SOLUTIONS MANUAL for Business Statistics: For Contemporary Decision Making, Canadian Edition 3rd Edi by StudyGuide

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

SOLUTIONS TO PROBLEMS IN CHAPTER 1: INTRODUCTION TO STATISTICS 1.1

Examples of data in business disciplines: accounting – revenue/sales, cost of goods, salary expense, depreciation, utility costs, taxes, equipment inventory, etc. finance - World bank bond rates, number of failed savings and loans companies, measured risk of common stocks, stock dividends, foreign exchange rate, liquidity rates for a single-family, etc. human resources - salaries, size of engineering staff, years experience, age of employees, years of education, etc. marketing - number of units sold, dollar sales volume, forecast sales, size of sales force, market share, measurement of consumer motivation, measurement of consumer frustration, measurement of brand preference, attitude measurement, measurement of consumer risk, etc. operations and supply chain management - the number, location, and network missions of suppliers, production facilities, distribution centers, warehouses, cross-docks, and customers; the quantity and location of inventory, including raw materials, work in process (WIP), and finished goods, etc. information systems - CPU time, size of memory, number of work stations, storage capacity, percent of professionals who are connected to a computer network, dollar assets of company computing, number of “hits” on the Internet, time spent on the Internet per day, percentage of people who use the Internet, retail dollars spent in e-commerce, etc. production - number of production runs per day, weight of a product; assembly time, number of defects per run, temperature in the plant, amount of inventory, turnaround time, etc. management - measurement of union participation, measurement of employer support, measurement of tendency to control, number of subordinates reporting to a manager, measurement of leadership style, etc.

1.2

Examples of data that can be gathered for decision-making purposes in business industries:

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

manufacturing - size of punched hole, number of rejects, amount of inventory, amount of production, number of production workers, etc. insurance - number of claims per month, average amount of life insurance per family head, life expectancy, cost of repairs for major auto collision, average medical costs incurred for a single female over 45 years of age, etc. travel - cost of airfare, number of miles traveled for ground transported vacations, number of nights away from home, size of traveling party, amount spent per day on vacation besides lodging, etc. retailing - inventory turnover ratio, sales volume, size of sales force, number of competitors within 2 miles of retail outlet, area of store, number of sales people, etc. communications - cost per minute, number of phones per office, miles of cable per customer headquarters, minutes per day of long distance usage, number of operators, time between calls, etc. computing - age of company hardware, cost of software, number of CAD/CAM stations, age of computer operators, measure to evaluate competing software packages, size of data base, etc. agriculture - number of farms per county, farm income, number of acres of corn per farm, wholesale price of a gallon of milk, number of livestock, grain storage capacity, etc. banking - size of deposit, number of failed banks, amount loaned to foreign banks, number of tellers per drive-in facility, average amount of withdrawal from automatic teller machine, federal reserve discount rate, etc. healthcare - number of patients per physician per day, average cost of hospital stay, average daily census of hospital, time spent waiting to see a physician, patient satisfaction, number of blood tests done per week. 1.3

Descriptive statistics in recorded music industry : 1)

RCA total sales of compact discs this week, number of artists under contract to a company at a given time.

total dollars spent on advertising last month to promote an album.

number of units produced in a day.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

number of retail outlets selling the company's products.

Inferential statistics in recorded music industry : 1)

Measure the average amount spent per month on recorded music for a few consumers, then use that figure to infer the average amount for the population.

Determination of market share for rap music by using the proportion of a randomly selected sample of 500 purchasers of recorded music who purchased rap music.

Determination of top ten single records by sampling the number of requests at a few radio stations.

Estimation of the average length of a single recording by using the average length of a sample of records.

The difference between descriptive and inferential statistics lies mainly in the usage of the data. These descriptive examples all gather data from every item in the population about which the description is being made. For example, RCA measures the sales on all its compact discs for a week and reports the total. In each of the inferential statistics examples, a sample of the population is taken and the population value is estimated or inferred from the sample. For example, it may be practically impossible to determine the proportion of buyers who prefer rap music. However, a random sample of buyers can be contacted and interviewed for music preference. The results can be inferred to population market share. 1.4

Descriptive statistics in manufacturing batteries to make better decisions : 1)

total number of worker hours per plant per week - help management understand labor costs, work allocation, productivity, etc.

company sales volume of batteries in a year - help management decide if the product is profitable, how much to advertise in coming year and compare to costs to determine profitability.

total amount of sulfuric acid purchased per month for use in battery production. - can be used by management to study wasted inventory, scrap, etc.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

total amount of defective and/or rejected units produced per day/week/month. - can be used by management to study wasted resources, need for labor training, redesign of assembly process, etc.

Inferential Statistics in manufacturing batteries to make decisions :

1.5

Take a sample of batteries and test them to determine the average shelf life - use the sample average to reach conclusions about all batteries of this type. Management can then make labeling and advertising claims. They can compare these figures to the shelflife of competing batteries.

Take a sample of battery consumers and determine how many batteries they purchase per year. Infer to the entire population management can use this information to estimate market potential and penetration.

Interview a random sample of production workers to determine attitude towards company management - management can use this survey results to ascertain employee morale and to direct efforts towards creating a more positive working environment which, hopefully, results in greater productivity.

Size of sale ($) per customer in men‟s formal wear. Either by taking a sample or using a census, management could compute the average sale in men‟s formal wear of a weekly period and compare the number to the same average taken a year ago or a month ago to determine if more is being purchased per customer. Other variables might include number of sales per hour, number of people entering the department per day, number of dress shirts sold per day, etc.

Number of employees working per day. This variable could indicate the day of the week (certain days have more or less sales), sales activity (how sales are doing overall), or even health of associates. Other variables might include percent of employees absent due to illness, average number of hours worked per week per employee, number of open positions, etc.

Inventory turnover rate. How fast are items in the store selling? Other variables might include reorder rate, percent of storage space utilized, number of stockouts per week, etc.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

1.6

Number of customers that enter the store per hour. This figure will vary by day, time of day, and season. Comparing figures on this variable from period to period can give some indication of sales trends which can help drive human resource planning, etc. Other variables might include amount of time spent per customer in the store per visit, distance that customers travel to shop in the store, number of referrals that customers make to other people annually, etc.

Percentage of people paying with cash. Percentage of people using credit cards. These can be used to expedite pay systems, investigate employee theft, calculate surcharges associated with credit cards, etc. Other variables might include average time per checkout, average wait time in pay line, etc.

Size of bill or tab. This variable is the total amount in dollars spent by a patron per visit to the restaurant. The bill or tab could be for an individual or a group and would include both food and beverages if they are all included in the bill. Of course, the measurement would be in dollars. This information could be very useful for the manager or owner to know the average size of a bill both in projecting out total revenues over a period or as a baseline before a marketing effort to increase sales.

Percentage of Capacity Filled. This variable could be measured at various intervals, times, and days of the week. The measurement would be calculated by taking the number of patrons in the restaurant at any one time divided by the total number of seats in the restaurant (capacity). From this, management could make staffing decisions for various times and days of the week. In addition, management could make decisions about when to expand, how much to advertise, and/or when to run specials.

Length of Stay. The measurement is how many minutes people are actually in the restaurant from the time they are assigned a table until they are leaving. From this, management could determine customer turnover rates which have capacity implications. That is, how many times in a day is an average table “turned over”. If people stay longer, do they spend more?

Number of Arrivals per 5-minute intervals. The measurement is how many customers arrive at the front door to be greeted by the maître„d in any given five-minute period. This figure will likely vary by day of the week, season of the year, and time of day.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Management can use this information for staffing decisions and planning. 1.7

a) b) c) d) e) f) g) h)

ratio ratio ordinal nominal ordinal ratio nominal ratio

1.8

a) b) c) d) e) f) g) h)

ordinal ratio nominal ratio ratio ratio nominal ordinal

1.9

The population for this study is the 900 electric contractors who purchased Mapletech‟s wire.

The sample is the randomly chosen group of 35 contractors.

The statistic is the average satisfaction score for the sample of thirty-five contractors.

The parameter is the average satisfaction score for all 900 electric contractors in the population.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Legal Notice Copyright

Copyright © 2020 by John Wiley & Sons Canada, Ltd. or related companies. All rights reserved. The data contained in these files are protected by copyright. This manual is furnished under licence and may be used only in accordance with the terms of such licence. The material provided herein may not be downloaded, reproduced, stored in a retrieval system, modified, made available on a network, used to create derivative works, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning, or otherwise without the prior written permission of John Wiley & Sons Canada, Ltd.

SOLUTIONS TO PROBLEMS IN CHAPTER 2

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Range = max-min = 29-(-12) = 41 Class width = 41/5 = 8.2 ≈ 9 The 5 class frequency distribution: Class Interval –15 - under –6 –6 - under 3 3 - under 12 12 - under 21 21 - under 30 Totals

Frequency 7 12 13 9 9 50

b) Class width = 41/10 = 4.1 ≈ 5. 10 class frequency distribution: Class Interval –15 - under –10 –10 - under –5 –5 - under 0 0 - under 5 5 - under 10 10 - under 15 15 - under 20 20 - under 25 25 - under 30 30 - under 35 Totals

Frequency 2 5 7 10 7 3 7 4 5 0 50

c) The ten class frequency distribution gives a more detailed breakdown of temperatures. It allows locating more accurately the temperatures with the greatest frequency. The class with the highest frequency (modal class), is 0 – under 5 class with a frequency of 10. The five class distribution collapses the intervals into broader classes making it appear that there are nearly equal frequencies in each class.

2.2 The range is 22 (61-39). We will use 8 classes. The class width will be 3. The frequency distribution is as shown below Class Interval

Frequency

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

39 - under 42 42 - under 45 45 - under 48 48 - under 51 51 - under 54 54 - under 57 57 - under 60 60 - under 63 Totals

3 5 20 21 26 11 11 3 100

The distribution reveals that 49 of the 100 boxes of raisins contain 51 or less raisin (39 under 51). It shows that there are three boxes that have 10 or more extra raisins (60 under 63) and three boxes that have 8-11 less raisins (39 – under 41) than the boxes are supposed to contain.

2.3 Class Cumulative Interval Frequency Frequency 0-5 6 5 - 10 8 10 - 15 17 15 - 20 23 20 - 25 18 25 - 30 10 30 - 35 4 TOTAL 86

Class

Relative

Midpoint

Frequency

2.5 7.5 12.5 17.5 22.5 27.5 32.5

6/86 = .0698 .0930 .1977 .2674 .2093 .1163 .0465 1.0000

6 14 31 54 72 82 86

The relative frequency tells us that it is most probable that a customer is in the 15 - 20 category (.2674). Over two thirds (.6744) of the customers are between 10 and 25 years of age.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

2.4 Class

Class

Relative

Cumulative Interval

Frequency

Midpoint

Frequency

0-2 2-4 4-6 6-8 8-10 TOTAL

218 207 56 11 8 500

1 3 5 7 9

.436 .414 .112 .022 .016 1.000

Frequency

2.5

2.6

218 425 481 492 500

Some examples of cumulative frequencies in business: sales for the fiscal year, costs for the fiscal year, spending for the fiscal year, inventory build-up, accumulation of workers during a hiring buildup, production output over a time period.

Histogram:

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Frequency Polygon:

Comment: The assembly times “pile up” near the middle of the graphs indicating that many of the assembly times are between 36 and 42 minutes.

2.7

Histogram:

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Frequency Polygon:

Comment: The histogram indicates that the number of calls per shift varies widely. However, the heavy numbers of calls per shift fall in the 50 to 80 range. Since these numbers occur quite frequently, staffing planning should be done with these number of calls in mind realizing from the rest of the graph that there may be shifts with as few as 10 to 20 calls.

2.8 Ogive:

Cummulative Frequency

60 50 40 30 20 10 0 4.5

7.5

10.5

13.5

16.5

19.5

Midpoint

2.9

STEM

LEAF

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

21 22 23 24 25 26 27

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

The stem and leaf plot indicates that sales prices vary quite a bit within the range of $212,000 and $273,000. It is evident from the stem and leaf plot that there is a strong grouping of prices in the five price ranges from the $220‟s through the $260‟s.

2.10

STEM

LEAF 1 2 3 4 5 6 7 8

3, 6, 7, 7, 7, 9, 9, 9 0, 3, 3, 5, 7, 8, 9, 9 2, 3, 4, 5, 7, 8, 8 1, 4, 5, 6, 6, 7, 7, 8, 8, 9 0, 1, 2, 2, 7, 8, 9 0, 1, 4, 5, 6, 7, 9 0, 7 0

The stem and leaf plot shows that the numbers of passengers per flight were relatively evenly distributed between the high teens through the sixties. Rarely was there a flight with at least 70 passengers. The category of 40's contained the most flights (10). 2.11 The histogram shows that there is only one airport with more than 105 million passengers and from the given problem information, we know that that airport is Atlanta‟s Hartsfield-Jackson International Airport which has more than 105 million passengers. 40% (12) of the top 30 airports have between 65 and 75 million passengers. The next largest grouping is between 55 and 65 million passengers in which there are six airports. 2.12 We assume that the class endpoints (10, 20, 30, …) are indicated on the horizontal axis and the marks between them represent the class midpoints; the cumulative frequencies are marked on the vertical axis. The ogive shows that out of 200, 50 pots contain less than 10 legal king crabs. About 90 selected pots (45%) have fewer than 20 legal king crabs, and about 120 (60%) contain less than 30 legal king crabs. However, 180 pots (90%) have less than 80 legal king crabs.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

2.13 Airlines

Number of passengers (in millions)

Delta

164.6

164.6/787.7= 0.209

United

140.4

0.178

Southwest

134.0

0.170

American

107.9

0.137

China Southern

86.5

0.110

Ryanair

79.6

0.101

Lufthansa

74.7

0.095

Totals

787.7

1.000

360

Proportion

Pie chart:

Degrees

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Bar chart:

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

2.14 Firm

Revenue ($U.S. millions)

Pfizer

67,809

67,809/291,153=0.233

Novartis

53,324

0.183

Merck & Co.

45,987

0.158

Bayer

44,200

0.152

GlaxoSmithKline

42,813

0.147

Johnson and Johnson

37,020

0.127

Totals

291,153

1.000

360

Proportion

Degrees

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Pie chart:

Bar chart:

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Underwriting Firm RBC Capital Markets TD Securities CIBC World Markets National Bank Financial BMO Capital Markets Total

Total Amount ($million) 29,172 25,193 19,875

Percent 26.8 23.1 18.3

17,653 16,958 108,851

16.2 15.6 100.0

Bar chart:

Total Amount ($million) BMO Capital Markets National Bank Financial CIBC World Markets TD Securities RBC Capital Markets 0

5,000

10,000 15,000 20,000 25,000 30,000 35,000

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Pie chart:

15.6 26.8 16.2

18.3

23.1

RBC Capital Markets

TD Secirities

CIBC World Markets

National Bank Financial

BMO Capital Markets

The proportion, sizes and color of the pie slices clearly shows that RBC Capital Markets raised the highest amount ($29,172 million, 26.8%) and BMO Capital Markets raised the lowest amount ($16,958 million, 15.8%). 2. 16 Destination

Revenue ($ millions)

Proportion Degrees

U.S.

1,697

0.7767

279.61

Mexico

269

0.1231

44.32

Japan

171

0.0783

28.19

South Korea

0.0128

4.61

Taiwan

0.0073

2.63

China

0.0018

0.65

Totals

2,185

1.0000

360.0

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Pie Chart:

2.17

Complaint Busy Signal Too long a Wait Could not get through Got Disconnected Transferred to the Wrong Person Poor Connection Total

Number

% of Total

420 184 85 37 10 8 744

56.45 24.73 11.42 4.97 1.34 1.08 99.99

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

2.18

Generally speaking, the tendency is that less fish caught and used for industrial products when more fish is caught and used for human food.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

2.19

Generally speaking, the tendency is that sales are higher when more money is spent on advertising.

2.20 It appears from the graph that as job satisfaction decreases, there is an increase in tardiness. Thus, there appears to be an inverse relationship between job satisfaction and tardiness. The scatter plot also shows that when employees are highly satisfied, the level of tardiness is low.

2.21 One-Way Commute Distance (in km) 0-3

4 - 10

More than 10

Total

Number of Annual

0-2

184

117

396

Non-vacation -Day

3-5

114

More than 5

Absences

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Total

119

231

182

532

There is a slight tendency for there to be a few more absences as plant workers commute further distances. Say, 6.6% of those who commute more than 10 km had more than 5 non-vacation absent days, as compared to 2.5% and 3% for those who commute 0-3 km and 4-10 km respectively. Comparing workers who travel 4-10 km to those who travel 03 km, there is about a 2:1 ratio in all three cells (0-2, 3-5, more than 5 non-vacation day absences) indicating that for these two categories (0-3 and 4-10), number of absences is essentially independent of commute distance.

2.22 Level of Education High School only

University Degree

Total

Acceptable

Unacceptable

Total

Rating of Service

There is a tendency that the lower the level of education, the more acceptable is the service. First of all, according to the table, the numbers of customers with different level of education are almost the same (11 with “high school only” and 14 with “university degree”). It appears that a much higher proportion of high school level education customers rate the service as “acceptable” than as “unacceptable” (9 to 2 ratio or about 4.5 times as many). On the other hand, more of the university degree educated customers rated the service as “unacceptable” than as “acceptable” (8 to 6 ratio). In addition, 60% or 9 customers out of 15 who rate the service as “acceptable” have only high school education, and 80% or 8 customers out of 10 who rate the service as “unacceptable” are university degree level educated.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

2.23 a.

Time Series plot of Year 1 Sales 9,500

Sales in Year 1

9,000 8,500 8,000 7,500 7,000 6,500 6,000

Month

This time-series plot for the year 1 shows up and down swings of sales until October with sales strongly increasing for November and December due to perhaps the holiday season. If true, we could be seeing possible seasonal effects of the holiday season followed by low sales in January.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Time Series Plot of Year 1 to Year 5 Sales 12,000 11,000

Sales

10,000 9,000 8,000 7,000 6,000

Month Year 1

Year 2

Year 3

Year 4

Year 5

When placed together on one time series plot, the graphs tend to show the same shape virtually every year. This underscores the notion that there is some seasonality in the data. Sales are down in January, April, June, and October. Sales are highest in December, increases every year from October through December, and then drops dramatically in January. c.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

With all five years of data shown chronologically, it is apparent that there is a repeating seasonal cycle within each year. In addition, however, there is a general trend of increasing sales over the five years.

2.24

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Time Series Plot of Coolsville 2310 2290

Exports

2270 2250 2230 2210 2190 2170 2150 1

Quarter

Examining the exports for Coolsville over these two years, there was an increase in exports from quarter 1 to quarter 2, followed by a decline in quarter 3 and 4 in year 1. In year 2, there was an increase in exports from quarter 1 to 2 followed by a decline in quarter 3 and other increase in quarter 4. Overall, there seems to be little or no trend in exports over time just a pattern of ups and downs. b.

Time Series Plot of City Exports over 8 Quarters 14,000 12,000

Exports

10,000 8,000 6,000 4,000 2,000 0 1

Quarter Coolsville

Genius Grove

Highland

Inner City

Metro City

New New York

Orbit City

Springfield

Many of the cities actually have relatively constant exports over these 8 quarters with few significant ups and downs. Metro City (second highest at the end) seems to be the most volatile of the cities in terms of exports. Genius Grove consistently has the highest exports. 2.25 Range = max-min = 57-18 = 39 Solutions Manual 1-27 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Class width = 39/6 = 6.5 ≈ 7 Class Interval

Frequencies

16 - under 23 23 - under 30 30 - under 37 37 - under 44 44 - under 51 51 - under 58 TOTAL

6 9 4 4 4 3 30

2.26

2.27

Class Interval

Frequency

Midpoint

Rel. Freq.

Cum. Freq.

20 - under 25 25 - under 30 30 - under 35 35 - under 40 40 - under 45 45 - under 50

17 20 16 15 8 6

22.5 27.5 32.5 37.5 42.5 47.5

.207 .244 .195 .183 .098 .073

17 37 53 68 76 82

Class Interval

Frequencies

50 - under 60 60 - under 70 70 - under 80 80 - under 90 90 - under 100 TOTAL

13 27 43 31 9 123

Histogram:

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Frequency Polygon:

Ogive:

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

2.28

STEM 28 29 30 31 32 33

2.29

LEAF 4, 6, 9 0, 4, 8 1, 6, 8, 9 1, 2, 4, 6, 7, 7 4, 4, 6

5 Label A B C D

Value 55 121 83 46

Proportion .180 .397 .272 .151

Degrees 65 143 98 54 TOTAL 1.000

305 360

Pie Chart:

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

2.30

Bar Graph: Category A B C D E

2.31

Problem

Frequency 7 12 14 5 19

Frequency

Percent of Total

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

1 2 3 4 5 6 7 8 9 10

673 29 108 379 73 564 12 402 54 202 2496

26.96 1.16 4.33 15.18 2.92 22.60 0.48 16.11 2.16 8.09

Pareto Chart:

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

0 0

2.33 Yellowknife Steel Company Range = max-min = 69-36 = 33 Class width = 33/8 = 4.125 ≈ 5 Class Interval 32 - under 37 37 - under 42 42 - under 47 47 - under 52 52 - under 57 57 - under 62 62 - under 67 67 - under 72 TOTAL

Frequency 1 4 12 11 14 5 2 1 50

The highest frequencies are between 42 and 57.

2.34 Class Cumulative Interval

Frequency

Class

Relative

Midpoint

Frequency

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

20 – 25 25 – 30 30 – 35 35 – 40 40 – 45 45 – 50 TOTAL

2.35

8 6 5 12 15 7 53

22.5 27.5 32.5 37.5 42.5 47.5

8/53 = .1509 .1132 .0943 .2264 .2830 .1321 .9999

8 14 19 31 46 53

Frequency Distribution: Class Interval 10 - under 20 20 - under 30 30 - under 40 40 - under 50 50 - under 60 60 - under 70 70 - under 80 80 - under 90

Frequency 2 3 9 7 12 9 6 2 50

Histogram:

Frequency Polygon:

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

The normal distribution appears to peak near the center and diminish towards the end intervals.

2.36

a. Histogram and a Frequency Polygon for Problem 2.34 Class Interval 20 – 25 25 – 30 30 – 35 35 – 40 40 – 45 45 – 50 TOTAL

Frequency 8 6 5 12 15 7 53

Cumulative Frequency 8 14 19 31 46 53

Histogram:

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Frequency Polygon:

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

b. Ogive:

Cumulative Frequency

50 40 30 20 10 0 22.5

27.5

32.5

37.5

42.5

47.5

Class Midpoint

2.37 Asking Price

Frequency

Cumulative Frequency

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

$ 80,000 - under $ 100,000 $ 100,000 - under $ 120,000 $ 120,000 - under $ 140,000 $ 140,000 - under $ 160,000 $ 160,000 - under $ 180,000 $ 180,000 - under $ 200,000

21 27 18 11 6 3 86

21 48 66 77 83 86

Histogram: 30

Frequency

0 90,000

110,000

130,000

150,000

170,000

190,000

Class Midpoints

Frequency Polygon:

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Frequency

0 90,000

110,000

130,000

150,000

170,000

190,000

Class Midpoints

Ogive: 100

Frequencies

0 80,000

100,000

120,000

140,000

160,000

180,000

200,000

Class Endpoints

2.38

Stem and Leaf Plot: STEM 1 2 3

LEAF 2, 3, 6, 7, 8, 8, 8, 9, 9 0, 3, 4, 5, 6, 7, 8 0, 1, 2, 2

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Comments: The stem and leaf plot shows that the travel times are relatively evenly spread out between 12 days and 32 days. It also shows that the most travel times fall in the 12 to 19 day interval followed by the 20 to 28 day interval. Only four of the travel times were thirty or more days; 18 days is the most frequently occurring travel time (occurred three times).

2.39 Price

Frequency

$1.75 - under $1.90 $1.90 - under $2.05 $2.05 - under $2.20 $2.20 - under $2.35 $2.35 - under $2.50 $2.50 - under $2.65 $2.65 - under $2.80

9 14 17 16 18 8 5 87

Cumulative Frequency 9 23 40 56 74 82 87

Histogram:

Frequency Polygon:

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Ogive:

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

2.40

Genre R&B Alternative Rap Country Soundtrack Metal Classical Latin TOTAL

Albums Sold 146.4 102.6 73.7 64.5 56.4 26.6 14.8 14.5

Proportion .29 .21 .15 .13 .11 .05 .03 .03 1.00

Degrees 104.4 75.6 54.0 46.8 39.6 18.0 10.8 10.8 360.0

Pie Chart:

Classical 3% Latin 3% Metal 5%

R&B 29% Soundtrack 11% Country 13%

Rap 15%

Alternative 21%

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

2.41

The higher is the exchange rate more Canadian travellers go to the U.S. and less the U.S. travellers do to Canada. Solutions Manual 1-43 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

2.42 Industrial Source

Total Emissions (tonnes)

Proportion

Degrees

Upstream petroleum

1,637,651

0.430

154.6

Aluminum

496,103

0.130

46.8

Wood

385,909

0.101

36.4

Nonferrous smelting and refining

354,907

0.093

33.5

Mining and rock quarrying

312,598

0.082

29.5

Pulp and paper

193,447

0.051

18.3

Downstream petroleum

145,192

0.038

13.7

Cement and concrete

128,856

0.034

12.2

Chemicals

82,202

0.022

7.8

Grain

75,632

0.020

7.1

Totals

3,812,497

1.0000

360.0

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

2.43 The Pareto chart indicates that fault in raw material causes 44.2% of the defects and becomes the major problem. According to the chart, 23.4% of the bottles were rejected because of incorrect thickness which can be identified as the second severe problem. The steepest slopes correspond to “fault in raw material”, “thickness”, and “broken handle” categories. They represent 84.8% causes of poor-quality bottles.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

2.44

STEM 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59

LEAF 12, 16, 24, 32, 99, 99 04, 28, 39, 46, 61, 88 20, 40, 59 12 53, 54 30, 34, 58 22, 34, 66, 78 63 48, 49, 90 66 21, 54, 57, 63, 91 38, 66, 66 31, 78 56 69 37, 50 31, 32, 58, 73 19, 23

2.45 STEM

LEAF

00, 68

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

93 94 95 96 97 98 99 100

01, 37, 44, 75 05, 37, 48, 60, 68 24, 55 02, 56, 70, 77 42, 60, 64 14, 30 22, 61, 75, 76, 90, 96 02, 10

2.46 The histogram shows that all household incomes fall between $30,000 and 140,000. Since the distribution is almost bell-shaped, the data are approximately normally distributed. The centre of the histogram is located left of $100,000 and indicates that the average household income of mall shoppers is about $90,000. The heights of the tallest three rectangles (around 30 observations each) show most of the mall shoppers have household income between $75,000 and $105,000. There are no outliers.

