlateGap

The scatter plot shows a positive linear relationship. (b)

12.9

For each % increase in alcohol content, there is an expected increase in quality of an estimated 0.5624.

Yˆ = −0.3529 + 0.5624 X = −0.3529 + 0.5624(10) = 5.2715

There appears to be a positive linear relationship between quality and % alcohol content. For each % increase in alcohol content, there is an expected increase in quality of an estimated 0.5624.

(a)

(b) (c)

b0 = 47674.55 and b1 = –370.351 When ticket price is zero merchandise sales are $4,7674.55 (note this is a nonsensical result as we do not observe ticket prices even close to zero). For each additional increase in ticket price of $1, merchandise sales are expected to decline by $370.35.

(d)

Yˆ = 47674.55 − 370.351(100) = $10639.48

12.10

75% of the variation in the dependent variable can be explained by the variation in the independent variable.

12.11

SST = SSR + SSE = 25 + 24 = 49 r2 = SSR/SST = 25/49 = 0.5102 So, 51.02% of the variation in the dependent variable can be explained by the variation in the independent variable.

12.12

r2 = 55/72 = 0.764 So, 76.4% of the variation in the dependent variable can be explained by the variation in the independent variable.

12.13

SST = SSR + SSE = 67 + 28 = 95

r2 = SSR/SST = 67/95 = 0.7053 So, 70.53% of the variation in the dependent variable can be explained by the variation in the independent variable. 12.14

Since SST = SSR + SSE and since SSE cannot be a negative number, SST must be at least as large as SSR.

12.15 (a)

r2 = SSR/SST = 28350388757982.2/444694949364.38 = 0.984556567 Therefore, 98.5% of the variation in the dependant variable can be explained by the variation in the independent variable.

SSE 444694949364.38 = = 222284.8 n-2 11- 2

(b)

SYX =

(c)

Based on (a) and (b) the model should be useful for predicting revenue.

12.16 (a)

r2 = SSR/SST = 10.98/28.81 = 0.3811 So, 38.11% of the variation in the dependent variable can be explained by the variation in the independent variable.

SSE 17.83 = = 1.024 n−2 19 − 2

(b)

SSE =

(c)

This model may be moderately useful in predicting tear ratings.

12.17 (a)

r2 = SSR/SST = 52879.56/65600.74 = 0.806 So, 80.6% of the variation in the dependent variable can be explained by the variation in the independent variable.

SSE 12721.18 = = 27.355 n-2 19 - 2

(b)

SYX =

(c)

This model is useful in predicting spending.

12.18 (a)

r2 = SSR/SST = 585.69/638.93 = 0.917 So, 91.7% of the variation in the dependent variable can be explained by the variation in the independent variable.

SSE 53.24 = = 2.024 n-2 15 - 2

(b)

SYX =

(c)

This model is quite useful in predicting spending.

12.19 (a)

r2 = SSR/SST = 21.87/64 = 0.3417 So, 34.17% of the variation in the dependent variable can be explained by the variation in the independent variable.

(b)

SYX =

SSE 42.13 = = 0.9369 n−2 50 − 2

This model is very useful in predicting wine quality. r2 = SSR/SST = 640535694.53/664387646.67 = 0.964 So, 96.4% of the variation in the dependent variable can be explained by the variation in the independent variable.

12.21

SSE 23851952.14 = = 1544.41 n−2 12 − 2

(b)

SYX =

(c)

This model is very useful in predicting merchandise sales. A residual analysis of the data indicates no apparent pattern. The assumptions of regression appear to be met.

12.22

(a) The residual plot reveals that the equal variance assumption is most likely violated. The linearity assumption may also have been violated. So a linear fit does not appear to be adequate.

Residual Plot 3 2.5 2 1.5 1 0.5 0 -0.5 -1 -1.5

sl a u d is e R

-4

-2

0 X

This is not a time series data, so you do not need to evaluate the independence assumption. The normal probability plot suggests that the errors might be right-skewed. 12.23 (a)

Normal Probability Plot 3 2.5 2 sl 1.5 a 1 u d is0.5 e R 0 -0.5 -1 -1.5 -2

-1

0 ZValue

There seems to be some non-linear relationship present as residuals increase for increasing values of energy.

The simple linear regression model seems to fit well.

(b)

The normal probability plot of the residuals indicates a departure from the normality assumption. There is a non-linear pattern. The normal distribution is not a good model for these data. 12.24 (a)

The residuals are well spread and show no particular pattern.

The simple linear regression model seems to fit well. (b)

The plot indicates the data is normally distributed.

12.25 (a)

The residuals are well spread and show no particular pattern.

The simple linear regression model seems to be appropriate.

(b)

There appears to be a slight positive skewness, but the data can be assumed to have normal distribution. 12.26 (a)

Residual Plot 3 2 1 0 -1 -2 -3

Based on the residual plot, there does not appear to be a pattern in the residual plot. The linearity and equal variance assumptions appear to be holding up.

Normal Probability Plot 3 2 1 0

Residual s

-1 -2 -3 -3

-2

-1

0 Z Value

12.27 (a)

Residuals seem to decrease as ticket price increases.

The simple linear regression model seems to fit well. (b)

The plot indicates the data is normally distributed.

12.28 (a)

The residuals fluctuate between positive followed by negative values. (b)

There appears to be strong negative autocorrelation among residuals.

12.29 (a)

(b) (c)

There appears to be some positive autocorrelation among the residuals. D = 0.5625 <dL = 1.08 There is evidence of positive autocorrelation between residuals. The data is positively autocorrelated.

12.30 (a) It is not necessary to calculate the Durbin–Watson statistics the data have collected for a single period of time (week 6 of the semester).

(b)

If one student’s spending was considered over a period of time and income varied over time (unlikely), calculation of the Durbin–Watson statistic would be necessary before proceeding with the least-squares regression.

12.31 (a)

b0 = 0.6795 and b1 = 0.6643

(b) (c)

Ŷ = 0.6795 + 0.6643 * 4 = 8.3369%

There is no apparent pattern visible. (d) (e) 12.32 (a) (b) (c)

D = 1.0804 < 1.32 and D=1.0804>0.88. The test results are inconclusive. There are some questions about the validity of the model.

b0 = 20.9362 and b1 = 0.6677

Ŷ = 20.9362 + 0.6677 * 650 = 9236

(d) (e) 12.33 (a) (b) (c)

(d) (e)

The variability of residuals seems to be increasing as income increases. D = 1.117 < 1.18. There is positive autocorrelation among the residuals. There are definite questions about the validity of the model.

b0 = 8.9873 and b1 = 0.5455

Ŷ = 8.9873+ 0.5455 * 23 = 21.5344

There is no apparent pattern visible. D = 1.8816 > 1.37. There is negative autocorrelation among the residuals. There are definite questions about the validity of the model.

There appears to be a positive relationship between the size of the crowd attending the game and number of meat pies sold. (b)

Ŷ = 32.2431+1.8293X , where X is attendance (‘000).

(c)

There is no apparent pattern visible. (d) (e) 12.35 (a) (b) (c) (d)

D = 1.4141 > 1.32. There is negative autocorrelation among the residuals. There are definite questions about the validity of the model.

H0 : b1 = 0

H1 : b1 ¹ 0

Test statistic: t = (3.4-0)/1.2 = 2.83 n = 34, df = 32, t32, = ±2.0369 Reject H0. There is sufficient evidence of a relationship between the two variables.

b1 - t25Sb £ b1 £ b1 + t25Sb 1

3.4 − 2.0369(1.2)  1  3.4 + 2.0369(1.2)

0.9557  1  5.8443 12.36 (a) (b) (c) (d) (e)

MSR = SSR/k = 34/1 = 34 MSE = SSE/(n – k – 1) = 25/(21-1-1) = 1.32 F = MSR/MSE = 34/1.32 = 25.76 F1,19 = 4.38 Reject H0. There is sufficient evidence of a relationship between the two variables. r2 = SSR/SST = 34/59 = 0.5763 r = –√0.5763 = –0.7591

H0 : r = 0 H1 : r ¹ 0

There is no correlation between X and Y There is Correlation between X and Y

df = 19 Critical values: t19 = ±2.0930 Decision rule: Reject H0 if

tcalc  2.0930 Test statistic:

r-r

1- r 2 n-2 −0.7591 = −5.08 1 − 0.5763 21 − 2

Since –5.08 < –2.0930, reject H0. There is sufficient evidence to conclude that there is a significant correlation between X and Y. 12.37 (a)

H0 : b1 = 0 H1 : b1 ¹ 0 b - b 1442.216 - 0 t= 1 1 = = 23.954 Sb 60.209 1

Critical values: t9 = ±2.2622 Decision rule: Reject H0 if |tcalc|> 2.2622 Since 23.954 > 2.262, reject H0. There is enough evidence to conclude that there is a linear relationship between energy generation and CO2 emissions. (b)

b1 - t9 Sb £ b1 £ b1 + t9 Sb 1

1442.216 - 2.2622(60.209) £ b £ 1442.216 + 2.2622(60.209) 1306.0113 £ b £ 1578.421 12.38 (a) C o e f f i c i e n t s Inte rce pt

Plat e Gap

H0 : b1 = 0

0 . 7 5 0 0 0 . 5 0 0 0

S t a n d a r d E r r o r 0 . 2 3 4 9 0 . 1 5 4 5

t S t a t

3 . 1 9 2 2 3 . 2 3 5 6

P v a l u e

0 . 0 0 5 3 0 . 0 0 4 9

0 . 2 5 4 3 0 . 1 7 4 0

L o w e r

U p p e r

9 5 %

1 . 2 4 5 7 0 . 8 2 6 0

H1 : b1 ¹ 0

0.5000 − 0 = 3.2356 0.1545

Critical values: t17 = ±2.1098 Decision rule: Reject H0 if |tcalc|> 2.1098 Since |3.2356| > 2.1098, reject H0. There is enough evidence to conclude that there is a linear relationship between plate gap and tear rating. (b)

b1 − t12 Sb1  1  b1 + t12 Sb1

0.5000 − 2.1098  0.1545  1  0.5000 + 2.1098  0.1545

0.1740  1  0.8260 12.39 (a)

H0 : b1 = 0 H1 : b1 ¹ 0 b - b 0.6676 - 0 t= 1 1 = = 8.406 Sb 0.0794 1

Critical values: t17 = ±2.1098 Decision rule: Reject H0 if |tcalc|> 2.1098 Since |8.406| > 2.1098, reject H0. There is enough evidence to conclude that there is a linear relationship between spending and income.

(b)

b1 - t17 Sb £ b1 £ b1 + t17 Sb 1

0.6676 - 2.1098(0.0794) £ b1 £ 0.6676 + 2.1098(0.0794) 0.50 £ b1 £ 0.84

12.40 (a)

H0 : b1 = 0 H1 : b1 ¹ 0 b - b -1.0568 - 0 t= 1 1 = = -11.959 Sb 0.0884 1

Critical values: t13 = ±2.1604 Decision rule: Reject H0 if |tcalc|> 2.1604 Since |11.959| > 2.1604, reject H0. There is enough evidence to conclude that there is a linear relationship between tutorial class size and average class mark. (b)

b1 - t13Sb £ b1 £ b1 + t13Sb 1

− 1.0568 − 2.1604 (0.0884 )  1  0.6676 + 2.1098 (0.0794 )

-1.25 £ b1 £ -0.87 12.41 (a)

C o e f f i c i e n t s 0 . 3 5 2 9 0 . 5 6 2 4

Inter cept

alco hol

H0 : b1 = 0

St an da rd Er ro r

1. 20 00

0 . 2 9 4 1 4 . 9 9 1 3

0. 11 27

H1 : b1 ¹ 0 b1 − 1 0.5624 − 0

Sb1

S t a t

0.1127

P v a l u e

L o w e r

U p p e r

9 5 %

0 . 7 7 0 0

2. 7 6 5 6

2 . 0 5 9 9

0 . 0 0 0 0

0. 3 3 5 9

0 . 7 8 9 0

= 4.9913

Decision rule: Reject H0 if |tcalc|> 2.0106 Since |4.9913| > 2.0106, reject H0. There is enough evidence to conclude that there is a linear relationship between alcohol content and wine quality. (b)

b1 − t48 Sb1  1  b1 + t48Sb1

0.5624 − 2.0106(0.1127)  1  0.5624 + 2.0106(0.1127)

0.3359  1  0.7890 12.42 (a)

H0 : b1 = 0 H1 : b1 ¹ 0 b1 − 1 −370.351 − 0

Sb1

22.600

= −16.39

Critical values: t10 = ±2.2281 Decision rule: Reject H0 if |tcalc|> 2.2281 Since |–16.39| > 2.2281, reject Ho. There is enough evidence to conclude that there is a linear relationship between ticket prices and merchandise sales. (b)

b1 − t10 Sb1  1  b1 + t10 Sb1 −370.351 − 2.2281 22.600  1  −370.351 − 2.2281 22.600 −420.705  1  −319.996

12.43 (a)

(a)

First weekend gross and the U. S. gross

r = 0.7284. There appears to be a strong positive linear relationship.

First weekend gross and the worldwide gross

r = 0.8233. There appears to be a strong positive linear relationship.

U. S. gross, and worldwide gross (b)

r = 0.9642 There appears to be a very strong positive linear relationship.

First weekend gross and the U. S. gross

Since tstat = 2.6042 > 2.4469 and p-value = 0.0404 < 0.05, reject H0. At the 0.05 level of significance, there is evidence of a linear relationship between first weekend sales and U.S. gross.

First weekend gross and the worldwide gross

Since tstat = 3.5532 > 2.4469 and p-value = 0.0120 < 0.05, reject H0. At the 0.05 level of significance, there is evidence of a linear relationship between first weekend sales and worldwide gross.

U. S. gross, and worldwide gross

Since tstat = 8.9061 > 2.4469 and p-value = 0.0001 < 0.05, reject H0. At the 0.05 level of significance, there is evidence of a linear relationship between U.S. gross and worldwide gross. 12.44 (a)

H0 : b1 = 0

H1 : b1 ¹ 0

Critical values: t10 = ±2.2281 Decision rule: Reject H0 if |tcalc|> 2.2281 Since |–12.002| > 2.2281, reject H0. There is enough evidence to conclude that there is a linear relationship between hours worked and childcare costs. (b)

b1 − t10 Sb1  1  b1 + t10 Sb1

−0.080 − 2.2281 0.007  1  −0.080 + 2.2281 0.007 −0.095  1  −0.065

12.45 (a)

r2 = SSR/SST = 480.1146/607.2143 = 0.7907 r = √0.7907 = 0.889

(b)

H0 : b1 = 0 H1 : b1 ¹ 0 b - b 0.0682 - 0 t= 1 1 = = 6.7327 Sb 0.0101 1

Critical values: t12 = ±2.1788 Decision rule: Reject H0 if |tcalc|> 2.1788 Since |6.7327| > 2.1788, reject H0. There is enough evidence to conclude that there is a linear relationship between the number of drivers tested and the number of drivers charged for drink-driving.

12.46

(b) 12.47 (a)

(a) Proctor & Gamble’s stock moves only 32% as much as the overall market and is much less volatile than the market. Dr. Pepper Snapple Group’s stock moves only 2% as much as the overall market in the opposite direction and is considered very nonvolatile compared to the market. The stock of Disney Company moves 7% more than the overall market and is considered a little more volatile than the market. Apple’s stock moves 69% as much as the overall market and is considered less volatile than the market. eBay’s stock moves 79% as much as the overall market and is considered almost as volatile as the market. Marriot’s stock moves 32% more than the overall market and is considered more volatile than the market. Investors can use the beta value as a measure of the volatility of a stock to assess its risk. r2 = SSR/SST = 5362.7489/6056.9583 = 0.8854 r = √0.8854 = 0.9409

(b)

H0 : b1 = 0 H1 : b1 ¹ 0 b - b 0.0043 - 0 t= 1 1 = = 13.0365 Sb 0.0003 1

Decision rule: Reject H0 if |tcalc|> 2.0739 Since |13.0365| > 2.0739, reject H0. There is enough evidence to conclude that there is a linear relationship between distance and the time taken to deliver. (c)

There is a strong relationship between distance and time taken to deliver.

(d)

Ŷ = 3.7941+ 0.0043(9250) = 43.48 hours

12.48 (a)

r2 = SSR/SST = 321763.524/727674 = 0.4422 r = √0.4422 = 0.6649

(b)

H0 : b1 = 0 H1 : b1 ¹ 0 b - b 27.3469 - 0 t= 1 1 = = 2.8155 Sb 9.7130 1

Critical values: t10 = ±2.2281 Decision rule: Reject H0 if |tcalc|> 2.2281 Since |2.8155| > 2.2281, reject H0. There is enough evidence to conclude that there is a linear relationship between the length of service and sales. (c)

There is a moderately strong relationship between length of service and sales.

12.49 (a)

Ŷi ∓ tn-2 SYX hi 1 ( X i - X )2 + n SSX Ŷi = 5+ 3X = 5+ 3 * 4 = 17 hi =

hi =

1 (4 − 3) 2 + = 0.0817 24 25

t33 = 2.0345 95% confidence interval for the mean: 17  2.0345  2.1 0.0817

15.7790  YX = 4  18.2210 (b)

95% prediction interval for an individual response: 17  2.0345  2.1 1 + 0.0817

11.5066  Y | X =4  22.4934 12.50 (a)

Ŷi ∓ tn-2 SYX hi 1 ( X i - X )2 + n SSX Yˆi = 5 + 12 X = 5 + 12*4 = 53 hi =

hi =

1 (12 − 2) 2 + = 5.1 t8 = 2.3060 10 20

95% confidence interval for the mean: 53± 2.3060 *1*

5.1

42.79 £ mY|X =12 £ 58.21 (b)

95% prediction interval for an individual response: 53± 2.3060 *1*

hi = (c)

12.51 (a)

1+ 5.1

1 (25 - 22.64) + = 0.080 47.30 £ YX =12 £ 58.70 14 627.21 2

The confidence interval for the mean in this exercise is narrower because the standard error of the estimate is smaller.

Ŷi ∓ tn-2 SYX hi 1 ( X i - X )2 + n SSX Yˆi = 0.7500 + 0.5000(1.8) = 1.6500 hi =

hi =

1 (1.8 − 0) 2 + = 0.1264 19 43.92

t17=2.1098 95% confidence interval for the mean: 1.6500  2.1098*1.024* 0.1264 Copyright © 2019 Pearson Australia (a division of Pearson Australia Group Pty Ltd) 9781488617249, Berenson, Basic Business Statistics 5e

0.8818  Y | X =1.8  2.4182 (b)

95% prediction interval for an individual response:

1.6500  2.1098*1.024* 1.1264 −0.6432  YX =1.8  3.9432 (c)

Part (b) provides an interval prediction for the individual response given a specific value of the independent variable, whereas part (a) provides an interval estimate for the mean value given a specific value of the independent variable. Due to more variation in predicting an individual value, prediction interval is wider.

12.52 (a)

Ŷi ∓ tn-2 SYX hi 1 ( X i - X )2 + n SSX Ŷi = -63,667.920 +1442.216 X = 1378,547.668 hi =

hi =

1 (1,000 -1,045.818)2 + = 0.0835 12 13630,087.636

t10 = 2.2281 95% confidence interval for the mean:

1378,547.668 ∓ 2.2281*222,284.840 0.0835

1235, 431.809  Y | X =1000  1521,663.526 (b)

95% prediction interval for an individual response:

1378,547.668 ∓ 2.2281*222,284.840 1.0835 863,011.688 < YX =1,000 < 1894,083.647 (c)

12.53 (a)

Ŷi ∓ tn-2 SYX hi hi =

1 ( X i - X )2 + n SSX

Ŷi = 91.1918 -1.0568 X = 70.0553 hi =

1 (20 -19.8)2 + = 0.0667 15 524.4

t10 = 2.1604

95% confidence interval for the mean:

70.0553 ∓ 2.1604 *2,02370 0.0667 68.926  Y | X =20  71.185 (b)

95% prediction interval for an individual response:

70.0553 ∓ 2.1604 *2,02370 1.0667 65.540 < YX =20 < 74.571 (c)

12.54 (a)

Ŷi ∓ tn-2 SYX hi hi =

1 ( X i - X )2 + n SSX

Ŷi = 20.936 - 0.668 X = 121.087 hi =

1 (150 -150.579)2 + = 0.0526 19 118,620.632

t17 = 2.1098 95% confidence interval for the mean: 121.087 ∓ 2.1098*27.355

107.846  Y | X =150  134.328

(b)

0.0526

95% prediction interval for an individual response:

121.087 ∓ 2.1098*27.355 1.0526 61.873 < YX =20 < 180.301 (c)

12.55 (a)

Ŷi ∓ tn-2 SYX hi hi =

1 ( X i - X )2 + n SSX

Ŷi = 6.039 + 0.107 X = 10.297 hi =

1 (40 - 53.65)2 + = 0.089 20 4,736.55

t18 = 2.1009 95% confidence interval for the mean: 10.297 ∓ 2.1009 *2.3578

8.816  Y | X =40  11.777

(b)

0.089

95% prediction interval for an individual response:

10.297 ∓ 2.1009 *2.3578 1.089 5.127 < YX =40 < 15.467 (c)

12.56 (a)

Ŷi ∓ tn-2 SYX hi hi =

1 ( X i - X )2 + n SSX

Ŷi = 47674.55 - 370.351(80) = 18046.49 hi =

1 (80 − 89) 2 + = 0.1007 12 4670

T10 = 2.2281 95% confidence interval for the mean:

18046.49 ± 2.2281*1544.408 * 0.1007 16954.64 £ mY|X =80 £ 19138.34 (b)

95% prediction interval for an individual response:

18046.49  2.2281*1544.408* 1.1007 14436.33  YX =80  21.656.65 (c)

12.58 SUMMARY OUTPUT Regression Statistics Multiple R

0.992248

R Square Adjusted R Square Standard Error

0.984557 0.982841 152.9326

Observations

ANOVA SS

Regression

df 1

13419592

573.772

Significance F

Residual

210495.3

23388.37

Total

13630088

1.84E-09

Coefficients

Standard Error

t Stat

P-value

Intercept

59.61517

61.81673

0.964386

0.360048

–80.224

199.4543

C02

0.000683

2.85E-05

23.95354

1.84E-09

0.000618

0.000747

Lower 95%

We pick up a statistically significant relationship. However, the direction of causality is incorrect. The estimated slope would be interpreted as indicating that an increase in CO 2 emissions leads to an increase in energy generation. 12.60

The Y intercept b0 represents the estimated mean value of Y when X equals 0. The slope of the line, b1, represents the estimated expected change in Y per unit change in X. It represents the estimated mean amount that Y changes (either positively or negatively) for a particular unit change in X.

12.61

The coefficient of determination is the square of the correlation coefficient.

12.62

The ANOVA table is used to calculate the coefficient of determination as well as being used to test the slope of the model with an F test.

12.63

The test statistic t follows a t distribution with n – 2 degrees of freedom while the test statistic F follows an F distribution with k and n – k – 1 degrees of freedom.

12.64

Unless a residual analysis is undertaken, you will not know whether the model fit is appropriate for the data. In addition, residual analysis can be used to check whether the assumptions of regression have been seriously violated.

12.65

Interpolation is predicting Y for a given value of X within a relevant range of X. Extrapolation is predicting beyond the given range.

12.66

The normality of error assumption can be evaluated by obtaining a histogram, boxplot and/or normal probability plot of the residuals. The homoscedasticity Copyright © 2019 Pearson Australia (a division of Pearson Australia Group Pty Ltd) 9781488617249, Berenson, Basic Business Statistics 5e

Upper 95%

assumption can be evaluated by plotting the residuals on the vertical axis and the X variable on the horizontal axis. The independence of errors assumption can be evaluated by plotting the residuals on the vertical axis and the time order variable on the horizontal axis. This assumption can also be evaluated by computing the Durbin–Watson statistic. 12.67

The unexplained variation would be equal to zero if the model perfectly predicted the data. This would result in a coefficient of determination of 1.

12.68

The confidence interval for the mean response estimates the mean response for a given X value. The prediction interval estimates the value for a single item or individual.

12.69 (a)

(b)

(e)

C St t P o a e n S v f d t a f ar a l i d t u c Er e i ro e r n t s Inter 6 85 7. 0 cept 8 4. 96 . 0 96 30 0 8 82 0 . 0 1 5 0 4 7 Twitt 0 0. 14 0 er . 00 .2 . Activ 0 35 53 0 ity 5 2 0 0 0 3 0 For each additional unit increase in Twitter activity, the mean receipts per theater will increase by an estimated $0.05. The estimated mean receipt per theater is $6808.10 when there is no Twitter activity.

Yˆ= b0 + b1X = 6808.1047+0.0503(100000) = $11,835.26

You should not use the model to predict the receipts for a movie that has a Twitter activity of 1,000,000 because 1,000,000 falls outside the domain of the independent variable and any prediction performed through extrapolation will not be reliable. r2 = 0.9760. So 97.60% of the variation in receipt per theater can be explained by the variation in Twitter activity.

(f) Residual Plot 3000 2000

Residuals

1000 0 -1000 -2000 -3000 -4000 0

200000

400000 X

600000

800000

(g)

(h) (i)

The residual plot does not reveal specific pattern. However, the sample size is too small for the residual analysis to be reliable. Scatter t = 14.2532 and p-value = Diagram 0.0000. Since p-value <  = 0.05 , reject H0. There is evidence 4.5 of a linear relationship between Twitter activity and receipts. )s 4 $10,015.85  Y | X =100,000  $13,654.67 ur3.5 $6,790.94  YX =100,000  $16,879.58 ( 3 e

Them of (a)-(h) suggest that Twitter activity is a useful predictor of receipts on the first 2.5 iT results weekend n 2 a movie opens. However, the sample size of 7 is too small for the prediction to be o it reliable. l e1.5 p m1 o C0.5 0

12.70

(a)

100

200 Invoice Processed

300

400

b0 = 0.4872, b1 = 0.0123

(b) 0.4872 is the portion of estimated mean completion time that is not affected by the number of invoices processed. When there is no invoice to process, the mean completion time is estimated to be 0.4872 hours. Of course, this is not a very meaningful interpretation in the context of the problem. For each additional invoice processed, the estimated mean completion time increases by 0.0123 hours. (c) (d)

Yˆ = 0.4872 + 0.0123X = 0.4872 + 0.0123(150) = 2.3304 r2 = 0.8623. 86.23% of the variation in completion time can be explained by the variation in the number of invoices processed.

(e)

Invoices Residual Plot 0.8 0.6 0.4 sl u a ids0.20 e R0.2 0.4 0.6 0.8

10 0

200 Invoice s

30 0

40 0

Invoices Residual Plot 0.8 0.6 0.4 sl 0.2 a u d is 0 e R-0.2 -0.4 -0.6 -0.8

13.79

(e) cont. (f)

(g)

20 Invoices

Based on a visual inspection of the graphs of the distribution of residuals and the residuals versus the number of invoices and time, there appears to be autocorrelation in the residuals. D = 0.69 < 1.37 = dL. There is evidence of positive autocorrelation. The model does not appear to be adequate. The number of invoices and, hence, the time needed to process them, tend to be high for a few days in a row during historically heavier shopping days or during advertised sales days. This could be the possible causes for positive autocorrelation. Due to the violation of the independence of errors assumption, the prediction made in (c) is very likely to be erroneous.

12.71 (a)

b0 = 0.3527 and b1 = 0.0117 (b)

When the crude oil price = $0/barrel, the price of petrol is $0.3527/litre. For each additional dollar of increase in the price of crude oil, the price of petrol increases by $0.0117.

(c)

Ŷ = 0.3527 + 0.0117 X Ŷ = 0.3527 + 0.0117(75) = $1.23

(d)

r2 = SSR/SST = 0.1044/0.1351 = 0.773 So, 77.3% of the variation in the dependent variable can be explained by the variation in the independent variable.

(e)

The residuals are well spread and show no particular pattern.

The simple linear regression model seems to fit well. (f)

H0 : b1 = 0 H1 : b1 ¹ 0 b - b 0.0117 - 0 t= 1 1 = = 5.8373 Sb 0.0020 1

Decision rule: Reject H0 if |tcalc|> 2.2281 Since |5.8373| > 2.2281, reject H0. There is sufficient evidence to conclude that there is a linear relationship between petrol and crude oil prices. (g)

Ŷi ∓ tn-2 SYX hi hi =

1 ( X i - X )2 + n SSX

Ŷi = 0.3527 + 0.01166 X = 5+ 3 * 4 = 1.2271 1 (75 - 81.75)2 + = 0.1426 12 768.25 t10 = 2.2281 hi =

95% confidence interval for the mean:

1.2271∓ 2.2281 * 0.0554 0.1426 1.1805 < YX =75 < 1.2737 (h)

Ŷi ∓ tn-2 SYX 1+ hi 95% prediction interval for an individual response:

1.2271∓ 2.2281 * 0.0554 1.1426 1.0953 < YX =4 < 1.3590 (i)

b1 - t10 Sb £ b1 £ b1 + t10 Sb 1

0.0117 - 2.2281(0.0020) £ b1 £ 0.0117 + 2.2281(0.0020) 0.007 £ b1 £ 0.016 12.72 (a) Regres sion Statisti cs M u lt i p l e R R S q u a r e A

0 . 9 1 4 3 0 . 8 3 6 0 0

d j u s t e d R S q u a r e S t a n d a r d E r r o r O b s e r v a ti o n s

. 8 3 0 1

7 8 . 2 8 8 4

3 0

A N O V A d f

R e g r e s s i o n

R e s i d u a

2 8

S S

8 7 4 5 4 6 . 7 0 7 2 1 7 1 6 1 4 .

M S

8 7 4 5 4 6 . 7 0 7 2 6 1 2 9 . 0 8

1 4 2 . 6 8 8 1

S i g n i f i c a n c e F 0 . 0 0 0 0

T o t a l

2 9

C o e f f i c i e n t s I n t e r c e p t R e v e n u e ( $ m il )

1 3 2 . 5 1 1 4 5 . 2 2 8 6

b0 = 132.5 114 b1 = (b)

2 5 9 5 1 0 4 6 1 6 0 . 9 6 6 7

0 7

S t a n d a r d E r r o r 5 5 . 5 7 6 7 0 . 4 3 7 7

t S t a t

P v a l u e

2 . 3 8 4 3

0 . 0 2 4 1

1 1 . 9 4 5 2

0 . 0 0 0 0

2 4 6 . 3 5 5 1 4 . 3 3 2 0

L o w e r

U p p e r

9 5 %

1 8 . 6 6 7 6 6 . 1 2 5 2

5.2286

For each additional million-dollars of revenue generated, the mean value of the franchise will increase by an estimated $5.2286 million. Literal interpretation of the intercept is not meaningful because an operating franchise cannot have zero revenue.

Yˆ= −132.5114 + 5.2286 X = −132.5114 + 5.2286 (150) = $ 651.7731millions

r2 = 0.8360. So, 83.60% of the variation in the value of an NBA franchise can be explained by the variation in its annual revenue.

(e) Residual Plot 250 200 150

Residuals

0 -50 -100

13.82

(e) cont. Normal Probability Plot 250 200

Residuals

150 100 50 Residuals

0 -50

-100 -150 -3

-2

-1

0 Z Value

The normal probability plot suggests possible departure from the normality assumption. = 11.9452 with a p-value that is approximately zero, reject H0 at the 5% level of t

( f )

S T A T

significance. There is evidence of a linear relationship between annual revenue and franchise value.

(g) =150

(i)

$613.6103 millions  Y | X =150  $689.9359 millions (h)

$486.9282 millions  YX

 $816.6180 millions The strength of the relationship between revenue and value is stronger for NBA franchises than for European soccer teams and Major League Baseball teams.

(a) Regres sion Statisti cs M u lt i p l e R R S q u a r e A dj u st e d R S q

0 . 9 1 0 4 0 . 8 2 8 8 0 . 8 1 9 3

u a r e S t a n d a r d E r r o r O b s e r v a ti o n s

4 0 1 . 9 3 8 1

2 0

A N O V A d f

R e g r e s si o n

R e si d u a l

1 8

T o t

1 9

S S

1 4 0 7 7 4 6 9 . 3 1 3 2 2 9 0 7 9 7 5 . 6 3 6 8 1 6 9

M S

1 4 0 7 7 4 6 9 . 3 1 3 2 1 6 1 5 5 4 . 2 0 2 0

8 7 . 1 3 7 7

S i g n i f i c a n c e F 0 . 0 0 0 0

a l

8 5 4 4 4 . 9 5 0 0 C o e f f i c i e n t s

I n t e r c e p t

7 2 5 . 0 1 2 4

R e v e n u e ( $ m ill io n s

5 . 5 4 5 2

S t a n d a r d E r r o r 2 0 2 . 4 0 9 4 0 . 5 9 4 0

t S t a t

P v a l u e

3 . 5 8 1 9

0 . 0 0 2 1

9 . 3 3 4 8

0 . 0 0 0 0

1 1 5 0 . 2 5 8 8 4 . 2 9 7 2

L o w e r

U p p e r

9 5 %

2 9 9 . 7 6 6 1 6 . 7 9 3 3

b0 = -725.0124, b1 = 5.5452 (b)

Literal interpretation of the intercept is meaningless since no team can have 0 revenues. For each additional million dollars increase in team revenues, the estimated mean franchise value increases by 5.5452 million dollars.

Yˆ= −725.0124 + 5.5452 X = −725.0124 + 5.5452 (150) = $106.7739 millions

r2 = 0.8288. So, 82.88% of the variation in the value of an European soccer franchise can be explained by the variation in its annual revenue.

13.83

(e) cont.

Residuals

Residual Plot 1200 1000 800 600 400 200 0 -200 -400 -600 -800 0

200

600

400 X

800

Residuals

Normal Probability Plot 1200 1000 800 600 400 200 0 -200 -400 -600 -800

Residuals

-2

(f)

-1

0 Z Value

Based on a visual inspection of the graphs of the distribution of the residuals versus revenues, the equal variance assumption appears to be violated. The normal probability plot suggests that the normality assumption might have been violated. The p-value is virtually zero, reject H 0 at the 5% level of significance. There is evidence of a linear relationship between annual revenue and franchise value.

-$163.8184 millions  Y | X =150  $377.3662 millions -$779.9617 millions  YX =150  $993.5095 millions (i) The strength of the relationship between revenue and value is stronger for NBA franchises than for European soccer teams and Major League (g) (h)

12.73 (a)

b0 = –3.856 and b1 = 0.113 (b)

For each fast food outlet, obesity rates increase by 0.113 percentage points.

(c)

Ŷ = -3.856 + 0.113(350) = 35.70

(d)

r2 = SSR/SST = 1011.895/1046.4 = 0.9670 So, 96.70% of the variation in the dependent variable can be explained by the variation in the independent variable.

(e)

The residuals are well spread and show no particular pattern.

The simple linear regression model seems to fit well. (f)

H0 : b1 = 0

H1 : b1 ¹ 0

0.113 − 0 t= = 15.317 0.0074

Critical values: t8 = ±2.306 Decision rule: Reject H0 if |tcalc|> 2.306 Since |15.317| > 2.306, reject H0. There is sufficient evidence to conclude that there is a linear relationship between obesity rates and the number of fast-food outlets in a region. (g)

Ŷi ∓ tn-2 SYX hi 1 ( X i - X )2 + n SSX Ŷ = -3.856 + 0.113(350) = 35.70

hi =

1 (350 - 409.3)2 + = 0.1444 10 79228.1

t8=2.306

95% confidence interval for the mean:

35.70 ± 2.306 * 2.0768 * 0.1444 33.88 £ mY|X =350 £ 37.52 (h)

Ŷi ∓ tn-2 SYX 1+ hi 95% prediction interval for an individual response:

35.70 ± 2.306 * 2.0768 * 1.1444 30.58 £ YX =350 £ 40.82 (i)

b1 - t10 Sb £ b1 £ b1 + t10 Sb 1

0.113 - 2.306(0.0074) £ b1 £ 0.113 + 2.306(0.0074) 0.0960 £ b1 £ 0.1300 (j) 12.74 (a)

Other independent variables that can be included in the regression are variables relating to exercise and other aspects of diet. b0 = 64.1888 and b1 = 1.3046

(b)

For each additional degree of increase in temperature, the number of golfers increases by 1.3046.

(c)

Ŷ = 64.1888 +1.3046 X Ŷ = 64.1888 +1.3046(25) = 96.81

(d)

r2 = SSR/SST = 907.0473/6280.9333 = 0.1444 So, 14.44% of the variation in the dependent variable can be explained by the variation in the independent variable.

(e)

The residuals show no particular pattern.

The simple linear regression model seems to fit reasonably well.

13.85 13.86 ( b )

12.75

(a)

The correlation between compensation and the investment return is 0.1719.

H 0

v s .

H1 :   0

(b)

The tSTAT value is 2.2615 with a p-value = 0.0250 < 0.05, reject H0 . The correlation between compensation and the investment return is statistically significant. The small correlation between compensation and stock performance was surprising (or maybe it should not have been!).

Chapter 13: Introduction to multiple regression Learning objectives After studying this chapter you should be able to: 1. construct a multiple regression model and analyse model output 2. differentiate between independent variables and decide which ones to include in the regression model, and determine which independent variables are more important in predicting a dependent variable 3. incorporate categorical and interactive variables in a regression model 4. detect collinearity 13.1

(a)

(b) 13.2

13.3

13.4

(a)

Holding constant the effect of X2, for each increase of one unit in X1, the response variable Y is estimated to increase a mean of 8 units. Holding constant the effect of X1, for each increase of one unit in X2, the response variable Y is estimated to decrease an average of 3 units. The Y intercept 10 is the estimate of the mean value if X1 and X2 are both 0.

(b)

Holding constant the effect of X2, for each increase of one unit in X1, the response variable Y is estimated to increase an average of 10 units. Holding constant the effect of X1, for each increase of one unit in X2, the response variable Y is estimated to decrease an average of 15 units. The Y intercept 100 is the estimate of the mean value of Y if X1 and X2 are both 0.

(a)

Yˆ = 0.02686 + 0.79116X 1 + 0.60484X 2

(b)

For a given measurement of the change in impact properties over time (holding constant the effect of X2), each increase in one unit in forefoot shock-absorbing capability, X1, is estimated to result in a mean increase in the long-term ability to absorb shock of 0.79116 units. For a given forefoot shock-absorbing ability (holding constant the effect of X1), each increase of one unit in measurement of the change in impact properties over time is estimated to result in a mean increase in the long-term ability to absorb shock by 0.60484 units.

(a)

Yˆ = 9.8156 + 1.47693X 1 + 0.10344X 2

(b)

Holding the effect of stock market rate, X2, constant, for each increase of 1% in unemployment rate, is estimated to result in a mean increase in the retirement rate by 1.477%. Holding the effect of unemployment rate, X1, constant, for each increase of 1% in the stock market return rate, is estimated to result in a mean increase in the retirement rate by 0.1034%. The interpretation of b0 has no practical meaning here because it would have been the estimated mean retirement rate when there were no unemployment rate and no stock market return rate. However, this does not allow for any other influence on retirement rate.

(c)

(d)

Yˆ = 9.8156 + 1.47693(6) + 0.10344(5) = 19.1944%

13.5

(e)

15.2758  Y|X  23.1130

(f)

6.4057  YX  31.98312

(a)

The predicted sign for b1 will be positive since the higher the GDP we would expect higher CO2 emissions. Similarly, the predicted sign for b2 will be positive; the higher the population density we would expect higher CO2 emissions.

(b)

Yˆ = 151.3935 + 204.6397 X1 − 0.1639 X 2

(c)

For a given population density, each increase of GDP US$1 trillion is estimated to result in a mean increase of CO2 emission of 204.6397 million metric tonnes. For a given GDP, each increase in population density of 1 person per kilometre is estimated to result in a mean decrease of 0.1639 million metric tonnes. The interpretation of b0 has no practical meaning here because it would have been the estimated mean CO2 emission for a country with zero GDP and population density.

(d)

(f)

Yˆ = 151.3935 + 204.6397(1) − 0.1639(50) = 347.8404 million metric tonnes −1677.83  Y | X  1030.133

(g)

−1677.83  YX  2373.51

(e)

13.6

PHStat output:

Interce pt

alcohol

chlorid es

C o e f f i c i e n t s 1 . 1 5 9 2 0 . 4 9 6 2 9 . 6 3

Sta nd ard Err or

1.27 19

0 . 9 1 1 4 4 . 5 3 7 8

0.10 94

3.68 18

S t a t

2 . 6 1

P v a l u e

0 . 3 6 6 7 0 . 0 0 0 0 0 . 0 1 1

3 1

6 4

Yˆ = 1.1592 + 0.4962 X1 − 9.63 31 X2

(b)

(c)

13.7

For a given amount of chlorides, each increase of one percent in alcohol is estimated to result in a mean increase in quality rating of 0.4962. For a given alcohol content, each increase of one unit in chlorides is estimated to result in the mean decrease in quality rating of 9.6331. The interpretation of b0 has no practical meaning here because it would have meant the estimated mean quality rating when a wine has 0 alcohol content and 0 amount of chlorides.

(d)

Yˆ = 1.1592 + 0.4962 (10 ) − 9.6331 ( .08 ) = 5.3510.

(e)

5.0635   Y | X  5.6386

(f) (g)

3.5484  YX  7.1536 The model uses both alcohol content (%) and the g) The model uses both alcohol content (%) and the amount of chlorides to predict wine quality. This may produce a better model than if only one of these independent variables is included.

(a)

Yˆ = 39.45 − 0.0003X1 + 0.1526 X 2

(b)

For a given CPI, each increase in GDP/capita of $1 is estimated to result in a mean decrease in the percentage of very happy citizens of 0.0003 percentage points. For a given GDP/capita, each one-point increase in CPI is estimated to result in a mean increase in the percentage of very happy citizens by 0.1526 percentage points. The interpretation of b0 has no practical meaning here because it would have been the estimated mean percentage of very happy citizens when GDP/capita and CPI were zero.

(c)

(e)

Yˆ = 0.3945 − 0.000003(35000) + 0.0015(75) = 40.03 37.28  Y | X  42.77

(f)

30.33  YX  49.72

(d)

13.8

(a) Yˆ = 156 .4 + 13 .081 X1 + 16 .795 X2 (b) For a given amount of newspaper advertising, each increase of $1000 in radio advertising is estimated to result in a mean increase in sales of $13,081. For a given amount of radio advertising, each increase of $1000 in newspaper advertising is estimated to result in the mean increase in sales of $16,795. (c)When there is no money spent on radio advertising and newspaper advertising, the estimated mean amount of sales is $156,430.44. (d) According to the results of (b), newspaper advertising is more effective as each increase of $1000 in newspaper advertising will result in a higher mean increase in sales than the same amount of increase in radio advertising. 13.9

(a)

MSR = SSR/k = 55/2 = 27.5 MSE = SSE/(n – k – 1) = 145/18 = 8.06

(b) F = MSR/MSE = 27.5/8.06 = 3.41 (c) F = 3.41 < FU(2,18) = 3.555. Do not reject H0. There is no evidence of significant linear relationship. (d)

R2 =

SSR 55 = = 0.2895 SST 190

28.95% of the variation in y is explained by the model (e)

n −1  21 − 1    2 Radj = 1 − (1 − R2 ) = 1 − (1 − 0.2895) = 0.2105  n − k − 1 21 − 2 − 1  

13.10 (a)

MSR = SSR/k = 185/2 = 92.5 MSE = SSE/(n-k-1) = 315/10 = 31.5 (b) F = MSR/MSE = 92.5/31.5 = 2.9365 (c) F = 2.9365 > FU(2,8) = 4.46. Do not reject H0. There is not sufficient evidence of a significant linear relationship. (d) (e)

13.11 (a)

(b) (c)

13.12 (a) (b) (c)

SSR 185 = = 0.37 SST 500 n −1  11 − 1    2 Radj = 1 − (1 − R 2 ) = 1 − (1 − 0.37) = 0.2125  n − k − 1 11 − 2 − 1  

R2 =

72% of the total variability in enjoyment can be explained by length of stay after adjusting for the number of predictors and sample size. 78% of the total variability in enjoyment can be explained by average income after adjusting for the number of predictors and sample size. 68% of the total variability in enjoyment can be explained by both length of stay and average income after adjusting for the number of predictors and sample size. 2 Model 2 is the best predictor of enjoyment because it has the highest adjusted R . The regression coefficients are needed to explain the relationships, especially the b slope coefficients. For example, if the beta of the length of stay is negative and the beta for average income is negative, it might be that the circle trams appeal most to the tourists with little time and not much money. F = 97.69 > FU(2,15-2-1) = 3.89. Reject H0. There is evidence of a significant linear relationship with at least one of independent variables. p-value is virtually zero. The probability of obtaining an F test statistic of 97.69 or larger is virtually zero if H0 is true.

R2 =

SSR 12.61020 = = 0.9421. So, 94.21% of the variation in the long-term SST 13.3847

ability to absorb shock can be explained by variation in forefoot absorbing capability and variation in midsole impact. (d) 13.13 (a) (b) (c)

n −1  15 − 1    2 Radj = 1 − (1 − R 2 ) = 1 − (1 − 0.9421) = 0.93245  n − k − 1 15 − 2 − 1   F = MSR/MSE = 6.39 < FU(2,11) = 3.98. Reject H0 and conclude there is evidence of a significant relationship. p-value = 0.014. The probability of obtaining an F test statistic of 6.39 or larger is 0.014 if H0 is true.

R2 =

SSR 9594740 = = 0.5374 . So, 53.74% of the variation in CO2 emissions can SST 17855076

(d) 13.14 (a) (b) (c)

n −1  14 − 1    2 Radj = 1 − (1 − R2 ) = 1 − (1 − 0.5374) = 0.4533  n − k − 1 14 − 2 − 1   F = MSR/MSE = 124.5622939/31.21739546 = 3.990157 > FU(2,12) = 3.89. Reject H0 and conclude that there is enough evidence of a significant linear relationship. p-value = 0.047. The probability of obtaining an F test statistic of 3.9902 or larger is 0.047 if H0 is true.

R2 =

SSR 249.12 = = 0.3994 . So, 39.94% of the variation in retirement rates can SST 623.73

be explained by the variation in unemployment rates and stock market returns. (d) 13.15 (a) (b) (c)

n −1  15 − 1    2 Radj = 1 − (1 − R 2 ) = 1 − (1 − 0.3994) = 0.2993  n − k − 1 15 − 2 − 1   F = MSR/MSE = 95.841/17.401= 5.508 < FU(2,10) = 4.10. Reject H0 and conclude that there is enough evidence of a significant linear relationship. p-value = 0.024. The probability of obtaining an F test statistic of 5.508 or larger is 0.024 if H0 is true.

R2 =

SSR 191.683 = = 0.5242 . So, 52.42% of the variation in very happy citizens SST 365.692

can be explained by GDP/capita and CPI. (d)

n −1  13 − 1    2 Radj = 1 − (1 − R2 ) = 1 − (1 − 0.5242) = 0.4290  n − k − 1 13 − 2 − 1  

13.16

(a)

Partial PHStat output: d f

S S

Regres sion

27.2 241

Residu al

4 7

36.7 759

Total

4 9

64.0 000

M S

1 3 . 6 1 2 0 0 . 7 8 2 5

Si g ni fic a nc e F 1 7 . 3 9 6 3

0 . 0 0 0 0

= M S R / M S E = 17.3963

FS T A T

Since p-value = 0.0000 < 0.05, reject H0. There is evidence of a significant linear relationship. p-value = 0.0000. The probability of obtaining an F test statistic of 17.3963 or larger is 0.0000 if H0 is true.

(b)

R2 =

S o ,

SSR 27.2241 = = 0.4254 SST 64 %

the variation in quality rating can be

o f explained by variation in the percentage of alcohol and variation in chorides.

(d)

n −1   2 Radj = 1 − (1 − R2 ) = 0.4009 n − k − 1 

1 3 . 1 7

M SR = SSR / k = 2, 028, 033 / 2 = 1, 014, 016 M SE = SSE /(n − k − 1) = 479, 759.9 / 19 = 25, 251

( a ) FSTAT

FSTAT

M S R

M S E



4 0

3 . 5 2 2

. 1 6

(b)



= 1,014 ,016 / 25 ,251

= 40 .16

. Reject H0. There is evidence of a significant linear

relationship. p-value < 0.001. The probability of obtaining an F test statistic of 40.16 or larger is less than 0.001 if H0 is true.

SSR 2028033 = = 0.8087 80.87% of the variation in sales can be explained by variation SST 2507793 in radio per advertising.

(c)

R2 =

(d)

n −1   2 Radj = 1 − (1 − R2 ) = 0.7886 n − k − 1 

13.18

From the plot residuals versus predicted Y, we can see that it does not show a pattern for different predicted values of Y and the model appears to be adequate.

There is no evidence of a pattern in the residual versus stock market return.

There is no evidence of a pattern in the residual against unemployment rate. 13.19 Excel output

All the plots above show random pattern and there is evidence that the model is adequate. Copyright © 2019 Pearson Australia (a division of Pearson Australia Group Pty Ltd) 9781488617249, Berenson, Basic Business Statistics 5e

13.20

Residual Vs. Fitted Value 3 2.5 2 1.5 1 0.5 0 -0.5 0 -1 -1.5 -2 -2.5

4.7969690126, 0.796990126

Fitted Values

Residual Plot for alcohol 3 2.5 2 1.5 1 0.5 0 -0.5 -1 -1.5 -2 -2.5 0

alcohol

Residual Plot for chlorides 3 2.5 2 1.5 1 0.5 0 -0.5 -1 -1.5 -2 -2.5 0

0.05

0.1

0.15 chlorides

0.2

0.25

0.3

The residual plots do not reveal any specific pattern.

13.21

All the plots above show random pattern and there is no evidence in the plot to suggest a non-linear relationship. There is evidence that the model is adequate.

13.22

R e s i dua l s V e r s us R a di o (r e s p o ns e is S a le s ) 300

200

100

-100

-200

-300 0

R a d io

R e s i dua l s V e r s us N e w s pa pe r (r e s p o ns e is S a le s ) 300

200

100

-100

-200

-300 0

Ne w s p a p e r

There appears to be a quadratic relationship in the plot of the residuals against the fitted value and both radio and newspaper advertising. Thus, quadratic terms for each of these explanatory models should be considered for inclusion in the model. The normal probability plot suggests that the distribution of the residuals is very close to a normal distribution. 13.23 (a) (b)

The slope of X2 in terms of t statistic is 4 which is larger than the slope of X1 in terms of t statistic which is 2.8. 95% confidence interval of  1 : b1  t n−k −1 S  , 7  2.018(2.5) 1

1.955  1  12 .045 (c)

13.24 (a)

For X1 : t = b1/Sb1 = 7/2.5 = 2.8 > t42 = 2.018 with 42 degrees of freedom for α = 0.05. Reject H0. There is evidence that the variable X1 contributes to a model already containing X2. For X2 : t = b2/Sb2 = 6/1.5 = 4 > t42 = 2.018 with 42 degrees of freedom for α = 0.05. Reject H0. There is evidence that the variable X2 contributes to a model already containing X1. On the basis of these results, both variables X1 and X2 should be included in the model. 95% confidence interval of  1 : b1  t n−k −1 S  , 0.79  2.17(0.063) 1

0.6533  1  0.926 (b)

For X1 : t = b1/Sb1 = 0.79/0.063 = 12.57 > t12 = 2.17 with 12 degrees of freedom for α = 0.05. Reject H0. There is evidence that the variable X1 contributes to a model already containing X2. For X2 : t = b2/Sb2 = 0.605/0.071 = 8.43 > t12 = 2.17 with 12 degrees of freedom for α = 0.05. Reject H0. There is evidence that the variable X2 contributes to a model already containing X1. On the basis of these results, both variables X1 and X2 should be included in the model.

13.25 (a)

95% confidence interval of  1 : b1  t n−k −1 S  , 1.477  2.17(0.566) 1

0.244  1  2.71 (b)

13.26 (a)

For X1 : t = b1/Sb1 = 1.477/0.566 = 2.61 > t12 = 2.17 with 12 degrees of freedom for α = 0.05. Reject H0. There is evidence that the variable X1 contributes to a model already containing X2. For X2 : t = b2/Sb2 = 0.103/0.891 = 0.1161 < t12 = 2.17 with 12 degrees of freedom for α = 0.05. Do not reject H0. There is not enough evidence that the variable X2 contributes to a model already containing X1. On the basis of these results, only variables X1 should be included in the model. 95% confidence interval of  1 : b1  t n−k −1 S  , 1

204.6397  2.201(57.4754) 78.1371  1  331.1422 (b)

For X1 : t = b1/Sb1 = 204.6397/57.4754 = 3.5605 > t11 = 2.201 with 11 degrees of freedom for α = 0.05. Reject H0. There is evidence that the variable X1 contributes to a model already containing X2. For X2 : t = b2/Sb2 = –0.1639/1.5509 = –0.1057 > t11 = -2.201 with 11 degrees of freedom for α = 0.05. Do not reject H0. There is not enough evidence that the variable X2 contributes to a model already containing X1. On the basis of these results, only variables X1 should be included in the model. 13.27

(a)

PHStat output:

Interce pt

alcohol

chlorid es

C o e f f i c i e n t s 1 . 1 5 9 2 0 . 4 9 6 2 9 .

Sta nd ard Err or

1.27 19

0 . 9 1 1 4 4 . 5 3 7 8

0.10 94

3.68 18

S t a t

2 .

P v a l u e

0 . 3 6 6 7 0 . 0 0 0 0 0 . 0

6 3 3 1

6 1 6 4

1 1 9

0.2762   1  0.7162 For X1 : t = b1/Sb1 = 0.4962/0.1094 = 4.5378 > t47 = 2.0117 with 47 degrees of freedom for α = 0.05. Reject H0. There is evidence that the X1 alcohol contributes to a model already containing X2 chlorides. For X2 : t = b2/Sb2 = -9.6331/3.6818 = –2.6164 < t47 = -2.0117 with 47 degrees of freedom for α = 0.05. Reject H0. There is evidence that the variable X2 chlorides contributes to a model already containing X1 alcohol.Both variables should be included in the model.

13.28 (a) (b)

95% confidence interval of  1 : b1  t n−k −1 S  , 1

−0.0003  2.2281(0.00009) −0.0005  1  −0.0001

For X1 : t = b1/Sb1 = -0.0003/0.00009 = –3.3174 < t10 = –2.2281 with 10 degrees of freedom for α = 0.05. Reject H0. There is evidence that the variable X1 contributes to a model already containing X2. For X2 : t = b2/Sb2 = 0.1526/0.0702 = 2.1729 < t10 = –2.2281 with 10 degrees of freedom for α = 0.05. Do not reject H0. There is not enough evidence that the variable X2 contributes to a model already containing X1. On the basis of these results, only X1 should be included in the model.

13.29 (a) 13.0807  2.093 (1.7594 )

9.398  1  16.763 (b) For X1 : t = b1/Sb1 = 13.0807/1.7594 = 7.43 > t19 = 2.093 with 19 degrees of freedom for α = 0.05. Reject H0. There is evidence that the X1 radio advertsising contributes to a model already containing X2 newspaper. For X2 : t = b2/Sb2 = 1.7649/0.379 = 4.66 > t19 = 2.093 with 19 degrees of freedom for α = 0.05. Reject H0. There is evidence that the variable X2 newspaper advertising contributes to a model already containing X1 radio advertising.Both variables should be included in the model.

13.30 (a)

For X1: SSR( X 1 | X 2 ) = SSR( X 1 andX 2 ) − SSR( X 2 ) = 60 − 25 = 35

SSR( X 1 | X 2 ) 35 = = 5.25  FU (1,18) = 4.41 with 1 and 18 degrees of MSE 120 / 18

freedom and α = 0.05. Reject H0. There is evidence that the variable X1 contributes to a model already containing X2. For X2: SSR( X 2 | X 1 ) = SSR( X 1 andX 2 ) − SSR( X 1 ) = 60 − 45 = 15

SSR( X 2 | X 1 ) 15 = = 2.25  FU (1,18) = 4.41 with 1 and 18 degrees of MSE 120 / 18

freedom and α = 0.05. Do not reject H0. There is no sufficient evidence that the variable X2 contributes to a model already containing X1. Since variable X2 does not significantly contribute to a model in the presence of X1, only variable X1 should be included and a simple linear regression model is developed. (b)

RY21, 2 =

SSR( X 1 | X 2 ) 35 = = 0.2258 SST − SSR( X 1andX 2 ) + SSR( X 1 | X 2 ) 180 − 60 + 35

Holding constant the effect of variable X2, 22.58% of the variation in Y can be explained by the variation in variable X1.

RY22,1 =

SSR( X 2 | X 1 ) 15 = = 0.1111 SST − SSR( X 1andX 2 ) + SSR( X 2 | X 1 ) 180 − 60 + 15

Holding constant the effect of variable X1, 11.11% of the variation in Y can be explained by the variation in variable X2. 13.31 (a)

For X1: SSR( X 1 | X 2 ) = SSR( X 1 andX 2 ) − SSR( X 2 ) = 234 − 97 = 137

SSR( X 1 | X 2 ) 137 = = 5.147  FU (1,13) = 4.67 with 1 and 13 degrees of SSE 346 /13

freedom and α = 0.05. Reject H0. There is evidence that the variable X1 contributes to a model already containing X2. For X2:

SSR( X 2 X1 ) = SSR( X1 and X 2 ) − SSR( X1 ) = 234 −124 = 110 SSR( X 2 | X 1 ) 110 F= = = 4.133  FU (1,13) = 4.67 SSE 346 /13 with 1 and 13 degrees of freedom and α = 0.05. Do not reject H0. There is no sufficient evidence that the variable X2 contributes to a model already containing X1. Since variable X2 does not significantly contribute to a model in the presence of X1, based on these results only variable X1 should be included and a simple linear regression model is developed. (b)

RY21,2 =

SSR( X 1 X 2 ) 137 = = 0.2836 SST − SSR( X 1 and X 2 ) + SSR( X 1 X 2 ) 580 − 234 + 137

Holding constant the effect of variable X2, 28.36% of the variation in Y can be explained by the variation in variable X1.

RY22,1 =

SSR( X 2 X 1 ) 110 = = 0.2412 SST − SSR( X 1 and X 2 ) + SSR( X 2 X 1 ) 580 − 234 + 110

Holding constant the effect of variable X1, 24.12% of the variation in Y can be explained by the variation in variable X2.

13.32 (a)

For

SSR( X1 X 2 ) = SSR( X1 and X 2 ) − SSR( X 2 ) = 9594740.095–

X1:

75100.326=9519639.769

SSR ( X 1 X 2 ) MSE

9519639.769 = 12.677 >FU(1,11) = 4.84 with 1 and 11 degrees 8260336.027 / 11

of freedom and α = 0.05. Reject H0. There is evidence that the variable X1 contributes to a model already containing X2. For X2: SSR( X 2 X1 ) = SSR( X1 and X ) − SSR( X1 ) = 9594740.095 – 9586357.683 = 2

8382.4125

SSR ( X 2 X 1 ) MSE

8382.4125 = 0.0112 < FU (1,11) = 4.84 with 1 and 11 degrees 8260336.027 / 11

of freedom and α = 0.05. Do not reject H0. There is no sufficient evidence that the variable X2 contributes to a model already containing X1. Since variable X2 does not significantly contribute to a model in the presence of X1, only variable X1 should be included and a simple linear regression model is developed. (b)

RY21,2 =

SSR( X 1 X 2 ) SST − SSR( X 1 and X 2 ) + SSR( X 1 X 2 )

9519639.769 = 0.5354 17855076.12 - 9594740.095 + 9519639.769

Holding constant the effect of variable X2, 53.54% of the variation in Y can be explained by the variation in variable X1.

RY22,1 =

SSR( X 2 X 1 ) SST − SSR( X 1 and X 2 ) + SSR( X 2 X 1 )

8382.4125 = 0.0010 17855076.12 - 9594740.095 + 8382.4125

Holding constant the effect of variable X1, 0.10% of the variation in Y can be explained by the variation in variable X2. 13.33 (a)

For X1: SSR( X 1 | X 2 ) = SSR( X 1 andX 2 ) − SSR( X 2 ) = 249 .125 − 36 .48 = 212 .65

SSR( X 1 | X 2 ) 212.65 = = 6.81  FU (1,13) = 4.67 MSE 31.22

with

and

degrees

freedom and α = 0.05. Reject H0. There is evidence that the variable X1 contributes to a model already containing X2. For X2: SSR( X 2 | X 1 ) = SSR( X 1 andX 2 ) − SSR( X 1 ) = 249 .125 − 248 .704 = 0.421

SSR( X 2 | X 1 ) 0.421 = = 0.013  FU (1,13) = 4.67 MSE 31.22

with

and

degrees

(b) RY 1, 2 = 2

SSR( X 1 | X 2 ) 212.65 = = 0.3621Holdin SST − SSR( X 1andX 2 ) + SSR( X 1 | X 2 ) 623.73 − 249.125 + 212.65

g constant the effect of variable X2, 36.21% of the variation in Y can be explained by the variation in variable X1.

RY22,1 =

SSR( X 2 | X 1 ) 0.421 = = 0.0011 SST − SSR( X 1andX 2 ) + SSR( X 2 | X 1 ) 623.73 − 249.125 + 0.421

Holding constant the effect of variable X1, 0.11% of the variation in Y can be explained by the variation in variable X2 13.34 (a)

For X1: SSR( X1 X 2 ) = SSR( X1 and X ) − SSR( X 2 ) = 191.683 – 0.189 = 191.494 2

SSR ( X 1 X 2 ) MSE

191.494 = 11.0048  FU (1,10) = 4.96 with 1 and 10 degrees of 17.401

freedom and α = 0.05. Reject H0. There is sufficient evidence that the variable X1 contributes to a model already containing X2. For X2: SSR( X 2 X1 ) = SSR( X1 and X ) − SSR( X1 ) = =365.692 – 109.526 = 82.157 2

SSR ( X 2 X 1 ) MSE

82.157 = = 4.72  FU (1,10) = 4.96 with 1 and 10 degrees of freedom 17.401

and α = 0.05. Do not reject H0. There is no sufficient evidence that the variable X2 contributes to a model already containing X1. Since variable X2 does not significantly contribute to a model in the presence of X1, based on these results only variable X1 should be included and a simple linear regression model is developed. (b)

RY22,1 =

SSR( X 1 X 2 ) 191.494 = = 0.5239 SST − SSR( X 1 and X 2 ) + SSR( X 1 X 2 ) 365.692 − 191.683 + 191.494

Holding constant the effect of variable X2, 52.39% of the variation in Y can be explained by the variation in variable X1.

RY22,1 =

SSR( X 2 X 1 ) 82.157 = = 0.3207 SST − SSR( X 1 and X 2 ) + SSR( X 2 X 1 ) 365.692 − 191.683 + 82.157

Holding constant the effect of variable X1, 32.07% of the variation in Y can be explained by the variation in variable X2. 13.35

(a) F=

For X1: SSR( X1 X 2 ) = SSR( X1 and X ) − SSR( X 2 ) = = 27.2241 - 11.1119 = 16.1122 2

SSR( X 1 X 2 ) MSE

16.1122 = 20.5916 0.7825

with 1 and 47 degrees of freedom, and pvalue = 0.0000. Reject H0. There is sufficient evidence that the variable percentage alcohol contributes to a model already containing chorides.

For X2: SSR( X 2 X1 ) = SSR( X1 and X ) − SSR( X1 ) = = 27.2241 - 21.8677 = 5.3564 2

SSR ( X 2 X 1 ) MSE

5.3564 = 0.68455 0.7825

with 1 and 47 degrees of freedom and p-value = 0.0119. Reject H0. There is enough evidence that the variable chorides contributes to a model already containing percentage alcohol. Since both percentage alcohol and chlorides make a significant contribution to the model in the presence of the other, the most appropriate regression model for this data set should include both percentage alcohol and chlorides.

( b )

RY22,1 =

SSR( X 1 X 2 ) SST − SSR( X 1 and X 2 ) + SSR( X 1 X 2 )

16.1122 64 − 27.2241 + 16.1122

= 0.3046. Holding constant the effect of chlorides, 30.46% of the variation in quality rating can be explained by the variation in percentage alcohol.

SSR( X 2 X 1 ) 5.3564 = == 0.1271. Holding constant the SST − SSR( X 1 and X 2 ) + SSR( X 2 X 1 ) 64 − 27.2241 + 5.3564 effect of percentage alcohol, 12.71% of the variation in quality rating can be explained by the variation in chlorides. RY22,1 =

13.36 (a) For X1: SSR( X1 X 2 ) = SSR( X1 and X 2 ) − SSR( X 2 ) = 2, 028, 033 − 632, 259.4 = 1, 395, 773.6 F=

SSR( X1 X 2 ) MSE

1395773.6 = 55.28 >F(1,19)=4.381 479 ,759 .9 /19

Reject H0. There is evidence that the variable X1 contributes to a model already containing X2.

For X2: For X2: SSR( X 2 X1 ) = SSR( X1 and X ) − SSR( X1 ) = = 2,028,033 – 1,216,940 = 811,093 2

SSR( X 2 X1 ) MSE

811,093 = 32.12 >FU(1,19) = 4.381 with 1 and 19 degrees of freedom 9. Reject H0. 479,759.9

There is enough evidence that the variable X2 contributes to a model already containing X1. Since both variables make a significant contribution to the model in the presence of the other, the most appropriate regression model for this data set should include both variables.

(b) RY21,2 =

SSR( X 1 X 2 ) SST − SSR( X 1 and X 2 ) + SSR ( X 1 X 2 )

1,395, 773.6 = 0.7442 2,507, 793 − 2, 028, 033 + 1,395, 773.6

Holding constant the effect of newspaper advertising, 74.42% of the variation in Y can be explained by the variation in radio advertising.

RY22,1 =

SSR( X 2 X 1 ) 811, 093 = = 0.6283 SST − SSR( X 1 and X 2 ) + SSR( X 2 X 1 ) 2,507, 793 − 2, 028, 033 + 811, 093

Holding constant the effect of radio advertising, 62.83% of the variation in Y can be explained by the variation in newspaper advertising.

13.37 (a) (b) (c) 13.38 (a)

(b)

13.39 (a) (b)

(c)

Holding constant the effect of X2, the estimated mean value of the dependent variable will increase by 5 units for each increase of one unit of X1. Holding constant the effect of X1, the presence of the condition represented by X2 = 1 is estimated to increase the mean value of the dependent variable by 0.5 units. t = 2.67 > t32 = 2.037. Reject H0. The presence of X2 makes a significant contribution to the model. First develop a multiple regression model using X1 as the variable amount spent on advertising and X2 a dummy variable with X2 = 1 if the music has air-time on ratio. If the dummy variable coefficient is significantly different from zero, you need to develop a model with the interaction term X1X2 to make sure that the coefficient of X1 is not significantly different if X2 = 0 or X2 = 1. If the music receives air-time on radio, the sales would be estimated to be 0.30 greater than had the same amount been spent on advertising, but without air-time on radio.

Yˆ = −6.456 − 199.1872X 1 + 69.4149X 2 X1 = Location (east = 0), X2 = Number of rooms Holding the effect of neighbourhood constant, for each additional room, the selling price is estimated to increase by a mean of $69,414 000. For a given number of rooms, a western city side is estimated to decrease the selling price over an eastern city location by $199,187.

Yˆ = −6.456 − 199.1872(0) + 69.4149(9) = $618,278 554.4951  YX=Xi  682.0602

428.005  Y|X=Xi  808.5503

(d)

The models appear to be adequate from residual analysis. (e) (f)

(g) (h)

F = 71.1155 > FU(2,17) = 3.59. Reject H0. There is evidence of a relationship between selling price and the two independent variables. For X1: t = –4.622 < –t17 = -2.110. Reject H0. Location makes a significant contribution and should be included in the model. For X2: t = 7.5143 > t17 = 2.110. Reject H0. Number of rooms makes a significant contribution and should be included in the model.

− 290 .1025  1  −108 .272 ,49 .9251   2  88 .9047 SSR R2 = = 0.9451 . So, 94.51% of the variation in selling price is explained by SST

variation in the two independent variables. (i)

2 Radj = 0.8932

(j)

RY21, 2 =

SSR( X 1 | X 2 ) = 0.5569 . Holding constant the effect of SST − SSR( X 1andX 2 ) + SSR( X 1 | X 2 )

number of rooms, 55.69% of selling price can be explained by location.

RY22,1 =

SSR( X 2 | X 1 ) = 0.7686 . Holding constant the effect SST − SSR( X 1andX 2 ) + SSR( X 2 | X 1 )

(k)

of location, 76.86% of selling price can be explained by number of rooms. The slope of selling price with number of rooms is the same regardless of whether the house is located in the east or west of Melbourne.

(l)

Yˆ = −354.5434 − 274.4005X 1 + 102.5660X 2 − 48.2717X 1 X 2 .

(m)

13.40 (a) (b)

For X1X2 the p-value = 0.01. Reject H0. It is likely that the size of houses and location interact, and the two independent variables are not independent of each other. The regression results indicate that the two variables model can be used as both independent variables do contribute to an explanation of selling price. However, there is a risk of lack of independence between the two independent variables and the model could be run with just number of rooms (the higher partial coefficient of determination).

Yˆ = 16.5613 + 0.1529X 1 − 6.9086X 2 Holding constant effect of traffic lights for each additional one-unit increase in traffic volume, accidents increase by 0.1529 units.

For a given number of traffic volume, an intersection with traffic lights is estimated to decrease the accidents over a no traffic lights location by a mean of 6.9 units. (c)

Yˆ = 16.5613 + 0.1529(18) − 6.9086(1) = 12.4053 1.3684  YX=Xi  2.4156

1.6880  Y|X=Xi  2.096 (d) (e) (f)

(g) (h)

The model appears to be adequate from residual analysis. F = 4.5799 > F2,12 = 3.89. Reject H0. There is evidence of a relationship between accidents and two independent variables. For X1: t = 1.0538 < 2.179. Do not reject H0. Likely that traffic volume does not contribute significantly to explaining accidents and could be excluded from the model. For X2: t = 2.6160 > 2.179. Reject H0. Traffic lights do make significant contribution to the explanation of accidents and should be included in the model.

− 0.1633  1  0.4691 ,−12 .6083   2  −1.2087 SSR R2 = = 0.6579 . So, 65.79% of the variation in accidents are explained by SST

variation in the two independent variables. (i)

2 Radj = 0.3384 . Note that the sample size is small so adjusted R2 is much lower than

R2. (j)

RY21, 2 =

SSR( X 1 | X 2 ) = 0.0847 . SST − SSR( X 1andX 2 ) + SSR( X 1 | X 2 )

Holding

traffic

lights

constant, 8.47% of the variation in accidents can be explained by traffic volume.

RY22,1 =

SSR( X 2 | X 1 ) = 0.3676 . SST − SSR( X 1andX 2 ) + SSR( X 2 | X 1 )

Holding

traffic

volume

(k)

constant 36.76% of the variation in accidents can be explained by traffic lights. The assumption for the slope on volume of traffic is that there is no change in the variable traffic lights.

(l)

Yˆ = 4.2182 + 0.6533X 1 + 8.3687X 2 − 0.6344X 1 X 2

(m)

For X1X2 the p-value = 0.0716. Do not reject H0 at 5% significance level. It is unlikely that traffic volume and traffic lights interact. The regression result indicates that the model should only contain one independent variable traffic lights. Traffic volume is not significantly related to accidents. Because of low partial R2 of 36.76% for traffic lights there must be some other variables to explain the variation in accidents that have not been measured. 13.41

PHStat output: Regression Statistics Mult iple R

R Squ are

0 . 5 0 6 8 0 . 2 5

Adju sted R Squ are Stan dard Erro r Obs erva tion s

6 8 0 . 2 4 1 5 1 . 0 5 0 9 1 0 0

ANO VA d f

S S

Regr essi on

3 7. 0 2 5 7

Resi dual

9 7

Tota l

9 9

1 0 7. 1 3 4 3 1 4 4. 1 6 0 0

C o e f f i c

S t a n d a r

M S

1 8 . 5 1 2 9 1 . 1 0 4 5

S i g n i f i c a n c e 1 6 . 7 6 1 7

t S t a t

P v a l u e

F 0 . 0 0 0 0

L o w e r

U p p e r

i e n t s Inter cept

0 . 9 3 4 2

alco hol

0 . 4 6 5 2 0 . 2 5 7 7

Type of Win e

d E r r o r 0. 8 7 7 0 0. 0 8 2 0 0. 2 1 0 2

5 %

1 . 0 6 5 2

0 . 2 8 9 4

5 . 6 7 6 2 1 . 2 2 5 8

0 . 0 0 0 0 0 . 2 2 3 2

0 . 8 0 6 4 0 . 3 0 2 5 0 . 6 7 4 9

5 %

2 . 6 7 4 7 0 . 6 2 7 8 0 . 1 5 9 5

(a)

Yˆ = 0.9342 + 0.4652 X − 0.25 77 X

(b)

Holding constant the effect of the type of wine, for each additional % increase in alcohol content, wine quality is estimated to increase by a mean of 0.4652. For a given amount of alcohol content, a white wine is estimated to have a 0.2577 higher mean quality than a red wine.

Yˆ = 0.9342 + 0.4652 (10 ) −

= 5.3283

0.2577 (1)

( c )

3.2196  YX = Xi  7.4371 5.0184  Y | X = Xi  5.6382 3 PHStat output: Residual Plot for alcohol 3 2 1 0 -1 -2 -3 -4 0

alcohol

Residual Plot for Type of Wine 3 2 1 0 -1 -2 -3 -4 0

0.2

0.4

0.6 0.8 Type of Wine

1.2

Normal Probability Plot 3 2 1 0 -1

Residual

-2 -3 -4

(d) ( e )

Based on a residual analysis, there is not any obvious pattern in the residual plots but the normal probability plot indicates departure from the normality assumption. = 16.7617 with a p-value = 0.0000. Reject H0. There is evidence of a relationship F S T A T

between quality and percentage of alcohol and the type of wine

(f)

= 5.6762 with a p-value = 0.0000. Reject H0. Alcohol content makes a

For X1:

tS T A T

significant contribution and should be included in the model. = -1.2258 with a p-value = 0.2232. Do not reject H0. The type of wine does F

o r X 2

: t S T A T

(g) (h) ( i )

not make a significant contribution and should not be included in the model. Only alcohol content should be kept in the model. 0.3025   1  0.6278, -0.6749   2  0.1595 The slope here takes into account the effect of the other predictor variable, type of wine, while the solution for Problem 13.4 did not. R2= 0.2568. So, 25.68% of the variation in quality can be explained by variation in

alcohol content and variation in the type of wine. 2 = 0.2415 Radj

( j )

0.2568 while R 2 = 0.3417 in Problem 13.16 (a).

RY21,2 = 0.2493. Holding constant the effect of wine type, 24.93% of the variation in

( k ) ( l )

(m) (n)

quality can be explained by variation in alcohol content. RY22,1 = 0.0153. Holding constant the effect of alcohol content, 1.53% of the variation in quality can be explained by variation in wine type. The slope of alcohol content is the same regardless of whether the wine is red or white. PHStat output: C o e f f i c i e

St an da rd Er ro r

t S t a t

P v a l u e

L o w e r

U p p e r

9 5 %

Intercept

alcohol

Type of Wine

alcohol X Type of Wine

Si nc e th et

n t s 1 . 6 7 8 0 0 . 3 9 4 7 2 . 0 3 0 9 0 . 1 6 7 8

1. 14 48

1 . 4 6 5 8

0 . 1 4 6 0

0. 10 76

3 . 6 6 6 7 1 . 1 4 9 4 1 . 0 1 0 7

0 . 0 0 0 4 0 . 2 5 3 3

1. 76 69

0. 16 60

for the significa nce of

0 . 3 1 4 7

0 . 5 9 4 4 0 . 1 8 1 0 5 . 5 3 8 2 0 . 1 6 1 7

3 . 9 5 0 3 0 . 6 0 8 3 1 . 4 7 6 4 0 . 4 9 7 3

has a p-value = 0.3147, do not reject H0.

X 2

STA T

T h e r e

(o) (p)

13.42 (a)

i s n o t

(b)

s a contribution to the model. The one-variable model should be used. Only the alcohol content is significant in predicting the wine quality.

Yˆ = 23.8016 − 0.5052X 1 − 2.955X 2 + 0.4528X 1 X 2 Where X1: Unemployment rate, X2: Stock market return, X1 X2 = interaction between unemployment rate and stock market return. For X1 X2: the p-value = 0.3187 > 0.05. Do not reject H0. There is not enough evidence that the interaction term makes a contribution to the model. Since there is not enough evidence of any interaction effect between unemployment rates and stock market return, the model in problem 13.4 should be used.

e v i 13.43 d e(a) Yˆ = −1293.3105 + 43.6600 X + 56.9335 X − 0.8430 X . n where X1 = radio advertisement, X2 = newspaper advertisement, X3 = X1 X2 c e X1X2: the p-value is 0.0018 < 0.05. Reject H0. There is enough evidence that the For interaction term makes a contribution to the model. t h a t t h e i n t e r a c t i o n t e r m m a k e

(b) Since there is enough evidence of an interaction effect between radio and newspaper advertisement, the model in this problem should be used. 13.44 (a)

(b)

13.45 (a)

(b)

Yˆ = 186.595 + 185.701X1 − 0.879 X 2 + 0.293 X1 X 2 Where X1: GDP, X2: population density, X1 X2 = interaction between GDP and population density For X1 X2: the p-value = 0.685 > 0.05. Do not reject H0. There is not enough evidence that the interaction term makes a contribution to the model. Since there is not enough evidence of any interaction effect between GDP and population density, the model in problem 13.5 should be used.