2.47 Family practice is most prevalent - followed by pediatrics . A virtual tie exists between ob/gyn, general surgery, anesthesiology, and psychiatry.

2.48 Stem-and-leaf plots can be thought of as histograms that are produced by the observations themselves, rather than by columns or bars representing the values. Since we have the observations themselves, stem-and-leaf plots often illustrate more detailed information than equivalent histograms. In this stem-and-leaf plot of the audits for 100 CPA firms, we can observe that the range is 30 audits (the largest value of 42 minus the smallest value of 12). The observations are concentrated on the top stems between 12-19 audits, with other concentrations for 26-29 audits and also for 34-35 audits. The stem-and-leaf plot does not show a clear central tendency, indeed, one could say that the plots is skewed since the density of observations is higher on the top stems.

2.49 Ogives are useful when the analyst wants to see running totals (or cumulative frequencies). Thus, ogives are useful when identifying steep increases in frequencies. In this ogive, particularly steep increases occur in November and December, indicating large Solutions Manual 1-47 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

jumps (the largest being in December) in monthly frequency totals, likely due to the holidays season. Notice also a steep increase in April, can you explain why this might be?

2.50 This is a time series, cross-sectional data. One can examine trends in representation using time series plots. Pie chart or bar chart can also be used to visualize representation for each year.

Representation

Time Series Plot of Political Party Representation Over 20 Years 43 41 39 37 35 33 31 29 27 25 1

Year Liberal

Conservative

Other

Conservative had greater representation in the first 4 year while the Liberal Party had greater representation from year 5 through year 12. Over the last 8 years the Conservative Party had the majority representation.

Representation in Year 20

Liberal

Conservative

Other

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

The sizes and color of the pie slices clearly shows that the Conservative Party had majority representation in year 20 followed by the Liberal Party. 2.51

Revenue (US$ billions)

Time Series Plot of Revenue for Coca-Cola Company Over 17 Years 60.0 50.0 40.0 30.0 20.0 10.0 0.0 1

Year

This time-series plot shows that the company enjoyed revenue growth from year 1 to year 15. Revenue appears to be declining after year 15.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Legal Notice Copyright

SOLUTIONS TO PROBLEMS IN CHAPTER 3: DESCRIPTIVE STATISTICS

3.1

Mean 17.3 44.5 µ = x/N = (333.6)/8 = 41.7 31.6 40.0 52.8 = x/n = (333.6)/8 = 41.7 38.8 30.1 78.5 (It is not stated in the problem whether the x = 333.6 data represent as population or a sample).

3.2

Mean 7 -2 5

µ = x/N = -12/12 = -1

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

9 0 -3 -6 -7 -4 -5 2 -8 x = -12

= x/n = -12/12 = -1

(It is not stated in the problem whether the data represent a population or a sample).

3.3 Median for values in 3.1 Arrange in ascending order: 17.3 , 30.1 , 31.6 , 38.8 , 40 , 44.5 , 52.8 , 78.5 There are 8 terms. Since there is an even number of terms, the median is the average of the two middle values, 38.8 and 40. The median =

n1 th 8 1 term = = 4.5th term. 2 2 The median is located halfway between the 4th and 5th terms or the average of 38.8 and 40. So, the median is 39.4. Using the formula, the median is located at the

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

3.4

Median Arrange terms in ascending order: 73, 167, 199, 213, 243, 345, 444, 524, 609, 682 There are 10 terms. Since there is an even number of terms, the median is the average of the middle two terms: (243 345) 588 Median = = 294  2 2 n1 th Using the formula, the median is located at the term 2 10  1 11 n = 10 therefore = 5.5th term.  2 2 The median is located halfway between the 5th and 6th terms. 5th term = 243

6th term = 345

Halfway between 243 and 345 is the median = 294 3.5

Mode Arrange terms in ascending order 2, 2, 3, 3, 4, 4, 4, 4, 5, 6, 7, 8, 8, 8, 9 There are 15 terms. Since there is an odd number of terms, the median is the middle number. Median = 4 n1 th Using the formula, the median is located at the term 2 n = 15, therefore (n+1)/2 = 8th term. The median is located at the 8th term. Median = 4 4 is the most frequently occurring value The mode = 4

3.6

Rearranging the data into ascending order: 11, 13, 16, 17, 18, 19, 20, 25, 27, 28, 29, 30, 32, 33, 34

P35 is located at the 5 + 1 = 6th term,

P35 = 19

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

P55 is located at the 8 + 1 = 9th term,

P55 = 27

Q1 = P25 but Q1 = P25 is located at the 3 + 1 = 4th term,

Q2 = Median but:

Q1 = 17

The median is located at the

Q2 = 25

Q3 = P75 but Q3 = P75 is located at the 11 + 1 = 12th term, Q3 = 30 3.7

Rearranging the data in ascending order: 80, 94, 97, 105, 107, 112, 116, 116, 118, 119, 120, 127, 128, 138, 138, 139, 142, 143, 144, 145, 150, 162, 171, 172 n = 24 For P20: Thus, P20 is located at the 4 + 1 = 5th term and

P20 = 107

For P47: Thus, P47 is located at the 11 + 1 = 12th term and

P47 = 127

For P83: Thus, P83 is located at the 19 + 1 = 20th term and

P83 = 145

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Q1 = P25 For P25: Since i is a whole number, Q1 is the average of the terms at 6th and 7th , and therefore Q1 = (112 + 116)/ 2 = 114 Q2 = Median The median is located at the: Thus, Q2 = (127 + 128)/ 2 = 127.5 Q3 = P75 For P75: Since i is a whole number, Q3 is the average of the terms at 18th and 19th and Q3 = (143 + 144)/ 2 = 143.5 3.8

Mean: ∑

The mean is 2,414.27 The median is located at the

= 8th term

Median = 2,357 If there are extreme values, the median has an advantage over the mean Q2 = Median = 2,357 For P63, P63 is located at the 9 + 1 = 10th term,

P63 = 2,534

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

P29 is located at the 4 + 1 = 5th term,

P29 = 1,952

As a measure of central tendency, identifying specific percentile values can help to determine where more (or fewer) of the data values lie, or where a specific data value lies. In this knowing where a certain bank lies within the percentile range may help to decide investment and/or lending decisions. 3.9

Rearranging the data into ascending order: 1,738,160 1,804,523 2,337,237 2,443,040 2,639,735 2,886,343 3,161,955 3,581,445 6,118,221 6,793,714 6,867,465 8,157,058 The median is located at the

= 6.5th term

Thus, Q2 = (2,886,343 + 3,161,955)/ 2 = 3,024,149 Q3 = P75 For P75: Thus, Q3 is average of the 9th and 10th terms and Q3 = (6,118,221 + 6,793,714)/ 2 = 6,455,967.5 For P20: P20 is located at the 2 + 1 = 3rd term,

P20 = 2,337,237

For P60: P60 is located at the 7 + 1 = 8th term,

P60 = 3,581,445

For P80: P80 is located at the 9 + 1 = 10th term,

P80 = 6,793,714

For P93: P93 is located at the 11 + 1 = 12th term,

P93 = 8,157,058

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

3.10

n = 17;

Mean =

The median is located at the

The mean is 3.588

= 9th term,

Median = 4

There are eight 4‟s, therefore the Mode = 4 Q3 = P75:

Q3 is located at the 13th term and P11:

P11 is located at the 2nd term and P35:

P35 = 3

P58 is located at the 10th term and P67:

P11 = 1

P35 is located at the 6th term and P58:

Q3 = 4

P58 = 4

P67 is located at the 12th term and

P67 = 4

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

3.11

x 6 2 4 9 1 3 5 x = 30

6-4.2857 =

x-µ =

 x -µ 1.7143 2.2857 0.2857 4.7143 3.2857 1.2857 0.7143 14.2857

(x-µ)2 2.9388 5.2244 0.0816 22.2246 10.7958 1.6530 0.5102 (x -µ)2 = 43.4284

a.) Range = 9 - 1 = 8 b.) M.A.D. =

2.0408

c.) 2 =

= 6.2041

d.)  =

= 2.4908

e.) Arranging the data in order:

1, 2, 3, 4, 5, 6, 9

Q1 = P25

i =

= 1.75

Q1 is located at the 2nd term, Q1 = 2

Q3 = P75:

i =

= 5.25

Q3 is located at the 6th term, Q3 = 6 IQR = Q3 - Q1 = 6 - 2 = 4 f.)

z =

= 0.69

z =

= -0.92

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

z =

= -0.11

z =

= 1.89

z =

= -1.32

z =

= -0.52

z =

= 0.29

3.12

x 4 3 0 5 2 9 4 5 x = 32

0 1 4 1 2 5 0 1 = 14

0 1 16 1 4 25 0 1 = 48

=4 a) Range = 9 - 0 = 9

b) M.A.D. =

c) s2 =

d) s =

= 1.75

= 6.8571

= 2.6186

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

e) Numbers in order: 0, 2, 3, 4, 4, 5, 5, 9 Q1 = P25

i =

= 2

Q1 is located at the average of the 2nd and 3rd terms,

Q1 =

2.5 Q3 = P75

i =

= 6

Q3 is located at the average of the 6th and 7th terms,

Q3 =

IQR = Q3 - Q1 = 5 - 2.5 = 2.5 3.13

a.) x 12 23 19 26 24 23 x = 127

 =

(x-µ) 12-21.167= -9.167 1.833 -2.167 4.833 2.833 1.833 (x -µ) = -0.002

(x -µ)2 84.034 3.360 4.696 23.358 8.026 3.360 2 (x -µ) = 126.834

= 21.167

 =

= 4.598

ORIGINAL

FORMULA b.) x 12 23 19 26 24 23 x = 127

x2 144 529 361 676 576 529 2 x = 2815

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

= 4.598

SHORT-CUT FORMULA

The short-cut formula is faster, but the original formula gives insight into the meaning of a standard deviation.

3.14

s2 = 433.9267 s = 20.8309

3.15

2 = 58,631.295  = 242.139

3.16 14, 15, 18, 19, 23, 24, 25, 27, 35, 37, 38, 39, 39, 40, 44, 46, 58, 59, 59, 70, 71, 73, 82, 84, 90 Q1 = P25

i =

= 6.25

P25 is located at the 7th term, and therefore, Q1 = 25 Q3 = P75

i =

= 18.75

P75 is located at the 19th term, and therefore, Q3 = 59 IQR = Q3 - Q1 = 59 - 25 = 34

3.17

.75

.84

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

.609

.902

3.18 Set 1:

= 14.2215 Set 2:

= 14.5344

CV1 =

= 21.71%

CV2 =

= 10.20%

3.19

7 5 10 12 9 8

1.833 3.833 1.167 3.167 0.167 0.833

3.361 14.694 1.361 10.028 0.028 0.694

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

14 3 11 13 8 6 106

5.167 5.833 2.167 4.167 0.833 2.833 32.000

26.694 34.028 4.694 17.361 0.694 8.028 121.665

= 8.833 a) MAD =

= 2.667

b) s2 =

= 11.06

c) s =

= 3.326

d) Rearranging terms in order:

3 5 6 7 8 8 9 10 11 12 13 14

Q1 = P25: i = (.25)(12) = 3 Q1 = the average of the 3rd and 4th terms:

Q1 = (6 + 7)/2 = 6.5

Q3 = P75: i = (.75)(12) = 9 Q3 = the average of the 9th and 10th terms:

Q3 = (11 + 12)/2 = 11.5

IQR = Q3 - Q1 = 11.5 – 6.5 = 5 e.) z =

= - 0.85

f.) CV = 3.20 n = 11

= 37.65% x 768 429 323 306 286 262 215

 x- 475.64 136.64 30.64 13.64 6.36 30.36 77.36

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

172 162 148 145 x = 3216

120.36 130.36 144.36 147.36 x-µ = 1313.08

µ = 292.36 x = 3216

x2 = 1,267,252

a.) Range = 768 - 145 = 623 b.) MAD =

= 119.37

c.) 2 =

= 29,728.23

d.)  =

= 172.42

e.) Q1 = P25:

i = .25(11) = 2.75

Q1 is located at the 3rd term and Q1 = 162 Q3 = P75:

i = .75(11) = 8.25

Q3 is located at the 9th term and Q3 = 323 IQR = Q3 - Q1 = 323 - 162 = 161 f.) xnestle = 172 z =

= -0.70

g.) CV = 3.21

µ = 125

= 58.98%

 = 12

68% of the values fall within: µ ± 1 = 125 ± 1(12) = 125 ± 12 between 113 and 137 95% of the values fall within: µ ± 2 = 125 ± 2(12) = 125 ± 24 Solutions Manual 1-63 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

between 101 and 149 99.7% of the values fall within: µ ± 3 = 125 ± 3(12) = 125 ± 36 between 89 and 161 3.22

=6

µ = 38 between 26 and 50:

x1 - µ = 50 - 38 = 12 x2 - µ = 26 - 38 = -12 = 2

= -2 Another suggested method From Chebyshev‟s theorem the interval will be ( -k , +k ) This means that - k = 26 38-6k = 26 and k = 2 Similarly + k , = 50 38 – 6k = 50 and k = -2 k = 2, and since the distribution is not normal, use Chebyshev‟s theorem: = .75 at least 75% of the values will fall between 26 and 50 between 14 and 62?

µ = 38

 =6

x1 - µ = 62 - 38 = 24 x2 - µ = 14 - 38 = -24 = 4

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

From Chebyshev‟s theorem the interval will be ( -k , This means that -k = 14 38-6k = 14 and k = 4 Similarly +k = 62 38 – 6k = 64 and k = -4

+k )

k=4 = .9375 at least 93.75% of the values fall between 14 and 62 between what two values do at least 89% of the values fall? 1-

=.89

.11 = .11 k2 = 1 k2 = k2 = 9.09 k = 3.015 With µ = 38,  = 6 and k = 3.015 at least 89% of the values fall within: µ ± 3.015 = 38 ± 3.015 (6) = 38 ± 18.09 Between 19.91 and 56.09

3.23

= .80

1 - .80 =

.20 =

and

.20k2 = 1

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

k2 = 5

and

k = 2.236

2.236 standard deviations

3.24

µ = 43. 68% of the values fall within µ + 1 . Thus, between the mean,

43, and one of the values, 46, is one standard deviation. Therefore, 1 = 46 - 43 = 3 99.7% of the values lie within µ + 3. Thus, between the mean, 43, and one of the values, 51, are three standard deviations. Therefore, 3 = 51 - 43 = 8

 = 2.67 µ = 28 and 77% of the values lie between 24 and 32 or + 4 from the mean: 1-

= .77

Solving for k: .23 =

and therefore,

.23k2 = 1

k2 = 4.3478 k = 2.085 2.085 = 4

 =

3.25

µ = 29

= 1.918

=4

Between 21 and 37 days: = -2 Standard Deviations Solutions Manual 1-66 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

= 2 Standard Deviations Since the distribution is normal, the empirical rule states that 95% of the values fall within µ ± 2 .

Exceed 37 days: Since 95% fall between 21 and 37 days, 5% fall outside this range. Since the normal distribution is symmetrical, 2½% fall below 21 and above 37. Thus, 2½% lie above the value of 37. Exceed 41 days: = 3 Standard deviations The empirical rule states that 99.7% of the values fall within µ ± 3 = 29 ± 3(4) = 29 ± 12. That is, 99.7% of the values will fall between 17 and 41 days. 0.3% will fall outside this range and half of this or .15% will lie above 41. Less than 25: µ = 29

= 4 = -1 Standard Deviation

According to the empirical rule, µ ± 1 contains 68% of the values. 29 ± 1(4) = 29 ± 4 Therefore, between 25 and 33 days, 68% of the values lie and 32% lie outside this range with ½(32%) = 16% less than 25.

3.26

City London Mexico City Tokyo Bangalore Bangkok

Expense 482 362 485 493 241

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Riyadh Lagos Cape Town Zurich Paris Guatemala City x = 4703

x2 = 2,158,469

485 446 315 546 608 240 n = 11

= 427.55

s = 121.54

Lagos: x = 446

Riyadh: x = 485

Bangkok: x = 241

3.27

mean = $35 median = $33 mode = $21 If mean > median > mode, then the data is skewed to the right The stock prices are skewed to the right. While many of the stock prices are at the cheaper end, a few extreme prices at the higher end pull the mean.

3.28

mean = 51 median = 54 mode = 59 If mean < median < mode, then the data is skewed to the left. The distribution is skewed to the left. More people are older but the most extreme ages are younger ages.

3.29 Sk =

= 0.726

Because the value of Sk is positive, the distribution is positively skewed.

3.30

n = 25

x = 600

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

= 24

s = 6.6521

Md = 23

Sk =

= 0.451

There is a slight skewness to the right

3.31

n = 24 Q1 = P25: i =

= 6

Thus, Q1 is located at the 6.5th term and Q1 = (497 + 503)/ 2 = 500 Median:

12.5th term

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Median = Q2 = (558 +559)/ 2 = 558.5 Q3 = P75: i =

= 18

Thus, Q3 is located at the 18.5th term and Q3 = (588 + 590)/ 2 = 589 IQR = 589 - 500 = 89 Inner Fences:

Q1 - 1.5 IQR = 500 - 1.5 (89) = 366.5 and Q3 + 1.5 IQR = 589 + 1.5 (89) = 722.5

Outer Fences: Q1 - 3.0 IQR = 500 - 3 (89) = 233 and Q3 + 3.0 IQR = 589 + 3 (89) = 856 The distribution is negatively skewed. There are no mild or extreme outliers.

3.32

n = 18 = 9.5th term

Median: Median = 74 Q1 = P25:

i =

= 4.5

Q1 = 5th term = 66

Q3 = P75:

i =

= 13.5

Q3 = 14th term = 90

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Therefore, IQR = Q3 - Q1 = 90 - 66 = 24 Inner Fences: Q1 - 1.5 IQR = 66 - 1.5 (24) = 30 Q3 + 1.5 IQR = 90 + 1.5 (24) = 126 Outer Fences: Q1 - 3.0 IQR = 66 - 3.0 (24) = -6 Q3 + 3.0 IQR = 90 + 3.0 (24) = 162

There are no extreme outliers. The only mild outlier is 21. The distribution is positively skewed since the median is nearer to Q1 than Q3. 3.33 The summary statistics is provided below Column1 Mean Standard Error

50.68 1.1867858

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Median Mode Standard Deviation Sample Variance Kurtosis Skewness Range Minimum Maximum Sum Count

51 44 11.8678583 140.8460606 -0.074269122 -0.084975432 57 21 78 5068 100

These descriptive statistics give us many insights into the data. The mean or average of the data is 50.68, the median or middle number of the data is 51. We notice that the mean and the median are quite close to each other, giving us evidence that the data appear to near normally distributed. Further study of the data reveals that even the mode is 44 and there are 8 observations with this value. The smallest value in the dataset is 21 and the largest value is 78. From these two values, the range can be computed (78 − 21) and is 57, as shown in the output. Both the kurtosis and skewness values are low. Since the data appear to be near normally distributed or at least bell-shaped, the empirical rule can be applied. With a mean of 50.68 and a standard deviation of 11.88 we can conclude that approximately 68% of the data lie between 38.80 and 62.56 (50.68 + 11.88) and that approximately 95% of the data lie between 26.32 and 74.44 (50.68 + 2(11.88)). Furthermore, z scores could be computed. 3.34 The mean or average of the Holiday Spending data is $1,142.14, the median or middle number of the data is $1,129.49. We notice that the mean is greater than the median giving us evidence that the data appear to be skewed to the right which is supported by the positive skewness. The mode of the data is $1,251.15. The smallest value in the dataset is $233.80 and the largest value is $2,097.50. From these two values, the range is $1,863.71, as shown in the output. Since the data appear to be not normally distributed or at least bell-shaped, Chebyshev‟s theorem could be used to determine ranges that are certain standard deviations from the mean.

3.35

Arranging the values in an ordered array: 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 4, 4, 5, 6, 8 Mean:

= 2.5

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Mode = 2 (There are eleven 2‟s) Median: There are n = 30 terms. = 15.5th position.

The median is located at

Median is the average of the 15th and 16th value. However, since these are both 2, the median is 2.

Range = 8 - 1 = 7

Q1 = P25:

i =

Q1 is the 8th term =

Q3 = P75:

= 7.5 1

i =

= 22.5

Q3 is the 23rd term = 3 IQR = Q3 - Q1 = 3 - 1 = 2

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

3.36 P10: i =

= 4 P10 = 4.5th term = 23

P80:

i =

= 32

P80 = 32.5th term = 49.5

Q1 = P25:

i =

= 10

P25 = 10.5th term = 27.5

Q3 = P75:

i =

= 30

P75 = 30.5th term = 47.5 IQR = Q3 - Q1 = 47.5 - 27.5 = 20 Range = 81 - 19 = 62

3.37

Mean:

µ =

The median is located at the

= 6,345.2

th value = 21/2 = 10.5th value

The median is the average of 5,414 and 5,563 = 5,488.5 P30:

i = (.30)(20) = 6 P30 is located at the average of the 6th and 7th terms

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

P30 = (4,507+4,541)/2 = 4,524 P60:

i = (.60)(20) = 12 P60 is located at the average of the 12th and 13th terms P60 = (6,101+6,498)/2 = 6,299.5

P90:

i = (.90)(20) = 18 P90 is located at the average of the 18th and 19th terms P90 = (9,863+11,019)/2 = 10,441

Q1 = P25:

i = (.25)(20) = 5 Q1 is located at the average of the 5th and 6th terms Q1 = (4,464+4,507)/2 = 4,485.5

Q3 = P75:

i = (.75)(20) = 15 Q3 is located at the average of the 15th and 16th terms Q3 = (6,796+8,687)/2 = 7,741.5

Range = 11,388 - 3,619 = 7,769 IQR = Q3 - Q1 = 7,741.5 - 4,485.5 = 3,256

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

x 262 497 335 351 467 44,587 10,984 1,783 193,194 1,374

68644 247009 112225 123201 218089 1988000569 120648256 3179089 37323921636 1887876

=253,834 39,438,406,594

n = 10

Mean:

= 253,834

39,438,406,594

µ = (x)/N = 253834/10 = 25,383.4

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Standard Deviation:  =

= :  = √

= 57,441.48

3.39 x2

x 12.5

156.25

9.7

94.09

6.4

40.96

5.3

28.09

4.4

19.36

4.1

16.81

3.9

15.21

3.6

12.96

3.5

12.25

3.2

10.24

2.9

8.41 414.63

= 59.5 N = 11

a.)

= 59.5

Mean: Median:

µ=

414.63

= 59.5/11 = 5.41 6th term

Median = Q2 = 4.1 Since there is no extreme value, the median has no advantage over the mean. Solutions Manual 1-77 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

b.)

Range = 12.5 – 2.9 = 9.6 Q1 = P25: i =

= 2.75

Thus, Q1 is located at the 3rd term and Q1 = 3.5. Q3 = P75: i =

= 8.25

Thus, Q3 is located at the 9th term and Q3 = 6.4 IQR = Q3 - Q1 = 6.4 – 3.5 = 2.9 c.)

Variance:

2 =

= 8.435

Standard Deviation:

 = d.)

= 2.904

Saudi Aramco: z =

= 2.44

Daily capacity for Saudi Aramco is 2.44 standard deviations over the mean daily capacity. ExxonMobil: z =

= -0.04

Daily capacity for ExxonMobil is 0.04 standard deviations under the mean daily capacity. e.)

Skewness: Sk =

= 1.35

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Because the value of Sk is positive, the distribution is positively skewed.

3.40 a.) Mean:

= 3.3071 7.5th term

Median:

Median = (3.76 + 3.91)/2 = 3.835 Mode: b.) Range:

No mode 6.20 – 0.36 = 5.84

Q1:

Located at the 4th term. Q1 = 0.71

Q3:

Located at the 11th term.

Q3 = 5.45

IQR = Q3 – Q1 = 5.45 – 0.71 = 4.74 x 0.78 6.12 4.16 0.71 0.59 6.09 2.14 6.20 3.91 0.63 0.36 5.45 5.40 3.76

2.5271 2.8129 0.8529 2.5971 2.7171 2.7829 1.1671 2.8929 0.6029 2.6771 2.9471 2.1429 2.0929 0.4529 = 29.2658

6.3862 7.9124 0.7274 6.7449 7.3826 7.7445 1.3621 8.3689 0.3635 7.1669 8.6854 4.5920 4.3802 0.2051 = 72.0221

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

MAD =

s2 =

= 2.0904

= 5.5402

= 2.3538

c.) Pearsonian coefficient of skewness: Sk =

= -0.6728

d.) Use Q1 = 0.71, Q2 = 3.835, Q3 = 5.45, IQR = 4.74 0.71– 1.5(4.74) = – 6.4 5.45 + 1.5(4.74) = 12.56 There are no mild outliers. Inner Fences:

Outer Fences: 0.71– 3(4.74) = – 13.51 5.45 + 3(4.74) = 19.67 There are no extreme outliers.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

3.41

CVX =

= 10.78%

CVY =

= 6.43%

Stock X has a greater relative variability.

3.42 µ = 7.5 mean.

Each of the numbers, 1 and 14, are 6.5 units away from the

From the Empirical Rule: 99.7% of the values lie in µ + 3 3 = 14 - 7.5 = 6.5

Solving for 3 = 6.5 for :

 = 2.167

Suppose that µ = 7.5,  = 1.7: 95% lie within µ + 2 = 7.5 + 2(1.7) = 7.5 + 3.4 Between 4.1 and 10.9 lie 95% of the values.

3.43

µ = 419,  = 27 a.) 68%:

µ + 1

419 + 27

95%:

µ + 2

419 + 2(27) 365 to 473

99.7%: µ + 3

419 + 3(27) 338 to 500

392 to 446

b.) Use Chebyshev‟s theorem: Each of the points, 359 and 479 is a distance of 60 from the mean, µ = 419. k = (distance from the mean)/ = 60/27 = 2.22 Proportion = 1 - 1/k2 = 1 - 1/(2.22)2 = .797 = 79.7% Solutions Manual 1-82 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

c.) Since x = 400, z =

= -0.704. This worker is in the lower

half of workers but within one standard deviation of the mean.

3.44 a.)

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

x Albenia Bulgaria Croatia Czech Republic

12,021 144,504,441 20,329 413,268,241 25,264 638,269,696 36,916 1,362,791,056

x =

x2 =

94,530 2,558,833,434

 =

= 23,632.5

 = √

√

= 9,011.84

b.) x2

x Hungary Poland Romania Bosnia and Herzegovina Sum

28,375 29,291 25,841 12,876

805,140,625 857,962,681 667,757,281 165,791,376

x =

x2 =

96,383 2,496,651,963

 =

 = √

= 24,095.80

√

= 6,599.84

c.) CV1 =

= 38.13%

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

CV2 =

= 27.39%

The first group has a larger coefficient of variation.