Yˆ = 29.08 + 0.0001X1 + 0.37 X 2 − 0.0000008X1 X 2 Where X1:GDP/capita, X2: CPI, X1 X2 = interaction between GDP/capita and CPI For X1 X2: the p-value = 0.0499 < 0.05. Reject H0. There is enough evidence that the interaction term makes a contribution to the model. Since there is enough evidence of an interaction effect between GDP/capita and CPI, the model including this interaction term should be used.

13.46 (a)

Interc ept

Profic iency

Classr oom

Onlin e

C o e f f i c i e n t s 6 3 . 9 8 1 3 1 . 1 2 5 8 2 2 . 2 8 8 7 8 .

St a n d ar d Er ro r

t S t a t

P v a l u e

16. 79 97

3 . 8 0 8 5

0 . 0 0 0 8

0.1 58 9

7 . 0 8 6 8 5 . 1 6 4 9

0 . 0 0 0 0 0 . 0 0 0 0

1 .

0 .

4.3 15 4

4.3 10

0 8 8 0

(b)

(c)

8 7 6 5

0 7 1 9

where X1 = proficiency exam, X2 = classroom dummy, X3 = online dummy Holding constant the effect of training method, for each point increase in proficiency exam score, the end-of-training exam score is estimated to increase by a mean of 1.1258 points. For a given proficiency exam score, the end-of-training exam score of a trainee who has been trained by the classroom method will have an estimated mean score that is 22.2887 points below a trainee that has been trained using the courseware app method. For a given proficiency exam score, the end-of-training exam score of a trainee who has been trained by the online method will have an estimated mean score that is 8.0880 points above a trainee that has been trained using the courseware app method Yˆ = −63.9813 + 1.1258 ( 100 ) = 48.5969

( d )

(e)

FSTAT = 31.77 with 3 and 26 degrees of freedom. The p-value is virtually 0. Reject H0 at 5% level of significance. There is evidence of a relationship between end-of-training exam score and the independent variables.

(f)

For X1: tSTAT = 7.0868 and the p-value is virtually 0. Reject H0. Proficiency exam score makes a significant contribution and should be included in the model. For X2: tSTAT = -5.1649 and the p-value is virtually 0. Reject H0. The classroom dummy makes a significant contribution and should be included in the model. and the p-value = 0.07186. Do not reject H0. There is not sufficient For

X3: t = 1 .8 7 6

(g)

evidence to conclude that there is a difference in the online method and the courseware app method on the mean end-of-training exam scores. Base on the above result, the regression model should use the proficiency exam score and the classroom dummy variable. 0.7992   1  1.4523 , −31.1591   2  −13.4182 , − 0.7719   3  16.9480

RY2123 = 0 .7 8 5 7 . 78.57% of the variation in the end-of-training exam score can be

( h )

explained by the proficiency exam score and the various training methods. ( i ) ( j )

2 = 0.7610 Radj

= 0 .6 5 8 9 . Holding constant the effect of training method, 65.89% of the

Y 1,23 variation

(k) (l) H 0:  4

in end-of-training exam score can be explained by variation in the proficiency exam score. RY22,13 = 0 .5 0 6 4 . Holding constant the effect of proficiency exam score, 50.64% of the variation in end-of-training exam score can be explained by the difference between classroom and courseware app methods. RY23,12 = 0 .1 1 9 3 . Holding constant the effect of proficiency exam score, 11.93% of the variation in end-of-training exam score can be explained by the difference between online and courseware app methods. The slope of end-of-training exam score with proficiency score is the same regardless of the training method. Let X4 = X1X2, X5 = X1X3. =

There is no interaction among X1 , X2 and X3.

 5

= 0 H 1 : At

least one of  4

a n d

is not zero.

 5

STAT

SSR ( X , X | X , X , X ) 5 1 2 3 = 4 = 1 1 2 3 4 M S , , , , E

( X

X 5

SSR ( X , X , X , X , X ) − SSR ( X , X , X )/ 2 2

M S E

(

4, X 5

= 0 . 8 1 13.47 2 2 . 13.48

T h 13.49 e p 13.50 -

v a l u e

(m)

H0. The interaction terms do not make a significant contribution to the model. The regression model should use the proficiency exam score and the classroom dummy variable.

VIF =

1 = 1.54 1 − 0.35

VIF =

1 = 2.0 1 − 0.5

Collinearity is potentially significant and you need to consider whether all the independent variables are needed.

1 = 1.215 1 − 0.1767 1 R22 = 0.1767 , VIF = = 1.215 1 − 0.1767 R12 = 0.1767 , VIF =

There is no reason to suspect the existence of collinearity.

1 = 1.0035 1 − 0.0035 1 = 1.0035 R22 = 0.0035 , VIF = 1 − 0.0035

13.51 =

, R1 = 0.0035 VIF =

0 . 4 6

There is no reason to suspect the existence of collinearity.

13.52

R12 = 0.759 , VIF =

0 . 0 5 13.53 . D o

1 = 4.15 1 − 0.759 1 R22 = 0.759 , VIF = = 4.15 1 − 0.759 There is no reason to suspect the existence of collinearity.

1 = 1.225 1 − 0.184 1 R22 = 0.184 , VIF = = 1.225 1 − 0.184 R12 = 0.184 , VIF =

There is no reason to suspect the existence of collinearity.

n o 13.54 t r e 13.55 j e c t

VIF =

1 = 1.466 1 − 0.3181

There is no reason to suspect the existence of collinearity.

In the case of the simple linear regression model, the slope b1 represents the change in the estimated mean of Y per unit change in X and does not take into account any other variables. In the multiple linear regression model, the slope b1 represents the change in the estimated mean of Y per unit change in X1, taking into account the effect of all the other independent variables. Copyright © 2019 Pearson Australia (a division of Pearson Australia Group Pty Ltd) 9781488617249, Berenson, Basic Business Statistics 5e

1 3 . 5 6 T e s13.58 t i n g t h e13.59 s i g n i13.60 f i c a13.61 n c e o f 13.62 t h e 13.63 (a) e (b) n t i r e (c) r e g r e s s

ion model involves a simultaneous test of whether any of the independent variables are significant. Testing the contribution of each independent variable tests the contribution of the independent variable after accounting for the effect of the other independent variables in the mo partial determination measures the proportion of variation in Y explained by a particular X variable holding constant the effect of other independent variables in the model. The coefficient of multiple determination measures the proportion of variation in Y explained by all the X variables included in the model. The coefficient of partial determination measures the proportion of variation in Y explained by a particular X variable holding constant the effect of other independent variables in the model. The coefficient of multiple determination measures the proportion of variation in Y explained by all the X variables included in the model.

To test for collinearity, we can use VIF values. If a set of independent variables is uncorrelated, each VIFj is equal to 1. If the set is highly correlated, then a VIFj might even exceed 10. Snee recommends using alternatives to least square regression if maximum VIFj exceeds 5. A dummy variable will be included to represent a categorical independent variable. One category is coded as 0 and the other category of the variable is coded as 1. You test whether the interaction term in the regression model makes a significant contribution to the regression model. If it makes a significant contribution, then it should be included in the model. We can’t interpret them separately since the two variables interact. The effect of an independent variable on the response variable Y is dependent on the value of a second variable. Start with an F test of the overall significance of the model. If you are able to reject this null hypothesis you should test the contribution of each individual x variable using t tests.

Yˆ = 16.0906 − 0.1437 X1 −1.0947 X 2 Holding constant the effect of video access, for each one km increase in distance from campus, lecture attendance decreases by 0.1437 classes per session. Holding constant the effect of distance to campus, for each access to video lecture recordings, lecture attendance decreases by 1.0947 classes per session.

Yˆ = 16.0906 − 0.1437(5) −1.0947(3) = 12.0880

(d)

a d e q u a t e f r o m t h e r e s i d u a l a n a l y s i s .

The model appears to be adequate from the residual analysis. (e) F = 34.09 > F2,9 =4.26. Reject H0. There is evidence of a relationship between lecture attendance and the two independent variables. (f) p-value is virtually zero. The probability of obtaining an F test statistic of 34.09 or larger is virtually zero if H0 is true. (g) R2 = 0.8834. So, 88.34% of the variation in lecture attendance is explained by the variation in the two independent variables. (h) 0.8575 Copyright © 2019 Pearson Australia (a division of Pearson Australia Group Pty Ltd) 9781488617249, Berenson, Basic Business Statistics 5e

(i) (j) (k)

For X1: t = –3.9087 < –2.2622. Reject H0. There is sufficient evidence that distance to campus Forˆ X2: t = –3.5972 < –2.2622. Reject H0. there is suficient evidence that video lecture access Y = 61.8354 + 3.8250 X − 4.2881 X For X1: p-value = 0.0036, which means the probability of obtaining a t statistic of –3.9087 or s For X2: p-value = 0.0058, which means the probability of obtaining a t statistic of –3.9087 or s VIF = 1.5248. The measure of collinearity is below 5 and indicates the two independent variab

1 3 . 6 4

( a ) where

X 1=

= opponent field goal %

field goal %, X2

(b)

For a given opponent field goal %, each increase of 1% in field goal % increases the estimated mean number of wins by 3.8250. For a given field goal %, each increase of 1% in opponent field goal % decreases the estimated mean number of wins by 4.2881.

(c)

Yˆ= 61.8354 + 3.8250 ( 45 ) − 4.2881 ( 44 ) = 45.2825

(d)

H 0: 1 =  2= 0

( e

H 1 : Not all  j = 0 for j = 1, 2

)

F = MSR/MSE = 34.0700 (f)

(g) (h) (i)

p-value is essentially 0 < 0.05. Reject H0 at 5% level of significance. There is evidence of a significant linear relationship between number of wins and the two explanatory variables. p-value is virtually 0. The probability of obtaining an F test statistic equal to or larger than 34.0700 is is true. virtually 0 if H 0 R2 = SSR/SST = 0.7162. So, 71.62% of the variation in number of wins can be explained by variation in field goal % for the team and opponent field goal %. 2 = 0.6952. Radj For X1: t STAT = b1 / s b = 4.4077 and p-value = 0.0001< 0.05, reject H0. There is evidence that the variable X1 contributes to a model already containing X2.

For X2: t STAT = b 2 / s b = -4.3061 and p-value = 0.0002 < 0.05, reject H0. There is

(j)

(k)

evidence that the variable X2 contributes to a model already containing X1. Both variables X1 and X2 should be included in the model. For X1: p-value = 0.0001. The probability of obtaining a t test statistic that differs from 0 by 4.4077 or more in either direction is 0.01% if X1 is insignificant. For X2: p-value is virtually 0. The probability of obtaining a t test statistic that differs from 0 by 4.3061 or more in either direction is 0.02% if X2 is insignificant. RY21,2 =0.4185. Holding constant opponent field goal %, 41.85% of the variation in number of wins can be explained by variation in field goal% for the team. RY22,1 = 0.4071. Holding constant the effect of field goal % for the team, 40.71% of the

variation in number of wins can be explained by variation in opponent field goal %.

13.65 (a)

Yˆ = 0.809 − 0.008X 1 + 0.019X 2

(b)

Holding constant the effect of unemployment rate for a 1% increase in the percentage of adults with high literacy, robbery rate decreases by 0.008 units. Holding constant the effect of percentage of adults with high literacy for a 1% increase in unemployment, robbery rate increases by 0.019 units.

(c)

Yˆ = 0.809 − 0.008(20) + 0.019(7) = 0.782

(d)

R2 = 0.048. So, 4.8% of the variation in robbery rate is explained by the variation in the two independent variables (unemployment and percentage of adults with high literacy).

(e)

(f)

There is some indication in the residual plot of a non-linear relationship. F = 0.301 < F(2,12) = 3.89. Do not reject H0. There is no evidence of a relationship between robbery rate and two independent variables (unemployment rate and percentage of adults with high literacy).

(g)

− 0.06  1  0.04 ,−0.083   2  0.122

(h)

For X1: t = –0.35 > –2.179. Do not reject H0. Percentage of adults with high literacy does not contribute significantly to an explanation of robbery rate. For X2: t = 0.418 < 2.179. Do not reject H0. It is likely that unemployment rate does not contribute to an explanation of sales. So, there must be other variables that better explained the robbery rate. VIF = 1.35. The measure of collinearity is below 5 and indicates the two independent variables are independent of each other.

(i)

13.66 Excel output: Regression Statistics Multi 0 ple R . 7 5 2 0 R 0 Squar . e 5 6 5 5 Adjus 0 ted R . Squar 4 e 7 8 5 Stand 0 ard . Error 9 Copyright © 2019 Pearson Australia (a division of Pearson Australia Group Pty Ltd) 9781488617249, Berenson, Basic Business Statistics 5e

Obser vation s

1 3 6 1 9

ANOV A d f

S S

Regre ssion

16. 29 08

Resid ual

1 5

12. 51 92

Total

1 8

28. 81 00

C o e f f i c i e n t s 1 8 . 6 9 1 5 0 .

Sta nd ar d Err or

Interc ept

Viscos ity

M S

5 . 4 3 0 3 0 . 8 3 4 6

6 . 5 0 6 3

P v a l u e

S t a t

Si g ni fi c a n c e F 0. 0 0 4 9

L o w e r

U p p e r

9 5 %

7.9 78 9

2 . 3 4 2 6

0 . 0 3 3 4

3 5. 6 9 8 2

1 . 6 8 4 8

0.0 08

1 .

0 .

Press ure

Plate Gap

0 1 2 1 0 . 0 8 4 4 0 . 5 0 0 0

0.0 41 4

0.1 37 9

4 8 1 7 2 . 0 4 1 5 3 . 6 2 7 1

1 5 9 1 0 . 0 5 9 2 0 . 0 0 2 5

0 0 5 3 0. 0 0 3 7 0. 2 0 6 2

0 2 9 6 0 . 1 7 2 6 0 . 7 9 3 8

The r 2 of the multiple regression is 0.5655. So 56.66% of the variation in tear rating can be explained by the variation of viscosity, pressure, and plate gap on the bag-sealing equipment. The F test statistic for the combined significant of viscosity, pressure, and plate gap on the bag- sealing equipment is 6.5063 with a p-value of 0.0049. Hence, at a 5% level of significance, there is enough evidence to conclude that viscosity, pressure, and plate gap on the bag-sealing equipment affect tear rating. The p-value of the t test for the significance of viscosity is 0.1591, which is larger than 5%. Hence, there is not sufficient evidence to conclude that viscosity affects tear rating holding constant the effect of pressure and plate gap on the bag-sealing equipment. The p-value of the t test for the significance of pressure is 0.0592, which is also larger than 5%. There is not enough evidence to conclude that pressure affects tear rating at 5% level of significance holding constant the effect of viscosity and plate gap on the bag-sealing equipment. The p-value of the t test for the significance of plate gap is 0.0025, which is smaller than 5%. There is enough evidence to conclude that plate gap affects tear rating at 5% level of significance holding constant the effect of viscosity and pressure. 13.67 (a)

Yˆ = −5603.64 + 20028.74 X 1 + 22075.85X 2

(b)

Holding constant the effect of economics major, for each one-point increase in WAM, the wage is estimated to increase in a mean of $20,028.74. For a given WAM, the wage of students with an economics major will have an estimated mean wage that is 22,075.85 above a wage of students with no economic major.

(c)

Yˆ = −5603.64 + 20028.74(3.5) + 22075.85(1) = $86572.80

(d) (e)

F = 47.99 > F(2,7) = 4.74. There is evidence of a significant relationship between wage and WAM and economics major. p-value = 0. The F statistic is statistically significant.

(f)

2 Radj = 0.9126

13.68 (a)

Yˆ = 0.5610 − 0.1633X 1 + 0.2614X 2

(b)

Holding constant the effect of growth in exports, for each one-unit increase in manufacturing growth there is a mean decrease of 0.1633 in GDP. Holding constant the effect of manufacturing growth for each one-unit increase in export growth, there is a mean increase of 0.2614 in GDP.

(c)

Yˆ = 0.5610 − 0.1633(1.2) + 0.2614(6.3) = 2.012

(d)

R2 = 0.3327. So, 33.27% of the variation in GDP is explained by the variation in the two independent variables (manufacturing growth and exports).

(e)

2 2 to be reliable. Radj = −0.009 . The sample size is too small for the Radj

(f)

(g)

13.69 (a)

For X1: t = –0.1710 > –3.5. Do not reject H0. It is likely that manufacturing growth does not contribute significantly to an explanation of GDP. For X2: t = 0.689 < 3.5. Do not reject H0. It is likely that export growth does not contribute to an explanation of GDP. The regression does not explain GDP growth using either of the independent variables, and the sample size is possibly too small. The model theory must be strong enough too allow for the small sample size in that group of countries to make up a significant proportion of the population countries in that group.

Yˆ = 56.7688 − 0.7770 X1 −1.5486 X 2

(b)

Holding constant the effect school life expectancy for each one-year increase in the average age of females having their first child, the percentage of youth not in employment, education or training (NEET) decreases by 0.78 percentage points. Holding constant the effect of mean birth age, for each one-year increase in school life expectancy, the percentage of youth NEET decreases by 1.55 years.

(c)

Yˆ = 56.7688 − 0.7770(25) −1.5486(15) = 14.12

(d)

F = 9.29 > F2,27 = 3.35. Reject H0. There is enough evidence of a significant relationship between the percentage of youth NEET and the two independent variables. R2 = 0.4077. So, 40.77% of the variation in percentage of youth NEET is explained by the variation in the two independent variables.

(e) (f)

2 Radj = 0.3342 0.3638

(g)

For X1: t = –1.4340 > –2.0518. Do not reject H0. It is likely that average age of first-child birth does not contribute significantly to an explanation of percentage of youth NEET. For X2: t = –1.5486 < –2.0518. Reject H0. It is likely that school life expectancy does contribute to an explanation of percentage of youth NEET at 5% significance level.

13.70 (a) Yˆ = −3.9152 + 0.0319 X1 + 4.2228 X 2

where X1 = amount of cubic feet moved and X2 =

number of pieces of large furniture b) Holding constant the number of pieces of large furniture, for each additional cubic foot moved, the mean labor hours are estimated to increase by 0.0319. Holding constant the amount of cubic feet moved, for each additional piece of large furniture, the mean labor hours are estimated to increase by 4.2228. (c) Yˆ = −3.9152 + 0.0319(500) + 4.2228(2) = 20.4926

(d)

Based on a residual analysis, the errors appear to be normally distributed. The equal cont. variance assumption might be violated because the variances appear to be larger around the center region of both independent variables. There might also be violation of the linearity assumption. A model with quadratic terms for both independent variables might be fitted. (e) FSTAT = 228.80, p-value is virtually 0. Since p-value < 0.05, reject H0. There is evidence of a significant relationship between labor hours and the two independent variables (the amount of cubic feet moved and the number of pieces of large furniture). (f) The p-value is virtually 0. The probability of obtaining a test statistic of 228.80 or greater is virtually 0 if there is no significant relationship between labor hours and the two independent variables (the amount of cubic feet moved and the number of pieces of large furniture). (g) RY212 = 0.9327. 93.27% of the variation in labor hours can be explained by variation in the amount of cubic feet moved and the number of pieces of large furniture. (h) R 2 = 0.9287 adj

(i)

(j)

For X1: tSTAT = 6.9339, p-value is virtually 0. Reject H0. The amount of cubic feet moved makes a significant contribution and should be included in the model. For X2: tSTAT = 4.6192, p-value is virtually 0. Reject H0. The number of pieces of large furniture makes a significant contribution and should be included in the model. Based on these results, the regression model with the two independent variables should be used. For X1: tSTAT = 6.9339, p-value is virtually 0. The probability of obtaining a sample that will yield

(k)

a test statistic farther away than 6.9339 is virtually 0 if the amount of cubic feet moved does not make a significant contribution holding the effect of the number of pieces of large furniture constant. For X2: tSTAT = 4.6192, p-value is virtually 0. The probability of obtaining a sample that will yield a test statistic farther away than 4.6192 is virtually 0 if the number of pieces of large furniture does not make a significant contribution holding the effect of the amount of cubic feet moved constant. 0.0413. We are 95% confident that the mean labor hours will increase by

0 .0226   1

somewhere between 0.0226 and 0.0413 for each additional cubic foot moved holding constant the number of pieces of large furniture. In Problem 13.44, we are 95% confident that the mean labor hours will increase by somewhere between 0.0439 and 0.0562 for each additional cubic foot moved regardless of the number of pieces of large furniture. = 0.5930. Holding constant the effect of the number of pieces of large furniture, 2

( l )

RY 1,2

RY22,1

(m)

59.3% of the variation in labor hours can be explained by variation in the amount of cubic feet moved. = 0.3927. Holding constant the effect of the amount of cubic feet moved, 39.27% of the variation in labor hours can be explained by variation in the number of pieces of large furniture. Both the number of cubic feet moved and the number of large pieces of furniture are useful in predicting the labor hours, but the cubic feet removed is more important.

13.71 Excel output: Regression Statistics Multi ple R

0 . 5 7 0 7 R 0 Squar . e 3 2 5 7 Adjus 0 ted R . Squar 2 e 6 1 5 Stand 1 ard 1 Error . 3 Copyright © 2019 Pearson Australia (a division of Pearson Australia Group Pty Ltd) 9781488617249, Berenson, Basic Business Statistics 5e

Obser vatio ns

6 2 1 2 4

ANOV A d f

S S

Regre ssion

13 09. 51 92

Resid ual

2 1

27 11. 03 79

Total

2 3

40 20. 55 71

M S

6 5 4 . 7 5 9 6 1 2 9 . 0 9 7 0

5 . 0 7 1 8

S i g n i f i c a n c e F 0. 0 1 6 0

C St t P L U o an o p e da S v w p f rd t a e e f Er a l r r i ro t u c r e 9 9 i 5 5 e % % n t s Interc 1 71. 0 0 1 ept 8 48 . . 1 6 Copyright © 2019 Pearson Australia (a division of Pearson Australia Group Pty Ltd) 9781488617249, Berenson, Basic Business Statistics 5e

Die Temp eratur e Die Diam eter

. 2 8 9 2

2 5 5 8

8 0 0 6

0 . 5 9 7 6 1 3 . 5 1 0 8

0.4 63 9

1 . 2 8 8 3 2 . 9 1 2 7

0 . 2 1 1 7 0 . 0 0 8 3

4.6 38 6

3 0. 3 7 1 8 0. 3 6 7 1 2 3. 1 5 7 2

6 . 9 5 0 2 1 . 5 6 2 2 3 . 8 6 4 4

The r 2 of the multiple regression is 0.3257. So 32.57% of the variation in unit density can be explained by the variation of die temperature and die diameter. The F test statistic for the combined significant of die temperature and die diameter is 5.0718 with a p-value of 0.0160. Hence, at a 5% level of significance, there is enough evidence to conclude that die temperature and die diameter affect unit density. The p-value of the t test for the significance of die temperature is 0.2117, which is larger than 5%. Hence, there is not sufficient evidence to conclude that die temperature affects unit density holding constant the effect of die diameter. The p-value of the t test for the significance of die diameter is 0.0083, which is smaller than 5%. There is enough evidence to conclude that die diameter affects unit density at 5% level of significance holding constant the effect of die temperature. Excel output after dropping die temperature: Regression Statistics Multi ple R

R Squa re

Adjus ted R Squa re

Stan dard Error

0 . 5 2 1 9 0 . 2 7 2 4 0 . 2 3 9 3 1 1 . 5

Obse rvati ons

3 1 2 2 4

ANO VA d f

S S

Regr essio n

10 95 .2 55 7

Resid ual

2 2

29 25 .3 01 4

Total

2 3

40 20 .5 57 1

C o e f f i c i e n t s 1 0 7 .

St an da rd Er ro r

Inter cept

16 .6 43 8

M S

1 0 9 5 . 2 5 5 7 1 3 2 . 9 6 8 2

8 . 2 3 7 0

t S t a t

6 . 4 8

S i g n if i c a n c e F 0 . 0 0 8 9

P v a l u e

0 . 0 0

7 3 . 4

L o w e r

U p p e r

9 5 %

1 4 2 .

Die Diam eter

9 2 6 7 1 3 . 5 1 0 8

4. 70 76

4 5

0 0

0 9 5

2 . 8 7 0 0

0 . 0 0 8 7

2 3 . 2 7 3 8

4 4 3 9 3 . 7 4 7 9

Die diameter still remains statistically significant at the 5% level of significance. Hence, only die cont. diameter needs to be used in the model. The residual plot suggests that the equal variance assumption is likely violated.

Re sid ual s

The normal probability plot and the boxplot both suggest that the normal distribution assumption might be violated. None of the observations have a Cook’s Di > F = 0.7155 with d.f. = 2 and 22. Hence, using the Studentized deleted residuals, hat matrix diagonal elements and Cook’s distance statistic together, there is insufficient evidence for removal of any observation from the model.

13.71 (a)

(b)

1 = 2917.00 2 = 3801 3 = 4384.5 4 = 8745 5 = 10064 The better choice would be 5. The problem in (a) is that the regression model is a constant linear growth trend and even if the values in (a) are within the range of the data, it would follow that sales would increase as both salary and commission increase. However, this may not be realistic in a real-world setting.

Chapter 14: Time-series forecasting and index numbers After studying this chapter you should be able to: 1. explain the importance of business forecasting 2. disaggregate the components of a time series 3. use moving averages and exponential smoothing 4. apply linear trend, quadratic trend and exponential trend time-series models 5. estimate the Holt–Winters forecasting model 6. use autoregressive models 7. choose an appropriate forecasting model 8. estimate a forecasting model for seasonal data 9. calculate various price indices 14.1

If the smoothed value for this year is $12 million, our forecast for next year is $12 million (using exponential smoothing).

14.2

(a) The first centred value in the smoothed series is the second year of the time series, 1956. (b) By using the 5-year moving averages, you are unable to calculate a moving average for the first two or last two values in the time series.

14.3

(a) (b)

14.4

The smoothed value of this series in 2006: The smoothed value of this series in 2007:

(a), (b), (c), (d)

14.5

(e)

The exponential smoothing with W = 0.5 assigns more weight to the more recent values and is better for forecasting, while the exponential smoothing with W = 0.25, which assigns more weight to more distance values, is better suited for eliminating unwanted cyclical and irregular variations.

(f)

Using ES(W=0.5) Yˆ2017 = E2016 = 14.28

(a), (b), (c), (d)

(e)

The exponential smoothing with W = 0.65 assigns more weight to the more recent values and is better for forecasting, while the exponential smoothing with W = 0.30, which assigns more weight to more distance values, is better suited for eliminating unwanted cyclical and irregular variations.

(f)

Yˆ2013 = E2012 = 642536.826 RM million

14.6

(a), (b), (c), (d)

14.7

(a), (b), (c), (e)

(d)

Yˆ2015 = E2014 = 16.33

(f)

The exponential smoothing with W = 0.50 assigns more weight to the more recent values and is better for forecasting, while the exponential smoothing with W = 0.25, which assigns more weight to more distance values, is better suited for eliminating unwanted cyclical and irregular variations.

14.8

(a) (b) (c) (d) (e)

14.9

(a) (b) (c) (d) (e)

The Y intercept b0 = 23.2 is the fitted trend value reflecting the total revenues (in millions of constant 1996 dollars) during the origin or base year 1992. The slope b1 = 4.8 indicates that the real total revenues are increasing at a rate of 4.8 million dollars per year. Year is 1997, X= 1997 – 1992 = 5 million dollars Year is 2011, X = 2011 – 1992 = 19 million dollars Year is 2014, x = 2014 – 1992 = 22 million dollars The Y intercept b0 = 1.2 is the fitted trend value reflecting the real total revenues (in billions of real constant 1998 dollars) during the origin or base year 1969. The slope b1 = 0.5 indicates that the real total revenues are increasing at a rate of 0.5 billion dollars per year. billion dollars Year is 2008, X = 2008 – 1969 = 39 billion dollars Year is 2010, X = 2010 – 1969 = 41 billion dollars

14.10 (a), (b), (c), (d)

Replacement Rate for Unemployment Benefits in Australia (1967 - 2007)

Replacement Rate

30 25 20 y = 0.2009x + 18.519

y = -0.0223x 2 + 1.0909x + 12.734 10 y = 17.991e 0.0103x 5 0 0

Year (from 1967)

(e)

The quadratic model is most appropriate since its fit is best for this sample of data.

14.11 (a), (b)

(c)

Yˆ2017 = 72807.78 + 3301.70(17) = 128936.649 $USmillion Yˆ = 72807.78 + 3301.70(18) = 132238.347 $USmillion

(d)

GDP is forecast to increase at a constant rate of $3301.70 $USmillion per year.

2018

14.12 (a), (b), (c), (d)

Welfare Expenditure as a percentage of GDP (1971 - 2001)

Expenditure (% of GDP)

18 16 14 12 10 8

y = 0.2365x + 7.7658

y = -0.0182x 2 + 0.7813x + 5.1329

y = 7.6623e 0.0239x

2 0 0

Year (from 1971)

(e) (f)

The quadratic model appears to be most appropriate. Year is 2002, X = 1971 – 2002 = 31

14.13 (a), (b), (c), (d) Linear Yˆi = 658.88 + 14.47 X i Quadratic Yˆi = 648.71 + 17.68 X i − 0.16 X i

Exponential log Yˆi = 2.82 + 0.0079 X i

(e)

Annual trend forecasts using linear, quadratic and exponential models derived in (b), (c) and (d), respectively.

YˆMay17

YˆNov17

Linear 962.78

Quadratic 949.25

Exponential 976.16

977.25

960.03

994.13

14.14 (a), (b), (c), (d)

Consumer Price Index (CPI) for Australia

September quarter 2006 to the September quarter 2011

CPI

185 180 175 170 y = 1.2075x + 154.52 165

y = 0.0022x 2 + 1.1641x + 154.65

160

y = 154.78e 0.0073x

155 150 0

Quarter (from Sep 2006)

(e) (f)

The linear model is most appropriate. Quarter is December 2011, X = 21

14.15 (a) (b) (c) (d)

Linear Yˆ = 178033.2 + 32256.08 X Quadratic Yˆ = 164846 + 37201.27 X − 290.893 X

Exponential log Yˆ = 5.3264 + 0.0344 X (e)

The quadratic exponential model provides the best fit (marginally), although there was no clear standout using the first differences, second differences and percentage differences criteria.

(f) 14.16 (a)

(b) (c)

14.17 (a)

(b) (c)

740219.412 RM million For time series I, the graph of Y vs X appears to be more linear than the graph of log Y vs X, so a linear model appears to be more appropriate. For time series II, the graph of log Y vs X appears to be more linear than the graph of Y vs X, so an exponential model appears to be more appropriate. Time series I: Time series II: X = 10 for year 2006 in all models. Forecasts for the year 2006: Time series I: Time series II: Based on a comparison of first differences, second differences and percentage differences for the time series, none appear ideal. The linear model is chosen for forecasts.

Yˆ = 50.25 + 3.37 X i . Yˆ = 50.25 + 3.37(25) = 134.41 (‘000) overseas arrivals

Yˆ = 50.25 + 3.37(26) = 137.78 (‘000) overseas arrivals 14.18 (a), (b), (c), (d)

(e)

Annual forecasts using linear, quadratic and exponential models derived in (b), (c) and (d), respectively. Linear

Quadratic

Exponential

2013

2951.66271

2659.23229

5228.97014

2014

3117.0402

2732.26333

6292.56262

(b) 14.20 i

152

N/A

146

152.000

-6.000

148

147.400

-5.020

135

137.214

-8.636

130

129.573

-7.939

124

123.290

-6.780

120

118.953

-5.070

113

113.265

-5.503

110

109.329

-4.406

104

104.277

-4.858

335

N/A

332

335.000

-3.000

325

131.400

-41.320

312

80.416

-18.461

298

71.991

-5.377

275

68.323

-1.809

267

66.703

-0.686

245

62.203

-1.037

223

56.833

-1.281

213

53.710

-0.881

14.21

14.22 (a)

U = 0.3, V = 0.3

Yˆ2017 = E2016 + 1*T2016 = 160880.184 + 1*(5563.81947) = 166444.0037 $USmillion Yˆ2018 = E2016 + 2*T2016 = 160880.184 + 2*(5563.81947) = 172007.8231$USmillion (b)

U = 0.7, V = 0.7

Yˆ2017 = E2016 + 1*T2016 = 157287.1679 + 1*(3326.434264) = 160613.6022 $USmillio n

Yˆ2018 = E2016 + 2*T2016 = 157287.1679 + 2*(3326.434264) = 163940.0364 $USmillion (c)U = 0.3, V = 0.7

Yˆ2017 = E2016 + 1*T2016 = 160506.4077 + 1*(4314.55151) = 164820.9592 $USmillion $USmillion Yˆ = E + 1*T = 160506.4077 + 2*(4314.55151) = 169135.5108 2018

2016

(d)

Given the historical movement of the time series, which suggests that there is a cyclical component in addition to the trend, a better projection model will be to give more weight to the more recent levels and trends. As a result, any recent changes in the level or trend of the time series that are caused by the cyclical component will be readily picked up in the projection. Hence, the projection in model (a) will be a better choice.

14.23 (a)

U = 0.3, V = 0.3 keep formula = 679623.427 + 1 * (21926.132) = 701549.559 U = 0.7, V = 0.7 =581528.463 + 1 * (26602.003) = 608130.466 U = 0.3, V = 0.7 =681953.452 + 1 * (27185.972) = 709139.424 Given the historical movement of the time series, which suggests that there is a cyclical component in addition to the upward trend, a better projection model will be to give more weight to the more recent levels and trends. As a result, any recent changes in the level or trend of the time series that are caused by the cyclical component will be readily picked up in the projection. Hence, the projection in model (a) will be a better choice. The forecasts in (a)–(c) are lower than those in Problem 14.12(c). The Holt– Winters method, with U = 0.3 and V = 0.3, gives more weight to the recent levels and trends of the time series, which is a minor downturn.

(b) (c) (d)

(e)

14.24 (a)

U = 0.3, V = 0.3

YˆMay 2017 = ENov 2016 + 1*TNov 2016 = 933.718 + 1*8.394 = 942.112 Yˆ =E + 2*T = 933.718 + 2*8.394 = 950.506 Nov 2017

(b)

Nov 2016

U = 0.7, V = 0.7

YˆMay 2017 = ENov 2016 + 1*TNov 2016 = 939.737 + 1*11.140 = 950.877 Yˆ =E + 2*T = 939.737 + 2*11.140 = 962.018 Nov 2017

(c)

Nov 2016

U = 0.3, V = 0.7

YˆMay 2017 = ENov 2016 + 1*TNov 2016 = 934.250 + 1*10.548 = 944.798 Copyright © 2019 Pearson Australia (a division of Pearson Australia Group Pty Ltd) 9781488617249, Berenson, Basic Business Statistics 5e

YˆNov 2017 = ENov 2016 + 2*TNov 2016 = 934.250 + 2*10.548 = 955.346 14.25 (a)

U = 0.3, V = 0.3

Yˆ2016 = E2015 + 1*T2015 = 116.803 + 1* −0.210 = 116.593

Yˆ2017 = E2015 + 2*T2015 = 116.803 + 2* −0.210 = 116.383 (b)

U = 0.7, V = 0.7

Yˆ2016 = E2015 + 1*T2015 = 123.167 + 1* −0.295 = 122.872 Yˆ2016 = E2015 + 2*T2015 = 123.167 + 2* −0.295 = 122.577 (c)

U = 0.3, V = 0.7

Yˆ2016 = E2015 + 1*T2015 = 117.768 + 1* −0.097 = 117.672

Yˆ2016 = E2015 + 2*T2015 = 117.768 + 2* −0.097 = 117.575 14.26 (a) Five of the 50 values in the series are lost with a fifth-order autoregressive model. Comparisons cannot be made for the first five observations. (b) Six. One for each of the five variables and one for the intercept. (c) The final five observations are needed to generate forecasts. (d) (e) 14.27

Reject H0. There is sufficient evidence that the third-order regression parameter is significantly different from zero. A third-order autoregressive model is appropriate.

14.28

14.29 (a)

(b)

Do not reject H0. There is not sufficient evidence that the third-order regression parameter is significantly different from zero. A third-order autoregressive model is not appropriate. Fit a second-order autoregressive model and test to see if it is appropriate.

14.30 (a)

Intercept YLag1 YLag2 YLag3

Coefficients

Standard Error

tStat

p-value

3559.475810 1.017722 0.092143 -0.113584

2093.863214 0.225135 0.319249 0.203664

1.699956 4.520499 0.288623 -0.557702

0.104640 0.000209 0.775840 0.583237

Since the p-value = 0.583237 is > 0.05 level of significance, the third-order term should be dropped. (b)

Intercept YLag1 YLag2

Coefficients

Standard Error

tStat

p-value

2639.398228 1.150426 -0.147511

1826.097790 0.195773 0.198118

1.445376 5.876339 -0.744561

0.162447 0.000007 0.464417

Since the p-value = 0.464417 is > 0.05 level of significance, the second-order term should be dropped. (c)

Intercept YLag1

Coefficients

Standard Error

tStat

p-value

1471.300348 1.015686

1830.772583 0.015700

0.803650 64.694714

0.429487 0.000000

Since the p-value is essentially zero, the first-order term is significant and should remain. (d)

The most appropriate model for forecasting is the first-order autoregressive model.

Yˆi = 1471.300348 + 1.015686Yi −1 Year 2017 2018

Forecasts

165479.864 169546.941

14.31 (a)

Intercept YLag1 YLag2 YLag3

Coefficients

Standard Error

tStat

p-value

51.474846 0.565407 0.261809 0.135807

17.114674 0.267859 0.298563 0.264002

3.007644 2.110840 0.876896 0.514416

0.009408 0.053254 0.395342 0.614986

Given the p-value = 0.614986 > 0.05, do not reject H0 that A3 = 0. Thirdorder term should be dropped. (b)

Intercept YLag1 YLag2

Coefficients

Standard Error

tStat

p-value

41.826324 0.630501 0.340717

12.863554 0.241006 0.237247

3.251537 2.616121 1.436128

0.005005 0.018717 0.170230

Given the p-value = 0.170230 > 0.05, do not reject H0 that A2 = 0. Secondorder term should be dropped. (c) Copyright © 2019 Pearson Australia (a division of Pearson Australia Group Pty Ltd) 9781488617249, Berenson, Basic Business Statistics 5e

Intercept YLag1

Coefficients

Standard Error

tStat

p-value

31.423610 0.977998

10.431118 0.013012

3.012487 75.163892

0.007479 0.000000

Given the p-value is virtually zero, reject H0 that A1 = 0. First-order autoregressive model is appropriate. 14.32 (a) Coefficients Intercept YLag1 YLag2 YLag3

3.3188 1.185985 -0.381923 0.18377

Standard Error 5.221899 0.248937 0.38344 0.239876

tStat

p-value

0.635554 4.764196 -0.996042 0.766103

0.535318 0.000302 0.336135 0.456338

Given the p-value = 0.456338 > 0.05, do not reject H0 that A3 = 0. Thirdorder term should be dropped.

(b) Coefficients Intercept YLag1 YLag2

2.090678 1.244529 -0.251326

Standard Error 4.807489 0.22615 0.233039

tStat

p-value

0.434879 5.503122 -1.078475

0.669459 0.000048 0.296813

Given the p-value = 0.296813 > 0.05, do not reject H0 that A2 = 0. Secondorder term should be dropped. (c) Coefficients Intercept YLag1

-1.744941 1.017656

Standard Error 4.582059 0.027587

tStat

p-value

-0.38082 36.889165

0.707794 0.000000

Given the p-value is virtually zero, reject H0 that A1 = 0. First-order autoregressive model is appropriate. (d)

The most appropriate model for forecasting is the first-order autoregressive model.

Year 2012

Forecasts 180.823

14.33 (a)

Intercept YLag1

Coefficients

Standard Error

P-value t Stat

7.493222 0.943778

6.190099 0.066542

1.210517 14.183270

0.238926 0.000000

Given the p-value = 0.000000 < 0.05, reject H0 that A1 = 0. First-order autoregressive model is appropriate. (b) Year 2016 2017

Forecasts

119.519639 120.293195

14.34 (a) The standard error of the estimate is 2.117. (b) The mean absolute deviation is 0.89.

14.35 (a) The standard error of the estimate is 0.869. (b) The mean absolute deviation is 3.633. 14.36 (a)

(b) (c) (d)

The residuals in the linear trend model show strings of consecutive positive and negative values. SYX = 39301.24 MAD = 31188.30 The residuals in the linear trend model show strings of consecutive positive and negative values. The linear trend model is inadequate in capturing the non-linear trend.

14.37 (a) Coded Year Residual Plot

2.2

Residuals

1.2 0.2 -0.8 -1.8 -2.8 0

Coded Year

Quadratic AR1 Holt-Winters

(b), (c)

(d)

Quadratic

AR1

Holt– Winters

SSE

18.28805772

16.29545022

21.24602288

SYX

0.823003931

0.776875275

0.921868166

MAD

0.581666667

0.591867877

0.697342676

On the basis of (a), (b), (c) and parsimony, it appears that the first-order autoregressive model is most appropriate with respect to the quadratic and Holt–Winters models. The residuals of this model are more randomly distributed, and it has the lowest values for the standard error of the estimate (SYX) and mean absolute deviation (MAD).

14.38 (a)

(b), (c) SSE SYX MAD (d)

Linear

AR1

Holt-Winters

907.886654 6.912561 5.628827

441.098989 4.950302 4.046780

43.53234 1.555141 1.330365

On the basis of (a), (b), (c) and parsimony, we would select the Holt-Winters forecasting model.

14.39 (a)

(b), (c)

SSE SYX MAD (d)

Linear

AR1

Holt-Winters

1623.52668 8.401673 6.019557

1513.32954 8.293835 6.068906

161.2834 7.012322 1.784890

On the basis Holt-Wintersfirst-order autoregressive forecasting model.

14.40 (a) This is the fitted value for January 2004 prior to adjustment by the January multiplier. (b) The estimated monthly compound growth rate is (b1 – 1)100% = 7.15% (c) The January values in the time series are estimated to require 51.36% increase above the value determined by the monthly growth rate. 14.41 Eleven dummy variables would be required to handle monthly data. 14.42 (a) This is the fitted value for January 2004 prior to adjustment by the January multiplier. (b) The estimated monthly compound growth rate is (b1 – 1)100% = 7.15% (c) The second-quarter values in the time series are estimated to require 51.19% increase above the value determined based on the quarterly compound growth rate.

14.43 (a)

(b)

(c)

(d)

log YˆQ 3,2017 = 4.5 + 0.03(54) + 0.17 = 6.29 YˆQ 3,2017 = 106.29 = 1949844.6

log YˆQ 4,2017 = 4.5 + 0.03(55) = 6.15 YˆQ 4,2017 = 106.15 = 1412537.545 log YˆQ1,2018 = 4.5 + 0.03(56) − 0.32 = 5.86 YˆQ1,2018 = 105.86 = 724435.9601 log YˆQ 2,2018 = 4.5 + 0.03(57) + 0.40 = 6.61 YˆQ 2,2018 = 106.61 = 4073802.778

14.44 (a)

Composite Stock Price Index

Standard & Poor's Composite Stock Price Index (Q1 1994 - Q2 2004) 1600 1400 1200 1000 800 600 400 200 0 0

Coded Quarter

(b)

2004: 2005:

(f) The estimated quarterly compound growth rate is (b1–1)100% = 2.10%. (g) The second-quarter values in the time series are estimated to have a mean 1.41% below the fourth-quarter values.

14.45 (a)

(b) Coefficien Standard ts Error

tStat

p-value

0.002576 0.001124

2.2918

0.0335

-0.0377

0.02175

-1.7332

0.09925

-0.03277

0.0218

-1.5007

0.1499

-0.07586

0.02175

-3.4877

0.00246

0.01972

54.437

0.0000

Constan 1.0734 t

log10 Yˆ = 1.0734 + 0.002576 X − 0.0377Q1 − 0.03277Q2 − 0.07586Q3 (c)

May 2014:

log10 Yˆ23 = 1.0734 + 0.002576(23) − 0.03277 = 1.0999

Yˆ23 = 101.0999 = 12.5866 (d)

August 2014:

log10 Yˆ24 = 1.0734 + 0.002576(24) − 0.07586 = 1.0594 Yˆ24 = 101.0594 = 11.4656 November 2014:

log10 Yˆ25 = 1.0734 + 0.002576(25) = 1.1378

Yˆ25 = 101.1378 = 13.7351

(e)

log b1 = 0.002576, b1 = 100.002576 = 1.00595 The quarterly compounded growth rate is 0.595%

(f)

log b3 = −0.03277, b3 = 10−0.03277 = 0.9273 The second-quarter values in the time series are estimated to have a mean 7.2683% below the fourth-quarter values.

14.46 (a)

(b)

There is no obvious quarterly seasonal pattern to the CPI data.

(d)

Standard tcalc Error

p-value

0.001583 1383.926738 0.000000

0.003143 0.000096 32.817277

0.000000

Sept

0.000719 0.001602 0.448774

0.659613

Dec

-0.001196 0.001681 -0.711324

0.487125

Mar

-0.000944 0.001673 -0.564233

0.580419

log b1 = 0.003143 b1 = 100.003142 = 1.007264 The estimated quarterly compound growth rate is (b1–1)100% = 0.07264%.

(e)

log b2 = 0.000719 b2 = 100.000719 = 1.001657 The September values in the time series are estimated to have a mean 0.1657% above the June values.

(f)

December 2011:

log yˆ 21 = 2.190097 + 0.003143(21) − 0.001196 = 2.254908 yˆ 21 = 179.849030 (g)

March 2012:

(h)

This is for students to look up.

log yˆ 22 = 2.190097 + 0.003143(22) − 0.000944 = 2.258303 yˆ 22 = 181.260499

14.47 (a)

Number ('000s)

Married Females aged 25-34 in labour force in Australia (Nov 2005 - Sep 2008) 710 700 690 680 670 660 650 640 630 620 610 0

Coded Month

There does not appear to be an obvious seasonal variation given the above time-series plot. (b) Parameter

Standard Error

tcalc

p-value

0.00088

0.000164

5.349

0.000025

0.003539 0.007437

0.475831

0.638886

0.005818 0.007429

0.783136

0.441897

– 0.007425 0.010233

–1.378114

0.182024

– 0.007425 0.001323

–0.178217

0.860184

– 0.007428 0.008143

–1.096357

0.284779

– 0.007434 0.008642

–1.162412

0.257523

– 0.007445 0.004827

–0.648443

0.52341

– 0.007458 0.004071

–0.545849

0.590662

– 0.007476 0.005618

–0.751431

0.460357

M10

– 0.007497 0.000188

–0.025029

0.980257

M11

0.011305 0.008585

1.316816

0.201453

Constant 2.801041 0.005777

484.823618 0.000000

(c) The estimated monthly compound growth rate is (b1–1)100% = 0.203%. (d) The November values in the time series are estimated to have a mean 0.818% above the October values.

The December values in the time series are estimated to have a mean 1.35% above the November values.

The January values in the time series are estimated to have a mean 2.33% below the December values.

The February values in the time series are estimated to have a mean 0.304% below the January values. The March values in the time series are estimated to have a mean 1.858% below the February values. The April values in the time series are estimated to have a mean 1.97% below the March values.

The May values in the time series are estimated to have a mean 1.11% below the April values.

The June values in the time series are estimated to have a mean 0.933% below the May values.

The July values in the time series are estimated to have a mean 1.29% below the June values.

The August values in the time series are estimated to have a mean 0.0433% below the July values. The September values in the time series are estimated to have a mean 2.637% above the August values. (e)

14.48 (a) Earnings could be affected by seasonal variation if affected by the regular spending patterns within the year. (b)

It is difficult to ascertain a seasonal pattern in the chart. (c)

log yˆi = 2.823497 + 0.007922 xi − 0.000625nov

(d)

log b1 = 0.007922. b1 = 100.007922 = 1.018407. The estimated biannual compound growth rate is (b1–1)100% = 1.8407%

(e)

log b2 = -0.000625. b2 = 10–0.000625 = 0.998562. The November values in the time series are estimated to have a mean 0.14378%% below the May estimate.

(f)

Forecast for May 2017: log yˆ 21 = 2.823497 + 0.007922(21)=2.989848

Here we include a dummay, as opposed to quarterly, as our data is available on a 6 monthly basis.

biannual

log yˆ 22 = 2.823497 + 0.007922(22) − 0.000625 = 993.44665 yˆ 22 = 993.44665 14.49 The price of the commodity in 2012 was 25% higher than in 1998. 14.50 (a)

With 2015 as the base year:

P2015 5 = (100) = 100 P2015 5 P 8 I 2015 = 2016 = (100) = 160 P2015 5 P 7 I 2015 = 2017 = (100) = 140 P2015 5 I 2015 =

(b)

With 2016 as the base year:

P2015 5 = (100) = 62.5 P2016 8 P 8 I 2015 = 2016 = (100) = 100 P2016 8 P 7 I 2015 = 2015 = (100) = 87.5 P2016 8 I 2015 =

 P =  P 3

2017 U

14.51 (a)

 P =  P  P =  P 3

(b)

2017 L

2017 i =1 i 3 2011 i =1 i 3

(c)

2017 P

2017 i =1 i 3 2011 i =1 i

(100) =

Qi2007

(100) = 2007

2017 i =1 i 3 2011 i =1 i

Qi2017

(100) = 2017

40 (100) = 102.56 39

327 (100) = 119.34 274 258 (100) = 125.85 205

14.52 (a), (b) cpi-u (base = 1982)

cpi-c (base = 1965)

cpi-c (base = 1990)

1965

32.51

100.00

24.04

1966

33.63

103.47

24.87

1967

34.54

106.25

25.54

1968

36.12

111.11

26.71

1969

37.92

116.67

28.05

1970

40.18

123.61

29.72

1971

41.99

129.17

31.05

1972

43.34

133.33

32.05

1973

46.05

141.67

34.06

1974

51.02

156.94

37.73

1975

55.76

171.53

41.24

1976

58.92

181.25

43.57

1977

62.75

193.06

46.41

1978

67.49

207.64

49.92

1979

75.17

231.25

55.59

1980

85.33

262.50

63.11

1981

94.13

289.58

69.62

1982

100.00

307.64

73.96

1983

103.16

317.36

76.29

1984

107.45

330.56

79.47

1985

111.29

342.36

82.30

1986

113.54

349.31

83.97

1987

117.61

361.81

86.98

1988

122.35

376.39

90.48

1989

128.44

395.14

94.99

1990

135.21

415.97

100.00

1991

141.08

434.03

104.34

1992

145.15

446.53

107.35

1993

149.44

459.72

110.52

1994

153.50

472.22

113.52

1995

157.79

485.42

116.69

1996

162.30

499.31

120.03

1997

166.14

511.11

122.87

1998

168.85

519.44

124.87

1999

172.46

530.56

127.55

2000

178.33

548.61

131.89

2001

183.30

563.89

135.56

2002

186.23

572.92

137.73

2003

190.52

586.11

140.90

2004

195.49

601.39

144.57

2005

202.26

622.22

149.58

2006

208.58

641.67

154.26

2007

214.67

660.42

158.76

2008

222.80

685.42

164.77

2009

222.12

683.33

164.27

2010

225.73

694.44

166.94

2011

232.96

716.67

172.29

2012

237.70

731.25

175.79

2013

241.08

741.67

178.30

2014

245.15

754.17

181.30

(c)

The price index using 1990 as the base year is more useful because it is closer to the present.

14.53 (a), (b)

(c)

The price index using 1990 as the base year is more useful because it is closer to the present and the DJIA has grown more than 1000% over the 23year period.

14.54 (a) Quarter Sep-06 Dec-06 Mar-07 Jun-07 Sep-07 Dec-07 Mar-08 Jun-08 Sep-08 Dec-08 Mar-09 Jun-09 Sep-09 Dec-09 Mar-10

CPI 155.7 155.5 155.6 157.5 158.6 160.1 162.2 164.6 166.5 166 166.2 167 168.6 169.5 171

Price Index (Base = Sep 06) 100.000 99.872 99.936 101.156 101.863 102.826 104.175 105.716 106.936 106.615 106.744 107.258 108.285 108.863 109.827

Jun-10 Sep-10 Dec-10 Mar-11 Jun-11 Sep-11 (b)

172.1 173.3 174 176.7 178.3 179.4

110.533 111.304 111.753 113.487 114.515 115.222

The CPI has significantly changed, an increase of 15.222% in the 5-year period from September quarter 2006 to September quarter 2011. The CPI has progressively increased during the period.

14.55 (a), (b)

year

price index

1999

73.0

2000

80.3

2001

79.0

2002

76.5

2003

78.0

2004

82.0

2005

89.9

2006

94.4

2007

94.7

2008

101.8

2009

87.7

2010

91.8

2011

99.5

2012

100.0

2013

97.3

(c)

price index (base = price index (base = 2000) 2013) 90.83237915 75.00719424 100 82.57759507 98.30736297 81.17985612 95.16104943 78.58170606 97.09513616 80.17882837 102.0983721 84.31038027 111.9256235 92.42548818 117.5337283 97.05652621 117.8535869 97.32065776 126.7486434 104.6659815 109.1178374 90.10688592 114.2878479 94.37615622 123.8898292 102.3052415 124.4586051 102.7749229 121.0982227 100

Both price indices are useful. The one using 2000 as the base year conveys a picture of how the CPI has grown since 2000 as a percentage of that year. The one using 2013 as the base year reveals what the CPI in prior years was as a percentage of the current level. Since the price index is usually used to compare the growth of price from some base year in the past, the price using 2000 as the base is more useful.

14.56 (a), (b) Year 1999 1999 1999 2000 2000 2000

Quarter Q2 Q3 Q4 Q1 Q2 Q3

CPI for Tradables 895 895 899 903 910 924

Price Index (Base=Q3'00) 96.861 96.861 97.294 97.727 98.485 100.000

Price Index (Base=Q1'05) 93.424 93.424 93.841 94.259 94.990 96.451

2000 2001 2001 2001 2001 2002 2002 2002 2002 2003 2003 2003 2003 2004 2004 2004 2004 2005 2005 2005 2005 2006 2006 2006 2006 2007 2007 2007 2007 2008 2008 2008 2008 2009 2009 2009 2009 2010 2010 2010 2010 2011 2011 2011

Q4 Q1 Q2 Q3 Q4 Q1 Q2 Q3 Q4 Q1 Q2 Q3 Q4 Q1 Q2 Q3 Q4 Q1 Q2 Q3 Q4 Q1 Q2 Q3 Q4 Q1 Q2 Q3 Q4 Q1 Q2 Q3 Q4 Q1 Q2 Q3 Q4 Q1 Q2 Q3 Q4 Q1 Q2 Q3

940 938 953 957 960 958 973 970 975 972 962 956 955 950 956 956 962 958 963 974 979 978 1000 1003 990 986 995 1000 1018 1020 1043 1063 1041 1037 1045 1062 1057 1058 1055 1065 1092 1097 1113 1114

101.732 101.515 103.139 103.571 103.896 103.680 105.303 104.978 105.519 105.195 104.113 103.463 103.355 102.814 103.463 103.463 104.113 103.680 104.221 105.411 105.952 105.844 108.225 108.550 107.143 106.710 107.684 108.225 110.173 110.390 112.879 115.043 112.662 112.229 113.095 114.935 114.394 114.502 114.177 115.260 118.182 118.723 120.455 120.563

98.121 97.912 99.478 99.896 100.209 100.000 101.566 101.253 101.775 101.461 100.418 99.791 99.687 99.165 99.791 99.791 100.418 100.000 100.522 101.670 102.192 102.088 104.384 104.697 103.340 102.923 103.862 104.384 106.263 106.472 108.873 110.960 108.664 108.246 109.081 110.856 110.334 110.438 110.125 111.169 113.987 114.509 116.180 116.284

(c)

The CPI for tradables in New Zealand is 20.563% higher in the third quarter of 2011 when compared to the third quarter of 2000.

(d)

The CPI for tradables in New Zealand is 3.68% higher in the third quarter of 2011 when compared to the first quarter of 2005.

(e)

The CPI values generally appear to increase consistently in approximately a linear trend. Thus, using the third quarter of 2000 as a base is appropriate.

(f)

CPI for Tradables NZ (Q2 '99- Q3 '11) 1200 1000

CPI

800 600 400 200 0 0

Coded Quarter

There appears to be a linear trend, as indicated by the time-series plot above. 14.57 (a)

Sydney

(b)

Melbourne

(c)

Prices in Melbourne for staple food items appear to have not increased as rapidly as those in Sydney from 2007 to 2017, although the difference is only 3.34%.

14.58 Forecasting methodology is important as a tool for planning for the future. 14.59 A time series is a set of numerical data obtained at regular periods over time. 14.60 Trend is the overall long-term tendency or impression of upward or downward movements. The cyclical component depicts the up-and-down swings or movements through the series. Any observed data that do not follow the trend curve modified by the cyclical component are indicative of the irregular or random component. When data are recorded monthly, or quarterly, an additional component called the seasonal factor is considered. 14.61 Moving averages take into account the result of a limited number of periods of time. Exponential smoothing takes into account all the time periods but gives increased weight to more recent time periods. 14.62 When the first differences are constant. 14.63 The linear trend model in this chapter has the time period as the X variable. 14.64 The Holt–Winters model extends the smoothing approach described in section 14.3 by including the future trend. 14.65 The different methods for choosing an appropriate forecasting model are residual analysis, the sum of squared deviations, the mean absolute deviation and parsimony.

14.66 The standard error of the estimate relies on the squared sum of the deviations, which gives increased weight to large differences. The mean absolute deviation is the mean of the absolute value of the deviations. 14.67 Forecasting for monthly or quarterly data uses an exponential trend model with dummy variables to represent either months or quarters. 14.68 An index number provides a measure of the value of an item (or group of items) at a particular point in time as a percentage of the item’s (or group of items’) value at another point in time. 14.69 A simple price index reflects the price of a single commodity in a given period of time over the price for that commodity at a particular point of time in the past. An aggregate price index reflects the percentage change in the price of a group of commodities in a given period of time over the price paid for that group of commodities at a particular point of time in the past. 14.70 Both the Laspeyres price index and Paasche price index are weight aggregate price indexes. The Laspeyres price index uses the consumption quantities associated with the base year as the weights while the Paasche price index uses the consumption quantities in the year of interest as the weights. 14.71 (a)

(b)

Yˆ = −2.5854 + 0.7197 X

(c)

Yˆ = 1.225 + 0.0847 X + 0.0172 X 2

(d)

log Yˆ = 0.1741 + 0.0374 X (e) Intercept YLag1

Coefficients Standard Error t Stat P-value 0.4887 0.1561 3.1301 0.0038 1.1725 0.1543 7.5975 0.0000

YLag2 YLag3

-0.8248 0.7121

0.2592 -3.1817 0.1833 3.8855

0.0033 0.0005

Test of A3: p-value = 0.0005< 0.05. Reject H0 that A3 = 0. Third-order term cannot be deleted. A third-order autoregressive model is appropriate (f)

(g)

SSE Syx MAD

Linear QuadraticExponential AR3 138.6905 9.4759 206.5249 7.7632 1.9628 0.5203 2.3952 0.5004 1.6711 0.3576 1.3060 0.3682

(h)

The residuals in the first three models show strings of consecutive positive and negative values. The autoregressive model performs well for the historical data and has a fairly random pattern of residuals. It also has the smallest values in the standard error of the estimate, MAD and SSE. Based on the principle of parsimony, the autoregressive model would probably be the best model for forecasting.

(i)

Yˆ2013 = 0.4887 + 1.1725Y2012 − 0.8248Y2011 + 0.7121Y2010 = $27.7445Billion

14.72 (a), (b), (c), (d)

Number of Employees

Labour Force Participation Rate of Male Older Workers (1967-2010) 120 100 80 60 y = -0.419x + 88.263

y = 88.23e-0.005x

y = 0.0239x2 - 1.4488x + 95.472

0 0

Coded Year

(e)

The quadratic model appears to fit the data best.

14.73 (a)

(b) (c)

There is a distinct linear growth trend in the series. There are several possible forecasting models that could be used. A linear trend model, or possibly an exponential trend (but this is not evident in the graph) or a Holt–Winters model. The Holt–Winters would need weighting for the more recent values as the earlier irregularity ceases towards the more recent time period.

(d)

r2 = 0.9445 b1 has a p-value of virtually zero, so reject H0 that . Overall the model is very good and it is unlikely that other models could improve much upon this choice. You may wish to see whether you can improve upon this model. 2007: 2008:

(e)

14.74 (a)

(b) (c) (d) (e)

Yˆ = 0.8695 + 0.0610 X Yˆ = 0.5365 + 0.2426 X − 0.0151X 2 log Yˆ = −0.0635 + 0.0228 X AR(3): Yˆ = 0.3107 + 0.3953Y + 7.479Y i −1

i −2

− 7.7638Yi −3

Test of A3: p-value = 0.2107 > 0.05. Do not reject H0 that A3 = 0. Third-order term can be deleted. AR(2): Yˆ = 0.5742 + 0.6466Yi −1 − 0.0698Yi −2 Test of A2: p-value = 0.8319> 0.05. Do not reject H0 that A2 = 0. Secondorder term can be deleted. AR(1): Yˆ = 0.4677 + 0.6511Yi −1 Test of A1: p-value =0.0092 < 0.05. Reject H0 that A1 = 0. A first-order autoregressive model is appropriate.

(f)

(g) linear

quadratic

exponential

AR1

SYX

0.29908118

0.22920703

0.10179095

0.2764989

MAD

0.22236686

0.16532698

0.23874359

0.16156055

(h)

All of the residual series display distinctive non-random patterns. The quadratic is chosen as having relatively low standard error and MAD.

(i)

Yˆ = 0.5365 + 0.2426(13) − 0.0151(169) = 1.13

14.75 (a)

(b)

Diversified Equity: Yˆ= 16.0848 +1.8976 X , where X = years relative to 1984 Intercept Coded Yr

Coefficients Standard Error 16.0848 3.8786 1.8976 0.2297

t Stat P-value 4.1471 0.0003 8.2618 0.0000

Stable-Value: Yˆ = 12.8522 + 0.2959 X , where X = years relative to 1984 Coefficients Standard Error t Stat P-value Intercept 12.8522 0.4671 27.5151 0.0000 Coded Yr 0.2959 0.0277 10.6974 0.0000 (c)

Diversified Equity: Yˆ= 3.7482 + 4.5411X − 0.0912X 2 , where X = years relative to 1984 Coefficients Standard Error t Stat P-value Intercept 3.7482 4.6364 0.8084 0.4259 Coded Yr 4.5411 0.7401 6.1359 0.0000 Coded Yr SQ -0.0912 0.0247 -3.6966 0.0010 Stable-Value: Yˆ = 10.3055 + 0.8416 X − 0.0188X 2 , where X = years relative to 1984 Coefficients Standard Error t Stat P-value Intercept 10.3055 0.0769 133.9758 0.0000 Coded Yr 0.8416 0.0123 68.5430 0.0000 Coded Yr SQ -0.0188 0.0004 -45.9961 0.0000 (d)

Diversified Equity: log10 Yˆ= 1.2404 + 0.0237 X , where X = years relative to

1984 Intercept Coded Yr

Coefficients Standard Error t Stat P-value 1.2404 0.0424 29.2269 0.0000 0.0237 0.0025 9.4244 0.0000

Stable-Value: log10 Yˆ = 1.1080 + 0.0082X , where X = years relative to 1984 Intercept Coded Yr (e)

Coefficients Standard Error t Stat P-value 1.1080 0.0145 76.5294 0.0000 0.0082 0.0009 9.5737 0.0000

Diversified equity AR(3) Coefficients Standard Error

t Stat

P-value

Intercept 7.8372 4.4294 1.7693 0.0901 YLag1 0.8938 0.2079 4.2992 0.0003 YLag2 -0.1715 0.2788 -0.6149 0.5446 YLag3 0.1456 0.2001 0.7277 0.4741 Test of A3: p-value = 0.4741> 0.05. Do not reject H0 that A3 = 0. Thirdorder term can be deleted. AR(2): Coefficients Standard Error t Stat P-value Intercept 7.5708 4.0137 1.8862 0.0709 YLag1 0.8949 0.2002 4.4704 0.0001 YLag2 -0.0286 0.1931 -0.1481 0.8835 Test of A2: p-value = 0.8835> 0.05. Do not reject H0 that A2 = 0. Second-order term can be deleted. AR(1): Coefficients Standard Error t Stat P-value Intercept 6.6774 3.6651 1.8219 0.0796 YLag1 0.8835 0.0776 11.3821 0.0000 Test of A1: p-value is virtually 0. Reject H0 that A1 = 0. A first-order autoregressive model is appropriate. Stable value AR(3) Coefficients Standard Error t Stat P-value Intercept 0.7751 0.4161 1.8631 0.0753 YLag1 1.4361 0.2632 5.4563 0.0000 YLag2 -0.3226 0.4497 -0.7173 0.4804 YLag3 -0.1538 0.2258 -0.6813 0.5025 Test of A3: p-value = 0.5025 > 0.05. Do not reject H0 that A3 = 0. Thirdorder term can be deleted. AR(2): Coefficients Standard Error t Stat P-value Intercept 0.8950 0.3492 2.5629 0.0168 YLag1 1.5440 0.1678 9.2029 0.0000 YLag2 -0.5901 0.1523 -3.8749 0.0007 Test of A2: p-value = 0.0007 < 0.05. Reject H0 that A2 = 0. A secondorder autoregressive model is appropriate. (f) Diversified equity

(g)

2011

2008

2005

2002

1999

1996

1993

1990

1987

1984

Stable-value

Diversified Equity:

Linear Quadratic Exponential AR1 3319.7859 2204.2269 5653.7558 1802.4609 10.8887 9.0354 14.2099 8.1705 8.3322 7.0083 10.1558 5.9621

SSE Syx MAD

Stable-Value: SSE Syx MAD

(g)

(h) (i)

Linear Quadrati cExponent AR2 48.1474 0.6067 66.1971 0.3774 1.3113 0.1499 1.5376 0.1229 1.1080 0.1109 1.2974 0.0873

(h) Diversified Equity: The residuals in three trend models show

strings of consecutive positive and negative values. There is no apparent pattern in the residuals of the autoregressive AR(1) model . The autoregressive model has the smallest values in the standard error of the estimate and MAD. Based on the principle of parsimony, the autoregressive model would probably be the best model for forecasting. Stable-Value: The residuals in the linear and exponential trend models show strings of consecutive positive and negative values. There is no apparent pattern in the residuals of the quadratic trend and autoregressive AR(2) model. The autoregressive model AR(2) has the smallest values in the standard error of the estimate and MAD. Based on the principle of parsimony, the autoregressive model would probably be the best model for forecasting. Diversified Equity: Yˆ2014 = 6.6774 + 0.8835Y2013 = 61.0404 Stable-value Yˆ2014 = 0.8950 + 1.5440Y2013 − 0.5901Y2012 = 18.9569 You would recommend Diversified Equity, which consists of investments that are primarily made in stocks, for a member of Teachers’ Retirement System of New York City because it has a higher return than Stable-Value over the past 29 years period

14.76 (a) (b) (c)

Chapter 15: Chi-square tests Learning objectives After studying this chapter you should be able to: 1. identify how and when to use the chi-square test for contingency tables 2. use the Marascuilo procedure for determining pairwise differences when evaluating more than two proportions 3. apply the chi-square test of independence 4. use the chi-square test to evaluate the goodness of fit of a set of data to a specific probability distribution 5. use the chi-square distribution to test for the population variance or standard deviation 15.1

(a) (b) (c)

For df = 12 and  = 0.01, 2 = 26.217 For df = 23 and  = 0.025, 2 = 38.076 For df = 24 and  = 0.05, 2 = 36.415

15.2

(a) (b) (c)

For df = 4 and  = 0.95, 2 = 0.711 For df = 26 and  = 0.975, 2 = 13.844 For df = 15 and  = 0.99, 2 = 5.229

15.3

(a), (b) A oi = 20 ei = 20 Chi-sq contrib= 0 oi = 30 ei = 30 Chi-sq contrib= 0 50

1 2 Total (c)

B oi = 30 ei = 30 Chi-sq contrib= 0 oi = 45 ei = 45 Chi-sq contrib= 0 75

Total 50 75 125

( f0 – fe )2 2 = 0 + 0 + 0 + 0 = 0. Since  STAT < 3.841, it is not fe All Cells

2  STAT = 

significant at the 5% level of significance with 1 df. 15.4

(a) A oi = 15 ei = 13.86

B oi = 24 ei = 25.14

Total 39

oi = 12 ei = 13.14

oi = 25 ei = 23.86

Total

(b)

Decision rule: If 

> 6.635, reject H0 at the 1% significance level with 1 df.

Test statistic:  STAT = 2

( f0 – fe )2 = 0.298833  fe All Cells

( f – f )2  fe All Cells

0 e Decision: Since  STAT == 0.29883 is not greater than the critical value of 2

6.635, it is not significant at the 1% level of significance. 15.5

(a)

Excel output: Observed Frequencies Gender

Prefer black cars?

Men

Women

Total

Yes

Total

100

Expected Frequencies Gender Prefer black cars?

Women

Total

Yes

Men 9

Total

100

Data Level of Significance

0.05

Number of Rows

Number of Columns

Degrees of Freedom

Results Critical Value Chi-Square Test Statistic

3.841459

p-Value

0.037314

4.336043

Reject the null hypothesis Expected frequency assumption is met.

H0:  1 =  2

H1:  1   2

where population: 1 = males, 2 = females

Decision rule: df = 1. If  STAT > 3.8415, reject H0 at 5% significance level. 2

Test statistic:  STAT = 4.3360 2

Conclusion: Since  STAT = 4.3360 is greater than the upper critical bound of 2

3.8415, reject H0. There is evidence to conclude that there is significant difference between the proportions of males and females who prefer black cars at the 5% level of significance. (b)

Excel output: Observed Frequencies Gender

Prefer black cars?

Women

Total

Yes

Men 52

148

180

328

Total

200

400

Women

Total

Expected Frequencies Gender Prefer black cars?

Men Yes

164

328

Total

200

400

Data Level of Significance

0.05

Number of Rows

Number of Columns

Degrees of Freedom

Results Critical Value Chi-Square Test Statistic

3.841459

p-Value

3.12E-05

17.34417

Reject the null hypothesis Expected frequency assumption is met. Copyright © 2019 Pearson Australia (a division of Pearson Australia Group Pty Ltd) 9781488617249, Berenson, Basic Business Statistics 5e

H0:  1 =  2

H1:  1   2

where population 1 = males, 2 = females

Decision rule: df = 1. If  STAT > 3.8415, reject H0 at 5% significance 2

level. Test statistic:  STAT = 17.3442 2

Conclusion: Since  STAT = 17.3442 is greater than the upper critical 2

bound of 3.8415, reject H0. There is sufficient evidence to conclude that there is significant difference between the proportions of males and females who prefer black cars at the 5% level of significance.

15.6

(c)

The chi-square test statistic is more likely to be significant for larger sample sizes, as demonstrated in the increase in the test statistic value between (a) and (b).

(a)

Excel output: Observed Frequencies Country of Origin Own shares? China Japan Yes 14 36 No 320 80 Total 334 116

Total 50 400 450

Expected Frequencies Country of Origin Own shares? China Japan Yes 37.1111 12.8889 No 296.8889 103.1111 Total 334 116

Total 50 400 450

Data Level of Significance Number of Rows Number of Columns Degrees of Freedom

0.05 2 2 1

Results Critical Value 3.8415 Chi-Square Test Statistic 62.8123 p-value 0.0000 Reject the null hypothesis H0:  1 =  2

H1:  1   2

where population: 1 = China, 2 = Japan

Decision rule: df = 1. If  STAT > 3.8415, reject H0 at 5% significance level. 2

Test statistic:  STAT = 62.8123 2

Conclusion: Since  STAT = 62.8123 is greater than the upper critical bound of 2

3.8415, reject H0. There is sufficient evidence to conclude that there is significant difference in share ownership between business executives in China and Japan at the 5% level of significance.

15.7

(b)

The p-value is 0.000. The probability of obtaining a test statistic of 62.8123 or larger when the null hypothesis is true is about 0.000.

(c)

We can conclude that there is overwhelming evidence in this sample to reject the null hypothesis, that business executives in both countries are equally likely to own shares on the stock market.

Excel output: Observed Frequencies Workers WhiteMode of Negotiation Blue-collar collar Collective Bargaining 112 23 Individual Contracts 53 25 Total 165 48

Total 135 78 213

Expected Frequencies Workers Mode of Negotiation Blue-collar White-collar Collective Bargaining 104.5775 30.4225 Individual Contracts 60.4225 17.5775 Total 165 48

Total 135 78 213

Data Level of Significance Number of Rows Number of Columns Degrees of Freedom

0.05 2 2 1

Results Critical Value 3.8415 Chi-Square Test Statistic 6.3840 p-value 0.0115 Reject the null hypothesis H0:  1 =  2

H1:  1   2 where population: 1 = blue-collar, 2 = white-collar

Decision rule: df = 1. If  STAT > 3.8415, reject H0 at 5% significance level. 2

Test statistic:  STAT = 6.3840 2

Conclusion: Since  STAT = 6.3840 is greater than the upper critical bound of 2

3.8415, reject H0. There is sufficient evidence to conclude that there is significant difference in negotiation mode preference between blue- and white-collar workers at the 5% level of significance. 15.8

(a)

Excel output: Observed Frequencies Tyres Punctured? Sandtrak Standard Yes 6 18 No 114 162 Total 120 180

Total 24 276 300

Expected Frequencies Tyres Punctured? Sandtrak Standard Yes 9.6000 14.4000 No 110.4000 165.6000 Total 120 180

Total 24 276 300

Data Level of Significance Number of Rows Number of Columns Degrees of Freedom

0.01 2 2 1

Results Critical Value 0.0002 Chi-Square Test Statistic 2.4457 p-value 0.1179 Reject the null hypothesis H0:  1 =  2

H1:  1   2 where population: 1 = Sandtrak, 2 = standard

Decision rule: df = 1. If  STAT > 0.0002, reject H0 at 99% significance level. 2

Test statistic:  STAT = 2.4457 2

Conclusion: Since  STAT = 2.4457 is higher than the lower critical bound of 2

0.0002, reject H0. There is sufficient evidence to conclude that there is significant difference in the puncture resistance between the SDsandtrak and standard tyre at the 1% level of significance. It can be concluded at the 1% level that the new SDsandtrak can be advertised to reduce the rate of punctures in off-road driving over the SD standard tyre.

15.9

(b)

The level of significance is chosen by the practitioner according to the situation and the potential consequences of failing to reject a false null hypothesis. In this case, driver and passenger safety are considered when choosing a large significance level.

(a)

Excel output: Observed Frequencies Location Response Adelaide Perth Yes 73 98 No 138 117 Total 211 215

Total 171 255 426

Expected Frequencies Location Response Adelaide Perth Yes 84.6972 86.3028 No 126.3028 128.6972 Total 211 215

Total 171 255 426

Data Level of Significance Number of Rows Number of Columns Degrees of Freedom

0.05 2 2 1

Results Critical Value 3.8415 Chi-Square Test Statistic 5.3473 p-value 0.0208 Reject the null hypothesis H0:  1 =  2

H1:  1   2

where population: 1 = Adelaide, 2 = Perth

Decision rule: df = 1. If  STAT > 3.8415, reject H0 at 5% significance level. 2

Test statistic:  STAT = 5.3473 2

Conclusion: Since  STAT = 5.3473 is greater than the upper critical bound of 2

3.8415, reject H0. There is sufficient evidence to conclude that there is significant difference in negotiation mode preference between the two treatment groups in the proportion of responses to the survey at the 5% level of significance. (b)

The p-value is 0.0208. The probability of obtaining a test statistic of 5.3473 or larger when the null hypothesis is true is about 0.0208.

(c)

Adelaide financial planners are incentivised to participate whereas Perth financial planners are not. This introduces a new element of bias to the study.