3.45 Mean Median Mode

$30,468 $26,721 $25,000

Since these three measures are not equal, the distribution is skewed. The distribution is skewed to the right because the mean is greater than the median. Often, the median is preferred in reporting income data because it yields information about the middle of the data while ignoring extremes.

3.46 a.)

Q1 = P25:

i =

= 5

Q1 = 5.5th term = (5,739.30 + 6,185.50)/2 = 5,962.40

Q3 = P75:

i =

= 15

Q3 = 15.5th term = (25,785.80 + 27,387.50)/2 = 26,586.65

Median:

= 10.5th term

Median = (10,015.60 + 13,663.40)/2 = 11,839.50

IQR = Q3 – Q1 = 26,586.65– 5,962.40 = 20,624.25 1.5 IQR = 30,936.375 ;

3.0 IQR = 61,872.75

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Inner Fences: Q1 – 1.5 IQR = 5,962.40 – 30,936.375= – 24,973.975 Q3 + 1.5 IQR = 26,586.65 + 30,936.375 = 57,523.025 Outer Fences: Q1 – 3.0 IQR = 5,962.40 – 61,872.75 = – 55,910.35 Q3 + 3.0 IQR = 26,586.65 + 61,872.75 = 88,459.40

b.) and d.) Since the median is nearer to Q1, the distribution is positively skewed. There are no mild/extreme outliers in the lower end. The value of 107, 575.20 is outside the inner and the outer fences. Thus, there is one extreme outlier in the upper end (Metro Vancouver, 107,575.20). In other words, there is one dominating, large port. Solutions Manual 1-86 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

3.47

Paris:

Since 1 - 1/k2 = .53, solving for k:

k = 1.459

The distance from µ = 349 to x = 381 and to x = 317 is 32 1.459 = 32

 = 21.93 Moscow:

Since 1 - 1/k2 = .83, solving for k:

k = 2.425

The distance from µ = 415 to x = 459 and to x = 371 is 44 2.425 = 44

 = 18.14 3.48 The Count shows that there were 100 companies in the study. The minimum profit of these 100 Canadian public companies is 201,134.00. This figure represents the company ranked last in these 100 companies. The maximum is 5,889,000 which represents the profit of the top company (Toronto-Dominion Bank). The mean profit of these top 100 companies is 1,063,117.59 and the median profit is 522,123.50. The gap between these two figures (mean and median) indicates that there is skewness in the data. Observing that the mean is greater than the median, the skewness is positive. This conclusion is reinforced by the fact that the skewness figure given in the output is 2.16 which indicates considerable positive skewness. It is likely that the largest companies have extremely large profits while the middle and bottom companies are more closely packed together. The range is 5,687,866.00 which is the difference between the minimum and maximum profit values. The standard deviation is quite large at 1,187,594.89. If we apply the empirical rule to these data using the mean and the standard deviation, we can see that there is about one standard deviation between the mean and the minimum value reinforcing the notion that the data are not normally distributed (because there should be at least three standard deviations of data on either side of the mean if the data are approximately normally distributed) and are likely to be positively skewed.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

3.49

First of all, the graph indicates that the data are very skewed to the right. This is supported by the output which yields a skewness value of 3.6214. This is a very high positive skewness number telling us that there are a few advertisers in the Hispanic market who are spending extremely large numbers on advertising while most of the other companies are spending moderate to less amounts. In fact, looking at the graph, there is a very high modal group of advertisers centering around the $ 5 million mark. The output reveals that the mean amount being spent by Hispanic advertisers is $ 7.8560 million while the median is $ 5.75 million. The gap between the mean and the median supports the notion that the data are positively skewed. The standard deviation of the advertising data is 5.8860. If we apply the empirical rule to these data using the mean and the standard deviation, we can see that there are less than one standard deviation between the mean and the minimum value reinforcing the notion that the data are not normally distributed (because there should be at least three standard deviations of data on either side of the mean if the data are approximately normally distributed) and are likely to be positively skewed. The “N” indicates that there were 50 companies in this data set. The minimum amount spent on advertising cultivating the Hispanic market by one of these companies is $ 3.25 million and the maximum spent is $ 40.00 million. Also, presented in the data are the first and third quartiles. Observe the gap between the third quartile and the maximum amount. The “range” of this top 25 percent is over $31 million (8.625 – 40.00) and this is where the skewness is occurring.

3.50

This Excel output represents the number of employees per company for a database of 46 of the largest employers with headquarters outside the United States. The minimum number of employees for companies in this group is 125,894 and the maximum is 382,000. This yields a range of 256,106 (382,000 – 125,894). The Sum tells us that there is a total of 8,433,048 employees in these 46 companies. The mean number of employees for this group is 183,327.1304 while the median number is 156,670. The gap between the mean and the median indicates that there is positive skewness. This is underscored by the fact that Skewness is given as 1.2996 which is a relatively high figure. The data are skewed to the right indicating that a few of the largest companies have a very high number of employees and near the bottom of the 46 companies, the numbers of employees are more closely compacted. The standard deviation is 64,302.4905. If we apply the empirical rule to these data using the mean and the standard deviation, we can see that there is less than one standard deviation between the mean and the minimum value reinforcing the notion that the data are not normally distributed (because there should be at least three standard deviations of data on either side of the mean if the data are

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

approximately normally distributed) and are likely to be positively skewed. The mode is 135,000 employees. Typically, in data set like this, there is no mode because it is unlikely that any two companies have the exact same number of employees. Since the raw data are not available to us, we can speculate that the mode most likely represents only two or three companies and has limited value here.

3.51

The N indicates that there are 25 companies in this database. The mean advertising expenditure is $ 772,702 thousands while the median is $ 613,823 thousands. The gap between the mean and the median indicates that there is positive skewness. While no skewness figure is given in the MINITAB output, the boxplot shows extreme outliers on the right or positive side of the data supporting the conclusion that there is positive skewness. Since the median is is closer to Q1 than Q3, the data are likely to be skewed to the right. The standard deviation is $ 436,067 thousand. Using this figure, it is possible to see that there is less than one standard deviation between the mean and the minimum ($ 445,958). If we apply the empirical rule to these data using the mean and the standard deviation, we can see that the data are not normally distributed (because there should be at least three standard deviations of data on either side of the mean if the data are approximately normally distributed) and are likely to be positively skewed. The values of the first quartile ($ 484,600 thousand) and the third quartile ($788,256 thousand) are also given and can be used to compute the interquartile deviation (middle 50% of the data), construct the boxplot, and indicate advertising expenditures for both the bottom and top 25 percent of companies.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Legal Notice Copyright

SOLUTIONS TO PROBLEMS IN CHAPTER 4

4.1

Enumeration of the six parts: D1, D2, D3, A4, A5, A6 D = Defective part A = Acceptable part

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

6C2

= 15 There are 15 elementary events in the sample space. Sample Space: D1 D2, D2 D3, D3 A5 D1 D3, D2 A4, D3 A6 D1 A4, D2 A5, A4 A5 D1 A5, D2 A6, A4 A6 D1 A6, D3 A4, A5 A6 There are 15 elementary events in the sample space and 9 of them have exactly 1 defective part The probability of selecting exactly one defect out of two is: 9/15 = .60

4.2

X = {1, 3, 5, 7, 8, 9}, Y = {2, 4, 7, 9}, and Z = {1, 2, 3, 4, 7} a) = {1, 2, 3, 4, 5, 7, 8, 9} b) = {7, 9} c) = {1, 3, 7} d) = {1, 2, 3, 4, 5, 7, 8, 9} e) = {7} f) = {1, 2, 3, 4, 5, 7, 8, 9} {1, 2, 3, 4, 7} = {1, 2,

3, 4, 7} g) h) i)

4.3 through 30,

X or Y = X Y and Z = Y

= {2, 4, 7} {7, 9} = {2, 4, 7, 9} Y = {1, 2, 3, 4, 5, 7, 8, 9} Z = {2, 4, 7}

If A = {2, 6, 12, 24} and the population is the positive even numbers A‟ = {4, 8, 10, 14, 16, 18, 20, 22, 26, 28, 30}

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

4.4 4.5

6(4)(3)(3) = 216 Enumeration of the six parts: D1, D2, A1, A2, A3, A4 D = Defective part A = Acceptable part Sample Space: D1 D2 A1, D1 D2 A2, D1 D2 A3, D1 D2 A4, D1 A1 A2, D1 A1 A3, D1 A1 A4, D1 A2 A3, D1 A2 A4, D1 A3 A4, D2 A1 A2, D2 A1 A3, D2 A1 A4, D2 A2 A3, D2 A2 A4, D2 A3 A4, A1 A2 A3, A1 A2 A4, A1 A3 A4, A2 A3 A4 Combinations are used to counting the sample space because sampling is done without replacement and the order of elements is not important. 6C3 =

= 20 (sample size)

Probability that one of three is defective is: 12/20 = 3/5 = .60 There are 20 elementary events in the sample space and 12 of them have exactly 1 defective part.

4.6

107 = 10,000,000 different numbers

4.7

a) 20C6 =

= 38,760

It is assumed here that 6 different (without replacement) employees are to be selected. b) 720

4.8

P(A) = .10, P(B) = .12, P(C) = .21, P(A

C) = .05

P(B

C) = .03

a) P(A

C) = P(A) + P(C) - P(A

C) = .10 + .21 - .05 = .26

b) P(B

C) = P(B) + P(C) - P(B

C) = .12 + .21 - .03 = .30

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

c) If A, B are mutually exclusive, P(A

B) = P(A) + P(B) = .10 + .12 =

.22 4.9 D

a) P(A  D) = P(A) + P(D) - P(A  D) = 25/60 + 23/60 - 5/60 = 43/60 = .7167 b) P(E  B) = P(E) + P(B) - P(E  B) = 16/60 + 20/60 - 6/60 = 30/60 = .5000

c) P(D  E) = P(D) + P(E) = 23/60 + 16/60 = 39/60 = .6500 d) P(C  F) = P(C) + P(F) - P(C  F) = 15/60 + 21/60 - 5/60 = 31/60

= .5167

4.10 E

.10

.03

.13

.04

.12

.16

.27

.06

.33

.31

.07

.38

.72

.28

1.00

a) P(A  F) = P(A) + P(F) - P(A  F) = .13 + .28 - .03 = .38 b) P(E  B) = P(E) + P(B) - P(E  B) = .72 + .16 - .04 = .84 c) P(B  C) = P(B) + P(C) =.16 + .33 = .49 d) P(E  F) = P(E) + P(F) = .72 + .28 = 1.00

4.11

A = event of having flown in an airplane at least once T = event of having ridden in a train at least once P(A) = .47

P(T) = .28

P (ridden either a train or an airplane) = Solutions Manual 1-93 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

P(A  T) = P(A) + P(T) - P(A  T) = .47 + .28 - P(A  T). We cannot solve this problem without knowing the probability of the intersection. We need to know the probability of the intersection of A and T, the proportion who have ridden both. 4.12

Let L = participating in the labour force and M =being married. P(L) = .75 P(M) = .78 P(M  L) = .61 a) P(M  L) = P(M) + P(L) - P(M  L) = .78 + .75 - .61 = .92 b) P(M  L but not both) = P(M  L) - P(M  L) = .92 - .61 = .31 c) P(NM  NL) = 1 - P(M  L) = 1 - .92 = .08 Note: the neither/nor event is solved for here by taking the complement of the union.

4.13

Let C = have cable TV Let T = have 2 or more TV sets P(C) = .67, P(T) = .74, P(C  T) = .55 a) P(C  T) = P(C) + P(T) - P(C  T) = .67 + .74 - .55 = .86 b) P(C  T but not both) = P(C  T) - P(C  T) = .86 - .55 = .31 c) P(NC  NT) = 1 - P(C  T) = 1 - .86 = .14 d) The special law of addition does not apply because P(C  T) is not

.0000. Possession of cable TV and 2 or more TV sets are not mutually exclusive.

4.14

Let T = review transcript F = consider faculty references P(T) = .54 P(F) = .44 P(T  F) = .35 a) P(F  T) = P(F) + P(T) - P(F  T) = .44 + .54 - .35 = .63 b) P(F  T but not both) = P (F  T) - P(F  T) = .63 - .35 = .28 c) P(NF  NT) = 1 - P(F T) = 1 - .63 = .37

d) Faculty References Y

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Transcript

.35

.19

.54

.09

.37

.46

.44

.56

1.00

Answer for a) .35 + .19 + .09 = .63 Answer for b) .19 + .09 = .28 Answer for c) .37

4.15 C

a) P(A  E) = 16/57 = .2807 b) P(D  B) = 3/57 = .0526 c) P(D  E) = .0000 d) P(A  B) = .0000

4.16 D

.12

.13

.08

.33

.18

.09

.04

.31

.06

.24

.06

.36

.46

.18

1.00

a) P(E  B) = .09 b) P(C  F) = .06 c) P(E  D) = .00

4.17

Let D = Defective part a) (without replacement) P(D1  D2) = P(D1)  P(D2 D1) =

= .0122

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

b) (with replacement) P(D1  D2) = P(D1)  P(D2) =

4.18

= .0144

Let U = urban, I = care for ill relatives P(U) = .83 P(I) = .15 P(I|U) = .11 a) P(U∩I) = P(U).P(I|U) = (.83)(.11) = .0913

b P(U∩NI) = P(U).P(NI|U) = (.83)(1-.11) = .7387 c) U I

Yes No

Yes .0913 (a.) .7387 (b.) 0.83

No .0587 .1113 0.17

.15 .85 1.00

P(U∩I) = .0913

P(NU∩I) = .15 - .0913 = .0887

P(U∩NI) = .7387

P(NU∩NI) = .1113

The answer is in the Yes-No cell (first row, second column): P(NU∩I) = .15 - .0913 = .0587

4.19

Let

T = employed in tourist industry,

U = under 35 years of age

P(T) = .11 P(U|T) = .51 P(U|NT) = .44 a) P(NT) = 1 - .11 = .89 b) P(T∩U) = P(T) P(U|T) = (.11)(.51) = .056 Solutions Manual 1-96 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

c) P(T∩NU) = P(T) P(NU|T) = (.11)(1 - .51) = .054 d) P(NT∩U) = P(NT) P(U|NT) = (.89)(.44) = .392 e) P(NT∩NU) = P(NT) P(NU|NT) = (.89)(1 - .44) = .498 P(NT∩not over 35) = P(NT∩U) = P(NT) P(U|NT) = (.89)(.44) = .392 (same as d) g) P(NT∩NU) = P(NT) P(NU|NT) = (.89)(1 - .44) = .498 (same as e) The matrix: U

Yes No

4.20

Let

Yes .056 .392 .448

D = Air Conditioner,

No .054 .498 .552

.11 .89 1.00

C = Tree on Property

P(D) = .60 P(C) = .81 P(C|D) = .91 a) b) P ( D ∪ C) = P(D) + P(C) – P(D∩C) = .60 + .81 - .546 = .864 c) d) P(ND∩NC) = 1 - P ( D ∪ C) = 1 - .864 = .136 e) P(ND∩C) = P(C) – P(D∩C) = .81 - .546 = .264 Matrix:

C .546

NC .054

.60

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

.264 .81

.136 .19

Let S = safety

Let A = age

.40 1.00

4.21

P(S) = .30 P(A) = .39 P(AS) = .87 a) P(S  NA) = P(S) P(NAS) but P(NAS) = 1 - P(AS) = 1 - .87 = .13 P(S  NA) = (.30)(.13) = .039 b) P(NS  NA) = 1 - P(S  A) = 1 - [P(S) + P(A) - P(S  A)] but P(S  A) = P(S)  P(AS) = (.30)(.87) = .261 P(NS  NA) = 1 - (.30 + .39 - .261) = .571 c) The matrix: A

4.22

Let

Yes No

Yes .261 .129

No .039 .571

.30 .70

.39

.61

1.00

L = Have energy saving lights,

T = truck or van or SUV

P(C) = .90 P(T) = .40 P(CT) = .38 a) P(CT) = P(C) + P(T) - P(CT) = .90 + .40 - .38 = .92 b) P(NCNT) = 1 - P(CT) = 1 - .92 = .08 c) P(NCT) = P(T) - P(CT) = .40 - .38 = .02 d) P(CNT) = P(C) - P(CT) = .90 - .38 = .52 Matrix:

T .38

NT .52

.90

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

.02 .40

.08 .60

.10 1.00

4.23 E

205

a) P(GA) = 8/35 = .2286 b) P(BF) = 17/74 = .2297 c) P(CE) = 21/65 = .3231 d) P(EG) = .0000

4.24 C

.36

.44

.80

.11

.09

.20

.47

.53

1.00

a) P(CA) = .36/.80 = .4500 b) P(BD) = .09/.53 = .1698 c) P(AB) = .0000

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

4.25 Calculator

Computer

Yes

Select a category from each variable and test P(V1V2) = P(V1). For example, P(Yes ComputerYes Calculator) = P(Yes Computer)? ? .8070  .6533 Since this is one example that the conditional does not equal the marginal in is matrix, the variable, computer, is not independent of the variable, calculator.

4.26

Let

O = built overseas,

3,337,380 1,716,803 1,620,577 444,070 325,421

F = sold in Year 1,

S = sold in Year 2

sold in Canada in Year 1 and Year 2 sold in Year 2 sold in Year 1 sold in Year 2 and built overseas sold in Year 1 and built overseas

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

4.27

Let E = Economy Let Q = Qualified P(E) = .46 P(Q) = .37 P(E  Q) = .15 a) P(EQ) = P(E  Q)/P(Q) = .15/.37 = .4054 b) P(QE) = P(E  Q)/P(E) = .15/.46 = .3261 c) P(QNE) = P(Q  NE)/P(NE) but P(Q  NE) = P(Q) - P(Q  E) = .37 - .15 = .22 P(NE) = 1 - P(E) = 1 - .46 = .54 P(QNE) = .22/.54 = .4074 d) P(NE  NQ) = 1 - P(E  Q) = 1 - [P(E) + P(Q) - P(E  Q)] = 1 - [.46 + .37 - .15] = 1 - (.68) = .32 The matrix: Q

4.28

Yes No

Let S = Same day shipping

Yes .15 .22

No .31 .32

.46 .54

.37

.63

1.00

Let L = Shopping Lover

P(S) = .80 P(L) = .24 P(SL) = ..61 a) P(S  L) = P(L)  P(SL) = (.24)(.61) = .1464 b) P(NS|L) = P(L  NS)/P(L) = (P(L) – P(S  L))/P(L) = (.24 - .1464)/(.24) = .360 c) c)

P(SNL) = = PS) – P(SL) = (.80 - .1464)) = .6536

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

d) P(NS|NL) = P(NS  NL)/P(NL) But P(NS  NL) = 1 - P(S  L) = 1 – [P(S) + P(L) - P(S  L)] = 1 – [.8+.24-.1464] = .1064 P(NS|NL) = .1064/(1-.24) = .140

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

4.29

Let

E = Entry Level

P(E)= .80 P(S) = .77

S = Social Media

P(S|E) = .83

a) P(S  E) = P(E)  P(SE) = (.80)(.83) = .664 P ( S ∪ E) = P(S) + P(E) – P(S∩E) = .77 + .80 - .664 = .906 c) P(NS|E) = 1 – P(S|E) = 1 - .83 = .17

d) P(S|NE) = P(S  NE)/P(NE) = (P(S) – P(S  E))/P(NE) = (.77 - .664)/(1 - .80) =.530 e) = P(NE|NS) = P(NS  NE)/P(NS) But P(NS  NE) = 1 - P ( S ∪ E) =1 – [ P(S) + P(E) – P(S∩E)] = 1 – [.77 + .80 - .664] = 1-.906 = .094 P(NE|NS) = .094/(1-.77) = .4087 4.30

Let R = agreed or strongly agreed that lack of role models was a barrier Let S = agreed or strongly agreed that gender-based stereotypes was a

barrier P(R) = .43

P(S) = .46

P(RS) = .77

P(not S) = .54

a.) P(not RS) = 1 - P(RS) = 1 - .77 = .23 b.) P(not SR) = P(not S  R)/P(R) but P(S  R) = P(S)  P(RS) = (.46)(.77) = .3542 so P(not S  R) = P(R) - P(S  R) = .43 - .3542 = .0758 Therefore, P(not SR) = (.0758)/(.43) = .1763 c.) P(not Rnot S) = P(not R  not S)/P(not S) but P(not R  not S) = P(not S) - P(not S  R) = .54 - .0758 = .4642 P(not Rnot S) = .4642/.54 = .8596 Solutions Manual 1-104 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

The matrix: S

4.31

Yes No

Yes .3542 .1058

No .0758 .4642

.43 .57

.46

.54

1.00

Let A = product produced on Machine A B = product produces on Machine B C = product produced on Machine C D = defective product P(A) = .10 P(B) = .40 P(C) = .50 P(DA) = .05 P(DB) = .12

Event

Prior

Ei A B C

Conditional

P(Ei) .10 .40 .50

P(DEi) .05 .12 .08

Revised Probability:

Joint P(D  Ei) .005 .048 .040 P(D)=.093

P(DC) = .08 Revised

.005/.093=.0538 .048/.093=.5161 .040/.093=.4301

P(AD) = .005/.093 = .0538 P(BD) = .048/.093 = .5161 P(CD) = .040/.093 = .4301

4.32

Let

P(A) = .30 P(IA) = .20

A = Alex fills the order B = Natasha fills the order C = Juan fills the order I = order filled incorrectly K = order filled correctly P(B) = .45 P(IB) = .12

P(C) = .25 P(IC) = .05

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

P(KA) = .80 P(KB) = .88

P(KC) = .95

a) P(B) = .45 b) P(KC) = 1 - P(IC) = 1 - .05 = .95 c) Event

Prior

Conditional

Joint

Ei A B C

P(Ei) .30 .45 .25

P(IEi) .20 .12 .05

P(I  Ei) .0600 .0540 .0125 P(I)=.1265

Revised P(EiI) .0600/.1265=.4743 .0540/.1265=.4269 .0125/.1265=.0988

Revised Probability: P(AI) = .0600/.1265 = .4743 P(BI) = .0540/.1265 = .4269 P(CI) = .0125/.1265 = .0988

d) Event

Prior

Conditional

Joint

Ei A B C

P(Ei) .30 .45 .25

P(KEi) .80 .88 .95

P(K  Ei) .2400 .3960 .2375 P(K)=.8735

Revised P(EiK) .2400/.8735=.2748 .3960/.8735=.4533 .2375/.8735=.2719

Revised Probability: P(AK) =.2748 P(BK) = .4533 P(CK) = .2719

4.33

Let

M = lawn treated by Maritime Lawn Service G = lawn treated by Greenchem V = very healthy lawn

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

P(M) = .72

P(G) = .28

P(VM) = .30

Event

Prior

Conditional

Joint

Ei M G

P(Ei) .72 .28

P(VEi) .30 .20

P(V  Ei) .216 .056 P(V)=.272

Revised Probability:

P(VG) = .20 Revised P(EiV) .216/.272=.7941 .056/.272=.2059

P(MV) = .216/.272 = .7941 P(GV) = .056/.272 = .2059

4.34

Let S = small

Let L = large

The prior probabilities are: P(S) = .70 P(TS) = .18 P(TL) = .82 Event

Prior

Conditional

Joint

Ei S L

P(Ei) .70 .30

P(TEi) .18 .82

P(T  Ei) .1260 .2460 P(T)=.3720

Revised Probability:

P(L) = .30

Revised P(EiT) .1260/.3720 = .3387 .2460/.3720 = .6613

P(ST) = .1260/.3720 = .3387 P(LT) = .2460/.3720 = .6613

37.2% offer training since P(T) = .3720.

4.35 Variable 1

Variable 2

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

a) P(E) = 40/95 = .42105 b) P(B  D) = P(B) + P(D) - P(B  D) = 20/95 + 55/95 - 15/95 = 60/95 = .63158 c) P(A  E) = 20/95 = .21053 d) P(BE) = 5/40 = .1250 e) P(A  B) = P(A) + P(B) = 30/95 + 20/95 = 50/95 = .52632 f) P(B  C) = .0000 (mutually exclusive) g) P(DC) = 30/45 = .66667 h) P(AB) =

P( A  B) .0000  = .0000 (A and B are mutually P( B) 20 / 95

exclusive)

4.36 Since

variables 1 and 2 are not independent.