15.10 (a) (b) (c) 15.11 (a)

df = (r – 1)(c – 1) = (2 – 1)(5 – 1) = 4 2 = 9.488  STAT 2  STAT = 13.277

Excel output for expected frequencies of each cell A 16.0000 19.0000 35

1 2 Total (b)

B 15.0857 17.9143 33

C 32.9143 39.0857 72

Total 64 76 140

( f0 – fe )2 = 2.945159 fe All Cells

2  STAT = 

This is not statistically significant at the 1% level. (c) 15.12 (a)

The Marascuilo procedure is only appropriate if we were to reject the null hypothesis. Excel output for expected frequencies of each cell

1 2 Tot al (b)

A 25.0000 25.0000

B 25.0000 25.0000

C 25.0000 25.0000

Total 75.0000 75.0000

150

( f0 – fe )2 = 4.000 fe All Cells

2  STAT = 

This is not statistically significant at the 5% level. 15.13 (a)

H0: p 1 = p 2 = p 3 = p 4 = p 5 H1: p 1 ¹ p 2 ¹ p 3 ¹ p 4 ¹ p 5

where population: 1 = Australia, 2 = New Zealand, 3= China, 4= Japan, 5=South Korea Decision rule: df = 5 – 1 = 4. If  STAT > 9.488, reject H0 at 5% significance 2

level. Test statistic:

( f0i – fei )2 ( 265 -100) ( 240 -100) (190 -100) ( 270 -100) ( 245-100) = + + + + =1048.5 fei 100 100 100 100 100 i 2

2 c STAT =å

Conclusion: Since  STAT = 1,049 is greater than the upper critical bound of 2

9.488, reject H0. There is sufficient evidence to conclude that there is

significant difference in the proportion of households who own more than 1 laptop computer at the 5% level of significance. (b)

The p-value is 0.000. The probability of obtaining a test statistic of 62.8123 or larger when the null hypothesis is true is about 0.000.

(c)

Excel output of the Marascuilo procedure: Marascuilo Procedure Group

Sample Proportions

0.53

0.48

0.38

0.54

0.49

MARASCUILO TABLE Absolute Critical Differences Range 0.13706290 0.05 5 0.14381877 0.15 5 0.13284457 0.01 1 0.13636891 0.04 1

0.1 0.06 0.97

0.14708038 6 0.13636891 1 0.13980443 5

0.11

0.14315753 6 0.14643387 6

0.05

0.13567136 8

0.16

Statistically Significant? No Yes No No

No No Yes

Yes No

Australia is significantly different from China; New Zealand is significantly different from South Korea; and China is significantly different from Japan.

15.14 (a)

Excel output: Observed Frequencies Age group Major shopping day Under 35 35-54 Saturday 48 56 A day other than Saturday 152 144 Total 200 200

Over 54 24 176 200

Total 128 472 600

Expected Frequencies Age group Major shopping day Under 35 35-54 Saturday 42.6667 42.6667 A day other than Saturday 157.3333 157.3333 Total 200 200

Over 54 42.6667 157.3333 200

Total 128.0000 472.0000 600

Data Level of Significance Number of Rows Number of Columns Degrees of Freedom

0.05 2 3 2

Results Critical Value Chi-Square Test Statistic

5.991465 16.525424 2.5796Ep-value 04 Reject the null hypothesis There is evidence of a significant difference between the age groups with respect to major grocery shopping day at the 5% significance level. (b)

The p-value is 0.0003. The probability of obtaining a sample that gives rise to a test statistic that is equal to or more than 16.5254 is 0.0003 if the null hypothesis is true.

Critical Range 0.1073 0.0959 0.0929

pj − pj' 0.04 0.16* 0.12*

There is a significant difference between the 35–54 and over 54 groups, and between the under 35 and over 54 groups. (d)

The stores can use this information to target their marketing to the specific group of shoppers on Saturday and the days other than Saturday.

15.15 (a)

H 0 : 1 =  2 =  3

H1 : at least one proportion differs

where population 1 = under 35, 2 = 35–54, 3 = over 54 Decision rule: df = (c – 1) = (3 – 1) = 2. If  STAT > 5.9915, reject H0. 2

Test statistic:  STAT = 4.1314 2

Decision: Since

2 = 4.1314 is less than the upper critical bound of  STAT

5.9915, do not reject H0. There is not enough evidence to show that there is a significant relationship between age and major grocery shopping day. (b)

p-value = 0.1267. The probability of obtaining a sample that gives rise to a test statistic that is equal to or more than 4.1314 is 0.12.67 if the null hypothesis is true. The larger the sample size, the more power the 

test has and, hence,

there is a higher likelihood of rejecting a false null hypothesis. 15.16 (a)

Excel output: Observed Frequencies Age group (yrs) Prefer lowcarb beer Yes No Total

18-25 45 32 77

26-45 45 43 88

46-65 23 31 54

>65 12 18 30

Total 125 124 249

Total 125.000 0 124.000 0 249

Expected Frequencies Age group (yrs) Prefer low-carb beer

18-25

26-45

46-65

>65

Yes

38.6546

44.1767

27.1084

15.0602

No Total

38.3454 77

43.8233 88

26.8916 54

14.9398 30

Data Level of Significance Number of Rows Number of Columns Degrees of Freedom Results Critical Value Chi-Square Test Statistic

0.05 2 4 3

7.814728

At the 5% level of significance there is not a statistically significant difference in the proportion of people of different ages who prefer low-carbohydrate beer. (b) 15.17 (a)

Since we were not able to reject the null hypothesis at this level the Marascuilo procedure is not relevant. Excel output:

ATAR < 90 ATAR > 90 Total

Arts 164 172 336

Observed Frequencies University Course Business Engineering 172 68 195 73 367 141

Science 120 139 259

Total 524 579 1103

Arts 159.62 176.38 336

Expected Values University Course Business Engineering 174.35 66.98 192.65 74.02 367 141

Science 123.04 135.96 259

Total 524 579 1103

Data Level of Significance Number of Rows Number of Columns Degrees of Freedom Results Critical Value

0.05 2 4 4

7.8147

Chi-Square Test Statistic

0.4619682 69 0.2152330 p-Value 8 Do not reject the null hypothesis (b)

The p-value is greater than 5%. The probability of obtaining a test statistic of 0.462 or larger when the null hypothesis is true is about 21.52%.

Observed Frequencies Country Free WiFi available?

Australia

China

Malaysia

Total

UK 38

252

Yes

248

Total

100

500

Expected Frequencies Country Free WiFi available?

Australia

China

Malaysia

Total

Yes

50.4

252

49.6

248

Total

100

500

Data Level of Significance

0.025

Number of Rows

Number of Columns

Degrees of Freedom

Results Critical Value Chi-Square Test Statistic p-Value

11.14329 15.08897 0.00452

Reject the null hypothesis (b)

The p-value is 0.00452. The probability of obtaining a test statistic of 15.08897 or larger when the null hypothesis is true is about 0.00452.

15.19 Excel output of the Marascuilo procedure: Group

Sample Proportions

0.3

0.4

Proportions |Group 1 - Group 2| |Group 1 - Group 3|

Absolute Differences

|Group 2 - Group 3|

0.1

Critical Range 0.02188766 2 0.02109147 9

0.1

0.02109147 9

Statistically Significant? No Yes

Yes

Travellers to North Queensland appear to be significantly different from travellers to Bali and Fiji at the 95% significance level. 15.20 Excel output: Observed Frequencies Country of Origin Political voting intention Australia Overseas Liberal 12 23 Labor 23 28 National 8 12 Other 4 8 Total 47 71

Total 35 51 20 12 118

Expected Frequencies Country of Origin Political voting intention Australia Overseas Liberal 13.9407 21.0593 Labor 20.3136 30.6864 National 7.9661 12.0339 Other 4.7797 7.2203 Total 47 71

Total 35 51 20 12 118

Data Level of Significance Number of Rows Number of Columns Degrees of Freedom

0.05 4 2 3

Results Critical Value 7.814728 Chi-Square Test Statistic 1.251070 p-value 7.4078E-01 Do not reject the null hypothesis There is not enough evidence to conclude that there is a statistically significant difference between political party voting intentions, at the 5% significance level. 15.21 df = (r – 1)(c – 1) = (5 – 1)(6 – 1) = 4*5 = 20 Copyright © 2019 Pearson Australia (a division of Pearson Australia Group Pty Ltd) 9781488617249, Berenson, Basic Business Statistics 5e

15.22 (a) (b) (c) (d) (e)

2 = 21.026 2 = 18.549 2 = 26.217 2 = 18.307 2 = 13.277

15.23 Excel output: Observed Frequencies Age group (yrs) Prefer low-carb beer Yes No Total

18-25 45 32 77

26-45 45 43 88

46-65 23 31 54

>65 12 18 30

Total 125 124 249

>65 15.0602 14.9398 30

Total 125.0000 124.0000 249

Expected Frequencies Age group (yrs) Prefer low-carb beer Yes No Total Data Level of Significance Number of Rows Number of Columns Degrees of Freedom Results Critical Value Chi-Square Test Statistic

18-25 38.6546 38.3454 77

26-45 44.1767 43.8233 88

46-65 27.1084 26.8916 54

0.05 2 4 3

7.814728

4.621503 2.0171Ep-value 01 Do not reject the null hypothesis is no relationship between age group and preference for low-carb beer. H 0 : There 1 = 2 = 3 H1 : There is a relationship between age group and preference for low-carb beer. Decision rule: df = (r – 1)(c – 1) = (2 – 1)(4 – 1) = 3. If  STAT > 7.815, reject H0. 2

Test statistic:  STAT = 4.622 2

Decision: Since  STAT = 4.622 is less than critical level of 7.815, do not reject H0 at 2

5% significance level. There is not enough evidence to support the alternative Copyright © 2019 Pearson Australia (a division of Pearson Australia Group Pty Ltd) 9781488617249, Berenson, Basic Business Statistics 5e

hypothesis that states that age group and preference for low-carbohydrate beer are dependent events. 15.24 (a)

Excel output: Observed Frequencies Stress Commuting time High Moderate Under 15 min 9 5 15-45 min 17 8 Over 45 min 18 6 Total 44 19

Low 18 28 7 53

Total 32 53 31 116

Expected Frequencies Stress Commuting time High Moderate Under 15 min 12.138 5.241 15-45 min 20.103 8.681 Over 45 min 11.759 5.078 Total 44 19

Low 14.621 24.216 14.164 53

Total 32 53 31 116

Data Level of Significance Number of Rows Number of Columns Degrees of Freedom

0.01 3 3 4

Results Critical Value 13.276704 Chi-Square Test Statistic 9.831141 p-value 4.3370E-02 Do not reject the null hypothesis is no relationship between stress and commuting time. H 0 : There 1 = 2 = 3 H1 : There is a relationship between stress and commuting time. Decision rule: df = (r – 1)(c – 1) = (3 – 1)(3 – 1) = 4 and =1%. If  STAT > 2

13.277, reject H0. Test statistic:  STAT = 9.831 2

Conclusion: Since

2 = 9.831 is less than the upper critical bound of  STAT

13.277, do not reject H0 at 1% significance level. There is not enough evidence to support the alternative hypothesis, which states that stress and commuting time are dependent events. (b)

Using a significance level of 0.05 would alter our decision rule and conclusion in the following way:

Decision rule: df = (r – 1)(c – 1) = (3 – 1)(3 – 1) = 4 and =5%. If  STAT > 2

9.488, reject H0. Conclusion: Since  STAT = 9.831 is greater than the upper critical bound of 2

9.488, reject H0 at 5% significance level. There is not enough evidence to say that stress and commuting time are independent events. 15.25 Excel output: Observed Frequencies Country of Origin Political voting intention Australia Overseas Liberal 12 23 Labor 23 28 National 8 12 Other 4 8 Total 47 71

Total 35 51 20 12 118

Expected Frequencies Country of Origin Political voting intention Australia Overseas Liberal 13.9407 21.0593 Labor 20.3136 30.6864 National 7.9661 12.0339 Other 4.7797 7.2203 Total 47 71

Total 35 51 20 12 118

Data Level of Significance Number of Rows Number of Columns Degrees of Freedom

0.01 4 2 3

Results Critical Value 11.344867 Chi-Square Test Statistic 1.251070 p-value 7.4078E-01 Do not reject the null hypothesis is no relationship between country of origin and voting intention. H 0 : There 1 = 2 = 3 H1 : There is a relationship between country of origin and voting intention. Decision rule: df = (r – 1)(c – 1) = (4 – 1)(2 – 1) = 3 and =1%. If  STAT > 11.345, 2

reject H0. Test statistic:  STAT = 1.251 2

Conclusion: Since  STAT = 1.251 is less than the upper critical bound of 11.345, do 2

not reject H0. At the 1% level of significance, there appears to be no relationship between country of origin and voting intention. Copyright © 2019 Pearson Australia (a division of Pearson Australia Group Pty Ltd) 9781488617249, Berenson, Basic Business Statistics 5e

15.26 Excel output: Observed Frequencies Country Preferred method

USA

Australia

China

Total

Face to face

824

509

667

545

2545

Phone, talk

111

127

191

499

Phone, text

154

414

118

120

142

400

Total

1106

894

1011

847

3858

Expected Frequencies Country Preferred method

USA

Australia

China

Total

Face to face

729.5931

589.7434

666.9246

558.739

2545

Phone, talk

143.0518

115.6314

130.7644

109.5524

499

Phone, text

118.6843

95.93468

108.4899

90.89114

414

114.6708

92.69051

104.8212

87.81752

400

Total

1106

894

1011

847

3858

Data Level of Significance

0.01

Number of Rows

Number of Columns

Degrees of Freedom

Results Critical Value Chi-Square Test Statistic

21.66599

p-Value

6.71E-48

246.0794

Reject the null hypothesis is no relationship between country and type of preferred communication H 0 : There 1 = 2 = 3 method. H1 : There is a relationship between country and type of preferred communication method. Decision rule: df = (r – 1)(c – 1) = (4 – 1)(4 – 1) = 9 and =1%. If  STAT > 21.666, 2

reject H0.

Test statistic:  STAT = 246.0794 2

Conclusion: Since

2 = 246.0794 is greater than the upper critical bound of  STAT

21.666, reject H0. There appears to be a highly significant relationship between country and type of preferred communication method (also supported by a p-value that is virtually zero). 15.27 (a)

Excel output: Observed Frequencies Level of return Managed fund type High Medium Maximum capital gain 25 41 Long-term growth 22 31 Growth and current income 33 41 Balanced income 35 39 Common stock 28 15 Total 143 167

Low 52 42 53 42 10 199

Total 118 95 127 116 53 509

Expected Frequencies Level of return Managed fund type High Medium Maximum capital gain 33.151 38.715 Long-term growth 26.690 31.169 Growth and current income 35.680 41.668 Balanced income 32.589 38.059 Common stock 14.890 17.389 Total 143.0000 167

Low 46.134 37.141 49.652 45.352 20.721 199

Total 118 95 127 116 53 509

Data Level of Significance Number of Rows Number of Columns Degrees of Freedom

0.05 5 3 8

Results Critical Value 15.507313 Chi-Square Test Statistic 22.650608 p-value 3.8430E-03 Reject the null hypothesis is no relationship between level of return and type of managed H 0 : There 1 = 2 = 3 fund. H1 : There is a relationship between level of return and type of managed fund. Decision rule: df = (r – 1)(c – 1) = (5 – 1)(3 – 1) = 8 and =5%. If  STAT > 2

15.507, reject H0. Test statistic:  STAT = 22.651 2

Conclusion: Since  STAT = 22.651 is greater than the upper critical bound of 2

15.507, reject H0. There appears to be a significant relationship between level of return and type of managed fund. (b) 15.28

This finding tells us that it matters (in terms of level of return) as to which type of managed fund an investor chooses. H0: The number of sick days in the past 12 months follows a Poisson distribution. H1: The number of sick days in the past 12 months does not follow a Poisson distribution.

Our estimate of  is 1.7 and this enables us to determine the expected probability distribution of the data. Number of sick days

Expected Probabilities

Expected Frequencies

0.19

0.31

152

0.26

127

0.15

0.06

0.02

0.01

1.00

483

Decision rule: df = k – p - 1 = 7 – 1 - 1 = 5 and  = 1%. If reject H0. Test statistic: 

2 STAT

> 11.071,

( f0 – fe )2 = 53.321 =  fe All Cells

Conclusion: Since = 53.321 is greater than the upper critical bound of 11.071, reject H0. The distribution of sick days does not appear to follow the Poisson distribution. 15.29

With a significance level of 0.01 and a population mean of 1.5, the distribution of sick days does not follow the Poisson distribution.

15.30

H0: The distribution of commercial mortgages approved follows a Poisson distribution. H1: The distribution of commercial mortgages approved does not follow a Poisson distribution.

Our estimate of  is 2.1 and this enables us to determine the expected probability distribution of the data. No. of commercial mortgages approved

Expected Probabilities

Expected Frequencies

0.12

0.26

0.27

0.19

0.10

0.04

0.01

0.00

1.00

104

Decision rule: df = k – p – 1 = 8 – 1 – 1 = 6 and  = 1%. If H0. Test statistic: 

2 STAT

> 16.812, reject

( f0 – fe )2 = 2.539 =  fe All Cells

Conclusion: Since = 2.539 is lower than the upper critical bound of 16.812, do not reject H0. The distribution of commercial mortgages approved appears to follow the Poisson distribution. 15.31 H0: The distribution of battery lives follows a normal distribution. H1: The distribution of battery lives does not follow a normal distribution. Life (in years)

Expected Probabilities

Expected Frequencies

0-under 1

0.07

1-under 2

0.29

146

2-under 3

0.40

201

3-under 4

0.19

4-under 5

0.03

5-under 6

0.00

Decision rule: df = k – p – 1 = 6 – 2 – 1 = 3 and  = 5%. If Test statistic:  STAT = 2

> 7.815, reject H0.

( f0 – fe ) = 195.73 fe All Cells



Conclusion: Since = 195.73 is greater than the upper critical bound of 7.815, reject H0. The distribution of battery lives does not appear to follow a normal distribution at the 5% level of significance. Copyright © 2019 Pearson Australia (a division of Pearson Australia Group Pty Ltd) 9781488617249, Berenson, Basic Business Statistics 5e

15.32 (a)

(b)

H0: The distribution of call length follows a normal distribution. H1: The distribution of call length does not follow a normal distribution. Length (in minutes)

Expected Probabilities

Expected Frequencies

0-under 5

0.0853

42.65

5-under 10

0.1923

96.15

10-under 15

0.2977

148.85

15-under 20

0.2587

129.35

20-under 25

0.1259

62.95

25-under 30

0.0344

17.20

Decision rule: df = k – p – 1 = 6 – 2 – 1 = 3 and  = 5%. If reject H0. Test statistic:  STAT = 2

> 7.815,

( f0 – fe )2 = 13.28  fe All Cells

Conclusion: Since = 13.28 is greater than the upper critical bound of 7.815, reject H0. The distribution of call lengths does not appear to follow a normal distribution at the 5% level of significance.

15.33 (a)

For df = 25 and a = 0.01, 

(b)

For df = 16 and a = 0.05, 

(c)

For df = 13 and a = 0.10, 

15.34 (a) (b) (c)

15.35



2 STAT

2 /2

= 10.520 and  = 6.908 and  = 5.892 and 

2 1− / 2

= 46.928. = 28.845. = 22.362.

For df = 23 and a = 0.01,   / 2 = 9.2604 and  1− / 2 = 44.1814. 2

For df = 19 and a = 0.05,   / 2 = 8.9065 and  1− / 2 = 32.8523. 2

For df = 15 and a = 0.10,   / 2 = 7.2609 and  1− / 2 = 24.9958. 2

(n – 1)  S 2

2

24  150 2 = = 54 100 2

15.36

2  STAT =

( n – 1)  S 2

2

15 10 2 = 10.417 12 2

15.37 df = n – 1 = 16 – 1 = 15 15.38 (a)

For df = 15 and a = 0.05, 

(b)

For df = 15 and a = 0.05, 

15.39 (a)

2 /2

= 6.262 and 

2 1− / 2

= 27.488.

= 7.261.

H1 : s ¹12 , do not reject H0 since the test statistic c = 10.417 falls 2 2 between the two critical bounds,  = 6.262 and  = 27.488.   2 If H1 :   12 , do not reject H0 since the test statistic c = 10.417 is greater 2

1− / 2

(b)

than the critical bound 7.261. 15.40 You must assume that the data in the population are normally distributed to be able to use the chi-square test of a population variance or standard deviation. If the data selected do not come from an approximately normally distributed population, particularly for small sample sizes, the accuracy of the test can be seriously affected. 15.41 (a)

H0: s £1.2 C. The standard deviation of the oven temperature has not increased above 1.20C. 0

H1: s >1.2 C. The standard deviation of the oven temperature has increased above 1.20C. 0

Decision rule: df = 29. If  STAT > 42.557, reject H0. 2

Test statistic:  STAT = 2

(n – 1)  S 2

2

29  2.12 = 88.813 1.2 2

Conclusion: Since the test statistic of  STAT = 88.813 is greater than the critical 2

bound of 42.557, reject H0. There is sufficient evidence to conclude that the standard deviation of the oven temperature has increased above 1.2 0C. (b) (c)

15.42 (a)

You must assume that the data in the population are normally distributed to be able to use the chi-square test of a population variance or standard deviation. p-value = 5.53 x 10–8 or 0.00000005. The probability that a sample is obtained whose standard deviation is equal to or larger than 2.1 0C when the null hypothesis is true is 5.53 x 10–8, a very small probability. Note: The p-value was found using Excel. H0: = $200. The standard deviation of the amount of auto repairs is equal to $200. H1: $200. The standard deviation of the amount of auto repairs is not equal to $200. Decision rule: df = 24. If  STAT < 12.401 or  STAT > 39.364, reject H0. 2

Test statistic:  STAT = 2

(n – 1)  S 2

2

24  237 .52 2 = 33.849 200 2

Decision: Since the test statistic of

(b) (c)

15.43 (a)

2 = 33.849 is between the critical  STAT

bounds of 12.401 and 39.364, do not reject H0. There is insufficient evidence to conclude that the standard deviation of the amount of auto repairs is not equal to $200. You must assume that the data in the population are normally distributed to be able to use the chi-square test of a population variance or standard deviation. p-value = 2(0.0874) = 0.1748. The probability of obtaining a sample whose standard deviation will give rise to a test statistic equal to or more extreme than 33.849 is 0.1748 when the null hypothesis is true. Note: The p-value was found using Excel. H0: = $32. The standard deviation of the monthly cost of gas usage within the local region is $32. H1: $32. The standard deviation of the monthly cost of gas usage within the local region differs from $32. Decision rule: df = 14. If  STAT < 6.571 or  STAT > 23.685, reject H0. 2

Test statistic:  stat = 2

( n −1) .S 2 = 14(29.25)2 = 11.6971 2

322

Conclusion: Since the test statistic of  STAT = 11.6971 is between the critical 2

(b) (c)

bounds of 6.571 and 23.685, do not reject H0. There is insufficient evidence to conclude that the standard deviation of the monthly cost of calls within the local calling region differs from $32. You must assume that the data in the population are normally distributed to be able to use the chi-square test of a population variance or standard deviation. p-value = 2(1 – 0.6306) = 0.7388. The probability of obtaining a test statistic equal to or more extreme than the result obtained from this sample data is 0.7388 if the standard deviation of the monthly cost of gas usage within the local region is $32. Note: Excel returns an upper-tail area of 0.6306 for  STAT = 11.6971. 2

But

since the sample standard deviation is smaller than the hypothesised value, the amount of area in the lower tail is (1 – 0.6306). That value is doubled to accommodate the two-tail hypotheses. 15.44 (a)

H0: = 0.135 cm. The standard deviation of the diameter of doorknobs is greater than or equal to 0.135 cm in the redesigned production process. H1: < 0.135 cm. The standard deviation of the diameter of doorknobs is less than 0.135 cm in the redesigned production process. Decision rule: df = 24. If  STAT < 13.848, reject H0. 2

Test statistic: c STAT = 2

(n –1)× S 2

24 × 0.1252 = 20.576 0.1352

Conclusion: Since the test statistic of

2 = 12.245 is greater than the  STAT

critical bound of 13.848, do not reject H0. There is insufficient evidence to conclude that the standard deviation of the diameter of doorknobs is less than 0.135 cm in the redesigned production process. Copyright © 2019 Pearson Australia (a division of Pearson Australia Group Pty Ltd) 9781488617249, Berenson, Basic Business Statistics 5e

(b) (c)

15.45 (a)

You must assume that the data in the population are normally distributed to be able to use the chi-square test of a population variance or standard deviation. p-value = 0.0664. The probability of obtaining a test statistic equal to or more extreme than the result obtained from this sample data is 6.64% if the population standard deviation is indeed no less than 0.135 cm. H0: = 5g. The standard deviation of the weight of smoked salmon is 5 g. H1: $12. The standard deviation of the weight of smoked salmon differs from 5 g. Decision rule: df = 29. If  STAT < 16.047 or  STAT > 45.722, reject H0. 2

Test statistic: Conclusion: Since the test statistic of  STAT = 51.92 is greater than the critical 2

(b) (c)

15.46 (a)

boundary of 45.722, reject H0. There is sufficient evidence to conclude that the standard deviation of salmon weights differs from 5 g at the 5% level of significance. You must assume that the data in the population are normally distributed to be able to use the chi-square test of a population variance or standard deviation. p-value = 2(1 – 0.994436) = 0.011129. The probability of obtaining a test statistic equal to or more extreme than the result obtained from this sample data is 0.011129 if the standard deviation of the weight of smoked salmon is 5 g. H0: s £ 2.5 ampere-hours. The standard deviation in the capacity of the battery is equal to 2.5 ampere-hours. H1: s > 2.5 ampere-hours. The standard deviation in the capacity of the battery differs from 2.5 ampere-hours. Decision rule: df = 19. If  STAT > 30.144, reject H0. 2

Test statistic:  STAT = 2

(n – 1)  S 2

2

19  2.6589 2 = 21.492 2.5 2

Conclusion: Since the test statistic of  STAT = 21.492 is less than the critical 2

bound of 30.144, do not reject H0. There is not sufficient evidence to conclude that the standard deviation in the capacity of a certain type of battery differs from 2.5 ampere-hours. (b)

You must assume that the data in the population are normally distributed to be able to use the chi-square test of a population variance or standard deviation.

(c)

p-value = 0.3103. The probability of obtaining a test statistic equal to or more extreme than the result obtained from this sample data is 0.3103 if the population standard deviation is indeed no greater than 2.5 ampere-hours.

15.47 The chi-square test for the difference between two proportions can be used only when the alternative hypothesis is two-tailed. Copyright © 2019 Pearson Australia (a division of Pearson Australia Group Pty Ltd) 9781488617249, Berenson, Basic Business Statistics 5e

15.48 The chi-square test can be used for c populations as long as all expected frequencies are at least one. 15.49 The chi-square test for independence can be used as long as all expected frequencies are at least one. 15.50 (a)

Excel output:

Ease of access Easy to find Difficult to find Very difficult to find Total

Ease of access Easy to find Difficult to find Very difficult to find Total Data Level of Significance Number of Rows Number of Columns Degrees of Freedom

Observed Frequencies Bank Comm. ANZ 77 70 22 30

Westpac 65 25

Total 212 77

10 110

8 98

29 318

Expected Frequencies Bank Comm. ANZ 73.333 73.333 26.635 26.635 10.031 10.031 110 110

Westpac 65.333 23.730 8.937 98

Total 212 77 29 318

11 110

0.05 3 3 4

Results 9.48772 9 1.82815 7 7.6733Ep-value 01 Do not reject the null hypothesis Critical Value Chi-Square Test Statistic

is no difference between the public perceptions of ease of access H 0 : There 1 = 2 = 3 to these three banks. H1 : There is a difference between the public perception of ease of access to these three banks. Conclusion: Given the Excel output above, there isn’t a significant difference between the public perception of ease of access to these three banks at the 5% significance level.

(b)

15.51 (a)

The comment in the question states that there should not be a difference in the perception of people about the ease of accessing different main brand service suppliers. Our findings support this. Excel output: Observed Frequencies Gender Intend to Buy Male Female Yes 122 91 No 79 49 Total 201 140

Total 213 128 341

Expected Frequencies Gender Intend to Buy Male Female Yes 125.5513 87.44868 No 75.44868 52.55132 Total 201 140

Total 213 128 341

Data Level of Significance Number of Rows Number of Columns Degrees of Freedom

0.05 2 2 1

Results Critical Value 3.841459 Chi-Square Test Statistic 0.651822 p-value 0.419462 Do not reject the null hypothesis H0:  1 =  2

H1:  1   2

where population: 1 = males, 2 = females

Conclusion: Since  STAT = 0.651822 is lower than the upper critical bound of 2

3.8415, do not reject H0. There is insufficient evidence to conclude that there is significant difference between the proportions of males and females who intend to buy the carpet cleaning product at the 5% significance level. (b)

15.52 (a)

The p-value is 0.419462. The probability of obtaining a test statistic of 0.651822 or larger when the null hypothesis is true is about 41.95%.

Excel output: Observed Frequencies Island Number of days of use North South

Total

<10 10 or more Total

86 45 131

161 89 250

Expected Frequencies Island Number of days of use North South <10 76.636 84.364 10 or more 42.364 46.636 Total 119 131

Total 161 89 250

Data Level of Significance Number of Rows Number of Columns Degrees of Freedom

75 44 119

0.05 2 2 1

Results Critical Value 3.8415 Chi-Square Test Statistic 0.18722 p-value 0.66524 Do not reject the null hypothesis H0:  1 =  2

H1:  1   2

where population: 1 = North, 2 = South

Conclusion: Since  STAT = 0.18722 is lower than the upper critical bound of 2

3.8415, do not reject H0. There is insufficient evidence to conclude that there is significant difference in the number of days of use between North and South islands at the 5% significance level. (b)

The p-value is 0.66524. The probability of obtaining a test statistic of 0.18722 or larger when the null hypothesis is true is about 66.52%.

15.53 Excel output: Observed Frequencies Country Windscreen damage Australia New Zealand Yes 23 32 No 387 448 Total 410 480

Total 55 835 890

Expected Frequencies Country Windscreen damage Australia New Zealand Yes 25.3371 29.6629 No 384.6629 450.3371 Total 410 480

Total 55 835 890

Data Level of Significance Number of Rows Number of Columns Degrees of Freedom

0.05 2 2 1

Results Critical Value 3.8415 Chi-Square Test Statistic 0.4260 p-value 0.5139 Do not reject the null hypothesis H0:  1 =  2

H1:  1   2 where population: 1 = Australia, 2 = New Zealand

Conclusion: Since  STAT = 0.4260 is lower than the upper critical bound of 2

3.8415, do not reject H0. There is insufficient evidence to conclude that there is significant difference in the proportion of cars returned with windscreen damage between Australia and New Zealand at the 5% significance level.

15.54 (a)

Excel output: Observed Frequencies Age group 35-54 Support increase in age pension 18-34 yrs yrs Yes 34 45 No 78 51 Total 112 96

>54 yrs 55 24 79

Total 134 153 287

Expected Frequencies Age group 35-54 Support increase in age pension 18-34 yrs yrs Yes 52.2927 44.8223 No 59.7073 51.1777 Total 112 96

>54 yrs 36.8850 42.1150 79

Total 134.0000 153.0000 287

Data Level of Significance Number of Rows Number of Columns Degrees of Freedom

0.05 2 3 2

Results Critical Value Chi-Square Test Statistic p-value

5.991465 28.693186 5.8797E07 Reject the null hypothesis

H0:

H1:

where population: 1 = 18–34 yrs, 2 = 35–54 yrs, 3 = > 54 yrs Conclusion: Since  STAT = 28.693 is greater than the upper critical bound of 2

5.991, reject H0. There is sufficient evidence to conclude that there is a significant difference between the age groups in the proportion of support at the 5% significance level. (b)

The p-value is virtually zero. The probability of obtaining a test statistic of 28.693 or larger when the null hypothesis is true is about 0.000.

(c)

Excel output of the Marascuilo procedure: Group

Sample Proportions

1 (18-34 yrs)

0.39

2 (35-54 yrs)

0.33

3 (>54 yrs)

0.28

MARASCUILO TABLE Statisticall y Significant ?

Proportions

Absolute Differences

|Group 1 - Group 2|

0.06

Critical Range 0.16346082 9

|Group 1 - Group 3|

0.11

0.16662284

|Group 2 - Group 3|

0.05

0.17033979 2

Even though we rejected the null hypothesis in part (a), the results from the Marascuilo procedure displayed in the table above indicate that there is not enough data to conclude any particular difference is significant. However, note that the comparison between age groups 1 and 3 comes closest to significance. Further tests with larger samples might actually show a statistically significant difference between these two age groups. 15.55 (a)

Excel output: Observed Frequencies Attitude Type of job Favour Neutral Hourly worker 108 46 Supervisor 18 12 Middle Management 35 14 Upper Management 24 7 Total 185 79

Oppose 71 30 26 9 136

Total 225 60 75 40 400

Expected Frequencies Attitude Type of job Favour Neutral Hourly worker 104.0625 44.4375 Supervisor 27.7500 11.8500 Middle Management 34.6875 14.8125 Upper Management 18.5000 7.9000 Total 185 79

Oppose 76.5000 20.4000 25.5000 13.6000 136

Total 225 60 75 40 400

Data Level of Significance Number of Rows Number of Columns Degrees of Freedom

0.05 4 3 6

Results Critical Value 12.591587 Chi-Square Test Statistic 11.895309 p-value 6.4346E-02 Do not reject the null hypothesis is no relationship between attitude and type of job. H 0 : There 1 = 2 = 3 H1 : There is a relationship between attitude and type of job. Decision rule: df = (r – 1)(c – 1) = (4 – 1)(3 – 1) = 6. If  STAT > 12.592, 2

reject H0. Test statistic:  STAT = 11.895 2

Conclusion: Since  STAT = 11.895 is less than the upper critical bound of 2

12.592, do not reject H0 at 5% significance level. The evidence suggests that there is no relationship between attitude towards self-managed work teams and type of job. (b)

Excel output: Observed Frequencies Attitude Type of job Favour Neutral Hourly worker 135 23 Supervisor 39 7 Middle Management 47 6 Upper Management 26 6 Total 247 42

Oppose 67 14 22 8 111

Total 225 60 75 40 400

Expected Frequencies Attitude Type of job Favour Neutral Hourly worker 138.9375 23.6250 Supervisor 37.0500 6.3000 Middle Management 46.3125 7.8750 Upper Management 24.7000 4.2000 Total 247 42

Oppose 62.4375 16.6500 20.8125 11.1000 111

Total 225 60 75 40 400

Data Level of Significance Number of Rows Number of Columns Degrees of Freedom

0.05 4 3 6

Results Critical Value 12.591587 Chi-Square Test Statistic 3.293706 p-value 7.7118E-01 Do not reject the null hypothesis Copyright © 2019 Pearson Australia (a division of Pearson Australia Group Pty Ltd) 9781488617249, Berenson, Basic Business Statistics 5e

(b)

Conclusion: Since

2 = 3.294 is less than the upper critical bound of  STAT

12.592, do not reject H0 at 5% significance level. The evidence suggests that there is no relationship between attitude towards time off without pay and type of job. 15.56 (a)

Excel output: Observed Frequencies Country Emphasis on banking? China India Yes 34 45 No 55 50 Total 89 95

Australia 48 51 99

Total 127 156 283

Expected Frequencies Country Emphasis on banking? China India Yes 39.9399 42.6325 No 49.0601 52.3675 Total 89 95

Australia 44.4276 54.5724 99

Total 127.0000 156.0000 283

Data Level of Significance Number of Rows Number of Columns Degrees of Freedom

0.95 2 3 2

Results Critical Value Chi-Square Test Statistic

0.102587 2.362197 3.0694Ep-value 01 Reject the null hypothesis

H 0 : 1 =  2 =  3 H1 :

where population 1 = China, 2 = India, 3 = Australia

Decision rule: df = (r – 1)(c – 1) = (2 –1)(3 – 1) = 2 and  = 95%. If  STAT 2

> 0.103, reject H0. Test statistic:  STAT = 2.362 2

Conclusion: Since  STAT = 2.362 is greater than the upper critical bound of 2

0.103, reject H0 at 95% significance level. The evidence suggests that there is a difference in the emphasis on an understanding of the banking system between these three countries.

(b)

Excel output of the Marascuilo procedure: Group

Sample Proportions

0.314487633

0.335689046

0.349823322 MARASCUILO TABLE

Proportions

Absolute Differences

Critical Range

Statistical ly Significan t?

|Group 1 - Group 2|

0.021201413

0.16904211

|Group 1 - Group 3|

0.035335689

0.168154744

|Group 2 - Group 3|

0.014134276

0.166814844

None of the differences between countries are statistically significant. (c)

The p-value is 0.03069. The probability of obtaining a test statistic of 2.362 or larger when the null hypothesis is true is about 3.069%.

(d)

Excel output: Observed Frequencies Country Corporate Accounting? China India Less than 20% 29 35 21-40% 50 47 Greater than 40% 10 13 Total 89 95

Australia 45 41 13 99

Total 109 138 36 283

Expected Frequencies Country Corporate Accounting? China India Less than 20% 34.279 36.590 21-40% 43.399 46.325 Greater than 40% 11.322 12.085 Total 89 95

Australia 38.131 48.276 12.594 99

Total 109 138 36 283

Data Level of Significance Number of Rows Number of Columns Degrees of Freedom

0.05 3 3 4

Results Critical Value 9.487729 Chi-Square Test Statistic 4.466557 p-value 3.4653E-01 Do not reject the null hypothesis (d)

is no relationship between the level of corporate accounting H 0 : There 1 = 2 = 3 desired in each country. H1 : There is a relationship between the level of corporate accounting desired in each country. Decision rule: df = (r – 1)(c – 1) = (3 – 1)(3 – 1) = 4 and  = 5%. If  STAT > 2

9.488, reject H0. Test statistic:  STAT = 4.467 2

Conclusion: Since

2 = 4.467 is less than the upper critical bound of  STAT

9.488, do not reject H0 at 5% significance level. The evidence suggests that there is no relationship between the level of corporate accounting desired in each country. (e)

The p-value is 0.03465. The probability of obtaining a test statistic of 4.467 or larger when the null hypothesis is true is about 3.465%.

(f)

Excel output: Observed Frequencies Country Market systems? China India Yes 59 50 No 30 45 Total 89 95

Australia 45 54 99

Total 154 129 283

Expected Frequencies Country China India Yes 48.4311 51.6961 No 40.5689 43.3039 Total 89 95

Australia 53.8728 45.1272 99

Total 154.0000 129.0000 283

Market systems?

Data Level of Significance Number of Rows Number of Columns Degrees of Freedom

0.025 2 3 2

Results Critical Value Chi-Square Test Statistic

7.377759 8.387754 1.5088Ep-value 02 Reject the null hypothesis

(f)

H 0 : 1 =  2 =  3 H1 :

where population 1 = China, 2 = India, 3 = Australia

Decision rule: df = (r – 1)(c – 1) = (2 – 1)(3 – 1) = 2 and  = 2.5%. If  STAT 2

> 7.478, reject H0. Test statistic:  STAT = 8.388 2

Conclusion: Since  STAT = 8.388 is greater than the upper critical bound of 2

7.478, reject H0 at 2.5% significance level. The findings suggest that there is evidence of a significant difference between the countries on whether comparative market systems should be a significant part of an economics senior subject. (g) 15.57 (a)

See part (b). None of the differences between countries are statistically significant. Further investigation is required. Excel output: Observed Frequencies Accommodation City

Pool

Barcelona

130 10 143 79 362

Lyon Melbourne London Total

No pool

Total

5887 912 1685 7644 16128

6017 922 1828 7723 16490

Expected Frequencies Accommodation City Barcelona Lyon Melbourne London Total

Pool

No pool

Total

132.0894 20.24039 40.12953 169.5407 362

5884.911 901.7596 1787.87 7553.459 16128

6017 922 1828 7723 16490

Level of Significance

0.01

Number of Rows

Number of Columns

Degrees of Freedom

Results Critical Value Chi-Square Test Statistic

11.34487

p-Value

5.22476E-70

324.3916

Reject the null hypothesis (a)

H 0 : 1 =  2 =  3 H1 : where population 1 = swimming pool, 2 = no swimming pool Decision rule: df = (r – 1)(c – 1) = (4 – 1)(2 – 1) = 3 and  = 1%. If  STAT > 2

11.3449, reject H0. Test statistic:  STAT = 324.3916 2

Conclusion: Since

2 = 324.3916 is more than the critical bound of  STAT

11.3449, reject H0 at 1% significance level. There is a significant difference between accommodation with swimming pool proportions in different countries. (b)

15.58 (a)

Absolute differences 0.010759 0.056622 0.011376 0.067382 0.000617 0.067998

Critical Range 0.01311 0.022076 0.007398 3.233808 0.01211 0.021503

Not significant Significant Significant Not significant Not significant Significant

Excel output:

Observed Frequencies Preference after ads Preference before ads Toyota Mazda Toyota 79 13 Mazda 11 97 Total 90 110

Total 92 108 200

Expected Frequencies Preference after ads Preference before ads Toyota Mazda Toyota 41.4 50.6 Mazda 48.6 59.4 Total 90 110

Total 92 108 200

Data Level of Significance Number of Rows Number of Columns Degrees of Freedom

0.05 2 2 1

Results Critical Value 3.8415 Chi-Square Test Statistic 114.9791 p-value 7.95E-27 Reject the null hypothesis

H 0 : 1 =  2 =  3 H1 : where population 1 = Toyota, 2 = Mazda Decision rule: df = (r – 1)(c – 1) = (2 – 1)(2 – 1) = 1 and  = 5%. If  STAT > 2

3.8415, reject H0. Test statistic:  STAT = 114.9791 2

Conclusion: Since  STAT = 114.9791 is greater than the upper critical bound 2

of 3.8415, reject H0 at 5% significance level. There is a significant difference in the proportion of respondents who prefer Toyota before and after viewing the ads. (b)

The p-value is approximately 0.000. The probability of obtaining a test statistic of 114.9791 or larger when the null hypothesis is true is about 0.000.

(c)

Table is derived from the margins of the first table to form the variable preference, ‘Before ad’ and ‘After ad’.

(d)

Excel output: Observed Frequencies

Preference after ads Toyota Mazda 92 108 90 110 182 218

Total 200 200 400

Expected Frequencies Preference after ads Preference Toyota Mazda Before 91 109 After 91 109 Total 208 192

Total 200 200 400

Preference Before After Total

Data Level of Significance Number of Rows Number of Columns Degrees of Freedom

0.05 2 2 1

Results Critical Value 3.841459 Chi-Square Test Statistic 0.040327 p-value 0.840843 Do not reject the null hypothesis According to the above information, there is not a significant difference in the proportion of respondents who prefer Toyota before and after viewing the ads. (e)

The p-value has no meaning because the null hypothesis was not rejected.

(f)

The difference in the results of (a) and (d) is due to the layout of the data in each table. Use the method in part (a) since the contingency table in part (d) does not make sense, given our research objectives.

15.59 (a)

Excel output: Observed Frequencies Income Level Subscriber Option High Low TV episodes 352 769 Movies 198 498 Total 550 1267

Total 1121 696 1817

Expected Frequencies Income Level Subscriber Option High Low

Total

TV episodes Movies Total Data Level of Significance Number of Rows Number of Columns Degrees of Freedom

339.3231 210.6769 550

781.6769 485.3231 1267

1121 696 1817

0.1 2 2 1

Results Critical Value 2.705543 Chi-Square Test Statistic 1.773126 p-value 0.182996 Do not reject the null hypothesis is no relationship between income level and subscriber option. H 0 : There 1 = 2 = 3 H1 : There is a relationship between income level and subscriber option.

Decision rule: df = (r – 1)(c – 1) = (2 – 1)(2 – 1) = 4 and  = 10%. If  STAT 2

> 2.706, reject H0. Test statistic:  STAT = 1.773 2

Conclusion: Since

2 = 1.773 is less than the upper critical bound of  STAT

2.706, do not reject H0 at 10% significance level. The evidence suggests that there is no relationship between the income level and subscriber download choice. (b) 15.60 (a)

The p-value is 0.182996. The p-value has no meaning because the null hypothesis was not rejected. Excel output: Observed Frequencies Gender Duration of current job Male Female Under 1 year 20.7 31.2 1 year and under 2 years 9.9 15.3 2 years and under 3 years 9.5 13 3 years and under 4 years 7.4 9.6 4 years and under 5 years 5.7 6 5 years and under 10 years 18.8 15.6 10 years and under 15 years 11.5 4.8 15 years and under 20 years 6 2.3 Over 20 years 10.5 2.2 Total 100 100

Total 51.9 25.2 22.5 17 11.7 34.4 16.3 8.3 12.7 200

Expected Frequencies Duration of current job Under 1 year 1 year and under 2 years 2 years and under 3 years 3 years and under 4 years 4 years and under 5 years 5 years and under 10 years 10 years and under 15 years 15 years and under 20 years Over 20 years Total

Results Chi-Square Test Statistic p-value (b)

Gender Male Female 25.95 25.96 12.6 12.6 11.25 11.25 8.5 8.5 5.85 5.85 17.2 17.2 8.15 8.15 4.15 4.15 6.35 6.35 100 100

Total 51.9 25.2 22.5 17 11.7 34.4 16.3 8.3 12.7 200

14.24373 0.07563

Excel output: Observed Frequencies Gender Multiple job holders Male Female Multiple jobs 3.3 97.5 Single jobs 96.7 2.5 Total 100 100

Total 100.8 99.2 200

Expected Frequencies Gender Multiple job holders Male Female Multiple jobs 50.4000 50.4000 Single jobs 49.6000 49.6000 Total 100 100

Total 100.8 99.2 200

Results Chi-Square Test Statistic p-value

177.4842 0.0000

(c) Excel output: Observed Frequencies Gender Difficulties reported Male Female Too young or too old 12.74 20.1 Unsuitable hours 3.94 5.1 Transport problems 9.74 3.2 Lacked training/education 13.74 2.2

Total 32.84 9.04 12.94 15.94

Insufficient work experience No vacancies in line of work No vacancies at all Total

18 25.8 25.6 100

33.54 45.64 50.23 200.17

Expected Frequencies Gender Difficulties reported Male Female Too young or too old 16.434 16.406 Unsuitable hours 4.524 4.516 Transport problems 6.475 6.465 Lacked training/education 7.977 7.963 Insufficient work experience 16.784 16.756 No vacancies in line of work 22.839 22.801 No vacancies at all 25.136 25.094 Total 100.1700 100

Total 33 9 13 16 34 46 50 200.17

Results Chi-Square Test Statistic p-value (d)

15.54 19.84 24.63 100

14.435610 7.1094E02

Excel output: Observed Frequencies Gender Duration of unemployment Male Female 1-4 weeks 11.5 12.9 5-13 weeks 7.5 17.5 14-26 weeks 9.5 23.2 Total 28.5 53.6

Total 24.4 25 32.7 82.1

Expected Frequencies Duration of unemployment 1-4 weeks 5-13 weeks 14-26 weeks Total Results Chi-Square Test Statistic p-value (e)

Gender Male Female 8.470 15.930 8.678 16.322 11.351 21.349 28.5 53.6

Total 24 25 33 82.1

2.367695 6.6847E-01

Excel output: Observed Frequencies Gender Birthplace of unemployed Male Female Australia 34.9 34.6 Outside of Australia 15.5 15.2

Total 69.5 30.7

Total

50.4

49.8

100.2

Gender Male Female 34.9581 34.5419 15.4419 15.2581 50.4 49.8

Total 69.5 30.7 100.2

Expected Frequencies Birthplace of unemployed Australia Outside of Australia Total Results Chi-Square Test Statistic p-value (f)

0.0006 0.9799

Ranked order of importance of each problem area analysed (according to pvalues shown above) are: 1. Birthplace of unemployed p-value: 0.9799 2. Duration of current job p-value: 0.07563 3. Duration of unemployment p-value: 0.0669 4. Difficulties reported p-value: 0.0071 5. Multiple job holder p-value: 0.0000

15.61 Numbers of persons using each social media tool have been rounded to the nearest integer. Excel output: Facebook

Observed Frequencies Column variable Row variable Male Female Total Facebook user 536 571 1107 Non Facebook user 53 18 71 Total 589 589 1178 Expected Frequencies Column variable Row variable Male Female Facebook user 553.5 553.5 Non Facebook user 35.5 35.5 Total 589 589

Data Level of Significance Number of Rows Number of

Total 1107 71 1178

0.05 2 2

Columns Degrees of Freedom

Results Critical Value 3.841459 Chi-Square Test Statistic 18.36012 p-Value 1.83E-05 Reject the null hypothesis Excel output: Instagram

Observed Frequencies Column variable Row variable Male Female Total Instagram user 295 241 536 Non Instagram user 294 348 642 Total 589 589 1178 Expected Frequencies Column variable Row variable Male Female Instagram user 268 268 Non Instagram user 321 321 Total 589 589 Data Level of Significance Number of Rows Number of Columns Degrees of Freedom

Results Critical Value Chi-Square Test Statistic p-Value

Total 536 642 1178

0.05 2 2 1

3.841459 9.982355 0.00158

Reject the null hypothesis Excel output:LinkedIn

Observed Frequencies Column variable Row variable Male Female Total LinkedIn user 130 82 212 Non LinkedIn user 459 507 966 Total 589 589 1178 Expected Frequencies Column variable Row variable Male Female LinkedIn user 106 106 Non LinkedIn user 483 483 Total 589 589 Data Level of Significance Number of Rows Number of Columns Degrees of Freedom

Total 212 966 1178

0.05 2 2 1

Results Critical Value 3.841459 Chi-Square Test Statistic 13.25302 p-Value 0.000272 Reject the null hypothesis

Excel output: Twitter

Observed Frequencies Column variable Row variable Male Female Total Twitter user 206 165 371 Copyright © 2019 Pearson Australia (a division of Pearson Australia Group Pty Ltd) 9781488617249, Berenson, Basic Business Statistics 5e

Non Twitter user Total

383 589

424 589

807 1178

Expected Frequencies Column variable Row variable Male Female Twitter user 185.5 185.5 Non Twitter user 403.5 403.5 Total 589 589

Total 371 807 1178

Data Level of Significance Number of Rows Number of Columns Degrees of Freedom

0.05 2 2 1

Results Critical Value 3.841459 Chi-Square Test Statistic 6.614021 p-Value 0.010118 Reject the null hypothesis 15.62–15.66 Team project. 15.62 (a) Objective Negative 8 10 18

Category Australian International Grand Total (b)

Non-negative 25 5 30

Grand Total 33 15 48

H0: There is no relationship between the category of investment fund and whether or not there is a negative monthly return. H1: There is a relationship between the category of an investment fund and whether or not there is a negative monthly return. Decision rule: If

, reject H0.

Test statistic:  = 7.9192 2

Decision: Since = 7.9192 is greater than the critical bound of 3.8415, reject H0. There is enough evidence to conclude at the 5% level of Copyright © 2019 Pearson Australia (a division of Pearson Australia Group Pty Ltd) 9781488617249, Berenson, Basic Business Statistics 5e

significance that there is a relationship between the category of an investment and whether or not there is a negative monthly return. 15.63 (a) Market Capitalisation High Medium 8 11 2 13 10 24

Fee No Yes Grand Total (b)

Low 2 5 7

Grand Total 20 21 41

H0: There is no relationship between the market capitalisation of an investment fund and whether or not there is an outperformance fee. H1: There is a relationship between the market capitalisation of an investment fund and whether or not there is an outperformance fee. Decision rule: If

, reject H0.

Test statistic:  = 5.0310 2

Decision: Since = 5.0310 is less than the critical bound of 5.9915, do not reject H0. There is not enough evidence to conclude at the 5% level of significance that there is a relationship between the market capitalisation of an investment fund and whether or not there is an outperformance fee. (Note: in this calculation three cells have expected frequencies below 5 but all expected frequencies are greater than 1.) 15.64 (a) a Traded volume High Low Grand Total (b)

High

Medium

Low

Grand Total

6 4 10

17 8 25

2 11 23

25 23 48

H0: There is no relationship between the market capitalisation of an investment fund and its traded volume. H1: There is a relationship between the market capitalisation of an investment fund and its traded volume. Decision rule: If

, reject H0.

Test statistic:  = 9.8044 2

Decision: Since = 9.8044 is greater than the critical bound of 5.9915, reject H0. There is enough evidence to conclude at the 5% level of significance that there is a relationship between the market capitalisation and traded volume.

15.65

 X = 4.8441 = 0.1242

 X − nX = 1.1445 − 39(0.1242) = 0.1195

n −1

H0: Three-year return follows a normal distribution. H1: Three-year return does not follow a normal distribution.

 k − p −1 =  2

( f0 − fe ) = 10.9213 2

 2crit =  27−2−1,0.05 =  24,0.05 = 9.4877 Since 10.9213 > 9.4877, reject H0. There is sufficient evidence to conclude that the distribution of three-year return does not follow a normal distribution. Note that variations in answers are possible according to the categories chosen. 15.66

 X = 65.4866 = 1.7699 n

 X − nX = 3845.016 − 37(1.7699) = 10.1777 S= 2

n −1

H0: Five-year return follows a normal distribution. H1: Five-year return does not follow a normal distribution.

 k − p −1 =  2

( f0 − fe ) = 3690.68 2

 2crit =  27−2−1,0.05 =  24,0.05 = 9.4877 Since 3690.68 > 9.48771, reject H0. There is sufficient evidence to conclude that the distribution of five-year return does not follow a normal distribution. Note the effect of the large outlier on expected frequencies. Variations in answers are possible according to the categories chosen.

Chapter 16: Multiple regression model building After studying this chapter you should be able to: 1. utilise quadratic terms in a regression model 2. calculate and use transformed variables in a regression model 3. examine the effect of each observation on the regression model 4. construct a regression model using either the stepwise or the best-subsets approach 5. recognise the many pitfalls involved in developing a multiple regression model

16.1

(a) (b)

Ŷ = 10+2X1i -0.5X1i 2 = 10+2(4)-0.5(4)2 = 10 tcalc = 2.35  t27 = 2.0518 with 27 degrees of freedom. Reject H0. The quadratic term is significant.

(c)

tcalc = 1.17  t27 = 2.0518 with 27 degrees of freedom. Do not reject H0. The quadratic term is

not significant. (d) 16.2

Ŷ = 10-3X -0.5X 2 = 10-3(2)-0.5(2)2 = 2

(a)

(b) (c)

2 Yˆ = -7.555+1.2717X - 0.0145X 2 Ŷ = -7.555+1.2717(90)-0.0145(90) = 224.358

(d)

A residual analysis indicates no strong pattern. (e)

(f)

H0 = 1 = 2 = 0 H1 = at least one j ≠ 0 F = 141.4596 > F2,25 = 3.39. Reject H0. There is a significant quadratic relationship between litres per 100 kilometres and speed. H0 = including the quadratic effect does not significantly improve the model (2 = 0). H1 = including the quadratic effect significantly improves the model (2 ≠ 0). tcalc = –16.6325 > t25 = 2.0595. Reject H0. The quadratic effect is significant. Therefore, the quadratic model is a better fit than the linear regression model.

(g)

(h)

The coefficient of multiple determination R2 represents the proportion of variation in Y that is explained by the variation in the independent variables. Therefore, 91.88% of the variation in litres per 100 kilometres is explained by the quadratic relationship between litres per 100 kilometres and speed.

é (n-1) ù 2 Radj = 1- ê(1- R2 ) ú (n- k -1) û ë é (27-1) ù 2 Radj = 1- ê(1- 0.9188) ú (27- 2-1) û ë 2 Radj = 1-0.088 = 0.912

16.3

(a) GPA 2 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9

Predicted HOCS 2.8600 3.0342 3.1948 3.3418 3.4752 3.5950 3.7012 3.7938 3.8728 3.9382

GPA

Predicted HOCS 3.9900 4.0282 4.0528 4.0638 4.0612 4.0450 4.0152 3.9718 3.9148 3.8442 3.7600

3 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9 4

(b) Predicted HOCS

HOC S

4.50 4.00 3.50 3.00 2.50 2.00 1.50 1.00 0.50 0.00 0

(c)

(d)

16.4

GP A

The curvilinear relationship suggests that HOCS increases at a decreasing rate. It reaches its maximum value of 4.0638 at GPA = 3.3 and declines after that as GPA continues to increase. An r2 of 0.07 and an adjusted r2 of 0.06 tell you that GPA has very low explanatory power in explaining the variation in HOCS. You can tell that the individual HOCS scores will have scattered quite widely around the curvilinear relationship plotted in (b) and discussed in (c).

(b)

Yˆ = 6.7380 + 2.4198 X − 0.0109 X 2

(c)

. Ŷ = 6.7380 + 2.4198(75) - 0.0109(5625) = 126.8039 = $126803.90

(d)

A residual analysis indicates no strong pattern. (e)

H0 = 1 = 2 = 0 H1 = at least one j ≠ 0 F = 225.6689 > F2,9 = 4.26. Reject H0. There is a significant quadratic relationship between client revenue and social marketing expenditure.

(f)

The p-value = virtually zero. It indicates that the probability of having an F test statistic of at least 225.6689 when 1 and 2 = 0 is virtually zero.

(g)

H0 = including the quadratic effect does not significantly improve the model (2 = 0). H1 = including the quadratic effect significantly improves the model (2 ≠ 0). tcalc = –4.5303 < t9 = 2.2622. Reject H0. The quadratic effect is significant and therefore the quadratic model is a better fit than the linear regression model.

(h)

The p-value = 0.0014. It indicates that the probability of having a t test statistic of less than –4.5303 when 2 = 0 is 0.0014.

(i)

The coefficient of multiple determination R2 represents the proportion of variation in Y that is explained by the variation in the independent variables. Therefore, 98.04% of the variation in client revenue is explained by the quadratic relationship between client revenue and social marketing expenditure.

(j)

é (n-1) ù 2 Radj = 1- ê(1- R2 ) ú (n- k -1) û ë é (12 -1) ù 2 Radj = 1- ê(1- 0.9804) = 0.9761 (12 - 2 -1) úû ë

16.5

(a)

Scatter Plot 1400000 1200000 1000000 800000 600000 400000 200000

(b)

0 0

50000

100000

150000

200000

Tourism Establishments

Regression Statistics Multiple R 0.8459 R Square 0.7155 Adjusted R Square 0.6918 Standard Error 173172.0845 Observations 27 ANOVA df Regression Residual Total

Intercept Tourism Establishments Tourism Establishments Sq

SS MS F Significance F 2 1809871444435.5500 904935722217.7750 30.1760 0.0000 24 719725700749.6360 29988570864.5682 26 2529597145185.1900

Coefficients 87251.7358 17.4863 -0.0001

Standard Error 41921.1371 2.9851 0.0000

Yˆ = 87251.7358 + 17.4863 X − 0 .0001 X

(c)

t Stat 2.0813 5.8578 -3.8046

P-value 0.0482 0.0000 0.0009

Lower 95% Upper 95% 730.7611 173772.7104 11.3254 23.6473 -0.0001 0.0000

2 Yˆ = 87251.7358 + 17.4863 ( 3000 ) − 0.0001 ( 3000 )

= 138972.3308

(d)

Residual

Normal Probability Plot 700000 600000 500000 400000 300000 200000 100000 0 -100000 -200000 -300000

Residual

-2

-1

0 Z Value

Residuals

Residual Plot for Tourism Establishments 700000 600000 500000 400000 300000 200000 100000 0 -100000 -200000 -300000 0

(e) (f) (g) (h)

(i)

50000 100000 150000 Tourism Establishments

200000

A residual analysis reveals potential violation of the equal variance assumption and the normality assumption. F = 30.1760 with a p-value of essentially 0. Reject H0. The overall model is significant. The p-value = 0.0000 indicates that the probability of having an Ftest statistic of at least 30.1760 when 1 = 0 and 2 = 0 is essentially 0. t = − 3.8046 with a p-value = 0.0009. Reject H0. The quadratic effect is significant. The p-value = 0.0009 indicates that the probability of having a ttest statistic with an absolute value of at least 3.8046 when 2 = 0 is 0.0009. r = 0 .7 1 5 5 . So, 71.55% of the variation in taxes can be explained by the quadratic relationship between the number of jobs generated in the travel and tourism industry in 2012 and the number of establishments. ST A T

ST A T

2 Y .1 2

(j)

2 adj

= 0.6918 . r

(k) There is a significant quadratic relationship between the number of jobs generated in the travel and tourism industry in 2012 and the number of establishments that provide overnight accommodation for tourists. 16.6

(a) (b) (c) (d)

Ŷ = 9.0356+0.85204 X1

Ŷ = 9.0356+0.85204 85 = 16.89

The residual analysis indicates a clear quadratic term. The model does not adequately fit the data.

t = 1.3537 >t 26 = 2.055. Fail to reject H0. The model does not provide a

significant

relationship. (e)

R2 = 0.0658. So, 6.58% of the variation on litres per 100 kilometres can be explained by

variation in the square root of highway speed. (f)

2 Radj = 0.0299

(g)

The quadratic regression model in 16.2 is superior to this model. Radj is far less than the

model for 16.2. 16.7

(a)

ln Yˆ = 2.3882 + 0.0045X1

(b)

ln Ŷ = 1.9708+0.0040(85) = 2.77

(c)

The residual analysis indicates a clear quadratic pattern. The model does not adequately fit the data. F = 1.218 > F1,26 = 4.23. Fail to reject H0. The model does not provide a significant relationship. R2 = 0.04475. 4.475% of the variation in litres per 100 kilometres can be explained by variation in the natural logarithm of speed. R2adj = 0.008. The quadratic regression model in 16.2 is superior to 16.6 and this model. R2adj = 0.008 is less than both the other models. The model in 16.2 with the highest R2adj is the most appropriate model to use.

(d) (e) (f) (g)

16.8

(a) (b) (c) (d) (e) (f) (g)

16.9

(a) (b) (c) (d) (e)

Ŷ = e2.77 = 15.97/100 km

ln Yˆ = 11.413 + 0.0000234 X ln Yˆ = 11.413 + 0.0000234(5000) = 11.53

Yˆ = e11.53 = 101794.52

The residual analysis does not indicate any clear patterns. t = 3.6164 > t25 = 2.0595. Reject H0. The model provides a significant relationship. R2 = 0.3435. 34.35% of the variation of the log of tourism employment is explained by the number of tourism establishments. R2adj = 0.3172. We cannot directly compare R2 values as there is a different dependent variable.

Yˆ = 6390.1439 + 2814.4438 X Yˆ = −10.3422 + 15.4973 5000 = 205401.37 The residual analysis does not indicate clear patterns. t = 7.556 > t25 = 2.0595. Reject H0. The model provides a significant relationship. R2 = 0.6955. So, 69.55% of the variation in tourism employment can be explained by variation in the square root of the number of tourism establishments.

(f)

2 Radj = 0.6833

(g)

This model here is the best, based upon R2adj and residual analysis.

16.10

ervation

Predicted Retirement

Residuals

Standard Residuals

h_i

Stud_t

Cook_d

t test

Hat test

Cook's test

18.0278

1.972205

0.381265

0.247177

0.40966

0.027171

23.72864

10.27136

1.985652

0.114677

2.366037

0.247231

Yes

21.94137

0.058628

0.011334

0.177088

0.011567

1.44E-05

19.29784

-3.29784

-0.63754

0.097828

-0.63167

0.020937

22.2517

9.748295

1.884533

0.088544

2.151245

0.162225

Yes

20.67133

-3.67133

-0.70974

0.075133

-0.69695

0.018962

21.83793

-1.83793

-0.35531

0.308384

-0.39815

0.034881

23.52175

-5.52175

-1.06746

0.09565

-1.0894

0.057114

19.40128

-0.40128

-0.07758

0.142931

-0.0776

0.000502

17.51057

6.489426

1.254531

0.344031

1.575389

0.539284

26.37217

-1.37217

-0.26527

0.194796

-0.27455

0.009061

33.86028

0.139721

0.027011

0.662903

0.043075

0.001824

Yes

16.44742

-2.44742

-0.47313

0.207118

-0.49697

0.031607

20.77477

-2.77477

-0.53642

0.082927

-0.5245

0.01216

22.35515

-7.35515

-1.42189

0.160812

-1.57932

0.197858

Since none of the observations are flagged by all three diagnostic tests, there is no strong evidence that any of them are influential. 16.11

ervation

Predicted emission

Residual

Stud_t

Cook_d

Hat test

t test

Cook’s test

1840.275412

2,425.76

0.181384

8.588

1.060474428

yes

3355.265466

-877.24

0.762805

-2.67

6.947020552

yes

561.3083952

438.87

0.141735

0.554

0.024675892

472.7739058

490.71

0.194705

0.643

0.048136377

1315.87294

-754.66

0.225743

-1.04

0.142795886

808.685565

-458.18

0.097207

-0.56

0.016670476

304.3976324

34.30

0.436311

0.053

0.001075796

yes

462.0410228

-220.34

0.155167

-0.28

0.007027675

295.0947749

-53.76

0.164843

-0.07

0.000454879

224.1418807

4.91

0.153256

0.006

3.42854E-06

610.6208019

-413.92

0.105474

-0.51

0.015036999

384.3232706

-193.56

0.122784

-0.24

0.003980501

230.8918401

-65.22

0.109291

-0.08

0.000390177

523.2870923

-357.67

0.149297

-0.45

0.017571869

No observation is flagged by all tests as being influential. Analysis re-performed without influential data. Since none of the observations are flagged by all three diagnostic tests, there is no strong evidence that any of the remaining data are influential. Copyright © 2019 Pearson Australia (a division of Pearson Australia Group Pty Ltd) 9781488617249, Berenson, Basic Business Statistics 5e

16.12

Predicted Observation Quality 1 5.525891816 2 6.398411466 3 4.893321595 4 4.988773819 5 6.759794591 6 5.231061475 7 5.0465727 8 5.170344626 9 5.73548282 10 5.082187321 11 5.407955826 12 5.656958661 13 5.844365403 14 5.560627192 15 5.090641717 16 5.736941803 17 5.743656983 18 5.208877214 19 5.988862606 20 5.677683938 21 6.187361479 22 5.016214276 23 5.97251428 24 5.012417064 25 6.927935039 26 6.668719318 27 5.0465727 28 5.182895741 29 5.989442345 30 4.905872709 31 7.480522118 32 6.10883732 33 5.611131155 34 5.738400787 35 6.17393112 36 5.219969344 37 5.221428328 38 5.368843499 39 6.256252491 40 5.116922696

Residual 0.474108 -0.39841 0.106678 0.011226 1.240205 0.768939 0.953427 -0.17034 0.264517 -0.08219 0.592044 -0.65696 0.155635 0.439373 0.909358 -0.73694 0.256343 -0.20888 -0.98886 0.322316 -1.18736 -0.01621 0.027486 0.987583 1.072065 0.331281 0.953427 -0.1829 1.010558 1.094127 -1.48052 -0.10884 0.388869 1.261599 -0.17393 -1.21997 -1.22143 -0.36884 -1.25625 -0.11692

h 0.030716 0.043837 0.062633 0.03463 0.092437 0.034376 0.034007 0.043049 0.02299 0.044375 0.02755 0.022129 0.023494 0.020114 0.18272 0.021154 0.026906 0.045917 0.034022 0.020884 0.036323 0.037124 0.026087 0.034452 0.112993 0.073121 0.034007 0.032502 0.03509 0.049036 0.189116 0.032774 0.046464 0.020728 0.034934 0.039621 0.033932 0.03117 0.039418 0.029223

student t 0.546126 -0.46165 0.124584 0.012917 1.506841 0.892074 1.110955 -0.19694 0.302827 -0.09505 0.682066 -0.75559 0.178108 0.503129 1.153121 -0.84849 0.294043 -0.2419 -1.1534 0.368773 -1.3954 -0.01868 0.031486 1.152131 1.310127 0.38963 1.110955 -0.21031 1.180119 1.290673 -1.93099 -0.12513 0.45117 1.474187 -0.20024 -1.43795 -1.43531 -0.42444 -1.48252 -0.13418

Cook D 0.004416 0.004455 0.000457 2.79E-06 0.091636 0.013003 0.019776 0.0008 0.001028 0.000192 0.006175 0.006106 0.000364 0.002482 0.099887 0.007345 0.001131 0.001283 0.021283 0.001387 0.032766 6.25E-06 1.26E-05 0.021497 0.084058 0.005155 0.019776 0.000695 0.022927 0.037601 0.274713 0.000248 0.0045 0.021091 0.000676 0.037741 0.032385 0.002712 0.039813 0.000256

t test no no no no no no no no no no no no no no no no no no no no no no no no no no no no no no no no no no no no no no no no

hat test no no no no no no no no no no no no no no yes no no no no no no no no no no no no no no no yes no no no no no no No No No

Cook test no no no no no no no no no no no no no no no no no no no no no no no no no no no no no no no no no no no no no no no no

41 42 43 44 45 46 47 48 49 50

6.087232798 -0.08723 0.036257 -0.10046 0.000177 4.675256923 -1.67526 0.081215 -2.06334 0.146614 4.655990629 -0.65599 0.085169 -0.78035 0.023591 5.016214276 -0.01621 0.037124 -0.01868 6.25E-06 6.17393112 -0.17393 0.034934 -0.20024 0.000676 5.97251428 -1.97251 0.026087 -2.39331 0.064902 5.53406598 2.465934 0.037801 3.122916 0.147094 3.450388285 -0.45039 0.573945 -0.78515 0.110941 7.682818203 0.317182 0.193676 0.399999 0.012936 4.796990126 -0.79699 0.057739 -0.93681 0.023514

no yes no no no yes yes no no no

No No No No No No No No Yes No

no no no no no no no no no no

No observation is flagged by all tests as being influential.

.13

vation

Stud_t

Cook_d

Hat test

t test

Cook’s test

0.39175

3.503247

1.774455

yes

0.344334

-0.33097

0.028452

0.114163

0.620158

0.023865

0.399828

-0.63806

0.130305

0.10992

0.651784

0.025163

0.133582

-0.103

0.000817

0.389995

2.75405

1.378803

yes

0.281909

-0.23319

0.010616

0.368244

-1.25789

0.398152

0.087988

-0.38193

0.006935

0.105201

-0.40399

0.00944

0.108234

-1.12388

0.068056

0.164938

-1.05384

0.098716

Since none of the observations are flagged by all three diagnostic tests, there is no strong evidence that any of them are influential. 16.14 Observation

Predicted Sales

Sales

Residual

studt

cook

hat test

t test

cook test

828.2415577

973

144.7584

0.292389

1.118013

0.13

828.2415577

1119

290.7584

0.292389

2.510081

0.54

yes

903.3294104

875

-28.3294

0.098132

-0.1879

0.00

903.3294104

625

-278.329

0.098132

-2.0356

0.15

1052.709206

910

-142.709

0.066755

-0.95154

0.03

1052.709206

971

-81.7092

0.066755

-0.53629

0.01

1202.089001

931

-271.089

0.060799

-1.92424

0.09

1099.539625

882

-217.54

0.055995

-1.48895

0.05

1099.539625

982

-117.54

0.055995

-0.7732

0.02

1500.848591

1628

127.1514

0.125148

0.872466

0.04

1500.848591

1577

76.15141

0.125148

0.515937

0.01

810.4644823

1044

233.5355

0.356373

2.018837

0.44

yes

810.4644823

914

103.5355

0.356373

0.826626

0.09

1295.749839

1329

33.25016

0.069023

0.217134

0.00

1295.749839

1330

34.25016

0.069023

0.223681

0.00

1445.129634

1405

-40.1296

0.080129

-0.26379

0.00

1445.129634

1436

-9.12963

0.080129

-0.05991

0.00

1594.509429

1521

-73.5094

0.116655

-0.49537

0.01

1594.509429

1741

146.4906

0.116655

1.006684

0.05

1743.889225

1866

122.1108

0.178602

0.864406

0.05

1743.889225

1717

-26.8892

0.178602

-0.18688

0.00

Since none of the observations are flagged by all three diagnostic tests, there is no strong evidence that any of them are influential. 16.15 (a)

For the model that includes independent variables A and B, the value of Cp exceeds 3, the number of parameters, so this model does not meet the criterion for further consideration. For the model that includes independent variables A and C, the value of Cp is less than or equal to 3, the number of parameters, so this model does meet the criterion for further consideration. For the model that includes independent variables A, B and C, the value of Cp is less than or equal to 4, the number of parameters, so this model does meet the criterion for further consideration.

(b)

The inclusion of variable C in the model appears to improve the model’s ability to explain variation in the dependent variable sufficiently to justify its inclusion in a model that contains only variables A and B.

16.16 (a)

(b) 16.17 (a)

CP =

(1 − Rk2 )( n − T ) (1 − 0.274 )( 26 − 7) − n − 2(k + 1) = − 26 − 2(2 + 1) 2 1 − RT 1 − 0.623

= 16.59 Cp overwhelmingly exceeds k + 1 = 3, the number of parameters (including the Y-intercept), so this model does not meet the criterion for further consideration as the best model. Let Y = sales, X1 = Wonderlic personnel test score, X2 = Strong–Campbell interest inventory test score, X3 = experience, X4 = 1 with a degree in electrical engineering; 0 otherwise. Based on a full regression model involving all of the variables: All VIFs are less than 5. So there is no reason to suspect collinearity between any pair of variables. The best-subset approach yields the following models to be considered:

Partial output from the best-subsets selection:

Model

5 4

Consider Std. Error This Model? 0.593228 0.552551222 11.74203 Yes 0.5922 0.562360664 11.6126 Yes

0.593217 0.563452639 11.59811 Yes

0.5922

X1X2X3X4 5 X1X2X4 3.1011 55 X2X3X4 3.0010 97 X2X4 1.1011 72

R2adj

0.572780469 11.47353 Yes

Partial PHStat2 output of the full regression model: Coefficients Intercept Wonder SC Experience Engineer Dummy

25.7683 -0.0134 1.3514 0.1682 7.2747

Standard Error 13.9537 0.4050 0.1947 0.5287 4.1011

t Stat

P-value

1.8467 -0.0331 6.9407 0.3180 1.7738

0.0722 0.9737 0.0000 0.7521 0.0837

Since the p-value for X1 and X3 are larger than 0.05, they do not have significant effect individually on sales. The best model should include both X2 and X4. PHStat2 output of the model with only X2 and X4: Regression Statistics Multiple R 0.7695 R2 0.5922 R2adj 0.5728 Standard Error 11.4735 Observations 45

ANOVA df Regression

Residual

Total

44 Coefficients

Intercept 26.8910 SC 1.3408 Engineer Dummy 7.2869

Significance F 8029.041 4014.520 30.4958 6.59784E-09 3 7 5528.958 131.6419 7 13558 Standard tcalc P-value Error 9.7718 2.7519 0.0087 0.1792 7.4824 0.0000 3.9857 1.8282 0.0746

The most appropriate model to predict sales is:

Ŷ = 26.8910 +1.3408 X 2 + 7.2869 X 4 (b)

With the exception of one single residual point at a value of –44.58 when SC = 54, there is no specific pattern in the residual plot. Influential analysis using the hat matrix elements, the deleted residuals and Cook’s distance statistic does not reveal any observation that needs to be deleted.

According to the finding in (a), the company only needs to administer the Strong–Campbell test. (c) (d) (e)

According to the model in (a), the variable X4 helps predict sales and, hence, the idea of only hiring electrical engineers should be supported. Prior selling experience (X3) does not help predict sales according to the model chosen in (a). The company only needs to administer the Strong–Campbell test to save time and money. It should hire only sales managers with an electrical engineering degree.

16.18

Let Y = Revenue, X1 = Number of partners, X2 = Number of Professionals, X3 = MAS (%), X4 = 1 for Southeast Region; 0 otherwise, X5 = 1 for Gulf Coast Region; 0 otherwise. Based on a full regression model involving all of the variables:

X1 and X2 have VIFs > 5 with X2 having the highest VIF = 6.9763. All VIFs < 5 after dropping X2. So there is no reason to suspect collinearity between any pair of the remaining variables. Let Y = Revenue, X1 = Number of partners, X2 = MAS (%), X3 = 1 for Southeast Region; 0 otherwise, X4 = 1 for Gulf Coast Region; 0 otherwise. The best-subset approach yields the following models to be considered: Partial PHStat output from the best-subsets selection: Model

k+1

R Square

Adj. R Square

Std. Error

Consider This Model? Copyright © 2019 Pearson Australia (a division of Pearson Australia Group Pty Ltd) 9781488617249, Berenson, Basic X1X2 2.1964 3 0.8237 0.8157 20.5374 Yes Business Statistics 5e X1X2X3 3.3349 4 0.8272 0.8151 20.5667 Yes X1X2X4 3.2422 4 0.8276 0.8155 20.5441 Yes

Partial PHStat output of the full regression model: Coefficients Standard Error Intercept -3.4970 6.5617 Number of Partners 1.3536 0.1063 MAS (%) 0.7215 0.2507 SE 3.9856 8.0979 GC -4.2658 7.3713

t Stat -0.5329 12.7379 2.8778 0.4922 -0.5787

P-value 0.5969 0.0000 0.0063 0.6252 0.5659

Since the p-value for X3 and X4 are larger than 0.05, they do not have significant effect cont. individually on revenue. X3 with the largest pvalue is dropped. Partial PHStat output of the regression model with X3 dropped: Intercept Number of Partners MAS (%) GC

Coefficients Standard Error -1.8107 5.5465 1.3705 0.0997 0.6994 0.2445 -6.1515 6.2416

t Stat P-value -0.3264 0.7457 13.7456 0.0000 2.8608 0.0065 -0.9856 0.3299

Since the p-value for X4 is larger than 0.05, it does not have significant effect individually on revenue. X4 with the largest p-value is dropped. Partial PHStat output of the regression model with X4 dropped: Coefficients Standard Error Intercept -4.1412 5.0156 Number of Partner s 1.3698 0.0997 MAS (%) 0.7090 0.2442

t Stat P-value -0.8257 0.4135 13.7439 0.0000 2.9031 0.0058

The best model should include both X1 and X2 PHStat output of the model with only X1 and X2: Regression Statistics Multiple R 0.9076 R Square 0.8237 Adjusted R Square 0.8157 Standard Error 20.5374 Observations 47 ANOVA df Regression Residual Total

Intercept Number of Partners MAS (%)

2 44 46

SS MS F Significance F 86692.4957 43346.2479 102.7682 0.0000 18558.6100 421.7866 105251.1057

Coefficients Standard Error -4.1412 5.0156 1.3698 0.0997 0.7090 0.2442

t Stat P-value -0.8257 0.4135 13.7439 0.0000 2.9031 0.0058

Lower 95% Upper 95% -14.2494 5.9671 1.1689 1.5707 0.2168 1.2012

The p-value of the t-test for the significance of individual independent variables are all < .05. t = 4.2369  t  = 1.6811 The first observation has a Studentized deleted residual *

with d.f. = 44, a hat matrix diagonal element hi = 0.5315  2 ( k + 1) / n = = 0.1277 2

and a Cook’s Di = 4.8999 > F =

with d.f. = 3 and 44.