4.37

a) P(F  A) = 7/78 = .08974 b) P(AB) =

P( A  B) .0000  = .0000 (A and B are mutually P( B) 22 / 78

exclusive)

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

c) P(B) = 22/78 = .28205 d) P(E  F) = .0000 (Mutually Exclusive) e) P(DB) = 8/22 = .36364 f) P(BD) = 8/21 = .38095 g) P(D  C) = P(D) + P(C) – P(D  C) = 21/78 + 25/78 – 10/78 = 36/78 = .46154 h) P(F) = 16/78 = .20513

4.38 Age(years)

Gender

<35

35-44

45-54

55-64

>65

Male

.11

.20

.19

.12

.16

.78

Female

.07

.08

.04

.02

.01

.22

.18

.28

.23

.14

.17

1.00

a) P(35-44) = .28 b) P(Woman  45-54) = .04 c) P(Man  35-44) = P(Man) + P(35-44) - P(Man  35-44) = .78 + .28 - .20 = .86 d) P(<35  55-64) = P(<35) + P(55-64) = .18 + .14 = .32 e) P(Woman45-54) = P(Woman  45-54)/P(45-54) = .04/.23= .1739 f) P(not Woman  not 55-64) = 1 – P(Woman  55-64) = = 1 – [P(Woman) + P(55-64) – P(Woman  55-64)] = 1 –[.22 + .14 - .02] = .66

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

4.39

Let T = thoroughness P(T) = .78 P(K) = .40

Let K = knowledge P(T  K) = .27

a) P(T  K) = P(T) + P(K) - P(T  K) = .78 + .40 - .27 = .91 b) P(NT NK) = 1 - P(T K) = 1 - .91 = .09 c) P(KT) = P(K  T)/P(T) = .27/.78 = .3462 d) The matrix: K

4.40

Yes No

Yes .27 .13

No .51 .09

.78 .22

.40

.60

1.00

Let R = retirement

Let L = life insurance

P(R) = .42

P(R  L) = .33

P(L) = .61

a) P(RL) = P(R  L)/P(L) = .33/.61 = .5410 b) P(LR) = P(R  L)/P(R) = .33/.42 = .7857 c) P(L  R) = P(L) + P(R) - P(L  R) = .61 + .42 - .33 = .70 d) P(R  NL) = P(R) - P(R  L) = .42 - .33 = .09 e) P(NLR) = P(NL  R)/P(R) = .09/.42 = .2143

The matrix: L Yes

Yes .33

No .09

.42

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

.28

.30

.58

.61

.39

1.00

4.41 Let T = technology driven, B = lives in British Columbia, M = lives in Manitoba

a) b) c) d)

4.42

Let M = MasterCard

A = American Express

P(M) = .30 P(A) = .20 P(V) = .25 P(M  A) = .08 P(V  M) = .12

V = Visa

P(A  V) = .06

a) P(V  A) = P(V) + P(A) - P(V  A) = .25 + .20 - .06 = .39 b) P(VM) = P(V  M)/P(M) = .12/.30 = .40 c) P(MV) = P(V  M)/P(V) = .12/.25 = .48 Solutions Manual 1-111 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

d) P(V) = P(VM)?? .25  .40 Possession of Visa is not independent of possession of MasterCard e) American Express is not mutually exclusive of Visa because P(A  V)  .0000 4.43

Let C = cell phone only Given: P(C) = .25

Let I = high-speed Internet

P(I) = .65

P(IC) = .80

a) P(C  I) = P(C)  P(IC) = (.25)(.80) = .20 b) P(C  I) = P(C) + P(I) - P(C  I) = .25 + .65 - .20 = .70 c) P(C  NI) = P(C)  P(NIC) Since P(IC) = .80, P(NIC)= 1 - P(I C) = 1 - .80 = .20 P(C  NI) = (.25)(.20) = .05 d) P(NC  NI) = 1 - P(C  I) = 1 - .70 = .30 e) P(NC  I) = P(I) - P(C  I) = .65 - .20 = .45 The joint probability table:

4.44

.20

.05

.25

.45

.30

.75

.65

.35

1.00

Let M = expect to save more R = expect to reduce debt NM = don't expect to save more NR = don't expect to reduce debt P(M) = .43 P(R) = .45 P(RM) = .81 P(NRM) = 1 - P(RM) = 1 - .81 = .19 P(NM) = 1 - P(M) = 1 - .43 = .57

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

P(NR) = 1 - P(R) = 1 - .45 = .55 a) P(M  R) = P(M)P(RM) = (.43)(.81) = .3483 b) P(M  R) = P(M) + P(R) - P(M  R) = .43 + .45 - .3483 = .5317 c) P(neither save nor reduce debt) = P(NM  NR)= 1 - P(M  R) = 1 - .5317 = .4683 d) P(M  NR) = P(M)P(NRM) = (.43)(.19) = .0817 Probability matrix for problem 4.44: Reduce

Save

4.45

Yes

.3483

.0817

.43

.1017

.4683

.57

.45

.55

1.00

Let R = read Let B = checked in the with boss P(R) = .40

P(B) = .34

P(BR) = .78

a) P(B  R) = P(R)P(BR) = (.40)(.78) = .312 b) . P(NR  NB) = 1 - P(R  B) But P(R  B) = P (R) + P (B) - P(B  R) = .40 + .34 - .312 = .428 P(NR  NB) = 1 - .428 = .572 c) P(RB) = P(R  B)/P(B) = (.312)/(.34) = .9176 d) P(NBR) = 1 - P(BR) = 1 - .78 = .22 e) P(NBNR) = P(NB  NR)/P(NR) but P(NR) = 1 - P(R) = 1 - .40 = .60 Solutions Manual 1-113 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

P(NBNR) = .572/.60 = .9533 f) Probability matrix for problem 4.45:

4.46

.312

.088

.40

.028

.572

.60

.34

.66

1.00

Let Q = keep quiet when they see co-worker misconduct Let C = call in sick when they are well P(Q) = .35 P(NQ) = 1 - .35 = .65 P(CQ) = .75 a) P(C  Q) = P(Q)P(CQ) = (.35)(.75) = .2625

P(QC) = .40

b) P(Q  C) = P(Q) + P(C) - P(C  Q) but P(C) must be solved for: P(C  Q) = P(C)  P(QC) .2625 = P(C) ( .40) Therefore, P(C) = .2625/.40 = .65625 and P(Q  C) = .35 + .65625 - .2625 = .74375

c) P(NQC) = P(NQ  C)/P(C) but P(NQ  C) = P(C) - P(C  Q) = .65625 - .2625 = .39375 Therefore, P(NQC) = P(NQ  C)/P(C) = .39375/.65625 = .60 d) P(NQ  NC) = 1 - P(Q  C) = 1 - .74375 = .25625 e) P(Q  NC) = P(Q) - P(Q  C) = .35 - .2625 = .0875 Probability matrix for problem 4.46:

C Y

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

.2625

.0875

.35

.39375

.25625

.65

.65625

.34375

1.00

4.47 Let

T = time to get through to a customer representative R = impolite customer reps C = customer service P = payment disputes B = incorrect billing H = incompetent complaint handling I = indifference to customers

a) Since P and B are mutually exclusive,

(mutually exclusive)

c) d) Since H and P are mutually exclusive, P(NH  NP) = 1 – (.08+.11) = .81

4.48

Let B = Believe plastic shopping bags should be banned Let R = Recycle aluminum cans P(B) = .54

P(R  B) = .41

P(RNB) = .60

a) P(R  NB) = P(NB  R) = P(NB) P(RNB) but P(NB) = 1 – P(B) = 1 - .54 = .46 and P(RNB) = .60. P(R  NB) = P(NB  R) = P(NB) P(RNB) = (.46)(.60) = .2760. b) P(R) = P(R  B) + P(R  NB) = .41 + .2760 = .6860.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

c) P(R  B) = P(R) + P(B) - P(R  B) = .6860 + .54 - .41 = .8160 d) P(NR  NB) = P(NR) + P(NB) - P(NR  NB) But P(NB) =.46, P(NR) = 1 - P(R) = 1 - .6860 = .3140, and P(NR  NB) = 1 - P(R  B) = 1 - .8160 = .1840, therefore P(NR  NB) = .3140 + .46 - .1840 = .5900 e) P(NBR) = P(NB  R)/ P(R) P(NB  R) = P(R  NB) = .2760 and P(R) = .6860, therefore P(NBR) = P(NB  R)/ P(R) = .2760/.6860 = .4023. Initial joint probability table for problem 4.48: B Y Y R

.41

N .54

.46

1.00

Multiply .60 by the .46 to get the intersection of R and not B = .2760. Fill in the rest of the cells. Final joint probability table for problem 4.48: B

4.49

.41

.276

.686

.13

.184

.314

.54

.46

1.000

Let M = Credit card Let S = Debit P(M) = .42 P(M  S) = .0000 P(NM  NS) = .47 a) P(M  NS) = P(M) - P(M  S) = .42 - .00 = .42 b) Because P(M  S) = .0000, P(M  S) = P(M) + P(S) Therefore, P(S) = P(M  S) - P(M)

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

but P(M  S) = 1 - P(NM  NS) = 1 - .47 = .53 Thus, P(S) = P(M  S) - P(M) = .53 - .42 = .11 c) P(SM) = P(S  M)/P(M) = .0000/.42 = .0000 d) P(NMNS) = P(NM  NS)/P(NS) = .47/.89 = .5281 where: P(NS) = 1 - P(S) = 1 - .11 = .89 Probability matrix for problem 4.49: S

4.50

.00

.42

.11

.47

.58

.11

.89

1.00

Let F = Flexible work Let V = gives time off for Volunteerism P(F) = .41 P(VNF) = .10 P(VF) = .60 from this, P(NF) = 1 - .41 = .59 a) P(F  V) = P(F) + P(V) - P(F  V) P(F) = .41 and P(F  V) = P(F)P(VF) = (.41)(.60) = .246 Find P(V) by using P(V) = P(F  V) + P(NF  V) but P(NF  V) = P(NF)P(VNF) = (.59)(.10) = .059 so, P(V) = P(F  V) + P(NF  V) = .246 + .059 = .305 and P(F  V) = P(F) + P(V) - P(F  V) = .41 + .305 - .246 = .469 b) P(F  NV) = P(F) - P(F  V) = .41 - .246 = .164 c) P(FNV) = P(F  NV)/P(NV) P(F  NV) = .164 P(NV) = 1 – P(V) = 1 - .305 = .695. P(FNV) = P(F  NV)/P(NV) = .164/.695 = .2360

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

d) P(NFV) = P(NF  V)/P(V) = .059/.305 = .1934 e) P(NF  NV) = P(NF) + P(NV) - P(NF  NV) P(NF) = .59 P(NV) = .695 Solve for P(NF  NV) = P(NV) – P(F  NV) = .695 - .164 = .531 P(NF  NV) = P(NF) + P(NV) - P(NF  NV) = .59 + .695 - .531 = .754

Probability matrix for problem 4.50: V

4.51

Event Ei K C J

.246

.164

.41

.059

.531

.59

.305

.695

1.00

Let K = Khan Let C = Chan Let J = Jackson Let B = blood test P(K) = .41 P(C) = .32 P(J) = .27 P(BK) = .05 P(BC) = .08 P(BJ) = .06

Prior P(Ei) .41 .32 .27

4.52

Conditional

Joint

Revised

P(BEi) .05 .08 .06

P(B  Ei) .0205 .0256 .0162 P(B) = .0623

P(EiB) .3291 .4109 .2600

Let R = regulations

T = tax burden

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

P(R) = .30

P(T) = .35

P(TR) = .71

a) P(R  T) = P(R)P(TR) = (.30)(.71) = .2130 b) P(R  T) = P(R) + P(T) - P(R  T) = .30 + .35 - .2130 = .4370 c) P(R  T) - P(R  T) = .4370 - .2130 = .2240 d) P(RT) = P(R  T)/P(T) = .2130/.35 = .6086 e) P(NRT) = 1 - P(RT) = 1 - .6086 = .3914 f) P(NRNT) = P(NR  NT)/P(NT) = [1 - P(R  T)]/P(NT) = (1 - .4370)/.65 = .8662

Probability matrix for problem 4.52: T

4.53

.213

.087

.30

.137

.563

.70

.35

.65

1.00

Let RB = Canadians who read a book or a magazine more than 10 hours

per week.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Event

E 0-24 25-34 35-44 >45 l

Prior Conditional Joint Revised P(RB  E) P(E) P(RB|E) P(E|RB) 0.283 0.11 0.03113 0.113684 0.138 0.24 0.03312 0.120951 0.132 0.27 0.03564 0.130154 0.446 0.39 0.17394 0.635212 P(RB) 0.999 =0.27383

4.54 Let GH = Good health P(GH) = .29

Let HM = Happy marriage P(HM) = .21

Let HF = Happy family

P(HF) = .40

a) P(HM  HF) = P(HM) + P(HF) - P(HM  HF) but P(HM  HF) = .0000 P(HM  HF) = P(HM) + P(HF) = .21 + .40 = .61

b) Since all categories are mutually exclusive, P(HM  HF  GH) = P(HM) + P(HF) + P(GH) = .21 + .40 + .29 = .9000 c) P(HF  GH) = .0000 The categories are mutually exclusive. The respondent could not select more than one answer. d) P(neither HF nor GH nor HM) = 1 - P(HM  HF  GH) = 1 - .9000 = = .1000

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Legal Notice Copyright

SOLUTIONS TO PROBLEMS IN CHAPTER 5: DISCRETE DISTRIBUTIONS

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

5.1 x P(x) x·P(x) (x – µ)2 2 (x – µ) ·P(x) 1 .238 .238 2.775556 0.6605823 2 .290 .580 0.443556 0.1286312 3 .177 .531 0.111556 0.0197454 4 .158 .632 1.779556 0.2811698 5 .137 .685 5.447556 0.7463152 µ = [x·P(x)] = 2.666 2 = [(x – µ)2·P(x)] = 1.8364439  = 1.8364439 = 1.355155

5.2 x P(x) (x – µ)2·P(x) 0 .103 1 .118 0.362201 2 .246 0.139114 3 .229 0.014084 4 .138 0.214936 5 .094 0.475029 6 .071 0.749015 7 .001 0.018046

x·P(x)

(x – µ)2

.118

3.069504

.492

0.565504

.687

0.061504

.552

1.557504

.470

5.053504

.426

10.549504

.007

18.045504

.000

µ =  [x·P(x)] = 2.752

2 = [(x – µ)2·P(x)] =

2.752496

 =

5.3 x (x – µ) ·P(x) 0 0.421324

2.752496 = 1.6591

P(x)

x·P(x)

(x – µ)2

.461

.000

0.913936

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

1 .285 .285 0.001936 0.000552 2 .129 .258 1.089936 0.140602 3 .087 .261 4.177936 0.363480 4 .038 .152 9.265936 0.352106 Expected number = µ = [x·P(x)]= 0.956 2 = [(x – µ)2·P(x)] = 1.278064  = 1.278064 = 1.1305

P(x)

x·P(x)

(x – µ)2

0 0.37791 1 0.01588 2 0.15705 3 0.26538 4 0.11752 5 0.02886 6 0.00000

.262

.000

1.4424

.393

0.0404

.246

.492

0.6384

.082

.246

3.2364

.015

.060

7.8344

.002

.010

14.4324

.000

23.0304

5.4 – µ) ·P(x)

2 =  [(x – µ)2·P(x)] =

µ =  [x·P(x)] = 1.201 0.96260

5.5

n=4 P(x=3) =

n=7

 =

.96260 = .98112

p = .10

q = .90

3 1 4C3(.10) (.90) = 4(.001)(.90) =

p = .80

.0036

q = .20

P(x=4) = 7C4(.80)4(.20)3 = 35(.4096)(.008) = .1147 c)

n = 10

p = .60

q = .40

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

P(x > 7) = P(x=7) + P(x=8) + P(x=9) + P(x=10) = 7 3 8 2 9 1 10C7(.60) (.40) + 10C8(.60) (.40) + 10C9(.60) (.40) 10 0

+10C10(.60) (.40) =

120(.0280)(.064) + 45(.0168)(.16) + 10(.0101)(.40) + 1(.0060)(1) = .2150 + .1209 + .0403 + .0060 = .3822

n = 12

p = .45

q = .55

P(5 < x < 7) = P(x=5) + P(x=6) + P(x=7) = 5 7 6 6 7 5 12C5(.45) (.55) + 12C6(.45) (.55) + 12C7(.45) (.55) =

792(.0185)(.0152) + 924(.0083)(.0277) + 792(.0037)(.0503) = .2225 + .2124 + .1489 = .5838

5.6

By Table A.2: a)

n = 20

p = .50

P(x=12) = .120

n = 20

p = .30

P(x > 8) = P(x=9) + P(x=10) + P(x=11) + ...+ P(x=20) = .065 + .031 + .012 + .004 + .001 + .000 = .113

n = 20

p = .70

P(x < 12) =

P(x=11) + P(x=10) + P(x=9) + ... + P(x=0) =

.065 + .031 + .012 + .004 + .001 + .000 = .113

n = 20

p = .90

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

P(x < 16) = P(x=16) + P(x=15) + P(x=14) + ...+ P(x=0) = .090 + .032 + .009 + .002 + .000 = .133

n = 15

p = .40

P(4 < x < 9) = P(x=4) + P(x=5) + P(x=6) + P(x=7) + P(x=8) + P(x=9) = .127 + .186 + .207 + .177 + .118 + .061 = .876

n = 10

p = .60

P(x > 7) = P(x=7) + P(x=8) + P(x=9) + P(x=10) = .215 + .121 + .040 + .006 = .382

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

5.7

n = 20

µ = np = 20(.70) = 14

 = b)

n  p  q  20(.70)(.30)  4.2 = 2.05

n = 70

p = .35

q = .65

µ = np = 70(.35) = 24.5

 = c)

n  p  q  70(.35)(.65)  15.925 = 3.99

n = 100

p = .50

q = .50

µ = np = 100(.50) = 50

 =

5.8

a) n = 6

n  p  q  100(.50)(.50)  25 = 5

p = .70

2 4

x 0 1 .060 3 .324 5 6

Prob .001 .010 .185 .303 .118

  4.2

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

n = 20

p = .50

x 0 1 .000 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

Prob .000 .000 .001 .005 .015 .037 .074 .120 .160 .176 .160 .120 .074 .037 .015 .005 .001 .000 .000 .000

  10

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

n=8

p = .80

x 0 1 2 3 4 5 6 7 8

Prob .000 .000 .001 .009 .046 .147 .294 .336 .168

  6.4

5.9

n = 20

p = .78

x = 14

P(x=14) = 20C14 (.78)14(.22)6 = 38,760(.030855)(.00011338) = .1356 b)

n = 20

p = .75

x = 20

P(x=20) = 20C20 (.75)20(.25)0 = (1)(.0031712)(1) = .0032 c)

n = 20

p = .70

x < 12

Use table A.2: P(x<12) = P(x=0) + P(x=1) + . . . + P(x=11)= .000 + .000 + .000 + .000 + .000 + .000 + .000 + .001 + .004 + .012 + .031 + .065 = .113 Solutions Manual 1-128 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

5.10

n = 16

p = .40

P(x > 9): from Table A.2: x 9 10 11 12 13 > 13

Prob .084 .039 .014 .004 .001 .000 .142

P(3 < x < 6): x 3 4 5 6

Prob .047 .101 .162 .198 .508

n = 13

p = .88

P(x = 10) = 13C10(.88)10(.12)3 = 286(.278500976)(.001728) = .1376 P(x = 13) = 13C13(.88)13(.12)0 = (1)(.1897906171)(1) = .1898 Expected Value = µ = np = 13(.88) = 11.44

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

5.11

n = 25

p = .60

a) x > 15 P(x > 15) = P(x = 15) + P(x = 16) + · · · + P(x = 25) Using Table A.2 n = 25, p = .60 x 15 16 17 18 19 20 21 22 >22

Prob .161 .151 .120 .080 .044 .020 .007 .002 .000 .585

b) x > 20 P(x > 20) = P(x = 21) + P(x = 22) + P(x = 23) + P(x = 24) + P(x = 25) = Using Table A.2 n = 25, p = .60 .007 + .002 + .000 + .000 + .000 = .009

c) P(x < 10) Using Table A.2 n = 25, p = .60 and x = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 x 9 8 7 <6

Prob. .009 .003 .001 .000 .013

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

5.12

n = 16

p = .50

x > 10

Using Table A.2, n = 16 and p = .50, x 11 12 13 14 15 16

Prob. .067 .028 .009 .002 .000 .000 .106

P(x > 10) = P(x=11) + P(x=12) + . . . + P(x=16) = .106

For n = 10

p = .87

x=6

P(x = 6) = 10C6 (.87)6(.13)4 = 210(.433626)(.00028561) = .0260 5.13

n = 15

p = .20

a) P(x = 5) = 15C5(.20)5(.80)10 = 3003(.00032)(.1073742) = .1032 b) P(x > 9): Using Table A.2 P(x = 10) + P(x = 11) + . . . + P(x = 15) = .000 + .000 + . . . + .000 = .000 c) P(x = 0) = 15C0(.20)0(.80)15 = (1)(1)(.035184) = .0352 d) P(4 < x < 7): Using Table A.2 P(x = 4) + P(x = 5) + P(x = 6) + P(x = 7) = .188 + .103 + .043 + .014 = .348 e)

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

The distribution is skewed to the right. Values of x near   3 have the highest probabilities. For the large values of x the probabilities are very small. = 5.14 n = 18 a)

p =.22

µ = 18(.22) = 3.96 (earn between $100,000 and

p = .32

µ = 18(.32) = 5.76 (earn $150,000 or more)

$149,999)

n = 18

p = .22

P(x > 8) = P(x = 8) + P(x = 9) + ...+P(x = 18) P(x = 8) = 18C8(.22)8(.78)10 = .0200 P(x = 9) = 18C9(.22)9(.78)9 = .0063 P(x = 10) = 18C10(.22)10(.78)8 = .0016 P(x = 11) = 18C11(.22)11(.78)7 = .0003 P(x = 12) = 18C12(.22)12(.78)6 = .0001 P(x) for x > 13 are negligible. Therefore P(x > 8) = .0200 + .0063 + .0016 + .0003 + .0001 = .0283 c) n = 18

p = .32

P(2 < x < 4) = P(x = 2) + P(x = 3) + P(x = 4) = 2 16 3 15 4 14 18C2(.32) (.68) + 18C3(.32) (.68) + 18C4(.32) (.68) =

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

.0327 + .0822 + .1450 = .2599 d) n = 18

p = .22

x=0

P(none earn between $100,000 and $149,999) = 18C0(.22)0(.78)18 = .01142 n = 18

p = .34

x=0

P(none earn $150,000 or more) = 18C0(.32)0(.68)18 = .00097 Since only 22% (compared to 32%) fall in the $100,000 to $149,999 category, it is more likely that none of the 18 CPFs would fall in this category.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

5.15

2.35  e  2.3 (64.36343)(.100259) a) P(x=5 = 2.3) = = .0538  5! 120 3.9 2  e  3.9 (15.21)(.020242) b) P(x=2 = 3.9) = = .1539  2! 2

c) P(x < 3 = 4.1) = P(x=3) + P(x=2) + P(x=1) + P(x=0) = 4.13  e  4.1 (68.921)(.016573) = .1904  3! 6 4.12  e  4.1 (16.81)(.016573) = .1393  2! 2 4.11  e  4.1 (4.1)(.016573) = .0679  1! 1 4.10  e  4.1 (1)(.016573) = .0166  0! 1

.1904 + .1393 + .0679 + .0166 = .4142 d) P(x=0 = 2.7) = 2.7 0  e  2.7 (1)(.06721)  = .0672 0! 1

e) P(x=1  = 5.4)= 5.4 1  e  5.4 (5.4)(.0045166)  = .0244 1! 1

f) P(4 < x < 8  = 4.4): P(x=5 = 4.4) + P(x=6 = 4.4) + P(x=7 = 4.4)= 4.4 5  e  4.4 4.4 6  e  4.4 4.4 7  e  4.4 + + = 6! 7! 5! (1649.1622)(.01227734) (7256.3139)(.01227734) + + 120 720 (31,927.781)(.01227734) 5040 Solutions Manual 1-134 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

5.16

= .1687 + .1237 + .0778 = .3702 a) P(x=6 = 3.8) = .0936 b) P(x>7 = 2.9): x 8 9 10 11 12

Prob .0068 .0022 .0006 .0002 .0000 .0098

c) P(3 < x < 9 = 4.2)= x 3 4 5 6 7 8 9

Prob .1852 .1944 .1633 .1143 .0686 .0360 .0168 .7786

d) P(x=0 = 1.9) = .1496 e) P(x < 6 = 2.9)= x 0 1 2 3 4 5 6

Prob .0550 .1596 .2314 .2237 .1622 .0940 .0455 .9714

f) P(5 < x < 8  = 5.7) = x 6 7 8

Prob .1594 .1298 .0925

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

.3817 5.17 a)  = 6.3

mean = 6.3 x 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

Standard deviation =

6.3 = 2.51

Prob .0018 .0116 .0364 .0765 .1205 .1519 .1595 .1435 .1130 .0791 .0498 .0285 .0150 .0073 .0033 .0014 .0005 .0002 .0001 .0000

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

b)  = 1.3

mean = 1.3 x 0 1 2 3 4 5 6 7 8 9

standard deviation =

1.3 = 1.14

Prob .2725 .3543 .2303 .0998 .0324 .0084 .0018 .0003 .0001 .0000

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

c)  = 8.9

mean = 8.9 x 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22

standard deviation =

8.9 = 2.98

Prob .0001 .0012 .0054 .0160 .0357 .0635 .0941 .1197 .1332 .1317 .1172 .0948 .0703 .0481 .0306 .0182 .0101 .0053 .0026 .0012 .0005 .0002 .0001

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

d)  = 0.6

mean = 0.6 x 0 1 2 3 4 5 6

standard deviation =

0.6 = .775

Prob .5488 .3293 .0988 .0198 .0030 .0004 .0000

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

5.18

 = 2.84 minutes a) P(x=6  = 2.8) = from Table A.3 .0407 b) P(x=0  = 2.8) = from Table A.3 .0608 c) Unable to meet demand if x > 44 minutes: x 5 6 7 8 9 10 11

Prob. .0872 .0407 .0163 .0057 .0018 .0005 .0001 .1523

There is a .1523 probability of being unable to meet the demand. Probability of meeting the demand = 1 – (.1523) = .8477 15.23% of the time a second window will need to be opened. d)  = 2.8 arrivals4 minutes P(x=3) arrivals2 minutes = ?? Lambda must be changed to the same interval (½ the size) New lambda=1.4 arrivals2 minutes P(x=3)=1.4) = from Table A.3 = .1128

P(x > 5 8 minutes) = ?? Lambda must be changed to the same interval (twice the size): New lambda = 5.6 arrivals8 minutes

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

P(x > 5   = 5.6): From Table A.3: 6

x 5 .1584 7 8 9 10 11 12 13 14 15 16 17

Alternate Solution Suggestion P(x > 5) = 1 – P(x<5) x Prob. 0 1 .0207 2 3 4

Prob. .1697 .1267 .0887 .0552 .0309 .0157 .0073 .0032 .0013 .0005 .0002 .0001 .6579

.0037 .0580 .1082 .1515 .3421

P(x > 5) = 1 – P(x<5) = 1 - .3421 = .6579

5.19

 = x/n = 126/36 = 3.5 Using Table A.3 a) P(x = 0) = .0302 b) P(x > 6) = P(x = 6) + P(x = 7) + . . . = .0771 + .0385 + .0169 + .0066 + .0023 + .0007 + .0002 + .0001 = .1424

Alternate Solution Suggestion P(x > 6) = 1 – P(x<6) = 1 – [ P(x=0)+p(x=1)+P(x=2)+…….+P9x=5)] = 1 – [.0302+.1057+.1850+.2158+.1888+.1322] =.1424 c) P(x < 4 10 minutes) Solutions Manual 1-141 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Double Lambda to  = 7.010 minutes P(x < 4) = P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3) = .0009 + .0064 + .0223 + .0521 = .0817 d) P(3 < x < 6 10 minutes)  = 7.0  10 minutes P(3 < x < 6) = P(x = 3) + P(x = 4) + P(x = 5) + P(x = 6) = .0521 + .0912 + .1277 + .1490 = .42 e) P(x = 8  15 minutes) Change Lambda for a 15 minute interval by multiplying the original Lambda by 3.  = 10.5  15 minutes P(x = 815 minutes) =

5.20

x  e  x!



(10.58 )(e 10.5 ) 8!

= .1009

 = 5.6 days3 weeks a) P(x=0  = 5.6): from Table A.3 = .0037 b) P(x=6  = 5.6): from Table A.3 = .1584 c) P(x > 15 = 5.6): x 15 16 17

Prob. .0005 .0002 .0001 .0008

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Because this probability is so low, if it actually occurred, the researcher would question the Lambda value as too low for this period. Perhaps the value of Lambda has changed because of an overall increase in pollution.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

5.21

 = 0.6 trips1 year a) P(x=0  = 0.6): from Table A.3 = .5488 b) P(x=1  = 0.6): from Table A.3 = .3293 c) P(x > 2  = 0.6): from Table A.3

x 2 3 4 .0004 6

Prob. .0988 .0198 .0030

.0000 .1220 Alternate Solution Suggestion P(x > 2) = 1 – P(x<2) = 1 – [ P(x=0)+p(x=1)] = 1 – [.5488+.3293] =.1220 d) P(x < 3 3 year period): The interval length has been increased (3 times) New Lambda =  = 1.8 trips3 years P(x < 3  = 1.8): from Table A.3

x 0 1 2 3

Prob. .1653 .2975 .2678 .1607 .8913

e) P(x=46 years): The interval has been increased (6 times) New Lambda =  = 3.6 trips6 years P(x=4 = 3.6): Solutions Manual 1-144 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

from Table A.3 = .1912 5.22

 = 1.2 collisions4 months a) P(x=0  = 1.2): from Table A.3 = .3012

b) P(x=22 months): The interval has been decreased (by ½) New Lambda =  = 0.6 collisions2 months P(x=2  = 0.6): from Table A.3 = .0988

c) P(x < 1 collision6 months): The interval length has been increased (by 1.5) New Lambda =  = 1.8 collisions6 months P(x < 1  = 1.8): from Table A.3

x 0 1

Prob. .1653 .2975 .4628

The result is likely to happen almost half the time (46.28%). Ship channel and weather conditions are about normal for this period. Safety awareness is about normal for this period. There is no compelling reason to reject the lambda value of 0.6 collisions per 4 months based on an outcome of 0 or 1 collisions per 6 months.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

5.23

 = 1.2 penscarton a) P(x=0   = 1.2): from Table A.3 = .3012 b) P(x > 8   = 1.2): from Table A.3 = .0000 c) P(x > 3   = 1.2): from Table A.3

x 4 5 6 .0002 8

Prob. .0260 .0062 .0012 .0000 .0336

Alternate Solution Suggestion P(x >3) = 1 – P(x< 3) = 1 – [ P(x=0)+p(x=1)+p(x=2)+p(x=3)] = 1 – [.3012+.3614+.2169+.0867] =.0338

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

5.24

n = 100,000

p = .00004

P(x > 7n = 100,000 p = .00004):  = µ = np = 100,000(.00004) = 4.0

Since n > 20 and np < 7, the Poisson approximation to this binomial problem is close enough. P(x > 7   = 4): Using Table A.3

x 7 8 9 10 11 12 13 14

Prob. .0595 .0298 .0132 .0053 .0019 .0006 .0002 .0001 .1106

x 11 12 13 14

Prob. .0019 .0006 .0002 .0001 .0028

P(x >10   = 4): Using Table A.3

Since getting more than 10 is a rare occurrence, this particular geographic region appears to have a higher average rate than other regions. An investigation of particular characteristics of this region might be warranted.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

5.25

p = .009

n = 200

Use the Poisson Distribution:  = np = 200(.009) = 1.8 a) P(x > 6) from Table A.3 = P(x = 6) + P(x = 7) + P(x = 8) + P(x = 9) + . . . = .0078 + .0020 + .0005 + .0001 = .0104 b) P(x > 10) = .0000 c) P(x = 0) = .1653 d) P(x < 5) = P(x = 0) + P(x = 1) + P(x = 2) + P( x = 3) + P(x = 4) = .1653 + .2975 + .2678 + .1607 + .0723 = .9636

5.26 If 99% see a doctor, then 1% do not see a doctor. Thus, p = .01 for this problem. n = 300,

p = .01,

 = n(p) = 300(.01) = 3

a) P(x = 5): Using  = 3 and Table A.3 = .1008 b) P(x < 4) = P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3) = .0498 + .1494 + .2240 + .2240 = .6472 c) The expected number = µ =  = 3

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

5.27

a) P(x = 3 N = 11, A = 8, n = 4) 8

C3 3 C1 (56)(3)  = .5091 330 11 C 4

b) P(x < 2)N = 15, A = 5, n = 6) P(x = 1) + P (x = 0) = 5

C1 10 C5 C  C (5)(252) (1)(210) + 5 0 10 6 =  C C 5005 5005 15 6 15 6

.2517 + .0420 = .2937 c) P(x=0 N = 9, A = 2, n = 3) 2

C 0 7 C3 (1)(35)  = .4167 84 9 C3

d) P(x > 4 N = 20, A = 5, n = 7) = P(x = 5) = 5

C 5 15 C 2 (1)(105) = = .0014 77520 20 C 7

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

5.28

N = 19

n=6

A = 11

a) P(x = 1 private) 11

C1 8 C5 (11)(56)  = .0227 C 27 , 132 19 6

b) P(x = 4 private) 11

C 4 8 C 2 (330)(28)  = .3406 27,132 19 C 6

c) P(x = 6 private) 11

C 6 8 C 0 (462)(1)  = .0170 27,132 19 C 6

d) P(x = 0 private) 11

C 0 8 C 6 (1)(28)  = .0010 27,132 19 C 6

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

5.29

N = 20

A=4 4

a) P(x = 0) =

b) P(x = 4) = 4

N = 20

C 0 16 C 4 (1)(1820) = = .3756 C 4845 20 4

C 4 16 C 0 (1)(1) = = .0002 4845 20 C 4 4

c) P(x = 2) =

5.30

n=4

C 2 16 C 2 (6)(120) = = .1486 4845 20 C 4

A = 16 white

a) P(x = 4 white) =

N - A = 4 red

n=5

C 4  4 C1 (1820)(4) = = .4696 15504 20 C 5

b) P(4 red ; x = 1 white) =

C1  4 C 4 (16)(1) = = .0010 15504 20 C 5

c) P(5 red) = .0000 The participant cannot draw 5 red beads if there are only 4 to draw from.

5.31

N = 10

n=4

a) A = 4 (British Columbia, Alberta, Saskatchewan, Manitoba) x = 2 P(x = 2) =

C 2 6 C 2 (6)(15)  = .4286 210 10 C 4

b) A = 4 (Nova Scotia, New Brunswick, Newfoundland and Labrador, Prince Edward Island)

x=0

P(x = 0) =

C 0 6 C 4 (1)(15)  = .0714 210 10 C 4

c) A = 4 (Ontario, British Columbia, Alberta, Quebec)

x=3

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

P(x = 3) =

C3 6 C1 (4)(6)  = .1143 210 10 C 4

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

5.32

N = 16 a) P(x = 0) =

A = 4 defective 4

n=3

C 0 12 C3 (1)(220)  = .3929 C 560 16 3

C3 12 C0 (4)(1)  = .0071 560 16 C 3 C  C c) P(x > 2) = P(x=2) + P(x=3) = 4 2 12 1 + .0071 (from part b.) = 16 C 3 b) P(x = 3) =

(6)(12) + .0071 = .1286 + .0071 = .1357 560

d) P(x < 1) = P(x=1) + P(x=0) = 4

C1 12 C 2 (4)(66) + .3929 (from part a.) = + .3929 = .4714 + .3929 C 560 16 3

= .8643

5.33

N = 18

A = 11 men

n=5

P(x < 1) = P(1) + P(0) = 11

C1 7 C 4 11 C 0 7 C 5 (11)(35) (1)(21) + = = .0449 + .0025 =  8568 8568 18 C 5 18 C 5

.0474 It is fairly unlikely that these results occur by chance. A researcher might want to further investigate this result to determine causes. Were officers selected based on leadership, years of service, dedication, prejudice, or some other reason?

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

5.34

a) P(x=4 n = 11 and p = .23) 4 7 11C4(.23) (.77) = 330(.0028)(.1605) =

.1482

b) P(x > 1n = 6 and p = .50) = 1 – P(x < 1) = 1 – P(x = 0) = 1 – [6C0(.50)0(.50)6] = 1 – [(1)(1)(.0156)] = .9844 c) P(x > 7 n = 9 and p = .85) = P(x = 8) + P(x = 9) = 8 1 9 0 9C8(.85) (.15) + 9C9(.85) (.15) =

(9)(.2725)(.15) + (1)(.2316)(1) = .3679 + .2316 = .5995 d) P(x < 3 n = 14 and p = .70) = P(x = 3) + P(x = 2) + P(x = 1) + P(x = 0) = 3 11 2 12 14C3(.70) (.30) + 14C2(.70) (.30) + 1 13 0 14 14C1(.70) (.30) + 14C0(.70) (.30) =

(364)(.3430)(.00000177) + (91)(.49)(.000000531)= (14)(.70)(.00000016) + (1)(1)(.000000048) = .0002 + .0000 + .0000 + .0000 = .0002

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

5.35

a) P(x = 14 n = 20 and p = .60) = .124 b) P(x < 5 n = 10 and p =.30) = P(x = 4) + P(x = 3) + P(x = 2) + P(x = 1) + P(x=0) = x 0 1 2 3 4

Prob. .028 .121 .233 .267 .200 .849

c) P(x > 12 n = 15 and p = .60) = P(x = 12) + P(x = 13) + P(x = 14) + P(x = 15) x 12 13 14 15

Prob. .063 .022 .005 .000 .090

d) P(x > 20 n = 25 and p = .40) = P(x = 21) + P(x = 22) + P(x = 23) + P(x = 24) + P(x=25) = x 21 22 23 24 25

Prob. .000 .000 .000 .000 .000 .000

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

5.36 a) P(x = 4 = 1.25) (125 . 4 )(e 1.25 ) (2.4414)(.2865) = .0291  4! 24

b) P(x < 1 = 6.37) = P(x = 1) + P(x = 0) = (6.37) 1 (e  6.37 ) (6.37 0 )(e  6.37 ) (6.37)(.0017) (1)(.0017)    1! 0! 1 1

.0109 + .0017 = .0126

c) P(x > 5 = 2.4) = P(x = 6) + P(x = 7) + ... = (2.4 6 )(e  2.4 ) (2.4 7 )(e  2.4 ) (2.4 8 )(e  2.4 ) (2.4 9 )(e  2.4 ) (2.4 10 )(e  2.4 )       6! 7! 8! 9! 10!

.0241 + .0083 + .0025 + .0007 + .0002 = .0358 for values x > 11 the probabilities are each .0000 when rounded off to 4 decimal places.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

5.37

a) P(x = 3 = 1.8) = .1607 b) P(x < 5 = 3.3) = P(x = 4) + P(x = 3) + P(x = 2) + P(x = 1) + P(x = 0) = x 0 1 2 3 4

Prob. .0369 .1217 .2008 .2209 .1823 .7626

c) P(x > 3  = 2.1) = x 3 4 5 6 7 8 9 10 11

Prob. .1890 .0992 .0417 .0146 .0044 .0011 .0003 .0001 .0000 .3504

Alternate Solution Suggestion P(x > 3) = 1 – P(x< 3) = 1 – [ P(x=0)+p(x=1)+p(x=2)] = 1 – [.1225+.2572+.2700] =.3503 d) P(2 < x < 5  = 4.2): P(x=3) + P(x=4) + P(x=5) = x 3 4 5

Prob. .1852 .1944 .1633 .5429

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

5.38

a) P(x = 3 N = 6, n = 4, A = 5) =

C3 1 C1 (10)(1)  = .6667 15 6 C4

b) P(x < 1 N = 10, n = 3, A = 5): P(x = 1) + P(x = 0) =

C1 5 C 2 5 C 0 5 C 3 (5)(10) (1)(10) + =  120 120 10 C 3 10 C 3

= .4167 + .0833 = .5000 c) P(x > 2  N = 13, n = 5, A = 3): P(x=2) + P(x=3) Note: only 3 x's in population 3

C 2 10 C 3 + 13 C 5

C3 10 C 2 (3)(120) (1)(45) = = .2797 + .0350 =  1287 1287 13 C 5

.3147

5.39

n = 25 p = .20 retired from Table A.2: P(x = 7) = .111 P(x > 10):

P(x = 10) + P(x = 11) + . . . + P(x = 25) = .012 + .004 + .001

= .017 Expected number of retired people = µ = np = 25(.20) = 5 n = 20 p = .40 mutual funds from Table A.2: P(x = 8) = .180 P(x < 6) = P(x = 0) + P(x = 1) + . . . + P(x = 5) = .000 + .000 + .003 +.012 + .035 + .075 = .125 P(x = 0) = .000 P(x > 12) = P(x = 12) + P(x = 13) + . . . + P(x = 20) = .036 + .015 + .005 + .001 = .057 The highest probability is .180 for x = 8 .

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Expected number of people invested in mutual funds = µ = n p = 20(.40) =8

5.40

 = 3.2 cars2 hours a) P(x=3 cars per 1 hour) = ? The interval has been decreased by ½. The new  = 1.6 cars1 hour. P(x = 3 = 1.6) = (from Table A.3) .1378

b) P(x = 0 cars per ½ hour) = ? The interval has been decreased by ¼ the original amount. The new  = 0.8 cars½ hour. P(x = 0  = 0.8) = (from Table A.3) .4493 c) P(x > 5  = 1.6) = (from Table A.3)

x 5 .0047 7 8

Prob. .0176 .0011 .0002 .0236

Either a rare event occurred or perhaps the long-run average, , has changed (increased).

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

5.41

N = 32

A = 10

a) P(x = 3) =

b) P(x = 6) =

c) P(x = 0) =

n = 12

C3  22 C9 (120)(497,420) = = .2644 225 , 792 , 840 C 32 12 C 6  22 C 6 (210)(74,613) = = .0694 225,792,840 32 C12 C 0  22 C12 (1)(646,646) = = .0029 225,792,840 32 C12

d) A = 22 P(7 < x < 9) =

C 7 10 C5 C  C C  C + 22 8 10 4 + 22 9 10 3 32 C12 32 C12 32 C12

(170,544)(252) (319,770)(210) (497,420)(120)   225,792,840 225,792,840 225,792,840

= .1903 + .2974 + .2644 = .7521

5.42

 = 1.4 defects1 lot

If x > 3, buyer rejects

If x < 3, buyer

accepts P(x < 3  = 1.4) = (from Table A.3) x 0 1 2 3

Prob. .2466 .3452 .2417 .1128 .9463

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

5.43

a) n = 20 and p = .176 The expected number = µ = np = (20)(.176) = 3.52

b) P(x < 1 n = 20 and p = .176) = P(x = 1) + P(x = 0) =

1 19 0 20 20C1(.176) (.824) + 20C0(.176) (.824)

= (20)(.176)(.02527) + (1)(1)(.02082) = .0890 +. 0208 = .1098 Since the probability is comparatively low, the population of your province may have a lower percentage of chronic heart conditions than those of other provinces.

5.44

a) P(x > 7 n = 10 and p = .69) = P(x = 8) + P(x = 9) + P(x = 10) = 8 2 9 1 10 0 10C8(.69) (.31) + 10C9(.69) (.31) + 10C10(.69) (.31) =

= .2222 +. 1099 +. 0245 = .3566 Expected number = µ = np = 10(.69) = 6.9

b) n = 15

p = .25 Expected number = µ = np = 15(.25) = 3.75

P(x=0 n = 15 and p = .25) = 0 15 15C0(.25) (.75)

c) n = 7

= .0134

p = .49

P(x = 7n = 7 and p = .49) = 7C7(.49)7(.51)0 = .0068 Probably the 49% figure is too low for this population since the probability of this occurrence is so low (.0068).

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

5.45

n = 12 a.) P(x = 0 long hours): 0 12 12C0(.20) (.80) = .0687

p = .20

b.) P(x > 6 long hours): p = .20 Using Table A.2: P(x = 6) + P(x = 7) + . . . + P(x = 12) = .016 + .003 + .001 = .020 c) P(x = 5 good financing): p = .25,

5 7 12C5(.25) (.75) =

.1032

d.) p = .19 (good plan), expected number = µ = n(p) = 12(.19) = 2.28

5.46

n = 100,000

p = .000014

Worked as a Poisson:

 = np = 100,000(.000014) = 1.4

a) P(x = 5): from Table A.3 = .0111 b) P(x = 0): from Table A.3 = .2466 c) P(x > 6): x 7 8

(from Table A.3)

Prob .0005 .0001 .0006

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

5.47

P(x < 3 n = 8 and p = .60):

From Table A.2: x 0 1 2 3

Prob. .001 .008 .041 .124 .174

17.4% of the time in a sample of eight, three or fewer customers are walk-ins by chance. Other reasons for such a low number of walk-ins might be that she is retaining more old customers than before or perhaps a new competitor is attracting walk-ins away from her.

5.48

n = 25

p = .63

a) P(x = 15 n = 25 and p = .63) = 25C15(.63)15(.37)10 = .1536

b) P(x > 20n = 25 and p = .63) = P(x = 21) + P(x = 22) + P(x = 23) + P(x = 24) + P(x = 25) = 21 4 22 3 23 2 25C21(.63) (.37) + 25C22(.63) (.37) + 25C23(.63) (.37) +

24 1 25C24(.63) (.37) +

25 0 25C25(.63) (.37) = .0145 + .0045 + .0010 + .0001 + .0000 = .0201

c) Since such a result would only occur 2% of the time by chance, it is likely that the analyst's list was not representative of the entire province of Manitoba or the 63% figure for the Manitoba census is not correct.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

5.49

 = 0.6 flats2,000 km P(x = 0  = 0.6) = (from Table A.3) .5488 P(x > 3  = 0.6) = (from Table A.3) x 3 4 5

Prob. .0198 .0030 .0004 .0232

Assume one trip is independent of the other. Let F = flat tire and NF = no flat tire P(NF1  NF2) = P(NF1)  P(NF2) but P(NF) = .5488 P(NF1  NF2) = (.5488)(.5488) = .3012

5.50

N = 20

n=8

a) P(x = 1 in Ontario)

A=9

= 9C1 *11C7 /20C8 = (9)(330)/125,970 = 0.0236 b) P(x = 4 in top 10) A = 10 =

10C4 *10C4 /20C8 = (210)(210)/125,970 = 0.3501

c) P(x = 0 in British Columbia)

A=3

C 0  22 C8 (1)(319,770)  = .2957 1,081,575 25 C8 3C0 *17C8 /20C8 = (1)(2210)/125,970 = 0.0175 3

P(x = 3 with N) = .0000 There are no newspapers located in Nova Scotia.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

5.51

N = 24

n=6

a) P(x = 6) =

b) P(x = 0) =

A=8

C 6 16 C 0 (28)(1)  = .0002 C 134 , 596 24 6 8

C 0 16 C 6 (1)(8008)  = .0595 134,596 24 C 6

c) The probability from part b) is greater as only 8 out of 24 commute from Airdrie. d) A = 16 (do not commute from Airdrie) P(x = 3) =

5.52

n = 25 p = .16

µ = 25(.16) = 4

C3 8 C3 (560)(56)  = .2330 134,596 24 C 6

Expected Value = µ = np = 25(.16) = 4

 =

n  p  q  25(.16)(.84) = 1.833

P(x > 8) = P(x = 9) + P(x = 10) + ... + P(x = 25) = .0086 + .0026 + .0007 + .0002 + .0000 + ... + .0000= .0121 The values for x > 8 are so far away from the expected (4) that they are very unlikely to occur. Solutions Manual 1-166 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

P(x = 10) = 25C10(.16)10(.84)15 = .0026 which is very unlikely. If this value (x = 10) actually occurred, one would doubt the validity of the p = .16 figure or one would have experienced a very rare event. 5.53

 = 2.4 calls1 minute a) P(x = 0 = 2.4) = (from Table A.3) .0907

b) Can handle x < 5 calls

Cannot handle x > 5 calls

P(x > 5 = 2.4) = (from Table A.3) x 6 7 8 9 10 11

Prob. .0241 .0083 .0025 .0007 .0002 .0000 .0358

c) P(x = 3 calls2 minutes) The interval has been increased 2 times. New Lambda:  = 4.8 calls2 minutes. from Table A.3: .1517

d) P(x < 1 calls15 seconds): The interval has been decreased by ¼. New Lambda =  = 0.6 calls15 seconds. P(x < 1 = 0.6) = (from Table A.3) P(x = 1) = .3293 P(x = 0) = .5488 .8781

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

5.54

p = .13 n = 50 > 20 and n  p  6.5  7 We can use the Poisson distribution to approximate the binomial

problem:  = np = (50)(.13) = 6.5 a) Expected number of households =  = μ = np = (50)(.13) = 6.5. b) P(x > 8 = 6.5) = ? P(x > 8 ) = P(x = 8) + P(x = 9) + . . . + P(x = 19) or P(x > 8 ) = 1 - P(x < 8) =1- ( P(x = 0) + P(x = 1) +. . . + P(x = 7) ) Using Table A.3:

x 0 1 2 3 4 5 6 7

Prob. .0015 .0098 .0318 .0688 .1118 .1454 .1575 .1462 .6728

P(x > 8 ) = 1 - P(x < 8) =1- .6728 = .3272

c) P(2 < x < 6 = 6.5) = ? P(2 < x < 6) = P(x = 2) + P(x = 3) + P(x = 4) + P(x = 5) + P(x = 6) Using Table A.3:

x 2 3 4 5 6

Prob. .0318 .0688 .1118 .1454 .1575 .5153

P(2 < x < 6 = 6.5) = .5153

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

5.55

n=8

p = .55

x = 2 women

2 6 8C2(.55) (.45) = (28)(.3025)(.0083038) = .0703

It is not very likely that a company would randomly hire 8 physicians and only two of them would be female.

5.56

N = 28 a) n = 5

P(x = 3enrolments > 18,000)

A = 24

24C3 *4C2 /28C5 = (12)(2024)/98,280 = .2471

b) n = 8 P(x < 2universities in Quebec)

A=5

P(x < 2) = P(x = 0) + P(x = 1) + P(x = 2) = = =5C0 *23C8 /28C5 +5C1 *23C7 /28C5 +5C2 *23C6 /23C8 = = 490314/3,108,105 + 1225785/3,108,105 +1009470/3,108,105 = .1575+.39438+.32479 = .8769 c) n = 5 . The list contains only four universities in British Columbia. So, the probability of selecting one university in B.C. from the list of 28 is 4/28. Thus, p = 4/28 and q = 24/28. The binomial formula yields the final answer: P(x = 2 n = 5, p = 3/28) = 5C2(4/28)2(24/28)3 = (10)(.0204)(.6297) = .1285 5.57

N = 14 n = 4 a) P(x = 4 N = 14, n = 4, A = 10 north side) 10

C 4  4 C 0 (210((1)  = .2098 1001 14 C 4

b) P(x = 4 N = 14, n = 4, A = 4 west) 4

C 4 10 C 0 (1)(1)  = .0010 1001 14 C 4

c) P(x = 2 N = 14, n = 4, A = 4 west) Solutions Manual 1-169 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

C 2 10 C 2 (6)(45)  = .2697 1001 14 C 4

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

5.58a)  = 3.051,000 P(x = 0) =

3.050  e 3.05 = .0474 0!

b) P(x = 62,000) : The number of doctors has been increased 2 times. New Lambda:  = 6.12,000 P(x = 6) =

6.16  e 6.1 (51,520.37436)(.00224287) = .1605  6! 720

c)  = 1.61,000 and  = 4.83,000 from Table A.3: P(x < 7) = P(x = 0) + P(x = 1) + . . . + P(x = 6) = .0082 + .0395 + .0948 + .1517 + .1820 + .1747 + .1398 = .7907

5.59

This is a binomial distribution with n = 15 and p = .36.

 = np = 15(.36) = 5.4  =

15(.36)(.64) = 1.86

The most likely values are near the mean, 5.4. Note from the printout that the most probable values are at x = 5 and x = 6 which are near the mean.

5.60

This printout contains the probabilities for various values of x from zero to eleven from a Poisson distribution with  = 2.78. Note that the highest probabilities are at x = 2 and x = 3 which are near the mean. The probability is slightly higher at x = 2 than at x = 3 even though x = 3 is nearer to the mean because of the “piling up” effect of x = 0. The distribution is skewed to the right, however this may change dependent upon the p value which is unknown..

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

5.61

This is a binomial distribution with n = 22 and p = .64. The mean is np = 22(.64) = 14.08 and the standard deviation is:

 =

n  p  q  22(.64)(.36) = 2.25

The x value with the highest peak on the graph is at x = 14 followed by x = 15 and x = 13 which are nearest to the mean. The distribution is skewed to the left.