Hence, using the Studentized deleted residuals, hat matrix diagonal elements and Cook’s distance statistic together, the first observation should be deleted from the data set.

PHStat output with the first observation deleted: cont. Regression Statistics Multiple R 0.7959 R Square 0.6334 Adjusted R Square 0.6164 Standard Error 17.4495 Observations 46 ANOVA df Regression Residual Total

Intercept Number of Partners MAS (%)

2 43 45

SS MS F Significance F 22621.7672 11310.8836 37.1478 0.0000 13092.7953 304.4836 35714.5625

Coefficients Standard Error 5.7798 4.8624 1.0089 0.1201 0.5402 0.2113

t Stat P-value 1.1887 0.2411 8.4006 0.0000 2.5565 0.0142

Lower 95% Upper 95% -4.0262 15.5858 0.7667 1.2512 0.1140 0.9663

The most appropriate model to predict sales is Yˆ = 5.7798 + 1.0089 X 1 + 0.5402 X 3

Residual Plot for Number of Partners 80

20 Residual

Residuals

Residual

Normal Probability Plot 80

20 0 -20

-20

-40

-40 -3

-2

-1

0 Z Value

50 100 Number of Partners

150

Residual Plot for MAS (%) 80 60

Residuals

40 20 0 -20 -40 0

30 MAS (%)

The normal probability plot suggests possibly departure from the normality assumption. However, because of the large sample size, the validity of the results is not seriously impacted. The residual plots do not reveal any specific pattern. 16.19 Using best subsets Best subsets analysis Intermediate Calculations R2T

0.399409

1 - R2T

0.600591

n-T

Model

k+1

R2adj

Std. Error

1.0135

0.3987

0.3525

5.3711

7.8115

0.0585

-0.0139

6.7211

X1X2

3.0000

0.3994

0.2993

5.5873

Using Stepwise df

Significance F

Regression

248.7041

8.6211

0.0116

Residual

375.0293

28.8484

Total

Intercept Unemployment Rate

623.7333

Coefficients

Standard Error

tcalc

Pvalue

Lower 95%

Upper 95%

10.5619

4.0923

2.5809

0.0228

1.7210

19.4028

1.4493

0.4936

2.9362

0.0116

0.3829

2.5157

Based upon best subsets and stepwise techniques we should include only the unemployment rate as an explanatory variable. 16.20 Using best subsets Best subsets analysis Intermediate Calculations R2T

0.968235

1 – R2T

0.031765

n–T

Model

k+1

R Square

Adj. Square

Std. Error

9.1995

0.9538

0.9513

43.0746

80.4280

0.8281

0.8190

83.0648

X1X2

3.0000

0.9682

0.9647

36.6819

Using stepwise Stepwise Regression Analysis Table of Results for General Stepwise price entered.

Significance F

Regression

727224.1267

391.9454

0.0000

Residual

35253.0161

1855.4219

Total

762477.1429

Coefficients

Standard Error

tcalc

P-value

Lower 95%

Upper 95%

Intercept

985.2239

29.7425

33.1251

0.0000

922.9721

1047.4758

price

-0.9649

0.0487

-19.7976

0.0000

-1.0670

-0.8629

Significance F

Regression

738257.0376

369128.5188

274.3305

0.0000

Residual

24220.1053

1345.5614

Total

762477.1429

Coefficients

Standard Error

tcalc

P-value

Lower 95%

Upper 95%

Intercept

735.9766

90.6539

8.1185

0.0000

545.5198

926.4334

price

-0.7539

0.0846

-8.9122

0.0000

-0.9316

-0.5762

income

0.2116

0.0739

2.8635

0.0103

0.0564

0.3669

income entered.

Based upon the output we should include both explanatory variables – price and income. 16.21 Using best subsets Best subsets analysis

Intermediate Calculations R2T

0.524165

1–R T

0.475835

n–T

Model

k+1

R2adj

Std. Error

5.7214

0.2995

0.2358

4.8257

12.0048

0.0005

-0.0903

5.7643

X1X2

3.0000

0.5242

0.4290

4.1714

Using stepwise Stepwise Analysis

Regression

Table of Results for General Stepwise No variables could be entered into the model. Based upon the output we should include both explanatory variables. 16.22 We can measure the influence of individual observations on the model by computing the Variance Inflationary Factor (VIF). 16.23 Stepwise regression and best-subsets regression both identify useful predictors during the exploratory stages of model building. Stepwise regression starts with one independent variable and adds variables until we reach a stage where no more significant variables can be added. In comparison, best-subsets regression evaluates all possible specifications, and we generally use the adjusted R2 or Cp to arrive at the preferred model. 16.24 The Cp is used to determine which combination of independent X variables is best to use in a multiple regression model. The Cp statistic, defined in Equation 16.10, measures the differences between a fitted regression model and a true model, along with random error. When a regression model with k independent variables contains only random differences from a true model, the mean value of Cp is k + 1. The goal is to find models whose Cp is close to or less than k + 1. 16.25 There are several ways of validating a regression model: • Collect new data and compare the results. • Compare the results of the regression model with previous results. • If the data set is large, split the data into two parts and cross-validate the results.

16.26 (a)

For the scatter diagram: X-axis to represent the independent variable ‘exchange rate’ while Y-axis to represent the dependent variable ‘investor confidence’.

(b)

Ŷ = 2.6761-0.1095X +0.0018X 2

(c)

Ŷ = 2.6761-0.1095(85)+0.0018(85)2 = 6.1391

(d)

H0 = 1 = 2 = 0 H1 = at least one j ≠ 0 F = 131.66 > F2,9 = 4.26. Reject H0. There is a significant quadratic relationship between investor confidence and exchange rates.

(e)

H0 = including the quadratic effect does not significantly improve the model (2 = 0). H1 = including the quadratic effect significantly improves the model (2 ≠ 0). tcalc = 2.1595 < t9 = 2.2622. Fail to reject H0. The quadratic effect is not significant. Therefore, the quadratic model is not a better fit than the linear regression model.

(f)

(g)

The coefficient of multiple determination R2 represents the proportion of variation in Y that is explained by the variation in the independent variables. Therefore, 96.69% of the variation in investor confidence is explained by the quadratic relationship between investor confidence and exchange rates.

é (n-1) ù 2 Radj = 1- ê(1- R2 ) ú (n- k -1) û ë é (12-1) ù 2 Radj = 1- ê(1- 0.9669) ú (12- 2-1) û ë 2 Radj = 1-0.0405 = 0.9595

Since the variable Rooms is the sum of Bathrooms, Bedrooms, Loft/Den and Finished Basement, it is removed from the list of potential independent variable. Including it will introduce perfect collinearity.

An analysis of the linear regression model using PHStat with all of the remaining seven possible independent variables revealed that none of the variables have VIF values in excess of 5.0. A best subsets regression produces the following potential models that have Cp values less than or equal to k+1. Model X1X2X3X4X5 X1X2X3X5X6 X1X2X3X4X5X6 X1X2X3X4X5X6X7

Cp 5.657125 5.78991 6.074978 8

k+1 6 6 7 8

R Square 0.532645 0.531509 0.546173 0.546814

Adj. R Square 0.490157971 0.488919347 0.49574803 0.486959627

Std. Error yes yes yes yes

where X1 = Hot tub (0 = No and 1 = Yes), X2 = Lake View (0 = No and 1 = Yes), X3 = Bathrooms, X4 = Bedrooms , X5 = Loft/Den (0 = No and 1 = Yes), X6 = Finished Basement (0 = No and 1 = Yes), X7 = Acres.

Looking at the p-values of the t statistics for each slope coefficient of the model that includes X1 through X7 reveals that Bathrooms, Bedrooms, Loft/Den, Finished Basement and Acres are not significant at 5% level of significance. Intercept Hot Tub Lake View Bathrooms Bedrooms Loft/Den Finished Basement Acres

Coefficients 56.86025281 83.33704829 188.1459004 44.97723054 31.7825021 68.68786861 42.40047126 8.214876036

Standard Error 75.83635651 39.54169465 46.55629756 28.05985879 24.30126221 34.62218517 35.03765953 30.00092343

t Stat 0.749775641 2.107574019 4.041255648 1.602902954 1.307853963 1.983926441 1.210139942 0.273820773

P-value 0.456705242 0.039815205 0.000172817 0.114899805 0.196567371 0.052453764 0.231594833 0.7852867

Dropping Acres which has the highest p-value, the new regression indicates that Bathrooms, Bedrooms and Finished Basement are still not significant. Intercept Hot Tub Lake View Bathrooms Bedrooms Loft/Den Finished Basement

Coefficients 63.04035916 82.87369204 190.4252197 44.42625163 31.82322517 68.8302055 43.67867801

Standard Error 71.77716439 39.16564269 45.41206354 27.74686825 24.09177116 34.3204956 34.42659968

t Stat 0.878278763 2.115979372 4.193273877 1.601126702 1.320916796 2.005513157 1.268747957

P-value 0.383683708 0.038973223 0.000102798 0.115183479 0.192099589 0.049930143 0.209972962

Dropping Finished Basement, which has the largest p-value, the new regression indicates cont. that Bathrooms and Bedrooms are still insignificant.

Intercept Hot Tub Lake View Bathrooms Bedrooms Loft/Den

Coefficients 33.50312632 98.1802218 181.7931221 52.45485906 36.99281947 79.72730905

Standard Error 68.27210075 37.46721666 45.14769931 27.1649841 23.87596476 33.41209396

t Stat 0.490729389 2.620430087 4.026630921 1.930973303 1.549374856 2.386181158

P-value 0.625570141 0.011330891 0.000174862 0.058647073 0.127027169 0.020493138

Dropping Bedrooms next, which has the largest p-value, the new regression indicates that all the remaining variables become significant at 5% level of significance. Intercept Hot Tub Lake View Bathrooms Loft/Den

Coefficients 102.0014472 89.12922256 183.8349004 77.69078822 76.80439638

Standard Error 52.67086383 37.4689494 45.68930995 22.01055013 33.77336974

t Stat 1.936582007 2.378748911 4.023586712 3.529706789 2.274111141

P-value 0.057848524 0.020806217 0.00017363 0.000839855 0.026811687

The best linear model is determined to be: Yˆ= 102.0014

+ 89.1292 X 1 + 183.8349 X 2 + 77.6908

X 3 + 76.8044 X .

The overall model has FSTAT = 14.7030 (4 and 56 degrees of freedom) with a p-value that is virtually 0. r2 = 0.5122, r = 0.4774. 2

adj

The 52nd observation has a Studentized deleted residual

= 5.7102  t t /2 = 1.6730 with d.f. = i

56, a hat matrix diagonal element hi = 0.1835  2 ( k + 1) / n = 0.1639 and a Cook’s Di = 0.9370 > F = 0.8809

with d.f. = 5 and 56.

Hence, using the Studentized deleted residuals, hat matrix diagonal elements and Cook’s distance statistic together, the 52nd observation should be deleted from the data set.

PHStat output after deleting the 52nd observation. cont. Regression Statistics Multiple R 0.7622 R Square 0.5810 Adjusted R Square 0.5505 Standard Error 76.9111 Observations 60 ANOVA df Regression Residual Total

4 55 59

SS MS F Significance F 451063.3243 112765.8311 19.0634 0.0000 325342.2590 5915.3138 776405.5833

Coefficients Standard Error 88.6802 42.1756 34.2776 31.4593 204.4737 36.7076 85.2307 17.6472 36.9265 27.8907

Intercept Hot Tub Lake View Bathrooms Loft/Den

t Stat P-value 2.1026 0.0401 1.0896 0.2806 5.5703 0.0000 4.8297 0.0000 1.3240 0.1910

Lower 95% Upper 95% 4.1584 173.2020 -28.7683 97.3235 130.9101 278.0374 49.8650 120.5965 -18.9677 92.8208

A residual analysis does not reveal any strong patterns and the normal probability plot does not suggest any departure from the normality Residual Plot for Hot Tub assumption. 250 200

Residuals

150 100 50 0 -50 -100 -150 0

0.2

0.4

0.6 Hot Tub

0.8

1.2

Residual Plot for Lake View 250 200

Residuals

150 100 50 0 -50 -100 -150 0

0.2

0.4

0.6 Lake View

0.8

1.2

Residual Plot for Bathrooms 250 200 150

Residuals

100 50 0 -50 -100 -150 0

2 3 Bathrooms

Residual Plot for Loft/Den 250 200 150

Residuals

100 50

Residual

0 -50

0.5

1.5

-100 -150

Loft/Den

Normal Probability Plot 250 200 150 100

Residual

50 Residual

0 -50

-100

16.28 -150Using best subsets -3 -2 analysis -1 0 Best subsets

Z Value

Intermediate Calculations

R2T

0.970118

1–R T

0.029882

n–T

Model

k+1

R2adj

Std Error

0.1637

0.9632

0.9586

177500.8425

40.3816

0.7228

0.6882

486916.0372

3.2951

0.9444

0.9375

217976.0978

151.2612

0.0602

-0.0573

896585.6861

X1X2

1.7114

0.9659

0.9561

182661.4553

X1X3

1.6629

0.9662

0.9565

181885.4082

X1X4

1.9200

0.9646

0.9545

185968.4955

X2X3

4.3839

0.9499

0.9356

221309.7260

X2X4

41.8160

0.7262

0.6480

517351.3096

X3X4

4.3188

0.9503

0.9361

220449.2488

X1X2X3

3.3110

0.9683

0.9524

190256.2955

X1X2X4

3.5184

0.9670

0.9505

193935.6927

X1X3X4

3.2829

0.9684

0.9526

189751.7029

X2X3X4

5.6387

0.9543

0.9315

228170.1253

X1X2X3X4

5.0000

0.9701

0.9462

202220.9133

Using stepwise Stepwise Regression Analysis Table of Results for General Stepwise price chicken ($/kg) entered. df

Significance F

Regression

6590515245923.2000

209.1792

0.0000

Residual

252052392626.8990

31506549078.3624

Total

6842567638550.1000

Intercept price chicken ($/kg)

Coefficients

Standard Error

t Stat

P-value

Lower 95%

Upper 95%

7371855.9427

248124.5325

29.7103

0.0000

6799679.7448

7944032.1406

-293600.0036

20300.0352

-14.4630

0.0000

-340411.9687

-246788.0385

No other variables could be entered into the model. Stepwise ends. The best model includes the price of chicken only as an explanatory variable. Copyright © 2016 Pearson Australia (a division of Pearson Australia Group Pty Ltd) 9781486018956 Berenson/Basic Business Statistics 4e

16.29

Glencove: Based on a full regression model involving all of the variables: All VIFs are less than 5. So there is no reason to suspect collinearity between any pair of variables. The best-subset approach yielded the following models to be Adj. R Model Cp k+1 R Square Std. Error Consider considered: This Square X1X2X3 X1X2X3X4 X1X2X3X5 X1X2X3X6 X1X2X3X4X5 X1X2X3X4X6 X1X2X3X5X6 X1X2X3X4X5X6

2.1558 4.1117 3.2400 3.8887 5.1832 5.7825 5.1038 7.0000

4 5 5 5 6 6 6 7

0.8424 0.8427 0.8484 0.8442 0.8488 0.8449 0.8493 0.8500

0.8242 0.8175 0.8241 0.8192 0.8173 0.8125 0.8179 0.8108

60.5007 61.6425 60.5180 61.3567 61.6903 62.4827 61.5846 62.7677

Model? Yes Yes Yes Yes Yes Yes Yes Yes

The stepwise regression approach reveals the following best model: Intercept Property Size (acres) House Size (square feet) Age

Coefficients Standard Error 260.6791 66.3288 362.8318 48.6233 0.1109 0.0228 -1.7543 0.5483

t Stat P-value 3.9301 0.0006 7.4621 0.0000 4.8682 0.0000 -3.1996 0.0036

The p-value of the individual slope coefficients indicate that all the remaining independent variables are significant individually. Combining the results of both approaches, the most appropriate multiple regression model for predicting fair market value in Glencove is Yˆ = 260.6791+362.8318 X 1 + 0.1109 X 2 − 1.7543 X 3

Chapter 17: Decision making

17.1

(a)

Opportunity loss table Optimum Event, Ei E1 E2

(b)

Action

Profit

B A

$100 $200

Alternative courses of action A B $50 $0

$0 $75

Decision tree

$50

$200

$100

$125

17.2

(a)

Opportunity loss table Optimum Event, Ei E1 E2 E3

(b)

Action

Profit

A A A

$50 $300 $500

Alternative courses of action A B $0 $0 $0

$40 $200 $300

Decision tree

$50

$300

$500

E3 17.3

(a)

$10 $100 $200

Possible levels of production for a small factory and payoffs for each level of production

(b) Level of Production (pairs of jeans/ Payoff year) 10,000 $100,000 – $200,000 = $100,000 20,000 $200,000 – $200,000 = $0 50,000 $500,000 – $200,000 = (b) Possible levels of production for a large factory and payoffs for each level of $300,000 Level of Production (pairs of jeans/ Payoff year) 10,000 $100,000 – $400,000 = – $300,000 20,000 $200,000 – $400,000 = – $200,000 50,000 $500,000 – $400,000 = $100,000 100,000 $1,000,000 – $400,000 = $600,000 (c) Payoff table Copyright © 2019 Pearson Australia (a division of Pearson Australia Group Pty Ltd) 9781488617249, Berenson, Basic Business Statistics 5e

production

(c)

Let

Event 1 (E1) = 10,000 pairs of jeans/yr Event 2 (E2) = 20,000 pairs of jeans/yr Event 3 (E3) = 50,000 pairs of jeans/yr, and Event 4 (E4) = 100,000 pairs of jeans/yr.

Event, Ei E1 E2 E3 E4 (d)

Alternative Courses of Action Small Factory Large Factory -$100,000 -$300,000 $0 -$200,000 $300,000 $100,000 $600,000

Decision tree E1

-$100,000

$300,000

-$300,000

-$200,000

$100,000

$600,000

Small factory

Large factory

(e)

Opportunity loss table Optimum Event, Ei

Action

Profit

E1 E2 E3 E4

Small Small Small Large

-$100,000 $0 $300,000 $600,000

Alternative courses of action Small Large $0 $0 $0 –

$200,000 $200,000 $200,000 $0

17.4

(a)

Payoff for each level of demand for company A and company B Payoffs Level of Demand (Books sold) 1,000 2,000

10,000

Payoff table indicating events and alternative courses of action Let

Event 1 (E1) = 1,000 books sold Event 2 (E2) = 2,000 books sold Event 3 (E3) = 5,000 books sold, and Event 4 (E4) = 100,000 books sold.

Event, Ei E1 E2 E3 E4 (c)

Company B

$10,000 + $2*1,000 = $12,000 $2000 + $4*1,000 = $6,000 $10,000 + $2*2,000 = $14,000 $2000 + $4*2,000 = $10,000 $10,000 + $2*5,000 = $20,000 $2000 + $4*5,000 = $22,000 $10,000 + $2*10,000 = $2000 + $4*10,000 = $30,000 $42,000

5,000

(b)

Company A

Alternative Courses of Action Company A Company B $12,000 $6,000 $14,000 $10,000 $20,000 $22,000 $30,000 $42,000

Decision tree

Company A

Company B

$12,000

$14,000

$20,000

$30,000

$6,000

$10,000

$22,000

$42,000

(d)

Opportunity loss table Optimum Event, Ei

Action

Profit

A A B B

$12,000 $14,000 $22,000 $42,000

E1 E2 E3 E4

17.5

(a)

Alternative courses of action Company Company B A $0 $6,000 $0 $4,000 $2,000 $0 $12,000 $0

Payoffs of purchases for each level of demand Payoffs (for each quantity of Christmas trees purchased)

Demand 100 (trees) 100 $75*100 – $40*100 = $3,500 200 $3,500

200 $75*100 – $40*200 – $5*100 = $-1,000 $75*200 – $40*200 = $7,000

500

$3,500

$7,000

1000

$3,500

$7,000

(b)

$75*100 – $40*500 – $5*400 = –$14,500 $75*200 – $40*500 – $5*300 = –$6,500 $75*500 – $40*500 = $17,500 $17,500

1000 $75*100 – $40*1000 – $5*900 = –$37,000 $75*200 – $40*1000 – $5*800 = –$29,000 $75*500 – $40*1000 $5*500 = $-5,000 $75*1000 – $40*1000 = $35,000

Payoff table indicating events and alternative courses of action Let

Event 1 (E1) = Event 2 (E2) = Event 3 (E3) = Event 4 (E4) =

Also,

Action A = Action B = Action C = Action D =

Event, Ei E1 E2 E3 E4

500

A $3,500 $3,500 $3,500 $3,500

100 trees sold 200 trees sold 500 trees sold, and 1,000 trees sold.

100 trees purchased 200 trees purchased 500 trees purchased 1,000 trees purchased Alternative Courses of Action B C -$1,000 $7,000 $7,000 $7,000

-$14,500 -$6,500 $17,500 $17,500

D -$37,000 -$29,000 -$5,000 $35,000

(c)

Decision tree

(d)

$3,500

-$1,000

$7,000

-$14,500

-$6,500

$17,500

-$37,000

-$29,000

-$5,000

$35,000

Opportunity loss table

Event, Ei E1 E2

Optimum Action Profit

Alternative courses of action B C D

A B

$3,500 $7,000

$0 $3,500

$4,500 $0

$18,000 $13,500

$40,500 $36,000

$17,500

$14,000

$35,000

$31,500

$10,500

$22,500

$28,000

$17,500

17.6

Excel output Probabilities & Payoffs Table: P

0.5

100

0.5

200

125

200 50

125 100

Max Min Statistics for:

Expected Monetary Value Variance

125

112.5

5625

156.25

Standard Deviation

12.5

Coefficient of Variation Return-to-Risk Ratio

0.6

0.1111 11 9

Opportunity Loss Table: Optimu m Action E1 B E2 A Expected Opportunity Loss

1.666667

Optimum

Alternatives

Profit

A 100 200

B 50 0

0 75

37.5

EVPI (a) (b) (c)

EMVA = 50(0.5) + 200(0.5) = 125 EMVB = 100(0.5) + 125(0.5) = 112.50 EOLA = 50(0.5) + 0(0.5) = 25 EOLB = 0(0.5) + 75(0.5) = 37.50 Perfect information would correctly forecast which event, 1 or 2, would occur. The value of perfect information is the increase in the expected value if you knew which of the events 1 or 2 would occur prior to making a decision between actions. It allows us to select the optimum action given a correct forecast. EMV with perfect information = 100(0.5) + 200(0.5) = 150 EVPI = EMV with perfect information – EMVA = 150 – 125 = 25

(d)

Based on (a) and (b) above, select action A because it has a higher expected monetary value (a) and a lower opportunity loss (b) than action B.

(e)

 A 2 = (50 – 125)2(0.5) + (200 – 125)2(0.5) = 5625

 A = 75

CVA = 75 100% = 60%

(f)

17.7

125  B 2 = (100 – 112.5)2(0.5) + (125 – 112.5)2(0.5) = 156.25 CVB = 12.5 100% =11.11% 112.5 Return-to-risk ratio for A = 125 =1.667 75 Return-to-risk ratio for B = 112.5 = 9.0 12.5

 B = 12.5

(g)

Based on (e) and (f), select action B because it has a lower coefficient of variation and a higher return-to-risk ratio.

(h)

The best decision depends on the decision criteria. In this case, expected monetary value leads to a different decision than the return-to-risk ratio.

PHStat2 output: Expected Monetary Value Probabilities & Payoffs Table: P A E1 0.8 50 E2 0.1 300 E3 0.1 500 Max 500 Min 50

B 10 100 200 200 10

Statistics for: A B Expected Monetary 120 38 Value Variance 21600 3636 Standard Deviation 146.969 60.2992 4 5 Coefficient of Variation 1.22474 1.58682 5 2 Return-to-Risk Ratio 0.81649 0.63019 7 Opportunity Loss Table: Optimum Optimum Action Profit E1 A 50 E2 A 300 E3 A 500 Expected Opportunity Loss

Alternatives A B 0 40 0 200 0 300 A B 0 82 EVPI

(a)

EMVA = 50(0.8) + 300(0.1) + 500(0.1) = 120 EMVB = 10(0.8) + 100(0.1) + 200(0.1) = 38

(b) (c) (d) (e)

EOLA = 0(0.8) + 0(0.1) + 0(0.1) = 0 EOLB = 40(0.8) + 200(0.1) + 300(0.1) = 82 EVPI = 0. The expected value of perfect information is zero because the optimum decision is action A across all three event states. Based on the results of (a) and (b), select action A because it has higher expected monetary value than action B and an opportunity loss of zero.  A 2 = (50 – 120)2(0.8) + (300 – 120)2(0.1) + (500 – 120)2(0.1) = 21,600  A = 146.97 CVA 146.97 (100% ) = 122.47% 120  B 2 = (10 – 38)2(0.8) + (100 – 38)2(0.1) + (200 – 38)2(0.1) = 3636  B = 60.299

CVB = (f)

(g) (h) (i)

17.8

17.9

60.299 (100%) = 158.68% 38

Return-to-risk ratio for A =

120 = 0.816 146.97

Return-to-risk ratio for B =

38 = 0.630 60.299

Based on the results of (e) and (f), select option A because it has a lower coefficient of variation and a higher return-to-risk ratio than action B. The recommendation to select action A is made consistently across both parts (d) and (g). No, the recommendation to select option A is independent of the probabilities whenever action A is the preferred choice across all event states.

(a)

Rate of return = $100 100% = 10%

(b)

CV = $25 100% = 25%

(c)

Return-to-risk ratio = $100 = 4.0

(a) (b)

EMV = 50(0.3) + 100(0.3) + 120(0.3) + 200(0.1) = 101  2 = (50 – 101)2(0.3) + (100 – 101)2(0.3) + (120 – 101)2(0.3)+ (200 – 101)2(0.1)  2= 1,869

$1,000

$100

$25

 = 43.23

(c)

CV = $101 = 42.80%

$43.23 (d)

Return-to-risk ratio =

$101 = 2.336 $43.23

17.10 Select stock A because it has a higher expected monetary value while it has the same standard deviation as stock B. 17.11 Select stock B because it has the same expected monetary value as stock A but a smaller standard deviation and therefore presumably less risk. 17.12 PHStat2 output: Expected Monetary Value Probabilities & Payoffs Table: P Cool weather Warm Weather

Sell hot dogs

Sell Ice cream

0.4

260

190

0.6

230

290

Statistics for: Expected Monetary Value

Sell hot dogs 242

250

Variance

216

2400

Sell Ice cream

Standard Deviation

14.69693846

48.98979486

Coefficient of Variation

0.060731151

0.195959179

Return to Risk Ratio

16.46601438

5.103103631

Optimum

Action Sell hot dogs Sell Ice cream

Profit

Alternatives Sell Ice Sell hot dogs cream

Opportunity Loss Table:

Cool weather Warm Weather

260

290

60 Sell hot dogs

Expected Opportunity Loss

70 0 Sell Ice cream

28 EVPI

(a)

EMV(hot dogs) = 260(0.4) + 230(0.6) = 242 EMV(ice cream) = 190(0.4) + 290(0.6) = 250

(b)

EOL(hot dogs) = 0(0.4) + 60(0.6) = 36 EOL(ice cream) = 70(0.4) + 0(0.6) = 28

(c)

EVPI is the maximum amount of money the vendor is willing to pay for the information about which event will occur.

(d)

Based on (a) and/or (b), choose to sell ice cream because you will earn a higher expected monetary value and incur a lower opportunity loss than choosing to sell hot dogs.

(e)

CV(hot dogs) =

14.6969 100% = 6.0731% 242

CV(ice cream) =

48.9898 100% = 19.5959% 250

(f)

Return-to-risk ratio for hot dogs =

242 = 16.4660 14.6969

Return-to-risk ratio for ice cream =

250 = 5.1030 48.9898

(g)

Based on (e) and (f) we would choose to sell hot dogs because it has the smaller coefficient of variation and the larger return-to-risk ratio.

(h)

Ignoring the variability of the payoff in (d), you will choose to sell ice cream. However, when risk, which is measured by standard deviation, is taken into consideration as in coefficient of variation or return-to-risk ratio, you will choose to sell hot dogs because it has the lower variability per unit of expected return or the higher expected return per unit of variability.

17.13 PHStat2 output: Expected Monetary Value Probabilities & Payoffs Table: Demand P (A) Produce 500 Sell 500 0.2 500 Sell 1000 0.4 500 Sell 2000 0.4 500 Max 500 Min 500

Supply (B) Produce 1000

Statistics for: Expected Monetary Value Variance Standard Deviation Coefficient of Variation Return-to-Risk Ratio

Buy 1000 Buy 2,000 800 600

Buy 500 500 0 0 0 #DIV/0!

0 1000 1000 1000 0

160000 400 0.5 2

1440000 1,200 2 0.5

Optimum Action

Optimum Profit

Sell 500 Buy 500 Sell 1000 Buy 1000 Sell 2000 Buy 2000

Expected Opportunity Loss

Alternatives Produce 500 Produce 1000 500 0 500 1000 500 0 2000 1500 1000 Produce Produce 500 1000 800 500

Produce 2000 1500 1000 0 Produce 2000 700

EVPI (a) (b)

See the table above. EMVA = 500(0.2) + 500(0.4) + 500(0.4) = 500 EMVB = 0(0.2) + 1,000(0.4) + 1,000(0.4) = 800 EMVC = – 1,000(0.2) + 0(0.4) + 2,000(0.4) = 600 Based on the expected monetary value, the company should produce 1,000 kg of tomatoes. This level of supply results in the highest expected monetary value (i.e. will expect to net $800).

(c)

 A 2 = (500 – 500)2(0.2) + (500 – 500)2(0.4) + (500 – 500)2(0.4) = 0 A = 0

 B 2 = (0 – 800)2(0.2) + (1,000 – 800)2(0.4) + (1,000 – 800)2(0.4) =

160,000  B = 400 2 2 2 2  C = (– 1,000 – 600) (0.2) + (0 – 600) (0.4) + (2,000 – 600) (0.4) = 1,440,000  C = 1,200 (d)

Opportunity loss table: Optimum Event, Ei

Action

Profit

E1 E2

A B

$500 $1,000

$0 $500

$500 $0

$1,500 $1,000

$2,000

$1,500

$1,000

EOLA EOLB EOLC (e)

Alternative Courses of Action

= 0(0.2) + 500(0.4) + 1,500(0.4) = 800 = 500(0.2) + 0(0.4) + 1,000(0.4) = 500 = 1,500(0.2) + 1,000(0.4) + 0(0.4) = 700

EMV with perfect information = 500(0.2) + 1,000(0.4) + 2,000(0.4) = 1,300 EVPI = EMV with perfect information – EMVB = 1,300 – 800 = 500 The company should not be willing to pay more than $500 for a perfect forecast.

(f)

CVA = 0 100% = 0% 500

CVB = 400 100% = 50% 800

CVC = 1,200 100% = 200% 600

(g)

Return-to-risk ratio for A = 500 = undefined 0

Return-to-risk ratio for B = 800 = 2.0 400

Return-to-risk ratio for C = 600 = 0.5 1,200

(h)

Based on (b) and (d), choose to supply 1,000 kg of tomatoes. Buying 1,000 kg has the highest expected monetary value ($800) and the lowest expected opportunity loss ($500).

(i)

There is no discrepancy between (b), (d), (f) and (g). In particular, part (g) shows action B to be the preferred option since it has the larger of the two return-to-risk ratios with defined solutions.

(j)

PHStat2 output: Probabilities & Payoffs Table: Demand P (A) Produc e 500 Sell 500 0.2 750 Sell 1000 0.4 750 Sell 2000 0.4 750 Max 750 Min 750 Statistics for: Expected Monetary Value Variance Standard Deviation Coefficient of Variation Return to Risk Ratio

Opportunity Loss Table: Optimu m Action Sell 500 Sell 1000

Buy 500 Buy

(B) Produc e 1000 250 1500 1500 1500 250

Buy 500 750

Buy 1000 1250

Buy 2000 1250

25000 0 500 0.4

22500 00 1500 1.2

#DIV/ 0!

2.5

0.8333 33

Optimu m Profit

Alternatives

0 0

750

Produce 500 0

Produce 1000 500

Produce 2000 1500

1500

750

1000

Sell 2000

1000 Buy 2000

3000

Expected Opportunity Loss

2250

1500

Buy 500 1200

Buy 1000 700

Buy 2000 700

EVPI

(a)

See table above.

(b)

Optimal number of kg of tomatoes the farm should produce is 1,000 or 2,000 kg as both options have EMV of $1,250.

(c)

 A 2 = (750 – 750)2(0.2) + (750 – 750)2(0.4) + (750 – 750)2(0.4) =

A = 0

 B 2 = (250 – 1,250)2(0.2) + (1,500 – 1,250)2(0.4)

+ (1,500 – 1,250)2(0.4) = 250,000 = 500 B 2 2 2  C = (– 750 – 1,250) (0.2) + (500 – 1,250) (0.4) + (3,000 – 1,250)2(0.4) = 2,250,000 (d)

EOLA = 0(0.2) + 750(0.4) + 2,250(0.4) = 1,200 EOLB = 500(0.2) + 0(0.4) + 1,500(0.4) = 700 EOLC = 1,500(0.2) + 1,000(0.4) + 0(0.4) = 700

(e)

EMV with perfect information = 750(0.2) + 1,500(0.4) + 3,000(0.4) = 1,950 EVPI = EMV, perfect information – EMVB or C = 1,950 – 1,250 = 700 The company should not be willing to pay more than $700 for a perfect forecast.

(f)

CVA = 0 100% = 0% 750

CVB = 500 100% = 40% 1,250

CVC = 1,500 100% = 120% 1,250

(g)

750 = undefined 0 Return-to-risk ratio for B = 1,250 = 2.5 Return-to-risk ratio for A =

500

Return-to-risk ratio for C = 1,250 = 0.833 1,500

(h)

Produce 1,000 or 2,000 kg of tomatoes, actions B or C. Buying 1,000 or 2,000 kg has the highest expected monetary value ($1,250) and the lowest expected opportunity loss ($700). But action B has the higher return-to-risk ratio and is the best choice with respect to the return-to-risk.

(k)

PHStat2 output: Probabilities & Payoffs Table: Demand P (A) Produce 500 Sell 500 0.4 500 Sell 1000 0.4 500 Sell 2000 0.2 500 Max 500 Min 500

(B) Produce 1000 0 1000 1000 1000 0

Statistics for: Buy 500 Buy 1000 Buy 2,000 Expected Monetary Value 500 600 0 Variance 0 240000 1200000 Standard Deviation 0 489.8979 1,095.445 Coefficient of Variation 0 0.816497 #DIV/0! Return-to-Risk Ratio #DIV/0! 1.224745 0 Opportunity Loss Table: Optimum Optimum Action Profit Sell 500 Buy 500 Sell 1000 Buy 1000 Sell 2000 Buy 2000

Expected Opportunity Loss

Alternatives Produce Produce Produce 500 1000 2000 500 0 500 1500 1000 500 0 1000 2000 1500 1000 0 Buy Buy 1000 Buy 500 2000 500 400 1000 EVPI

(a)

See table above.

(b)

Optimal number of kg of tomatoes the farm should produce is 1,000 kg as this action has the highest EMV ($600).

(c)

 A 2 = (500 – 500)2(0.4) + (500 – 500)2(0.4) + (500 – 500)2(0.2) =0 A = 0  B 2 = (0 – 600)2(0.4) + (1,000 – 600)2(0.4) + (1,000 – 600)2(0.2) = 240,000  B = 489.90 2 2 2 2  C = (–1,000 – 0) (0.4) + (0 – 0) (0.4) + (2,000 – 0) (0.2) =

1,200,000  C = 1,095.45

(d)

Opportunity loss table: Optimum Event, Ei

Alternative Courses of Action

Action

Profit

E1 E2

A B

$500 $1,000

$0 $500

$500 $0

$1,500 $1,000

$2,000

$1,500

$1,000

EOLA = 0(0.4) + 500(0.4) + 1,500(0.2) = 500 EOLB = 500(0.4) + 0(0.4) + 1,000(0.2) = 400 EOLC = 1,500(0.4) + 1,000(0.4) + 0(0.2) = 1,000 (e)

EMV with perfect information = 500(0.4) + 1,000(0.4) + 2,000(0.2) = 1,000 EVPI = EMV, perfect information – EMVB = 1,000 – 600 = 400 The company should not be willing to pay more than $400 for a perfect forecast.

(f)

CVA = 0 100% = 0% 500

CVB = 489.90 100% = 81.65% 600

CVC = 1,095.45 100% = undefined (g)

Return-to-risk ratio for A = 500 = undefined 0 600 = 1.22 489.90 0 Return-to-risk ratio for C = =0 1,095.45

Return-to-risk ratio for B =

(h)

Produce 1,000 kg of tomatoes, action B. Buying 1,000 kg has the highest expected monetary value ($600) and the lowest expected opportunity loss ($400). Action B has the higher return-to-risk ratio and is the best choice with respect to the return-to-risk.

(i)

Although the values of EMV, EOL and  are affected by $.50/kg changes in the profit and by shifts in the probability with which events occur, the recommendation for action B remains unaffected.

17.14 PHStat2 output: Probabilities & Payoffs Table:

Economy declines No change Economy expands Max

P 0.3 0.5 0.2

Investment Selection A B C 500 –2000 –7000 1000 2000

2000 5000

–1000 20000

2000

5000

20000

Min

500

–2000

–7000

Statistics for: Expected Monetary Value Variance

A 1050

B 1400

C 1400

27250 0 522.01 53 0.4971 57 2.0114 35

62400 00 2497.9 99 1.7842 85 0.5604 49

93240 000 9656.0 86 6.8972 04 0.1449 86

Standard Deviation Coefficient of Variation Return-to-Risk Ratio

Opportunity Loss Table: Optimu m Action Economy A declines No change B Economy C expands Expected Opportunity Loss

(a)

Optimu m Profit 500 2000 20000

Alternatives A 0

B 2500

C 7500

1000 18000

0 15000

3000 0

A 4100

B 3750

C 3750

EVPI

EMVA = 500(0.3) + 1,000(0.5) + 2,000(0.2) = 1,050 EMVB = – 2,000(0.3) + 2,000(0.5) + 5,000(0.2) = 1,400 EMVC = – 7,000(0.3) – 1,000(0.5) + 20,000(0.2) = 1,400 The best investment according to the expected monetary value (EMV) criterion is either B or C since both have the highest net return ($1,400).

(b)

 A 2 = (500 – 1,050)2(0.3) + (1,000 – 1,050)2(0.5) + (2,000 – 1,050)2(0.2) = 272,500  A = 522.02  B 2 = (– 2,000 – 1,400)2(0.3) + (2,000 – 1,400)2(0.5) + (5,000 – 1,400)2(0.2) = 6,240,000  B = 2,498.00

C

= (– 7,000 – 1,400)2(0.3) + (– 1,000 – 1,400)2(0.5) + (20,000 –

1,400)2(0.2) = 93,240,000  C = 9656.09 (c)

EOLA = 0(0.3) + 1,000(0.5) + 18,000(0.2) = 4,100 EOLB = 2,500(0.3) + 0(0.5) + 15,000(0.2) = 3,750

EOLC = 7,500(0.3) + 3,000(0.5) + 0(0.2) = 3,750 (d)

(e)

EMV with perfect information = 500(0.3) + 2,000(0.5) + 20,000(0.2) = 5,150 EVPI = EMV with perfect information – EMVB or C = 5,150 – 1,400 = 3,750 The investor should not be willing to pay more than $3,750 for a perfect forecast. CVA = 522.02 100% = 49.72% 1050

CVB = 2498.00 100% = 178.43% 1400

CVC = 9656.09 100% = 689.72% 1400

(f)

Return-to-risk ratio for A = 1050 = 2.01 522.02

Return-to-risk ratio for B = 1400 = 0.56 Return-to-risk ratio for C =

2498 1400 = 0.14 9656.09

(g)

Based on (e) and (f), choose investment option A since it minimises the coefficient of variation and maximises the investor’s return-to-risk.

(h)

The investor’s decision will vary depending on the criterion that it is based on. In part (a), the best investment option is either B or C since these will result in the highest net return. However, in terms of risk, option A is least risky with the lowest coefficient of variation of 49.72% and the highest return-to-risk ratio (2.01).

(i)

(a) Max EMV

(i) 0.1, 0.6, 0.3 C: 4,700

(ii) 0.1, 0.3, 0.6 C: 11,000

(iv) 0.6, 0.3, 0.1 A: 800

A: 36.64% A: 2.7294

(iii) 0.4, 0.4, 0.2 A or B: 800  A : 548  B : 2,683 A: 4,000 or B: 4,000 A: 54.77% A: 1.8257

(b)  Max EMV

 C : 10,169

 C : 11,145

C: 2,550

C: 1,650

A: 40.99% A: 2.4398 Choose A

Choose A

Different: (a) C (g) A

Different: (a) A or B (g) A

Same: A

 A : 458

A: 2,100

A: 57.28% A: 1.7457

17. 15 PHStat2 output: Probabilities & Payoffs Table: P (A) Large factory Sell 10000 0.1 – 300000 Sell 20000 0.4 – 200000 Sell 50000 0.2 100000 Sell 100000 0.3 600000 Max 600000 Min – 300000

(B) Small factory – 100000 0 300000 300000 300000 – 100000

Statistics for:

Large Small factory factory Expected Monetary 90000 140000 Value Variance 1.27E+1 2.64E+1 1 0 Standard Deviation 356230. 162480. 3 8 Coefficient of Variation 3.95811 1.16057 4 7 Return-to-Risk Ratio 0.25264 0.86164 6 Opportunity Loss Table: Optimum Optimum Action Profit Sell 10000 Small factory Sell 20000 Small factory Sell 50000 Small factory Sell 100000 Large factory

Expected Opportunity Loss

Alternatives Large Small factory factory –100000 200000 0 0

200000

300000

200000

600000

300000

Large Small factory factory 140000 90000 EVPI

(a)

See the table above. EMVA = – 300,000(0.1) + – 200,000(0.4) + 100,000(0.2) + 600,000(0.3) = 90,000 EMVB = – 100,000(0.1) + 0(0.4) + 300,000(0.2) + 300,000(0.3) = 140,000

(b)

See the table above. EOLA = 200,000(0.1) + 200,000(0.4) + 200,000(0.2) + 0(0.3) = 140,000 EOLB = 0(0.1) + 0(0.4) + 0(0.2) + 300,000(0.3) = 90,000

(c)

EMV with perfect information = – 100,000(0.1) + 0(0.4) + 300,000(0.2) + 600,000(0.3) = 230,000 EVPI = EMV, perfect information – EMVB = 230,000 – 140,000 = 90,000 The company should not be willing to pay more than $90,000 for a perfect forecast.

(d)

The company should build a small factory to maximise expected monetary value ($140,000) and minimise expected opportunity loss ($90,000). So, based on the results in (a) and (b), the company should choose to build a small factory.

(e)

CVA = 356,230 100% = 395.81%

(f)

Return-to-risk ratio for A = 90,000 = 0.2526

CVB = 162, 481 100% = 116.06% 140,000

90,000

356,230

Return-to-risk ratio for B = 140,000 = 0.8616 162, 481

(g)

To minimise risk and maximise the return-to-risk, the company should decide to build a small plant, based on the results in (e) and (f).

(h)

There are no discrepancies.

(i)

PHStat2 output: Probabilities & Payoffs Table: P (A) Large factory Sell 10000 0.4 –300000

Sell 20000 Sell 50000

0.2 0.2

–200000 100000

Sell 100000

0.2

600000

Max

600000

Min

–300000

Statistics for:

Large factory –20000

Expected Monetary Value Variance

1.18E+11

(B) Small factory – 10000 0 0 30000 0 30000 0 30000 0 – 10000 0 Small factory 80000 3.36E+ 10

Standard Deviation

342928.6

Coefficient of Variation Return-to-Risk Ratio

–17.1464

Opportunity Loss Table: Optimu m Action Sell 10000 Sell 20000 Sell 50000 Sell 100000

Small factory Small factory Small factory Large factory

–0.05832

Optimum

18330 3 2.2912 88 0.4364 36

Alternatives

Profit -100000

Large factory 200000

Small factory 0

200000

300000

200000

600000

300000

Large factory 16000 0

Small factory 60000

Expected Opportunity Loss

EVPI (a) EMVA = – 300,000(0.4) + – 200,000(0.2) + 100,000(0.2) + 600,000(0.2) = -20,000 EMVB = – 100,000(0.4) + 0(0.2) + 300,000(0.2) + 300,000(0.2) = 80,000 (b)

EOLA = 200,000(0.4) + 200,000(0.2) + 200,000(0.2) + 0(0.2) = 160,000 EOLB = 0(0.4) + 0(0.2) + 0(0.2) + 300,000(0.2) = 60, 000

(c)

EMV with perfect information = – 100,000(0.4) + 0(0.2) + 300,000(0.2) + 600,000(0.2) = 140,000 EVPI = EMV, perfect information – EMVB = 140,000 – 80,000 = 60,000 Under these conditions, the company should not be willing to pay more than $60,000 for a perfect forecast.

(d)

The company should build a small factory to maximise expected monetary value ($80,000) and minimise expected opportunity loss ($60,000).

(e)

CVA = 342, 929 ×100% = -17.15%

(f)

-20, 000

CVB = 183, 303 ×100% = 2.29%

80, 000

Return-to-risk ratio for A = 80, 000 = –0.05832 183, 303

(g) (j)

Return-to-risk ratio for B = 80, 000 = 0.436436 183, 303 To minimise risk and maximise the return-to-risk, the company should decide to build a small plant.

There are no discrepancies. The company’s decision is not affected by the changed probabilities.

17.16

PHStat2 output: Probabilities & Payoffs Table: P A B Demand 1000 0.45 12000 6000 Demand 2000 0.2 14000 10000 Demand 5000 0.15 20000 22000 Demand 10000 0.1 30000 42000 Demand 50000 0.1 110000 202000 Max 110000 202000 Min 12000 6000 Statistics for: A B Expected Monetary 25200 32400 Value Variance 8.29E+0 3.32E+0 8 9 Standard Deviation 28791.6 57583.3 7 3 Coefficient of Variation 1.14252 1.77726 6 3 Return-to-Risk Ratio 0.87525 0.56266 3 3 Opportunity Loss Table: Optimum Optimum Action Profit Demand 1000 A 12000 Demand 2000 A 14000 Demand 5000 B 22000 Demand 10000 B 42000 Demand 50000 B 202000 Expected Opportunity Loss (a)

EMVA EMVB

(b)

EOLA

Alternatives A B 0 6000 0 4000 2000 0 12000 0 92000 0 A B 10700 3500

EVPI = 12,000(0.45) + 14,000(0.2) + 20,000(0.15) + 30,000(0.1) + 110,000(0.1) = 25,200 = 6,000(0.45) + 10,000(0.2) + 22,000(0.15) + 42,000(0.1) + 202,000(0.1) = 32,400 = 0(0.45) + 0(0.2) + 2,000(0.15) + 12,000(0.1) + 92,000(0.1) = 10,700

EOLB = 6,000(0.45) + 4,000(0.2) + 0(0.15) + 0(0.1) + 0(0.1)

(e)

= 3,500 EMV with perfect information = 12,000(0.45) + 14,000(0.2) + 22,000(0.15) + 42,000(0.1) + 202,000(0.1) = 35,900 EVPI = EMV, perfect information – EMVB = 35,900 – 32,400 = 3,500 The author should not be willing to pay more than $3,500 for a perfect forecast. Sign with company B to maximise the expected monetary value ($32,400) and minimise the expected opportunity loss ($3,500). CVA = 28, 792 100% =114.25% CVB = 57,583 100% = 177.73%

(f)

Return-to-risk ratio for A = 25,200 = 0.8752

(c)

(d)

32,400

25,200

28,792

Return-to-risk ratio for B = 32,400 = 0.5627 57,583

(g) (h) (i)

Signing with company A will minimise the author’s risk and yield the higher return-to-risk. Company B has a higher EMV than A, but choosing company B also entails more risk and has a lower return-to-risk ratio than A. (a), (b), (c), (e), (f) (h) See the table below. Probabilities & Payoffs Table: P A B Demand 0.3 12000 6000 1000 Demand 0.2 14000 10000 2000 Demand 0.2 20000 22000 5000 Demand 0.1 30000 42000 10000 Demand 0.2 11000 20200 50000 0 0 Max 11000 20200 0 0 Min 14000 10000 Statistics for: (a) Expected Monetary Value Variance

A 35400

B 52800

(e) Coefficient of Variation (f) Return-to-Risk Ratio

1.42E+ 09 37672. 8 1.0642 03 0.9396 7

5.68E+ 09 75345. 6 1.427

Opportunity Loss Table: Optimu m Action

Optimu m Profit

Standard Deviation

0.7007 71

Alternatives A

Demand 1000 Demand 2000 Demand 5000 Demand 10000 Demand 50000

A A B B B

12000 14000 22000 42000 202000

(b) Expected Opportunity Loss

0 0 2000 12000 92000 A 20000

6000 4000 0 0 0 B 2600 (c) EVPI

(d)

Signing with company B will maximise EMV and minimise EOL. See (a) and (b) in table above. (g) Signing with Company A will minimise the author’s risk and yield the higher return-to-risk. (h) Company B has a higher EMV than A, but choosing company B also entails more risk and has a lower return-to-risk ratio than A. The author’s decision is not affected by the changed probabilities. 17.17 PHStat2 output:

Probabilities & Payoffs Table: P Sell 100 Sell 200 Sell 500 Sell 1000

0.2 0.5 0.2 0.1

Statistics for: Expected Monetary Value Variance Standard Deviation Coefficient of Variation Return to Risk Ratio Opportunity Loss Table: Optimum

Sell 100 Sell 200 Sell 500 Sell 1000

Action Buy 100 Buy 200 Buy 500 Buy 1000

Buy 100 3500 3500 3500 3500

Buy 200 -1000 7000 7000 7000

Buy 500 -14500 -6500 17500 17500

Buy 1000 -37000 -29000 -5000 35000

Buy 100

Buy 200

Buy 500

Buy 1000

3500 5400 -900 -19400 0 10240000 1.54E+08 4.45E+08 0 3200 12419.34 21105.45 0 0.592593 -13.7993 -1.08791 #DIV/0! 1.6875 -0.07247 -0.91919

Optimum Profit 3500 7000 17500 35000

Alternatives Buy 100 0 3500 14000 31500

Buy 200 4500 0 10500 28000

Buy 500 18000 13500 0 17500

Buy 1000 40500 36000 22500 0

Expected Opportunity Loss

(a) (b) (c)

(d) (e) (f) (g) (h) (i)

Buy 100 7700

Buy 200 5800 EVPI

Buy 500 12100

Buy 1000 30600

See table above. See table above. EMV with perfect information = 3,500(0.2) + 7,000(0.5) + 17,500(0.2) + 35,000(0.1) = 11,200 EVPI = EMV, perfect information – EMVB = 11,200 – 5,400 = $5,800 We should not be willing to pay more than $5,800 for a perfect forecast. The scout group should buy 200 trees, action B, to maximise expected monetary value ($5,400) and minimise expected opportunity loss ($5,800). See table above. See table above. To minimise risk and maximise the return-to-risk, the scout group should decide to buy 200 trees, action B. There are no discrepancies. (a), (b), (c), (e), (f) see table below.

Probabilities & Payoffs Table: P Sell 100 Sell 200 Sell 500 Sell 1000

0.4 0.2 0.2 0.2

Statistics for: Expected Monetary Value Variance Standard Deviation Coefficient of Variation Return to Risk Ratio

Buy 100 3500 3500 3500 3500

Buy 200 -1000 7000 7000 7000

Buy 500 -14500 -6500 17500 17500

Buy 1000 -37000 -29000 -5000 35000

Buy 100

Buy 200

Buy 500

Buy 1000

3500 3800 -100 -14600 0 15360000 2.15E+08 7.53E+08 0 3919.184 14664.24 27434.29 0 1.031364 -146.642 -1.87906 #DIV/0! 0.96959 -0.00682 -0.53218

17.18 (a)

(b) (c) (d) (e) (f)

Optimum

Action Buy 100 Buy 200 Buy 500 Buy 1000

Profit

Alternatives

3500 7000 17500 35000

Buy 100 0 3500 14000 31500

Buy 200 4500 0 10500 28000

Buy 500 18000 13500 0 17500

Expected Opportunity Loss

Buy 100 9800

Buy 200 9500 EVPI

Buy 500 13400

Sell 100 Sell 200 Sell 500 Sell 1000

(i)

Optimum

Buy 1000 40500 36000 22500 0 Buy 1000 27900

(c) We are not willing to pay more than $13,300 for a perfect forecast. (d) Choose to purchase 200 trees to maximise EMV and minimise EOL. (g) To minimise risk and maximise return-to-risk, buy 200 trees, action B. (h) No discrepancies. The change in probabilities did not result in a different decision. Under these conditions, the scout group should buy 200 trees.

P(E1 | F) =

P(F | E1 ) P(E1 ) 0.6(0.5) = 0.6 = P(F | E1 ) P(E1 ) + P(F | E2 )  P(E2 ) 0.6(0.5) + 0.4(0.5)

P(E2 | F) =1 – P(E1 | F) = 1 – 0.6 = 0.4 EMVA = (0.6)(50) + (0.4)(200) = 110 EMVB = (0.6)(100) + (0.4)(125) = 110 EOLA = (0.6)(50) + (0.4)(0) = 30 EOLB = (0.6)(0) + (0.4)(75) = 30 EVPI = (0.6)(100) + (0.4)(200) = 30 You should not be willing to pay more than $30 for a perfect forecast. Both have the same EMV and the same EOL.  2A = (0.6)(60)2 + (0.4)(90)2 = 5400  A = 73.4847  B = 12.2474  B2 = (0.6)(10)2 + (0.4)(15)2 = 150 12.2474 73.4847 CV = CV = 100% = 11.1% 100% = 66.8% A

110

110 = 1.497 73.4847 110 Return-to-risk ratio for B = = 8.981 12.2474

110

(g)

Return-to-risk ratio for A =

(h) (i)

Action B has a better return-to-risk ratio. Both have the same EMV, but action B has a better return-to-risk ratio.

17.19 (a)

P( E 1 | F ) = =

P( F | E1 )  P( E1 ) P( F | E1 )  P( E1 ) + P( F | E2 )  P( E2 ) + P ( F | E3 )  P ( E3 )

0.2(0.8) = 0.667 or 2/3 0.2(0.8) + 0.4(0.1) + 0.4(0.1)

P(E2 | F) =

0.4(0.1) = 0.167 or 1/6 0.2(0.8) + 0.4(0.1) + 0.4(0.1)

(e) (f)

P(E3 | F) = 1 – P(E1 | F) – P(E2 | F) = 1 – 0.667 – 0.167 = 0.167 or 1/6 EMVA = (0.667)(50) + (0.167)(300) + (0.167)(500) = 166.95 EMVB = (0.667)(10) + (0.167)(100) + (0.167)(200) = 56.77 EOLA = (0.667)(0) + (0.167)(0) + (0.167)(0) = 0 EOLB = (0.667)(40) + (0.167)(200) + (0.167)(300) = 110.18 EVPI = 0 You should not be willing to pay any money for a perfect forecast. Action A has a higher EMV and is better for all events.  2A = (0.667)( 13677.3025) + (0.167)(17702.3025) + (0.167)(110922.3025) = 30603.0698  A = 174.937  B2 = (0.667)(2187.4329) + (0.167)(1868.8329) + (0.167)(20514.8329) = 5197.090  B = 72.091

(g)

174.937 72.091 CVB =  100% = 126.99%  100% = 104.78% 166.95 56.77 166.95 Return-to-risk ratio for A = = 0.954 174.937 56.77 Return-to-risk ratio for B = = 0.787 72.091

(h) (i)

Action A has a better return-to-risk ratio. Both support action A.

17.20 (a)

P(forecast cool | cool weather) = 0.80 P(forecast warm | warm weather) = 0.70

(b) (c) (d)

CVA =

Forecast Cool 0.8

0.32

Cool 0.4

Warm

Forecast Warm 0.08 0.2 Forecast Cool 0.18 0.3

0.6 Forecast Warm 0.42 0.7

Cool Warm Totals

Forecast Cool 0.32 0.18 0.5

Forecast Warm 0.08 0.42 0.5

Totals 0.4 0.6

Revised probabilities: P(cool | forecast cool) = 0.32 = 0.64 0.5

P(warm | forecast cool) = 0.18 = 0.36 0.5

Cool 0.64

0.32

Warm 0.36 Cool 0.16

0.18

Warm 0.84

0.42

Forecast cool 0.5

Forecast warm 0.5

(b)

0.08

EMV(hot dogs) = 260(0.64) + 230(0.36) = 249.2 EMV(ice cream) = 190(0.64) + 290(0.36) = 226 EOL(hot dogs) = 0(0.64) + 60(0.36) = 21.6 EOL(ice cream) = 70(0.64) + 0(0.36) = 44.8 EMV with perfect information = 260(0.64) + 290(0.36) = 270.8 EVPI = EMV, perfect information – EMVA = 270.8 – 249.2 = 21.6 The vendor should not be willing to pay more than $21.6 for a perfect forecast of the weather. The vendor should sell hot dogs to maximise value and minimise loss.

14.4 .100% = 5.78% 249.2 48 .100% = 21.24% CV(ice cream) = 226 249.2 Return-to-risk ratio for soft drinks = = 17.31 14.4 226 Return-to-risk ratio for ice cream = = 4.71 48 CV(hot dogs) =

(c)

17.21 (a)

Based on these revised probabilities, the vendor’s decision changes because of the increased likelihood of cool weather given a forecast for cool. Under these conditions, they should sell hot dogs to maximise the expected monetary value and also to minimise the expected opportunity loss, as well as minimising risk and maximising return. P(rosy | decline) = 0.2 P(rosy | no change) = 0.4 P(rosy | expanding) = 0.7

Decline No Change Expanding Totals

Forecast Rosy Gloomy 0.06 0.24 0.20 0.30 0.14 0.4

0.06 0.60

Totals 0.3 0.5 0.2

Rosy 0.2

0.06

Decl ine 0.3 Gl oomy 0.8 Rosy 0.4 N o Change 0.5

Gl oomy 0.6 Rosy

0.24 0.20

0.30 0.14

0.7 Expandi ng 0.2 Gl oomy 0.3

0.06

Given a gloomy forecast, revised conditional probabilities are: P(decline | gloomy) = .24 = 0.40 P(no change | gloomy) = .30 = 0.50 .60

.60

P(expanding | gloomy) = .06 = 0.10 .60

(b) Payoff table, given gloomy forecast: Pr 0.4 0.5 0.1

Decline No Change Expanding Max Min EMV

2  CV Return-torisk

A 500 1,000 2,000 2,000 500 900 190,000

B – 2,000 2,000 5,000 5,000 -2,000 700 5,610,000

435.89 48.43%

2,368.54 338.36%

2.0647

0.2955

C – 7,000 –1,000 20,000 20,000 -7,000 – 1,300 58,410,00 0 7,642.64 – 587.90% – 0.1701

Opportunity loss table: Profit of Optimum Optimum Event Action Action 1 A 500 2 B 2,000 3 C 20,000

Alternative Courses of Action A B C 0 2,500 7,500 1,000 0 3,000 18,000 15,000 0

EOLA = 0(0.4) + 1,000(0.5) + 18,000(0.1) = 2,300 EOLB = 2,500(0.4) + 0(0.5) + 15,000(0.1) = 2,500 EOLC = 7,500(0.4) + 3,000(0.5) + 0(0.1) = 4,500 EMV with perfect information = 500(0.4) + 2,000(0.5) + 20,000(0.1) = 3,200 EVPI = 3,200 – 900 = 2,300

The investor should not be willing to pay more than $2,300 for a perfect forecast. (c)

17.22 (a)

Under the new conditions, action A optimises the expected monetary value, minimises the coefficient of variation and maximises the investor’s return-to-risk. The probability of decline has increased, which lowered the expected monetary value of actions B and C. P(favourable | 1,000) = 0.01 P(favourable | 2,000) = 0.01 P(favourable | 5,000) = 0.25 P(favourable | 10,000) = 0.60 P(favourable | 50,000) = 0.99 P(favourable and 1,000) P(favourable and 2,000) P(favourable and 5,000) P(favourable and 10,000) P(favourable and 50,000) Joint probability table: Favourable 1,000 2,000 5,000 10,000 50,000 Totals

0.0045 0.0020 0.0375 0.0600 0.0990 0.2030

= 0.01(0.45) = 0.0045 = 0.01(0.20) = 0.0020 = 0.25(0.15) = 0.0375 = 0.60(0.10) = 0.0600 = 0.99(0.10) = 0.0990

Unfavourabl e 0.4455 0.1980 0.1125 0.0400 0.0010 0.7970

Totals 0.45 0.20 0.15 0.10 0.10

Given an unfavourable review, the revised conditional probabilities are: P(1,000 | unfavourable) = 0.4455/0.7970 = 0.5590 P(2,000 | unfavourable) = 0.1980/0.7970 = 0.2484 P(5,000 | unfavourable) = 0.1125/0.7970 = 0.1412 P(10,000 | unfavourable) = 0.0400/0.7970 = 0.0502 P(50,000 | unfavourable) = 0.0010/0.7970 = 0.0013 (b)

Payoff table, given unfavourable review:

1,000 2,000 5,000 10,000 50,000

Pr 0.5590 0.2484 0.1412 0.0502 0.0013 EMV

2  CV Return-torisk

A 12,000 14,000 20,000 30,000 110,000 14,658.60 31,719,333.50 5,631.99 38.42% 2.6027

B 6,000 10,000 22,000 42,000 202,000 11,315.4 126877326.67 11263.98 99.55% 1.0046

Opportunity loss table: Pr Event 1 0.5590 Event 2 0.2484 Event 3 0.1412 Event 4 0.0502 Event 5 0.0013 EOL (c)

A 0 0 2,000 12,000 92,000 1,004.40

B 6,000 4,000 0 0 0 4,347.60

The author’s decision is affected by the changed probabilities. Under the new circumstances, signing with company A maximises the expected monetary value ($14,658.60), minimises the expected opportunity loss ($1,004.40), minimises risk with a smaller coefficient of variation and yields a higher return-to-risk than choosing company B.

17.23 Answers will vary. 17.24 The expected monetary value (risk neutral) criteria is inappropriate in circumstances where each incremental change of profit or loss does not have the same value as the previous amount of profits attained or losses incurred. 17.25 Events are the actual states of the world that can occur. Alternative courses of action represent the choices of the decision maker. 17.26 A payoff table presents the alternatives in a tabular format, while the decision tree organises the alternatives and events visually. 17.27 The opportunity loss is the difference between the highest possible profit for an event and the actual profit obtained for an action taken. 17.28 Since it is the difference between the highest possible profit for an event and the actual profit obtained for an action taken, it can never be negative. 17.29 Expected monetary value represents the mean profit of an alternative course of action. Expected opportunity loss represents the mean opportunity loss of the alternative course of action as compared to the action that would be taken if you knew the event that was going to occur. 17.30 The expected value of perfect information represents the maximum amount you would pay to obtain perfect information. It represents the alternative course of action with the smallest expected opportunity loss. It is also equal to the expected profit under certainty minus the expected monetary value of the best alternative course of action. 17.31 The expected value of perfect information equals the expected profit under certainty minus the expected monetary value of the best alternative course of action.

17.32 Expected monetary value measures the mean return or profit of an alternative course of action over the long run without regard for the variability in the payoffs under different events. The return-to-risk ratio considers the variability in the payoffs in evaluating which alternative course of action should be chosen. 17.33 Bayes’ theorem uses conditional probabilities to revise the probability of an event in the light of new information. 17.34 A risk averter attempts to reduce risk, while a risk seeker looks for increased return, usually associated with greater risk. 17.35 Under many circumstances in the business world, the assumption that each incremental change of profit or loss has the same value as the previous amount of profits attained or losses incurred is not valid. Utilities should be used instead of payoffs under such differential evaluation of incremental profits or losses. 17.36 PhStat output:

Probabilities & Payoffs Table: Demand P Buy 6000 Sell 6000 0.1 4800 Sell 8000 0.5 4800 Sell 10,000 0.3 4800 Sell 12,000 0.1 4800

Buy 8000 3440 6400 6400 6400

Buy 10,000 2080 5040 8000 8000

Buy 12,000 720 3680 6640 9600

Buy 8000

Buy 10,000

Buy 12,000

6104 788544 888 0.145478 6.873874

5928 3592256 1895.3248 0.3197242 3.1276962

4864 5607424 2368 0.4868421 2.0540541

Optimum Profit 4800 6400

Buy 6000 0 1600

Alternatives Buy 8000 Buy 10,000 1360 2720 0 1360

8000

3200

1600

1360

9600

4800 Buy 6000 2240

3200 Buy 8000 936 EVPI

1600 Buy 10,000 1112

0 Buy 12,000 2176

Statistics for: Buy 6000 Expected Monetary Value 4800 Variance 0 Standard Deviation 0 Coefficient of Variation 0 Return to Risk Ratio #DIV/0! Opportunity Loss Table: Optimum Action Sell 6000 Buy 6000 Sell 8000 Buy 8000 Buy Sell 10,000 10,000 Buy Sell 12,000 12,000

Expected Opportunity Loss

Buy 12,000 4080 2720

(a) (b)

See table above. 1 2 A

3 4

1 2

3 4

1 2

3 4

1 2

3 4

4,800 4,800 4,800 4,800 3,440 6,440 6,440 6,440 2,080 0 5,040 8,000 8,000 72 0 3,680 6,640 9,600

(j)

See table above. See table above. The maximum the supermarket chain should be willing to spend for perfect information is $936. Based on the results of (c) and (d), purchase 8,000 loaves (action B) since this option maximises EMV and minimises EOL. See table above. See table above. Based on (g) and (h), purchase 8,000 loaves (action B) since this option is associated with the least risk and has the highest return-to-risk ratio. There are no discrepancies.

(k)

PhStat output:

(f) (g) (h) (i)

Commented [JW1]: The first number in the last group of four above should be 720 ( not 72) but I can’t edit it to fit in the box or expand the box. Commented [AC2R1]: Production please can we have this fixed, as I can’t edit either.

Probabilities & Payoffs Table: Demand Sell 6000 Sell 8000 Sell 10,000 Sell 12,000

P 0.3 0.4 0.2 0.1

Statistics for: Expected Monetary Value Variance Standard Deviation Coefficient of Variation Return to Risk Ratio

Buy 6000 4800 4800 4800 4800

Buy 8000 3440 6400 6400 6400

Buy 10,000 2080 5040 8000 8000

Buy 12,000 720 3680 6640 9600

Buy 6000

Buy 8000

Buy 10,000

Buy 12,000

4800 5512 5040 3976 0 1839936 5256960 7797824 0 1356.442 2292.8061 2792.4584 0 0.246089 0.4549219 0.7023286 #DIV/0! 4.063571 2.1981797 1.423835

Opportunity Loss Table: Optimum

Optimum

Action Buy 6000 Buy 8000 Buy 10,000 Buy 12,000

Profit 4800 6400 8000 9600

Sell 6000 Sell 8000 Sell 10,000 Sell 12,000

Expected Opportunity Loss

17.37 (a), (d), (g)

Buy 6000 0 1600 3200 4800 Buy 6000 1760

Alternatives Buy Buy 8000 10,000 1360 2720 0 1360 1600 0 3200 1600 Buy Buy 8000 10,000 1048 1520 EVPI

Buy 12,000 4080 2720 1360 0 Buy 12,000 2584

See table above. See table above. See table above. Based on the results of (c) and (d), purchase 8,000 loaves (action B) since this option maximises EMV and minimises EOL. See table above. See table above. Based on the results of (g) and (h), purchase 8,000 loaves (action B) since this option is associated with the least risk and highest return-to-risk ratio. There are no discrepancies. The supermarket chain’s decision has not been affected by the changed probabilities. Payoff table:

Even t 1 2 3

Pr 0.40 0.30 0.30 EMV

50 100 200



CV Return-torisk

A: Install – 50,000 50,000 250,000 70,000 124,900 178.43% 0.5604

B: Do Not Install 0 0 0 0 0 undefined undefined

(b) 1 A

-50,000

50,000

250,000

1 0 B

(c), (e)

Opportunity loss table: Event 1 2 3

50 100 200

Pr 0.40 0.30 0.30 EOL

(f) (h) (i)

A: Install 50,000 0 0 20,000

B: Do Not Install 0 50,000 250,000 90,000

EVPI = $20,000. The owner of the home heating oil delivery company should not be willing to pay more than $20,000 for a perfect forecast. To maximise the expected monetary value and minimise the expected opportunity loss, the owner should offer solar heating. Payoff table:

50 100 200

Pr 0.40 0.30 0.30 EMV



CV Return-torisk

A: Install – 100,000 0 200,000 20,000 124,900 624.50% 0.1601

B: Do Not Install 0 0 0 0 0 undefined undefined

Opportunity loss table: Pr 0.40 0.30 0.30

50 100 200

EOL

A: Install 100,000 0 0 40,000

B: Do Not Install 0 0 200,000 60,000

EVPI = $40,000. The owner of the home heating oil delivery company should not be willing to pay more than $40,000 for a perfect forecast. Although individual values are different, the owner’s decision is not affected by the altered start-up costs. 17.38 (a)

-4,000,000

1,000,000

5,000,000

1 0 B

(c), (f) Payoff table: Event 1 2 3

Weak Moderat e Strong

Pr 0.3 0.6

New –4,000,000 1,000,000

Old

0.1

5,000,000 –100,000 2,808,914 –2,808.94% –0.0356

0 0 0 undefined undefined

EMV

 CV Return-torisk Opportunity loss table: Pr New Weak 0.3 4,000,000 Moderat 0.6 0 e Strong 0.1 0 EOL 1,200,000

0 0

(b), (d), (e)

Old 0 1,000,000 5,000,000 1,100,000

EVPI = $1,100,000. The product manager should not be willing to pay more than $1,100,000 for a perfect forecast. (g)

The product manager should continue to use the old packaging to maximise expected monetary value and to minimise expected opportunity loss and risk.

(h)

(c), (f)

Payoff table: Weak Moderat e Strong

Pr 0.6 0.3

New –4,000,000 1,000,000

Old

0.1

5,000,000 –1,600,000 3,136,877 –196.05% –0.5101

0 0 0 undefined undefined

EMV



CV Return-torisk (b), (d), (e) Opportunity loss table: Pr Weak 0.6 Moderat 0.3 e Strong 0.1 EOL

0 0

New 4,000,000 0

Old 0 1,000,000

0 2,400,000

5,000,000 800,000

EVPI = $800,000. The product manager should not be willing to pay more than $800,000 for a perfect forecast. (g)

The product manager should continue to use the old packaging to maximise expected monetary value and to minimise expected opportunity loss and risk.

(h)

(c), (f) Payoff table: Weak Moderat e Strong

Pr 0.1 0.3

New –4,000,000 1,000,000

0.6

5,000,000 2,900,000 2,913,760.457 100.47% 0.9953

0 0 0 undefined undefined

New 4,000,000 0

Old

EMV

 CV Return-torisk (b), (d), (e) Opportunity loss table: Pr Weak 0.1 Moderat 0.3 e Strong 0.6 EOL

0 400,000

Old 0 0

0 1,000,000 5,000,000 3,300,000

EVPI = $400,000. The product manager should not be willing to pay more than $400,000 for a perfect forecast.

(g)

The product manager should use the new packaging to maximise expected monetary value and to minimise expected opportunity loss.

(i)

P(Sales decreased | weak response) = 0.6 P(Sales stayed same | weak response) = 0.3 P(Sales increased | weak response) = 0.1 P(Sales decreased | moderate response) = 0.2 P(Sales stayed same | moderate response) = 0.4 P(Sales increased | moderate response) = 0.4 P(Sales decreased | strong response) = 0.05 P(Sales stayed same | strong response) = 0.35 P(Sales increased | strong response) = 0.6 P(Sales decreased and weak response) = 0.6(0.3) = 0.18 P(Sales stayed same and weak response) = 0.3(0.3) = 0.09 P(Sales increased and weak response) = 0.1(0.3) = 0.03 P(Sales decreased and moderate response) = 0.2(0.6) = 0.12 P(Sales stayed same and moderate response) = 0.4(0.6) = 0.24 P(Sales increased and moderate response) = 0.4(0.6) = 0.24 P(Sales decreased and strong response) = 0.05(0.1) = 0.005 P(Sales stayed same and strong response) = 0.35(0.1) = 0.035 P(Sales increased and strong response) = 0.6(0.1) = 0.06 Joint probability table: Sales Sales Sales Pr Decrease Stay Same Increase Weak 0.3 0.180 0.090 0.030 Moderat 0.6 0.120 0.240 0.240 e Strong 0.1 0.005 0.035 0.060 Total 0.305 0.365 0.330

(j)

Given the sales stayed the same, the revised conditional probabilities are: P(weak response | sales stayed same) = .09 = 0.2466 .365

P(moderate response | sales stayed same) = .24 = 0.6575 .365

P(strong response | sales stayed same) = .035 = 0.0959 .365

(k)

(c), (f)

Payoff table: Weak Moderat e Strong

Pr 0.2466 0.6575

New –4,000,000 1,000,000

0.0959 EMV

5,000,000 150,600 2,641,575.219 1,754.03% 0.0570

 CV Return-torisk (b), (d), (e)

Opportunity loss table: Pr New Weak 0.2466 4,000,000

Old 0 0 0 0 0 undefined undefined

Old 0

Moderat e Strong

0.6575 0.0959 EOL

1,000,000

0 986,400

5,000,000 1,137,000

EVPI = $986,400. The product manager should not be willing to pay more than $986,400 for a perfect forecast. (g)

(l)

The product manager should use the new packaging to maximise expected monetary value and to minimise expected opportunity loss.

Given the sales decreased, the revised conditional probabilities are: P(weak response | sales decreased) = .18 = 0.5902 .305

P(moderate response | sales decreased) = .12 = 0.3934 .305

P(strong response | sales decreased) = .005 = 0.0164 .305

(m)

(c), (f)

Payoff table: Weak Moderat e Strong

Pr 0.5902 0.3934

New – 4,000,000 1,000,000

0.0164 EMV

5,000,000 – 1,885,400 2,586,864.287 – 137.21% – 0.7288

0 0 0 undefined undefined

Pr 0.5902 0.3934

New 4,000,000 0

Old 0 1,000,000

0.0164 EOL

0 2,360,800

5,000,000 475,400



CV Return-torisk (b), (d), (e)

Old 0 0

Opportunity loss table:

Weak Moderat e Strong

EVPI = $475,400. The product manager should not be willing to pay more than $475,400 for a perfect forecast. (g)

The product manager should continue to use the old packaging to maximise expected monetary value and minimise expected opportunity loss.

17.39

Commented [JW3]: Please insert the solution from the US 13th edition Q 20.40, I’ve included this seperatly as I can’t copy and edit the answer into the file.

Chapter 18: Statistical applications in quality and productivity management After studying this chapter you should be able to: 1

distinguish between common cause and special cause variation

select appropriate control charts for categorical and numerical data

construct and interpret attribute and variable control charts

measure the capability of a process (a)

Proportion of nonconformances largest on day 5, smallest on day 3.

Proportion

18.1

0.3 0.2 0.1 0 0

Day (b)

(c)

n = 100,

p = 1.48/10 = 0.148,

LCL = p − 3

p(1 − p) 0.148(1 − 0.148) = 0.148 − 3 = 0.04147 , n 100

UCL = p + 3

p(1 − p) 0.148(1 − 0.148) = 0.148 + 3 = 0.25453 n 100

Proportions are within control limits, so there do not appear to be any special causes of variation.

(a)

Proportion of nonconformances largest on day 4, smallest on day 3.

Proportion

18.2

0.3 0.2 0.1 0 0

Day

(b)

18.3

n = 1036/10 = 103.6, p = 148/1036 = 0.142857,

LCL = p − 3

p(1 − p ) 0.142857(1 − 0.142857) = 0.142857 − 3 = 0.039719 n 103.6

UCL = p + 3

p(1 − p ) 0.142857(1 − 0.142857) = 0.142857 + 3 = 0.245995 n 103.6

(c)

Proportions are within control limits, so there do not appear to be any special causes of variation.

(a)

n = 125,

p = 146/3875 = 0.0377,

LCL = p − 3

p(1 − p) 0.0377(1 − 0.0377) = 0.0377 − 3 = −0.0134 < 0, so the n 125

lower control limit does not exist.

UCL = p + 3

p(1 − p) 0.0377(1 − 0.0377) = 0.0377 + 3 = 0.0888 n 125

p Chart 0.12 0.1 UCL

Proportion

0.08 0.06 0.04

pBar

0.02 0 0

-0.02 X

18.4

(b)

The proportion of transmissions with errors on day 23 is substantially out of control. Possible causes of this value should be investigated.

(a)

n = 500,

p = 761/16000 = 0.0476

LCL = p − 3

p(1 − p) 0.0476(1 − 0.0476) = 0.0476 − 3 = 0.0190 > 0 n 500

UCL = p + 3

p(1 − p) 0.0476(1 − 0.0476) = 0.0476 + 3 = 0.0761 n 500

p Chart 0.08

UCL

0.07

Proportion

0.06 0.05

pBar

0.04 0.03 0.02

LCL

0.01 0 0

(b)

Since the individual points are distributed around

p without any pattern and

all the points are within the control limits, the process is in a state of statistical control.

18.5

(a)

n = 102.5667, p = 0.308742, LCL = 0.171895, UCL = 0.44559 p Chart 0.6 0.5

Proportion

UCL 0.4 pBar

0.3 0.2

LCL

0.1 0 0

X (b)

Yes, the process gives an out-of-control signal because the proportions fall outside of the control limits on four of the 30 days.

(c)

n = 103.1923, p = 0.297428, LCL = 0.162428, UCL = 0.432428

18.6

(a)

n = 113345/22 = 5152.0455, p = 1460/113345 = 0.01288 LCL = p − 3

p(1 − p) 0.01288(1 − 0.01288) = 0.01288 − 3 = 0.00817 n 5152.0455

UCL = p + 3

p(1 − p) 0.01288(1 − 0.01288) = 0.01288+3 = 0.01759 n 5152.0455

p Chart 0.02 0.018

UCL

Proportion

0.016 0.014

pBar

0.012 0.01

LCL

0.008 0.006 0.004 0.002 0 0

X The proportion of unacceptable cans is below the LCL on day 4. There is evidence of a pattern over time, since the last eight points are all above the mean and most of the earlier points are below the mean. Thus, the special causes that might be contributing to this pattern should be investigated before any change in the system of operation is contemplated. (b)

Once special causes have been eliminated and the process is stable, Deming’s fourteen points should be implemented to improve the system. They might also look at day 4 to see if they could identify and exploit the special cause that led to such a low proportion of defects on that day.

18.7

(a)

p Chart 0.16

0.14 Proportion

0.12 0.1 0.08

pBar

0.06

UCL

0.04 0.02 0

20 X

p = 0.0422, LCL = does not exist, UCL = 0.0848. Points 9, 16, 20, 31 and 36 are above the UCL. Yes, the process gives an out-of-control signal because the proportions fall outside of the control limits on five of the 30 days. (b)

18.8

First, the reasons for the special cause variation would need to be determined and local corrective action taken. Once special causes have been eliminated and the process is stable, Deming’s fourteen points should be implemented to improve the system.

(a)

p Chart 0.08 0.07

Proportion

0.06 0.05 0.04

UCL

0.03

pBar

0.02

LCL

0.01 0 0

15 X

p = 0.0317, LCL = 0.0126, UCL = 0.0508. Points 9, 10, 25 and 30 are above the UCL. Points 13 and 22 are below the LCL. The process appears to be out of control. (b)

18.9

Answers will vary.

18.10 Answers will vary. 18.11 (a)

Excel output:

Number of Nonconformities per unit

c Chart 12 10 8 UCL

LCL 2 0 0

c = /10 = 3.8,

6 Time period

LCL = c − 3 c = 3.8 − 3 3.8  0 , LCL does not exist.

UCL = c + 3 c = 3.8 + 3 3.8 = 9.648077 (b)

18.12 (a)

There do not appear to be special causes of variation, as there are no points outside the control limits and no discernable pattern.

c = 115/10 = 11.5, LCL = c − 3 c = 11 .5 − 3 11 .5 = 1.32651 UCL = c + 3 c = 11 .5 + 3 11 .5 = 21 .67349

Number of Nonconformities per unit

c Chart 30 25 20 UCL

LCL 5 0 0

6 Time period

c = 115/10 = 11.5, LCL = c − 3 c = 11 .5 − 3 11 .5 = 1.32651 UCL = c + 3 c = 11 .5 + 3 11 .5 = 21 .67349 (b)

18.13 (a)

Yes, the number of nonconformities per unit for time period 1 is above the upper control limit.

c = 155/24 = 6.458, LCL = c − 3 c = 6.458 − 3 6.458  0 , LCL does not exist. UCL = c + 3 c = 6.458 + 3 6.458 = 14 .082 The process appears to be in control since there are no points outside the upper control limit and there is no pattern in the results over time.

Noncomformances

16 14 12 10 8 6 4 2 0 0

Day (b)

The value of 12 is within the control limits, so that it should be identified as a source of common-cause variation. Thus, no action should be taken concerning this value. If the value were 20 instead of 12, c would be 6.792 and UCL would be 14.61. In this situation, a value of 20 would be substantially above the UCL and action should be taken to explain the special cause of variation.

(c)

The process needs to be studied and potentially changed using principles of Six Sigma® management and/or Deming’s 14 points for management.

18.14 (a)

The twelve errors committed by Gina appear to be much higher than all others; hence the bank manager is likely to single out Gina to explain her performance.

(b)

c = 5.5, UCL = 12.56, LCL does not exist. The number of errors is in a state of statistical control since none of the tellers are outside the UCL. Copyright © 2019 Pearson Australia (a division of Pearson Australia Group Pty Ltd) 9781488617249, Berenson, Basic Business Statistics 5e

(c)

Since Gina is within the control limits, she is operating within the system, and should not be singled out for further scrutiny.

(d)

The process needs to be studied and potentially changed using principles of Six Sigma® management and/or Deming’s 14 points for management.

18.15 (a) c-Chart 14

Error NumberofofErrors Number

12 10

Number Errors Number of Errors Centre Center UCL UCL

8 6 4 2 0 0

Teller

(b)

Data collection would be delayed until the start-up period for the hospital had passed.

(c)

For example, the severity of the illness or the age of the patients from month to month could be contributing factors.

18.16 (a)

June 2016 2.933… fatal crashes per day June 2017 3.333 fatal crashes per day

(b)

June 2017

Number of Fatal Crashes per day

c - Chart June 2017 9 8 7 6

5 4 3

2 1

0 0

Day Number of Fatal Crashes

LCL

Center

UCL

Intermediate Calculations Sum of Nonconformities Number of Units Sampled Cbar Preliminary Lower Control Limit c Chart Control Limits Lower Control Limit Center Upper Control Limit

100 30 3.3333 -2.1439 0.0000 3.3333 8.8106

June 2016

Number of Fatal Crashes per day

c-Chart June 2016 9 8 7 6 5

4 3 2 1 0

Number of Fatal Crashes

Day

LCL

Intermediate Calculations Sum of Nonconformities Number of Units Sampled Cbar Preliminary Lower Control Limit c Chart Control Limits Lower Control Limit Center Upper Control Limit

Center

30 UCL

88 30 2.9333333 -2.2048 0.0000 2.9333 8.0714

(c)

Since the number of fatal crashes on 10 June 2016 is lower than the UCL it appears this high value is explained by chance variation. That is, this value is not caused by special cause variation.

(d)

There is no evidence of a pattern over each month, however, there appears to be a slight increase in the number of fatalities per day. The pattern over the entire year should be investigated, so that any special cause variation which has led to this apparent increase can be identified to enable it to be eliminated.

18.17 (a)

Mean number recorded homicide and related offences per month 14.708…

(b)

Number of Homicide and Related Offences per month

c Chart: Number of Homicide and Related Offences 30 25 20 15

10 5 0 Jun-15

Sep-15

Dec-15

Mar-16

Homicide and Related Offences

Jul-16

Oct-16

LCL

Center

c Chart Summary Intermediate Calculations Sum of Nonconformities Number of Units Sampled Cbar Preliminary Lower Control Limit c Chart Control Limits Lower Control Limit Center Upper Control Limit (c)

Jan-17

May-17

UCL

353 24 14.708333 3.2029 3.2029 14.7083 26.2138

The 28 homicide and related offences in May 2016 is higher than the UCL, this may be due to special cause variation – possibly drug or gang related.

18.18 (a)

d2 = 2.059

(b)

d3 = 0.88

(c)

D3 = 0

(d)

D4 = 2.282

(e)

A2 = 0.729

18.19 (a) (b)

R = 3.97, R chart: UCL = 9.06; LCL does not exist According to the R chart below, the process appears to be in control with all points lying inside the control limits without any pattern and no evidence of special cause variation.

Sample Range

R Chart 10 9 8 7 6 5 4 3 2 1 0

UCL R Bar LCL

Day

(c)

X = 13.95, X chart: UCL = 16.84; LCL = 11.06

(d)

X Chart

Sample Mean

20 15 UCL

X BarBar 5

LCL

0 0

5 Day

There seems to be evidence of special cause variation in (c) with the sample mean on day 7 lying outside the statistical control limits.

18.20 (a)

Data Sample/Subgroup Size 4 R Chart Intermediate Calculations RBar 3.34 D3 Factor 0 D4 Factor 2.282 R Chart Control Limits Lower Control Limit 0 Center 3.34 Upper Control Limit 7.62188 XBar Chart Intemediate Calculations Average of Subgroup Averages 5.91625 A2 Factor 0.729 A2 Factor * RBar 2.43486 XBar Chart Control Limits Lower Control Limit 3.48139 Center 5.91625 Upper Control Limit 8.35111 Control charts for range and the mean (respectively)

Control Chart Waiting Time 9 Range UCL

8 7

Sample Range

LCL-R 6 Center-R

5 4

UCL-R RBar

3 2 1 0

LCL 0

Day

R = 3.40 LCL = 0

UCL = 7.6218…

Control Chart Waiting Time 9 UCL

Sample Mean

6 XBar

XBar

5 4

LCL-X LCL

3 Center-X 2 UCL-X

1 0 0

Day

X = 5.91625 LCL = 3.4813… (b)

18.21 (a)

UCL = 8.3511…

The process appears to be in control since all points are inside the control limits without any pattern and no evidence of special cause variation. Control charts for the range and the mean (respectively)

A 1 Help Desk Number of Calls 3 Data 4 Sample/Subgroup Size 11 R Chart Control Limits 12 Lower Control Limit 13 Center 14 Upper Control Limit 21 XBar Chart Control Limits 22 Lower Control Limit 23 Center 24 Upper Control Limit

5 0 56.8 120.0752 23.2264 56 88.7736

R Chart: Help Desk Number of Calls

140

Range Number of Calls

120

UCL

100

RBar

LCL 0

Day

X-bar Chart - Help Desk Number of Calls 100

Sample Mean: Number of Calls

UCL

80 70 60

XBar

50 40 30 LCL

20 10 0 0

Day (b)

The process appears to be in control since there are no points outside the control limits and no evidence of a pattern in the range chart, and there are

no points outside the control limits and no evidence of a pattern in the X chart. 18.22 (a) and (b) A B 1 Spring Water Magnesium Content 3 Data 4 Sample/Subgroup Size 4 6 R Chart Intermediate Calculations 7 RBar 0.061333333 8 D3 Factor 0 9 D4 Factor 2.282 11 R Chart Control Limits 12 Lower Control Limit 0 13 Center 0.061333333 14 Upper Control Limit 0.139962667 16 XBar Chart Intemediate Calculations 17 Average of Subgroup Averages 1.59953125 18 A2 Factor 0.729 19 A2 Factor * RBar 0.044712 21 XBar Chart Control Limits 22 Lower Control Limit 1.55481925 23 Center 1.59953125 24 Upper Control Limit 1.64424325

Range Magnesium Content (mg/l)

R Chart: Spring Water Magnesium Content

0.16 0.14

UCL

0.12 0.1 0.08 RBar

0.06 0.04 0.02 0 0

LCL 25

Hour

X-Bar Chart: Spring Water Magnesium Content

Sample Mean: Magnesium Content (mg/l)

1.65 1.64 1.63 1.62 1.61 1.6 1.59 1.58 1.57 1.56 1.55

(c)

UCL

XBar

LCL 0

Hour

The process appears to be in control since there are no points outside the lower and upper control limits of both the R chart and X chart, and there is no pattern in the results over time.

18.23 (a) and (b)

A 1 3 4 6 7 8 9 11 12 13 14 16 17 18 19 21 22 23 24

Strength Rope Data Sample/Subgroup Size 3 R Chart Intermediate Calculations RBar 52.85 D3 Factor 0 D4 Factor 2.575 R Chart Control Limits Lower Control Limit 0 Center 52.85 Upper Control Limit 136.08875 XBar Chart Intemediate Calculations Average of Subgroup Averages 609.1777778 A2 Factor 1.023 A2 Factor * RBar 54.06555 XBar Chart Control Limits Lower Control Limit 555.1122278 Center 609.1777778 Upper Control Limit 663.2433278

R Chart: Strength Rope

Sample Range: Strength (mPa)

160 140

UCL

120 100 80 60

RBar

40 20 0 0

LCL 25

Hour

Sample Mean: Strenth (mPa)

X-bar: Strength Rope 680 UCL

660 640 620

XBar

600 580 560

LCL

540 0

Hour (c)

The process appears to be in a state of statistical control since there are no points outside the control limits and there is no evidence of a pattern in the range or mean charts.

18.24 (a) Control charts for the range and the mean (respectively)

Sample Range

R Chart 20 18 16 14 12 10 8 6 4 2 0 -1

UCL R Bar LCL

Day

X Chart 30

Sample Mean

25 20 UCL

X BarBar 10

LCL

5 0 0

(b)

10 Day

The process does not appear to be in a state of statistical control since there are points outside the control limits for the mean chart.

18.25 (a)

Some possible sources of common-cause variation can be the fluctuation in temperature, humidity and geographical disturbances of the environment in which the machines operate. A machine that operates in an earthquake zone could experience chance variation during even undetectable quakes.

(b)

If a machine operates in a location near a subway line, the vibration from the transit trains can cause systematic assignable cause of variation.

(c)

Control charts for the range and the mean (respectively)

Control Chart Calculations Data Sample/Subgroup Size 4 R Chart Intermediate Calculations RBar 0.2248 D3 Factor 0 D4 Factor 2.282 R Chart Control Limits Lower Control Limit 0 Center 0.2248 Upper Control Limit 0.5129936 XBar Chart Intemediate Calculations Average of Subgroup Averages 5.509 A2 Factor 0.729 A2 Factor * RBar 0.1638792 XBar Chart Control Limits Lower Control Limit 5.3451208

Control Chart - Weight 0.6 UCL

Sample Range

0.5 0.4 0.3

Range

0.2

RBar LCL-R Center-R

0.1

UCL-R 0

LCL 0

Sample

Control Chart Weight 5.70

UCL 5.65

Sample Mean

5.60 5.55

XBar XBar

5.50

LCL-X

5.45

Center-X

5.40

UCL-X

5.35

LCL

5.30 0

(d)

18.26 (a)

The process appears to be in a state of statistical control since there are no points outside the control limits and there is no evidence of a pattern in the range or mean charts. Control charts for the range and the mean (respectively)

R Chart 1.2

Sample Range

1 0.8 UCL

0.6

R Bar

0.4

LCL 0.2 0 0

10 Time

X Chart 91.000

Sample Mean

90.800 90.600 90.400 UCL 90.200

X BarBar

90.000

LCL

89.800 89.600 0

(b)

18.27 (a) (b)

10 Time

The process does not appear to be in a state of statistical control since several points on both charts lie outside the control limits. Furthermore, there is evidence of a declining pattern in the range and mean after the halfway point. Possible special causes of variation should be investigated to explain this trend.

Estimate of the population mean of all X values = X = 20 Estimate

the

population

standard

deviation

all

values

2 R / d2 = = 0.9713 2.059

18.28 (a)

Estimate of the population mean = X = 100 Estimate of the population standard deviation = R / d 2 =

3.386 =2 1.693

102 − 100   98 − 100 P ( 98  X  102 ) = P  Z  = 0.6827 2 2   (b)

107.5 − 100   93 − 100 P ( 93  X  107.5) = P  Z  = .9997 2 2  

(c)

93.8 − 100   P ( X  93.8) = P  Z   = .9990 2  

(d)

110 − 100   P ( X  110) = P  Z   1 2  

18.29 (a)

Cp =

USL − LSL 102 − 98 = = 0.3333 6 ( 2) 6 ( R / d2 ) X − LSL 100 − 98 = = 0.3333 3( 2) 3 ( R / d2 )

CPL =

CPU =

USL − X 102 − 100 = = 0.3333 3( 2) 3 ( R / d2 )

Cpk = min(CPL, CPU ) = 0.3333 (b)

Cp =

USL − LSL 107.5 − 93 = = 1.2083 6 ( 2) 6 ( R / d2 ) X − LSL 100 − 93 = = 1.1667 3( 2) 3 ( R / d2 )

CPL =

CPU =

USL − X 107.5 − 100 = = 1.25 3( 2) 3 ( R / d2 )

Cpk = min(CPL, CPU ) = 1.1667

18.30 (a)

Estimate of the population mean = X = 1.6 Estimate of the population standard deviation = R / d 2 =

0.06 = 0.0291 2.059

æ 1.5-1.6 1.7 -1.6 ö P 1.5 < X <1.7 = P ç <Z< ÷ = 0.99942 0.0291 ø è 0.0291

(

(b)

Cp =

)

USL - LSL 1.7 -1.5 = = 1.1455 6 R / d2 6 0.0291

CPL =

(

)

(

)

X - LSL 1.6 -1.5 = = 1.1455 3 R / d2 3 0.0291

CPU =

(

) (

)

USL - X 1.7 -1.6 = = 1.1455 3 R / d2 3 0.0291

(

) (

)

C pk = min(CPL,CPU ) =1.1455

18.31 (a)

Estimate of the population mean = X = 609.18 Estimate of the population standard deviation = R / d 2 =

52.85 = 31.217 1.693

æ 550 - 609.18 ö P X > 550 = P ç Z > ÷ = 0.9713 31.217 ø è

(

(b)

CPL =

)

X - LSL 609.18 - 550 = = 0.6319 3 R / d2 3 31.217

(

)

(

)

C pk = min(CPL,CPU ) = 0.6319

18.32 (a)

Estimate of the population mean = X = 5.51 Estimate of the population standard deviation = R / d 2 =

0.22 = 0.1068 2.059

5.8 − 5.51   5.2 − 5.51 P ( 5.2  X  5.8 ) = P  Z  0.1068   0.1068 = P(−2.90  Z  2.72) = 0.9948 (b)

18.33 (a)

According to the estimate in (a), only 99.5% of the tea-bags will have weight fall between 5.2 grams and 5.8 grams. The process is, therefore, incapable of meeting the 99.7% goal.

Estimate of the population mean = X = 5.91625 Estimate of the population standard deviation = R / d 2 =

3.34 = 1.622 2.059

5 − 5.91625   P ( X  5) = P  Z   = P(Z  −0.56) = 0.2877 1.622   (b)

The process is not capable of reaching the goal with a 99% requirement and will not be capable with an even more stringent criterion of 99.7%.

18.34 Chance or common causes of variation represent the inherent variability that exists in a system. These consist of the numerous small causes of variability that operate randomly or by chance. Special or assignable causes of variation represent large fluctuations or patterns in the data that are not inherent to a process. These fluctuations are often caused by changes in a system that represent either problems to be fixed or opportunities to exploit.

18.35 Find the reasons for the special causes and take corrective action to prevent their occurrence in the future or exploit them if they improve the process. 18.36 When only common causes of variation are present, it is up to management to change the system. 18.37 The p chart is an attribute control chart. It can be used when sampled items are classified according to whether they conform or do not conform to operationally defined requirements. It is based on the proportion of nonconforming items in a sample. 18.38 Attribute control charts are used for categorical or discrete data such as the number of nonconformances. Variables control charts are used for numerical variables and are based on statistics such as the mean and standard deviation. 18.39 Since the range is used to obtain the control limits of the chart for the mean, the range needs to be in a state of statistical control. Thus, the range and mean charts are used together. 18.40 From the red bead experiment you learned that variation is an inherent part of any process, that workers work within a system over which they have little control, that it is the system that primarily determines their performance, and that only management can change the system. 18.41 The p chart is used to plot the proportion of items in a sample that are in a category of interest. The c chart on the other hand is used to plot the number of items in a sample that are in a category of interest. 18.42 Process potential measures the potential of a process in satisfying production specification limits or customer satisfaction but does not take into account the actual performance of the process; process performance refers to the actual performance of the process in satisfying production specification limits. 18.43 If a process has a Cp = 1.5 and a Cpk = 0.8, it indicates that the process has the potential of meeting production specification limits but fails to meet the specification limits in actual performance. The process should be investigated and adjusted to increase either the CPU or CPL or both. 18.44 Capability analysis is not performed on out-of-control processes because out-ofcontrol processes do not allow one to predict their capability. They are considered incapable of meeting specifications and, therefore, incapable of satisfying the production requirement.

18.45 (a) and (b)

1 3 4 6 7 8 9 11 12 13 14 16 17 18 19 21 22 23 24

A R Chart: Cat Food

Data Sample/Subgroup Size 5 R Chart Intermediate Calculations RBar 0.029266667 D3 Factor 0 D4 Factor 2.114 R Chart Control Limits Lower Control Limit 0 Center 0.029266667 Upper Control Limit 0.061869733 XBar Chart Intemediate Calculations Average of Subgroup Averages 1.517513333 A2 Factor 0.577 A2 Factor * RBar 0.016886867 XBar Chart Control Limits Lower Control Limit 1.500626467 Center 1.517513333 Upper Control Limit 1.5344002

R Chart: Cat Food 0.07 UCL

0.06

Sample Range

0.05 0.04 0.03

RBar

0.02 0.01

LCL 0

Time Interval

X-Bar Chart: Cat Food 1.540 1.535

UCL

Sample Mean

1.530 1.525 1.520 XBar 1.515 1.510 1.505 LCL

1.500 1.495 0

Time Interval (c)

All of the points are within the control limits, and there are no patterns in the R chart. All the points in the X chart are also within the control limits, and there are no patterns, so the process is in control. However, there does seem to be an upward trend, so would suggest that the process is watched carefully to see if this continues.

(d)

Let X = weight of filled can, kg

X = 1.51751K , d2 = 2.326

R = 0.029266K ,

1.48 − 1.51751K   P(X  1.48) = P  Z   = P ( Z  −2.98) = 0.9986 0.029266K / 2.326   (e) (f)

CPL =

X − LSL

(

3 R / d2

)

1.51751K − 1.48 = 0.9938K 3 ( 0.029266K / 2.326 )

The process is in control. From (d) we can conclude that since 99.86% of the items will fall above the lower specification limit, the process is capable of meeting the 99.7% goal. However, since CPL is slightly less than 1, the potential of the process may not be capable of meeting the 99.7% requirement.

18.46 (a) and (b) 1 3 4 11 12 13 14 21 22 23 24

A Online Test 1

Data Sample/Subgroup Size 5 R Chart Control Limits Lower Control Limit 0 Center 3.785714286 Upper Control Limit 8.003 XBar Chart Control Limits Lower Control Limit 4.444214286 Center 6.628571429 Upper Control Limit 8.812928571

Online Test 1 9

Sample Range - Test Mark

UCL

7 6 5 4

RBar

3 2 1

LCL

Day

X-Bar Chart: Online Test 1

Sample Mean - Test Mark

10 9

UCL

8 7

XBar

6 5 LCL

4 3 2 1 0

Day

1 3 4 11 12 13 14 21 22 23 24

A Online Test 4

Data Sample/Subgroup Size 5 R Chart Control Limits Lower Control Limit 0 Center 4.571428571 Upper Control Limit 9.664 XBar Chart Control Limits Lower Control Limit 3.848 Center 6.485714286 Upper Control Limit 9.123428571

R Chart: Online Test 4

Sample Range - Test Mark

12 10

UCL

8 6 RBar

4 2 0

LCL 0

Day

X-Bar Chart: Online Test 4 10

UCL

Sample Mean - Test Mark

9 8 7

XBar

6 5 4

LCL

2 1 0 0

Day (c)

All of the points are within the control limits, and there are no patterns in the R charts. All the points in the X charts are also within the control limits, and there is no pattern in the X chart for test 1. So the process is in control. However, there is a pattern in the X chart for test 4, with the mean generally decreasing, so this is a special cause of variation. As test 4 is the last of the online tests, it could be that weaker students or students who do not need the marks are not doing the test until later while good students are doing the test earlier to get it out of the way, so that they can concentrate on the final

exam. From this explanation of the special cause of variation, and as the pattern shows no advantage to students doing the test later, the lecturer need not be concerned about the validity of the test marks. 18.47 (a)

p Chart

Proportion

0.6 0.5

UCL

0.4

pBar

0.3

LCL

0.2 0.1 0 0

Day p = 0.391, LCL = 0.301, UCL = 0.480. (b)

The process is out of statistical control. The proportion of investigations that are closed is below the LCL on days 2 and 16 and is above the UCL on days 22 and 23.

(c)

Special causes of variation should be investigated and eliminated. Next, process knowledge should be improved to increase the proportion of investigations closed the same day. When the proportions are above the UCL, this is in the direction the process needs to go. Therefore, when investigating special causes for days 22 and 23, consideration of exploiting these special causes should be given.

18.48 (a)

Control chart for the data

p Chart 0.3

Proportion

0.25 0.2 UCL

0.15

pBar 0.1

LCL

0.05

0 0

15 X

(b)

The process does not appear to be in control since point 4 is above the UCL and therefore outside the control limits.

(c)

18.49 Answers will vary. 18.50 Answers will vary. 18.51 (a) p Chart Summary 0.18 0.16

UCL

Proportion

0.14 0.12

pBar

0.10 0.08

LCL

0.06

0.04 0.02 0.00 0

p Chart Summary Intermediate Calculations Sum of Subgroup Sizes 22740 Number of Subgroups Taken 30 Average Sample/Subgroup Size 758 Average Proportion of Nonconforming Items 0.1091469 Three Standard Deviations 0.0339778 Preliminary Lower Control Limit 0.0751691 p Chart Control Limits Lower Control Limit 0.075169 Center 0.109147 Upper Control Limit 0.143125 Process is not in control, days 9, 26 and 30 are all above UCL. (b)

18.52 (a)

Reasons for the special cause variation need to be determined and eliminated. Once special causes have been eliminated and the process is in control, Deeming’s fourteen points or the DMAIC model should be implemented to improve the process. Mean number of incidents per week is 3.0

(b)

Number of Incidents per Week

C Chart for Fire Bridgade 9 8 7 6 5 4 3 2 1 0 0

Week Incidents

LCL

Center

UCL

c Chart Summary Intermediate Calculations Sum of Nonconformities Number of Units Sampled Cbar Preliminary Lower Control Limit c Chart Control Limits Lower Control Limit Center Upper Control Limit

156 52 3 -2.1962 0.0000 3.0000 8.1962

(c)

There are no points outside the control limits. However, there is evidence of a possible pattern over time, with the first eight weeks all below the mean. Thus the special causes which may be contributing to this pattern should be investigated.

(d)

Since the seven incidents attended in Weeks 15 and 41 are below the upper control limit, they can be explained by common causes of variation.

(e)

After identifying any special causes which may have contributed to the first eight weeks being below the average, the fire brigade can use a c-chart to monitor the process in real time, to help identify any potential special causes of variation such as increased arson, severe drought, or holiday-related activities or other emergencies such as searches, storms and floods.

18.53 (a)

Control Chart Calculations Data Sample/Subgroup Size 5 R Chart Intermediate Calculations RBar 271.5666667 D3 Factor 0 D4 Factor 2.114 R Chart Control Limits Lower Control Limit 0 Center 271.5666667 Upper Control Limit 574.0919333 XBar Chart Intemediate Calculations Average of Subgroup Averages 198.6666667 A2 Factor 0.577 A2 Factor * RBar 156.6939667 XBar Chart Control Limits Lower Control Limit 41.9727 Center 198.6666667 Upper Control Limit 355.3606333

Control Chart for Range

700

Sample Range

600

UCL

500 400 300

RBar

200 100

0 0

15 Week

LCL 30

Control Chart for Mean 400 UCL

350

Sample Mean

300 250 200

XBar

150

100 50

LCL

0 0

(b)

15 Week

The process appears to be in control since there are no points outside the lower and upper control limits of both the R-chart and X chart, and there is no pattern in the results over time.

18.54(a)

Control Chart Calculations Data Sample/Subgroup Size 5 R Chart Control Limits Lower Control Limit 0 Center 0.363333333 Upper Control Limit 0.768086667 XBar Chart Control Limits Lower Control Limit 999.83369 Center 1000.043333 Upper Control Limit 1000.252977

Control Chart Range - Bench Length 0.9 0.8

UCL

Sample Range

0.7 0.6 0.5 0.4

RBar

0.3 0.2 0.1 0 0

LCL 30

Day

Sample Mean

Control Chart Mean - Bench Length UCL

1,000.25 1,000.20 1,000.15 1,000.10 1,000.05 1,000.00 999.95 999.90 999.85 999.80

XBar

LCL 0

Day

(b)

The process appears to be in control since there are no points outside the lower and upper control limits of both the R-chart and X chart, and there is no pattern in the results over time. Estimate of the population mean = X = 1000.04333...

(c)

Estimate of the population standard deviation = R /d2 = (i)

0.36333... = 0.15620... 2.326

1002.5 − 1000.0433...   997.5 − 1000.0433... P ( 997.5  X  1002.5) = P  Z  0.1562... 0.1562...   = P(−16.28  Z  15.72) =1 Therefore, 100% of benchtops manufactured will be within the specification limits.

(ii)

Cp =

USL − LSL 1002.5 − 997.5 = = 5.334... 6 ( R /d2 ) 6 ( 0.15620... )

CPL =

X − LSL 1000.0433... − 997.5 = = 5.427... 3 ( R /d2 ) 3 ( 0.15620... )

CPU =

USL − X 1002.5 − 1000.0433... = = 5.242... 3 ( R /d2 ) 3 ( 0.15620... )

Cpk = min(CPL,CPU) = 5.24 Since these capability indices are all above 5 the actual and potential process is exceeding the company’s requirements. (d) (i)

Suppose specification limits are reduced to 1000mm ± 1.0mm

1001 − 1000.0433...   999 − 1000.0433... P ( 999  X  1001) = P  Z  0.1562... 0.1562...   = P(−6.68  Z  6.12) 1 Therefore, 100% of benchtops manufactured will be within the new specification limits.

(ii)

Cp =

USL − LSL 1001 − 999 = = 2.1339... 6 ( R /d2 ) 6 ( 0.15620... )

CPL =

X − LSL 1000.0433... − 999 = = 2.226... 3 ( R /d2 ) 3 ( 0.15620... )

CPU =

USL − X 1001 − 1000.0433... = = 2.041... 3 ( R /d2 ) 3 ( 0.15620... )

Cpk = min(CPL,CPU) = 2.04 Since these capability indices are all above 2 the actual and potential process is capable of meeting the company’s requirements if the specification limits are reduced to 1000mm ± 1.0mm.

18.55 (a)

Control Chart Calculations Data Sample/Subgroup Size 5 R Chart Intermediate Calculations RBar 0.2424 D3 Factor 0 D4 Factor 2.114 R Chart Control Limits Lower Control Limit 0 Center 0.2424 Upper Control Limit 0.5124336 XBar Chart Intemediate Calculations Average of Subgroup Averages 1.00032 A2 Factor 0.577 A2 Factor * RBar 0.1398648 XBar Chart Control Limits Lower Control Limit 0.8604552 Center 1.00032 Upper Control Limit 1.1401848 Control Chart Range - Upload Speed

0.6 UCL

Sample Range

0.5 0.4 0.3

RBar

0.2 0.1 0 0

LCL 25

Day

Control Chart Mean - Upload Speed 1.2 1.2

UCL

Sample Mean

1.1 1.1 1.0

XBar

1.0 0.9 LCL

0.9 0.8 0

15 Day

(b)

The process appears to be in control since there are no points outside the lower and upper control limits of both the R-chart and X chart, and there is no pattern in the results over time.

(c)

Estimate of the population mean = X = 1.0032 Estimate of the population standard deviation = R /d2 =

(i)

0.2424 = 0.1042... 2.326

1.2 − 1.00032   P ( X  1.2 ) = P  Z   0.1042...   = P(Z  1.92) = 0.9726 Therefore, approximately 97.3% of uploads are within specification.

(ii) (d)

CPU =

USL − X 1.2 − 1.00032 = = 0.6386... 3 ( R /d2 ) 3 ( 0.1042... )

Since only 97.3% of uploads are currently meeting AMS requirements and CPU = 0.63 < 1 the process is not capable of meeting AMS’s requirements. Since the process is in control management will need to change the process to improve upload speed by reducing common cause variation.

Chapter 19: Further non-parametric tests After studying this chapter you should be able to: 1.

choose and conduct a selection of powerful non-parametric tests

19.1

(a)

H 0 : 1   2

H1 :  1   2

where population 1= brand A prior change, 2 = brand

A after change Decision rule: If Zcalc < –1.645, reject H0. Test statistic: Z =

B −C 20 − 43 = = −2.898 B+C 20 + 43

Conclusion: Since Zcalc = –2.898 is less than –1.645, reject H0 at the 5% significance level. There is enough evidence to conclude that the proportion of beer drinkers who prefer brand A is lower prior to the change in brewing technique.

19.2

(b)

p-value = 0.0019. The probability of obtaining a data set which gives rise to a test statistic smaller than –2.898 is 0.19% if the null hypothesis is true.

(a)

H 0 : 1 −  2

≥0

H1 :  1 −  2 > 0 where population 1= candidate A prior debate, 2= candidate A after debate Decision rule: If Z calc > 1.645, reject H0. Test statistic: Z =

25 -10 25 +10

B -C B +C

21- 36

= -1.987 keep formula

21+ 36

= 2.236

Conclusion: Since Zcalc = 2.236 > 1.645, reject H0 at the 5% significance level. There is enough evidence to conclude that the proportion of voters who prefer candidate A is lower prior to the debate.

19.3

(b)

p-value = 0.0.125. The probability of obtaining a data set that gives rise to a test statistic at least or more extreme than the one calculated here is 0.0125% if the null hypothesis is true.

(a)

H 0 : 1   2

H1 :  1   2 where population 1= prefer brand A before, 2= prefer brand A after Decision rule: If Zcalc < –1.645, reject H0. Test statistic: Z =

B −C 5 − 15 = = −2.236 B+C 5 + 15

Conclusion: Since Zcalc = –2.236 is less than –1.645, reject H0 at the 5% significance level. There is enough evidence to conclude that the proportion who prefer brand A is lower before the advertising than after.

19.4

(b)

p-value = 0.025. The probability of obtaining a data set that gives rise to a test statistic at least or more extreme than the one calculated here is 2.5% if the null hypothesis is true.

(a)

H 0 : 1   2

H1 :  1   2

where population 1 = satisfied last year, 2 = satisfied

this year Decision rule: If Zcalc < –1.645, reject H0. Test statistic: Z =

B −C 8 − 25 = = −2.959 B+C 8 + 25

Conclusion: Since Zcalc = –2.959 is less than –1.645, reject H0 at the 5% significance level. There is enough evidence to conclude that the satisfaction was lower last year prior to the introduction of a new catering company compared to this year.

19.5

(b)

p-value = 0.00135. The probability of obtaining a data set that gives rise to a test statistic smaller than –2.959 is 0.135% if the null hypothesis is true.

(a)

H 0 : 1   2

H1 :  1   2 where population 1 = absenteeism (< 5 days) year 1, 2 = absenteeism (< 5 days year 2) Decision rule: If Zcalc < –1.645, reject H0.

Test statistic: Z

B -C B +C

4 - 25 4 + 25

= -3.900

Conclusion: Since Zcalc = –3.900 is less than –1.645, reject H0 at the 5% significance level. There is enough evidence to conclude that the proportion of employees absent less than 5 days was lower in year 1 than in year 2.

19.6

19.7

19.8

19.9

(b)

p-value = 0.00005. The probability of obtaining a data set that gives rise to a test statistic smaller than –3.900 is 0.005% if the null hypothesis is true.

(a)

The lower and upper critical values are 24 and 51, respectively.

(b)

The lower and upper critical values are 22 and 53, respectively.

(c)

The lower and upper critical values are 18 and 57, respectively.

(d)

As the level of significance  gets smaller, the width of the nonrejection region gets wider.

(a)

The upper critical value is 51.

(b)

The upper critical value is 53.

(c)

The upper critical value is 55.

(d)

The upper critical value is 57.

(a)

The lower critical value is 24.

(b)

The lower critical value is 22.

(c)

The lower critical value is 20.

(d)

The lower critical value is 18.

T1 = 4 + 1 + 8 + 2 + 5 + 10 + 11 = 41

19.10 The lower and upper critical values are 40 and 79, respectively. 19.11 Decision: Since T1 = 41 is between the critical bounds of 40 and 79, do not reject H0 at the 5% significance level. 19.12 (a)

The ranks for sample 1 are 1, 2, 4, 5 and 10, respectively. The ranks for sample 2 are 3, 6.5, 6.5, 8, 9 and 11, respectively.

(b)

T1 = 1 + 2 + 4 + 5 + 10 = 22

(c)

T2 = 3 + 6.5 + 6.5 + 8 + 9 + 11 = 44

(d)

T1 + T2 =

n(n +1) 11(12) = = 66 T1 + T2 = 22 + 44 = 66 2 2

19.13 The lower critical value is 20. 19.14 Decision: Since T1 = 22 is greater than the lower critical bound of 20, do not reject H0 at the 5% significance level. 19.15 H0: M1 = M2

where populations 1 = group A, 2 = group B

H1: M1 ≠M2 T1 = 84 from Table E.8 critical values 62 and 109. Conclusion: Since T1 = 84 is between the lower and upper critical values of 62 and 109, do not reject H0 at the 5% significance level. There is insufficient evidence of a difference in the median complaints between the two groups. 19.16 (a)

H0: M1 = M2

where population 1 = LIRR, 2 = NJT

H1: M1 ≠ M2

Wilcoxon Rank Sum Test Data Level of Significance 0.05 Population 1 Sample Sample Size 10 Sum of Ranks 141 Population 2 Sample Sample Size 12 Sum of Ranks 112 Intermediate Calculations Total Sample Size n 22 T 1 Test Statistic 141 T 1 Mean 115 Standard Error of T 1 15.1658 Z Test Statistic 1.7143892 Two-Tail Test Lower Critical Value -1.9600 Upper Critical Value 1.9600 p -Value 0.0865 Do not reject the null hypothesis Decision: Since Z = 1.714 is between the critical values of −2.58 and 2.58, do not reject H0. There is not enough evidence to conclude that there is any difference in the median tendencies to be late. (b)

Assume populations have similar variability and shape.

(c)

There is insufficient evidence to conclude that there is any significant difference in the median tendencies of the railroads to be late.

19.17 (a)

H0: M1 = M2 where population 1 = computer-assisted individualbased, 2 = team-based program Median assembly time in seconds is the same for employees trained in a computer-assisted, individual-based program and those trained in a team-based program. H1: M1 ≠ M2 median assembly time in seconds is different for employees trained in a computer-assisted, individual-based program and those trained in a team-based program. Decision rule: If Z < –1.96 or Z > 1.96, reject H0.

Test statistic: T1 = 379

n1n2 ( n + 1) 21 21 43 = 39.7524 = 12 12

T = 1

n ( n + 1) 21 ( 43) = = 451.5, 2 2

T = 1

T1 – mT

379–451.5 = – 1.82 39.7524

Conclusion: Since Zcalc = –1.82 is between the critical bounds of ± 1.96, do not reject H0. There is not enough evidence to conclude that the median assembly time in seconds is different for employees trained in a computer-assisted, individual-based program and those trained in a team-based program. (b)

Assume populations have similar variability and shape.

(c)

Using the pooled-variance t test you reject the null hypothesis and conclude that mean assembly time in seconds is different for the two sets of trainees. Using the separate-variance t test the test statistic is between the critical values and you are unable to conclude there was any difference in the mean assembly times. In this test, the test statistic using the Wilcoxon rank sum test with large-sample Z approximation is also between the critical values and you fail to reject the null hypothesis here as well.

19.18 (a)

H0: M1 ≥ M2

where population 1 = females, 2 = males

H1: M1 < M2 Decision rule: If Z < –1.645, reject H0. Wilcoxon Rank Sum Test Data Level of Significance

0.05

Population 1 Sample Sample Size

Sum of Ranks

247

Population 2 Sample Sample Size

Sum of Ranks

573

Z Test Statistic

-3.316736

Lower-Tail Test Lower Critical Value

-1.6449

p-Value

0.0005

Reject the null hypothesis Conclusion: Since Zcalc = –3.317 is less than the lower critical bound of –1.645, reject H0. There is enough evidence to conclude that the median salary is less for females. (b)

Assume populations have similar variability and shape

(c)

Obtain the same conclusion using either a pooled variance or separate variance t-test.

19.19 (a)

H0: M1 = M2

where population 1 = bank 1, 2 = bank 2

H1: M1 ≠ M2 Wilcoxon Rank Sum Test Data Level of Significance

0.05

Population 1 Sample Sample Size

Sum of Ranks

153

Population 2 Sample Sample Size

Sum of Ranks

312

Intermediate Calculations Total Sample Size n

T1 Test Statistic

153

T1 Mean

232.5

Standard Error of T1

24.10913

Z Test Statistic

-3.29751

Two-Tailed Test Lower Critical Value

-1.95996

Upper Critical Value

1.959961

p-value

0.000976

Reject the null hypothesis

Decision rule: If Z < –1.96 or Z > 1.96, reject H0. Conclusion: Since Zcalc = –3.2975 is less than the lower critical bound of –1.96, reject H0. There is enough evidence to conclude that the median waiting time between the two branches is different. (b)

Assume populations have similar variability and shape.

19.20 (a)

H0: M1 = M2

where population 1 = office I, 2 = office II

H1: M1 ≠ M2 Wilcoxon Rank Sum Test Data Level of Significance

0.05

Population 1 Sample Sample Size

Sum of Ranks

434.5

Population 2 Sample Sample Size

Sum of Ranks

385.5

Intermediate Calculations Total Sample Size n

T1 Test Statistic

434.5

T1 Mean

410

Standard Error of T1

36.96846

Z Test Statistic

0.662727

Two-Tailed Test Lower Critical Value

-1.95996

Upper Critical Value

1.959961

p-value

0.507505

Do not reject the null hypothesis

Decision rule: If Z < –1.96 or Z > 1.96, reject H0. Conclusion: Since Zcalc = 0.6627 is between the lower and upper critical bounds of ±1.96, do not reject H0. There is not enough evidence to conclude that the median time to clear the problems between the two offices is different. (b)

Assume populations have similar variability and shape.

(c)

Using both the pooled-variance t test and the separate-variance t test in you do not reject the null hypothesis; you conclude that there is not enough evidence to show that the mean waiting time between the two branches is different. In this test using the Wilcoxon rank sum test with large-sample Z approximation, you also do not reject the null hypothesis; you conclude that there is not enough evidence to show that the median waiting time between the two branches is different.

19.21 (a)

WL = 13, WU = 53

(b)

WL = 10, WU = 56

(c)

WL = 7, WU = 59

(d)

WL = 5, WU = 61

19.22 (a)

WU = 53

(b)

WU = 56

(c)

WU = 59

(d)

WU = 61

19.23 (a)

WL = 13

(b)

WL = 10

(c)

WL = 7

(d)

WL = 5

19.24 Observation (Di)

abs(Di)

Sign of Di

Rank

Signed R

R(+)

4.8

1.7

4.5

-1.2

1.2

-2

11.1

-0.8

0.8

-3

2.3

-2.0

2.0

-1

0.0

Discard

14.8

7.8

1.7

4.5

W = S i=1 Ri n'

(+ )

= 60

19.25 n = 12,  = 0.05, WU = 61 '

19.26 Since W = 60 < WU = 61, do not reject H 0 at the 5% significance level.

19.27 Observation (Di)

abs(Di)

Sign Di

5.0

6.5

of Rank

Signed R

R(+)

6.5

7.5

4.0

11.0

-8.0

8.0

-1

2.5

-2.5

2.5

-2

1.0

12.0

6.5

7.5

10.0

9.0

W = Si=1 Ri n'

(+)

= 75

19.28 n = 12,  = 0.05, WU = 61 '

19.29 Since W = 75 > WU = 61, reject H 0 at the 5% significance level.

19.30 H0: MD ≥ 0

H1: MD < 0

where population 1 = before, 2 = after

Observation (Di)

abs(Di)

Sign of Di

Rank

Signed R

R(+)

-63

-4

-111

111

-2

-12

-5

-114

114

-1

-98

-3

-2

-6

Differences:

Before – After n'

( +)

Test statistic: W = i=1Ri

Decision rule: Since n = 6 and  = 0.05, if W < 2 reject H0.

Conclusion: Since W = 0 is less than the lower critical value 2, reject H 0 at the 5% significance level. There is evidence of an increase in the rate of drinks dispensed. 19.31 H0: MD ≤ 0 Woolworths, 2 = Coles

H1: MD > 0

where

population

Item Woolworths Coles Difference (Di) abs(Di) Sign of Di Lamb Chops $6.49 $6.50 -0.01 0.01 Fruit Juice $2.40 $2.37 0.03 0.03 + Apples $4.80 $4.50 0.30 0.30 + Lettuce $3.00 $3.50 -0.50 0.50 Vegemite $6.00 $6.60 -0.60 0.60 Oats $5.80 $5.00 0.80 0.80 + Wholemeal Bread $2.80 $2.00 0.80 0.80 + Milk $2.00 $1.20 0.80 0.80 + Rump Steak $21.00 $20.00 1.00 1.00 + Eggs $4.10 $3.00 1.10 1.10 + Differences:

Rank 1 2 3 4 5 7 7 7 9 10

Woolworths – Coles

Test statistic: W =

R = 2 + 3 + 7 + 7 + 7 + 9 + 10 = 45 i=1

( +) i

Decision rule: Since n = 10 and  = 0.01, if W > 50, reject H0. Conclusion: Since W = 45 is less than the upper critical value of 50, do not reject H0 at the 1% significance level. The sample does not provide evidence that the median price is higher at Woolworths than at Coles. 19.32 H0: MD = 0 where population 1 = bookstore, 2 = Amazon H1: MD ≠ 0 '

n = 15

WL = 16 , WU = 104

W = Ri( + ) = 54 i=1

Since W = 54 falls between WL = 16 and WU = 104 , do not reject H 0 . There

is insufficient evidence of a difference in the median price of textbooks between the local bookstore and Amazon.com

Bookshop

Amazon

|Di|

52.22

57.34

-5.12

5.12

52.74

44.47

8.27

39.04

41.48

-2.44

2.44

101.28

73.72

27.56

37.45

42.04

-4.59

4.59

113.41

95.38

18.03

109.72

119.8

-10.08

10.08

101.28

62.48

38.8

29.49

32.43

-2.94

2.94

70.07

74.43

-4.36

4.36

83.87

83.81

0.06

23.21

26.48

-3.27

3.27

72.8

73.48

-0.68

0.68

17.41

20.98

-3.57

3.57

37.72

40.43

-2.71

2.71

19.33 H0: MD = 0

Ri(+)

11 14 13 15

where population 1 = before, 2 = after

H1: MD ≠0

Before 78 88 72 88 61 89 60 Differences:

After 77 89 74 86 84 62 94

|D| 1 -1 -2 2 -23 27 -34

R 1 1 2 2 23 27 34

R+ 1.5 1.5 4.5 4.5 33.0 34.0 35.0 630.0

1.5 0 0 4.5 0 34 0 156.0

Before – After assume distribution approximately symmetric

Test statistic: W = 156 n’ = 35

n'(n'+ 1) 4 Z= n'(n'+ 1)(2n'+ 1) 24 35  36 156 − 4 = 35  36  71 24 156 − 315 = 3727.5 = −2.604... W−

Decision rule:  = 0.05, if Z <

−1.96 or Z > 1.96 reject H0.

Conclusion: Since Z = −2.60 < −1.96 reject H0 . There is sufficient evidence of a difference in the median performance ratings before and after workshops. 19.34 H0: MD ≥0 H1: MD < 0 Assume distribution of D = Thursday – Friday is approximately symmetric.

Station Thursday Friday 10 133 133 5 138 142 3 114 123 1 132 143 4 142 156 8 138 155 9 142 163 7 119 145 2 123 155 6 128 165 n’ = 9 W =

D 0 -4 -9 -11 -14 -17 -21 -26 -32 -37

|D| 0 4 9 11 14 17 21 26 32 37

R+

1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0

0 0 0 0 0 0 0 0 0

R( ) = 0 i =1

Decision rule: Since n’ = 9 and  = 0.01, if W < 3, reject H0. Conclusion: Since W = 0 is less than the lower critical value, 3, reject H 0 at the 1% significance level. There is sufficient evidence that the median petrol price is more expensive on a public holiday.

19.35

For the 0.01 level of significance and 5 degrees of freedom,  U = 15.086 .

19.36

(a)

Decision rule: If H >  U = 15.086 , reject H0.

(b)

Decision: Since Hcalc = 13.77 is less than the critical bound of 15.086, do not reject H0.

19.37 H0: MA = MB = MC H1: At least one of the medians differs. Assume populations have similar variability and shape. Decision rule: If H U = 9.210, reject H0. 2

Test statistic: H = 0.64 Decision: Since Hcalc = 0.64 is less than the critical value of 9.210, do not reject H0. There is insufficient evidence to show any real difference in the median reaction times for the three learning methods. Kruskal-Wallis Rank Test for Differences in Medians Data Level of Significance 0.01 Group Sample Size Sum of Ranks Mean Ranks 1 9 104 11.5555556 Intermediate Calculations 2 8 106 13.25 Sum of Squared Ranks/Sample Size 4259.403 3 8 115 14.375 Sum of Sample Sizes 25 Number of Groups 3 Test Result H Test Statistic 0.6351 Critical Value 9.2103 p-Value 0.7279 Do not reject the null hypothesis

19.38 H0: MSydney = MMelbourne = MAuckland= MSingapore H1: At least one of the medians differs. Assume populations have similar variability and shape. Decision rule: If H U = 7.815, reject H0. 2

Test statistic: H = 8.88 Decision: Since Hcalc = 8.88 is greater than the critical value of 7.815 reject H0. At the 5% level of significance there is evidence of a significant difference in the median sales between the four cities.

Kruskal-Wallis Rank Test for Differences in Medians Data Level of Significance 0.05 Group Sample Size Sum of Ranks Mean Ranks Sydney 5 62.5 12.5 Intermediate Calculations Auckland 5 21 4.2 Sum of Squared Ranks/Sample Size 2515.9 Melbourne 5 52.5 10.5 Sum of Sample Sizes 20 Singapore 5 74 14.8 Number of Groups 4 Test Result H Test Statistic 8.8829 Critical Value 7.8147 p-Value 0.0309 Reject the null hypothesis

19.39 H0: MFront = MMiddle = MRear H1: at least one of the medians differs. Assume populations have similar variability and shape. Decision Rule: Reject H0 if H > .05,2 = 5.991. 2

Test statistic: H = 11.792… c T2   12  12  j H=  (1960.5833) − 3 (19 ) = 11.79  − 3 (n + 1) =    n (n + 1) j=1 nj   18 (19 ) 

Decision: Since Hcalc = 11.82 is greater than the critical value of 5.991, reject H0. There is sufficient evidence to show there is a difference in the median sales of pet toys among the three store aisle locations.

Sampl e

Value

Rank

Middle

1.4

Middle

1.6

Middle

1.8

Middle

Front

Middle

Rear

88.5

57.5

Tj^2

7832.25

625

3306.25

Rear

2.2

Tj^2/nj

1305.375

104.1667

551.0417

Middle

2.4

Rear

2.8

7.5

Rear

2.8

7.5

Middle

3.2

Front

10.5

Rear

10.5

Rear

4.6

Front

5.4

Rear

Front

6.2

Front

7.2

Front

8.6

Kruskal-Wallis Rank Test for Differences in Medians Data Level of Significance 0.05 Group Front Intermediate Calculations Middle Sum of Squared Ranks/Sample Size 1960.583 Rear Sum of Sample Sizes 18 Number of Groups 3 Test Result H Test Statistic 11.7924 Critical Value 5.9915 p-Value 0.0027 Reject the null hypothesis

Sample Size

Sum of Ranks Mean Ranks 6 88.5 14.75 6 25 4.166666667 6 57.5 9.583333333

19.40 PhStat output: Kruskal-Wallis Rank Test for Differences in Medians Data Level of Significance

0.05

Intermediate Calculations Sum of Squared Ranks/Sample Size

7660.1

Sum of Sample Sizes

Number of Groups

Test Result H Test Statistic

5.84

Critical Value

5.991465

p-Value

0.053934

Do not reject the null hypothesis H0: M1 = M2 = M3 H1: at least one of the medians differs. Since the p-value = 0.0539 > 0.05, do not reject H0. There is no evidence of a difference in the median sick leave taken of the age groups.

19.41

H0: MKroger = MGlad = MHefty = MTuffstuff H1: At least one of the medians differs. Assume populations have similar variability and shape.

Kruskal-Wallis Rank Test for Differences in Medians Data Level of Significance Intermediate Calculations Sum of Squared Ranks/Sample Size Sum of Sample Sizes Number of Groups Test Result H Test Statistic Critical Value p-Value Reject the null hypothesis

0.05

19852.2 40 4

Group Sample Size Sum of Ranks Mean Ranks KROGER 10 254 25.4 GLAD 10 270 27 HEFTY 10 241 24.1 TUFFSTUFF 10 55 5.5

22.2600 7.8147 0.0001

Since the p-value is virtually zero, reject H0. There is sufficient evidence of a difference in the median strength of the four brands of garbage bags. 19.42 For df = 6 – 1 = 5 and  = 0.01,

U2 = 16.812

19.43 (a) (b)

Decision rule: Reject H0 when FR > U = 16.812. 2

FR = 11.56 is less than the upper critical value of 16.812, therefore, do not reject H0 at the 1% significance level.

19.44 Friedman rank test for Rating by Brand blocked by Expert

H0 : MA = MB = MC = MD H1 : not all medians are the equal. Assume populations have similar variability and shape.

Ranks Group 1 Group 2 Group 3 Group 4 2 4 3 1 3.5 3.5 2 1 2 4 3 1 2 4 3 1 2.5 4 2.5 1 3 4 1.5 1.5 4 3 1 2 3 4 2 1 3 4 2 1 25 34.5 20 10.5 625 1190.25 400 110.25

Block 1 Block 2 Block 3 Block 4 Block 5 Block 6 Block 7 Block 8 Block 9 Rank Totals Rank Totals squared Data Level of significance 0.05 Intermediate Calculations Number of blocks 9 Number of groups 4 First intermediate term 0.066667 Sum of rank total squares 2325.5 Second intermediate term 135 Degrees of freedom 3 Test Result Friedman Test Statistic 20.03333 Critical Value 7.8147 p -Value 0.0002 Reject the null hypothesis

Since the p-value is virtually zero, reject H0 at 0.05 level of significance. There is evidence of a difference in the median ratings of the four brands of Colombian coffee.

19.45 H0 : MA = MB = MC = MD

H1 : Not all medians are equal.

Assume populations have similar variability and shape. Rank Bazooka

Bubbletape

Bubbleyum

Bubblicious

2.5

1.5

R.j

(R.j)^2

169

144

Test statistic: FR =

c 12 12 R2j − 3r ( c + 1) = ( 426 ) − 3( 4 )( 5) = 3.9  rc ( c + 1) j=1 ( 4 )( 4 )( 5)

Upper critical value: U = 0.05,3 = 7.8147 2

Decision:

Since FR = 3.9 < 7.81, do not reject H0. There is insufficient evidence of a difference in the median diameter of the bubbles produced by the different brands.

19.46 Friedman rank test for price by store blocked by item. Assume populations have similar variability and shape.

H0 : MWoolworths = MColes = MAldis = MIGA H1 : not all medians are equal.

Item Milk Eggs Fruit Juice Lettuce Lamb Chops Vegemite Apples Oats Rump Steak Wholemeal Bread

Woolworths Coles 4.0 2.0 4.0 1.0 4.0 3.0 3.0 4.0 2.0 3.0 2.0 4.0 4.0 3.0 4.0 3.0 4.0 3.0 3.0 1.0 Ri 34 27 Ri^2 1,156 729

Aldis 1.0 3.0 1.0 1.0 1.0 1.0 2.0 2.0 2.0 2.0 16 256

IGA 3.0 2.0 2.0 2.0 4.0 3.0 1.0 1.0 1.0 4.0 23 529

100 2,670

Test statistic

FR =

c 12 12 R2j − 3r ( c + 1) = (2670 ) − 3(10 )( 5) = 10.2  rc ( c + 1) j=1 (10 )( 4 )( 5)

Upper critical value: U = 0.05,3 = 7.8147 2

Since FR = 10.2 > 7.81, reject H0. There is evidence of a difference in the median prices for these kitchen staples at the four stores.

Ranks Woolworths Milk 4 Eggs 4 Fruit Juice 4 Lettuce 3 Lamb Chops 2 Apples 2 Oats 4 Rump Steak 4 Rump Steak 4 Wholemeal Bread 3 Rank Totals 34 Rank Totals squared 1156 Data Level of significance 0.05 Intermediate Calculations Number of blocks 10 Number of groups 4 First intermediate term 0.06 Sum of rank total squares 2670 Second intermediate term 150 Degrees of freedom 3 Test Result Friedman Test Statistic 10.2 Critical Value 7.8147 p -Value 0.0169 Reject the null hypothesis

Coles

Aldis

2 1 3 4 3 4 3 3 3 1 27 729

1 3 1 1 1 1 2 2 2 2 16 256

IGA 3 2 2 2 4 3 1 1 1 4 23 529

19.47 The McNemar test should be used when determining whether there is evidence of a difference between the proportions of two related samples.

19.48 The Wilcoxon rank sum test should be used when you are unable to assume that each of two independent populations is normally distributed. 19.49 The Wilcoxon signed ranks test should be used when comparing the median difference between matched items or repeated measurements of the same item. 19.50 The Kruskal–Wallis test should be used if you cannot assume that the populations are normally distributed. 19.51 The Friedman rank test should be used to test for the difference in c medians in a randomised block design when data collected are only in rank form within each block or when normality cannot be assumed. 19.52

McNemar Test

(a)

H0 : 1 = 2 H1 : 1  2

where 1 = before, 2 = after

Decision rule: If Z < –1.96 or Z > 1.96, reject H0. Test statistic: Z STAT =

B−C 23 − 15 = = 1.297... B+C 23 + 15

Conclusion: Since Z = 1.30 < 1.96 the upper critical value do not reject H0 at the 5% significance level. At the 5% level of significance there is no evidence of a difference in the proportion of respondents who prefer Ford before and after viewing the ads. (b)

19.53

p-value = 0.1936. The probability of obtaining a data set that gives rise to a test statistic that differs from 0 by 1.66 or more in either direction is 9.7% if there is not a difference in the proportion of respondents who prefer Ford before and after viewing the ads.

McNemar Test (a)

H0 : 1 − 2 ≥0 H1 : 1 − 2 < 0 where 1 = before, 2 = after Decision rule: If Z < –1.28, reject H0. Test statistic: Z =

Decision:

3 − 16 B−C = = −2.9824 3 + 16 B+C

Since Z = –2.9824 is less than the lower critical value of – 1.28, reject H0 at the 10% significance level. There is enough evidence of an increase in the proportion of people supporting athletics after the campaign.

(b)

p-value = 0.0014. The probability of obtaining a data set that gives rise to a test statistic that differs from 0 by –2.98 or more is 0.14% if the null hypothesis is true.

19.54 Friedman rank test for Return by Fund blocked by Year Let

A = Australian Shares G = Multi-sector Growth

B = Multi-sector Balanced H = Multi-sector High Growth

I = International Shares Assume populations have similar variability and shape.

H0 : MA = MI = MB = MG = MH

H1 : not all medians are equal.

Return for year ending 30 June % 2012 2013 2014 2015 2016 2017 R R^2

Australian Shares

International Multi-Sector Multi-Sector Multi-Sector Shares Balanced Growth High Growth

1 4 2 1 2 2 12 144

3 5 5 5 1 5 24 576

5 1 1 2 5 1 15 225

4 2 3 3 4 3 19 361

2 3 4 4 3 4 20 400

90 1706

Test statistic:

FR =

c 12 12 R2j − 3r ( c + 1) = (1706 ) − 3(6 )(6 ) = 5.733...  rc ( c + 1) j=1 ( 6 )( 5)( 6 )

Upper critical value: U = 0.05,4 = 9.488 2

Since FR = 5.73 < 9.488 do not reject H0 at the 0.05 level of significance. There is insufficient evidence of a difference in the median returns for the five funds.

Ranks Australian Shares

International Multi-Sector Multi-Sector Multi-Sector Shares Balanced Growth High Growth 1 3 5 4 2 4 5 1 2 3 2 5 1 3 4 1 5 2 3 4 2 1 5 4 3 2 5 1 3 4 12 24 15 19 20 144 576 225 361 400

2012 2013 2014 2015 2016 2017 Rank Totals Rank Totals squared Data Level of significance 0.05 Intermediate Calculations Number of blocks 6 Number of groups 5 First intermediate term 0.066666667 Sum of rank total squares 1706 Second intermediate term 108 Degrees of freedom 4 Test Result Friedman Test Statistic 5.733333333 Critical Value 9.4877 p -Value 0.2200 Do not reject the null hypothesis 19.55 Wilcoxon signed ranks test H0: MD = 0 H1: MD ≠ 0

where D = Opening Price – Closing Price assume distribution of D approximately symmetric Test statistic: W = 57 n’ = 13 Decision rule: n’ = 13 and  = 0.05 from Table E.9 WL = 17 , WU = 74 . Reject H0 if W < 17 or W > 74

Singapore United States Malaysia South Korea Canada New Zealand China Australia Indonesia Japan Taiwan Hong Kong India

Open 3320.67 6373.33 1781.65 2404.68 15267.40 7771.57 3262.08 5824.50 5755.04 20062.65 10598.59 27684.60 32341.05

TTest Statistic: W =

Close D 3318.08 2.59 6370.46 2.87 1777.94 3.71 2394.73 9.95 15256.40 11.00 7782.72 -11.15 3281.87 -19.79 5795.70 28.80 5810.56 -55.52 19996.01 66.64 10470.38 128.21 27854.91 -170.31 32014.19 326.86

|D| 2.59 2.87 3.71 9.95 11.00 11.15 19.79 28.80 55.52 66.64 128.21 170.31 326.86

R 1 2 3 4 5 6 7 8 9 10 11 12 13

R+ 1 2 3 4 5

8 10 11 13 57

R ( ) = 57 +

i=1

Conclusion: Since 17 < W = 57 <

74 do not reject H0 . There is insufficient

evidence of a difference in the median opening and closing prices for the selected indices. 19.56 Friedman rank test for Contestant blocked by Reviewer Assume populations have similar variability and shape

H0 : MA = MB = Mc = MD H1 : not all medians are equal.

Reviewer 1 2 3 4 5 6 R R^2

A 2.0 3.0 2.5 2.5 2.0 1.5 13.5 182.25

Test statistic: FR =

Contestent B C 1.0 4.0 1.5 4.0 1.0 4.0 2.5 1.0 1.0 4.0 3.5 1.5 10.5 18.5 110.25 342.25

D 3.0 1.5 2.5 4.0 3.0 3.5 17.5 306.25

60 941

c 12 12 R2j − 3r ( c + 1) = ( 941) − 3(6 )( 5) = 4.1  rc ( c + 1) j=1 ( 6 )( 4 )( 5)

Upper critical value: U = 0.05,3 = 7.815 2

Since FR = 4.1 < 7.815 do not reject H0 at the 0.05 level of significance. There is insufficient evidence of any difference between the scores for the different contestants

Ranks A 1 2 2 3 3 2.5 4 2.5 5 2 6 1.5 Rank Totals 13.5 Rank Totals squared 182.25 Data Level of significance 0.05 Intermediate Calculations Number of blocks 6 Number of groups 4 First intermediate term 0.1 Sum of rank total squares 941 Second intermediate term 90 Degrees of freedom 3 Test Result Friedman Test Statistic 4.1 Critical Value 7.8147 p -Value 0.2509 Do not reject the null hypothesis

1 1.5 1 2.5 1 3.5 10.5 110.25

4 4 4 1 4 1.5 18.5 342.25

3 1.5 2.5 4 3 3.5 17.5 306.25

19.57 Kruskal-Wallis Rank Test Assume populations have similar variability and shape H0: MA = MB = MC = MD H1: at least one of the medians differs. Decision rule:

If H U = 7.815, reject H0. 2

Test statistic: H = 11.91 Decision:

Since Hcalc = 11.91 is greater than the critical value of 7.815, reject H0. There is sufficient evidence to show there is a significant difference in median battery life between the toys.

Kruskal-Wallis Rank Test for Differences in Medians Data Level of Significance 0.05 Group Sample Size Sum of Ranks Mean Ranks Toy A 5 73 14.6 Intermediate Calculations Toy B 5 76 15.2 Sum of Squared Ranks/Sample Size 2622 Toy C 5 39 7.8 Sum of Sample Sizes 20 Toy D 5 22 4.4 Number of Groups 4 Test Result H Test Statistic 11.9143 Critical Value 7.8147 p-Value 0.0077 Reject the null hypothesis 19.58 Kruskal-Wallis Rank Test H0: MBryon Bay = MCairns = MGold Coast = MNoosa = MPort Douglas H1: at least one of the medians differs. Assume populations have similar variability and shape. Decision rule:

If H U = 7.779, reject H0. 2

Test statistic:

T T^2

Bryon Bay 16.0 18.0 4.5 12.5 10.5

Cairns Gold Coast Noosa 9.0 6.0 14.5 2.0 10.5 7.5 12.5 14.5 24.5 23.0 24.5 20.5 3.0 17.0 19.0

61.5 3782.25

49.5 2450.25

72.5 5256.25

86.0 7396.00

Port Douglas 7.5 4.5 22.0 1.0 20.5 55.5 3080.25

325 21965

c T2   12  12 21965  j H=    − 3 (n + 1 ) =   − 3 ( 26 ) = 3.101...  5   n (n + 1) j=1 nj   25 ( 26 )

Decision: Since Hcalc = 3.10 < 7.779 do not reject H0. There is no evidence to show that there is a significant difference in the median occupancy rates in the five locations.

Kruskal-Wallis Rank Test for Differences in Medians Data Level of Significance 0.1 Group Sample Size Sum of Ranks Mean Ranks Bryon Bay 5 61.5 12.3 Intermediate Calculations Cairns 5 49.5 9.9 Sum of Squared Ranks/Sample Size 4393 Gold Coast 5 72.5 14.5 Sum of Sample Sizes 25 Noosa 5 86 17.2 Number of Groups 5 Port Douglas 5 55.5 11.1 Test Result H Test Statistic 3.1015 Critical Value 7.7794 p-Value 0.5410 Do not reject the null hypothesis 19.59 Wilcoxon signed ranks test H0: MD = 0

where D = Primary – Secondary

H1: MD > 0 Assume the distribution of differences, D, is approximately symmetric. Decision rule: n’ = 16 and  = 0.05 from Table E.9 WU = 101 . Reject H0 if W > 101

Material Primary Secondary Supplier Supplier 12 $21.00 $20.98 16 $37.05 $37.00 9 $1.99 $2.20 6 $95.00 $94.49 2 $48.00 $47.00 7 $34.00 $35.00 14 $19.55 $21.00 3 $31.00 $32.99 5 $92.00 $94.00 8 $235.00 $233.00 15 $26.00 $24.00 13 $15.00 $12.95 10 $102.05 $99.99 11 $114.60 $112.50 4 $83.00 $77.00 1 $55.10 $45.00

Test Statistic: W =

D 0.02 0.05 -0.21 0.51 1.00 -1.00 -1.45 -1.99 -2.00 2.00 2.00 2.05 2.06 2.10 6.00 10.10

|D| 0.02 0.05 0.21 0.51 1.00 1.00 1.45 1.99 2.00 2.00 2.00 2.05 2.06 2.10 6.00 10.10

R 1 2 3 4 5.5 5.5 7 8 10 10 10 12 13 14 15 16 W

R+ 1.0 2.0 4.0 5.5

10.0 10.0 12.0 13.0 14.0 15.0 16.0 102.5

R ( ) = 102.5 i=1

Conclusion: Since W =102.5 > 101 reject H0 . At the 5% level of significance can conclude that primary supplier has higher median prices. Alternatively, test statistic W = 102.5, n’ = 16

n'(n'+ 1) 4 Z= n'(n'+ 1)(2n'+ 1) 24 16  17 102.5 − 4 = 16  17  33 24 102.5 − 68 = 374 = 1.7839... W−

Decision rule:  = 0.05, if Z > 1.6449 reject H0. Conclusion:

Since Z = 1.78 > 1.6649 reject H0 . At the 5% level of significance there sufficient evidence that primary supplier has higher median prices.

19.60 Let Test

X = ranks airline X

O = ranks other airlines

H0 : MX  MO

H1 : MX  MO Level of significance:  = 0.05 Data ranked and independent so need to use a Wilcoxon rank sum test with test statistic

TX sum of the ranks for airline X as nX = 12 , nO = 18 > 10 can use the normal approximation

nX (n + 1) 2 nXnO (n + 1) 12

TX −

Critical value and decision rule: z 0.05 = 1.6449 Reject H if the calculated value of the test statistic z  1.645 otherwise do not 0 reject H 0

Calculations:

Airline X Rank X 1 2 3 4 4 4 4 5 5 6 6 7

2.5 6.0 10.5 16.0 16.0 16.0 16.0 20.5 20.5 25.0 25.0 29.0

203.0 TX = 203

Conclusion:

Other airlines Rank O 1 2.5 1 2.5 1 2.5 2 6.0 2 6.0 3 10.5 3 10.5 3 10.5 3 10.5 3 10.5 4 16.0 5 20.5 5 20.5 6 25.0 6 25.0 6 25.0 7 29.0 7 29.0 262.0

12  31 2 = 203 − 186 = 0.719... Z= 12 18  31 558 12 203 −

As 0.719 < 1.6449 do not reject H 0

Therefore, we conclude, at a 5% level of significance, that there is no evidence that Airline X is more highly rated than other airlines.

Wilcoxon Rank Sum Test Data Level of Significance 0.05 Population 1 Sample Sample Size 12 Sum of Ranks 203 Population 2 Sample Sample Size 18 Sum of Ranks 262 Intermediate Calculations Total Sample Size n 30 T 1 Test Statistic 203 T 1 Mean 186 Standard Error of T 1 23.6220 Z Test Statistic 0.7196674 Upper-Tail Test Upper Critical Value 1.6449 p -Value 0.2359 Do not reject the null hypothesis 19.61 Wilcoxon Rank Sum Test Let

B = ranks before campaign D = ranks during campaign

Assume populations have similar variability and shape Test

H0 : MB  MD H1 : MB  MD

 = 0.05 nB = nD = 10 from Table E.8 lower critical value TB  82

82, reject H 0 if

Calculations

Before campaign 8.9 9.0 9.1 9.3 9.6 12.0 12.3 12.3 17.3 18.8

Rank 1.0 2.0 3.0 5.0 6.5 10.5 12.5 12.5 16.0 18.0

TB =

87.0

During campaign 9.2 9.6 11.0 11.7 12.0 13.9 14.3 18.1 22.9 22.9

Rank 4.0 6.5 8.0 9.0 10.5 14.0 15.0 17.0 19.5 19.5 123.0

Conclusion:

As 87 > 82 do not reject H 0

Therefore, we conclude, at a 5% level of significance, that there is no evidence that sales increased during the campaign. Alternatively

10  21 87 − 105 2 Z= = = −1.3606... 10 10  21 175 12 87 −

Conclusion:

Since -1.36 > -1.6449 do not reject H 0

Therefore, we conclude, at a 5% level of significance, that there is no evidence that sales increased during the campaign.

Wilcoxon Rank Sum Test Data Level of Significance 0.05 Population 1 Sample Sample Size 10 Sum of Ranks 87 Population 2 Sample Sample Size 10 Sum of Ranks 123 Intermediate Calculations Total Sample Size n 20 T 1 Test Statistic 87 T 1 Mean 105 Standard Error of T 1 13.2288 Z Test Statistic -1.3606721 Lower-Tail Test Lower Critical Value -1.6449 p -Value 0.0868 Do not reject the null hypothesis Case Study - Tasman University Student Survey (a)(i) & (c)

Bachelor of Business

Let M = Male students and F = Female students Assume M and F have similar variability and shape. Testing for difference of medians for two independent populations so use Wilcoxon sum rank test For each variable test

H0 : MM = MF

H1 :MM  MF Use level of significance  = 0.05 Decision rule: reject null hypothesis if Z < -1.96 or Z > 1.96 Age

Wilcoxon Rank Sum Test Age by Gender Data Level of Significance 0.05 Female Sample Sample Size 33 Sum of Ranks 1009 Male Sample Sample Size 29 Sum of Ranks 944 Intermediate Calculations Total Sample Size n 62 T 1 Test Statistic 944 T 1 Mean 913.5 Standard Error of T 1 70.8819 Z Test Statistic 0.430292938 Two-Tail Test Lower Critical Value -1.9600 Upper Critical Value 1.9600 p -Value 0.6670 Do not reject the null hypothesis There is no evidence of any difference between males and females in median age. WAM

Wilcoxon Rank Sum Test WAM by Gender Data Level of Significance 0.05 Female Sample Size 33 Sum of Ranks 1070 Male Sample Size 29 Sum of Ranks 883 Z Test Statistic -0.430292938 Two-Tail Test Lower Critical Value -1.9600 Upper Critical Value 1.9600 p -Value 0.6670 Do not reject the null hypothesis Copyright © 2019 Pearson Australia (a division of Pearson Australia Group Pty Ltd) 9781488617249, Berenson, Basic Business Statistics 5e

There is no evidence of any difference between males and females in median WAM. Expected Salary

Wilcoxon Rank Sum Test - Expected Salary Data Level of Significance 0.05 Female Sample Size 33 Sum of Ranks 1049.5 Male Sample Size 29 Sum of Ranks 903.5 Z Test Statistic -0.141079652 Two-Tail Test Lower Critical Value -1.9600 Upper Critical Value 1.9600 p -Value 0.8878 Do not reject the null hypothesis There is no evidence of any difference between males and females in median expected starting salary. Textbook Cost

Wilcoxon Rank Sum Test Textbook Cost Data Level of Significance 0.05 Female Sample Size 33 Sum of Ranks 966 Male Sample Size 29 Sum of Ranks 987 Z Test Statistic 1.036935441 Two-Tail Test Lower Critical Value -1.9600 Upper Critical Value 1.9600 p -Value 0.2998 Do not reject the null hypothesis There is no evidence of any difference between males and females in median textbook cost.

Social Networking

Wilcoxon Rank Sum Test Social Networking Data Level of Significance 0.05 Female Sample Size 33 Sum of Ranks 998 Male Sample Size 29 Sum of Ranks 955 Z Test Statistic 0.585480555 Two-Tail Test Lower Critical Value -1.9600 Upper Critical Value 1.9600 p -Value 0.5582 Do not reject the null hypothesis There is no evidence of any difference between males and females in median number of social networking sites registered for. Number of Text Messages

Wilcoxon Rank Sum Test Text Messages Data Level of Significance 0.05 Female Sample Size 33 Sum of Ranks 1015 Male Sample Size 29 Sum of Ranks 938 Z Test Statistic 0.345645147 Two-Tail Test Lower Critical Value -1.9600 Upper Critical Value 1.9600 p -Value 0.7296 Do not reject the null hypothesis There is no evidence of any difference between males and females in median number of text messages sent in a typical week.