5.62

This is the graph of a Poisson Distribution with  = 1.784. Note the high probabilities at x = 1 and x = 2 which are nearest to the mean. Note also

that the probabilities for values of x > 8 are near to zero because they are so far away from the mean or expected value. The distribution is skewed to the right.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Legal Notice Copyright

SOLUTIONS TO PROBLEMS IN CHAPTER 6: CONTINUOUS DISTRIBUTIONS

6.1 a = 200

b = 240

1 1 1   = .025, we have b  a 240  200 40 0.025, 200  x  240 f x     0, for all other values

a) Since

b)  =

a  b 200  240  = 220 2 2

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

 =

b  a 240  200 40 = 11.547   12 12 12

c) P(x > 230) =

240  230 10 = .250  240  200 40

d) P(205 < x < 220) =

e) P(x < 225) =

220  205 15 = .375  240  200 40

225  200 25 = .625  240  200 40

6.2 a = 8

b = 21 1 1 1   = .0769, we have a) Since b  a 21  8 13 0.0769, 8  x  21 f x    0, for all other values 

b)  =

a  b 8  21 29   = 14.5 2 2 2

 =

b  a 21  8 13   = 3.7528 12 12 12

c) P(10 < x < 17) =

17  10 7  = .5385 21  8 13

d) P(x < 22) = 1.0000 e) P(x > 7) = 1.0000

6.3 a = 2.80

b = 3.14

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

 =

a  b 2.80  3.14 = 2.97  2 2

 =

b  a 3.14  2.80 = 0.098  12 12

P(3.00 < x < 3.10) =

b = 355.8   355.0

6.4 a = 354.0 Height =

3.10  3.00 = 0.2941 3.14  2.80

1 1 1   = 0.5556 b  a 355.8  354.0 1.8

P(x > 355.2) =

355.8  355.2 0.6  = .3333 355.8  354.0 1.8

P(354.3 < x < 355.2) =

6.5 µ = 2,100

 =

a = 400

ba 12

Height =



355.2  354.3 0.9  = .5000 355.8  354.0 1.8

b = 3,800

3,800  400 12

= 981.5

1 1  = .000294 b  a 3,800  400

P(x > 3,000) =

3,800  3,000 800  = .2353 3,800  400 3,400

P(x > 4,000) = .0000 P(700 < x < 1,500) =

1,500  700 800  = .2353 3,800  400 3,400

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

6.6 a) P(z > 1.96): Table A.5 value for z = 1.96:

.4750

P(z > 1.96) = .5000 – .4750 = .0250

b) P (z < 0.73): Table A.5 value for z = 0.73:

.2673

P(z < 0.73) = .5000 + .2673 = .7673

c) P(–1.46 < z < 2.84): Table A.5 value for z = 2.84: Table A.5 value for z = 1.46:

.4977 .4279

P(–1.46 < z < 2.84) = .4977 + 4279 = .9256

d) P(–2.67 < z < 1.08): Table A.5 value for z = 2.67: Table A.5 value for z = 1.08:

.4962 .3599

P(–2.67 < z < 1.08) = .4962 + .3599 = .8561 e) P (–2.05 < z < –.87): Table A.5 value for z = 2.05: .4798 Table A.5 value for z = 0.87: .3078 P(–2.05 < z < –.87) = .4798 – .3078 = .1720

6.7 a)

P(x < 635 µ = 604,  = 56.8):

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

z =

x





635  604 = 0.55 56.8

Table A.5 value for z = 0.55:

.2088

P(x < 635) = .2088 + .5000 = .7088 b) P(x < 20 µ = 48,  = 12): z =

x





20  48 = –2.33 12

Table A.5 value for z = 2.33:

.4901

P(x < 20) = .5000 – .4901 = .0099 c) P(100 < x < 150 µ = 111,  = 33.8): z =

x





150  111 = 1.15 33.8

Table A.5 value for z = 1.15: z =

x





.3749

100  111 = –0.33 33.8

Table A.5 value for z = 0.33:

.1293

P(100 < x < 150) = .3749 + .1293 = .5042 d) P(250 < x < 255 µ = 264,  = 10.9): z =

x





250  264 = –1.28 10.9

Table A.5 value for z = 1.28: z =

x





.3997

255  264 = –0.83 10.9

Table A.5 value for z = 0.83:

.2967

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

P(250 < x < 255) = .3997 – .2967 = .1030 e) P(x > 35 µ = 37,  = 4.35): z =

x



35  37 = –0.46 4.35



Table A.5 value for z = 0.46:

.1772

P(x > 35) = .1772 + .5000 = .6772 f) P(x > 170 µ = 156,  = 11.4): z =

x





170  156 = 1.23 11.4

Table A.5 value for z = 1.23:

.3907

P(x > 170) = .5000 – .3907 = .1093

6.8 µ = 6.7

 = 1.2

a) P(x > 5.2): z =

x





5.2  6.7 = –1.25 1.2

area between x = 5.2 and µ = 6.7 from table A.5 is .3944 P(x > 5.2) = .3944 + .5000 = .8944

b) P(x < 4): z =

x





4  6.7 = –2.25 1.2

from table A.5, area = .4878 P(x < 4) = .5000 – .4878 = .0122 Solutions Manual 1-179 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

c) P(7.6 < x < 9.4): z =

x





9.4  6.7 = 2.25 1.2

from table A.5, area = .4878 z =

x





7.6  6.7 = 0.75 1.2

from table A.5, area = .2734 P(7.6 < x < 9.4) = .4878 – .2734 = .2144

 = 11.35

6.9 µ = 73 a) P(x > 100): z =

x





100  73 = 2.38 11.35

from Table A.5, the value for z = 2.38 is .4913 P(x > 100) = P(z > 2.38) = .5000 – .4913 = .0087

b) P(60 < x < 83): z =

z =

x

 x





60  73 = –1.15 11.35



83  73 = 0.88 11.35

from Table A.5, the value for z = 1.15 is .3749 and for z = 0.88 is .3106

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

P(60 < x < 83) = P(–1.15 < z < 0.88) =.3749 + .3106 = .6855

c) P(80 < x < 90): z =

z =

x

 x





80  73 = 0.62 11.35



90  73 = 1.50 11.35

from Table A.5, the value for z = 0.62 is .2324 from Table A.5, the value for z = 1.50 is .4332 P(80 < x < 90) = P(0.62 < z < 1.50) =.4332 – .2324 = .2008

d) P(x < 55): z =

x





55  73 = –1.59 11.35

from Table A.5, the value for z = 1.59 is .4441 P(x < 55) = P(z < –1.59) = .5000 – .4441 = .0559

 = $725

6.10 µ = $1,332

a) P(x > $2,000): z =

x





2,000  1,332 = 0.92 725

from Table A.5, the z = 0.92 yields:

.3212

P(x > $2,000) = .5000 – .3212 = .1788

b) P(owes money) = P(x < 0): z =

x





0  1332 = –1.84 725

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

from Table A.5, the z = 1.84 yields: .4671 P(x < 0) = .5000 – .4671 = .0329

c) P($100 < x < $700): z =

x





100  1,332 = –1.70 725

from Table A.5, the z = 1.70 yields: z =

x





.4554

700  1332 = –0.87 725

from Table A.5, the z = 0.87 yields:

.3078

P($100 < x < $700) = .4554 – .3078 = .1476

6.11  = $30,000

 = $9,000

a) P($15,000 < x < $45,000): z =

x





45,000  30,000 = 1.67 9,000

From Table A.5, z = 1.67 yields: z =

x





.4525

15,000  30,000 = –1.67 9,000

From Table A.5, z = 1.67 yields:

.4525

P($15,000 < x < $45,000) = .4525 + .4525 = .9050

b) P(x > $50,000): z =

x





50,000  30,000 = 2.22 9,000

From Table A.5, z = 2.22 yields: .4868 Solutions Manual 1-182 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

P(x > $50,000) = .5000 – .4868 = .0132

c) P($5,000 < x < $20,000): z =

x





5,000  30,000 = –2.78 9,000

From Table A.5, z = 2.78 yields: z =

x





.4973

20,000  30,000 = –1.11 9,000

From Table A.5, z = 1.11 yields

.3665

P($5,000 < x < $20,000) = .4973 – .3665 = .1308

d) Since 90.82% of the values are greater than x = $7,000, x = $7,000 is in the lower half of the distribution and .9082 – .5000 = .4082 lie between x and µ. From Table A.5, z = 1.33 is associated with an area of .4082. x Solving for : z =



–1.33 =

7,000  30,000



 = 17,293.23

e)  = $9,000. If 79.95% of the costs are less than $33,000, x = $33,000 is in the upper half of the distribution and .7995 – .5000 = .2995 of the values lie between $33,000 and the mean. From Table A.5, an area of .2995 is associated with z = 0.84 Solutions Manual 1-183 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Solving for µ:

z =

0.84 =

x

 33,000   9,000

µ = $25,440

6.12

µ = 200,

 = 47

Determine x

a) 60% of the values are greater than x: Since 50% of the values are greater than the mean, µ = 200, 10% or .1000 lie between x and the mean. From Table A.5, the z value associated with an area of .1000 is z = –0.25. The z value is negative since x is below the mean. Substituting z = –0.25, µ = 200, and  = 47 into the formula and solving for x: z =

–0.25 =

x

 x  200 47

x = 188.25

b) x is less than 17% of the values. Since x is only less than 17% of the values, 33% (.5000 – .1700) or .3300 lie between x and the mean. Table A.5 yields a z value of 0.95 for an area of .3300. Using this z = 0.95, µ = 200, and  = 47, x can be solved for: z =

x



Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

0.95 =

x  200 47

x = 244.65

c) 22% of the values are less than x. Since 22% of the values lie below x, 28% lie between x and the mean (.5000 – .2200). Table A.5 yields a z of –0.77 for an area of .2800. Using the z value of –0.77, µ = 200, and  = 47, x can be solved for: z =

–0.77 =

x

 x  200 47

x = 163.81

d) x is greater than 55% of the values. Since x is greater than 55% of the values, 5% (.0500) lie between x and the mean. From Table A.5, a z value of 0.13 is associated with an area of .05. Using z = 0.13, µ = 200, and  = 47, x can be solved for: z =

0.13 =

x

 x  200 47

x = 206.11

6.13

 = 625. If 73.89% of the values are greater than 1,700, then 23.89%

or .2389 lie between 1,700 and the mean, µ. The z value associated with .2389 is –0.64 since the 1,700 is below the mean. Solutions Manual 1-185 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Using z = –0.64, x = 1,700, and  = 625, µ can be solved for: z =

–0.64 =

x

 1,700   625

µ = 2,100 µ = 2,258 and  = 625. Since 31.56% are greater than x, 18.44% or .1844 (.5000 – .3156) lie between x and µ = 2,258. From Table A.5, a z value of 0.48 is associated with .1844 area under the normal curve. Using µ = 2,258,  = 625, and z = 0.48, x can be solved for: z =

0.48 =

x

 x  2,258 625

x = 2,558

6.14 µ = 21.3

 = 6.7

Since only 3% of the values are greater than x, x is above the mean. In addition, 47% of the area or .4700 of the values lie between x and the mean. The z score associated with this area is 1.88. Solving for x: z = 1.88 =

x



x  21.3 6.7

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

6.15 = 6.2. Since 62.5% of values are greater than 21, x = 21 is in the lower half of the distribution and .1250 (.6250 - .5000) lie between x and the mean. Table A.5 yields a z = 0.32 for an area of .1255 (closest value to .1250). Solving for : z =

x



-1.984 = 21 - 

 = 22.98 minutes

6.16 µ = 9.3 Since 22.45% are greater than 18.7, x = 18.7 is in the upper half of the distribution and .2755 (.5000 – .2245) lie between x and the mean. Table A.5 yields a z = 0.76 for an area of .2755. Solving for : z =

0.76 =

x

 18.7  9.3



 = 12.4

6.17 a) P(x < 16 n = 30 and p = .70) µ = np = 30(.70) = 21

 =

n  p  q  30(.70)(.30) = 2.51 µ ± 3 = 21 ± 3(2.51) = 21 ± 7.53 (13.47 to 28.53) does lie between 0 and 30. Solutions Manual 1-187 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

P(x < 16.5µ = 21 and  = 2.51) b) P(10 < x < 20 n = 25 and p = .50) µ = np = 25(.50) = 12.5

 =

n  p  q  25(.50)(.50) = 2.5 µ ± 3 = 12.5 ± 3(2.5) = 12.5 ± 7.5 (5 to 20) does lie between 0 and 25. P(10.5 < x < 20.5µ = 12.5 and  = 2.5)

c) P(x = 22 n = 40 and p = .60) µ = np = 40(.60) = 24

 =

n  p  q  40(.60)(.40) = 3.10 µ ± 3 = 24 ± 3(3.10) = 24 ± 9.30 (14.70 to 33.30) does lie between 0 and 40. P(21.5 < x < 22.5µ = 24 and  = 3.10)

d) P(x > 14 n = 16 and p = .45) µ = np = 16(.45) = 7.2

 =

n  p  q  16(.45)(.55) = 1.99 µ ± 3 = 7.2 ± 3(1.99) = 7.2 ± 5.97 (1.23 to 13.17) does lie between 0 and 16. P(x > 14.5µ = 7.2 and  = 1.99)

6.18 a) n = 8 and p = .50

 =

µ = np = 8(.50) = 4

n  p  q  8(.50)(.50) = 1.414

µ ± 3 = 4 ± 3(1.414) = 4 ± 4.242 Solutions Manual 1-188 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

(–0.242 to 8.242) does not lie between 0 and 8. Do not use the normal distribution to approximate this problem. b) n = 18 and p = .80

 =

µ = np = 18(.80) = 14.4

n  p  q  18(.80)(.20) = 1.697

µ ± 3 = 14.4 ± 3(1.697) = 14.4 ± 5.091 (9.309 to 19.491) does not lie between 0 and 18. Do not use the normal distribution to approximate this problem. c) n = 12 and p = .30 µ = np = 12(.30) = 3.6

 =

n  p  q  12(.30)(.70) = 1.587

µ ± 3 = 3.6 ± 3(1.587) = 3.6 ± 4.761 (–1.161 to 8.361) does not lie between 0 and 12. Do not use the normal distribution to approximate this problem. d) n = 30 and p = .75 µ = np = 30(.75) = 22.5  = n  p  q  30(.75)(.25) = 2.37 µ ± 3 = 22.5 ± 3(2.37) = 22.5 ± 7.11 (15.39 to 29.61) does lie between 0 and 30. The problem can be approximated by the normal curve. e) n = 14 and p = .50

 =

µ = np = 14(.50) = 7

n  p  q  14(.50)(.50) = 1.87

µ ± 3 = 7 ± 3(1.87) = 7 ± 5.61 (1.39 to 12.61) does lie between 0 and 14. The problem can be approximated by the normal curve.

6.19 a) P(x = 8n = 25 and p = .40)

µ = np = 25(.40) = 10

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

 =

n  p  q  25(.40)(.60) = 2.449

µ ± 3 = 10 ± 3(2.449) = 10 ± 7.347 (2.653 to 17.347) lies between 0 and 25. Approximation by the normal curve is sufficient. P(7.5 < x < 8.5µ = 10 and  = 2.449): z =

7.5  10 = -1.02 2.449

From Table A.5, area = .3461 z =

8.5  10 = -0.61 2.449

From Table A.5, area = .2291 P(7.5 < x < 8.5) = .3461 - .2291 = .1170 From Table A.2 (binomial tables) = .120

b) P(x > 13n = 20 and p = .60)

 =

µ = np = 20(.60) = 12

n  p  q  20(.60)(.40) = 2.19

µ ± 3 = 12 ± 3(2.19) = 12 ± 6.57 (5.43 to 18.57) lies between 0 and 20. Approximation by the normal curve is sufficient. P(x > 12.5µ = 12 and  = 2.19): z =

x





12.5  12 = 0.23 2.19

From Table A.5, area = .0910 P(x > 12.5) = .5000 -.0910 = .4090

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

From Table A.2 (binomial tables) = .415 c) P(x = 7n = 15 and p = .50) µ = np = 15(.50) = 7.5

 =

n  p  q  15(.50)(.50) = 1.9365

µ ± 3 = 7.5 ± 3(1.9365) = 7.5 ± 5.81 (1.69 to 13.31) lies between 0 and 15. Approximation by the normal curve is sufficient. P(6.5 < x < 7.5µ = 7.5 and  = 1.9365): z =

x





6.5  7.5 = -0.52 1.9365

From Table A.5, area = .1985 From Table A.2 (binomial tables) = .196

d) P(x < 3n = 10 and p =.70):

 =

µ = np = 10(.70) = 7

n  p  q  10(.70)(.30)

µ ± 3 = 7 ± 3(1.449) = 7 ± 4.347 (2.653 to 11.347) does not lie between 0 and 10. The normal curve is not a good approximation to this problem.

6.20

P(x < 40 n = 120 and p = .37):

 =

µ = np = 120(.37) = 44.4

n  p  q  120(.37)(.63) = 5.29

µ + 3 = 28.53 to 60.27 does lie between 0 and 120.

It is okay to use the normal distribution to approximate thi Correcting for continuity: x = 39.5 Solutions Manual 1-191 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

z =

39.5  44.4 = –0.93 5.29

from Table A.5, the area of z = 0.93 is .3238 P(x < 39.5) = P(z < –0.93) =.5000 – .3238 = .1762

6.21 P(x < 35 n = 70 and p = .59):

µ = n(p) = 70(.59) = 41.3  = n  p  q  70(.59)(.41) = 4.115

µ + 3 = 41.3 + 3(4.115) 28.955 to 53.645 , does lie between 0 and 70. Correcting for continuity: x = 34.5 z =

34.5  41.3 = –1.65 4.115

from table A.5, area of z = 1.65 is .4505 P(x < 34.5) = P(z < –1.65) =.5000 – .4505 = .0495

6.22 For parts a) and b), n = 300

p = .53

µ = 300(.53) = 159

 =

n  p  q  300(.53)(.47) = 8.645

Test: µ + 3 = 159 + 3(8.645) = 133.065 to 184.935 Solutions Manual 1-192 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

which lies between 0 and 300. It is okay to use the normal distribution as an approximation on parts a) and b). a) P(x > 175 transmission) correcting for continuity: x = 175.5 z =

175.5  159 = 1.91 8.645

from A.5, the area for z = 1.91 is .4719 P(x > 175.5) = P(z > 1.91) =.5000 – .4719 = .0281

b) P(165 < x < 170 transmission) correcting for continuity: x = 164.5; x = 170.5 z =

170.5  159 164.5  159 = 1.33 and z = = 0.64 8.645 8.645

from A.5, the area for z = 1.33 is .4082 the area for z = 0.64 is .2389 P(164.5 < x < 170.5) = P(0.64 < z < 1.33) =.4082 – .2389 = .1693

For parts c) and d):

n = 300

 = 300(.60) = 180

p = .60

 =

n  p  q  300(.60)(.40) = 8.485

Test: µ + 3 = 180 + 3(8.485) = 180 + 25.455 154.545 to 205.455 lies between 0 and 300 It is okay to use the normal distribution to approximate

c) P(155 < x < 170 personnel): correcting for continuity: x = 154.5; x = 170.5

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

z =

170.5  180 154.5  180 = -1.12 and z = = -3.01 8.485 8.485

from A.5, the area for z = -1.12 is .3686 the area for z = -3.01 is .4987 P(154.5 < x < 170.5) = P(-3.01 < z < -1.12) = .4987 – .3686 = .1301

d) P(x < 200 personnel): correcting for continuity: x = 199.5 z =

199.5  180 = 2.30 8.485

from A.5, the area for z = 2.30 is .4893 P(x < 199.5) = P(z < 2.30) = .5000 + .4893 = .9893

6.23

p = .23 n = 130 µ = n(p) = 130(.23) = 29.9

 =

= 4.7982

µ ± 3 = 29.9 ± 3(4.7982) = 29.9 ± 14.3946 (15.5054 to 44.2946) lies between 0 and 130. Approximation by the normal curve is sufficient.

a) P(x > 35): z =

Correct for continuity: x = 35.5 = 1.17

from table A.5, area = .2910 P(x > 35.5) = P(z > 1.17) =.5000 – .3790 = .1210 b) P(24 < x < 34):

Correct for continuity: 23.5 to 34.5

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

z =

= -1.33 and z =

= 0.96

from table A.5, area for z = 0.96 is .3315 area for z = 1.33 is .4082 P(25.5 < x < 35.5) = P(-1.42 < z < 0.61) =.4082 + .3315 = .7397

c) P(x < 19):

correct for continuity:

x = 18.5

= –2.38

z =

from table A.5, area for z = 2.38 is .4913 P(x < 19.5) = P(z < –2.63) = .5000 – .4913 = .0087

d) P(x = 29): z =

correct for continuity: = – 0.29 and

z =

28.5 to 29.5

= – 0.08

from table A.5, area for 0.9 = .1141 area for 0.08 = .0319 P(29.5 < x < 30.5) = P(– 0.61 < x < – 0.40) = 0.1141 – .0319 = .0822

6.24 n = 95 a) P(44 < x < 52) agree with direct investments, p = .52 By the normal distribution: µ = n(p) = 95(.52) = 49.4

 =

n  p  q  95(.52)(.48) = 4.87

test: µ + 3 = 49.4 + 3(4.87) = 49.4 + 14.61 0 < 34.79 to 64.01 < 95 test passed Correct for continuity: 43.5 to 52.5 Solutions Manual 1-195 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

z =

43.5  49.4 = -1.21 4.87

from table A.5, area = .3869 z =

52.5  49.4 = 0.64 4.87

from table A.5, area = .2389 P(44 < x < 52) = .3869 + .2389 = .6258

b) P(x > 56): correcting for continuity, x = 56.5 z =

56.5  49.4 = 1.46 4.87

from table A.5, area = .4279 P(x > 56) = .5000 - .4279 = .0721

c) Joint Venture: p = .70, n = 95 By the normal distribution: µ = n(p) = 95(.70) = 66.5

 =

n  p  q  95(.70)(.30) = 4.47

test: 66.5 + 3(4.47) = 66.5 + 13.41 0 < 53.09 to 79.91 < 95

test passed

P(x < 60): correcting for continuity: x = 59.5 z =

59.5  66.5 = -1.57 4.47

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

from table A.5, area = .4418 P(x < 60) = .5000 – .4418 = .0582

d) P(55 < x < 62): correcting for continuity: 54.5 to 62.5 z =

54.5  66.5 = –2.68 4.47

from table A.5, area = .4963 z =

62.5  66.5 = –0.89 4.47

from table A.5, area = .3133 P(55 < x < 62) = .4963 – .3133 = .1830

6.25

a)  = 0.1 x 0 1 2 3 4 5 6 7 8 9 10

y .1000 .0905 .0819 .0741 .0670 .0607 .0549 .0497 .0449 .0407 .0368

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

b)  = 0.3 x 0 1 2 3 4 5 6 7 8 9

y .3000 .2222 .1646 .1220 .0904 .0669 .0496 .0367 .0272 .0202

c)  = 0.8 x 0 1 2

y .8000 .3595 .1615

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

3 4 5 6 7 8 9

.0726 .0326 .0147 .0066 .0030 .0013 .0006

d)  = 3.0 x 0 1 2 3 4 5

y 3.0000 .1494 .0074 .0004 .0000 .0000

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

6.26 a)  = 3.25

µ=



 =





1 = 0.31 3.25



1 = 0.31 3.25

b)  = 0.7

µ =



 =





1 = 1.43 .7



1 = 1.43 .7

c)  = 1.1

µ =



 =





1 = 0.91 1.1



1 = 0.91 1.1

d)  = 6.0

 =  =

 1





1 = 0.17 6



1 = 0.17 6

6.27 a) P(x > 5  = 1.35) = for x0 = 5:

P(x) = e-x = e-1.35(5) = e-6.75 = .0012

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

b) P(x < 3  = 0.68) = 1 – P(x > 3  = .68) = for x0 = 3:

1 – e-x = 1 – e-0.68(3) = 1 – e –2.04 = 1 – .1300 = .8700

c) P(x > 4  = 1.7) for x0 = 4:

P(x) = e-x = e-1.7(4) = e-6.8 = .0011

d) P(x < 6  = 0.80) = 1 – P(x > 6  = 0.80) for x0 = 6:

P(x) = e-x = e-0.80(6) = e-4.8 = .0082

P(x < 6  = 0.80) = 1 – .0082 = .9918

6.28 µ = 23 sec.  =

1 = .0435 per second 

a) P(x > 1 min = .0435/sec.) Change  to minutes:

 = .0435(60) = 2.61 per min.

P(x > 1 min = 2.61/min) = for x0 = 1:

P(x) = e-x = e-2.61(1) = .0735

b)  = .0435/sec Change  to minutes:

 = (.0435)(60) = 2.61 min

P(x > 3  = 2.61/min) = for x0 = 3:

P(x) = e-x = e-2.61(3) = e-7.83 = .0004

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

6.29  = 2.44/min. a) P(x > 10 min  = 2.44/min) = Let x0 = 10,

e-x = e-2.44(10) = e-24.4 = .0000

b) P(x > 5 min  = 2.44/min) = Let x0 = 5,

e-x = e-2.44(5) = e-12.20 = .0000

c) P(x > 1 min  = 2.44/min) = Let x0 = 1,

e-x = e-2.44(1) = e-2.44 = .0872

d) Expected time = µ =





1 min. = .41 min = 24.6 sec. 2.44

6.30  = 1.12 planes/hr. a) µ =





1 = .89 hr. = 53.4 min. 1.12

b) P(x > 2 hrs  = 1.12 planes/hr.) = Let x0 = 2,

e-x = e-1.12(2) = e-2.24 = .1065

c) P(x < 10 min  = 1.12/hr.) = 1 – P(x > 10 min = 1.12/hr.) Change  to 1.12/60 min. = .01867/min. 1 - P(x > 10 min  = .01867/min) = Let x0 = 10,

1 – e-x = 1 – e-.01867(10) = 1 – e-.1867 = 1 – .8297 =

.1703

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

6.31  = 8.83/ 1,000 passengers Average number: µ =





1 = 0.11325 or (0.11325)(1,000) = about 8.83

113 Passengers

P(x > 500): Let x0 = 500/1,000 passengers = .5 e-x = e-8.83(.5) = e-4.415 = .01209 P(x < 200) =1 – P(x > 200) Let x0 = 200/1,000 passengers = .2 e-x = e-8.83(.2) = e-1.766 = .1710 P(x < 200) = 1 – .1710 = .8290

6.32 µ = 20 years

 = x0 1 2 3

1 = .05/year 20 Prob(x > x0)=e-x .9512 .9048 .8607

If the foundation is guaranteed for 2 years, based on past history, 90.48% of the foundations will last at least 2 years without major repair and only 9.52% will require a major repair before 2 years.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

6.33

 = 2/month Average number of time between rain = µ =

 = µ = 15 days





1 month = 15 days 2

P(x < 2 days  = 2/month):

2 = .067/day 30 P(x < 2 days  = .067/day) = Change  to days:

 =

1 – P(x > 2 days  = .067/day) 1 – e-x = 1 – e-.067(2) = 1 – .8746 = .1254

let x0 = 2,

6.34 a = 6

b = 14

1 1 1   = .125 b  a 14  6 8 6  x  14 0.125, f x    for all other valu es  0,

 =

a  b 6  14  = 10 2 2

 =

b  a 14  6 8   = 2.309 12 12 12

P(x > 11) =

14  11 3  = .375 14  6 8

P(7 < x < 12) =

12  7 5  = .625 14  6 8

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

6.35 a) P(x < 21 µ = 25 and  = 4): z =

x



21 25 = –1.00 4



From Table A.5, area = .3413 P(x < 21) = .5000 –.3413 = .1587 b) P(x > 77 µ = 50 and  = 9): z =

x





77  50 = 3.00 9

From Table A.5, area = .4987 P(x > 77) = .5000 –.4987 = .0013 c) P(x > 47 µ = 50 and  = 6): z =

x





47  50 = –0.50 6

From Table A.5, area = .1915 P(x > 47) = .5000 + .1915 = .6915 d) P(13 < x < 29 µ = 23 and  = 4): z =

x





13  23 = –2.50 4

From Table A.5, area = .4938 z =

x





29  23 = 1.50 4

From Table A.5, area = .4332 P(13 < x < 29) = .4938 + 4332 = .9270 Solutions Manual 1-205 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

e) P(x > 105 µ = 90 and  = 2.86): z =

x





105  90 = 5.24 2.86

From Table A.5, area = .5000 P(x > 105) = .5000 – .5000 = .0000

6.36 a) P(x = 12 n = 25 and p = .60): µ = np = 25(.60) = 15

=

n  p  q  25(.60)(.40) = 2.45

µ ± 3 = 15 ± 3(2.45) = 15 ± 7.35 (7.65 to 22.35) lies between 0 and 25. The normal curve approximation is sufficient. P(11.5 < x < 12.5 µ = 15 and  = 2.45): z=

x

 x





11.5  15 = –1.43 2.45

From Table A.5, area = .4236



12.5  15 = –1.02 2.45

From Table A.5, area = .3461

P(11.5 < x < 12.5) = .4236 – .3461 = .0775 From Table A.2, P(x = 12) = .076

b) P(x > 5 n = 15 and p = .50): µ = np = 15(.50) = 7.5

=

n  p  q  15(.50)(.50) = 1.94

µ ± 3 = 7.5 ± 3(1.94) = 7.5 ± 5.82

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

(1.68 to 13.32) lies between 0 and 15. The normal curve approximation is sufficient. P(x > 5.5µ = 7.5 and = l.94) z =