Wealth

Wilcoxon Rank Sum Test Wealth Data Level of Significance 0.05 Female Sample Size 33 Sum of Ranks 795.5 Male Sample Size 29 Sum of Ranks 1157.5 Intermediate Calculations Total Sample Size n 62 T 1 Test Statistic 1157.5 T 1 Mean 913.5 Standard Error of T 1 70.8819 Z Test Statistic 3.4423435 Two-Tail Test Lower Critical Value -1.9600 Upper Critical Value 1.9600 p -Value 0.0006 Reject the null hypothesis There is no evidence of any difference between males and females in median wealth required to feel rich. (a)(ii) & (c)

Bachelor of Business

Let 1 = Other, 2 = Management, 3 = Information Systems, 4 = Economics/Finance, 5 = Undecided, 6 = International Business, 7 = Retail/Marketing, 8 = Accounting Assume that the eight populations have similar variability and shape. Testing for difference of medians for eight independent populations so use KruskalWallis rank test For each variable test

H0 : M1 = M2 = M3 = M4 = M5 = M6 = M7 = M8 H1 : At least one median not equal Use level of significance  = 0.05 Decision rule: reject null hypothesis if H   0.05,7 = 14.067 2

Age

Kruskal-Wallis Rank Test for Differences in Medians: Age Data Level of Significance 0.05 Group Sample Size Sum of Ranks Mean Ranks Other 7 225.5 32.2142857 Intermediate Calculations Management 10 360.5 36.05 Sum of Squared Ranks/Sample Size 62954.29 Information Systems 4 62.5 15.625 Sum of Sample Sizes 62 Economics/Finance 11 377 34.2727273 Number of Groups 8 Undecided 3 83.5 27.8333333 International Business 6 204 34 Test Result Retailing/Marketing 14 414.5 29.6071429 H Test Statistic 4.4080 Accounting 7 225.5 32.2142857 Critical Value 14.0671 p-Value 0.7318 Do not reject the null hypothesis There is no evidence of any difference based on major in median age. WAM

Kruskal-Wallis Rank Test for Differences in Medians: WAM Data Level of Significance 0.05 Group Sample Size Sum of Ranks Mean Ranks Other 7 124.5 17.7857143 Intermediate Calculations Management 10 385.5 38.55 Sum of Squared Ranks/Sample Size 65214.01 Information Systems 4 149 37.25 Sum of Sample Sizes 62 Economics/Finance 11 303.5 27.5909091 Number of Groups 8 Undecided 3 45.5 15.1666667 International Business 6 179.5 29.9166667 Test Result Retailing/Marketing 14 544.5 38.8928571 H Test Statistic 11.3503 Accounting 7 221 31.5714286 Critical Value 14.0671 p-Value 0.1241 Do not reject the null hypothesis There is no evidence of any difference based on major in median WAM.

Expected Salary

Kruskal-Wallis Rank Test for Differences in Medians: Expected Salary Data Level of Significance 0.05 Group Sample Size Sum of Ranks Mean Ra Other 7 238 Intermediate Calculations Management 10 291.5 2 Sum of Squared Ranks/Sample Size 64147.5 Information Systems 4 124 Sum of Sample Sizes 62 Economics/Finance 11 444 40.3636 Number of Groups 8 Undecided 3 96.5 32.1666 International Business 6 111.5 18.5833 Test Result Retailing/Marketing 14 376.5 26.8928 H Test Statistic 8.0738 Accounting 7 271 38.7142 Critical Value 14.0671 p-Value 0.3261 Do not reject the null hypothesis There is no evidence of any difference based on major in median expected starting salary. Textbook Cost

Kruskal-Wallis Rank Test for Differences in Medians: Textbook Cost Data Level of Significance 0.05 Group Sample Size Sum of Ranks Mean Ra Other 7 163.5 23.3571 Intermediate Calculations Management 10 343.5 3 Sum of Squared Ranks/Sample Size 62413.2 Information Systems 4 131 3 Sum of Sample Sizes 62 Economics/Finance 11 347.5 31.5909 Number of Groups 8 Undecided 3 97 32.3333 International Business 6 225 Test Result Retailing/Marketing 14 407.5 29.1071 H Test Statistic 2.7456 Accounting 7 238 Critical Value 14.0671 p-Value 0.9075 Do not reject the null hypothesis There is no evidence of any difference based on major in median textbook cost.

Social Networking

Kruskal-Wallis Rank Test for Differences in Medians: Social Networking Data Level of Significance 0.05 Group Sample Size Sum of Ranks Mean Ranks Other 7 230 32.8571429 Intermediate Calculations Management 10 295.5 29.55 Sum of Squared Ranks/Sample Size 62584.8 Information Systems 4 147.5 36.875 Sum of Sample Sizes 62 Economics/Finance 11 388 35.2727273 Number of Groups 8 Undecided 3 112 37.3333333 International Business 6 210 35 Test Result Retailing/Marketing 14 352 25.1428571 H Test Statistic 3.2728 Accounting 7 218 31.1428571 Critical Value 14.0671 p-Value 0.8587 Do not reject the null hypothesis There is no evidence of any difference based on major in median number of social networking sites registered for. Number of Text Messages

Kruskal-Wallis Rank Test for Differences in Medians: Text Messages Data Level of Significance 0.05 Group Sample Size Sum of Ranks Mean Ranks Other 7 240.5 34.3571429 Intermediate Calculations Management 10 295 29.5 Sum of Squared Ranks/Sample Size 63293.3 Information Systems 4 156.5 39.125 Sum of Sample Sizes 62 Economics/Finance 11 286.5 26.0454545 Number of Groups 8 Undecided 3 102.5 34.1666667 International Business 6 126.5 21.0833333 Test Result Retailing/Marketing 14 474.5 33.8928571 H Test Statistic 5.4496 Accounting 7 271 38.7142857 Critical Value 14.0671 p-Value 0.6053 Do not reject the null hypothesis There is no evidence of any difference based on major in median number of text messages sent in a typical week.

Wealth

Kruskal-Wallis Rank Test for Differences in Medians: Wealth Data Level of Significance 0.05 Group Sample Size Sum of Ranks Mean Ran Other 7 240.5 34.35714 Intermediate Calculations Management 10 368.5 36 Sum of Squared Ranks/Sample Size 62845.5 Information Systems 4 99 24 Sum of Sample Sizes 62 Economics/Finance 11 376 34.18181 Number of Groups 8 Undecided 3 124.5 4 International Business 6 161 26.83333 Test Result Retailing/Marketing 14 391 27.92857 H Test Statistic 4.0736 Accounting 7 192.5 2 Critical Value 14.0671 p-Value 0.7713 Do not reject the null hypothesis There is no evidence of any difference based on major in median wealth required to feel rich. (b)(i) & (c)

Master of Business Administration

H0 : MF = MM H1 :MF  MM Use level of significance  = 0.10 Decision rule: reject null hypothesis if Z < -1.6449 or Z > 1.6449

Age Wilcoxon Rank Sum Test: Age Data Level of Significance 0.1 Male Sample Size 25 Sum of Ranks 585.5 Female Sample Size 19 Sum of Ranks 404.5 Z Test Statistic -0.544961 Two-Tail Test Lower Critical Value -1.6449 Upper Critical Value 1.6449 p -Value 0.5858 Do not reject the null hypothesis There is no evidence of any difference between males and females in median age. MBA WAM

Wilcoxon Rank Sum Test: MBA WAM Data Level of Significance 0.1 Male Sample Size 25 Sum of Ranks 536 Female Sample Size 19 Sum of Ranks 454 Z Test Statistic 0.6278898 Two-Tail Test Lower Critical Value -1.6449 Upper Critical Value 1.6449 p -Value 0.5301 Do not reject the null hypothesis There is no evidence of any difference between males and females in median MBA WAM.

Undergraduate WAM

Wilcoxon Rank Sum Test: Undergraduate WAM Data Level of Significance 0.1 Male Sample Size 25 Sum of Ranks 490 Female Sample Size 19 Sum of Ranks 500 Z Test Statistic 1.717811746 Two-Tail Test Lower Critical Value -1.6449 Upper Critical Value 1.6449 p -Value 0.0858 Reject the null hypothesis At the 10% level of significance can conclude that there is a difference between males and females in median Undergraduate WAM. Expected Salary

Wilcoxon Rank Sum Test: Expected Salary Data Level of Significance 0.1 Male Sample Size 25 Sum of Ranks 580.5 Female Sample Size 19 Sum of Ranks 409.5 Z Test Statistic -0.426491192 Two-Tail Test Lower Critical Value -1.6449 Upper Critical Value 1.6449 p -Value 0.6697 Do not reject the null hypothesis There is no evidence of any difference between males and females in median expected starting salary.

Textbook Cost

Wilcoxon Rank Sum Test: Textbook Cost Data Level of Significance 0.1 Male Sample Size 25 Sum of Ranks 526 Female Sample Size 19 Sum of Ranks 464 Z Test Statistic 0.864829362 Two-Tail Test Lower Critical Value -1.6449 Upper Critical Value 1.6449 p -Value 0.3871 Do not reject the null hypothesis There is no evidence of any difference between males and females in median textbook cost. Number of Text Messages

Wilcoxon Rank Sum Test: Text Messages Data Level of Significance 0.1 Male Sample Size 25 Sum of Ranks 531.5 Female Sample Size 19 Sum of Ranks 458.5 Z Test Statistic 0.734512608 Two-Tail Test Lower Critical Value -1.6449 Upper Critical Value 1.6449 p -Value 0.4626 Do not reject the null hypothesis There is no evidence of any difference between males and females in median number of text messages sent in a typical week.

Wealth

Wilcoxon Rank Sum Test: Wealth Data Level of Significance 0.1 Male Sample Size 25 Sum of Ranks 564 Female Sample Size 19 Sum of Ranks 426 Z Test Statistic -0.0355409 Two-Tail Test Lower Critical Value -1.6449 Upper Critical Value 1.6449 p -Value 0.9716 Do not reject the null hypothesis There is no evidence of any difference between males and females in median wealth required to feel rich. (b)(ii) & (c)

Master of Business Administration

Let 1 = Other, 2 = Biological Sciences, 3 = Business, 4 = Engineering Assume that the four populations have similar variability and shape. Testing for difference of medians for four independent populations so use KruskalWallis rank test For each variable test

H0 : M1 = M2 = M3 = M4

H1 : At least one median not equal Use level of significance  = 0.10 Decision rule: reject null hypothesis if H   0.1,3 = 6.251 2

Age

Kruskal-Wallis Rank Test for Differences in Medians: Age Data Level of Significance 0.1 Group Sample Size Sum of Ranks Mean Ranks Other 13 299 23 Intermediate Calculations Biological Sciences 3 62 20.6666667 Sum of Squared Ranks/Sample Size 22604.66667 Business 25 532.5 21.3 Sum of Sample Sizes 44 Engineering 3 96.5 32.1666667 Number of Groups 4 Test Result H Test Statistic 1.9980 Critical Value 6.2514 p-Value 0.5728 Do not reject the null hypothesis There is no evidence of any difference based on undergraduate major in median age. Undergraduate WAM

Kruskal-Wallis Rank Test for Differences in Medians: Undergraduate WAM Data Level of Significance 0.1 Group Sample Size Sum of Ranks Mean Ranks Other 13 289.5 22.2692308 Intermediate Calculations Biological Sciences 3 66 22 Sum of Squared Ranks/Sample Size 22441.53 Business 25 587.5 23.5 Sum of Sample Sizes 44 Engineering 3 47 15.6666667 Number of Groups 4 Test Result H Test Statistic 1.0092 Critical Value 6.2514 p-Value 0.7990 Do not reject the null hypothesis There is no evidence of any difference based on undergraduate major in median undergraduate WAM.

MBA WAM

Kruskal-Wallis Rank Test for Differences in Medians: MBA WAM Data Level of Significance 0.1 Group Sample Size Sum of Ranks Mean Rank Other 13 304.5 23.423076 Intermediate Calculations Biological Sciences 3 44 14.666666 Sum of Squared Ranks/Sample Size 22476.05 Business 25 571 22.8 Sum of Sample Sizes 44 Engineering 3 70.5 23. Number of Groups 4 Test Result H Test Statistic 1.2185 Critical Value 6.2514 p-Value 0.7486 Do not reject the null hypothesis There is no evidence of any difference based on undergraduate major in median MBA WAM. Expected Salary

Kruskal-Wallis Rank Test for Differences in Medians: Expected Salary Data Level of Significance 0.1 Group Sample Size Sum of Ranks Mean Ranks Other 13 328 25.2307692 Intermediate Calculations Biological Sciences 3 61.5 20.5 Sum of Squared Ranks/Sample Size 22495.27 Business 25 521.5 20.86 Sum of Sample Sizes 44 Engineering 3 79 26.3333333 Number of Groups 4 Test Result H Test Statistic 1.3349 Critical Value 6.2514 p-Value 0.7209 Do not reject the null hypothesis There is no evidence of any difference based on undergraduate major in median expected salary upon graduation.

Textbook Cost

Kruskal-Wallis Rank Test for Differences in Medians: Textbook Cost Data Level of Significance 0.1 Group Sample Size Sum of Ranks Mean Ranks Other 13 285 21.9230769 Intermediate Calculations Biological Sciences 3 64 21.3333333 Sum of Squared Ranks/Sample Size 22305.1 Business 25 565.5 22.62 Sum of Sample Sizes 44 Engineering 3 75.5 25.1666667 Number of Groups 4 Test Result H Test Statistic 0.1824 Critical Value 6.2514 p-Value 0.9804 Do not reject the null hypothesis There is no evidence of any difference based on undergraduate major in median textbook cost. Number of Text Messages

Kruskal-Wallis Rank Test for Differences in Medians: Number of Text Messages Data Level of Significance 0.1 Group Sample Size Sum of Ranks Mean Ranks Other 13 289 22.2307692 Intermediate Calculations Biological Sciences 3 80.5 26.8333333 Sum of Squared Ranks/Sample Size 22335.6 Business 25 553.5 22.14 Sum of Sample Sizes 44 Engineering 3 67 22.3333333 Number of Groups 4 Test Result H Test Statistic 0.3673 Critical Value 6.2514 p-Value 0.9469 Do not reject the null hypothesis There is no evidence of any difference based on undergraduate major in median number of text messages sent in a typical week.

Wealth

Kruskal-Wallis Rank Test for Differences in Medians: Wealth Data Level of Significance 0.1 Group Sample Size Sum of Ranks Mean Rank Other 13 286 2 Intermediate Calculations Biological Sciences 3 89 29.666666 Sum of Squared Ranks/Sample Size 22455.63 Business 25 555.5 22.2 Sum of Sample Sizes 44 Engineering 3 59.5 19.833333 Number of Groups 4 Test Result H Test Statistic 1.0947 Critical Value 6.2514 p-Value 0.7784 Do not reject the null hypothesis There is no evidence of any difference based on undergraduate major in median wealth required to feel rich. (b)(iii) & (c) Master of Business Administration Let 1 = Economics/Finance, 2 = Other, 3 = Management, Retail/Marketing, 5 = Information Systems, 6 = International Business.

Assume that the seven populations have the similar variability and shape. Testing for difference of medians for seven independent populations so use KruskalWallis rank test. For each variable test

H0 : M1 = M2 = M3 = M4 = M5 = M6 = M7 H1 : At least one median not equal Use level of significance  = 0.10 Decision rule: reject null hypothesis if H   0.1,6 = 10.645 2

Age

Kruskal-Wallis Rank Test for Differences in Medians: Age Data Level of Significance 0.1 Group Sample Size Sum of Ranks Mean Ranks Economics/Finance 14 265.5 18.9642857 Intermediate Calculations Other 5 127 25.4 Sum of Squared Ranks/Sample Size 22683.69 Management 9 202.5 22.5 Sum of Sample Sizes 44 Retailing/Marketing 7 191 27.2857143 Number of Groups 7 Accounting 5 117 23.4 Information Systems 2 38.5 19.25 Test Result International Business 2 48.5 24.25 H Test Statistic 2.4769 Critical Value 10.6446 p-Value 0.8710 Do not reject the null hypothesis There is no evidence of any difference based on MBA major in median age. Undergraduate WAM

Kruskal-Wallis Rank Test for Differences in Medians: Undergraduate WAM Data Level of Significance 0.1 Group Sample Size Sum of Ranks Mean Ranks Economics/Finance 14 344.5 24.6071429 Intermediate Calculations Other 5 67 13.4 Sum of Squared Ranks/Sample Size 23447.74 Management 9 213.5 23.7222222 Sum of Sample Sizes 44 Retailing/Marketing 7 151 21.5714286 Number of Groups 7 Accounting 5 128 25.6 Information Systems 2 68 34 Test Result International Business 2 18 9 H Test Statistic 7.1075 Critical Value 10.6446 p-Value 0.3110 Do not reject the null hypothesis There is no evidence of any difference based on MBA major in median undergraduate WAM.

MBA WAM

Kruskal-Wallis Rank Test for Differences in Medians: MBA WAM Data Level of Significance 0.1 Group Sample Size Sum of Ranks Mean R Economics/Finance 14 327.5 23.392 Intermediate Calculations Other 5 81.5 Sum of Squared Ranks/Sample Size 23591.47 Management 9 197.5 21.944 Sum of Sample Sizes 44 Retailing/Marketing 7 165 23.571 Number of Groups 7 Accounting 5 145.5 Information Systems 2 65 Test Result International Business 2 8 H Test Statistic 7.9786 Critical Value 10.6446 p-Value 0.2397 Do not reject the null hypothesis There is no evidence of any difference based on MBA major in median MBA WAM. Expected Salary

Kruskal-Wallis Rank Test for Differences in Medians: Expected Salary Data Level of Significance 0.1 Group Sample Size Sum of Ranks Mean R Economics/Finance 14 227 16.214 Intermediate Calculations Other 5 174.5 Sum of Squared Ranks/Sample Size 23667.68 Management 9 210.5 23.388 Sum of Sample Sizes 44 Retailing/Marketing 7 166 23.714 Number of Groups 7 Accounting 5 108 Information Systems 2 51.5 2 Test Result International Business 2 52.5 2 H Test Statistic 8.4405 Critical Value 10.6446 p-Value 0.2076 Do not reject the null hypothesis There is no evidence of any difference based on MBA major in median expected salary upon graduation.

Textbook Cost

Kruskal-Wallis Rank Test for Differences in Medians: Textbook Cost Data Level of Significance 0.1 Group Sample Size Sum of Ranks Mean Ranks Economics/Finance 14 309 22.0714286 Intermediate Calculations Other 5 85.5 17.1 Sum of Squared Ranks/Sample Size 23747.47 Management 9 194.5 21.6111111 Sum of Sample Sizes 44 Retailing/Marketing 7 115 16.4285714 Number of Groups 7 Accounting 5 149 29.8 Information Systems 2 53 26.5 Test Result International Business 2 84 42 H Test Statistic 8.9241 Critical Value 10.6446 p-Value 0.1779 Do not reject the null hypothesis There is no evidence of any difference based on MBA major in median textbook cost. Number of Text Messages

Kruskal-Wallis Rank Test for Differences in Medians: Number of Text Messages Data Level of Significance 0.1 Group Sample Size Sum of Ranks Mean Ranks Economics/Finance 14 344.5 24.6071429 Intermediate Calculations Other 5 175.5 35.1 Sum of Squared Ranks/Sample Size 23980.93 Management 9 146.5 16.2777778 Sum of Sample Sizes 44 Retailing/Marketing 7 173.5 24.7857143 Number of Groups 7 Accounting 5 91 18.2 Information Systems 2 18 9 Test Result International Business 2 41 20.5 H Test Statistic 10.3389 Critical Value 10.6446 p-Value 0.1111 Do not reject the null hypothesis There is no evidence of any difference based on MBA major in median number of text messages sent in a typical week.

Wealth

Kruskal-Wallis Rank Test for Differences in Medians: Wealth Data Level of Significance 0.1 Group Sample Size Sum of Ranks Mean R Economics/Finance 14 314 22.428 Intermediate Calculations Other 5 128 Sum of Squared Ranks/Sample Size 22372.3 Management 9 197 21.888 Sum of Sample Sizes 44 Retailing/Marketing 7 155 22.142 Number of Groups 7 Accounting 5 99.5 Information Systems 2 47.5 Test Result International Business 2 49 H Test Statistic 0.5897 Critical Value 10.6446 p-Value 0.9966 Do not reject the null hypothesis There is no evidence of any difference based on MBA major in median wealth required to feel rich. (b)(iv) & (c)

Master of Business Administration

Let 1 = Employed Fulltime, 2 = Not Employed, 3 = Employed Part-time Assume that the three populations have the similar variability and shape. Testing for difference of medians for three independent populations so use KruskalWallis rank test. For each variable test

H0 : M1 = M2 = M3 H1 : At least one median not equal Use level of significance  = 0.10 Decision rule: reject null hypothesis if H   0.1,2 = 4.605 2

Age

Kruskal-Wallis Rank Test for Differences in Medians: Age Data Level of Significance 0.1 Group Sample Size Sum of Ranks Mean Ranks Full-Time 28 712.5 25.4464286 Intermediate Calculations Not Employed 9 207 23 Sum of Squared Ranks/Sample Size 23601.62 Part-Time 7 70.5 10.0714286 Sum of Sample Sizes 44 Number of Groups 3 Test Result H Test Statistic 8.0401 Critical Value 4.6052 p-Value 0.0180 Reject the null hypothesis At the 10% level of significance there is evidence of a difference based on employment status in median age. Undergraduate WAM

Kruskal-Wallis Rank Test for Differences in Medians: Undergraduate WAM Data Level of Significance 0.1 Group Sample Size Sum of Ranks Mean Ranks Full-Time 28 542.5 19.375 Intermediate Calculations Not Employed 9 259.5 28.8333333 Sum of Squared Ranks/Sample Size 23042.33 Part-Time 7 188 26.8571429 Sum of Sample Sizes 44 Number of Groups 3 Test Result H Test Statistic 4.6505 Critical Value 4.6052 p-Value 0.0978 Reject the null hypothesis At the 10% level of significance there is evidence of a difference based on employment status in median undergraduate WAM.

MBA WAM

Kruskal-Wallis Rank Test for Differences in Medians: MBA WAM Data Level of Significance 0.1 Group Sample Size Sum of Ranks Mean Ranks Full-Time 28 609.5 21.7678571 Intermediate Calculations Not Employed 9 245 27.2222222 Sum of Squared Ranks/Sample Size 22559.85 Part-Time 7 135.5 19.3571429 Sum of Sample Sizes 44 Number of Groups 3 Test Result H Test Statistic 1.7263 Critical Value 4.6052 p-Value 0.4218 Do not reject the null hypothesis There is no evidence of any difference based on employment status in median MBA WAM. Expected Salary

Kruskal-Wallis Rank Test for Differences in Medians: Expected Salary Data Level of Significance 0.1 Group Sample Size Sum of Ranks Mean Ranks Full-Time 28 772 27.5714286 Intermediate Calculations Not Employed 9 145.5 16.1666667 Sum of Squared Ranks/Sample Size 24388.29 Part-Time 7 72.5 10.3571429 Sum of Sample Sizes 44 Number of Groups 3 Test Result H Test Statistic 12.8078 Critical Value 4.6052 p-Value 0.0017 Reject the null hypothesis There is evidence of a difference based on employment status in median expected salary upon graduation.

Textbook Cost

Kruskal-Wallis Rank Test for Differences in Medians: Textbook Cost Data Level of Significance 0.1 Group Sample Size Sum of Ranks Mean Ranks Full-Time 28 609 21.75 Intermediate Calculations Not Employed 9 234.5 26.0555556 Sum of Squared Ranks/Sample Size 22421.81 Part-Time 7 146.5 20.9285714 Sum of Sample Sizes 44 Number of Groups 3 Test Result H Test Statistic 0.8898 Critical Value 4.6052 p-Value 0.6409 Do not reject the null hypothesis There is no evidence of any difference based on employment status in median textbook cost. Number of Text Messages

Kruskal-Wallis Rank Test for Differences in Medians: Text Messages Data Level of Significance 0.1 Group Sample Size Sum of Ranks Mean Ranks Full-Time 28 648.5 23.1607143 Intermediate Calculations Not Employed 9 220.5 24.5 Sum of Squared Ranks/Sample Size 22513.54 Part-Time 7 121 17.2857143 Sum of Sample Sizes 44 Number of Groups 3 Test Result H Test Statistic 1.4457 Critical Value 4.6052 p-Value 0.4854 Do not reject the null hypothesis There is no evidence of any difference based on employment status in median number of text messages sent in a typical week.

Wealth

Kruskal-Wallis Rank Test for Differences in Medians: Wealth Data Level of Significance 0.1 Group Sample Size Sum of Ranks Mean Ranks Full-Time 28 679 24.25 Intermediate Calculations Not Employed 9 155 17.2222222 Sum of Squared Ranks/Sample Size 22611.77 Part-Time 7 156 22.2857143 Sum of Sample Sizes 44 Number of Groups 3 Test Result H Test Statistic 2.0410 Critical Value 4.6052 p-Value 0.3604 Do not reject the null hypothesis There is no evidence of any difference based on MBA major in median wealth required to feel rich. (b)(iv) & (c) Undergraduate and MBA WAM relate to the same student so paired data. Let D = Undergraduate WAM – MBA WAM Assume the distribution of differences, D, is approximately symmetric. Wilcoxon signed ranks test Test:

H0: MD ≤ 0 H1: MD > 0

n'(n'+ 1) 4 Test statistic Z = n'(n'+ 1)(2n'+ 1) 24 W−

Decision rule:  = 0.10, if Z > 1.2816 reject H0

Calculation

Undergraduate MBA WAM WAM 70.1 70.6 83.6 84.5 68.4 69.5 72.7 71.5 89.3 67.1 92.4 70.2 92.6 67.7

D -0.5 -0.9 -1.1 1.2 22.2 22.2 24.9

|D| 0.5 0.9 1.1 1.2 22.2 22.2 24.9

R+

1.0 2.0 3.0 4.0 42.5 42.5 44.0 W=

4.0 42.5 42.5 44.0 885.5

n’ = 44 and W = 885.5

44  45 4 Z= 44  45  89 24 888.5 − 495.0 = 7342.5 = 4.592... 888.5 −

Conclusion: Since Z = 4.592 > 1.2816 reject H0 . Can conclude that there evidence the median undergraduate WAM is higher than median MBA WAM Case Study - Safe-As Houses Real Estate (a) Regional City 1 State A (a)(i) & (d) Let H = House price and U = Unit price Assume H and U have similar variability and shape. Testing for difference of medians for two independent populations so use Wilcoxon sum rank test Test

H0 : MU  MH

H1 : MU  MH Use level of significance  = 0.05 Decision rule: reject null hypothesis if Z < −1.6449

Wilcoxon Rank Sum Test: Price Data Level of Significance 0.05 Unit Sample Size 41 Sum of Ranks 1093 House Sample Size 84 Sum of Ranks 6782 Intermediate Calculations Total Sample Size n 125 T 1 Test Statistic 1093 T 1 Mean 2583 Standard Error of T 1 190.1631 Z Test Statistic -7.8353797 Lower-Tail Test Lower Critical Value -1.6449 p -Value 0.0000 Reject the null hypothesis Since p-value approximately zero can conclude that in Regional City 1, State A, houses have higher median prices than units. (a)(ii) & (d) Let H = House internal area and U = Unit internal area Assume H and U have similar variability and shape. Testing for difference of medians for two independent populations so use Wilcoxon sum rank test Test

H0 : MU  MH H1 : MU  MH

Use level of significance  = 0.05 Decision rule: reject null hypothesis if Z < −1.6449

Wilcoxon Rank Sum Test: Internal Area Data Level of Significance 0.05 Unit Sample Size 41 Sum of Ranks 924.5 House Sample Size 84 Sum of Ranks 6950.5 Intermediate Calculations Total Sample Size n 125 T 1 Test Statistic 924.5 T 1 Mean 2583 Standard Error of T 1 190.1631 Z Test Statistic -8.72146124 Lower-Tail Test Lower Critical Value -1.6449 p -Value 0.0000 Reject the null hypothesis Since p-value approximately zero can conclude that in Regional City 1, State A, houses have higher internal areas than units. (a)(iii) & (d) Let I = Internal Area based on Number of Bedrooms = 1, 2, 3, 4, 5, 6 and 9 Assume that the seven populations have similar variability and shape. Testing for difference of medians for seven independent populations so use Kruskal-Wallis rank test Test

H0 : M1 = M2 = M3 = M4 = M5 = M6 = M7 H1 : At least one median not equal

Use level of significance  = 0.05 Decision rule: reject null hypothesis if H   0.05,6 = 12.592 2

Kruskal-Wallis Rank Test for Differences in Medians: Internal Area Data Level of Significance 0.05 Bedrooms Sample Size Sum of Ranks Mean Ranks 1 1 1 1 Intermediate Calculations 2 37 844 22.8108108 Sum of Squared Ranks/Sample Size 632162.1 3 49 3008 61.3877551 Sum of Sample Sizes 125 4 29 2938 101.310345 Number of Groups 7 5 7 835 119.285714 6 1 124 124 Test Result 9 1 125 125 H Test Statistic 103.6473 Critical Value 12.5916 p-Value 0.0000 Reject the null hypothesis Since p-value approximately zero can conclude that in Regional City 1, State A, there is a difference in median internal area based on the number of bedrooms. (a)(iv) & (d) Let P = Price based on Number of Bedrooms = 1, 2, 3, 4, 5, 6 and 9 Assume that the seven populations have similar variability and shape. Testing for difference of medians for seven independent populations so use Kruskal-Wallis rank test Test

H0 : M1 = M2 = M3 = M4 = M5 = M6 = M7

H1 : At least one median not equal Use level of significance

= 0.05

Decision rule: reject null hypothesis if H   0.05,6 = 12.592 2

Kruskal-Wallis Rank Test for Differences in Medians: Price Data Level of Significance 0.05 Bedrooms Sample Size Sum of Ranks Mean Ranks 1 1 5 5 Intermediate Calculations 2 37 1046.5 28.2837838 Sum of Squared Ranks/Sample Size 590960 3 49 3099.5 63.255102 Sum of Sample Sizes 125 4 29 2801.5 96.6034483 Number of Groups 7 5 7 714 102 6 1 98 98 Test Result 9 1 110.5 110.5 H Test Statistic 72.2552 Critical Value 12.5916 p-Value 0.0000 Reject the null hypothesis Since p-value approximately zero can conclude that in Regional City 1, State A, there is a difference in median price based on the number of bedrooms. Copyright © 2019 Pearson Australia (a division of Pearson Australia Group Pty Ltd) 9781488617249, Berenson, Basic Business Statistics 5e

(a)(v) & (d) Let I = Internal Area based on Number of Bathrooms = 1, 2 and 3 Assume that the populations have similar variability and shape. Testing for difference of medians for three independent populations so use Kruskal-Wallis rank test Test

H0 : M1 = M2 = M3 H1 : At least one median not equal

Use level of significance  = 0.05 Decision rule: reject null hypothesis if H   0.05,2 = 5.991 2

Kruskal-Wallis Rank Test for Differences in Medians: Internal Area Data Level of Significance 0.05 Bathrooms Sample Size Sum of Ranks Mean Ranks 1 82 3814 46.5121951 Intermediate Calculations 2 39 3610.5 92.5769231 Sum of Squared Ranks/Sample Size 562384.1 3 4 450.5 112.625 Sum of Sample Sizes 125 Number of Groups 3 Test Result H Test Statistic 50.4831 Critical Value 5.9915 p-Value 0.0000 Reject the null hypothesis Since p-value approximately zero can conclude that in Regional City 1, State A, there is a difference in median internal area based on the number of bathrooms. (a)(vi) & (d) Let P = Price based on Number of Bathrooms = 1, 2 and 3 Assume that the populations have similar variability and shape. Testing for difference of medians for three independent populations so use Kruskal-Wallis rank test Test

H0 : M1 = M2 = M3 H1 : At least one median not equal

Use level of significance  = 0.05 Decision rule: reject null hypothesis if H   0.05,2 = 5.991 2

Kruskal-Wallis Rank Test for Differences in Medians: Price Data Level of Significance 0.05 Bathrooms Sample Size Sum of Ranks Mean Rank 1 82 3767 45.939024 Intermediate Calculations 2 39 3653.5 93.679487 Sum of Squared Ranks/Sample Size 566952.9 3 4 454.5 113.62 Sum of Sample Sizes 125 Number of Groups 3 Test Result H Test Statistic 53.9641 Critical Value 5.9915 p-Value 0.0000 Reject the null hypothesis Since p-value approximately zero can conclude that in Regional City 1, State A, there is a difference in median price based on the number of bathrooms. (b) Coastal City 1 State A (b)(i) & (d) Let H = House price and U = Unit price Assume H and U have similar variability and shape. Testing for difference of medians for two independent populations so use Wilcoxon sum rank test Test

H0 : MU  MH H1 : MU  MH

Use level of significance  = 0.01 Decision rule: reject null hypothesis if Z < −2.3263

Wilcoxon Rank Sum Test: Price Data Level of Significance 0.01 Unit Sample Size 53 Sum of Ranks 2204 House Sample Size 72 Sum of Ranks 5671 Intermediate Calculations Total Sample Size n 125 T 1 Test Statistic 2204 T 1 Mean 3339 Standard Error of T 1 200.1699 Z Test Statistic -5.6701824 Lower-Tail Test Lower Critical Value -2.3263 p -Value 0.0000 Reject the null hypothesis Since p-value approximately zero can conclude that in Coastal City 1, State A, houses have higher median prices than units. (a)(ii) & (d) Let H = House internal area and U = Unit internal area Assume H and U have similar variability and shape. Testing for difference of medians for two independent populations so use Wilcoxon sum rank test Test:

H0 : MU  MH H1 : MU  MH

Use level of significance  = 0.01 Decision rule: reject null hypothesis if Z < −2.3263

Wilcoxon Rank Sum Test: Internal Area Data Level of Significance 0.01 Unit Sample Size 53 Sum of Ranks 1559 House Sample Size 72 Sum of Ranks 6316 Intermediate Calculations Total Sample Size n 125 T 1 Test Statistic 1559 T 1 Mean 3339 Standard Error of T 1 200.1699 Z Test Statistic -8.89244463 Lower-Tail Test Lower Critical Value -2.3263 p -Value 0.0000 Reject the null hypothesis Since p-value approximately zero can conclude that in Coastal City 1, State A, houses have higher internal areas than units. (a)(iii) & (d) Let I = Internal Area based on Number of Bedrooms = 1, 2, 3, 4, 5, 6 and 7 Assume that the seven populations have similar variability and shape. Testing for difference of medians for seven independent populations so use Kruskal-Wallis rank test Test

H0 : M1 = M2 = M3 = M4 = M5 = M6 = M7 H1 : At least one median not equal

Use level of significance  = 0.01 Decision rule: reject null hypothesis if H   0.01,6 = 16.812 2

Kruskal-Wallis Rank Test for Differences in Medians: Internal Area Data Level of Significance 0.01 Bedrooms Sample Size Sum of Ranks Mean Ranks 1 4 10 2.5 Intermediate Calculations 2 28 583 20.8214286 Sum of Squared Ranks/Sample Size 637797.7 3 43 2309 53.6976744 Sum of Sample Sizes 125 4 35 3212.5 91.7857143 Number of Groups 7 5 10 1151.5 115.15 6 4 484 121 Test Result 7 1 125 125 H Test Statistic 107.9411 Critical Value 16.8119 p-Value 0.0000 Reject the null hypothesis Since p-value approximately zero can conclude that in Coastal City 1, State A, there is a difference in median internal area based on the number of bedrooms. (a)(iv) & (d) Let P = Price based on Number of Bedrooms = 1, 2, 3, 4, 5, 6 and 7 Assume that the seven populations have similar variability and shape. Testing for difference of medians for seven independent populations so use Kruskal-Wallis rank test Test

H0 : M1 = M2 = M3 = M4 = M5 = M6 = M7

H1 : At least one median not equal Use level of significance  = 0.01 Decision rule: reject null hypothesis if H   0.01,6 = 16.812 2

Kruskal-Wallis Rank Test for Differences in Medians: Price Data Level of Significance 0.01 Bedrooms Sample Size Sum of Ranks Mean Ranks 1 4 137.5 34.375 Intermediate Calculations 2 28 651.5 23.2678571 Sum of Squared Ranks/Sample Size 599866 3 43 2323 54.0232558 Sum of Sample Sizes 125 4 35 3342.5 95.5 Number of Groups 7 5 10 911 91.1 6 4 390.5 97.625 Test Result 7 1 119 119 H Test Statistic 79.0408 Critical Value 16.8119 p-Value 0.0000 Reject the null hypothesis

Since p-value approximately zero can conclude that in Coastal City 1, State A, there is a difference in median price based on the number of bedrooms. (a)(v) & (d) Let I = Internal area based on Number of Bathrooms = 1, 2, 3 and 4 Assume that the four populations have similar variability and shape. Testing for difference of medians for four independent populations so use Kruskal-Wallis rank test Test

H0 : M1 = M2 = M3 = M4 H1 : At least one median not equal

Use level of significance

= 0.10

Decision rule: reject null hypothesis if H   0.01,3 = 11.345 2

Kruskal-Wallis Rank Test for Differences in Medians: Price Data Level of Significance 0.01 Bathrooms Sample Size Sum of Ranks Mean Ranks 1 53 1740 32.8301887 Intermediate Calculations 2 55 4323.5 78.6090909 Sum of Squared Ranks/Sample Size 591523.5 3 13 1317.5 101.346154 Sum of Sample Sizes 125 4 4 494 123.5 Number of Groups 4 Test Result H Test Statistic 72.6846 Critical Value 11.3449 p-Value 0.0000 Reject the null hypothesis Since p-value approximately zero can conclude that in Coastal City 1, State A, there is a difference in median internal area based on the number of bathrooms. (a)(vi) & (d) Let P = Price based on Number of Bathrooms = 1, 2, 3 and 4 Assume that the four populations have similar variability and shape. Testing for difference of medians for four independent populations so use Kruskal-Wallis rank test Test

H0 : M1 = M2 = M3 = M4 H1 : At least one median not equal

Use level of significance  = 0.10 Decision rule: reject null hypothesis if H   0.01,3 = 11.345 2

Kruskal-Wallis Rank Test for Differences in Medians: Price Data Level of Significance 0.01 Bathrooms Sample Size Sum of Ranks Mean Ranks 1 53 1738 32.7924528 Intermediate Calculations 2 55 4436.5 80.6636364 Sum of Squared Ranks/Sample Size 585490.6 3 13 1260 96.9230769 Sum of Sample Sizes 125 4 4 440.5 110.125 Number of Groups 4 Test Result H Test Statistic 68.0881 Critical Value 11.3449 p-Value 0.0000 Reject the null hypothesis Since p-value approximately zero can conclude that in Coastal City 1, State A, there is a difference in median price based on the number of bathrooms. (c) Regional City 1 and Coastal City 1 State A (c)(i) & (d) Let C = Price Coastal City 1 and R = Price Regional City 1 Assume C and R have similar variability and shape. Testing for difference of medians for two independent populations so use Wilcoxon sum rank test Test

H0 : Mc = MR H1 : MC  MR

Use level of significance  = 0.1 Decision rule: reject null hypothesis if Z < −1.6449 or Z > 1.6449

Wilcoxon Rank Sum Test: Price Data Level of Significance 0.1 Coastal City 1 Sample Size 125 Sum of Ranks 19111.5 Regional City 1 Sample Size 125 Sum of Ranks 12263.5 Intermediate Calculations Total Sample Size n 250 T 1 Test Statistic 19111.5 T 1 Mean 15687.5 Standard Error of T 1 571.6843 Z Test Statistic 5.9893198 Two-Tail Test Lower Critical Value -1.6449 Upper Critical Value 1.6449 p -Value 0.0000 Reject the null hypothesis Since p-value approximately zero can conclude that there is a difference in prices between Coastal City 1 and Regional City 1 in State A.

(c)(ii) & (d) Let C = Internal Area Coastal City 1 and R = Internal Area Regional City 1 Assume C and R have similar variability and shape. Testing for difference of medians for two independent populations so use Wilcoxon sum rank test Test

H0 : Mc = MR

H1 : MC  MR Use level of significance  = 0.1 Decision rule: reject null hypothesis if Z < −1.6449 or Z > 1.6449

Wilcoxon Rank Sum Test: Internal Area Data Level of Significance 0.1 Coastal City 1 Sample Size 125 Sum of Ranks 16786 Coastal City 2 Sample Size 125 Sum of Ranks 14589 Intermediate Calculations Total Sample Size n 250 T 1 Test Statistic 16786 T 1 Mean 15687.5 Standard Error of T 1 571.6843 Z Test Statistic 1.921515139 Two-Tail Test Lower Critical Value -1.6449 Upper Critical Value 1.6449 p -Value 0.0547 Reject the null hypothesis Since p-value = 0.0547 reject the null hypothesis. Therefore, at the 10% level of significance can conclude that there is a difference in internal area between Coastal City 1 and Regional City 1 in State A.

Chapter 20: Business analytics 20.1

(a)

JMP output:

2 (b) The r for the classification tree model is 0.573. The first split is for the 23 customers who experienced delivery time difference < 4.8. Among customers who had previously stayed at the hotel, those who experience delivery time difference < 4 are more likely to be satisfied. Copyright © 2019 Pearson Australia (a division of Pearson Australia Group Pty Ltd) 9781488617249, Berenson, Basic Business Statistics 5e

20.2

(a)

(b)

JMP output:

2 The r for the classification tree model is 0.434. The first split is for the 8 customers who called 50 or more times. Among customers who called fewer than 50 times, those who called at least seven times and visited two or more times are more likely to be churned.

20.3

(a)

Partial JMP output:

(b)

2 The r for the classification tree model is 0.260. Based on the JMP output, the highest probability of a fraudulent claim is from a filed stolen vehicle claim that is not a new policy and has a claims/year >=0.75 and < 0.81.

(c)-(d) Because half the data will be used for a validation sample, the results will differ depending on which values are in the training sample and which are in the validation sample. 20.4 Because half the data will be used for a validation sample, the results will differ depending on which values are in the training sample and which are in the validation sample.

20.5

(a)

(b)

Partial JMP output:

2 The r of the regression tree model is 0.799. The first split occurs at $45 thousands expenditure on radio advertisement. With a radio advertisement of less than $45 thousands, the mean sales is $944.5 thousands. With a radio advertisement of at least $45 thousands, the mean sales is $1459 thousands. The second split occurs at $35 thousands expenditure on newspaper advertisement. With a newspaper advertisement of less than $35 thousands, the mean sales is $1243 thousands while the mean sales is $1675 thousands with a newspaper advertisement of at least $35 thousands.

20.6

(a)

(b)

2 The r for the regression tree model is 0.373. The first split is based on a plate gap of 1.8. For those bags with a plate gap less than 1.8, the mean tear is 0.3107. For those bags with a plate gap at least 1.8, the mean tear is 1.98. For those bags with a plate gap less than 0.0, the mean tear is 0.06. For those bags with a plate gap less than 1.8 but greater than 0, the mean tear is 0.45. Thus, you would recommend that a plate gap of less than 0 be used to minimize tears in the bag.

20.7 Because half the data will be used for a validation sample, the results will differ depending on which values are in the training sample and which are in the validation sample.

20.8

(a)

(b)

JMP output:

2 The r for the regression tree model is 0.789. The first split is based on 831 square feet. Moves of at least 831 sq. ft. have a mean moving time of 51.1875 hours. Moves of less than 831 square feet have a mean moving time of 22.6071 hours. Among moves of less than 831 sq. ft., moves of less than 486 sq. ft., have a mean moving time of 15.7955 hours. Moves of less than 344 sq. ft. have a mean moving time of 12.75 hours. Moves of between 344 and 486 sq. ft. have a mean moving time of 18.3333 hours. Moves of between 486 and 830 sq. ft. have a mean moving time of 27.0147 hours. Moves between 486 and 599 sq. ft. have a mean moving time of 24.825 hours. Moves between 600 and 830 have a mean moving time of 30.1429 hours. Moves between 557 and 599 sq. ft. have a mean moving time of 24.05 hours. Moves between 486 and 557 sq. ft have a mean moving time of 25.6 hours.

20.9 – 20.16 Because some of the data will be used for a validation sample, the results will differ depending on which values are in the training sample and which are in the validation sample. 20.17(a)

(b)

JMP output:

The first two titles to cluster are Order of the Phoenix and Half-Blood Prince followed by Chamber of Secrets and Prisoner of Azkaban. At the two cluster level, one cluster contains Deathly Hallows Part II alone and the other cluster contains the other titles. The first cluster of Deathly Hallows Part II is unique in having the highest first weekend gross, the U.S. gross, and the worldwide gross not at all similar to the other titles.

20.18 (a)

(b)

JMP output:

The first two cereals to cluster are Wheaties and Nature’s Path Organic Multigrain Flakes followed by Post Shredded Wheat Vanilla Almond and Kellogg’s Mini Wheats. At the two cluster level, one cluster contains Post Shredded Wheat Vanilla Almond and Kellogg’s Mini Wheats and the other cluster contains the other five cereals. The first cluster appears to contain brands with similar amount of calories, carbohydrates, and sugar.

20.19 (a)

(b)

Partial JMP output:

The first two protein foods to cluster are “lamb leg roast” and “chicken, no skin, roast” followed by “Pork loin roast” and “chicken, with skin, roast”. At the two cluster level, one cluster contains “liver, fried” and the other cluster contains the other protein foods. The first cluster of “liver, fried” is unique in having the highest cholesterol not similar in other protein foods.

(C)

(d) (e)

Beside the unique cluster of “liver, fried”, the seafoods with the execption of salmon, mackerel and tuna are clustered together while the remaining nonseafoods form another cluster. The results in (a) and (c) are the same and, hence, the conclusions are the same.

20.20 (a)

(b)

Partial JMP output:

The first two countries to cluster are Egypt and Jordan followed by Lithuania and Poland. At the two cluster level one cluster is France, Germany, Spain, United Kingdom, Japan, Israel, and the United States and the other cluster contains the remaining countries. The first cluster appears to contain the western European countries and the United States and Israel.

20.21 (a)

Partial JMP output:

(a)

(b)

Since the stress statistic is 0.0124 in one dimension, 0.00007 in two dimensions and essentially zero in three dimensions, it is reasonable to try to interpret a two-dimensional mapping of the titles. A vertical line separates Deathly Hallows Part II from the rest of the titles.

20.22 (a)

JMP output: 0.35

Stress Value

0.3 0.25 0.2 0.15 0.1 0.05 0

Number of Dimensions

Stress Values 1 0.31469932 7 2 0.13078311 3 3 0.09731535 7 4 0.00911931 2 5 0.11642666 2

Since the stress value shows steep decline for one-dimension and twodimension, two- dimensional MDS is investigated.

Dimension 2

(a)

(b)

Since the stress statistic is 0.0973 in three dimensions, 0.1308 in two dimensions, and 0.3147 in one dimension, it is reasonable to try to interpret a two-dimensional mapping of the cereals. Looking at a 45º rotation, one dimension separates Post Shredded Wheat Vanilla Almond and Kellogg’s Mini Wheats based on their higher calorie and sugar content. A second dimension does not seem to be interpretable. In addition, All Bran, which has lower calories and higher sugar is separated from the other cereals.

20.23 (a)

JMP output:

(a)

(b)

Since the stress statistic is essentially 0 for two, three and higher dimensions, and 0.0006 for one dimension, it is reasonable to try to first try interpret a two-dimensional mapping of the foods. “Liver, fried” is obviously separated from the others.

20.24 (a)

JMP output: 0.55

Stress Value

0.5 0.45 0.4 0.35 0.3 0.25 0.2

Number of Dimensions

Stress Values 1 0.53228397 2 0.37868642 3 0.31505665 4 0.27729418 5 0.23542606

Dimension 2

(a)

(b)

Since the stress statistic is 0.2773 in four dimensions, 0.3151 in three dimensions, 0.3787 in two dimensions, and 0.5323 in one dimension, it is reasonable to try to first try interpret a two-dimensional mapping of the countries. There does not seem to be a clear interpretation of the dimensions along the lines of GDP and social media usage. Pakistan seems very separated from the other countries with Indonesia on the other side of the graph. Russia and Lithuania are close as are Mexico and Spain, and Israel and Egypt.

20.25 While gauges have been a popular choice in business, it consumes a lot of visual space in a dashboard. Most information design specialists prefer bullet graphs because those graphs foster the direct comparison of each measurement. 20.26 Sparklines summarize time-series as small, compact graphs designed to appear as part of a table (or a written passage) while time-series plots are stand along grahic displays. 20.27 A treemap is used to help visualize two variables, one of which must be categorical. Treemaps are especially useful when categories can be grouped to form a multilevel hierarchy or tree. Copyright © 2019 Pearson Australia (a division of Pearson Australia Group Pty Ltd) 9781488617249, Berenson, Basic Business Statistics 5e

20.28 Classification trees and regression trees are decision trees that split data into groups based on the values of independent or explanatory variables. Classificaiton trees use categorical dependent variable while regression trees use numerical dependent variables. 20.29 While classification and regression tree models are decisions tress that can be used to determine which values of a specific independent variable are useful in predicting the dependent varibles, neural net models is a powerful and flexible data mining technique that does not use a suppied model as a starting point in prediction, classification, and clustering problem. 20.30 Cluster analysis seeks to classify data into a sequence of groupings such that objects in each group are more alike other objects in their group than they are to objects found in other groups. Multidimensional scaling visualizes objects in a two or more dimensional space, or map, with the goal of discovering patterns of similarities or dissimilarities among the objects. 20.31 (a)

Bullet Graph for Calories Weinhard's Amber Light Sierra Nevada Anniversary Ale

O'Doul's Miller Genuine Draft 1 Leinenkugel Amber Light Flying Dog Raging Bitch Budweiser Select 55 American Kight 0

100

150

200

250

(b)

Bullet Graph for Calories Weinhard's Amber Light Sierra Nevada Anniversary Ale O'Doul's Miller Genuine Draft 1 Leinenkugel Amber Light Flying Dog Raging Bitch

Budweiser Select 55 American Kight 0.0

5.0

10.0

15.0

20.0

25.0

(c) Majority (more than 50%) of the beers have number of calories that fall below 160 per 12 ounces. Majority (more than 50%) of the beers have number of carbohydrates (in grams) that fall below 14 grams per 12 ounces. (d)

Gauges will consume a lot of visual space in a dashboard while the bullet graphs foster the direct comparison of each measurement.

20.32 (a) Yea r Currency (Value of $1 U. S.)

2002 2003 2004 2005 2006 2007 2008 2009 2010

Canada

1.57

1.29 1.20 1.17

1.17 0.99

1.21 1.05 1.00

English Pound

0.67

0.56 0.52 0.58

0.51 0.50

0.69 0.62 0.64

Euro

0.96

0.79 0.74 0.85

0.76 0.69

0.72 0.70 0.75

Year Currency (Value of $1 U. S.)

2008 2009 2010 2011 2012

Canada

1.21

1.05 1.00 0.98

0.99

English Pound

0.69

0.62 0.64 0.64

0.62

Euro

0.72

0.70 0.75 0.77

0.76

(b) The sparklines show differences in the values of the U. S. dollar in terms of the Canadian dollar, the English pound, and the Euro over the time period 2002 – 2012. The value of the U. S. dollar in terms of the Canadian dollar declined drastically from 2002 to 2007, but has remained steady since 2009. The value of the U. S. dollar in terms of the English pound has remained relatively steady between 2002 and 2012. The value of the U. S. dollar in terms of the Euro declined between 2002 and 2007 (with an increase in 2005) but has remained steady since 2008. 20.33 Because half the data will be used for a validation sample, the results will differ depending on which values are in the training sample and which are in the validation sample. 20.34 Because half the data will be used for a validation sample, the results will differ depending on which values are in the training sample and which are in the validation sample.

20.35 (a)

(b)

JMP output:

The r

of the regression tree model is 0.808. The first split occurs at a program cost of

$100,389.00. With a program cost of less than $100.389.00, the mean starting salary is $87,134.65. With a program cost of at least $100,389.00, the mean starting salary is $114,500.1. Another split occurs at a program cost of $50,639.00. The last split occurs at a percentage job offer of 93% for program cost of at least $50,639.00 but less than $100,389.00. With a percentage job offer of less than 93%, the mean starting salary is $87.614.25 while the mean starting salary is $95,209.143 with a percentage job offer of at least 93%. For a program cost of at least $100,389.00, another split occurs at a program cost of $117,930.00. With a program cost of at least $100,389.00 but less than $117,930, the mean starting salary is $110,027.20 while the mean starting salary is $118,973.00 with a program cost of at least $117,930.00. (c)-(f) Because half the data will be used for a validation sample, the results of a regression tree will differ depending on which values are in the training sample and which are in the validation sample. 20.36 Because half the data will be used for a validation sample, the results will differ depending on which values are in the training sample and which are in the validation sample.

20.37 (a)

(b)

JMP output:

2 The r of the regression tree model is 0.665. The first split occurs at a property size of 0.4608 acres. With a property size of at least 0.4608 acres, the mean fair market value is $678.7 thousands. With a property size of less than 0.4608 acres, the mean fair market value is $434.15 thousands. The second split occurs at garage size of 2 within property size of less than 0.4608 acres. At garage size of at least 2, the mean fair market value is $522.12 thousands. At garage size of less than 2, the mean fair market value is $396.9 thousands. The next split occurs at 1.5 bath rooms within less than 2 garages. With less than 1.5 bath rooms, the mean fair market value is $334.18 thousands while the mean fair market value is $425.85 thousands with at least 1.5 bath rooms. The last split occurs at 1 garage within at least 1.5 baths. For less than 1 garage, the mean fair market value is $406.38 thousands while the mean fair market value is $457 thousands with at least 1 but less than 2 garages.

(c)-(f) Because half the data will be used for a validation sample, the results of a regression tree will differ depending on which values are in the training sample and which are in the validation sample. Copyright © 2019 Pearson Australia (a division of Pearson Australia Group Pty Ltd) 9781488617249, Berenson, Basic Business Statistics 5e

20.38 (a)

JMP output:

(b)

JMP output:

(b)

(c)

From the hierarchical clustering output in part (a), the first two foods to cluster are Cantonese and American, followed by French and Mandarin, followed by Spanish and Greek. At the two cluster level, the first cluster includes Japanese, French, Mandarin, Szechuan, and Mexican. The second cluster includes Cantonese, American, Spanish, Greek, and Italian. Since the stress statistic is 0.0468 in four dimensions, 0.1164 in three dimensions, 0.2339 in two dimensions, and 0.4079 in one dimension, it is reasonable to try to first try interpret a two-dimensional mapping of the foods. There does not seem to be a clear interpretation of the dimensions along the lines of the three scales. The two spicy foods, Mexican and Szechuan are close to each other as are French and Greek, and Japanese and American. Italian is separated by itself as is Spanish.

20.39 (a)

Partial JMP output:

(b)

JMP output:

(b)

(c) Based on clustering analysis, the first two teams to cluster are Detroit and Arizona followed by Oakland and San Francisco. At the two cluster level, one cluster contains Houston, Chicago Cubs, Colorado, Minnesota, Kansas City, Cleveland, Toronto and Boston while the other cluster contains the other teams. The clustering order is slightly different based on multidimensional scaling analysis. The first two teams to cluster are Kansas City and Cleveland followed by New York Mets and Pittsburgh. At the two cluster level, one cluster contains Colorado, Minnesota, Kansas City, Cleveland, Milwaukee, Boston, Toronto, Houston, Miami and Chicago Cubs while the other cluster contains the other teams. The stress value has large drops moving from one-dimension to two-dimensions and three- dimensions and levels off after that. Investigating a three-dimension scatterplot reveals that Colorado is uniquely different from the remaining teams.

Chapter 21: Data analysis: The big picture Problem 21.5 These are sample solutions for Toyota Corollas for sale in New South Wales. Other graphs, statistical inference techniques and regression models may also be appropriate. Furthermore, other questions can be asked and answered using this data. Question: For two and three-year-old Toyota Corollas for sale in NSW what is the minimum and maximum price and the average price? Provide an estimated price range for a two or three year old used car

Two and Three Year Old Toyota Corolla NSW 25

Frequency

20 15 10 5 0 11,000 13,000 15,000 17,000 19,000 21,000 23,000 25,000

Price, $ This mound shaped histogram with a slight right skew, indicates that prices of two and three year old Corollas in the sample varied from approximately $12,000 to $24,000, with the majority between $16,000 and $18,000.

Toyota Corolla NSW Mean $17,531 Standard Deviation $2,216 Range $10,613 Interquartile Range $2,350 Count 42

Five-Number Summary Minimum $12,888 First Quartile $16,500 Median $16,990 Third Quartile $18,850 Maximum $23,501

The cheapest car in the sample was $12,888 and most expensive $23501.

Toyota

The median, indicates that 50% of two and three year old Corollas in the sample had prices less than $16,990 and 50% more than. The mean, is slightly more than the median at $17,531. As the mean is affected by the maximum price, $23,501, above $22,000 and the median is not, the median is the better measure of the average price to use. Therefore, the average price of two and three year old Corollas can be considered to be approximately $17,000. The standard deviation in the above table, indicates that on average prices vary from the mean price by approximately $2216. The five number summary above divides the sample into quarters, with 25% of prices being below the first quartile, $16,500, and 25% of prices above the third quartile, $18,850. Therefore, the middle 50% of prices vary by $2350. That is, between $16,500 and $18,850. The above suggests that there is a good selection of card to choose from within the range of $16,000 to $19,000. Question: Suppose a buyer will only consider white cars, since they believe white cars are safer as they are more visible. Will this limit their choice of cars for sale? Answer by calculating “What proportion of Toyota Corollas in NSW cars are white?” using a 90% confidence level for a proportion Let then

= proportion of Corollas for sale which are white

P = sample proportion of Corollas for sale which are white, samples size 121

Sample proportion p =

51 = 0.4214... 121

As the sample size is large (np = 51 and n(1 p) = 70 > 5) the sampling distribution of P is approximately normal. Therefore, a confidence interval using the z-distribution for a single proportion can be used. Excel output confidence interval for a proportion.

Confidence Interval Estimate for the Proportion White Toyota Corolla NSW Data Sample Size 121 Number of Successes 51 Confidence Level 90% Intermediate Calculations Sample Proportion 0.421487603 Z Value -1.6449 Standard Error of the Proportion 0.0449 Interval Half Width 0.0738 Confidence Interval Interval Lower Limit 0.3476 Interval Upper Limit 0.4953 From this output it can be stated that we are 90% confident that between 34.8% and 49.5% of Corollas for sale in New South Wales are white.

Therefore, choice should not be severely limited if search is restricted to white cars. Question Suppose a buyer wishes to purchase a two and three year old car with less than 50,000kms. Will this limit their choice of cars for sale? Answer using a hypothesis test for a single mean to test if on average the odometer reading of two and three year Toyota Corollas for sale in NSW is less than 50,000? Let

X = Odometer reading two and three year old Toyota Corollas for sale in NSW

then

X = Mean odometer reading two and three year old Corollas for sale in NSW, samples size 42

and

= mean odometer reading two and three year old Toyota Corollas for sale in NSW

The boxplot below, indicates that the population of odometer readings is mound shaped and approximately symmetric. Two and Three Year Old Toyota Corolla NSW

15000

25000

35000

The population standard deviation sample standard deviation.

45000 55000 Odometer Reading

65000

75000

85000

, is unknown so needs to be estimated by s, the

As sample size is large and from the boxplot we can assume X is not extremely skewed a ttest may be used. Test

H0 :   50,000 (at least 50,000 km) H1 :   50,000 (less than 50,000 km)

Use level of significance of 1% Excel output lower tail t-test at a 1% level of significance.

t Test for Hypothesis of the Mean Data Null Hypothesis = Level of Significance Sample Size Sample Mean Sample Standard Deviation

50000 0.01 42 47966.2381 14405.82248

Intermediate Calculations Standard Error of the Mean 2222.8667 Degrees of Freedom 41 t Test Statistic -0.9149 Lower-Tail Test Lower Critical Value -2.4208 p -Value 0.1828 Do not reject the null hypothesis As p-value = 0.18 > 0.01 level of significance, or z = the null hypothesis H0 :   50,000

Calculations Area

0.9149 > -2.3263, do not reject

Alternatively, since p-value is 0.18 is large, the null hypothesis will not be rejected at any level of significance. Therefore, can conclude at the 1% level of significance the sample data provides no evidence that the mean odometer reading of two and three year old Corollas for sale in New South Wales is less than 50,000. This result indicates that choice may be limited if restricting a search in NSW to two and three year old Corollas with odometer readings less than 50,000 kms. Question Is there is a difference in price between cars for sale privately and those for sale by a used car dealer? Answer using a hypothesis test for difference of two independent means. As cars for sale through a used car dealer and privately are separate/unrelated cars, these are independent samples. Let

XD/P = Price of Corollas for sale in NSW, Dealer/Private

then

D/P = Mean Price, Dealer/Private

X D and X P are independent, with nD = 89 , nP = 32 and D /P unknown. The boxplots below indicate that the distribution of prices for Corollas for sale in NSW by used car dealers and privately are mound shaped and approximately symmetric.

Toyota Corolla New South Wales

Private

Dealer

5000

10000

15000

20000

25000

30000

Price, $

Population standard deviations, D and P are unknown so must be estimated by the corresponding sample standard deviations s D and s P . As sample sizes nD = 89 , nP = 32 are large and X D and X P are not extremely skewed the ttest for difference of two independent means can be used to test for a difference in the average price of Corollas for sale privately and through a used car dealer. The boxplots indicate that prices of Corollas for sale by dealers are more varied than those for sale privately so we cannot assume equal variances. Therefore an independent two-tail separate variance t-test can be used to test for a difference in the average price of Corollas, for sale privately and through a used car dealer. Test

H0 : D = P

(No difference)

H1 : D  P

(Difference)

Use level of significance of 5% Excel output of independent two-tail separate variance t-test Separate-Variances t Test (assumes unequal population variances) Data Hypothesized Difference Level of Significance Population 1 Sample - Dealer Sample Size Sample Mean Sample Standard Deviation Population 2 Sample Private Sample Size Sample Mean Sample Standard Deviation t Test Statistic Two-Tail Test Lower Critical Value Upper Critical Value p -Value Reject the null hypothesis

0 0.05 89 15773.61798 4840.6660 32 11741.1875 4132.1592 4.5172 -1.9983 1.9983 0.0000

Since p-value ≈ 0 reject the null hypothesis H0 : D = P at any level of significance. Therefore, accept H1 : D  P Therefore, we can conclude at the 5% level of significance that there is a difference in average price for Corollas for sale in NSW for sale privately and through a used car dealer. Question How does the value of a Toyota Corolla depreciate in value? Answer by developing simple and multiple regression models, and choosing one which best fits data. Note: Will only consider linear models, however, a non-linear model may fit the data better. Also only consider Age, Odometer Reading and Transmission as possible independent variables, could choose other variables as possible independent variables. Let X/X1 = Age (Independent Variable) X2 = Odometer Reading (Independent Variable) X3 = Transmission (Independent Dummy Variable Automatic = 1 and Manual = 0) Y = Price (Dependent Variable) Simple Linear Regression model

Toyota Corolla For Sale In NSW $35,000 $30,000

y = -1274.9x + 20987 R² = 0.7696

Price

$25,000

R = -0.8773

$20,000 $15,000

$10,000 $5,000 $0 0

Age (years)

Scatter plot indicates a strong, negative and approximately linear relationship between price and age of Corollas for sale in New South Wales.

Price = 20986.63 − 1274.85  Age r=

0.90211…

r2 = 0.81397

 81.4%

Regression Statistics Multiple R 0.892141428 R Square 0.795916327 Adjusted R Square 0.790683412 Standard Error 2277.646469 Observations 121 Coefficients Standard Error t Stat P-value Intercept 20084.99685 669.0781818 30.01890869 8.42E-57 Age -962.9370105 108.237724 -8.896500913 8.54E-15 Odometer (kms) -0.022171302 0.00750942 -2.952465277 0.003811 Transmission 1200.431883 546.937395 2.194825028 0.030151

Price = 20085 − 963  Age − 0.0222  Odometer + 1200  Transmission Where Transmission =

1 Automatic   0 Manual

Manual

Price = 20085 − 963  Age − 0.0222  Odometer Automatic

Price = 21285 − 963  Age − 0.0222  Odometer Test

H0 : Age = 0 (Age not significant)

H1 : Age  0 (Age significant) And

H0 : Odometer = 0 (Odometer not significant) H1 : Odometer  0 (Odometer significant)

And

H0 : Transmission = 0 (Transmission not significant)

H1 : Transmission  0 (Transmission significant) If use level of significance of 1% As p − value(X1 / Age) = 0.0000  0.01 reject null hypothesis As p − value(X2 /Odometer) = 0.0038  0.01 reject null hypothesis As p − value(X3 / Transmission) = 0.0301  0.01 do not reject null hypothesis Since the p-value for both Age and Odometer are small the null hypothesis would be rejected at any level of significance and the alternative hypothesis accepted. Therefore, both Age and

Kilometres make a significant contribution to the model and should be included in the linear model. Since p-value for Transmission is greater than the level of significance (0.01) the null hypothesis is not rejected at the 1% level of significance. Therefore, at the 1% level of significance Transmission does not make a significant contribution to the model and not be included in the best model. Therefore, include Age and Odometer but not Transmission in a new multiple regression model to predict price from age and kilometres travelled.

Price = 21300 − 991  Age − 0.0239  Odometer from

Regression Statistics Multiple R 0.887419609 R Square 0.787513562 Adjusted R Square 0.783912097 Standard Error 2314.193765 Observations 121 Coefficients Standard Error t Stat P-value Intercept 21298.94751 382.5377209 55.6780321 1.47E-86 Age -990.6388509 109.2242649 -9.06976899 3.16E-15 Odometer (kms) -0.023918756 0.007586912 -3.15263378 0.002052 The coefficient of Age in this equation indicates that that for the same kilometres travelled each additional year older a Toyota Corolla is its estimated price will decrease by approximately $991. While the coefficient of Odometer indicates that for the same age additional 10,000 kilometres travelled the estimated price will decrease by approximately $239. Therefore, from this model the estimated depreciation of a Toyota Corolla is approximately $1000 per year in addition to a depreciation of $239 for every 10,000 kilometres travelled. The constant in the equation gives the estimated price when age and odometer reading are zero. That is, the estimated price of a new Toyota Corolla is approximately $21,300. The equation to predict price is fairly accurate since from the scatter plot and the multiple correlation coefficient the relationship between price and the independent variables, age and odometer reading is strong. In particular, the coefficient of determination indicates that approximately 78.8% of the variation in price is explained by its relationship to age and number of kilometres travelled. If use level of significance of 5% As p − value(X1 / Age) = 0.0000  0.05 reject null hypothesis As p − value(X2 /Odometer) = 0.0038  0.05 reject null hypothesis As p − value(X3 / Transmission) = 0.0301  0.05 reject null hypothesis Since the p-value for both Age and Odometer are small the null hypothesis would be rejected at any level of significance and the alternative hypothesis accepted. Therefore, at any level

of significance both Age and Kilometres make a significant contribution to the model and should be included in the best model. Since p-value for Transmission is less than the level of significance (0.05) the null hypothesis is rejected at the 5% level of significance. Therefore, at the 5% level of significance Transmission makes a significant contribution to the model and should be included in the best model. At the 5% level of significance the best model includes all three independent variables. Therefore, the best model is given by the following two equations, one predicting price for automatic Corollas from age and kilometres travelled and the other for manual Corollas. Manual

Price = 20085 − 963  Age − 0.0222  Odometer Automatic

Price = 21285 − 963  Age − 0.0222  Odometer In either equation the coefficient of Age, indicates that that for the same kilometres travelled and transmission each additional year older a Toyota Corolla is its estimated price will decrease by approximately $950. The coefficient of Odometer in either equation indicates that for the same age and transmission each additional 10,000 kilometres travelled the estimated price will decrease by approximately $222. Therefore, from this model the estimated depreciation of a Toyota Corolla is approximately $963 per year in addition to a depreciation of $222 for every 10,000 kilometres travelled. The constant in both equations gives the estimated price when age and odometer reading are zero. That is, the estimated price of a new automatic or manual Toyota Corolla is approximately $20085 or $21285. The difference between these constants, $1200, indicates that for the same age and odometer reading the price of an automatic Corolla is approximately $1200 more than that of a manual. The equations predict price are fairly accurate since from the scatter plot and the multiple correlation coefficient the relationship between price and the independent variables is strong. In particular, the coefficient of determination indicates that approximately 78.6% of the variation in price is explained by its relationship to age, number of kilometres travelled and transmission type Problem 21.10 Here we are assuming a 5 % level of significance. The chi-square test for difference in proportions of Volkswagen Golf cars for sale shows: • For Age Chi test

observed

AGE

QLD 2 years and under

Total

3-7 years 8 years and over

Total

108

expected QLD 2 years and under 3-7 years 8 years and over

Chi square calc

UK 17

26.5

10.5

8.460969293

rows

columns degrees of freedom

2 2

0.05 critical value

5.991464547

p-value

0.014545339

Since p=0.0145 < 0.05 we reject the null hypothesis and conclude that at the 5% level of significance there is a difference in the proportions of cars of three age groups for sale in Queensland and the UK. • For Kilometres KMS

observed QLD

< 30,000 30,000 100,000

>100,000

108

< 30,000 30,000 100,000

>100,000

Chi square calc

15.03016591

expected

0.05 critical value

5.991464547

p-value

0.000544805

Since p=0.0005 < 0.05 we reject the null hypothesis and conclude that at the 5% level of significance there is a difference in the proportions of cars of three kilometres driven groups for sale in Queensland and the UK. • For Price Level Price

observed QLD

<15,000

15,00-20,000

>20,000

108

<15,000

16.5

15,00-20,000

12.5

expected

>20,000

Chi square calc

23.04484848

0.05 critical value

5.991464547

p-value

9.90546E-06

Since p=0.0000099 < 0.05 we reject the null hypothesis and conclude that at the 5% level of significance there is a difference in the proportions of cars of various price levels for sale in Queensland and the UK. • For Seller Type

Chi test

observed

SELLER

QLD

Total

Dealer

Private

Total

108

expected QLD

Dealer

41.5

Private

12.5

Chi square calc

15.04193

rows

columns degrees of freedom

2 1

0.05 critical value

3.841459

p-value

0.000105

Since p=0.000105 < 0.05 we reject the null hypothesis and conclude that at the 5% level of significance there is a difference in the proportions of cars for sale by dealers and private sale in Queensland and the UK. • For Colour

Colour

observed QLD

Total

White

Other

Total

108

expected QLD

White

10.5

Other

43.5

Chi square calc

1.477833

0.05 critical value

3.841459

p-value 0.224114 Since p=0.224114> 0.05 we cannot reject the null hypothesis and conclude that at the 5% level of significance we are not able to find a difference in the proportions of cars of white and non-white cars for sale in Queensland and the UK. • For transmission Transmission

observed QLD

Total

Automatic

Manual

Total

108

expected QLD

Automatic

Manual

Chi square calc

25.92614

0.05 critical value

3.841459

p-value 3.55E-07 Since p=0.00000036 < 0.05 we reject the null hypothesis and conclude that at the 5% level of significance there is a difference in the proportions of automatic and manual cars for sale in Queensland and the UK.

End of Part 1 problems A.1

A.2

A.3 (a)

Superannuation Fund 3 years 5 years

Conservative 7.72% 9.33%

Balanced 8.44% 11.02%

Growth 8.90% 12.13%

High Growth 9.60% 13.24%

(b)

The average rate of return is higher over the previous 5 years than the previous 3 years due to lower returns in 2015 to 2017 compared to 2013 and 2014. The growth and in high growth funds have higher returns over both three and five years.

A.4 (a) and (c)

Magnesium (mg/l) Frequency 1.50 to < 1.52 0 1.52 to < 1.54 3 1.54 to < 1.56 5 1.56 to < 1.58 13 1.58 to < 1.60 27 1.60 to < 1.62 25 1.62 to < 1.64 15 1.64 to < 1.66 5 1.66 to < 1.68 3 Total 96

Percent Cum. FrequencyCum. Percent 0.0% 0 0.0% 3.1% 3 3.1% 5.2% 8 8.3% 13.5% 21 21.9% 28.1% 48 50.0% 26.0% 73 76.0% 15.6% 88 91.7% 5.2% 93 96.9% 3.1% 96 100.0% 100.0%

(b) Natural Australian Spring Water

Frequency

25 20 15 10 5 0

1.5

1.52

1.54

1.56

1.58

1.6

1.62

1.64

1.66

1.68

1.7

Magnesium (mg/l)

Natural Australian Spring Water 30%

Percent

25% 20% 15% 10% 5% 0% 1.51

1.53

1.55

1.57

1.59

1.61

1.63

1.65

1.67

1.69

Magnesium (mg/l)

(c)

Percent

Natural Australian Spring Water

100% 90% 80% 70% 60% 50% 40% 30% 20% 10% 0% 1.52

1.54

1.56

1.58

1.60

1.62

1.64

1.66

1.68

Magnesium (mg/l)

(d) and (e)

Magnesium (mg/l) Mean

1.5995

Standard Error

0.0031

Median

1.5990

Mode

1.6150

Standard Deviation

0.030036102

Sample Variance

0.000902167

Coefficient of Variation

1.9%

First Quartile

1.5800

Third Quartile

1.6190

Range

0.155

Minimum

1.522

Maximum

1.677

(f) Australian Natural Spring Water

Magnesium (mg/l)

1.5

1.52

1.54

1.56

1.58

1.6

1.62

1.64

1.66

1.68

1.7

For 50% of bottles the magnesium content is between 1.58 and 1.62 mg/l. Magnesium content is symmetric about the median 1.60 mg/l. (g)

The distribution of magnesium (mg/l) is approximately symmetric and mound shaped, centred on the stated 1.60 mg/l. To 1 decimal place the average (mean and median) magnesium content is the stated 1.6 mg/l, as are the first and third quartile. Furthermore, as the coefficient of variation is small it can be concluded that there is little variation in the data. Therefore, it seems that the stated magnesium content is correct for this day’s production.

A.5 (a) Reports may vary over time. Examples of categorical variables might be: spending categories (Fashion, Department, Media etc); states (NSW, VIC, QLD etc); regions (Metro, Regional) (b) A numerical variable might be growth in spending. (c) Growth in spending is continuous. A.6 (a)

Ordered Arrays

August 0

100

140

158

194

236

237

250

253

264

287

September

(b)

105

106

184

185

186

187

210

226

242

259

PhStat Stem-and-leaf displays

August

September

Stem unit: 10

0 0003456678 1 0033899 2 124589 3 025668 4 0366 5 3 6 7 356 8 77 9 9 10 0 11 12 13 14 0 15 8 16 17 18 19 4 20 21 22 23 6 7 24 25 0 3 26 4 27 28 7

0 044899 1 125 2 12333579 3 02259 4 4567 5 133578 6 14568 7 25 8 3 9 10 5 6 11 12 13 14 15 16 17 18 4 5 6 7 19 20 21 0 22 6 23 24 2 25 9

(c) August Frequency 0.0 to < 25.0 20 25.0 to < 50.0 13 50.0 to < 75.0 2 75.0 to < 100.0 5 100.0 to < 125.0 1 125.0 to < 150.0 1 150.0 to < 175.0 1 175.0 to < 200.0 1 200.0 to < 225.0 0 225.0 to < 250.0 2 250.0 to < 275.0 3 275.0 to 300.0 1 Total 50

Percent 40.0% 26.0% 4.0% 10.0% 2.0% 2.0% 2.0% 2.0% 0.0% 4.0% 6.0% 2.0% 100.0%

Cum. Frequency Cum. Percent 20 40.0% 33 66.0% 35 70.0% 40 80.0% 41 82.0% 42 84.0% 43 86.0% 44 88.0% 44 88.0% 46 92.0% 49 98.0% 50 100.0%

September Frequency Percent 0.0 to < 25.0 14 28.0% 25.0 to < 50.0 12 24.0% 50.0 to < 75.0 12 24.0% 75.0 to < 100.0 2 4.0% 100.0 to < 125.0 2 4.0% 125.0 to < 150.0 0 0.0% 150.0 to < 175.0 0 0.0% 175.0 to < 200.0 4 8.0% 200.0 to < 225.0 1 2.0% 225.0 to < 250.0 2 4.0% 250.0 to 275.0 1 2.0% Total 50 100.0%

Cum. Frequency Cum. Percent 14 28.0% 26 52.0% 38 76.0% 40 80.0% 42 84.0% 42 84.0% 42 84.0% 46 92.0% 47 94.0% 49 98.0% 50 100.0%

(d) Histogram Frequency of Web Access

Frequency

15 10 5 0

100

125

150

175

200

225

250

275

300

August

Histogram Frequency of Web Access

Frequency

12 10 8 6 4 2 0 0

100

125

150

175

200

225

250

275

300

September

Polygon - Monthly Web Access

40%

30%

August 20%

September

10%

-25

100

125

150

175

200

225

250

275

300

325

Number of Times Site Accessed

(e) Ogive - Monthly Web Access

100% 90% 80% 70% 60% 50% 40% 30% 20% 10% 0%

August September

100

125

150

175

200

225

250

275

300

Number of Times Site Accessed

(f) and (g) Sample Data: Arithmetic mean Median Mode Minimum First Quartile Third Quartile Maximum Range Inter Quartile Range Variance (Sample) Standard Deviation (Sample) Coefficient of Variation (Sample)

(h)

August September 66.82 67.72 33.5 46.5 0 23 0 0 13 23 87 72 287 259 287 259 71.25 48 6618.518 4641.634 81.354 68.13 121.75% 100.60%

Both distributions are highly variable, but with few students accessing the website more than 100 times a month. But the majority of students access the website less than 40 times in August and less than 70 in September. In general students seem to be accessing the website more in September than in August. While the mean number of accesses is similar for August and September, the medians are not. In August 50% of students accessed the site more than 33 times while in September 50% had accessed it more than 46 times. The number of times a student accessed the site is more variable in August than in September, as the range, interquartile range, variance and standard deviation are all larger for August than September.

A.7 (a)

From solution of problem A.6, the five-number summaries are: August

33.5

287

September

46.5

259

(b) Monthly Web Access

September

August

(c)

100

120

140

160

180

200

220

240

260

280

300

Both distributions are positively skewed, with a long right whisker. However, the number of times a student accessed the site is more variable in August than in September.

A.8 Pie charts also appropriate graphs. (a) and (b) Web Access Saturday Friday Thursday Wednesday

September

Tuesday

August

Monday Sunday 0

200

400

600

800

1000

Frequency

Pattern similar for the two months but not the same, during August access peaked on Tuesday, with significant access on Monday and Friday. While during September access peaked on Monday, with significant access on Tuesday and Friday. For both months there was limited access during the weekend. Web Access Time of Day

24:00 23:00 22:00

21:00 20:00 19:00 18:00 17:00 16:00 15:00 14:00

September

13:00 12:00

August

11:00 10:00 9:00 8:00 7:00 6:00 5:00 4:00 3:00 2:00 1:00 0

100

200

300

400

500

600

Frequency

Pattern similar for the two months but not identical. For both months most common hour to access the website was between 12 noon and 1 pm. The majority of accesses happened between 8 am and 7 pm with very few during the evening. (c)

Monday morning at 8 am would be a good time to post an announcement.

A.9 All cars sold by top 20, so population data Top 20 sales for either February 2016 or February 2017 are AUDI

BMW

FORD

HOLDEN

HONDA

HYUNDAI

JEEP

KIA

LAND ROVER

LEXUS

MAZDA

MERCEDES-BENZ

MITSUBISHI

NISSAN

SKODA

SSANGYONG

SUBARU

SUZUKI

TOYOTA

VOLKSWAGEN

(a)

Make AUDI BMW FORD HOLDEN HONDA HYUNDAI JEEP KIA LAND ROVER LEXUS MAZDA MERCEDES-BENZ MITSUBISHI NISSAN SKODA SSANGYONG SUBARU SUZUKI TOYOTA VOLKSWAGEN Mean Standard Deviation Variance

Sales of New Cars - Top 20 February February 2017 Z-Scores 2016 176 -0.7954 137 160 -0.8568 193 611 0.8729 604 654 1.0378 645 373 -0.0399 292 606 0.8537 470 56 -1.2556 100 513 0.4970 407 93 -1.1137 64 62 -1.2326 59 755 1.4251 719 245 -0.5308 164 547 0.6274 413 346 -0.1434 484 104 -1.0715 102 93 -1.1137 95 305 -0.3007 208 624 0.9227 362 990 2.3263 915 355 -0.1089 309 383.40 337.10 260.7519

237.5548

67991.5400

56432.2900

Z-Scores -0.8423 -0.6066 1.1235 1.2961 -0.1899 0.5594 -0.9981 0.2942 -1.1496 -1.1707 1.6076 -0.7287 0.3195 0.6184 -0.9897 -1.0191 -0.5435 0.1048 2.4327 -0.1183

(b)

Within 1 standard deviation of mean 2 standard deviation of mean 3 standard deviation of mean (c)

February 2017 Number Percentage 12 60.0% 19 95.0% 20 100.0%

February 2016 Number Percentage 13 65.0% 19 95.0% 20 100.0%

The February sales data for both years are fairly consistent with the empirical rule, with slightly less than 68% within one standard deviation of the mean.

A.10

A.11

A.12 (a)&(b)

New Zealand and United States Exchange Rates 1.4 1.3 1.2

Rate per Australain Dollar

1.1 1.0 0.9 0.8 0.7

USD

0.6

NZD

0.5 0.4 0.3

0.2 0.1 0.0 Jan 10

Sep 10 May 11 Jan 12

Sep 12 May 13 Jan 14

Sep 14 May 15 Jan 16

Sep 16 May 17

The Australian dollar has depreciated or decreased in value against both the US and NZ dollar since May 2011. However, the value of the Australian dollar has been recently steady, between $1 and $1.10 NZ dollars since January 2014 and between 70 and 80 US cents since May 2015.

Chinese Yuan Renminbi 8

CNY per Australian Dollar

7 6

5 4 3 2

1 0 Jan 10

Sep 10 May 11 Jan 12

Sep 12 May 13 Jan 14

Sep 14 May 15 Jan 16

Sep 16 May 17

The Australian dollar has depreciated or decreased in value against the Chinese yuan renminbi since May 2011 falling to a low of approximately 4.5 yuan per Australian dollar, however, the value of the Australian dollar increased slightly since to just over 5 yuan.

Yen per Australain Dollar

Japanese Yen 110 100 90 80 70 60 50 40 30 20 10 0 Jan 10

Apr 11

Jul 12

Sep 13

Dec 14

Mar 16

May 17

The Australian dollar increased in value from 80 yen to between 90 and 100 yen on 2012 and 2014. However, it has since decreased in value back to just over 80 yen.

(c)

Scatter Plot 8

7 7

CNY

6 6

5 5 4 0.65

0.75

0.85

0.95

1.05

1.15

0.95

1.05

1.15

USD

Scatter Plot 105 100

JPY

95 90

85 80 75

70 0.65

0.75

0.85 USD

Scatter Plot 1.40 1.35

1.30

NZD

1.25 1.20

1.15 1.10 1.05

1.00 0.95 0.65

0.75

0.85

0.95

1.05

1.15

5.5

6.0

USD

Scatter Plot 1.20 1.18

1.16 1.14 NZD

1.12 1.10 1.08

1.06 1.04 1.02

1.00 4.0

4.5

5.0 CNY

JPY

Scatter Plot 102 100 98 96 94 92 90 88 86 84 82 80 78 76 74 4.4

4.6

4.8

5.0

5.2 CNY

5.4

5.6

5.8

6.0

Scatter Plot

1.40 1.35 1.30

NZD

1.25 1.20

1.15 1.10 1.05 1.00 74

98 100

JPY

(d)

Correlation Coefficients

Correlation Coefficient USD CNY JPY NZD (e)

USD 1.0000 0.9602 0.0289 0.8211

CNY 0.9602 1.0000 -0.1652 0.8923

JPY 0.0289 -0.1652 1.0000 -0.3464

NZD 0.8211 0.8923 -0.3464 1.0000

From the scatter plots and correlation coefficients there is a strong relationship between the US dollar and Chinese yuan renminbi in terms of the Australian dollar. There is also a moderate positive relationship between the New Zealand dollar and either the Chinese yuan renminbi or the US dollar in terms of the Australian dollar. There is no apparent relationship between the Japanese yen and any of the currencies in terms of the Australian dollar.

A.13 (a)

Contingency Table Total % Level Professor Associate Profesor Senior Lecturer Lecturer Associate Lecturer Total

Gender Female Male Total 3.93% 6.34% 10.27% 4.83% 7.25% 12.08% 11.18% 15.71% 26.89% 22.36% 17.52% 39.88% 6.95% 3.93% 10.88% 49.24% 50.76% 100.00%

Contingency Table Row Total % Level Professor Associate Profesor Senior Lecturer Lecturer Associate Lecturer Total

Gender Female Male 38.24% 61.76% 40.00% 60.00% 41.57% 58.43% 56.06% 43.94% 63.89% 36.11% 49.24% 50.76%

Total 100.00% 100.00% 100.00% 100.00% 100.00% 100.00%

Contingency Table Column Total % Level Professor Associate Profesor Senior Lecturer Lecturer Associate Lecturer Total

Gender Female Male 7.98% 12.50% 9.82% 14.29% 22.70% 30.95% 45.40% 34.52% 14.11% 7.74% 100.00% 100.00%

Total 10.27% 12.08% 26.89% 39.88% 10.88% 100.00%

Gender and Level Associate Lecturer Lecturer Senior Lecturer Associate Profesor

Professor 0%

10% Male

15%

20%

25%

Female

Gender and Level

Male

Female

Associate Lecturer

(b)

10% Lecturer

Senior Lecturer

15%

20%

Associate Profesor

25% Professor

While there are approximately equal numbers of male and female academic staff, male staff are generally at a higher level than female staff. For example, the majority of senior lecturers and above are male and the majority of lecturers and associate lecturers are female.

Female

Male

Total

$122,847 $23,694

$130,748 $24,577

$126,857 $24,467

(d)

Annual expenditure on academic salaries is approximately 42 million dollars. (126,857 × 331 = $41,989,667)

(f)

Female salaries are generally lower than and slightly less varied than male salaries at this university.

A.14 (a)

Contingency Table Total Percentage Search Items X-Ray Items Identified Yes No Yes 7.2% 2.4% No 2.8% 87.6% Total 10.0% 90.0%

Total 9.6% 90.4% 100.0%

Contingency Table Row Percentage Search Items X-Ray Items Identified Found Yes No Yes 75.0% 25.0% No 3.1% 96.9%

Total 100.0% 100.0%

Contingency Table Column Percentage Search Items X-Ray Items Identified Found Yes No Yes 72.0% 2.7% No 28.0% 97.3% Total 100.0% 100.0%

Mail Centre - illegal or threatening items

X-Ray Items Not Identified

Search Items Not Found

Search Items Found

X-Ray Items Identified

20%

40%

60%

80%

100%

120%

Mail Centre - illegal or threatening items

Search Items Not Found

X-Ray Items Not Identified

X-Ray Items Identified

Search Items Found

(b)

20%

40%

60%

80%

100%

120%

The X-ray is identifying 75% of the items of interest so it can be considered worthwhile, especially as 97.3% of items not identified as illegal or threatening items are found to be not illegal or threatening. However, the mail centre should probably continue to randomly search a proportion of the mail in the hope of identifying a proportion of the 25% of illegal or threatening mail not identified by X-ray.

A.15

(b)

A.16 By violating the rule that the axis should start at zero, this ad effectively communicates that BCU’s interest rate is less than that of the other lenders. A.17 (a)

Foods Never Eaten Cheese Cream Dairy Products Eggs Fish Seafood Any Meat Chicken/Poultry Pork/Ham Red Meat Sugar Wheat Products Eat all foods Total Number of Respondents

Men 236 623 131 175 123 166 111 126 234 159 1095 187 7299 10000

Women 219 917 196 279 266 268 353 368 495 247 897 380 7878 10000

Total 455 1540 327 454 389 434 464 494 729 406 1992 567 15177 20000

Men Women Total 2.36% 2.19% 2.28% 6.23% 9.17% 7.70% 1.31% 1.96% 1.64% 1.75% 2.79% 2.27% 1.23% 2.66% 1.95% 1.66% 2.68% 2.17% 1.11% 3.53% 2.32% 1.26% 3.68% 2.47% 2.34% 4.95% 3.65% 1.59% 2.47% 2.03% 10.95% 8.97% 9.96% 1.87% 3.80% 2.84% 72.99% 78.78% 75.89% 100.00% 100.00% 100.00%

Foods Never Eaten Wheat Products Sugar Red Meat Pork/Ham Chicken/Poultry Any Meat Seafood Fish Eggs Dairy Products Cream Cheese

4% Total

6% Women

10%

12%

Men

Of the participants, approximately 9% of women and 6% of men never eat cream and 11% of men and 9% of women never eat sugar. Over 95% eat some meat and dairy products.

As the categories are not mutually exclusive, a pie chart is not appropriate for this data.

A.18 (a)

Time to Resolve TS Problem

50% 45% 40% 35% 30% 25% 20% 15% 10% 5% 0%

Students Staff

11 13 Time (Minutes)

(b) & (c)

Time to Solve TS Problem (Minutes) Mean Standard Error Median Mode Standard Deviation Sample Variance Kurtosis Skewness Coefficient of Variation Sum Count

Minimum First Quartile Median Third Quartile Maximum Range Interquartile Range

Time Students Time Staff 4.5 3.7 0.747637 0.688635 3.05 2.05 1.5 1.5 4.72847 4.35531 22.35846 18.96872 5.07316 8.09229 2.27729 2.72591 105.08% 117.71% 180 148 40 40 Time to Resolve Problem (minutes) Students Staff 0.8 0.5 1.5 1.5 3.05 2.05 4.5 4.2 20.9 21.7 20.1 21.2 19.4 20.2

(d) Time to Resolve Problem

Staff

Students

Minutes

(e)

25% of problems are resolved in less than 1.5 minutes, and 75% in less than 4.5 minutes. 50% of problems reported by students are solved within 3.05 minutes with a mean of 4.5 minutes. While 50% of problems reported by staff are resolved in 2.05 minutes with a mean of 3.7 minutes. From the histogram almost all problems are resolved in less than 8 minutes. However, there are a few problems reported which take longer to resolve, with a maximum time of 21.7 minutes for this sample. The time to resolve a TS problem is skewed to the right for both students and staff. However, the time to resolve TS problems reported by staff are generally less, but more varied, than the time to resolve TS problems reported by students.

A.19

A.20 (a)

Number of Alana Resolved Calls Freq 10 to < 20 3 20 to < 30 5 30 to < 40 1 40 to < 50 2 50 to < 60 3 60 to < 70 1 70 to < 80 1 80 to < 90 2 90 to < 100 2 100 to < 110 0 Total 20

% Cum. Freq Cum. % 15.0% 3 15.0% 25.0% 8 40.0% 5.0% 9 45.0% 10.0% 11 55.0% 15.0% 14 70.0% 5.0% 15 75.0% 5.0% 16 80.0% 10.0% 18 90.0% 10.0% 20 100.0% 0.0% 20 100.0% 100.0%

Byung Freq 0 2 4 2 1 2 1 5 3 0 20

Percent Cum. Freq Cum. % 0.0% 0 0.0% 10.0% 2 10.0% 20.0% 6 30.0% 10.0% 8 40.0% 5.0% 9 45.0% 10.0% 11 55.0% 5.0% 12 60.0% 25.0% 17 85.0% 15.0% 20 100.0% 0.0% 20 100.0% 100.0%

Number of Claire Resolved Calls Freq 10 to < 20 1 20 to < 30 3 30 to < 40 2 40 to < 50 3 50 to < 60 2 60 to < 70 2 70 to < 80 1 80 to < 90 3 90 to < 100 2 100 to < 110 1 Total 20

% Cum. Freq Cum. % 5.0% 1 5.0% 15.0% 4 20.0% 10.0% 6 30.0% 15.0% 9 45.0% 10.0% 11 55.0% 10.0% 13 65.0% 5.0% 14 70.0% 15.0% 17 85.0% 10.0% 19 95.0% 5.0% 20 100.0% 100.0%

David Freq 1 2 5 4 1 2 3 2 0 0 20

Percent Cum. Freq Cum. % 5.0% 1 5.0% 10.0% 3 15.0% 25.0% 8 40.0% 20.0% 12 60.0% 5.0% 13 65.0% 10.0% 15 75.0% 15.0% 18 90.0% 10.0% 20 100.0% 0.0% 20 100.0% 0.0% 20 100.0% 100.0%

Number of Eru Resolved Calls Freq 10 to < 20 1 20 to < 30 2 30 to < 40 0 40 to < 50 4 50 to < 60 3 60 to < 70 0 70 to < 80 3 80 to < 90 3 90 to < 100 3 100 to < 110 1 Total

% Cum. Freq Cum. % 5.0% 1 5.0% 10.0% 3 15.0% 0.0% 3 15.0% 20.0% 7 35.0% 15.0% 10 50.0% 0.0% 10 50.0% 15.0% 13 65.0% 15.0% 16 80.0% 15.0% 19 95.0% 5.0% 20 100.0%

(b) HELP_DESK - ALANA

Frequency

5 4 3

2 1 0 0

100

110

100

110

100

110

Number of Resolved Calls per Shift

HELP_DESK - BYUNG

Frequency

5 4 3 2 1 0 0

Number of Resolved Calls per Shift

HELP_DESK - CLAIRE

Frequency

5 4 3 2 1 0 0

Number of Resolved Calls per Shift

HELP_DESK - DAVID

Frequency

5 4 3 2 1 0 0

100

110

100

110

Number of Resolved Calls per Shift

HELP_DESK - ERU

Frequency

5 4 3 2 1 0 0

Number of Resolved Calls per Shift

25% Alana

20% Byung

15%

Claire

10%

David Eru

5% 0% 5

105

115

Number of Resolved Calls

(d) Help Desk

100% 90% 80% 70% 60% 50% 40% 30% 20% 10% 0%

Alana Byung Claire David Eru

100

110

Number of Resolved Calls

(e) and (f)

Sample Data: Number of Data Points Minimum Maximum Total Arithmetic mean Median Mode First Quartile Third Quartile Range Inter Quartile Range Variance (Sample) Standard Deviation (Sample) Coefficient of Variation (Sample)

Alana

Byung Claire David Eru 20 20 20 20 20 13 25 19 18 24 93 99 106 83 93 939 1246 1159 996 1260 46.95 62.30 57.95 49.80 63.00 43 63 54 44 62.5 21 83 Nil Multi Multi 21 38 36 37 50 78 87 83 70 80 80 74 87 65 69 57 49 47 33 30 743.103 612.432 739.629 414.905 425.158 27.2599 24.7474 27.1961 20.3692 20.6194 58.1% 39.7% 46.9% 40.9% 32.7%

(g) Helpdesk

Eru David Claire Byung

Alana

110

Number of Resolved Calls per Shift

(h)

There is no pattern to the number of calls a staff member resolves in an eight-hour shift. This can vary from 13 to 106. Probably the number of resolved calls is a factor of the type of call, with some taking a few minutes to resolve and others far longer. On average Byung and Eru resolve the most calls per shift and Alana the least. This may be a factor of experience if Alana is newly appointed to the position. The number of resolved calls varies least for David and Eru and the most for Alana.

A.21 (a)

Age last birthday (years) 0-4 5-9 10-14 15-19 20-24 25-29 30-34 35-39 40-44 45-49 50-54 55-59 60-64 65-69 70-74 75-79 80-84 85-89 90 to 94 95 to 100 Over 100

Percentage at 30 June 2016

Percentage at 30 June 2013

Male 2016 Female 2016 Persons 2016

Male 2013 Female 2013 Persons 2013

6.71% 6.55% 6.18% 6.34% 7.19% 7.45% 7.42% 6.70% 6.69% 6.63% 6.35% 6.06% 5.38% 4.87% 3.66% 2.62% 1.70% 1.02% 0.39% 0.08% 0.01% 100.0%

6.27% 6.14% 5.78% 5.97% 6.73% 7.31% 7.37% 6.66% 6.74% 6.71% 6.44% 6.20% 5.51% 4.95% 3.79% 2.81% 2.11% 1.52% 0.76% 0.20% 0.03% 100.0%

6.49% 6.34% 5.98% 6.15% 6.96% 7.38% 7.40% 6.68% 6.72% 6.67% 6.39% 6.13% 5.44% 4.91% 3.73% 2.71% 1.91% 1.27% 0.58% 0.14% 0.02% 100.0%

6.77% 6.49% 6.22% 6.54% 7.27% 7.59% 7.23% 6.72% 7.16% 6.58% 6.66% 5.98% 5.34% 4.67% 3.33% 2.41% 1.68% 0.96% 0.33% 0.06% 0.01% 100.0%

6.35% 6.09% 5.87% 6.14% 6.92% 7.36% 7.10% 6.70% 7.22% 6.63% 6.74% 6.07% 5.39% 4.69% 3.43% 2.69% 2.17% 1.53% 0.71% 0.17% 0.03% 100.0%

6.56% 6.29% 6.05% 6.34% 7.10% 7.47% 7.16% 6.71% 7.19% 6.61% 6.70% 6.02% 5.36% 4.68% 3.38% 2.55% 1.93% 1.25% 0.52% 0.12% 0.02% 100.0%

Cummulative Percentage at 30 June 2016 Cummulative Percentage at 30 June 2013 Age last birthday (years) 0-4 5-9 10-14 15-19 20-24 25-29 30-34 35-39 40-44 45-49 50-54 55-59 60-64 65-69 70-74 75-79 80-84 85-89 90 to 94 95 to 100 Over 100 (b)

Males 6.71% 13.26% 19.44% 25.78% 32.98% 40.43% 47.85% 54.55% 61.24% 67.87% 74.22% 80.28% 85.66% 90.52% 94.19% 96.80% 98.50% 99.53% 99.91% 99.99% 100.00%

Females 6.27% 12.41% 18.19% 24.16% 30.89% 38.20% 45.57% 52.23% 58.97% 65.68% 72.12% 78.32% 83.83% 88.78% 92.56% 95.37% 97.49% 99.01% 99.77% 99.97% 100.00%

Persons 6.49% 12.83% 18.81% 24.97% 31.92% 39.31% 46.70% 53.38% 60.10% 66.77% 73.16% 79.29% 84.74% 89.65% 93.37% 96.08% 97.99% 99.27% 99.84% 99.98% 100.00%

Males Females Persons 6.77% 6.35% 6.56% 13.27% 12.45% 12.85% 19.49% 18.32% 18.90% 26.03% 24.46% 25.24% 33.30% 31.38% 32.34% 40.89% 38.74% 39.81% 48.12% 45.84% 46.97% 54.84% 52.53% 53.68% 62.00% 59.76% 60.87% 68.59% 66.39% 67.48% 75.25% 73.13% 74.18% 81.22% 79.20% 80.21% 86.56% 84.59% 85.57% 91.22% 89.28% 90.25% 94.55% 92.70% 93.62% 96.96% 95.39% 96.17% 98.64% 97.56% 98.10% 99.60% 99.09% 99.35% 99.94% 99.80% 99.87% 99.99% 99.97% 99.98% 100.00% 100.00% 100.00%

Frequency and percentage histograms, polygons and ogives are all appropriate.

Age Distribution 8% 7% 6% 5% 4% 3% 2% 1% 0% 0

5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 105

Age last birthday (years) Male 2016

Male 2013

Age Distribution 8% 7% 6% 5% 4% 3% 2% 1% 0% 0

5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 105

Age last birthday (years) Female 2016

Female 2013

Age Distribution 8% 7% 6% 5% 4% 3% 2% 1% 0% 0

5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 105

Age last birthday (years) Persons 2016

Persons 2013

There is no apparent change in age distribution between June 2013 and June 2016. However there is a higher proportion of children under 5 than between 10 and 14, indicating that there has been a slight increase in birth rate in recent years. The proportion in each age group after 50 is decreasing

Age Distribution 8% 7% 6% 5% 4% 3% 2% 1% 0% 0

5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 105

Age last birthday (years) Male 2016

Female 2016

Age Distribution 8% 7% 6% 5% 4% 3% 2% 1% 0% 0

5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 105

Age last birthday (years) Male 2013

Female 2013

From the graph we can see that males outnumber females until approximately the age of 35. Then until the age of approximately 70, there are approximately the same number of males and females have approximately the same number, however after the age of 70, females outnumber males. (c)

To 30 June

2013

2016

Persons

Mean

37.899…

38.266…

Standard deviation (population)

22.819…

23.023…

Note: these are population parameters and are approximations calculated from the frequency distribution with the mid-point of the last class (100 and over) estimated to be 102. These statistics show that the Australian population is growing older and more varied in age. A.22 (a)

Error -0.1 to < -0.05 -0.05 to < 0.0 0.0 to < 0.05 0.05 to < 0.1 0.1 to 0.15 Total

Frequency 13 36 36 14 1 100

Percent Cum. Frequency Cum. Percent 13% 13 13% 36% 49 49% 36% 85 85% 14% 99 99% 1% 100 100% 100%

(b) Histogram

40 35

Frequency

30 25 20 15 10 5 0 -0.15

-0.1

-0.05

0.05

0.1

0.15

0.1

0.15

0.2

Error in mm

Percentage %

Percentage Polygon

40 35 30 25 20 15 10 5 0 -0.15

-0.1

-0.05

0.05

0.2

Error mm

(c)

Cumulative Percentage

Cum. Percent Polygon

100% 90% 80% 70% 60% 50% 40% 30% 20% 10% 0% -0.125

-0.075

-0.025

0.025

0.075

0.125

Error mm

(d)

As there is only one part out of the sample of 100 that does not meet the requirements, the steel mill is doing a good job.

A.23 (a), (b) and (c)

Weekly Sales Regional City Forgive Chocolates Stem unit: 100 kgs 1 335 2 278 3 125999 4 0235567889 5 155689 6 001333469 7 2367789 8 0667 9 115 10 4 Forgive Chocolates Weekly Sales 0 kgs to < 2000 kgs 200 kgs to < 400 kgs 400 kgs to < 600 kgs 600 kgs to < 800 kgs 800 kgs to < 1000 kgs 1000 kgs to < 1200 kgs

Frequency 3 10 16 16 6 1

Percentage 5.8% 19.2% 30.8% 30.8% 11.5% 1.9%

Cumulative % 5.8% 25.0% 55.8% 86.5% 98.1% 100.0%

Forgive Chocolates

18 16

Frequency

14 12 10

8 6 4 2 0 0

200

400

600

800

1000

1200

Weekly Sales, kg

Percentage Polygon - Forgive Chocolates

35% 30% 25% 20% 15% 10% 5% 0% -200

200

400

600

800

1000

1200

1400

Weekly Sales, Kgs

Percentage Polygon - Forgive Chocolates

100%

90% 80% 70%

60% 50% 40%

30% 20% 10%

0% 0

200

400

600

800

1000

1200

Weekly Sales, Kgs

Can conclude that weekly sales of Forgive chocolates: •

range from approximately 0 kg to less than 1200 kg

•

are concentrated between 400 kg and 800 kg

•

are rarely below 200 kg or above 1000 kg

(d) Scatter Diagram - Forgive Chocolates

$12,000 $10,000

Cost

$8,000 $6,000 $4,000 $2,000 $0 0

2000

4000

6000

8000

10000

12000

Weekly Quantity Sold, kgs

Strong positive linear relationship between quantity sold and cost. (e)

As have weekly sales for all 52 weeks in the year this is population data.

Population Data Mean Variance (Pop) Standard Deviaiton (Pop) (f)

Forgive Rejoice 564.873 495.638 47778.799 66121.428 218.584 257.141

Weekly sales in the previous year were higher and less varied for Forgive chocolates than for Rejoice chocolates. However, as the population standard deviation is large for both products, we can conclude that the weekly quantity sold for each product was highly variable.

(g) and (h)

Forgive Within 1 standard deviation Within 2 standard deviations Within 3 standard deviations

Lower Value 346.29 127.71 -90.88

Upper Value 783.46 1002.04 1220.62

Number 35 51 52

Percentage 67.31% 98.08% 100.00%

Rejoice Within 1 standard deviation Within 2 standard deviations Within 3 standard deviations

Lower Value 238.50 -18.64 -275.78

Upper Value 752.78 1009.92 1267.06

Number 41 50 52

Percentage 78.85% 96.15% 100.00%

Last year, for the regional city store, sales were on average 564.9 kg and 495.6 kg for Forgive and Rejoice chocolates respectively. Furthermore, in a typical week sales were between 346.3 kg and 783.5 kg for Forgive chocolates and between 238.5 kg and 752.8 kg for Rejoice chocolates. The percentage of weeks with sales within one, two, and three standard deviations of the mean approximately follows the empirical rule for Forgive chocolates. Therefore, distribution of the quantity sold weekly of Forgive chocolates may be mound shaped Copyright © 2019 Pearson Australia (a division of Pearson Australia Group Pty Ltd) 9781488617249, Berenson, Basic Business Statistics 5e

with most weeks’ sales being close to the mean of 564.9 kg, and a few weeks having very low or high sales. However, Rejoice chocolates have a larger percentage of weeks within one standard deviation than that given by the empirical rule so may not be mound shaped. As all values are within three standard deviations of the mean there are no outliers. However, there is one possible outlier for Forgive, sales of 1039.2 kg in week 1 (New Year), and two for Rejoice, sales of 1031.4 kg and 1056.3 kg in weeks 49 and 51 (leading up to Christmas). As you would expect sales of Rejoice chocolates to increase over the Christmas period, and possibly Forgive chocolates are needed after New Year’s parties, these quantities are probably not outliers. (i)

Forgive

r = 0.9620K

Rejoice:

r = 0.9701K

For both products there is a very strong positive linear relationship between weekly quantity sold and associated costs.

r = −0.3420K There is a weak negative linear relationship between Forgive and Rejoice chocolates. When sales are high for one, sales tend to be low for the other. (j) Sweets-4-U Forgive

Rejoice

Weekly Quantity Sold, kg

1200 1000 800 600 400

200 0 1

Week

For Forgive chocolates there is no overall pattern in weekly sales during the year. Thus, management is unable to predict which weeks will have high sales and which will have low. However, there is a definite pattern in sales for Rejoice chocolates, with sales generally about 400 kg a week, with increases to approximately 1000 kg a week close to Valentines’ day, Easter and the weeks leading up to Christmas.

A.24

A.25

New Zealand

9.0

Offical Cash Rate %

8.0

7.0 6.0 5.0 4.0 3.0 2.0 0.0

1.0

2.0

3.0

4.0

5.0

6.0

Annual CPI %

rNew Zealand = 0.360K Australia 8.0

Target Cash Rate, %

7.0 6.0 5.0 4.0 3.0 2.0 0.0

1.0

2.0

3.0

4.0

5.0

6.0

7.0

Annual CPI, %

rAustralia = 0.5055K In both countries there is an extremely weak positive linear relationship between interest rates and inflation as measured by change in CPI.

A.26 (a)

Annual Change Real Gross Domestic Product 4.0

2.0

Mar-17

Mar-16

Mar-15

Mar-14

Mar-13

Mar-12

Mar-11

Mar-10

Mar-09

Mar-08

Mar-07

Mar-06

Mar-05

Mar-04

Mar-03

Mar-02

-2.0

Mar-01

0.0

Mar-00

Annual Averge % Change

6.0

-4.0 -6.0 NZ

(b)

Australia

USA

Japan

Other than Australia all countries had negative growth in GDP between 2008 and 2010. In general Japan’s GDP growth rate is less than that of the other countries. Australia has the least varied change in GDP with rate of growth generally between 2 and 4%.

A.27 (a)&(c)

Solar Power Generated per Day 25

0 15 Apr 15 May 14 Jun 14 Jul 13 Aug 12 Sep 12 Oct 11 Nov 11 Dec 10 Jan 09 Feb 11 Mar

The solar power generated per day varies from less than 5KW to more than 20KW. As expected in general there is less power generated per day in the winter months than during summer. However, this is extremely variable, due to changes in the weather.

Total Electricity Used Per Day 35 30 25

20 15

10 5 0 15 Apr 15 May 14 Jun 14 Jul 13 Aug 12 Sep 12 Oct 11 Nov 11 Dec 10 Jan 09 Feb 11 Mar

Total electricity use varies from less than 5 KW per day, probably periods when Alex and Tyler were away from home to over 25KW on a few days, probably due to air conditioning use during heat waves. Generally use is between 15 and 25 KW June to August or between 10 and 20KW September to May. Electricity Generation, Use and Destination

Fed into Grid Total Electricity Use Grid Use Solar Use Solar Generate

KW per Day

The amount of power generated by the solar panels is skewed to the left, while the total electricity use is approximately symmetric, as is the amount both used from and fed into the grid.

(b) and (c)

Five-Number Summary

Minimum First Quartile Median Third Quartile Maximum Range Interquartile Range

Solar Generate KW 2 12 16 18 22 20 6

Solar Generate KW Mean Standard Error Median Mode Standard Deviation Sample Variance Kurtosis Skewness Coefficcient of Variation Count

14.852174 0.229804 16 19 4.268415 18.219363 -0.047013 -0.652419 28.74% 345

Fed into Grid, KW 0 7 10 13 21 21 6

Solar Use, KW 5.095652 0.148008 5 4 2.749124 7.557685 2.287017 1.171238 53.95% 345

Use from Solar, KW 1 3 5 6 17 16 3

Total Electricity Use, KW 2 11 14 17.5 34 32 6.5

Use from Grid, KW 1 7 9 11 21 20 4

Grid Use, KW

Total Electricity Use, KW

Fed into Grid, KW

9.118841 0.214728 9 7 3.988402 15.907347 0.573945 0.308477 43.74% 345

14.214493 0.291523 14 13 5.414807 29.320138 0.517167 -0.118009 38.09% 345

9.756522 0.230281 10 11 4.277289 18.295197 -0.540388 -0.010537 43.84% 345

On 50% of days the solar panels generate at least 16KW of power, with a mean of 14.85KW. Of the electricity generated by the solar panels Alex and Tyler use an average of 5.10KW per day with on average 9.75 KW fed into the grid. A.28 (a) There were two populations. The first was all marketing executives and staff and their agents who worked for US organisations with 500 or more employees. The second was all IT staff who worked on mobile marketing for US organisations with 500 or more employees. (b)There were samples of 254 marketing executives and staff and 50 IT staff and analysts. (c)A parameter of interest is the proportion of marketing decision makers who are prioritising mobile apps. (d)The sample proportion of marketing decision makers who prioritise mobile apps was found to be 0.4. A.29 Before accepting the results of the poll you might want to know, for example: What was the radio station? From which demographic group does it draw its listeners? Copyright © 2019 Pearson Australia (a division of Pearson Australia Group Pty Ltd) 9781488617249, Berenson, Basic Business Statistics 5e

How were listeners selected for the poll? Did they self select by responding to a request to phone or send an SMS to the station? Did a polling company ring a large number of people and only record the responses of those who listen to the radio station? What was the operational definition of driving used?Did it include situations such as being stopped at traffic lights or in a traffic jam? Was the frequency of mobile use while driving mentioned? What other questions were asked? Were they clear, accurate, unbiased and valid? What was the response rate? A.30 (a) Row 16: 2323 6737 5131 8888 1718 0654 6832 4647 6510 4877 Row 17: 4579 4269 2615 1308 2455 7830 5550 5852 5514 7182 Row 18: 0989 3205 0514 2256 8514 4642 7567 8896 2977 8822 Row 19: 5438 2745 9891 4991 4523 6847 9276 8646 1628 3554 Row 20: 9475 0899 2337 0892 0048 8033 6945 9826 9403 6858 Row 21: 7029 7341 3553 1403 3340 4205 0823 4144 1048 2949 Row 22: 8515 7479 5432 9792 6575 5760 0408 8112 2507 3742 Row 23: 1110 0023 4012 8607 4697 9664 4894 3928 7072 5815 Row 24: 3687 1507 7530 5925 7143 1738 1688 5625 8533 5041 Row 25: 2391 3483 5763 3081 6090 5169 0546 Note: All sequences above 5000 are discarded. There were no repeating sequences. (b) 089 189 289 389 489 589 689 789 889 989 1089 1189 1289 1389 1489 1589 1689 1789 1889 1989 2089 2189 2289 2389 2489 2589 2689 2789 2889 2989 3089 3189 3289 3389 3489 3589 3689 3789 3889 3989 4089 4189 4289 4389 4489 4589 4689 4789 4889 4989 (c)With the single exception of invoice #0989, the invoices selected in the simple random sample are not the same as those selected in the systematic sample. It would be highly unlikely that a random process would select the same units as a systematic process.

End of Part 2 problems B.1 (a) P(unacceptable) unacceptable)

= P(machine I AND unacceptable) + P(machine II AND = 0.01 + 0.025 = 0.035

(b) P(machine I OR acceptable) = P(machine I) + P(acceptable) – P(machine I AND acceptable) = 0.5 + 0.965 – 0.49 = 0.1065 (c) P(unacceptable | machine I) = 0.01/0.5 = 0.02 (d) P(machine I | unacceptable) = 0.01/0.035 =0.2857… (e) The conditions are switched. Part (c) is an example of a posteriori probability or P(A|B) and part (d) is an example of a priori (Bayesian) probability or P(B|A). B.2 (a)(i) 𝑃(1.90 < 𝑋 < 2.00) = 𝑃(

1.90−2 0.05

<𝑍<

2.00−2 0.05

)

= 𝑃(−2.00 < 𝑍 < 0) = 0.4772 (ii)

𝑃(1.90 < 𝑋 < 2.10) = 𝑃(

1.90 − 2 2.10 − 2 <𝑍< ) 0.05 0.05

= 𝑃(−2.00 < 𝑍 < 2.00) = 0.9772 − 0.0228 = 0.9544 (iii)𝑃(𝑋 < 1.90) + 𝑃(𝑋 > 2.10) = 1 − 𝑃(1.90 < 𝑋 < 2.00) = 0.0456 (b) P( X  A) = P( Z  −2.33) = 0.99 𝐴 = 2.00 − 2.33(0.05) = 1.8835 (c) P( A  X  B ) = P( −2.58  Z  2.58 ) = 0.99

A = 2.00 − 2.58( 0.05) = 1.8710; B = 2.00 + 2.58( 0.05) = 21290 . ;

Commented [JW1]: Removed b in front of X

B.3 (a) (i) 𝑃(1.90 < 𝑋 < 2.00) = 𝑃(

1.90 − 2.02 2.00 − 2.02 <𝑍< ) 0.05 0.05

= 𝑃(−2.40 < 𝑍 < −0.40)

Commented [JW2]: Changed sign

= 0.3446 − 0.0082 = 0.3364 (ii)

P (1.90  X  2.10) = 𝑃(

1.90 − 2.02 2.10 − 2.02 <𝑍< ) 0.05 0.05

= 𝑃(−2.40 < 𝑍 < 1.60) = 0.9452 − 0.0082 = 0.9370 (iii) P (X  1.90) + P (X  2.10)

= 1 − P (1.90  X  2.00) = 0.0630 (b) 𝑃(𝑋 > 𝐴) = 𝑃(𝑍 > −2.33) = 0.99 𝐴 = 2.02 − 2.33(0.05) = 1.9035 (c) 𝑃(𝐴 < 𝑋 < 𝐵) = 𝑃(−2.58 < 𝑍 < 2.58) = 0.99 𝐴 = 2.02 − 2.58(0.05) = 1.8910 ; 𝐵 = 2.02 + 2.58(0.05) = 2.1490 B.4 (a) 27 = 128 (b) 67 = 279 936 (c) There are two mutually exclusive and collectively exhaustive outcomes in (a) and six in (b). B.5 1

𝜆

Let X = inter-arrival time, minutes. X is exponential with 𝜇 = = 6 minutes, so 𝜆 = minute

per

(a) P(X < 2) = 1 − 𝑒 −(6)×2 = 1 − 0.71653 … = 0.28346 … 1

(b) P(X > 10) = 𝑒 −(6)×10 = 0.18887 … (c) P ( 4  X  6 ) = P(X  6) − P(X  4)

= (1 − e−(1/ 6)6 ) − (1 − e−(1/ 6)4 ) = 0.6321

− 0.4865

= 0.14553

B.6

(a)(i)

Electricity Mean 147.06 Standard Error 4.48183727 Median 148.5 Mode 130 Standard Deviation 31.6913753 Sample Variance 1004.34327 Five-Number Summary Kurtosis -0.5441632 Minimum 82 Skewness 0.01584564 First Quartile 127 Range 131 Median 148.5 Minimum 82 Third Quartile 168 Maximum 213 Maximum 213 Sum 7353 Interquartile Range 41 Count 50 Range 131 Boxplot - Electricity Cost July

Electricity

$80

$100

$120

$140

$160

$180

$200

$220

From the boxplot electricity cost for July appears to be mound shaped and approximately symmetrical. The mean is very close to the median. The interquartile range is very close to 1.33 standard deviations. The range is about $50 below 6 times the standard deviation. In general, the distribution of the data appears to closely resemble a normal distribution. (ii) The normal probability plot below confirms that the data appear to be approximately normally distributed.

Electricity Cost 250

Electricity

200 150 100 50 0 -3

-2

-1

Z Value

(b) Let

X = Electricity Cost July, $

Given X is assumed normal with mean  = $147 and standard deviation  = $31.70 (i)

(ii)

120 − 147   P(X  120) = P  Z   31.70   = P(Z  −0.85) = 0.1977

120

147

− 0.85

160 − 147   100 − 147 P(100  X  160) = P  Z  31.70   31.70 = P(−1.48  Z  0.41) = P(Z  0.41) − P(Z  −1.48) = 0.6591 − 0.0694 = 0.5897

(iii)

100

147

160

−1.48

0.41

225 − 147   P ( X  225) = P  Z   31.70   = P(Z  2.46) = 1 − P(Z  2.46) = 1 − 0.9931 = 0.0069

147

225

2.46

P(X > x) = P(Z > z ) = 0.1

P(X ≤ x) = P(Z ≤ z) = 0.9

147

Find z such that P(Z ≤ z) = 0.9, from tables z = 1.28 gives closest area of 0.8997 Therefore for 10% of two-bedroom apartments the electricity cost for July is above $187.58 since X =  + z = 147 + (1.28×31.70) = 147 + 40.576 = $187.576

(d)

Want x1 and x2 such that P(x1 < X < x2) = 0.95 so P(X ≤ x1) = 0.025 and P(X ≥ x2) = 0.025

P(x1 < X < x2) = P(-z < Z < z) = 0.95 P(X ≤ x1) = P(Z ≤ -z) = 0.025

P(X≥ x2) = P(Z ≥ z) = 0.025

147

-z

Find z such that P(Z ≤ -z) = 0.025, from tables -z = -1.96 gives area of 0.025. By symmetry z = 1.96, that is, P(Z ≥ 1.96) = 0.025 Therefore cost of electricity for the middle 95% of two-bedroom apartments is between $84.87 and $209.13 since X =  + z = 147 + (-1.96×31.70) = 147 – 62.132 = $84.868 X =  + z = 147 + (1.96×31.70) = 147 + 62.132 = $209.132

Normal Probabilities Common Data Mean 147 Standard Deviation 31.7 Probability for X <= X Value 120 Z Value -0.851735 P(X<=120) 0.1972 Probability for X > X Value 225 Z Value 2.4605678 P(X>225) 0.0069 Probability for X<120 or X >225 P(X<120 or X >225) 0.2041 Probability for a Range From X Value 100 To X Value 160 Z Value for 100 -1.48265 Z Value for 160 0.4100946 P(X<=100) 0.0691 P(X<=160) 0.6591 P(100<=X<=160) 0.5900 Find X and Z Given Cum. Pctage. Cumulative Percentage 90.00% Z Value 1.2816 X Value 187.6252 Find X Values Given a Percentage Percentage 95.00% Z Value -1.96 Lower X Value 84.87 Upper X Value 209.13 B.7 Let X = number who use a credit card. The X has binomial distribution with n = 15 and p = 0.55. (a) (𝑖)𝑃(𝑋 = 0) =

15! 0!(15−0)!

(0.55)0 (1 − 0.55)15−0 = 0.0000 …

(𝑖𝑖)𝑃(𝑋 = 5) =

15! (0.55)5 (1 − 0.55)15−5 = 0.05146 … 0! (15 − 5)!

(𝑖𝑖𝑖)𝑃(𝑋 > 2)

= 1 − (𝑃(𝑋 = 0) + 𝑃(𝑋 = 1) + 𝑃(𝑋 = 2)) = 1 − (0.000006 + 0.000115 + 0.000986) = 0.998893

(b) 𝐸(𝑋) = 15 × 0.55 = 8.25 𝑆𝐷(𝑋) = √15 × 0.55 × 0.45 = 1.92678 … Copyright © 2019 Pearson Australia (a division of Pearson Australia Group Pty Ltd) 9781488617249, Berenson, Basic Business Statistics 5e

B.8

    0.9048 − 0.92  (a) P(p < 0.9048) = P  Z   ( 0.92 )( 0.08)   42   = P ( Z < -0.3631) = 0.3594     0.86 − 0.92 0.94 − 0.92  (b) P( 0.86 < p < 94) = P  Z   (0.92) ( 0.08 ) (0.92)(0.08)    50 50   = P ( -1.5639 < Z < 0.5213) = 0.6985 – 0.0594 = 0.6391 B.9 Excel output: Data Sample size

No. of successes in population

Population size

Hypergeometric Probabilities Table X

P(X)

0.08525

0.26616

0.33270

0.21664

0.08001

0.01707

2.041E-03

1.241E-04

2.908E-06

Let X = number of defective pots (a) (𝑖)𝑃(𝑋 = 8) = 2.908 × 10−6 (𝑖𝑖)𝑃(𝑋 = 0) = 0.08525 (𝑖𝑖𝑖)𝑃(𝑋 ≥ 1) = 1 − 0.08525 = 0.91475

(b) Excel output: Data Sample size

No. of successes in population

Population size

Hypergeometric Probabilities Table X

P(X)

0.32093

0.41918

0.20540

0.04833

0.00581

3.507E-04

9.742E-06

1.012E-07

2.259E-10

𝑃(𝑋 = 8) = 2.259 × 10−10 𝑃(𝑋 = 0) = 0.32093 𝑃(𝑋 ≥ 1) = 1 − 0.32093 = 0.67907 B.10 (a) ‘Is listed over 90 days’, ‘asking price under $200,000’ etc. (b) ‘Is listed over 90 days and asking price under $200,000’, ‘Is listed over 90 days or asking price under $200,000’ etc. (c) ‘Asking price $200,000 or higher’. (d) A house may have an asking price under $200,000 and be listed more than 90 days before it is sold. (e) 𝑃(𝑙𝑖𝑠𝑡𝑒𝑑 > 90 𝑑𝑎𝑦𝑠 |𝑎𝑠𝑘𝑖𝑛𝑔 𝑝𝑟𝑖𝑐𝑒 < $200, 000) = (f) 𝑃(𝑎𝑠𝑘𝑖𝑛𝑔 𝑝𝑟𝑖𝑐𝑒 < $200, 000 |𝑙𝑖𝑠𝑡𝑒𝑑 > 90 𝑑𝑎𝑦𝑠) =

10 100 10 190

= 0.10 = 0.0526 …

(g) The conditional events are reversed; in (a) we know the asking price, while in (b) we know the number of days until sold. (h) Not independent as P(listed | asking price < $200,000) = 0.10 and P(listed > 90 days) = 190/800 = 0.2375. (i) (𝑖)𝑃(𝑙𝑖𝑠𝑡𝑒𝑑 > 90 𝑑𝑎𝑦𝑠) =

190 800

= 0.2375

(𝑖𝑖)(𝑃(𝑎𝑠𝑘𝑖𝑛𝑔 𝑝𝑟𝑖𝑐𝑒 ≥ $400, 000) =

50 = 0.0625 800

(𝑖𝑖𝑖)𝑃(𝑎𝑠𝑘𝑖𝑛𝑔 𝑝𝑟𝑖𝑐𝑒 ≥ $400, 000 𝒂𝒏𝒅 𝑙𝑖𝑠𝑡𝑒𝑑 > 90 𝑑𝑎𝑦𝑠) =

10 = 0.0125 800

(𝑖𝑣)𝑃(𝑎𝑠𝑘𝑖𝑛𝑔 𝑝𝑟𝑖𝑐𝑒 ≥ $400, 000 𝒐𝒓 𝑙𝑖𝑠𝑡𝑒𝑑 > 90 𝑑𝑎𝑦𝑠) =

50 + 190 − 10 = 0.2875 800

(j) The probability of ‘initial asking price is over $400,000’ or ‘listed over 90 days before being sold’ includes the probability of ‘initial asking price is over $400,000’ and ‘listed over 90 days before being sold’. In (i) and (iii) both events must happen while in (i) and (iv) only one of the events needs to happen. B.11 (a) (i) 𝐸(𝑋) = $71; 𝐸(𝑌) = 97 (ii) 𝜎𝑋 = 61.88; 𝜎𝑌 = 84.27 (𝑖𝑖𝑖)𝜎𝑋𝑌 = 5113 (b) Share Y gives the investor a higher expected return than share X, but also has a higher standard deviation. A risk-averse investor should invest in share X, but an investor willing to sustain a higher risk can expect a higher return from share Y. B.12 (a)(𝑖)𝐸(𝑃) = $89.20; 𝜎𝑃 = 77.28 𝐶𝑉 =

𝜎𝑃 77.28 = × 100 = 86.64% 𝐸(𝑃) 89.20

(𝑖𝑖)𝐸(𝑃) = $84.00; 𝜎𝑃 = 72.73 𝐶𝑉 =

𝜎𝑃 72.73 = × 100 = 86.58% 𝐸(𝑃) 84.00

(𝑖𝑖𝑖)𝐸(𝑃) = $78.80; 𝜎𝑃 = 68.28 𝐶𝑉 =

𝜎𝑃 68.28 = × 100 = 86.65% 𝐸(𝑃) 78.80

(b) Based on the results of (i)–(iii), you should recommend a portfolio with 50% of share X and 50% of share Y because it has the lowest risk per unit average return as measured by the coefficient of variation. B.13 (a) 𝑃(𝑋 < 20) =

20−10 25−10

= 0.6667

(b) 𝑃(20 < 𝑋 < 22) =

22−10

25−18

(d) 𝜇 =

10+25 2

= 0.1333

25−10

= 0.4667

25−10

(25−10)2

= 17.5, 𝜎 = √

= 4.3301

B.14 (a)

‘Student is studying full time.’

(b)

‘Student is studying part time and is not employed.’

(c)

‘Not employed full time’, alternatively ‘either employed part time or not employed’.

(d)

A student can be employed full time and also be studying full time.

(e)

(i) P(employed) = P(employed full-time) + P(employed part-time)

652 482 + 1500 1500 1134 = 1500 = 0.756

(ii) P(employed and studying part-time) = (iii) P(employed or studying part-time) = (f)

558 190 748 + = = 0.4986... 1500 1500 1500

1134 664 748 1050 + − = = 0.70 1500 1500 1500 1500

The probability of ‘studying part time or is employed’ includes the probability of ‘studying part time and not employed’ as well as ‘studying full time and employed’ and ‘studying part time and employed’.

B.15 (a) 𝑃(𝑋 ≤ 0.05) = 1 − 𝑒 −(15)(0.05) = 0.5276 (b) 𝑃(𝑋 ≤ 0.25) = 0.9765 (c) If 𝜆 = 25, 𝑃(𝑋 ≤ 0.05) = 0.7135, 𝑃(𝑋 ≤ 0.25) = 0.9981 B.16

(a)

630 − 600.5   P( X  630) = P  Z   76 / 52   = P( Z > 2.7990) = 1 – 0.9974 = 0.0026

(b) 575 − 600.5   P( X  575) = P  Z   76 / 52   = P( Z  −2.4195) = 0.0078 B.17 45 C7 =

45! 45! = = 43,379,620 7!(45 - 7)! 7!38!

𝑃(𝑓𝑖𝑟𝑠𝑡 𝑑𝑖𝑣𝑖𝑠𝑖𝑜𝑛) =

1 = 2.20 … × 10−8 45,379,620

B.18 (a)

P(identified X-ray) =

50 = 0.1 = 10% 500

(b)

P(found in search | identified X-ray) =

36 = 0.72 50

(c)

P(identified X-ray | found in search) =

36 = 0.75 48

(d)

P(not identified X-ray and not found in search) =

438 = 0.876 = 87.6% 500

B.19 Let X = number of packages with illegal or threatening items X is binomial with n = 10 and p = 0.096 (a) (𝑖) 𝑝 = 0.096, 𝑛 = 10, 𝑃(𝑋 = 2) = 10𝐶2 (0.096)2 (1 − 0.096)8 = 0.1850

Commented [JW3]: There should not be boxes showing above 10 in each part of this answer

(𝑖𝑖) 𝑝 = 0.096, 𝑛 = 10, 𝑃(𝑋 = 0) = 10𝐶0 (0.096)0 (1 − 0.096)10 = 0.3645 (iii) 𝑝 = 0.096, 𝑛 = 10, 𝑃(𝑋 ≥ 1) = 1 − 𝑃(𝑋 = 0) = 0.6355 (𝑖𝑣) 𝑝 = 0.096, 𝑛 = 10, 𝑃(𝑋 > 5) = 𝑃(𝑋 = 6) + 𝑃(𝑋 = 7) + ⋯ + 𝑃(𝑋 = 10) = 10𝐶6 (0.096)6 (1 − 0.096)4 + 10𝐶7 (0.096)7 (1 − 0.096)3 + ⋯ + 10𝐶10 (0.096)10 (1 − 0.096)0 = 0.00011 (b) 𝐸(𝑋) = 𝑛 × 𝑝 = 10 × 0.096 = 0.96, 𝜎 = √𝑛 × 𝑝 × (1 − 𝑝) = √10 × 0.096 × (1 − 0.096) = 0.9315 B.20 (a)(i)

P(female) =

163 = 0.49244 331 97 = 0.57738 168

(ii)

P( senior lecturer|male) =

(iii)

P(female and associate lecturer) =

(iv)

P(female|professor) =

(v)

P(associate professor) =

23 = 0.06948 331

13 = 0.38235 34 40 = 0.12084 331

(b) No, as conditional probability is not equal to unconditional probability. For example,

P(female|professor)  P(female)

B.21 Hypergeometric with N = 163 and n = 5 (a)

A = 52 (M = male)

 52 111    5 0  P(M = 5) =   = 0.00288... 163     5  (b)

A = 34 (P = Professor)

 34 129     0 5  P(P = 0) =   = 0.30537 K 163     5  (c)

A = 66 (F = Female)

 66  97   66  97   66  97           3 2 4 1 5 0 P(F  3) =    +    +    = 0.32387 K 163  163  163         5   5   5  (d)

A = 34 (P = Professor)

 34 129     1 4  P(P = 1) =   = 0.41531K 163     5  (e)

A = 40 (A = Associate Professor)

 40 123     3 2  P(A = 3) =   = 0.08224 K 163     5  B.22 (a) 𝑃(𝑡𝑖𝑚𝑒 𝑡𝑖𝑙𝑙 𝑛𝑒𝑥𝑡 𝑖𝑛𝑗𝑢𝑟𝑦 ≤ 10) = 1 − 𝑒 −(0.1)(10) = 0.6321 (b) 𝑃(𝑡𝑖𝑚𝑒 𝑡𝑖𝑙𝑙 𝑛𝑒𝑥𝑡 𝑖𝑛𝑗𝑢𝑟𝑦 ≤ 5) = 0.3935 (c) 𝑃(𝑡𝑖𝑚𝑒 𝑡𝑖𝑙𝑙 𝑛𝑒𝑥𝑡 𝑖𝑛𝑗𝑢𝑟𝑦 ≤ 1) = 0.0952

B.23 Contingency table Male

Female

Single

207

225

432

Married

309

459

768

516

684

1200

(a) 𝑃(𝑚𝑎𝑙𝑒) =

516 1200

(b) 𝑃(𝑠𝑖𝑛𝑔𝑙𝑒) =

= 0.43

432 1200

= 0.36

309 1200 225 1200

= 0.2575 = 0.1875

B.24 X

P(x)

X*P(X)

(X-𝝁)2

(X-𝝁)2*P(X)

0.10

0.00

0.40

0.20

0.45

0.90

0.00

0.15

0.45

0.15

0.05

0.20

0.05

0.25

0.45

(a) Mean = 2.00, Variance = 1.40 (b) Standard deviation = 1.18321596 B.25 (a)(i) Let X = number of customers joining a queue in a minute X is a Poisson random variable with

P(X = 4) = (ii)

=

108 = 1.8 customers per minute. 60

e−1.81.84 = 0.0723K 4!

P(X  1) = 1 − P(X = 0) = 1 − 0.165 = 0.835

(b)(i) Let Y = number of customers joining a queue in 5 minutes Y is a Poisson random variable with  =

P(Y = 10) = (ii)

108 = 9 customers per 5 minutes. 12

e−9 910 = 0.11858K 10!

P(Y  10) = 1 − P(Y  10) = 1 − 0.5874 = 0.4126 Poisson Probabilities Data Average/Expected number of successes: 9 Poisson Probabilities Table X P(X) P(<=X) P(<X) 0 0.000123 0.000123 0.000000 1 0.001111 0.001234 0.000123 2 0.004998 0.006232 0.001234 3 0.014994 0.021226 0.006232 4 0.033737 0.054964 0.021226 5 0.060727 0.115691 0.054964 6 0.091090 0.206781 0.115691 7 0.117116 0.323897 0.206781 8 0.131756 0.455653 0.323897 9 0.131756 0.587408 0.455653 10 0.118580 0.705988 0.587408

B.26 Bayes’ Theorem Let

A = technician A, B = technicianB

S = Problem solved

P(A) = P(B) = 0.5

Given

(a)

P(S | A) = 0.95

P(S' | A) = 0.05

P(S | B) = 0.85

P(S' | B) = 0.15

P(S | A)P(A) P(S | A)P(A) + P(S | B)P(B) 0.95  0.5 = ( 0.95  0.5) + ( 0.85  0.5)

P(A | S) =

0.475 0.9 = 0.5277K =

(b)

P(S) = 0.9

(a)

Let W = number of wins in game = 0,1,2 binomial with p = 0.01 , n = 2

B.27

(b)

X = number of wins = 0,1,2,K ,40 binomial with p = 0.01 , n = 40

P(player wins at least once) = 1 − P(X = 0) = 1 − 0.6689K = 0.3310K (c)

For the table of 5 players, let

Y = number win at least once = 0,1,2, K ,5

binomial with

n = 5 , p = 0.331

(approximate)

P(Y = 4) = 5 C4 ( 0.331) ( 0.669 ) = 0.0401K 4

(d)

P(Biff does not win but 4 friends do)  (0.331)

(e)

Let G = amount won in game Win

Line

House

(0.669)1 = 0.0080  Both

Neither

$10

$20

$30

P(G)

0.0099

0.0001

0.9801

(f)

E(G) = (10  0.0099) + ( 20  0.0099) + ( 30  0.0001) + ( 0  0.9801) = 0.3

(g)

2 G = 102  0.0099 + 202  0.0099 + 302  0.0001 + 02  0.9801 − 0.32

(

) (

)

= 4.95

G = 4.95 = 2.2248K (h)

P = Profit, $ E(P) = 20  E(G) − 8 = 20  0.3 − 8 = −$2

B.28 (n = 200, p = 0.4. 𝜇 = np = 80, np = 80 ≥ 5 and n(1-p) = 120 ≥ 5 𝜎 = √𝑛𝑝(1 − 𝑝) = 6.9282 (a) 𝑃((𝑋 ≥ 75) ≃ 𝑃(𝑋 ≥ 74.5) = 𝑃(𝑍 ≥ −0.7939) = 0.7864 (b) 𝑃(𝑋 ≤ 70) ≃ 𝑃(𝑋 ≤ 70.5) = 𝑃(𝑍 ≤ −1.3712) = 0.0852 (c) 𝑃(70 ≤ 𝑋 ≤ 75) ≃ 𝑃(69.5 ≤ 𝑋 ≤ 75.5) = 𝑃(−1.5155 ≤ 𝑍 ≤ −0.6495) = 0.1932 B.29 (a) The assumptions needed are: i. The probability that a golfer loses a golf ball in a given interval in a game is constant. ii. The probability that a golfer loses more than one golf ball in a very small interval is 0. iii. The probability that a golfer loses a golf ball is independent from interval to interval.

(b) (𝑖)𝜆 = 3.8, 𝑃(𝑋 ≥ 1) = 1 − 𝑃(𝑋 = 0) = 1 − 0.0224 = 0.9776 (𝑖𝑖)𝜆 = 3.8, 𝑃(𝑋 < 3) = 𝑃(𝑋 = 0) + 𝑃(𝑋 = 1) + 𝑃(𝑋 = 2) = 0.0224 + 0.0850 + 0.1615 = 0.2689 (𝑖𝑖𝑖)𝜆 = 3.8, 𝑃(𝑋 > 5) = 1 − 𝑃(𝑋 ≤ 5) = 1 − 0.8152 = 0.1848 B.30

(a) np = 300(0.274) = 82.2 > 5, n(1-p) = 300(0.726) = 217.8 > 5. Therefore it is appropriate to use the normal distribution.    0.3 − 0.274   P(p > 0.3) = P  Z   0.274(0.726)    300   = P( Z > 1.0097) = 1 - 0.8438 = 0.1562    0.31 − 0.346 0.36 − 0.346   Z (b) P(0.31 < p <0.36) = P   0.346(0.654) 0.346(0.654)    300 300   = P (-1.3108 < Z < 0.5098) = 0.6950 – 0.0951 = 0.5999    0.23 − 0.256   (c) P(p < 0.23) = P  Z   0.256(0.744)    300   = P( Z < -1.0319) = 0.1515 𝑃B.31 7

(a) 𝑃(2 𝑟𝑖𝑔ℎ𝑡 ℎ𝑎𝑛𝑑𝑒𝑑 𝑔𝑙𝑜𝑣𝑒𝑠) = × =

7 12

= 0.5833

(b) 𝑃(1 𝑟𝑖𝑔ℎ𝑡 ℎ𝑎𝑛𝑑𝑒𝑑 𝑎𝑛𝑑 1 𝑙𝑒𝑓𝑡 ℎ𝑎𝑛𝑑𝑒𝑑 𝑔𝑙𝑜𝑣𝑒) =

7 9

× + × =

7 18

= 0.3889

(d) (a) 𝑃(2 𝑟𝑖𝑔ℎ𝑡 ℎ𝑎𝑛𝑑𝑒𝑑 𝑔𝑙𝑜𝑣𝑒𝑠) = × =

= 0.6049,

(b) 𝑃(1 𝑟𝑖𝑔ℎ𝑡ℎ𝑎𝑛𝑑𝑒𝑑 𝑎𝑛𝑑 1 𝑙𝑒𝑓𝑡 ℎ𝑎𝑛𝑑𝑒𝑑 𝑔𝑙𝑜𝑣𝑒) = × + × = 0.3457 B.32 (a) The 60% figure is most likely obtained by examining records of previous years, both number of customers and sales and, hence, is an empirical classical probability. If the stall owner estimates the figure then it should be classified as a subjective probability.

(b) Let X = number of showbags sold. Then X is binomial with n = 100 and p = 0.6. 𝐸(𝑋) = 100 × 0.6 = 60 𝜎 = √100 × 0.6 × 0.4 = 4.8989 … (c) Expected revenue = 12 × 𝐸(𝑋) = 12 × 60 = $720 (d) The assumptions needed are i.

The 100 visitors are each equally likely to buy a showbag.

ii. Each visitor’s decision to buy or not to buy a show bag is independent of the decision of any other visitor. B.33 (a) (i)–(iv) Portion of the PHStat output:

(b) The probability is essentially zero that all ten automatically opt to talk to a live operator. Hence, the 40% figure given in the article does not appear to apply to this particular system. B.34 (a)

Let X = number of years S&P 500 finish higher given higher in first 5 days Can use either binomial distribution exact or sampling distribution of sample proportion, P approximate Assume X is binomial with n = 59 (i)

If p = 0.50 and n = 59, P(X  41) = 0.001896...

(ii)

If p = 0.70 and n = 59, P(X  41) = 0.5967

(iii)

If p = 0.90 and n = 59, P(X  41) = 0.9999

Alternatively, using sampling distribution of sample proportion, P approximate Let  = proportion of years S&P 500 finish higher given higher in first 5 days Sample proportion p = (i)

41 = 0.6949... 59

If  = 0.50 and n = 59,

   41 0.6949... − 0.5      P  P  |  = 0.5  = P Z  59 0.5  0.5       59   = P(Z  2.99) = 0.0014 (ii)

If  = 0.70 and n = 59,

   41 0.6949... − 0.7      P  P  |  = 0.7  = P Z  59 0.7  0.3       59   = P(Z  −0.09) = 0.5359 (iii)

If  = 0.90 and n = 59,

   41 0.6949... − 0.9      P  P  |  = 0.9  = P Z  59 0.9  0.1       59   = P(Z  −5.25)  1.00 (b)

Based on the results in (a), the probability that the Standard & Poor’s 500 Index will increase if there is an early gain in the first five trading days of the year is closer to 0.7 than to either 0.5 (too low with a probability of 0.0019 or 0.0014 annual increase in at least 41 out of 59 years) or 0.9 (too high with a probability of approximately 1 annual increase in at least 41 out of 59 years).

B.35 (a) NI = Net increase, NO = No Increase in employment, TE = Telecommunications industry, OI = Other industries P(NI | TE) = 53%; P(TE) = (264/1764) = 15%; P(NI | OI) = 43%; P(OI) = (1764-264)/1764 = 85%

P(NI|TE) =

P(NI|TE)P(TE) 0.53  0.15 0.0795 = = = 17.9% P(NI|TE)P(TE) + P(NI|OI)P(OI) (0.53  0.15) + (0.43  0.85) 0.445

Commented [JW4]: I added spaces before each of the second Ps to make more easily read

(b) P(not from telecomm. nor expects an increase) = Sample Space

855 1764

= 48.5%

Event B

Event A

telecom

non telecom

Totals

140

645

785

increase no increase

124

855

979

Totals

264

1500

1764

B.36 (a) 8𝐶2 =

8! 2!(8−2)!

8! 2!6!

(b) 𝑃(𝑤𝑖𝑛 𝑞𝑢𝑖𝑛𝑒𝑙𝑙𝑎) =

Commented [JW5]: remove box above 18

= 28 1

= 0.0357 …

B.37 Let X = number of defective items Then X is binomial with n = 25 and p = 0.001 (a) (𝑖)𝑃(𝑋 = 2) =

25! 0!(25−2)!

(0.001)2 (1 − 0.999)25−2 = 0.00029 …

(𝑖𝑖)𝑃(𝑋 ≤ 1) = 𝑃(𝑋 = 0) + (𝑃𝑋 = 1) = 0.9752 … + 0.0244 … = 0.9997 … (𝑖𝑖𝑖)𝑃(𝑋 ≥ 2) = 1 − (𝑃(𝑋 = 0) + 𝑃(𝑋 = 1)) = 1 − 0.9997 … = 0.000295 (b) 𝐸(𝑋) = 25 × 0.001 = 0.025 𝜎 = √25 × 0.001 × 0.999 = 0.1580 … B.38 (a) 𝑃(𝑋 = 0) = 0.0907 (b) 𝑃(𝑋 = 1) = 0.2177 (c) 𝑃(𝑋 ≥ 2) = 0.6916 (d) 𝑃(𝑋 < 3) = 0.5697 B.39

n! = 6! = 720

B.40 (a) B = Base Construction Co. enters a bid O = Olive Construction Co. wins the contract 𝑃(𝐵′ |𝑂) =

𝑃(𝑂|𝐵′ ) ∙ 𝑃(𝐵′ ) 0.5 ∙ 0.3 = = 0.4615 𝑃(𝑂|𝐵 ) ∙ 𝑃(𝐵′ ) + 𝑃(𝑂|𝐵) ∙ 𝑃(𝐵) 0.5 ∙ 0.3 + 0.25 ∙ 0.7 ′

(b) 𝑃(𝑂) = 0.175 + 0.15 = 0.325 B.41

Given 𝜆 = 7.03, (a) 𝑃(𝑋 = 0) = 0.0009 (b) 𝑃(𝑋 ≥ 1) = 1 − 𝑃(𝑋 = 0) = 0.9991 (c) 𝑃(𝑋 ≥ 2) = 1 − 𝑃(𝑋 ≤ 1) = 1 − (0.0009 + 0.0064) = 0.9927

B.42 Let

A = Defect – incorrect fill amount B = Defect – incorrectly sealed

Given P(A) = 0.005, P(B) = 0.001, P(A and B) = 0.00002 (a)

P(A or B) = P(A) + P(B) − P(A and B) = 0.005 + 0.001 − 0.00002 = 0.00598

(b)

P(Not A or Not B) = 1 − P(A or B) = 1 − 0.00598 = 0.99402

(c)

(d)

P(B | A) =

P(A and B) 0.00002 = = 0.004 P(A) 0.005 A

Not A

Total

0.002

0.098

0.1

Not B

0.498

99.402

99.9

Total

0.5

99.5

100

Let X = number of bottles with incorrect fill amount As bottles randomly chosen, X is binomial with n = 20 and p = 0.005

20! 0.0051 (1 − 0.005)19 = 0.090915... 1!(20 − 1)!

(i)

P(X = 1) =

(ii)

P(X  1) = 1 − P(0) 20! 0.0050 0.99520 0!20! = 1 − 0.904610... =1−

= 0.09538... (iii)

P(X  2) = P(0) + P(1) + P(2) = 0.904610... + 0.090915... + 0.004340... = 0.999866...

Binomial Probabilities Data Sample size 20 Probability of an event of interest 0.005 Statistics Mean 0.1 Variance 0.0995 Standard deviation 0.3154 Binomial Probabilities Table X 0 1 2 3 (e)

P(X) 0.9046 0.0909 0.0043 0.0001

P(<=X) 0.9046 0.9955 0.9999 1.0000

P(<X) 0.0000 0.9046 0.9955 0.9999

P(>X) 0.0954 0.0045 0.0001 0.0000

P(>=X) 1.0000 0.0954 0.0045 0.0001

Let Y = number of bottles incorrectly sealed As bottles randomly chosen, Y is binomial with n = 100 and p = 0.001 Expected number bottles incorrectly sealed  = np = 100 × 0.001 = 0.1

B.43 (a) (𝑖)𝑃(𝑋̅ > 3) = 𝑃(𝑍 > −1.00) = 1 − 0.1587 = 0.8413 (𝑖𝑖)𝑃(𝑍 < 1.04) = 0.85,

𝑋̅ = 3.10 + 1.04(0.1) = 3.204 𝑚𝑖𝑛𝑢𝑡𝑒𝑠

(b) To be able to use the standard normal distribution as an approximation for the area under the curve, we must assume that the population is symmetrically distributed such that the central limit theorem will likely hold for samples of n = 16. (c) 𝑃(𝑋̅ < 𝐴) = 𝑃(𝑍 < 1.04) = 0.85 𝑋̅ = 3.10 + 1.04(0.05) = 3.152 B.44 (a) Waiting times seem to be exponential. This is expected, as most will wait a short time with only a few waiting longer times.

(b) Serving times seem to be exponential. This is expected, as most will cluster around the mean value with a few being either very quick or very long.

Serving time

Mean

6.10

60.00

Standard deviation

5.6509

13.3697

Assume X = waiting time, minutes is exponential with mean 𝜇 = 1/𝜆 = 6.1, so 𝜆 = 1/6.1 Assume Y = serving time, minutes is normal with mean 𝜇 = 60 𝜎 = 13.3697 5

(d) 𝑃(𝑋 < 5) = 1 − 𝑒 −6.1 = 0.5594 10

(e) 𝑃(𝑋 > 10) = 1 − 𝑒 −6.1 = 0.1941 (f) 𝑃(𝑌 < 60) = 0.5 (g) 𝑃(𝑌 > 90) = 𝑃(𝑍 > 2.24) = 1 − 0.9875 = 0.0125 B.45 (a)

Let M = number of deaths, motorcyclists per 5 million VKT M is Poisson with  =

6.47  5 = 0.3235 per 5 million VKT 100

e−0.3235 0.32350 = 0.723611 0!

(i)

P(M = 0) =

(ii)

P(M  1) = 1 − 0.72361= 0.27638

(iii)

P(M = 1) =

(iv)

P(M  2) = P(0) + P(1) + P(2)

e−0.3235 0.32351 = 0.23408 1!

= 0.72361... + 0.23408

+ 0.03786...

= 0.99556...

Data Mean/Expected number of events of interest: POISSON.DIST Probabilities Table X P(X) P(<=X) 0 0.7236 0.7236 1 0.2341 0.9577 2 0.0379 0.9956 3 0.0041 0.9996 4 0.0003 1.0000 (b)

0.3235 P(<X) 0.0000 0.7236 0.9577 0.9956 0.9996

P(>X) P(>=X) 0.2764 1.0000 0.0423 0.2764 0.0044 0.0423 0.0004 0.0044 0.0000 0.0004

Let C = number of deaths, car occupants per 300 million VKT C is Poisson with  = 0.35  3 = 1.05 per 300 million VKT

e−1.05 1.050 = 0.34993 0!

(i)

P(C = 0) =

(ii)

P(C  1) = 1 − 0.34993 e

−1.05

= 0.65006

1.05 = 0.36743 1!

(iii)

P(C = 1) =

(iv)

P(M  2) = P(0) + P(1) + P(2) = 0.34993... + 0.36743

+ 0.19290...

= 0.91027... Data Mean/Expected number of events of interest: POISSON.DIST Probabilities Table X P(X) P(<=X) 0 0.3499 0.3499 1 0.3674 0.7174 2 0.1929 0.9103 3 0.0675 0.9778 4 0.0177 0.9955

1.05 P(<X) 0.0000 0.3499 0.7174 0.9103 0.9778

P(>X) P(>=X) 0.6501 1.0000 0.2826 0.6501 0.0897 0.2826 0.0222 0.0897 0.0045 0.0222

B.46 Let Di = number with a disability age group i

D i is binomial with n = 15, p(15−24) = 0.082 , p(45−54) = 0.164 , p90 = 0.854 (a) (b)

P ( D(45−54) = 0 ) = 0.06809... ) ( P ( D(15−24)  1) = 1 − 0.27710... = 0.72289... P D(15− 24) = 0 = 0.2771...

(

P D

(45−54)

)

 1 = 1 − 0.06809... = 0.93190...

) ( P ( D(15− 24)  5) = 0.000793...

) ( P ( D(45−54)  5) = 0.02544...

P D(15− 24) = 5 = 0.00473...

P D(45−54) = 5 = 0.05940...

P (D90 = 0 )  0

P (D90  1)  1

(

)

P D90 = 5 = 0.000006...

)

P D90  5 = 0.9999...

B.47

X = number of calls answered

Let

binomial

with

n = 100 ,

p = 0.8

therefore

 = np = 100  0.8 = 80 and  = npq = 100  0.8  0.2 = 16 = 4 Using the normal approximation to the binomial (a)(i)

50.5 − 80   PB (X  50)  PN (X  50.5) = P  Z   = P(Z  −7.375)  1 4  

(ii) P (70  X  90)  P (69.5  X  90.5) B N

= P( −2.63  Z  2.63) = 0.9957 − 0.0043 = 0.9914 (iii) (b)

PB (X  75)  PN (X  74.5) = P(Z  −1.38) = 0.0838 PB (X  50)  1 PB (70  X  90) = 0.99160K , PB (X  75) = 0.08747K

B.48 Let X = download time, seconds X normal with  = 0.9 ,  = 0.3 (a) P(X  1) = P(Z  0.33) = 0.6293 (b) P(X  0.5) = P(Z  −1.33) = 1 − 0.0918 = 0.9082 (c) P(0.5  X  1.5) = P(−1.33  Z  2) = 0.9772 − 0.0918 = 0.8854 (d) P(X  2) = P(Z  3.67) = 1 − 0.99982 = 0.00018 (e) P(X  0.6) = P(Z  −1) = 0.1587 (f)

P(1.0  X  1.5) = P(0.33  Z  2) = 0.9772 − 0.6293 = 0.3479

B.49 Let X = time to next unplanned outage, years X exponential with  = 4 per year (a)(i)

P(X  1) = e−4(1/12) = 0.7165K

(ii)

P(X  3) = e−4(3/12) = 0.3678K

(iii)

P(X  6) = 1 − e−4(6 /12) = 0.8646K

(b)

3 months or 0.25 year

Let Y = number unplanned outages in year Poisson with  = 4 per year 12

(

)

P Y12 = 3 = 0.1954 (ii)

Let

Y6 = number unplanned outages in 6 months Poisson with  = 2 in 6-

months

(

)

P Y6  6 = 0.0045 (iii)

Let Y = number unplanned outages in 1 month Poisson with  = per month 1 3

(

)

(

) (

)

P Y1  2 = P Y1 = 0 + P Y1 = 1 = 0.7165K + 0.2388K = 0.9553K B.50

0.026 = 0.0722 0.36

(a)

P( $136,600|Not Enough) =

(b)

P( $35,700|Enough) =

(c)

P( $35,700 and Enough) = 0.096

(d)

P(Enough|  $35,700) =

(e)

P(Not Enough| $136,600) =

(f)

P( $136,600 and Not Enough) = 0.026

0.096 = 0.15 0.64

0.544 = 0.68 0.80

0.334 = 0.4175 0.80

<$35,700

$35,700 to $136,599

≥$136,600

Total

Enough

9.6

37.0

17.4

64.0

Not Enough

10.4

23.0

2.6

36.0

Total

20.0

60.0

20.0

100.0

B.51

(a)(i)&(b)

All Ords 2016-2017 90 80

Frequency

70 60 50 40 30 20 10

1650

1450

1550

1250

1350

1050

1150

850

950

650

750

450

550

250

350

150

Volume (millions)

All Ords 2016-2017

Volume (millions)

200

400

600

800

1000

1200

1400

1600

Both the Histogram and the Box-and-Whisker plot show that the daily volume traded in 2016-2017 financial year is mound shaped with a slight right skew. (a)(ii)&(b)

Volume (millions) Mean 810.94 Standard Error 10.8832 Median 786.688 Mode #N/A Standard Deviation 173.107 Sample Variance 29966.1 Kurtosis 4.10027 Skewness 1.05365 Range 1347.16 Minimum 187.923 Maximum 1535.08 Sum 205168 Count 253

Five-Number Summary Minimum 187.923 First Quartile 713.859 Median 786.688 Third Quartile 876.115 Maximum 1535.08 . Interquartile Range 162.256

•

The mean of 811 million is greater than the median of 787 million, as the distribution is slightly skewed to the right.

•

The interquartile range of 162 million is approximately 0.9 of a standard deviation; therefore more data is clustered closer to the mean than in a normal distribution.

•

The range of 1347 is equal to 7.9 standard deviations, indicating that the distribution is more spread out than a normal distribution.

Therefore, the properties of a normal distribution are not met. (a)(iii)&(b) All Ords 2016-2017 1800 1600

Volume (millions)

1400 1200 1000

800 600 400 200 0 -3

-2

-1

0 Z Value

The normal probability plot is not a straight line indicating that the distribution is not normal. Therefore, can conclude that the daily volume traded in 2016-2017 financial year does not appear to be normal. Hence, if the distribution of the number of shares traded daily is similar for the 2008 the normal probabilities calculated in Problem 6.12 are only approximate values. Copyright © 2019 Pearson Australia (a division of Pearson Australia Group Pty Ltd) 9781488617249, Berenson, Basic Business Statistics 5e

B.52

Answers will vary.