5.5  7.5 = –1.03 1.94

From Table A.5, area = .3485 P(x > 5.5) = .5000 + .3485 = .8485 Using table A.2, P(x > 5) = .849

c) P(x < 3n = 10 and p = .50): µ = np = 10(.50) = 5

=

n  p  q  10(.50)(.50) = 1.58

µ ± 3 = 5 ± 3(1.58) = 5 ± 4.74 (0.26 to 9.74) lies between 0 and 10. The normal curve approximation is sufficient. P(x < 3.5µ = 5 and  = l.58): z =

3.5  5 = –0.95 1.58

From Table A.5, area = .3289 P(x < 3.5) = .5000 – .3289 = .1711 Using table A.2, P(x < 3) = .172

d) P(x > 8n = 15 and p = .40): µ = np = 15(.40) = 6

=

n  p  q  15(.40)(.60) = 1.90

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

(0.3 to 11.7) lies between 0 and 15. The normal curve approximation is sufficient. P(x > 7.5µ = 6 and  = l.9): z =

7.5  6 = 0.79 1.9

From Table A.5, area = .2852 P(x > 7.5) = .5000 – .2852 = .2148 Using table A.2, P(x > 8) = .212

6.37 a) P(x > 3  = 1.3): let x0 = 3 P(x > 3  = 1.3) = e-x = e-1.3(3) = e-3.9 = .0202 b) P(x < 2  = 2.0): Let x0 = 2 P(x < 2  = 2.0) = 1 – P(x > 2  = 2.0) = 1 – e-x = 1 – e-2(2) = 1 – e-4 = 1 - .0183 = .9817 c) P(1 < x < 3  = 1.65): P(x > 1  = 1.65): Let x0 = 1 e-x = e-1.65(1) = e-1.65 = .1920 P(x > 3 = 1.65): Let x0 = 3 Solutions Manual 1-208 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

e-x = e-1.65(3) = e-4.95 = .0071 P(1 < x < 3) = P(x > 1) - P(x > 3) = .1920 - .0071 = .1849

d) P(x > 2 = 0.405): Let x0 = 2 e-x = e-(.405)(2) = e-.81 = .4449

6.38 µ = 42.2 12% more than 48.

x = 48

Area between x and µ is .50 – .12 = .38 z associated with an area of .3800 is z = 1.175 Solving for : z =

1.175 =

 =

x

 48  42.2



5.8 = 4.936 1.175

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

6.39

p = 1/4 = .25

n = 150

P(x > 50): µ = 150(.25) = 37.5

 = 150(.25)(.75) = 5.303  ± 3 = 37.5 ± 3(5.303) = 37.5 ± 15.909 (21.591 to 53.409) lies between 0 and 150. The normal curve approximation is sufficient. P(x > 50.5µ = 37.5 and  = 5.303): z =

50.5  37.5 = 2.45 5.303

Area associated with z = 2.45 is .4929 P(x > 50) = .5000 – .4929 = .0071

6.40  = 1 customer/20 minutes µ = 1/  = 1 a) 1 hour interval x0 = 3 because 1 hour = 3(20 minute intervals) P(x > 3) = e-x = e-1(3) = e-3 = .0498

b) 10 to 30 minutes x0 = .5, x0 = 1.5 P(x > .5) = e-x = e-1(.5) = e-.5 = .6065 P(x > 1.5) = e-x = e-1(1.5) = e-1.5 = .2231 P(10 to 30 minutes) = .6065 - .2231 = .3834

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

c) less than 5 minutes x0 = 5/20 = .25 P(x > .25) = e-x = e-1(.25) = e-.25 = .7788 P(x < .25) = 1 – .7788 = .2212

6.41

µ = 237

 = 54

a) P(x < 150):

from Table A.5, area for z = -1.61 is .4463 P(x < 150) = P(z < -1.61) =.5000 - .4463 = .0537

b) P(x > 400):

from Table A.5, area for z = 3.02 is .4987 P(x > 400) = P(z > 3.02) = .5000 - .4987 = .0013

c) P(120 < x < 185):

from Table A.5, area for z = 2.17 is .4850 area for z = 0.96 is .3315

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

P(120 < x < 185) = P(-2.17 < z < -0.96) =.4850 - .3315 = .1535

6.42

 = 83 Since only 3% = .0300 of the values are greater than 2,655(million), x =

2,655 lies in the upper tail of the distribution. .5000 - .0300 = .4700 of the values lie between 2,655 and the mean. Table A.5 yields a z = 1.88 for an area of .4700. Using z = 1.88, x = 2,655,  = 83, µ can be solved for. z =

1.88 =

x

 2,655   83

 = 2,498.96 million

6.43 a = 18

b = 65

P(25 < x < 50) =

 =

50  25 25  = .5319 65  18 47

a  b 65  18  = 41.5 2 2

Height =

1 1 1   = .0213 b  a 65  18 47

6.44  = 1.8 per 15 seconds

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

a)  =





1 = .5556(15 sec.) = 8.33 sec. 1.8

b) For x > 25 sec. use x0 = 25/15 = 1.67 P(x > 25 sec.) = e-1.6667(1.8) = .0498

c) For x < 5 sec. use x0 = 5/15 = 1/3 P(x < 5 sec.) = 1 – P(x > 5 sec.) =1 – e-(1/3)(1.8) = 1 – .5488 = .4512 d) P(x > 1 min.): x0 = 1 min. = 60/15 = 4 P(x > 1 min.) = e-4(1.8) = .0007

6.45

µ = 833

 = 84

a) P(x > 900): z =

x





900  833 = 0.80 84

from Table A.5, the area for z = 0.80 is .2881 P(x > 900) = .5000 – .2881 = .2119

b) P(800 < x < 1,000): z =

z =

x

 x





800  833 = –0.39 84



1,000  833 = 1.99 84

from Table A.5, the area for z = 0.39 is .1517 the area for z = 1.99 is .4767

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

P(800 < x < 1,000) = .1517 + .4767 = .6284

c) P(725 < x < 825): z =

z =

x

 x





725  833 = –1.29 84



825  833 = –0.10 84

from Table A.5, the area for z = 1.29 is .4015 the area for z = 0.10 is .0398 P(725 < x < 825) = .4015 – .0398 = .3617

d) P(x < 600): z =

x





600  833 = –2.77 84

from Table A.5, the area for z = 2.77 is .4972 P(x < 600) = .5000 – .4972 = .0028

6.46 n = 60

p = .24

µ = np = 60(.24) = 14.4

=

n  p  q  60(.24)(.76) = 3.308

test: µ + 3 = 14.4 + 3(3.308) = 14.4 + 9.924 = 4.476 , 24.324 Since 4.476 to 24.324 lies between 0 and 60, the normal distribution can be used to approximate this problem.

P(x > 17):

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

correcting for continuity: x = 16.5 z =

x





16.5  14.4 = 0.63 3.308

from Table A.5, the area for z = 0.63 is .2357 P(x > 17) = .5000 – .2357 = .2643

P(x > 22): correcting for continuity: x = 22.5 z =

x





22.5  14.4 = 2.45 3.308

from Table A.5, the area for z = 2.45 is .4929 P(x > 22) = .5000 – .4929 = .0071

P(8 < x < 12): correcting for continuity: x = 7.5 and x = 12.5 z =

x





12.5  14.4 = –0.57 3.308

z =

x





7.5  14.4 = –2.09 3.308

from Table A.5, the area for z = 0.57 is .2157 the area for z = 2.09 is .4817 P(8 < x < 12) = .4817 – .2157 = .2660

6.47

µ = 81,832

 = 7,712

a) P(x > 90,700): z =

= 1.15

from Table A.5, the area for z = 1.15 is .3749 Solutions Manual 1-215 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

P(x > 90,700) = P(z > 0.75) = .5000 – .3749 = .1251

b) P(x < 72,500): z =

= -1.21

from Table A.5, the area for z = 1.21 is .3869 P(x < 72,500) = P(z < –1.61) = .5000 – .3869 = .1131 c) P(x > 63,500): z =

= -2.38

from Table A.5, the area for z = 2.38 is .4913 P(x > 63,500) = P(z > –2.77) = .5000 + .4913 = .9913

d) P(70,800 < x < 85,600): z =

= -1.43

z =

= 0.49

from Table A.5, the area for z = 1.43 is .4236 the area for z = 0.09 is .1879 P(70,800 < x < 85,600) = P(–1.83 < x < 0.09) = .4236 + .1879 = .6115

6.48

µ = 9 minutes  = 1/µ = 1/9 = .1111/minute = .1111(60)/hour  = 6.67/hour P(x < 5 minutes  = .1111/minute) =

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

1 – P(x > 5 minutes  =.1111/minute): Let x0 = 5 P(x > 5 minutes  = .1111/minute) = e-x = e-.1111(5) = e-.5555 = .5738 P(x < 5 minutes) = 1 - P(x > 5 minutes) = 1 – .5738 = .4262 6.49  = 142

 = 10.3

a) P(x < 110): z =

x





110  142 = –3.11 10.3

From Table A.5, area = .4991 P(x < 110) = .5000 – .4991 = .0009

b) P(x > 130): z =

x





130  142 = –1.17 10.3

From Table A.5, area = .3790 P(x > 80) = .5000 + .3790 = .8790

c) P(145 < x < 160): z =

x





145  142 = 0.29 10.3

From Table A.5, area = .1141 z =

x





160  142 = 1.75 10.3

From Table A.5, area = .0.4599 P(145 < x < 160) = .4599 – .1141 = .3458 Solutions Manual 1-217 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

6.50 n = 200, p = .81 expected number = µ = n(p) = 200(.81) = 162 µ = 162

 =

n  p  q  200(.81)(.19) = 5.548

µ + 3 = 162 + 3(5.548) lie between 0 and 200, the normalcy test is passed

P(150 < x < 155): correction for continuity: 150.5 to 154.5 z =

150.5  162 = –2.07 5.548

from table A.5, area = .4808 z =

154.5  162 = –1.35 5.548

from table A.5, area = .4115 P(150.5 < x < 154.5) = P(–2.07 < x < –1.35) =.4808 – .4115 = .0693

P(x > 158): correcting for continuity, x = 158.5 z =

158.5  162 = –0.63 5.548

from table A.5, area = .2357 P(x > 158.5) = P(z > –0.63) =.2357 + .5000 = .7357

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

P(x < 144): correcting for continuity, x = 143.5 z =

143.5  162 = –3.33 5.548

from table A.5, area = .4996 P(x < 143.5) = P(z < –3.33) =.5000 – .4996 = .0004 6.51 n = 150 p = .71 µ = np = 150(.71) = 106.5

 =

n  p  q  150(.71)(.29) = 5.5574 µ + 3 = 106.5 + 3(5.5574) lie between 0 and 150, the normalcy test is passed a) P(x < 105): correcting for continuity: x = 104.5 z =

x





104.5  106.5 = –0.36 5.5574

from Table A.5, the area for z = 0.36 is .1406 P(x < 104.5) = P(z < –0.36) =.5000 – .1406 = .3594

b) P(110 < x < 120): correcting for continuity: x = 109.5, x = 120.5 z =

109.5  106.5 = 0.54 5.5574

z =

120.5  106.5 = 2.52 5.5574

from Table A.5, the area for z = 0.54 is .2054 the area for z = 2.52 is .4941 P(109.5 < x < 120.5) = P(0.54 < z < 2.52) = .4941 – .2054 = .2887 Solutions Manual 1-219 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

c) P(x > 95): correcting for continuity: x = 95.5 z =

95.5  106.5 = –1.98 5.5574

from Table A.5, the area for 1.98 is .4761 P(x > 95.5) = P(z > –1.98) =.4761 + .5000 = .9761 ab 6.52  = = 22.45 2 a + b = 2(22.45) = 44.90 b = 44.90 – a Height =

1 = 8.94 ba

1 = 8.94b – 8.94a Substituting b from above, 1 = 8.94(44.90 – a) – 8.94a 1 = 401.406 – 8.94a – 8.94a 1 = 401.406 – 17.88a 17.88a = 400.406 a = 22.39 and

b = 44.90 – 22.39 = 22.51

6.53 µ = 85,200 60% are between 75,600 and 94,800 94,800 –85,200 = 9,600 75,600 – 85,200 = – 9,600

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

The 60% can be split into 30% and 30% because the two x values are equal distance from the mean. The z value most closely associated with .3000 area is 0.84 z =

.84 =

x

 94,800  85,200



 = 11,428.57 6.54

n = 75

p = .81 prices

p = .44 products

a) Expected Value:  = np = 75(.81) = 60.75 seeking price information b) Expected Value:  = np = 75(.44) = 33 seeking information about products offered.

c) P(x > 67 prices)  = np = 75(.81) = 60.75 3.3974

and

 =

n  p  q  75(.81)(.19) =

Tests: µ + 3 = 60.75 + 3(3.3974) = 60.75 + 10.192 50.558 to 70.942 lies between 0 and 75. It is okay to use the normal distribution to approximate this problem. correcting for continuity: x = 66.5 z =

66.5  60.75 = 1.69 3.3974

from Table A.5, the area for z = 1.69 is .4545 P(x > 66.5) = P(z > 1.69) = .5000 – .4545 = .0455

d) P(x < 23 products):  = np = 75(.44) = 33

and

 =

n  p  q  75(.44)(.56) =

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

µ + 3 = 33 + 3(4.2988) = 33 + 12.8964 20.1036 to 45.8964 lies between 0 and 75. It is okay to use the normal distribution to approximate this problem. correcting for continuity: x = 22.5 z =

22.5  33 = –2.44 4.2988

from Table A.5, the area for z = 2.44 is .4927 P(x < 22.5) = P(z < –2.44) =.5000 – .4927 = .0073

6.55  = 3 hurricanes5 months P(x > 1 month  = 3 hurricanes per 5 months): Since x and  are for different intervals, change Lambda =  = 3/ 5 months = 0.6 per month. P(x > 1 month  = 0.6 per month): Let x0 = 1 P(x > 1) = e-x = e-0.6(1) = e-0.6 = .5488

P(x < 2 weeks):

2 weeks = 0.5 month.

P(x < 0.5 month  = 0.6 per month) = 1 – P(x > 0.5 month  = 0.6 per month) But P(x > 0.5 month  = 0.6 per month): Let x0 = 0.5 P(x > 0.5) = e-x = e-0.6(.5) = e-0.30 = .7408 P(x < 0.5 month) = 1 – P(x > 0.5 month) = 1 – .7408 = .2592

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Average time = Expected time = µ = 1/ = 1/0.6 = 1.67 months

6.56 n = 50

p = .80

 = np = 50(.80) = 40  =

n  p  q  50(.80)(.20) = 2.828

Test: µ + 3 = 40 +3(2.828) = 40 + 8.484 31.516 to 48.484 lies between 0 and 50. It is okay to use the normal distribution to approximate this binomial problem. P(x < 35): z =

correcting for continuity: x = 34.5

34.5  40 = –1.94 2.828

from Table A.5, the area for z = 1.94 is .4738 P(x < 35) = .5000 – .4738 = .0262

The expected value = µ = 40

P(42 < x < 47): correction for continuity: x = 41.5 z =

41.5  40 = 0.53 2.828

z =

x = 47.5

47.5  40 = 2.65 2.828

from Table A.5, the area for z = 0.53 is .2019 the area for z = 2.65 is .4960 P(42 < x < 47) = .4960 – .2019 = .2941

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

 = 175

6.57 µ = 2,087

If 20% of the operating costs are less than x, then 30% lie between x and µ. z.30 = –.84 z =

x



–.84 =

x  2,087 175

x = 1940

If 65% of the operating costs are more than x, then 15% lie between x and µ z.15 = –0.39 z =

–.39 =

x

 x  2,087 175

x = 2018.75

If x is more than 85% of the operating costs, then 35% lie between x and µ. z.35 = 1.04 z =

1.04 =

x

 x  2,087 175

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

6.58  = 0.8 personminute P(x > 1 minute  = 0.8 minute): Let x0 = 1 P(x > 1) = e-x = e-.8(1) = e-.8 = .4493 P(x > 2.5 Minutes  = 0.8 per minute): Let x0 = 2.5 P(x > 2.5) = e-x = e-0.8(2.5) = e-2 = .1353

6.59 µ = 302,190

 = 7,308

P(x > 315,000): z =

315,000  302,190 = 1.75 7,308

from table A.5 the area for z = 1.75 is .4599 P(x > 315,000) = P(z > 1.75) = .5000 – .4599 = .0401

P(x < 286,000): z =

286,000  302,190 = –2.22 7,308

from table A.5 the area for z = 2.22 is .4868 P(x < 286,000) = P(z < –2.22) = .5000 – .4868 = .0132

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

6.60  = 2.2 calls30 secs. Expected time between calls = µ = 1/ = 1/(2.2) = .4545(30 sec.) = 13.64 sec. P(x > 1 min.  = 2.2 calls per 30 secs.): Since Lambda and x are for different intervals, Change Lambda to:

 = 4.4 calls/1 min.

P(x > 1 min  = 4.4 calls/1 min.): For x0 = 1:

e-x = e-4.4(1) = .0123

P(x > 2 min.  = 4.4 calls/1 min.): For x0 = 2: e-x = e-4.4(2) = e-8.8 = .0002 6.61 This is a uniform distribution with a = 11 and b = 32. The mean is (11 + 32)/2 = 21.5 and the standard deviation is (32 - 11)/ 12 = 6.06. Almost 81% of the time there are less than or equal to 28 sales associates working. One hundred percent of the time there are less than or equal to 34 sales associates working and never more than 34. About 23.8% of the time there are 16 or fewer sales associates working. There are 21 or fewer sales associates working about 48% of the time.

6.62 The weight of the rods is normally distributed with a mean of 227 mg and a standard deviation of 2.3 mg. The probability that a rod weighs less than or equal to 220 mg is .0012, less than or equal to 225 mg is .1923, less than or equal to 227 is .5000 (since 227 is the mean), less than 231 mg is .9590, and less than or equal to 238 mg is 1.000. Solutions Manual 1-226 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

6.63 The lengths of cell phone calls are normally distributed with a mean of 2.35 minutes and a standard deviation of .11 minutes. Almost 99% of the calls are less than or equal to 2.60 minutes, almost 82% are less than or equal to 2.45 minutes, over 32% are less than 2.3 minutes, and almost none are less than 2 minutes.

6.64 The exponential distribution has  = 4.51 per 10 minutes and µ = 1/4.51 = .22173 of 10 minutes or 2.2173 minutes. The probability that there is less than .1 or 1 minute between arrivals is .3630. The probability that there is less than .2 or 2 minutes between arrivals is .5942. The probability that there is less than .5 or 5 minutes between arrivals is .8951. The probability that there is less than 1 or 10 minutes between arrivals is .9890. It is almost certain that there will be less than 2.4 or 24 minutes between arrivals.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Legal Notice Copyright

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

SOLUTIONS TO PROBLEMS IN CHAPTER 7: SAMPLING AND SAMPLING DISTRIBUTIONS

7.1

7.4

i. ii.

A union membership list for the company. A list of all employees of the company.

i. ii.

Residence Telephone Directory for Edmonton, Alberta. Utility company list of all customers.

i. ii.

Airline company list of phone and mail purchasers of tickets from the airline during the past six months. A list of frequent flyer club members for the airline.

i. ii.

List of boat manufacturer's employees. List of members of a boat owners association.

i. ii.

Cable company telephone directory. Membership list of cable management association.

Size of motel (rooms), age of motel, geographic location.

Gender, age, education, social class, ethnicity.

Size of operation (number of bottled drinks per month), number of employees, number of different types of drinks bottled at that location, geographic location.

Size of operation (sq.ft.), geographic location, age of facility, type of process used.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

7.5

Under 21 years of age, 21 to 39 years of age, 40 to 55 years of age, over 55 years of age.

Under $1,000,000 sales per year, $1,000,000 to $4,999,999 sales per year, $5,000,000 to $19,999,999 sales per year, $20,000,000 to $49,999,999 per year, $50,000,000 to $99,999,999 per year, $100,000,000 or over per year.

Less than 2,000 sq. ft., 2,000 to 4,999 sq. ft., 5,000 to 9,999 sq. ft., 10,000 or over sq. ft.

Atlantic Provinces, Central Canada, the Prairies, the West Coast, the North.

Government worker, teacher, lawyer, physician, engineer, business person, police officer, fire fighter, computer worker.

f) Manufacturing, finance, communications, health care, retailing, chemical, transportation.

7.6

n = N/k = 100,000/200 = 500

7.7

N = nk = 7511 = 825

7.8

k = N/n = 3,500/175 = 20 Start between 1 and 20. The human resource department probably has a list of company employees which can be used for the frame. Also, there might be a company phone directory available.

7.9

i. ii.

Municipalities Metropolitan areas

i. ii.

Provinces (beside which the oil wells lie) Companies that own the wells

i. ii.

Provinces Municipalities

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

7.10

Go to the prosecutor's office and observe the apparent activity of various prosecutors at work. Select some who are very busy and some who seem to be less active. Select some men and some women. Select some who appear to be older and some who are younger. Select prosecutors with different ethnic backgrounds.

7.11

Go to a conference where some of the Report on Business Magazine Top 1000 executives attend. Approach those executives who appear to be friendly and approachable.

7.12

Suppose 40% of the sample is to be people who presently own a personal computer and 60% with people who do not. Go to a computer show at the city's conference center and start interviewing people. Suppose you get enough people who own personal computers but not enough interviews with those who do not. Go to a mall and start interviewing people. Screen out personal computer owners. Interview non personal computer owners until you meet the 60% quota.

7.13

µ = 50,

 = 10,

n = 64

a) P( > 52): z =

= 1.6

from Table A.5, Prob. = .4452 P(

> 52) = .5000 – .4452 = .0548

b) P( < 51): z =

= 0.80

from Table A.5 prob. = .2881

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

P( < 51) = .5000 + .2881 = .7881

c) P( < 47): = –2.40

z =

from Table A.5 prob. = .4918 P( < 47) = .5000 – .4918 =

d) P(48.5 < z =

.0082

< 52.4): = –1.20

from Table A.5 prob. = .3849 z =

= 1.92

from Table A.5 prob. = .4726 P(48.5 <

< 52.4) = .3849 + .4726 = .8575

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

e) P(50.6 <

< 51.3):

z =

= 0.48

from Table A.5, prob. = .1844 z =

from Table A.5, prob. = .3508 P(50.6 <

7.14

< 51.3) = .3508 – .1844 = .1644

µ = 23.45 a) n = 10, P(

 = 3.8 > 22): = –1.21

z =

from Table A.5, prob. = .3869 P(

> 22) = .3869 + .5000 = .8869

b) n = 4, P( z =

> 26): = 1.34

from Table A.5, prob. = .4099 P(

> 26) = .5000 – .4099 =

.0901

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

7.15

n = 36

µ = 278

P( < 280) = .86

.3600 of the area lies between = 280 and µ = 278. This probability is associated with z = 1.08 from Table A.5. Solving for  : z =

1.08 =

7.16

1.08

= 2

=

= 11.11

n = 81

 = 12

> 300) = .18

.5000 – .1800 = .3200 and from Table A.5, z.3200 = 0.92 Solving for µ: z =

0.92 =

0.92

= 300 – 

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

µ = 300 – 1.2267 = 298.77

7.17 a) N = 1,000 P(

n = 60

µ = 75

=6

< 76.5):

z =

= 2.00

from Table A.5, prob. = .4772 P( < 76.5) = .4772 + .5000 = .9772 b) N = 90

n = 36

P(107 <

< 107.7):

 = 3.46

µ = 108

= –2.23

z =

from Table A.5, prob. = .4871 = –0.67

z =

from Table A.5, prob. = .2486 P(107 <

< 107.7) = .4871 – .2486 =

c) N = 250 P(

n = 100

µ = 35.6

.2385

 = 4.89

> 36):

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

z =

= 1.05

from Table A.5, prob. = .3531 > 36) = .5000 – .3531 = .1469

d) N = 5,000 P(

n = 60

µ = 125

 = 13.4

< 123): = –1.16

z =

from Table A.5, prob. = .3770 < 123) = .5000 – .3770 = .1230

7.18 µ = 77.8 a) P(

 = 20

n = 38

< 76):

z = z = ( -µ)/( /√n) = (76-77.8)/(20/√38) = -0.55 from table A.5, area = 0.2088 P(

< 36) = .5000 – .2088 = .2912

b) P(77 <

< 82):

z = ( -µ)/( /√n) = (82-77.8)/(20/√38) = 1.29 from table A.5, area = .4015 z = ( -µ)/( /√n) = (77-77.8)/(20/√38) = -0.25 from table A.5, area = .0987 Solutions Manual 1-236 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

P(77 <

c) P(

< 82) = .4015 + .0987 = .5002

< 85):

z = ( -µ)/( /√n) = (85-77.8)/(20/√38) = 2.22 from table A.5, area = .4868 P( < 85) = .5000 + .4868 = .9868

d) P(75 <

< 76):

z = ( -µ)/( /√n) = (75-77.8)/(20/√38) = -0.86 from table A.5, area = .3051

z = ( -µ)/( /√n) = (76-77.8)/(20/√38) = -0.55 from table A.5, area = .2088 P(37 <

7.19

< 38) = .3051 – .2088 = .0963

N = 1,500 P(

n = 100

µ = 277,000

 = 10,500

> $285,000):

z = ( -µ)/( /√n)( √N-n/N-1) = (285,000 – 277,000)/(10,500/√100)( √(1500-100)/(15001)) = 7.88 from Table A.5, prob. = .5000 P(

> $285,000) = .5000 – .5000 = .0000

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

7.20

 = $21.45

µ = $65.12

n = 45

) = .2300

P( lies between and µ) is .5000 – .2300 = .2700 from Table A.5, z.2700 = 0.74 Solving for

z =

0.74 =

2.366 =

7.21

– 65.12

µ = 27  = 6.7 a) P(

and

= 65.12 + 2.366 = $67.49

n = 42

> 29):

z = ( -µ)/( /√n) = (29 - 27)/(6.7/√42) = 1.93 from Table A.5, the area for z = 1.93 is .4732 P( > 27) = .5000 – .4732 = .0268 b) P(

< 23):

z = ( -µ)/( /√n) = (23 - 27)/(6.7/√42) = -3.87 from Table A.5, the area for z = 3.87 is .49997 P( c) P(

< 23) = .5000 – .49997 = .0.00003 < 20):

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

z = ( -µ)/( /√n) = (20 - 27)/(6.7/√42) = -6.77 from Table A.5, the area for z = 6.77 is .5000 P(

< 20) = .5000 – .5000 = .0000

Because the event is almost improbable, if it actually occurred, the researcher would question the given population parameters. d) 71% of the values are greater than 25. Therefore, 21% of all sample means are between 26 and the population mean, µ = 27. The z value associated with the 21% of the area is – 0.55: z.21 = – 0.55 z =