B.53 Let X = Number of problems newly purchased 2017 Brand H (a)

Need to assume that

(i)

Probability of a problem occurring in a newly purchased vehicle brand H, is the same for all newly purchased vehicles Brand H.

(ii)

Number of problems occurring in a newly purchased Brand H is independent of the number occurring in any other newly purchased vehicle Brand H.

(ii)

The probability that two or more problems occur in an area of a newly purchased Brand H approaches zero as the area gets smaller.

These assumptions seem reasonable. Assume X has a Poisson distribution with mean  = 1.27 problems (b)(i) (ii)

P(X = 0) =

e−1.271.270 = 0.28083 0!

P(X  2) = P(0) + P(1) + P(3) = 0.28083... + 0.35665... + 0.22647... = 0.86396...

(c)

An operational definition of a ‘problem’ could be any repair or adjustment taking at least an hour to fix. Different operational definitions could result in significantly different number of problems occurring, hence to the interpretation of the survey results.

B.54 Let Y = Number of problems newly purchased 2018 Brand H Assume Y has a Poisson distribution with mean  = 1.04 problems (a)(i)

P(Y = 0) = 0.35345...

(ii)

P(Y  2) = 0.91219...

(b)

As 2017 Brand H has a higher average number of problems per vehicle the probability that a randomly selected 2018 Brand H having no or at most two problems is lower than that of a randomly selected 2017 Brand H.

B.55

Number of days excess paid 3 3 = = = 0.001 Number of days driving 300  10 3000

(a)

Probability =

(b)

Let X = Number of days where insurance excess paid in 27 days Assume X is binomial with p = 0.001, n = 27

(i)

P(X = 0) =

27! 0.0010  0.99927 = 0.9733480... 0!  27!

(ii)

P(X = 1) =

27! 0.0011  0.99926 = 0.0263067... 1!  27!

(iii)

P(X = 2) =

27! 0.0012  0.99925 = 0.000342... 2!  25!

(iv)

P(X  3) = 1 − P(X  3) = 1 − 0.999997... = 0.0000028...  0

(v)

Mean number of days insurance excess paid is 0.027

= np = 27×0.001 =

Therefore expected payout is 0.027×2000 = $54.00

Binomial Probabilities Data Sample size 27 Probability of an event of interest 0.001 Statistics Mean 0.027 Variance 0.0270 Standard deviation 0.1642 Binomial Probabilities Table X 0 1 2 3 4 (c)

P(X) 0.9733 0.0263 0.0003 0.0000 0.0000

P(<=X) 0.9733 0.9997 1.0000 1.0000 1.0000

P(<X) 0.0000 0.9733 0.9997 1.0000 1.0000

P(>X) 0.0267 0.0003 0.0000 0.0000 0.0000

P(>=X) 1.0000 0.0267 0.0003 0.0000 0.0000

Let Y = Number of incidences where insurance excess paid in 27 days Assume Y is Poisson with  = 0.01  27 = 0.027 (i)

P(Y = 0) =

e−0.027 0.0270 = 0.9733612... 0!

(ii)

P(Y = 1) =

e−0.027 0.0271 = 0.026280... 1!

(iii)

P(Y = 2) =

e−0.027 0.0272 = 0.000354... 2!

(iv)

P(Y  3) = 1 − P(Y  3) = 1 − 0.9999967... = 0.0000032...  0

(v)

Mean number of incidences requiring insurance excess paid is  =  = 0.027. Therefore expected payout is 0.027×2000 = $54.00

POISSON.DIST Probabilities Data Mean/Expected number of events of interest: POISSON.DIST Probabilities Table X P(X) P(<=X) 0 0.9734 0.9734 1 0.0263 0.9996 2 0.0004 1.0000 3 0.0000 1.0000 (d)

0.027 P(<X) 0.0000 0.9734 0.9996 1.0000

P(>X) P(>=X) 0.0266 1.0000 0.0004 0.0266 0.0000 0.0004 0.0000 0.0000

Using either binomial or Poisson distribution the expected payout is $54.00. The cost of the policy is $18.40×27 = $496.80. Therefore, Jay made the right decision as the expected loss on the policy is $442.80 ($54.00 - $496.80 = −$442.80).

(f)

To break-even expected payout would be $496.80, the cost of the policy. That is, mean number of incidences requiring insurance excess paid in 27 days is 0.2484 since

2000 = 496.80 =

496.80 = 0.2484 2000

That is, probability of probability per day of driving that Jay will have to pay an insurance excess for the policy to break-even is 0.0092

 0.2484  or 27.6    27 

(0.0092×3000) incidences requiring insurance excess to be paid in 10 years. B.56 (a)

Let X =

Actual Rebuild Cost − Estimated Rebuild Cost Estimated Rebuild Cost

Assume X is normal with mean of zero. Given P(-0.1 < X < 0.1) = 0.80 so P(Z ≤ -z) = 0.1 and P(Z ≥ z) = 0.1

P(-0.1 < X < 0.1) = P(-z < Z < z) = 0.80 P(X ≤ -0.1) = P(Z ≤ -z) = 0.1 P(≥ X) = P(Z ≥ z) = 0.1

-0.1

0.1

-z

X Z

Find z such that P(Z ≤ -z) = 0.1, from tables -z = -1.28 gives area of 0.1003. By symmetry z = 1.28, that is, P(Z ≥ 1.28) = 0.1003 Therefore,

Z = −1.28 −0.1 − 0 = −1.28  −0.1 = −1.28 −0.1 = −1.28 0.078125 =  (b) Assume X is normal with mean of zero and standard deviation of 0.078125.

−0.3 − 0   P(X  −0.3) = P  Z   0.078125   = P(Z  −3.84) = 0.00006 (c)

If the online calculator is not biased in estimating the rebuild cost you would expect the difference between the estimated rebuild cost and actual rebuild cost to be approximately normal with a mean of zero. From calculation in Part (b) the probability of a rebuild cost at most 30% less than that given by the calculator is only 6 in 10,000. That is, extremely unlikely. Therefore, either the rebuild amount offered is not sufficient to rebuild the house or the online calculator is over-estimating the rebuild cost. In either case Sam and Jo probably have a case to ask for an increased rebuild amount. Given that you would expect building costs to be skewed to the right, if X is not normal would expect X to be right skewed. That is, the probability of actual rebuild costs being at most 30% less than those estimated by the calculator are likely to be reduced.

B.57 (a)

The percentages reported by the ABS are best classified as empirical classical probabilities as they are obtained from observed data.

(b)

The assumptions that must be made so that this random variable is distributed as a binomial random variable:

(c)

•

The sample consists of a fixed number of ‘observations’ (n = 10).

•

Each person can be classified into one of two mutually exclusive and collectively exhaustive categories (i.e. attended an opera or a musical OR did not attend an opera or a musical).

•

The probability of attending the opera or a musical is constant across all people.

•

The outcome for any one person is independent from any other. Mean

 = np = 10  0.148 = 1.48

Standard deviation

 = np(1 − p) = 10  0.148  0.852 = 1.26096 = 1.1229...

B.58 Let X = Number of people who have attended a musical or opera in the past year. Assume X is binomial with n = 10 and p = 0.148 (a)

P(X = 0) = 0.20155…

(b)

P(X = 10) = 0.00000…

(c)

P(X  5) = P(6) + P(7) + P(8) + P(9) + P(10) = 0.00116... + 0.00011... + 0.000007... + 0.000000... + 0.000000... = 0.001286...

(d)

P(X  8) = 0.0000078...  0

Data Sample size Probability of an event of interest Statistics Mean Variance Standard deviation Binomial Probabilities Table

10 0.148 1.48 1.2610 1.1229 X 0 1 2 3 4 5 6 7 8 9 10

P(X) 0.2016 0.3501 0.2737 0.1268 0.0385 0.0080 0.0012 0.0001 0.0000 0.0000 0.0000

P(<=X) 0.2016 0.5517 0.8254 0.9521 0.9907 0.9987 0.9999 1.0000 1.0000 1.0000 1.0000

P(<X) 0.0000 0.2016 0.5517 0.8254 0.9521 0.9907 0.9987 0.9999 1.0000 1.0000 1.0000

P(>X) 0.7984 0.4483 0.1746 0.0479 0.0093 0.0013 0.0001 0.0000 0.0000 0.0000 0.0000

P(>=X) 1.0000 0.7984 0.4483 0.1746 0.0479 0.0093 0.0013 0.0001 0.0000 0.0000 0.0000

B.59 Let Y = Number of people who have attended a cinema in the past year. Assume Y is binomial with n = 10 and p = 0.663 (a)

P(Y = 0) = 0.000018…

(b)

P(Y = 10) = 0.01641…

(c)

P(Y  5) = P(6) + P(7) + P(8) + P(9) + P(10) = 0.23004... + 0.25862... + 0.19080... + 0.08341... + 0.01641... = 0.77929...

(d)

P(Y  8) = 0.29062...

Binomial Probabilities Data Sample size Probability of an event of interest Statistics Mean Variance Standard deviation Binomial Probabilities Table

10 0.663 6.63 2.2343 1.4948 X 0 1 2 3 4 5 6 7 8

(b)

P(X) 0.0000 0.0004 0.0033 0.0173 0.0594 0.1403 0.2300 0.2586 0.1908

P(<=X) 0.0000 0.0004 0.0037 0.0209 0.0804 0.2207 0.4507 0.7094 0.9002

P(<X) 0.0000 0.0000 0.0004 0.0037 0.0209 0.0804 0.2207 0.4507 0.7094

P(>X) 1.0000 0.9996 0.9963 0.9791 0.9196 0.7793 0.5493 0.2906 0.0998

P(>=X) 1.0000 1.0000 0.9996 0.9963 0.9791 0.9196 0.7793 0.5493 0.2906

See the above calculations. Since probability of an Australian aged 15 years and over attending a cinema is greater than the probability of attending a musical or opera the probabilities of at least for (a)(ii) to (iv) in B.59 are higher than the corresponding probabilities in B.59 (b) to (d).

B.60 (a)(i)

𝑃(𝑑𝑒𝑠𝑠𝑒𝑟𝑡) =

114 600

= 0.19 114+158−52

(ii)

𝑃(𝑑𝑒𝑠𝑠𝑒𝑟𝑡 𝑂𝑅 𝑠ℎ𝑒𝑙𝑙𝑓𝑖𝑠ℎ) =

(iii)

𝑃(𝑓𝑒𝑚𝑎𝑙𝑒 𝐴𝑁𝐷 𝑛𝑜 𝑑𝑒𝑠𝑠𝑒𝑟𝑡) =

(iv)

600

𝑃(𝑓𝑒𝑚𝑎𝑙𝑒 𝑂𝑅 𝑛𝑜 𝑑𝑒𝑠𝑠𝑒𝑟𝑡) = 208

208 600

220 600

= 0.3667

= 0.3467

240+486−208 600

518 600

= 0.8633

(b)

𝑃(𝑛𝑜 𝑑𝑒𝑠𝑠𝑒𝑟𝑡|𝑓𝑒𝑚𝑎𝑙𝑒) =

(c)

If gender and ordering dessert are statistically independent, then the following identity should hold true:

240

= 0.8667

𝑃(𝑛𝑜 𝑑𝑒𝑠𝑠𝑒𝑟𝑡) = 𝑃(𝑛𝑜 𝑑𝑒𝑠𝑠𝑒𝑟𝑡|𝑓𝑒𝑚𝑎𝑙𝑒) 0.4 ≠ 0.8667 Gender and ordering dessert are not statistically independent random variables. (d)

If ordering shellfish and ordering dessert are statistically independent, then the following identity should hold true: 𝑃(𝑠ℎ𝑒𝑙𝑙𝑓𝑖𝑠ℎ) = 𝑃(𝑠ℎ𝑒𝑙𝑙𝑓𝑖𝑠ℎ|𝑑𝑒𝑠𝑠𝑒𝑟𝑡) 0.4561 ≠ 0.2633

B.61 (a)

(b)

X = time until next flood exponential with  =

1 = 0.1 per year 10

(i)

P(X  12) = e−0.112 = 0.30119 K

(ii)

P(X  5) = 1 − e−0.15 = 1 − 0.60653 K = 0.393469...

(iii)

P ( X  20) = e−0.120 = 0.13533K

(iv)

P(X  2) = 1 − e−0.12 = 1 − 0.81873 K = 0.18126...

(v)

P ( X  10) = e−0.110 = 0.36787 K

Y = time until next flood exponential with  =

1 = 0.05 per year 20

(i)

P(Y  12) = e−0.0512 = 0.54881K

(ii)

P(Y  5) = 1 − e−0.055 = 1 − 0.77880 K = 0.22119...

(iii)

P ( Y  20) = e−0.0520 = 0.36787K

(iv)

P(Y  2) = 1 − e−0.052 = 1 − 0.90483 K = 0.095162...

(v)

P ( Y  10) = e−0.0510 = 0.60653K

End of Part 3 problems C.1 (a) 𝐻0 : 𝜋 ≤ 0.098 𝐻1 : 𝜋 > 0.098 Decision rule: If Z > 1.645, reject 𝐻0 . 𝑝−𝜋 0.12−0.098 Test statistic: 𝑍 = = = 1.0465 √

𝜋(1−𝜋) 𝑛

√

(0.098(1−0.098) 200

Decision: Since Zcalc = 1.065 is below the critical bound of 1.645, do not reject 𝐻0 . There is not enough evidence to show that the sales rate has increased at the 5% significance level. (b) Decision rule at the 0.1 level of significance would be: If Z > 1.28, reject 𝐻0 . Decision: Since Zcalc = 1.065 is below the critical bound of 1.28, do not reject 𝐻0 . There is not enough evidence to show that the sales rate has increased at the 10% level of significance. C.2 𝑝(1−𝑝)

(a) 𝑝 + 𝑍 ∙ √

𝑛

=√

𝑁−𝑛 𝑁−1

= 0.024 + 1.645 ∙ √

0.024(1−0.024) 500

∙√

5000−500 5000−1

= 0.0347

(b) With 95% level of confidence, the auditor can conclude that the rate of noncompliance is less than 0.0347, which is less than the 0.05 tolerable exception rate for internal control, and, hence, the internal control compliance is adequate. C.3

H 0 : 1 − 2  0

H1 : 1 − 2  0 Where 1 represents the mean average weekly labour costs for June and  2 represents the mean average weekly labour costs for July. Decision rule: reject H0 if tcalc > 1.746 (assuming equal variances) Test statistic: tcalc = 0.001 Decision: Do not reject H0, there is no evidence to support that average weekly labour costs declined from June to July 2017.

t-Test: Two-Sample Assuming Equal Variances

Mean Variance Observations Pooled Variance

June July 40201.4444 40179.1111 1824855479 1792724889 9 9 1808790184

Hypothesized Mean Difference df t Stat P(T<=t) one-tail t Critical one-tail P(T<=t) two-tail t Critical two-tail

0 16 0.00111395 0.49956248 1.74588368 0.99912497 2.1199053

C.4 (a)

X  t /2,n−1

s 3.0554 = 7.1538  2.0595 n 26

(a) (5.92,8.39) (b) (b)You can be 95% confident that the population mean number of ‘foreign’ ATM (c) transactions per month is somewhere between 5.92 and 8.39 (d) (e) C.7 (a) H0: μ=180 (f) The mean amount of money withdrawn from the ATM machine per (g) customer transaction over a weekend is $180. (h) H1: μ > 180 (i) The mean amount of money withdrawn from the ATM machine per (j) customer transaction over a weekend is greater than $180. (k) Decision rule: If Z > 1.645, reject H0.

X −  187 − 180 = = 1.68  / n 25 / 36

(l) Test statistic: (m) Decision: Since Zcalc = 1.68 is greater than Z = 1.645, reject H0. There is (n) enough evidence to conclude that the mean amount of money withdrawn (o) from ATM machines per customer transaction over a weekend is more (p) than $180. (q) (b) Decision rule: If p-value < 0.05, reject H0. (r) From (a) the test statistic: Z (s) p-value = P(Z > 1.68) = 0.0465< 0.05, therefore reject H0. (t) There is enough evidence to conclude that the mean amount of money withdrawn (u) from ATM machines per customer transaction over a weekend is more (v) than $180. (w) (c) When the null hypothesis is true, the probability of obtaining a (x) sample whose mean is $187 or greater is 0.0465. (y) (d) The conclusions are the same. C.5 (a)

𝐻0 : 𝜎12 − 𝜎22 ≤ 0

𝐻1 : 𝜎12 − 𝜎22 > 0 where population 1 = Australia’s exchange rate, 2 = New Zealand’s exchange rate Decision rule: If F > 2.01, reject H0. 𝑆2

210.2

𝑆2

176.4

Test statistic: 𝐹 = 12 =

= 1.19

Decision: Since Fcalc = 1.19 is less than the critical bound of FU = 2.01, do not reject H0. There is not enough evidence to conclude that the volatility of Australia’s exchange rate is greater than that of New Zealand. (b) In order to use the F distribution, we must assume that both populations are independent of each other and that they are each normally distributed. C.6 (a) 𝐻0 : 𝜇 ≤ 400 The mean life of the batteries is not more than 400 hours. 𝐻1 : 𝜇 > 400 The mean life of batteries is more than 400 hours. Decision: Since t < 1.7823, do not reject H0. There is not enough evidence to conclude that the mean life of the batteries is more than 400 hours. (b) p-value = 0.1164. If the population mean life is indeed no more than 400 hours, the probability of obtaining a sample of 13 batteries that results in a sample mean of 473.46 or more is 11.64%. (c) Allowing for only 5% probability of making a type I error, the manufacturer should not say in advertisements that these batteries should last more than 400 hours. They can make the claim that these batteries should last more than 400 hours only if they are willing to raise the level of significance to more than 0.1164. (d) (a) The mean life of batteries is not more than 400 hours. The mean life of batteries is more than 400 hours. Decision rule: Reject if t > 1.7823 𝑋̅−𝜇

Test statistic: 𝑆

⁄ √𝑛

550.38−400

= 315.33

⁄ √13

= −1.7195

Decision: Since t < 1.7823, do not reject H0. There is not enough evidence to conclude that the mean life of the batteries is more than 400 hours. (b) p-value = 0.0556. If the population mean life is indeed no more than 400 hours, the probability of obtaining a sample of 13 batteries that result in a sample mean of 550.38 or more is 5.56%. (c) Allowing for only 5% probability of making a type I error, the manufacturer should not say on advertisements that these batteries should last more than 400 hours. They can make the claim that these batteries should last more than 400 hours only if they are willing to raise the level of significance to more than 0.0556. The extremely larger value of 1342 raises the sample mean and sample standard deviation and, hence, results in a higher measured t statistic of 1.7195. This is, however, still not enough to offset the lower hours in the remaining sample to the degree that the null hypothesis can be rejected. C.7 (a) 𝐻0 : 𝜇 ≤ $260. The mean amount of money withdrawn from the ATM machines per customer transaction over a weekend is $260. 𝐻1 : 𝜇 > $260. The mean amount of money withdrawn from the ATM machines per customer transaction over a weekend is greater than $260. Decision rule: If Z > 1.645, reject H0.

𝑋̅−𝜇

Test statistic: 𝑍 = 𝜎

⁄ √𝑛

$272−$260 = 2.40 $30 ⁄ √36

Decision: Since Zcalc = 2.40 is greater than 1.645, reject H0. There is enough evidence to conclude that the mean amount of money withdrawn from the ATM machines per customer transaction over a weekend is greater than $160. (b) Decision rule: If p-value < 0.05, reject H0. 𝑋̅−𝜇

Test statistic: 𝑍 = 𝜎

⁄ √𝑛

$272−$260 = 2.40 $30⁄ √36

p-value = 1.0 – 0.9918 = 0.0082 Decision: Since p-value = 0.0082 is less than 𝛼 = 0.05, reject H0. There is enough evidence to conclude that the mean amount of money withdrawn from the ATM machines per customer transaction over a weekend is greater than $260. (c) When the null hypothesis is true, the probability of obtaining a sample whose mean is $272 or greater is 0.0082. (d) The conclusions are the same. C.8 H0: 𝜋1 = 𝜋2 (a) 𝑝̅ =

𝑍=

66 150

H1: 𝜋1 ≠ 𝜋2 38 100

104 250

= 0.4160

(0.44 − 0.38) √0.4160(1 − 0.4160)( 1 + 1 ) 150 100

0.06 √0.242944(0.016667)

= 0.9429

Z critical = 1.96 Since Zcalc = 0.92429 is less than 1.96 there is insufficient evidence that the proportions are different. (b) p-value = 0.3457. There is a 34.57% chance of finding a Z value greater than or equal to 0.9429. C.9 (a) Equal variances: H0: 𝜇𝑢𝑛𝑓𝑙𝑎𝑤𝑒𝑑 ≥ 𝜇𝑓𝑙𝑎𝑤𝑒𝑑 H0: 𝜇𝑢𝑛𝑓𝑙𝑎𝑤𝑒𝑑 < 𝜇𝑓𝑙𝑎𝑤𝑒𝑑 The one tail p-value = 0.0026 which is less than 0.05, therefore reject the null hypothesis. (b) Unequal variances: H0: 𝜇𝑢𝑛𝑓𝑙𝑎𝑤𝑒𝑑 ≥ 𝜇𝑓𝑙𝑎𝑤𝑒𝑑 H0: 𝜇𝑢𝑛𝑓𝑙𝑎𝑤𝑒𝑑 < 𝜇𝑓𝑙𝑎𝑤𝑒𝑑 The one tail p-value = 0.000074 which is less than 0.05, therefore reject the null hypothesis. (c) Both (a) and (b) give the same result. The mean crack of the unflawed components is significantly less than that of the flawed components. C.10 Copyright © 2019 Pearson Australia (a division of Pearson Australia Group Pty Ltd) 9781488617249, Berenson, Basic Business Statistics 5e

(a) 𝑝 =

130 250

= 0.52

𝑝±𝑍∙√ (b) 𝑝 =

395 500

𝑝(1 − 𝑝) 0.52(1 − 0.52) = 0.52 ± 1.96√ = 0.46 ≤ 𝜋 ≤ 0.58 𝑛 250

= 0.79

𝑝±𝑍∙√

𝑝(1 − 𝑝) 0.79(1 − 0.79) = 0.79 ± 1.96√ = 0.75 ≤ 𝜋 ≤ 0.83 𝑛 500

(c) The interval for the proportion of people with a disability who are now in the labour force is wider. This is because the sample size is smaller. Less information corresponds to a decreased precision in the estimate of the population proportion, hence a wider confidence interval. C.11 (a) 𝐻0 : 𝜇 = 4.00 𝐿. The mean amount of paint per can is 4 litres. 𝐻0 : 𝜇 ≠ 4.00 𝐿. The mean amount of paint per can differs from 4 litres. Decision rule: Reject 𝐻0 if Z < –2.58 or Z > +2.58. Test statistic: 𝑍 =

𝑋̅ −𝜇

𝜎/√𝑛

3.98−4.00

= 0.08

⁄ √50

= −1.77

Decision: Since Zcalc = –1.77 is between the critical bounds of ± 2.58, do not reject H0. There is not enough evidence to conclude that the mean amount of paint per one-gallon can differs from 4 litres. (b) p-value = 2(0.0384) = 0.0768 Interpretation: The probability of getting a sample of 50 cans that will yield a mean amount that is farther away from the hypothesised population mean than this sample is 0.0768. (c) 𝑋 ± 𝑍 ∙

𝜎 √𝑛

= 3.98 ± 2.58 ∙

0.08 √50

= {3.9508 ≤ 𝜇 ≤ 4.0092}

(d) Same decision. The confidence interval includes the hypothesised value of 4.00. C.12 The design of the experiment satisfies the randomness and independence requirement for a one-way F test. The normal probability plots do not suggest severe departure from the normality assumption. To test homogeneity of variance, you perform a Levene’s test. 𝐻0 : 𝜎12 = 𝜎22 = 𝜎32 𝐻1 : 𝑁𝑜𝑡 𝑎𝑙𝑙 𝜎𝑗2 𝑎𝑟𝑒 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒.

Decision: Since the p-value = 0.89 > 0.05, do not reject H0. There is not sufficient evidence to show a difference in the variances in puzzle solution time between the three consumption levels. Hence, a one-way F test is appropriate. 𝐻0 : 𝜇1 = 𝜇2 = 𝜇3 𝐻1 : 𝑁𝑜𝑡 𝑎𝑙𝑙 𝜇𝑗 𝑎𝑟𝑒 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒. Decision rule: If p-value > 0.05, reject H0.

Test statistic: F = 81.915 Decision: Since the p-value is virtually zero, reject H0. There is sufficient evidence to show a difference in the mean puzzle solution time between three consumption levels. PHStat2 output for the Tukey-Kramer procedure:

At 5% level of significance, there is sufficient evidence of significant different between all the pairs. The order of the consumption levels from the shortest to the longest mean puzzle solution time is no alcohol, 1 drink and 5 drinks.

C.13 (a) 𝐻0 : 𝜇1 − 𝜇2 ≤ 0 𝐻0 : 𝜇1 − 𝜇2 > 0

Where 1 is the mean income of a male aged care worker and  2 is the mean income of a female aged care worker. Decision rule: df = 57 (assuming equal variances). If tcalc > 1.672, reject H0. Test statistic: (𝑛1 − 1) ∙ 𝑆12 + (𝑛2 − 1) ∙ 𝑆22 (11) ∙ 13.352 + (46) ∙ 9.422 𝑆𝑝2 = = = 106.006 (𝑛1 − 1) + (𝑛2 − 1) 11 + 46 𝑡=

(𝑋̅1 − 𝑋̅2 ) − (𝜇1 − 𝜇2 )

(567.87 − 546.76) − 0

= 6.34 1 1 √106.006( 1 + 1 ) √𝑆𝑝2 ( + ) 𝑛1 𝑛2 12 47 Decision: Since tcalc = 6.34 is greater than the upper critical bound of 1.672, reject H0. There is enough evidence to conclude that the mean income of a male aged care worker exceeds that of a female.

(b) In order to use the pooled-variance t test, you need to assume that the populations are normally distributed with equal variances. C.14 (a) 𝑋̅ ± 𝑡 ∙

𝑆 √𝑛

= −0.00575 ± 1.9842 ∙

0.042396 √100

= −0.0142 ≤ 𝜇 ≤ 0.0027

(b) The population distribution needs to be normally distributed. However, with a sample of 100, the t distribution can still be used as a result of the Central Limit Theorem even if the population distribution is not normal. (c)

Both the normal probability plot and the box-and-whisker plot suggest that the distribution is skewed to the right. (d) We are 95% confident that the mean difference between the actual length of the steel part and the specified length of the steel part is between –0.9142 mm and 0.0027 mm, which is narrower than the ±0.125 mm requirement. The steel mill is doing a good job at meeting the requirement. This is consistent with the finding in problem A.22. C.15 (a) 𝐻0 : 𝜇𝑅 = 𝜇𝐺 = 𝜇𝐻 = 𝜇 𝑇 𝐻1 : 𝐴t least one mean is different Copyright © 2019 Pearson Australia (a division of Pearson Australia Group Pty Ltd) 9781488617249, Berenson, Basic Business Statistics 5e

F critical = F0.05, 3.36 = 2.8663 Since the F = 45.5722 > 2.8663 and the p-value is approximately zero we reject H0. There is sufficient evidence to conclude at the 5% level of significance that there is a difference in the mean strength of the four brands of bags. (b)

From the Tukey–Kramer procedure there is a difference between the mean strength of Roger’s and Tuffer, Gard and Tuffer, and Hefty and Tuffer.

(c)

𝐻0 : 𝜎𝑅2 = 𝜎𝐺2 = 𝜎𝐻2 = 𝜎𝑇2 𝐻1 : 𝐴t least one variance is different.

Once again F critical = 2.8663 Since F = 1.4545 < 2.8663 and p-value = 0.2432 > 0.05, we do not reject H0 at the 5% level of significance. There is not enough evidence to conclude that the variances in strength among the four brands are different. (d) From the results in (a) and (b), Tuffer has the lowest mean strength and should be avoided. C.16 (a) 𝑛 =

𝑍 2 𝜋(1−𝜋)

(b) 𝑛 =

𝑍 2 𝜋(1−𝜋)

(d) 𝑛 =

𝑍 2 𝜋(1−𝜋)

𝑒2 𝑒2 𝑒2 𝑒2

= = = =

1.962 (0.35)(1−0.35) (0.04)2

= 546.2068

(2.5758)2 (0.35)(1−0.35) (0.04)2 1.962 (0.35)(1−0.35) (0.02)2

= 943.4009

= 2,184.8270

(2.5758)2 (0.35)(1−0.35) (0.02)2

= 3,773.6034

Use n = 547 Use n = 944 Use n = 2,185 Use n = 3,774

(e) Holding everything else constant, the higher the confidence level desired or the lower the acceptable sampling error, the larger the sample size needed. C.17 (a) 𝐻0 : 𝜇 = 5.5 𝐻1 : 𝜇 ≠ 5.5 Decision rule: Reject if |t| > 2.680 df = 49 𝑋̅−𝜇

Test statistic: 𝑍 = 𝜎

⁄ √𝑛

5.5014−5.5

= 0.1058

⁄ √50

= 0.0935

Decision: Since |t| < 2.680, do not reject H0. There is not enough evidence to conclude that the mean amount of tea per bag is different from 5.5 grams. (b) 𝑋 ± 𝑡 ∙

𝑆 √𝑛

= 5.5014 ± 2.6800 ∙

0.1058 √50

= 5.46 ≤ 𝜇 ≤ 5.54

With 99% confidence, you can conclude that the population mean amount of tea per bag is somewhere between 5.46 and 5.54 grams. (c) The conclusions are the same.

C.18

H 0 : 1   2

H1 :  1   2

p1 = 0.5897 p2 = 0.5185 p= Z=

46 + 70 = 0.5446 78 + 135 0.5897 − 0.5185 1   1 0.5446 ( 0.4554 )  +   78 135 

= 1.0056

Since Z = 1.0052 < 1.645, we cannot reject the null hypothesis at the 5% level. That is, it is possible that the proportion of students who rated teaching satisfaction highly is no higher for lecturers who have completed the Tertiary Certificate of Teaching. C.19 (a)

H 0 :  1   2 H1 :  1   2 𝑝1 =

p= Z=

71 87 = 0.6339, 𝑝2 = = 0.7768 112 112

71 + 87 = 0.7054 224 0.6339 − 0.7768

1   1 0.7054(0.2946)  +   112 112 

= −2.3457

Since Z = –2.3457 < –1.645, you reject H0. Student attendance is higher when rolls are kept than when not kept. Note that p2 is greater than p1, there is more attendance when rolls are kept. (b)

The p-value is 0.0096. The probability of obtaining a test statistic of –2.3457 or smaller when the null hypothesis is true is less than 1%.

End of Part 4 problems D.1 (a) 𝑌̂ = 7.718 − 1.525𝑥 Where x = the unemployment rate (b) When the unemployment rate is zero the inflation rate is 7.518% Interpreting this value is not meaningful, as the the unemployment rate is never equal to zero. For every one percentage point increase in the unemployment rate, the inflation rate is predicted to decrease by 1.525 percentage points. (c) 𝑌̂ = 7.718 − 1.525(3.5) = 2.182% (d) No, an unemployment rate of 10% is outside the range of the X data (for the sample in this data set). (e) r2 = 0.0772; approximately 7.72% of the variation in the inflation variable is explained by the variation in the unemployment rate.

SUMMARY OUTPUT Regression Statistics Multiple R 0.277869164 R Square 0.077211272 Adjusted R Square -0.025320809 Standard Error 1.321043109 Observations 11 ANOVA df Regression Residual Total

1 9 10

SS 1.314180191 15.70639407 17.02057426

MS F Significance F 1.314180191 0.753045013 0.408056089 1.745154897

Intercept unemployment rate

Coefficients Standard Error t Stat P-value Lower 95% Upper 95% 7.518075343 5.747319884 1.308101079 0.223260145 -5.483265499 20.5194162 -1.524504142 1.756783081 -0.867781662 0.408056089 -5.498623572 2.44961529

(f) D.2 (a) The 95% confidence interval is from $50,190 to $60,850. (b) The 95% prediction interval is from $33,130 to $77,910. (c) The prediction interval gives the interval for income for a specific level of education, whereas the confidence interval gives the interval for the mean income for all levels of education with index of 12 years. D.3 (a) 𝑌̂ = 226,223 − 1,866.3𝑥 (b) 𝑌̂ = 226,223 − 1,866.3(100) = 39,593 The mean overseas arrivals when the exchange rate is 100 is approximately 39,593. (c) 15000 10000

Residuals

5000 0 Mar-06

Oct-06

Apr-07

Nov-07

Jun-08

Dec-08

Jul-09

Jan-10

Aug-10

Feb-11

Sep-11

-5000 -10000 -15000

Time period

(d) D = 2.03 > 1.45. There is evidence of negative autocorrelation among the residuals.

(e) The plot of the residuals versus time period indicates that negative residuals tend to occur. A non-linear model might be more appropriate. The evidence of negative autocorrelation is a reason to question the validity of the model. D.4 (a) b0 = 24.83 b1=0.14 (b) The delivery person spends a fixed time of 24.83 minutes irrespective of the number of cases delivered plus 0.14 minutes for each case delivered. (c) Delivery time = 24.83 + 0.14(150) = 45.83 (d) No, because 500 is not within the data range used in this problem. (e) r2 = 0.97. 97% of the variation in delivery time can be explained by the variations in the number of cases delivered. (f)

The residual plot does not show any particular pattern, indicating that no assumptions such as normality have been violated. (g) The p-value of the slope coefficient, b1, of the independent variable cases is less than 0.05 (in fact almost zero at 2.15212E-15) indicating that there is strong evidence of a linear relationship between delivery time and the number of cases delivered. (h) 95% confidence interval estimate of delivery time for 150 cases: Lower limit 44.88 Upper limit 46.80 (i) 95% prediction interval estimate of delivery time for 150 cases:

Lower limit: 41.56 Upper limit: 50.12 (j) 95% confidence interval estimate of the population slope: Lower limit: 0.128 Upper limit: 0.152 (k) Delivery cost of each customer = {24.83 + 0.14 × (No. of cases delivered)} × Total cost/Total delivery time D.5 (a) Yˆ = 14.626 + 0.276 X (b) A student has a predicted mark of 14.626 if they do not spend any time using online content. For each additional minute using online content a student expects marks to increase by 0.276. (c)

Yˆ = 14.626 + 0.276(100) = 42.226 marks

(d) No, we do not observe any sample data near x = 500. Therefore, we would have to assume the estimated relationship would hold (extrapolate) for a wider range of data. (e) 30

Residuals

20 10 0 0

100

150

200

250

-10 -20

online

The residual plot shows a possibly upside-down U shape, indicating that we may have violated the assumption of linearity. (f) Disregard, there is no time aspect to this analysis. (g) Disregard, there is no time aspect to this analysis. (h) The p-value of the online slope coefficient is virtually zero. This indicates there is strong evidence of a linear relationship between online participation and student grades. (i) 95% confidence interval estimate of grades for 85 minutes of online content:

Lower limit 32.03 Upper limit 44.14 (j) 95% prediction interval estimate of grades for 85 minutes of online content: Lower limit: 13.88 Upper limit: 62.29 (k) 95% confidence interval estimate of the population slope: Lower limit: 0.179 Upper limit: 0.373 D.6 (a) 𝑌̂ = 0.9254𝑥 − 1.0356 b0 = –1.0356 b1=0.0.9254 (b) r2 = 0.72996. 73% of the variation in the customer satisfaction index variable is explained by the variation in employee satisfaction index. (c) 𝑌̂ = 0.9254(75) − 1.0356 = 68.3694 (d) 30 20

Residuals

10 0

100

120

-10 -20 -30

Employee satisfaction index

The residual plot shows a movement from positive to negative and back again to positive, indicating a violation of the assumptions of linear regression.

D.7 (a) Searching Not searching 𝑝̅ =

NSW 11 189

Qld 15 185

Tas 14 186

Vic 13 187

WA 16 184

11 + 15 + 14 + 13 + 16 69 = = 0.069 200 + 200 + 200 + 200 + 200 1000

Searching – NSW: 𝑝̅ = 0.069, 𝑛1 = 200, 𝑓𝑒 = 13.8 Searching – Qld: 𝑝̅ = 0.069, 𝑛2 = 200, 𝑓𝑒 = 13.8 Searching – Tas: 𝑝̅ = 0.069, 𝑛3 = 200, 𝑓𝑒 = 13.8 Searching – Vic: 𝑝̅ = 0.069, 𝑛4 = 200, 𝑓𝑒 = 13.8 Searching – WA: 𝑝̅ = 0.069, 𝑛5 = 200, 𝑓𝑒 = 13.8 Not Searching – NSW: 𝑝̅ = 0.931, 𝑛1 = 200, 𝑓𝑒 = 186.2 Not Searching – Qld: 𝑝̅ = 0.931, 𝑛2 = 200, 𝑓𝑒 = 186.2 Not Searching – Tas: 𝑝̅ = 0.931, 𝑛3 = 200, 𝑓𝑒 = 186.2 Not Searching – Vic: 𝑝̅ = 0.931, 𝑛4 = 200, 𝑓𝑒 = 186.2 Not Searching – WA: 𝑝̅ = 0.931, 𝑛5 = 200, 𝑓𝑒 = 186.2 fo 11 15 14 13 16 189 185 186 187 184

fe 13.8 13.8 13.8 13.8 13.8 186.2 186.2 186.2 186.2 186.2

(fo-fe) -2.8 1.2 0.2 -0.8 2.2 2.8 -1.2 -0.2 0.8 -2.2

(fo-fe)2 7.84 1.44 0.04 0.64 4.84 7.84 1.44 0.04 0.64 4.84

(fo-fe)2/fe 0.5681 0.1043 0.0029 0.0464 0.3507 0.0421 0.0078 0.0002 0.0034 0.0260 1.1519

degrees of freedom = (2 – 1)(5 – 1)=4 At the 0.05 significance level, the chi-square critical value is 9.488. 1.1519< 9.488, do not reject H0. We are not able find a significant difference in the proportion of people seeking new employment between the states.

(b) p-value = 0.8860 > 0.05, do not reject H0. (c) The Marascuilo Procedure is not appropriate as the null hypothesis was not rejected.

D.8

The critical chi-square value with (4-1)(4-1)=9 degrees of freedom is 16.9190. Since the chi-square test statistic 23.6679 > 16.9190, reject null hypothesis of independence. There is a significant relationship between political party and preferred head of state.

D.9 (a) 𝑌̂ = −40.80(5) + 0.33(7800) + 4421.88 = 6791.88 6090.90≤Yx = xi≤ 7541.55, 4112.62≤μY |X = Xi ≤ 9519.82 (b) The residual plot that examines the pattern of residuals versus the predicted values of Y shows a positive correlation indicating that there may be evidence of a non-linear effect in at least one of the independent variables. The second and third residual plots, however, show no pattern between the residuals and each of the independent variables. Further investigation is warranted. Thus, the model appears inadequate from the residual analysis. (c) For X1: |t | = b1/Sb1 = 0.33/0.23 = 1.43 < t0.05,11 = 2.201 with 11 degrees of freedom for 𝛼 = 0.05. Do not reject H0. There is insufficient evidence that the variable X1 contributes to a model already containing X2. For X2: |t | = b1/Sb1 = |–40.80/167.75| = 0.24 < t0.05,11 = 2.201 with 11 degrees of freedom for 𝛼 = 0.05. Do not reject H0. There is insufficient evidence that the variable X1 contributes to a model already containing X2. (d) -0.176 ≤ 𝛽1 ≤ 0.836, –410.01≤ 𝛽2 ≤ 328.42 (e) radj2=0.0545 (f) r2Y1.2= 0.1965. Holding constant the effect of beef supply, 19.65% of the variation in lamb supply can be explained by variation in price of feed. r2Y2.1= 0.5451. Holding constant the effect of feed price, 54.51% of the variation in lamb supply can be explained by variation in supply of beef. D.10 (a) 𝑌̂ = 55.9584 + 0.1478𝑋1 − 2.1664𝑋2 Holding constant the effect of hand feeding for each one-unit increase in rainfall the proportion increases by 0.1478 units. Holding constant the effect of rainfall for each one-unit increase in hand feeding the proportion decreases by 2.1664 units. 𝑌̂ = 55.9584 − 0.1478(180) − 2.1664(0) = 82.5537 78.30≤Yx = x1≤ 86.81, 70.56≤μY |X = Xi ≤ 94.52 (b) The model appears adequate from the residual analysis.

(c) F = 23.2409 > F2,11 = 3.98. Reject H0. There is evidence of a relationship between the proportion of clip at 75 mm and rainfall and hand feeding. (d) 0.0501 ≤ 𝛽1 ≤ 0.2439, –12.9079 ≤ 𝛽2 ≤ 8.5751 (e) r2=0.8915. So, 89.15% of the proportion of clip at 75 mm is explained by the variation in rainfall and hand feeding. (f) r2adj= 0.7607 (g) r2Y1.2= 0.4753. Holding constant the effect of beef supply, 47.53% of the variation in proportion can be explained by variation in rainfall. r2Y2.1= 0.0158. Holding constant the effect of rainfall, 1.58% of the variation in proportion can be explained by variation in hand feeding. (h) 𝑌̂ = 54.9628 + 0.1534𝑋1 − 0.5198𝑋2 − 0.0136𝑋1 . 𝑋2 For X1. X2, the p-value is 0.8888. Cannot reject H0 at the 5% significance level. It is likely that there is some interaction between rainfall (X1) and hand feeding (X2). D.11

𝑉𝐼𝐹 =

1 1−0.75

D.12 (a), (b), (c), (e)

Year 2000 2001 2002 2003 2004 2005 2006 2007 2008 2009 2010 2011

Petrol consumption (L) 89 123 102 112 87 213 112 123 101 109 99 103

MA(3) 104.67 112.33 100.33 137.33 137.33 149.33 112.00 111.00 103.00 103.67

ES(W=0.50) 89.00 106.00 104.00 108.00 97.50 155.25 133.63 128.31 114.66 111.83 105.41 104.21

ES(W=0.25) 89.00 97.50 98.63 101.97 98.23 126.92 123.19 123.14 117.61 115.46 111.34 109.26

250 Petrol consumption (L) MA(3) Petrol Consumption (L)

200

ES(W=0.50) ES(W=0.25)

150

100

0 2000

2001

2002

2003

2004

2005

2006

2007

2008

2009

2010

2011

Year

(d) The exponentially smoothed forecast for 2010 is 105.41. (f) The W = 0.25 line smooths the data more severely and removes the peaks and troughs. However, the W = 0.5 model fits the data more closely as it weights more recent values more heavily. D.13 (a) X = 0 (b) X = 10 (c) X = 31 (d) X = 34

D.14 Male full-time employed professionals ('000)

1000 900 800 y = 0.047x - 988.28

700 600 500 400 300 200 100

ug -9 6 Fe b9 A 7 ug -9 7 Fe b98 A ug -9 8 Fe b99 A ug -9 9 Fe b00 A ug -0 0 Fe b01 A ug -0 1 Fe b0 A 2 ug -0 2 Fe b03 A ug -0 3 Fe b04 A ug -0 4 Fe b0 A 5 ug -0 5 Fe b06 A ug -0 6 Fe b07 A ug -0 7 Fe b08 A ug -0 8

(a)

Time period

(b) 𝑌̂ = 4.2895𝑋 − 664.79 where X = quarters relative to August 1996. ̂ (c) 𝑋 = 50, 𝑌 = 4.2895(50) + 664.79 = 879.265 𝑋 = 51, 𝑌̂ = 4.2895(51) + 664.79 = 883.5545 (d) Probably not as there are too many current factors that influence employment that are not captured in a simple historical series (therefore model) such as this one. D.15 (a)

(b) Linear trend: 𝑌̂ = 77470.842 + 939.509𝑋, where X is relative to Feb 2016. (c) Quadratic trend: 𝑌̂ = 71964.351 + 3004.443𝑋 − 121.467𝑋 2 . (d) Exponential trend: 𝑙𝑜𝑔10 𝑌̂ = 4.883 + 0.005𝑋.

2016 Feb 2016 Mar 2016 Apr 2016 May 2016 Jun 2016 Jul 2016 Aug 2016 Sep 2016 Oct 2016 Nov

visitors 1st diff 2nd diff % diff 61,900 78,701 16,801 27.1421648 93,320 14,619 -2,182 18.5753675 64,175 -29,145 -43,764 -31.231247 77,805 13,630 42,775 21.2388002 103,689 25,884 12,254 33.2677848 78,888 -24,801 -50,685 -23.918641 92,211 13,323 38,124 16.8885002 85,395 -6,816 -20,139 -7.3917429 73,308 -12,087 -5,271 -14.154224

2016 Dec 2017 Jan 2017 Feb 2017 Mar 2017 Apr 2017 May 2017 Jun 2017 Jul

100,849 115,928 71,416 85,038 93,924 71,967 84,833 104,873

27,541 15,079 -44,512 13,622 8,886 -21,957 12,866 20,040

39,628 -12,462 -59,591 58,134 -4,736 -30,843 34,823 7,174

37.5688874 14.952057 -38.396246 19.0741571 10.4494461 -23.377412 17.8776384 23.6228826

(e) Investigating the first, second and percentage differences does not suggest any particular trend model is more appropriate than the others. D.16 (a) U= 0.3, V= 0.3

2016 Feb 2016 Mar 2016 Apr 2016 May 2016 Jun 2016 Jul 2016 Aug 2016 Sep 2016 Oct 2016 Nov 2016 Dec 2017 Jan 2017 Feb 2017 Mar 2017 Apr

Visitors E T 61,900 78,701 78,701 16,801 93,320 93974.6 15731.82 64,175 77834.426 -6578.5758 77,805 75840.2551 -3369.4924 103,689 94323.5288 11927.4439 78,888 87096.8918 -1480.4127 92,211 90232.6437 1750.90252 85,395 87371.5639 -1477.4851 73,308 77083.8236 -7644.6637 100,849 91426.048 7746.15793 115,928 110901.262 15956.497 71,416 88048.5276 -11209.965 85,038 82578.1689 -7192.2406 93,924 88362.5785 1891.41456

2017 May 2017 Jun 2017 Jul

71,967 84,833 104,873

77453.0979 80498.2658 97563.8359

-7069.212 10.8538869 11949.1553

(b) U= 0.7, V= 0.7

2016 Feb 2016 Mar 2016 Apr 2016 May 2016 Jun 2016 Jul 2016 Aug 2016 Sep 2016 Oct 2016 Nov 2016 Dec 2017 Jan 2017 Feb 2017 Mar 2017 Apr 2017 May 2017 Jun 2017 Jul

Visitors E T 61,900 78,701 78,701 16,801 93,320 94847.4 16604.62 64,175 97268.914 12349.6882 77,805 100074.522 9486.464 103,689 107799.39 8957.9853 78,888 105396.563 5549.74154 92,211 105325.713 3863.56416 85,395 102050.994 1722.07923 73,308 94633.5512 -1019.7774 100,849 95784.3417 -368.60701 115,928 101569.414 1477.49687 71,416 93557.6378 -1369.2851 85,038 90043.2469 -2012.8169 93,924 89798.501 -1482.3956 71,967 83411.3738 -2953.8151 84,833 81770.1911 -2560.0253 104,873 86909.016 -250.37027

2016 Feb 2016 Mar 2016 Apr 2016 May 2016 Jun 2016 Jul 2016 Aug 2016 Sep 2016 Oct 2016 Nov 2016 Dec 2017 Jan 2017 Feb 2017 Mar 2017 Apr 2017 May 2017 Jun 2017 Jul

Visitors E T 61,900 78,701 78,701 16,801 93,320 93974.6 16342.78 64,175 78017.714 6652.8802 77,805 79864.6783 5211.10542 103,689 98105.0351 9119.88085 78,888 87389.0748 3169.1285 92,211 91715.161 3516.21581 85,395 88345.913 1450.57668 73,308 78254.5469 -2012.0062 100,849 93467.0622 3155.35028 115,928 110136.324 7209.52365 71,416 85194.9542 -2435.7443 85,038 84354.363 -1957.1984 93,924 90465.9494 463.437051 71,967 77655.7159 -3518.6641 84,833 81624.2155 -1272.515 104,873 97516.6102 3876.9579

D.17

Coefficients Intercept Lag1 Lag2 Lag3

90178.706 -0.1123567 -0.3781981 0.45706277

Standard Error 49254.3054 0.27026226 0.25103751 0.24828069

t Stat

P-value

1.83087966 -0.4157321 -1.5065403 1.84091144

0.09431682 0.68560073 0.16009531 0.09274289

(a) Since the p-value = 0.09 > 0.05, do not reject H0 that A3 = 0. Third-order term can be deleted.

Coefficients Intercept Lag1 Lag2

149886.312 -0.2951109 -0.4418441

Standard Error 31525.439 0.25388762 0.23270794

t Stat

P-value

4.75445597 -1.1623683 -1.8987067

0.0003764 0.26598107 0.08002366

(b) Since the p-value = 0.08 > 0.05, do not reject H0 that A2 = 0. Second-order term can be deleted. (c)

Coefficients Intercept X Variable 1

97864.7992 -0.1307301

Standard Error 21241.4935 0.24852355

t Stat

P-value

4.60724662 -0.526027

0.00034203 0.60655996

Since the p-value = 0.606 > 0.05, do not reject H0 that A1 = 0. A first-order autoregressive model is also not appropriate. (d)

Due to the findings from parts a-c, a forecast is not appropriate with the AR models.

D.18 (a), (b), (c)

2016 Feb 2016 Mar 2016 Apr 2016 May 2016 Jun 2016 Jul 2016 Aug 2016 Sep 2016 Oct 2016 Nov 2016 Dec 2017 Jan 2017 Feb 2017 Mar 2017 Apr 2017 May 2017 Jun 2017 Jul SSE Syx MAD

Visitors linear residual quadratic residual exponential residual HW residual 61,900 77470.8421 -15570.842 71964.3509 -10064.351 76415.4845 14,515 78,701 78410.3509 290.649123 74847.3271 3853.67286 77298.3233 -1,403 78,701 0 93,320 79349.8596 13970.1404 77487.37 15832.63 78191.3617 -15,129 93,975 655 64,175 80289.3684 -16114.368 79884.4794 -15709.479 79094.7174 14,920 78,018 13,843 77,805 81228.8772 -3423.8772 82038.6553 -4233.6553 80008.5098 2,204 79,865 2,060 103,689 82168.386 21520.614 83949.8978 19739.1022 80932.8594 -22,756 98,105 -5,584 78,888 83107.8947 -4219.8947 85618.2069 -6730.2069 81867.888 2,980 87,389 8,501 92,211 84047.4035 8163.59649 87043.5826 5167.41744 82813.7192 -9,397 91,715 -496 85,395 84986.9123 408.087719 88226.0248 -2831.0248 83770.4778 -1,625 88,346 2,951 73,308 85926.4211 -12618.421 89165.5335 -15857.534 84738.2899 11,430 78,255 4,947 100,849 86865.9298 13983.0702 89862.1089 10986.8911 85717.2832 -15,132 93,467 -7,382 115,928 87805.4386 28122.5614 90315.7508 25612.2492 86707.587 -29,220 110,136 -5,792 71,416 88744.9474 -17328.947 90526.4592 -19110.459 87709.3319 16,293 85,195 13,779 85,038 89684.4561 -4646.4561 90494.2343 -5456.2343 88722.6502 3,685 84,354 -684 93,924 90623.9649 3300.03509 90219.0759 3704.92415 89747.6754 -4,176 90,466 -3,458 71,967 91563.4737 -19596.474 89700.984 -17733.984 90784.5429 18,818 77,656 5,689 84,833 92502.9825 -7669.9825 88939.9587 -4106.9587 91833.3895 7,000 81,624 -3,209 104,873 93442.4912 11430.5088 87936 16937 92894.3535 -11,979 97,517 -7,356 427656877 580155912 3343202799 720,258,344 14380.4551 14505.7708 14,455 6,929 11243.2515 11314.8763 11258.9535 5081.381

(d) The Holt-Winters model has the lowest MAD and SSE values and is the better model to use for forecasting. D.19

(a) 𝑌̂ = 0.0026𝑥1 + 136.08𝑥2 − 47.51 The constant b0 = –47.51. A customer who is earning no income and does not use a credit card will spend negative $47.51 at this restaurant. This coefficient estimate however is invalid as the scenario of a customer earning no income is unlikely (e.g. social welfare). Furthermore, x1 = 0 is outside the domain of this data set and therefore should not be interpreted. The coefficient b1 = 136.08. This tells us that if a customer were to pay with the company credit card, they would spend $136.08 more at the restaurant than if they weren’t using the credit card, holding all other factors constant. The coefficient b2 = 0.0026. This tells us that for every additional dollar in income that a customer in this sample is earning, their spending at the restaurant increases by $0.0026, holding all other factors constant. (b) For 𝛽1 : t = 0.00257/0.000426 = 6.033. Reject H0 at the 0.05 level of significance and 7 degrees of freedom. This coefficient is statistically significant to the model at 5% significance level. For 𝛽2 : t = 136.08/27.74 = 4.91. Reject H0 at the 0.05 level of significance and 7 degrees of freedom. This coefficient is statistically significant to the model at 5% significance level. For 𝛽0 : t = –47.509/28.682 = –1.656. Do not reject H0 at the 0.05 level of significance and 7 degrees of freedom. There is insufficient evidence to demonstrate statistical significance of this coefficient to the model at 5% significance level. (c) 𝑌̂ = 136.08(1) + 0.0026(75000) − 47.51= 276.07 (d) The model appears adequate from the residual analysis. D.20 (a) 𝑌̂ = −0.00203(450) + 0.78778(35) − 5.155 = 21.504 The second-hand sales price of a car with 450,000 km and a new car sales price of $35,000 is $21,504. 19.15≤Yx = x1≤ 23.85, 16.24≤μY |X = Xi ≤ 26.77 (b) The model appears adequate from the residual analysis. (c) For 𝛽1 : t = –0.00203/0.003511 = –0.58. Do not reject H0 at the 0.05 level of significance and 8 degrees of freedom. There is insufficient evidence to demonstrate statistical significance of this coefficient to the model at 5% significance level. For 𝛽2 : t = 0.7878/0.06656 = 11.84. Reject H0 at the 0.05 level of significance and 8 degrees of freedom. This coefficient is statistically significant to the model at 5% significance level. (d) 95% CI for 𝛽1 = {−0.00203 ± 2.3060 × 0.003511} = {−0.010, 0.006} 95% CI for 𝛽2 = {0.7878 ± 2.3060 × 0.06656} = {0.635, 0.941} (e) r2adj= 0.9594

(f) r2Y1.2= 0.9674. Holding constant the effect of mileage, 96.74% of the variation in second-hand price can be explained by variation in new car price. r2Y2.1= 0.9245. Holding constant the effect of new car price, 92.45% of the variation in second-hand car price can be explained by variation in mileage. D.21 (a)

(b)

𝑙𝑜𝑔𝑌̂ = 0.1272 + 0.003𝑋 − 0.061𝑀1 + 0.0109𝑀2 + 0.377𝑀3 + 0.0513𝑀4 + 0.0544𝑀5 + 0.0534𝑀6 + 0.0418𝑀7 + 0.0396𝑀8 + 0.0583𝑀9 + 0.0376𝑀10 + 0.0241𝑀11 (c) logb1 = 0.0003. b1 = 100.0003= 1.000715. The estimated monthly compound growth rate is (b1 – 1)100% = 0.0715%. (d)

(e) Beer prices are lower from November to February and higher from April to July with a surge in September. D.22 (a) Sydney 1564×72+137×77+246×68×256×70+609×93+75×58 765×72+108×77+216×68+220×70+662×93+59×58

218792 158472

= 1.3574 = 138.06%

Melbourne 1525 × 68 + 175 × 75 + 254 × 69 + 268 × 73 + 611 × 72 + 75 × 60 202407 = = 1.3794 = 137.94% 844 × 68 + 111 × 75 + 201 × 69 + 275 × 73 + 608 × 72 + 55 × 60 146737 (b) Sydney 1564×60+137×83+246×72×256×64+609×84+75×55 765×60+108×83+216×72+220×64+662×84+59×55

194588 143349

= 1.3574 = 135.74%

Melbourne 1525 × 58 + 175 × 82 + 254 × 70 + 268 × 60 + 611 × 75 + 75 × 50 186235 = = 1.3596 = 135.96% 844 × 58 + 111 × 82 + 201 × 70 + 275 × 60 + 608 × 75 + 55 × 50 136974 (c) The Paasche index uses weights in the current year 2012. The Laspeyres index uses weights in the base year 2002. Since the basket of items are staple food items, the quantity used may not vary on the basis of price so that either index may be useful and the difference between them is not great. D.23 (a) Year 2001 2003 2005 2007 2009 2011

Commodore Simple Price Index 2001 34500 100.00 35450 102.75 37650 109.13 39075 113.26 40650 117.83 42675 123.70

Falcon 35600 36000 37540 38650 39500 41500

Simple Price Index 2001 100.00 101.12 105.45 108.57 110.96 116.57

WRX 45000 46750 48750 49000 51350 53000

Simple Price Index 2001 100.00 103.89 108.33 108.89 114.11 117.78

Kluger 55000 56750 58900 62450 63000 65400

Simple Price Index 2001 100.00 103.18 107.09 113.55 114.55 118.91

(b)

Year 2001 2003 2005 2007 2009 2011

Commodore Simple Price Index 2003 34500 97.32 35450 100.00 37650 106.21 39075 110.23 40650 114.67 42675 120.38

Falcon 35600 36000 37540 38650 39500 41500

Simple Price Index 2003 98.89 100.00 104.28 107.36 109.72 115.28

WRX 45000 46750 48750 49000 51350 53000

Simple Price Index 2003 96.26 100.00 104.28 104.81 109.84 113.37

Kluger 55000 56750 58900 62450 63000 65400

Simple Price Index 2003 96.92 100.00 103.79 110.04 111.01 115.24

Unweighted Aggregate Price indices 100.00 102.85 107.49 111.21 114.34 119.09

(d) 2011 Laspeyres price index = (e) 2011 Paasche price index =

146385000

122375000 200920550 167957600

= 1.1962 = 119.62%

= 1.1963 = 119.63%

D.24 (a) t1 =

b b1 5 10 = = 2.5 , t2 = 2 = = 1.25 Sb2 Sb1 2 8

X1 has the biggest t statistic

b1  tn − k −1Sb1

(b) 2.5  2.0739* 2

0.4261  1  4.5739 (c)

H 0 : 1 = 0

H 0 : 2 = 0

H1 : 1  0

H1 :  2  0

Decision rule: reject H0 if |tcalc| > 2.0739 t1 = 2.5 , reject H0 t2 = 1.25, do not reject H0

Keep X1, X2 can be discarded D.25 There appears to be a ‘U’ shape to this residual pattern indicating that the assumption of linearity may be violated. D.26 H0: The number of domestic sales per week follows a Poison distribution. H1: The number of domestic sales per week does not follow a Poison distribution.

mj f j

Number Sold Number of Weeks 0 11 1 49 2 32 3 22 4 15 5 13 6 8 7 4 8 2 Total 156

0 49 64 66 60 65 48 28 16 396

P( X )

( f0 − fe ) / fe

0.078988 0.200508 0.254490 0.215338 0.136657 0.069380 0.029353 0.010644 0.003378

12.3221 31.27918 39.7005 33.59273 21.31846 10.82322 4.579054 1.660536 0.526901

0.141855 10.03951 1.493625 4.000608 1.872694 0.437797 2.55574 3.295978 4.118461 27.95626

 mj fj

X = j =1

where

( f − f ) 2 27.956  k − p −1 = k 0 e = fe 2

396 = 156

 =

2.538462

crit

7 , 0.05

= 14.067

Since 27.956 > 14.067, reject H0. There is sufficient evidence to conclude that the distribution of domestic sales does not follow a Poisson distribution at 5% level of significance. D.27 (a) 𝑋̅ − 27.47, 𝑠 = 14.65 (b)

Classes 1<10

𝑿 10

̅ 𝑿−𝑿 -17.47

𝒁 -1.19

Area Below 0.1170

p(x) Area in Class 0.1170

fx 42.12

10<20 20<30 30<40 40<50 50<60

20 30 40 50 60

fo 52 69 82 74 61 22

-7.47 2.53 12.53 22.53 32.53 fe 42.12 67.68 45.90 85.54 47.92 17.50

-0.51 0.17 0.86 1.54 2.22 (fo-fe) 9.88 1.32 36.10 -11.54 13.08 4.50

0.3050 0.4325 0.1949 0.0618 0.0132

0.1880 0.1275 -0.2376 -0.1331 -0.0486

(fo-fe)2 97.6144 1.7424 1303.2100 133.1716 171.0864 20.2500

67.68 45.90 85.54 47.92 17.50

(fo-fe)2/fe 2.3175 0.0257 28.3924 1.5568 3.5703 1.1571 37.0198

df = 7 – 2 – 1 = 4. At 0.05 significance critical chi-square at 4 df = 9.488. Since 37.02 > 9.88, reject H0, the distance kicked does not form a normal distribution. D.28 <Note the question has been changed but the answer is the same. However an extra bracket has been removed in the first equation. It is now shown correctly below.> (a) chi-square =

(30−1)(6)2 52

= 41.76

There insufficient evidence the standard deviation has changed. (b) Assume the data is normally distributed. (c) p-value= 0.059 > 0.05, cannot reject H0.

D.29

(a) 𝐻0 :  1 =  2

 

2 𝐻1 : 1 2 Decision rule: Reject 𝜒𝑐𝑎𝑙𝑐 if it is greater than 3.841 (given 𝛼 = 0.05, 𝜈 = 1)

Test statistic: 𝜒 2 =

(71−79)2 79

(87−79)2 79

(41−33)2 33

(25−33)2 33

= 5.4990

Decision: There is sufficient evidence to reject H0 at the 5% significance level. Therefore, the proportion of students attending class tutorial is not equal when class rolls are kept. (b) p-value= 𝑃(𝜒 2 ≥ 5.4990) ≈ 0.0190 The probability of obtaining a test statistic equal to or more extreme than 5.4990 is approximately 1.9%.

D.30 (a) 60,000,000 50,000,000 40,000,000 30,000,000 20,000,000 10,000,000

Mar-14

Dec-13

Sep-13

Jun-13

Mar-13

Dec-12

Sep-12

Jun-12

Mar-12

Dec-11

Sep-11

Jun-11

Mar-11

Dec-10

Sep-10

Jun-10

Mar-10

log ŷ = 6.9155+ 0.0466X + 0.1331Q + 0.0909Q - 0.0030Q

1 2 3 (b) (c) Antilog(0.0466) = 1.1133, iPhone sales increase by 11.33% per quarter. (d) Antilog(0.1331) = 1.3586, sales increase by 35.86% in March quarter compared to December quarter. Antilog(0.0909) = 1.2329, sales increase by 23.29% in June quarter compared to December quarter. Antilog(–0.0030) = 0.9931, sales decrease by 0.69% in September quarter compared to December quarter.

(e)

log yˆ = 6.9155 + 0.0466(16) + 0.1331(1) + 0.0909(0) − 0.0030(0) = 7.7944 yˆ = 62291634

End of Part 5 problems E.1 (a) and (b) Wine Bottles Data Sample/Subgroup Size

R Chart Intermediate Calculations RBar 3.483333333 D3 Factor 0 D4 Factor 2.004 R Chart Control Limits Lower Control Limit 0 Center 3.483333333 Upper Control Limit 6.9806 XBar Chart Intemediate Calculations Average of Subgroup Averages 750.80625 A2 Factor 0.483 A2 Factor * RBar 1.68245 XBar Chart Control Limits Lower Control Limit 749.1238 Center 750.80625 Upper Control Limit 752.4887

R-Chart: Wine Bottles 8 7

UCL

Sample Range: Volume mls

6 5 4

RBar

3 2 1 0

LCL 25

Hour

Sample Mean - Volume mls

X-bar Chart: Wine Bottles 753.0 752.5 752.0 751.5 751.0 750.5 750.0 749.5 749.0 748.5

UCL

XBar

LCL 0

Hour (c) All of the points are within the control limits, and there are no patterns in the R chart. All the points in the X chart are also within the control limits, and there are no patterns, so the process is in control. (d) Let X = Volume of filled bottle, ml

R = 3.48333K , X = 750.806 K , d2 = 2.534

753 − 750.806 K   747 − 750.806 K P(747  X  753) = P  Z  3.48333K / 2.534   3.48333K / 2.534 = P ( −2.78  Z  1.60 ) = 0.9425 (e)

Cp =

USL − LSL

(

6 R / d2

CPL = CPU =

)

X − LSL

(

3 R / d2

)

USL − X

(

3 R / d2

)

753 − 747 = 0.727 K 6 ( 3.48333K / 2.534 )

750.806K − 747 = 0.9229K 3 ( 3.48333K / 2.534 )

753 − 750.806K = 0.5320K 3 ( 3.48333K / 2.534 )

Cpk = Min(CPL,CPU) = Min(0.923,0.532) = 0.532 (f)

The process is in control. However, from (d) and (e) we can conclude that the process is not capable of meeting the 99.7% goal. Management needs to investigate ways of improving the process so that the specification goals can be met. In particular, the common causes of variation need to be investigated and, by changing and improving the process, some of these causes may be reduced.

E.2 (a) 1 2

-50,000 60,000 130,000

4 1 2

300,000 0 0 0

(b), (d) Opportunity loss table: Event

A: Garden Service

Pr 1 2 3 4

Very low Low Moderate High

0.2 0.5 0.2 0.1 EOL

50,000 0 0 0 10,000

B: No Garden Service 0 60,000 130,000 300,000 86,000

(c), (f) Payoff table: Event Pr 1 2 3 4

Very low Low Moderate High

0.2 0.5 0.2 0.1 EMV  CV Return-torisk

A: Garden Service – 50,000 60,000 130,000 300,000 76,000 94,361.01 124.16% 0.8054

B: No Garden Service 0 0 0 0 0 0 undefined undefined

(e) EVPI = $10,000. The entrepreneur should not be willing to pay more than $10,000 for a perfect forecast. (g) To maximise the expected monetary value and minimise expected opportunity loss, the entrepreneur should provide gardening service. (h) Given 3 successes out of 20, the binomial probabilities and their related revised conditional probabilities are:

(i)

Very low Low Moderate High

Pr 0.2 0.5 0.2 0.1

Binomial Probabilities 0.2054 0.0011 0.2054 0.1901 0.6019

Revised Conditional Probabilities 0.2054/0.6019 = 0.3412 0.0011/0.6019 = 0.0018 0.2054/0.6019 = 0.3412 0.1901/0.6019 = 0.3158

(i)(c), (f) Payoff table: Pr Very low Low Moderate High

0.3412 0.0018 0.3412 0.3158 EMV  CV Return-torisk

A: Garden Service – 50,000 60,000 130,000 300,000 122,159.8 141,884.9 116.15% 0.8610

B: No Garden Service 0 0 0 0 0 0 undefined undefined

(b), (d) Opportunity loss table: B: No Garden Service Very low 0.3412 50,000 0 Low 0.0018 0 60,000 Moderate 0.3412 0 130,000 High 0.3158 0 300,000 EOL 17,062.63 139,222.5 (e) EVPI = $17,062.63. The entrepreneur should not be willing to pay more than $17,062.63 for a perfect forecast. (g) To maximise the expected monetary value and minimise expected opportunity loss, the entrepreneur should provide gardening service. Pr

A: Garden Service

E.3 (a) p Chart 1 0.9 UCL 0.8

Proportion

0.7 pBar

0.6 0.5 0.4

LCL

0.3 0.2 0.1 0 0

1 Missed Goal Shots 2 3 Intermediate Calculations 4 Sum of Subgroup Sizes 4000 5 Number of Subgroups Taken 40 6 Average Sample/Subgroup Size 100 7 Average Proportion of Nonconforming Items 0.24825 8 Three Standard Deviations 0.12959928 9 10 p Chart Control Limits 11 Lower Control Limit 0.11865072 12 Center 0.24825 13 Upper Control Limit 0.37784928 Although none of the points is outside either the LCL or UCL, there is a clear pattern over time with higher values occurring in the first half of the sequence and lower values occurring toward the end of the sequence. (b) This would explain the pattern in the results over time, with the different shooting technique impacting adversely on her performance (c) The control chart would have been developed using the first 20 days’ data only.

E.4 (a) 1 2

35 350 700

1750

1 2

750 750 750 750

(b), (d) Opportunity loss table:

Very low Low Moderate High

Pr 0.25 0.25 0.25 0.25 EOL

A: Do Not insure 715 400 50 0 20

B: Insure 0 0 100 1000 100

c), (e), (f) Payoff table:*

Event 1 2 3 4

Very low Low Moderate High

Pr 0.25 0.25 0.25 0.25 EMV



CV Return-torisk

A: Do Not Insure 35 350 700 1750 708.75 645.55 91.08% 1.0979

B: Insure 750 750 750 750 750 0 0 undefined

*Note: The payoff here is cost and not profit. The opportunity cost is therefore calculated as the difference between the payoff and the minimum in the same row. (e) EVPI = $20. The manufacturer should not be willing to pay more than $20 for the information about which event will occur. (g) We want to minimise the expected monetary value because it is a cost. To minimise the expected monetary value, call the mechanic. (h) Given 2 successes out of 15, the binomial probabilities and their related revised conditional probabilities are:

(i) Pr 0.01 0.05 0.10 0.20

Very low Low Moderate High

Binomial Probabilities 0.0092 0.1348 0.2669 0.2309 0.6418

Revised Conditional Probabilities 0.0092/0.6418 = 0.0143 0.1348/0.6418 = 0.2100 0.2669/0.6418 = 0.4159 0.2309/0.6418 = 0.3598

(j) (c), (e), (f) Payoff table:

Very low Low Moderate High

Pr 0.0143 0.2100 0.4159 0.3598 EMV



CV Return-torisk

A: Do Not Call Mechanic 20 100 200 400 248.3860 121 48.68% 2.0544

B: Call Mechanic 100 100 100 100 100 0 0 undefined

(b), (d) Opportunity loss table: A: Do Not B: Pr Call Mechanic Call Mechanic Very low 0.0143 80 0 Low 0.2100 0 0 Moderate 0.4159 0 100 High 0.3598 0 300 EOL 1.1440 149.53 (e) EVPI = $1.14. The manufacturer should not be willing to pay more than $1.14 for the information about which event will occur. (g) We want to minimise the expected monetary value because it is a cost. To minimise the expected monetary value, call the mechanic. E.5 Processing time Control chart for the range:

R = 3.597, LCL = 0.802, UCL = 6.392 Control chart for the mean:

X =2.2653, LCL = 1.1575, UCL = 3.3732

Processing time can be considered to be in control in terms of the mean and the range since there is no strong pattern in either chart and there are no points that are outside the control limits. Proportion of rework in the laboratory

p = 0.04737, LCL = 0.02721, UCL = 0.06752

Days 6 and 29 are above the UCL. Thus, the special causes that might be contributing to these values should be investigated before any change in the system of operation is contemplated. Then Deming’s fourteen points can be applied to improve the system. Number of daily admissions

c = 26.8, LCL = 11.2694, UCL = 42.3306 The number of daily admissions can be considered to be in control since there is no strong pattern in the chart and there are no points that are outside the control limits. E.6

Ratings for cricket and tennis relate to the same city so paired data. Let D = Difference in Ratings Cricket - Tennis Assume the distribution of differences, D, is approximately symmetric. Wilcoxon signed ranks test Test:

H0: MD ≤ 0

H1: MD > 0 Test statistic Decision rule: n’ = 10,  = 0.05 from Table E.9 critical value is 45 so if W > 45 reject H0 Calculation W = 32 < 45 so do not reject null hypothesis. Conclude at the 5% level of significance that the sample provides no evidence that cricket ratings are higher.

E.7

Same applicant is tested at home then retested in workplace so paired data. Let D = Difference in rank Home Test – Workplace Test Assume the distribution of differences, D, is approximately symmetric. Wilcoxon signed ranks test Test:

H0: MD = 0 H1: MD ≠ 0

Test statistic Decision rule: n’ = 12,  = 0.05 from Table E.9 critical values is 13 and 65 so if W < 13 or W > 65 reject H0. Calculation 13 < W = 40 < 45 so do not reject null hypothesis. Conclude at the 5% level of significance that the sample provides no evidence of a difference in median rankings between the two tests. E.8

Kruskal–Wallis Rank Test:

Let SC = Shopping Centre, MR = Main Road, S = Suburban H0: MSC = MMR = MS H1: not all medians are equal Test statistic, H follows a chi-squared distribution, so the critical value with 17 degrees of freedom and a 0.05 significance level is 27.587. Do not reject the null hypothesis at 5% significance level. Conclude that there is no evidence of a difference in median sales between the store locations.

E.9 (a)

Friedman rank test for ranking by applicant blocked by Panel Member Assume populations have similar variability and shape. not all medians are equal.

Upper critical value: Since = 17.37 > 7.815 reject H0 at the 0.05 level of significance. There is sufficient evidence of a difference in the median ratings for the four applicants.

(b)

Answer to (a) applies to (b) also. The assumption made in (b) does not change our testing approach since the Friedman rank test can be used when analysing a randomised block design.

E.10

(a)

p-chart for proportion of dissatisfied students As there are no points outside the control limits the process is in control.

(b)

p-chart for proportion of students satisfied with unit. Process is not in control, need to investigate the causes of the low, below lower control limit, student satisfaction in Semester 6.

E.11

Let

A/B = Day to sell property, neighbourhood A/B

As properties are in different/separate neighbourhoods independent populations. Testing for a difference in medians of two independent populations so use Wilcoxon Rank Sum Test. Test Level of significance: Test Statistics: sum of the ranks for neighbourhood A as , can use the normal approximation Critical value and decision rule: Reject if the calculated value of the test statistic Z < -1.960 or Z > 1.960 otherwise do not reject Alternatively from Table E.9 critical values 82 and 128 so reject if TA < 82 or TA > 128, otherwise do not reject Calculations:

Conclusion:

Since -1.209 > -1.960 (alternatively 82 < TA = 89 < 128) do not reject

Therefore, conclude, at a 5% level of significance, that there is no evidence that there is a difference in the median number of days a property takes to sell between the two neighbourhoods? E.12

Let Mon = Number of lunchtime customers Monday etc Assume that the seven populations have similar variability and shape. Testing for difference of medians for five independent populations so use Kruskal-Wallis rank test

Use level of significance  = 0.01 Decision rule:

reject null hypothesis if

Test statistic:

Decision: Since Hcalc = 5.64 < 13.277 do not reject H0. There is no evidence to show that the median number of customers differs during the week.

E.13

(a)

(b)

Process is not in control, need to investigate the causes of the high, above upper control limit, number of dissatisfied customers on Day 8 – Monday Week 2.

E.14

Let

1/2 = Student satisfaction Semester 1/2

As generally different students attempting unit in each semester independent populations. Testing for a difference in medians of two independent populations so use Wilcoxon Rank Sum Test. Test Level of significance: Test Statistics:

sum of the ranks for Semester 1

as , can use the normal approximation Critical value and decision rule: Reject if the calculated value of the test statistic Z < -2.3263 otherwise do not reject PhStat Output: Conclusion: Since Z = -2.2376 > -2.3263 (alternatively p-value = 0.0126 > 0.01) do not reject Therefore, we conclude, at a 1% level of significance, that there is no evidence that the changes have resulted in higher student satisfaction in Semester 2. E.15

Testing for difference in two related proportions so McNemar Test

(a)

≥0 <0

where 1 = before, 2 = after

Decision rule:  = 0.1 if Z < –1.2816 reject H0. Test statistic: Decision: Since Z = –2.334 is less than the lower critical value of –1.2816, reject H0 at the 10% significance level. At the 10% level of significance can conclude that the proportion of coffee drinkers who prefer brand A was lower at the beginning of the advertising campaign than at the end of the advertising campaign.

(b)

p-value = 0.0099. The probability of obtaining a data set that gives rise to a test statistic that differs from 0 by –2.33 or more is 0.99% if the null hypothesis is true.

E.16 Friedman rank test for Car blocked by Reviewer

Assume populations have similar variability and shape not all medians are equal.

Test statistic: Upper critical value: Since = 2.01 < 11.345 do not reject H0 at the 0.01 level of significance. There is no evidence of any difference in the satisfaction ratings for the different brands. E.17

Testing for difference in two related proportions so McNemar Test

(a)

≥0 <0

where 1 = before, 2 = after

Decision rule:  = 0.01 if Z < –2.3263 reject H0. Test statistic: Decision: Since Z = –3.023 is less than the lower critical value of −2.3263, reject H0 at the 1% significance level. At the 1% level of significance can conclude that the proportion of students attending a tutorial is higher when students are aware that tutorial attendance records are being kept (b)

p-value = 0.00126. The probability of obtaining a data set that gives rise to a test statistic that differs from 0 by –3.02 or more is 0.0012 if the null hypothesis is true.

E.18 (a)