– 0.55 =

 = 11.78

7.22

p = .25 a) n = 110

< .21): = – 0.97

z =

from Table A.5, prob. = .3340 P( < .21) = .5000 – .3340 = .1660 b) n = 33

> .24):

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

= – 0.13

z =

from Table A.5, prob. = .0517 P(

> .24) = .5000 + .0517 = .5517

c) n = 59

P(.24 <

< .27): = – 0.18

z =

from Table A.5, prob. = .0714 z =

= 0.35

from Table A.5, prob. = .1368 P(.24 < d) n = 80

< .27) = .0714 + .1368 = .2082 P( > .30):

z =

= 1.03

from Table A.5, prob. = .3485 P(

> .30) = .5000 – .3485 = .1515

e) n = 800 z =

> .30): = 3.27

from Table A.5, prob. = .4995

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

7.23

> .30) = .5000 – .4995 = .0005

p = .58 a) P(

n = 660 > .60):

z =

= 1.04

from table A.5, area = .3508 P( > .60) = .5000 – .3508 = .1492 b) P(.55 <

< .65):

z =

= 3.64

from table A.5, area = .4998 = – 1.56

z =

from table A.5, area = .4406 P(.55 < c) P(

< .65) = .4998 + .4406 = .9404

> .57): = – 0.52

z =

from table A.5, area = .1985 P(

> .57) = .1985 + .5000 = .6985

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

d) P(.53 <

< .56): = – 1.04

z =

= –

2.60

from table A.5, area for z = 1.04 is .3508 from table A.5, area for z = 2.60 is .4953 < .56) = .4953 – .3508 = .1445

P(.53 < e) P(

< .48): = – 5.21

z =

from table A.5, area = .5000 P(

7.24

< .48) = .5000 – .5000 = .0000

p = .40

> .35) = .8000

P(.35 <

< .40) = .8000 – .5000 = .3000

from Table A.5, z.3000 = – 0.84 Solving for n: z =

– 0.84 =

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

8.23 = n = 67.73  68

7.25

p = .28 P(

n = 140

) = .3000

< .28) = .5000 – .3000 = .2000

from Table A.5, z.2000 = – 0.52 Solving for

: z =

– 0.52 =

– .02 =

– .28

= .28 – .02 = .26

7.26

P(x > 150): n = 600 =

p = .40

x = 150

= .25

z =

= – 7.5

from table A.5, area = .5000

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

P(x > 150) = P(

7.27

p = .36

> .25) = P(z > – 7.5) = .5000 +.5000 = 1.0000

n = 200

a) P(x < 90): =

= .45

z =

= 2.65

from Table A.5, the area for z = 2.65 is .4960 P(x < 90) = P(

< .45) = P( z < 2.65) = .5000 + .4960 = .9960

b) P(x > 100): =

= .50

z =

= 4.12

from Table A.5, the area for z = 4.12 is .49997 P(x > 100) = P(

> .50) = P( z > 4.12) =.50000 – .49997 = .00003

c) P(x > 80): =

z =

= .40

= 1.18

from Table A.5, the area for z = 1.18 is .3810 Solutions Manual 1-244 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

P(x > 80) = P(

7.28

p = .19 a) P(

< .40) = P( z > 1.18) =.5000 – .3810 = .1190

n = 950

> .25):

z =

= 4.71

from Table A.5, area = .5000 P(

> .25) = .5000 – .5000 = .0000

b) P(.15 <

< .20):

z =

= – 3.14

z =

= 0.79

from Table A.5, area for z = 3.14 is .4992 from Table A.5, area for z = 0.79 is .2852 P(.15 <

< .20) = .4992 + .2852 = .7844

c) P(133 < x < 171): =

= .14

P(.14 <

< .18):

= .18

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

= – 3.93

z =

= –

z =

0.79 from Table A.5, the area for z = 3.93 is .49997 the area for z = 0.79 is .2852 P(133 < x < 171) = P(.14 < = .49997 – .2852 = .21477

7.29

< .18) = P(– 3.93 < z < – 0.79) =

µ = 76,  = 14 Because the sample size in a) - c) is greater than 30, the central limit theorem

can be used. a) n = 35,

z =

> 79): = 1.27

from table A.5, area = .3980 P(

> 79) = .5000 – .3980 = .1020

b) n = 140, P(74 <

< 77): = – 1.69

z =

= 0.85

from table A.5, area for z = 1.69 is .4545 from table A.5, area for z = 0.85 is .3023 P(74 <

c) n = 219,

< 77) = .4545 + .3023 = .7568

< 76.5):

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

z =

= 0.53

from table A.5, area = .2019

7.30

P( < 76.5) = .5000 + .2019 = .7019 p = .46 a) n = 60 P(.41 <

< .53):

z =

= 1.09

from table A.5, area = .3621 = – 0.78

z =

from table A.5, area = .2823 P(.41 < b) n = 458 z =

< .53) = .3621 + .2823 = .6444 P( < .40):

p  p  pq n

.40 .46 = – 2.58 (.46)(.54) 458

from table A.5, area = .4951 P( p < .40) = .5000 – .4951 = .0049 c) n = 1350 z =

P( > .49): = 2.21

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

from table A.5, area = .4864 P( > .49) = .5000 – .4864 = .0136

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

7.31

7.32

Under 18 18 – 25 26 – 50 51 – 65 over 65

p = .55

n = 600

= .497

250(.22) = 55 250(.18) = 45 250(.36) = 90 250(.10) = 25 250(.14) = 35 n = 250

x = 298

P( < .497): z =

= – 2.61

from Table A.5, Prob. = .4955 P( < .497) = .5000 – .4955 = .0045 No, the probability of obtaining these sample results by chance from a population that supports the candidate with 55% of the vote is extremely low (.0045). This is such an unlikely chance sample result that it would cause the researcher to probably reject her claim of 55% of the vote.

7.33

a) Roster of production employees secured from the human resources department of the company. b) Safeway store records kept at the headquarters of their British Columbia division or merged files of store records from regional offices across the province. c) Membership list of Newfoundland lobster catchers association.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

7.34

µ = $ 21,755  = $ 650 The sample size is n = 30, but it is being drawn from a finite population of N =

120. P(

< 21,500):

from Table A.5, the area for z = 2.47 is .4932 P(

7.35

< 21,500) = .5000 – .4932 = .0068

Number the employees from 0001 to 1250. Randomly sample from the random number table until 60 different usable numbers are obtained. You cannot use numbers from 1251 to 9999.

7.36 µ = $125 P(

n = 32

= $110

2 = $525

> $110): = – 3.70

z =

from Table A.5, Prob.= .4998 P(

> $110) = .5000 + .4998 = .9998

> $135):

z =

= 2.47

from Table A.5, Prob.= .4932

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

> $135) = .5000 – .4932 = .0068

P($120 <

< $130):

z =

= – 1.23

z =

= 1.23

from Table A.5, Prob.= .3907 P($120 <

< $130) = .3907 + .3907 = .7814

7.37 n = 1,100 a) x > 810,

p = .73

z =

= 0.48

from table A.5, area = .1844 P(x > 810) = P( b) x < 1,030, =

> 0.7364) =P(z > 0.48) = .5000 – .1844 = .3156

p = .96, = .9364

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

= –3.99

z =

from table A.5, area = .49997 P(x < 1,030) = P(

< .9364) =P(z < –3.99) =. .5000 – .49997 = .00003

c) p = .85 P(.82 <

< .84): = – 2.79

z =

from table A.5, area = .4974 z =

= – 0.93

from table A.5, area = .3238 P(.82 < < .84) = .4974 – .3238 = .1736

7.38

1) 2) 3) 4) 5) 6) 7) 8) 9)

The managers from some of the companies you are interested in studying do not belong to the Board of Trade. The membership list of the Board of Trade is not up-to-date. You are not interested in studying managers from some of the companies belonging to the Board of Trade. The wrong questions are asked. The manager incorrectly interprets a question. The assistant accidentally marks the wrong answer. The wrong statistical test is used to analyze the data. An error is made in statistical calculations. The statistical results are misinterpreted.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

7.39

Divide Canada into nonoverlapping geographic areas. Randomly select a few areas. Take a random sample of employees from each selected Loblaws supermarket in each selected area.

7.40

N = 12,080

n = 300

k = N/n = 12080/300 = 40.27 Select every 40th outlet to assure n > 300 outlets. Use a table of random numbers to select a value between 1 and 40 as a starting point.

7.41

p = .54

n = 565

a) P(x > 339): =

= .60

z =

= 2.86

from Table A.5, the area for z = 2.86 is .4979 P(x > 339) = P(

> 0.60) = P(z > 2.86) = .5000 – .4979 = .0021

b) P(x > 288): = z =

= .5097 = – 1.45

from Table A.5, the area for z = 1.45 is .4265 P(x > 288) = P(

> 0.5097) = P(z > – 1.45) = .5000 + .4265 = .9265

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

c) P(

< .50): = –1.91

z =

from Table A.5, the area for z = 1.91 is .4719 P(

7.42

< .50) = P(z < – 1.91) = .5000 – .4719 = .0281

µ = $850  = $100 Because the sample size n = 50 is greater than 30, the central limit theorem can be used. P(

< $830):

z = ( -µ)/( /√n) = (830 - 850)/(100/√50) = -1.41 from Table A.5, Prob.=.4207 P(x < $530) = .5000 – .4207 = .0793

7.43

µ = 16,621 n = 51  = 3,500 Because the sample size n = 51 is greater than 30, the central limit theorem can be

used. a) P(

> 18,000):

z = ( -µ)/( /√n) = (18,000 – 16,621)/(3,500/√51) = 2.81 from Table A.5, Prob. = .4975 P(

> 18,000) = .5000 – .4975 = .0025

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

b) P(

> 17,500):

z = ( -µ)/( /√n) = (17,500 – 16,621)/(3,500/√51) = 1.79 from Table A.5, Prob.= .4633 P(

> 17,500) = .5000 – .4633 = .0367

c) P(17,000 <

< 18,000):

z = ( -µ)/( /√n) = (17,000 – 16,621)/(3,500/√51) = 0.77 z = ( -µ)/( /√n) = (18,000 – 16,621)/(3,500/√51) = 2.81 from Table A.5, Prob. for z = 0.77 is .2794 from Table A.5, Prob. for z = 2.81 is .4975 P(17,000 <

< 18,000)= .4975 – .2794 = .2181

d) P( < 16,000): z = ( -µ)/( /√n) = (16,000 – 16,621)/(3,500/√51) = -1.27 from Table A.5, Prob.= .3980 P( e) P(

< 16,000) = .5000 – .3980 = .1020 < 15,000):

z = ( -µ)/( /√n) = (15,000 – 16,621)/(3,500/√51) = -3.31 from Table A.5, Prob.= .4995 P( 7.45

< 15,000) = .50000 – .4995 = .0005

p = .46

n = 300

a) P(114 < x < 153):

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

= .38

= .51

= –2.78

= 1.74

from Table A.5, the area for z = 2.78 is .4973 the area for z = 1.74 is .4591 P(114 < x < 153) = P(–2.78 < z < 1.74) = .4973 + .4591 = .9564 b) P(

> .51):

z =

= 1.74

from Table A.5, the area for z = 1.74 is .4591 P( > .51) = .5000 – .4591 = .0409 c) p = .46

n = 800

z =

P( > .51): = 2.84

from Table A.5, the area for z = 2.84 is .4977 P( > .51) = .5000 – .4977 = .0023 The probability in this case is smaller than in question b) , because the sample sizes in b) and c) are different. 7.46

n = 140 (women)

P(x > 35):

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

= .25

p = .22 (women who do volunteer work)

z =

= 0.86

from Table A.5, the area for z = 0.86 is .3051 P(x > 35) = .5000 – .3051 = .1949 P(x < 21): =

= .15

= – 2.00

z =

from Table A.5, the area for z = 2.00 is .4772 P(x < 21) = .5000 – .4772 = .0228 n = 300 (men and women) P(.18 < z =

p = .20 (all people who do volunteer work)

< .25): = – 0.87

from Table A.5, the area for z = 0.87 is .3078 z =

= 2.17

from Table A.5, the area for z = 2.17 is .4850 P(.18 <

< .25) = .3078 + .4850 = .7928

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

7.47

By taking a sample, there is potential for obtaining more detailed information. More time can be spent with each employee. Probing questions can be asked. There is more time for trust to be built between employee and interviewer resulting in the potential for more honest, open answers. With a census, data is usually more general and easier to analyze because it is in a more standard format. Decision-makers are sometimes more comfortable with a census because everyone is included and there is no sampling error. A census appears to be a better political device because the CEO can claim that everyone in the company has had input.

7.48

p = .75

n = 150

= .80

x = 120

P( > .80): z =

= 1.41

from Table A.5, the area for z = 1.41 is .4207 P(

> .80) = .5000 – .4207 = .0793

7.49 Alberta: n = 40 P(25 < z =

µ = $ 26.50

=$3

< 26): = – 3.16

z = ( -µ)/( /√n) = (26 – 26.50)/(3/√40) = -1.05 from Table A.5, the area for z = 3.16 is .4992 the area for z = -1.05 is .3531

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

P(21 <

< 22) = .4992 – .3531 = .1461

Ontario: n = 35 P(

µ = $ 24.69

 = $3

> 27):

z = ( -µ)/( /√n) = (27 – 24.69)/(3/√35) = 4.56 from Table A.5, the area for z = 4.56 is .5000 P(

> 27) = .5000 – .5000 = .0000

British Columbia: n = 50 P(

µ = $ 26.47

 =$3

< 24.90):

z = ( -µ)/( /√n) = (24.90 – 26.47)/(3/√50) = -3.70 from Table A.5, the area for z = 3.70 is .5.000 P( < 18.90) = .5000 – .5000 = .0000

7.50

a) b) c) d)

7.51

Age, Ethnicity, Religion, Geographic Region, Occupation, UrbanSuburban-Rural, Party Affiliation, Gender Age, Ethnicity, Gender, Geographic Region, Economic Class Age, Ethnicity, Gender, Economic Class, Education Age, Ethnicity, Gender, Economic Class, Geographic Location

µ = $280 P( z =

n = 60

 = $50

> $270): = – 1.55

from Table A.5 the area for z = 1.55 is .4394 Solutions Manual 1-259 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

> $273) = .5000 + .4394 = .9394

Legal Notice Copyright

SOLUTIONS TO PROBLEMS IN CHAPTER 8: STATISTICAL INFERENCE: ESTIMATION FOR SINGLE POPULATIONS

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

8.1 a)

x = 25  = 3.5 95% Confidence xz

 n

x = 3.419 90% C.I. xz

= 25 + 1.96

x = 119.6 98% Confidence xz



 n

x = 56.7 80% C.I. xz

 n

n = 60 z.025 = 1.96 3.5 = 25 + 0.89 = 24.11 < µ < 25.89 60

 = 23.89

n = 75 z.01 = 2.33

= 119.6 + 2.33

23.89 75

 = 0.974

= 3.419 + 1.645

= 119.6 ± 6.43 = 113.17 < µ < 126.03

n = 32 z.05 = 1.645

0.974 = 3.419 ± .283 = 3.136 < µ < 3.702 32

 = 12.1 z.10 = 1.28

N = 500

12.1 500  47 N n = 56.7 + 1.28 N 1 47 500  1

n = 47

56.7 ± 2.15 = 54.55 < µ < 58.85

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

8.2 n = 36 95% C.I. xz

 n

8.3 n = 81 90% C.I. xz

 n

x = 90.4 94% C.I.

 n

8.5 n = 39 96% C.I. xz

= 211 ± 1.96

23 36

 n

211 ± 7.51 = 203.49 < µ < 218.51

 = 5.89

x = 47

z.05=1.645 = 47 ± 1.645

5.89 = 47 ± 1.08 = 45.92 < µ < 48.08 81

2 = 49

8.4 n = 70

xz

x = 211  = 23 z.025 = 1.96

x = 90.4

Point Estimate z.03 = 1.88 = 90.4 ± 1.88

49 = 90.4 ± 1.57 = 88.83 < µ < 91.97 70

x = 66

 = 11

N n 11 = 66 ± 2.05 N 1 39

200  39 = 200  1

N = 200 z.02 = 2.05

66 ± 3.25 = 62.75 < µ < 69.25

x = 66

Point Estimate

8.6 n = 120 99% C.I.

x = 18.72 z.005 = 2.575

 = 0.8735

x = 18.72 Point Estimate xz

 n

= 18.72 ± 2.575

0.8735 = 18.72 ± .21 = 18.51 < µ < 18.93 120

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

8.7 N = 1,500

x = 5.3 years

n = 187

x = 5.3 years

Point Estimate

95% C.I.

z.025 = 1.96

xz

 n

 = 1.28 years

N n 1.28 1,500  187 = 5.3 ± 1.96 = 1,500  1 N 1 187

5.3 ± .17 = 5.13 < µ < 5.47

8.8 n = 24 90% C.I. xz

 n

8.9 n = 36 98% C.I. xz

 n

 = 3.23

x = 5.625 z.05 = 1.645 = 5.625 ± 1.645

3.23 24

 = 1.17

x = 3.306 z.01 = 2.33 = 3.306 ± 2.33

1.17 36

= 3.306 ± .454 = 2.852 < µ < 3.760

 = .113

x = 2.139

8.10 n = 36

= 5.625 ± 1.085 = 4.540 < µ < 6.710

x = 2.139 Point Estimate 90% C.I. xz

 n

z.05 = 1.645 = 2.139 ± 1.645

(.113) = 2.139 ± .031 = 2.108 < µ < 2.170 36

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

8.11 95% confidence interval

x = 24.511 xz

 n

n = 45

= 5.124

= 24.511 + 1.96

z.025 = 1.96

5.124 45

24.511 + 1.497 = 23.014 <  < 26.008

x = 24.511 Point Estimate Error of the Interval = 1.497 We are 95% confident that the population mean commuting time in Saskatoon is between 23.01 and 26.01 minutes. It is smaller than ,27.4 minutes, the average travel time to work in Kitchener.

8.12 The point estimate is 0.9600. The assumed standard deviation is 0.1400. Sample size:

n = 41

Margin of error of the estimate: 1.002853 – 0.9600 = 0.042853  0.140000 z 2  z 2  0.042853 n 41

z 2 

0.042853 41  1.96 0.140000

95% level of confidence Confidence interval:

0.917147 < µ < 1.002853

x = 45.62

8.13 n = 13

s = 5.694

df = 13 – 1 = 12

95% Confidence Interval and /2=.025 t.025,12 = 2.179 xt

s n

= 45.62 ± 2.179

5.694 = 45.62 ± 3.44 = 42.18 < µ < 49.06 13

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

x = 319.17

8.14 n = 12

s = 9.104

df = 12 – 1 = 11

90% confidence interval

/2 = .05 xt

s n

t.05,11 = 1.796 = 319.17 ± (1.796)

9.104 = 319.17 ± 4.72 = 314.45 < µ < 323.89 12

x = 128.4

8.15 n = 41

s = 20.6

df = 41 – 1 = 40

98% Confidence Interval /2 = .01 t.01,40 = 2.423 xt

s n

= 128.4 ± 2.423

20.6 41

= 128.4 ± 7.80 = 120.6 < µ < 136.2

x = 128.4 Point Estimate

x = 2.364

8.16 n = 15

s2 = 0.81

df = 15 – 1 = 14

90% Confidence interval /2 = .05 t.05,14 = 1.761 xt

s n

= 2.364 ± 1.761

8.17 n = 25

x = 16.088

0.81 = 2.364 ± .409 = 1.955 < µ < 2.773 15

s = .817

df = 25 – 1 = 24

99% Confidence Interval

/2 = .005 t.005,24 = 2.797

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

xt

s n

= 16.088 ± 2.797

(.817)

= 16.088 ± .457 = 15.631 < µ < 16.545

x = 16.088 Point Estimate

x = 1,192

8.18 n = 51

s = 279

98% CI and /2 = .01 xt

s n

df = n – 1 = 50

t.01,50 = 2.403

= 1,192 + 2.403

279 51

= 1,192 + 93.880 = 1,098.12 <  < 1,285.88

The figure given by Runzheimer International falls within the confidence interval. Therefore, there is no reason to reject the Runzheimer figure as different from what we are getting based on the sample.

8.19 n = 20

df = 19

x = 2.36116

95% CI

t.025,19 = 2.093

s = 0.19721

2.36116 + 2.093

0.1972 = 2.36116 + 0.0923 = 2.26886 <  < 2.45346 20

Point Estimate = 2.36116 Error = 0.0923

x = 5.264

8.20 n = 28

90% Confidence Interval

s = 2.043

df = 28 – 1 = 27

/2 = .05

t.05,27 = 1.703 xt

s n

= 5.264 ± 1.703

2.043 28

= 5.264 + .658 = 4.606 < µ < 5.922

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

x = 49.8

8.21 n = 10

95% Confidence xt

s n

/2 = .025

= 49.8 ± 2.262

8.22 n = 14

s = 18.22

df = 10 – 1 = 9

t.025,9 = 2.262

18.22 = 49.8 + 13.03 = 36.77 < µ < 62.83 10

/2 = .01

98% confidence

df = 13

t.01,13 = 2.650 from data:

x = 152.16

confidence interval:

xt

s = 14.42 s n

= 152.16 + 2.650

14.42 = 14

152.16 + 10.21 = 141.95 <  < 162.37 The point estimate is 152.16

8.23

n = 41

df = 41 – 1 = 40

/2 = .005

99% confidence

t.005,40 = 2.704 from data:

x = 11.10

confidence interval:

xt

s = 8.45 s n

= 11.10 + 2.704

8.45 41

11.10 + 3.57 = 7.53 <  < 14.67

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

8.24

The point estimate is x which is 25.4134 hours. The sample size is 26 skiffs. The confidence level is 98%. The confidence interval is: s s or 22.8124 <  < 28.0145 x t    x t n n The margin of error of the confidence interval is 28.0145 – 25.4134 = 2.601. We can also calculate the margin of error using the formula: t 2

8.25

s n

n = 44

pˆ  z

n = 95

pˆ  z

 2.485

5.3369 26

 2.601 .

99% C.I.

z.005 = 2.575

(.51)(.49) pˆ  qˆ = .51 ± 2.575 = .51 ± .194 = .316 < p< .704 n 44

p̂ = .82

95% C.I.

z.025 = 1.96

(.82)(.18) pˆ  qˆ = .82 ± 1.96 = .82 ± .043 = .777 < p < .863 n 300

n = 1,150

pˆ  z

p̂ =.51

n = 300

pˆ  z

 t 0.01, 25

p̂ = .48

90% C.I.

z.05 = 1.645

pˆ  qˆ (.48)(.52) = .48 ± 1.645 = .48 ± .024 = .456 < p < .504 1,150 n

p̂ = .32

88% C.I.

z.06 = 1.555

(.32)(.68) pˆ  qˆ = .32 ± 1.555 = .32 ± .074 = .246 < p < .394 n 95

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

8.26

n = 116 p̂ =

8.27

z.005 = 2.575

(.49)(.51) pˆ  qˆ = .49 ± 2.575 = .49 ± .12 = .37 < p < .61 n 116

n = 800

x = 479

97% C.I.

z.015 = 2.17

p̂ =

x 479 = .60  n 800

pˆ  z

(.60)(.40) pˆ  qˆ = .60 ± 2.17 = .60 ± .038 = .562 < p < .638 n 800

n = 240

x = 106

85% C.I.

z.075 = 1.44

p̂ =

x 106 = .44  n 240

pˆ  z

(.44)(.56) pˆ  qˆ = .44 ± 1.44 = .44 ± .046 = .394 < p < .486 n 240

n = 60

x = 21

90% C.I.

z.05 = 1.645

p̂ =

x 21 = .35  n 60

pˆ  z

(.35)(.65) pˆ  qˆ = .35 ± 1.645 = .35 ± .10 = .25 < p < .45 n 60

n = 85 p̂ =

99% C.I.

x 57 = .49  n 116

pˆ  z

x = 57

x = 40

90% C.I.

z.05 = 1.645

x 40 = .47  n 85

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

pˆ  z

(.47)(.53) pˆ  qˆ = .47 ± 1.645 = .47 ± .09 = .38 < p < .56 n 85

95% C.I.

pˆ  z

(.47)(.53) pˆ  qˆ = .47 ± 1.96 = .47 ± .11 = .36 < p < .58 n 85

99% C.I.

pˆ  z

z.025 = 1.96

z.005 = 2.575

(.47)(.53) pˆ  qˆ = .47 ± 2.575 = .47 ± .14 = .33 < p < .61 n 85

All other things being constant, as the confidence increased, the width of the interval increased.

8.28

n = 1,003

pˆ  z

99% CI

z.005 = 2.575

pˆ  qˆ (.255)(.745) = .255 + 2.575 = .255 + .035 = .220 < p < .290 1,003 n

n = 10,000

pˆ  z

p̂ = .255

99% CI

z.005 = 2.575

pˆ  qˆ (.255)(.745) = .255 + 2.575 = .255 + .011 = .244 < p < .266 10,000 n

The confidence interval constructed using n = 1,003 is wider than the confidence interval constructed using n = 10,000. One might conclude that, all other things being constant, increasing the sample size reduces the width of the confidence interval.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

8.29

n = 560

p̂ = .47

n = 560

8.30

pˆ  z

8.31

p̂ = .28

90% CI

z.05 = 1.645

(.28)(.72) pˆ  qˆ = .28 + 1.645 = .28 + .0312 = .2488 < p < .3112 n 560

n = 1,250 p̂ =

z.025 = 1.96

(.47)(.53) pˆ  qˆ = .47 + 1.96 = .47 + .0413 = .4287 < p < .5113 n 560

pˆ  z

95% CI

x = 997

98% C.I.

z.01 = 2.33

x 997 = .80  n 1,250

pˆ  qˆ (.80)(.20) = .80 ± 2.33 = .80 ± .026 = .774 < p < .826 1,250 n

n = 3,481 p̂ =

x = 927 x 927  = .266 n 3,481

p̂ = .266 Point Estimate

99% C.I.

pˆ  z

z.005 = 2.575

pˆ  qˆ (.266)(.734) = .266 + 2.575 = .266 ± .019 = 3,481 n

.247 < p < .285

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

8.32

8.33

n = 89

x = 48

8.35

z.075 = 1.44

p̂ =

x 48 = .54  n 89

pˆ  z

(.54)(.46) pˆ  qˆ = .54 ± 1.44 = .54 ± .076 = .464 < p < .616 n 89

p̂ = .63

n = 672

95% Confidence

z.025 = 1.96

(.63)(.37) pˆ  qˆ = .63 + 1.96 = .63 + .0365 = .5935 < p < .6665 n 672

pˆ  z

8.34

85% C.I.

n = 275

x = 121

98% confidence

z.01 = 2.33

p̂ =

x 121 = .44  n 275

pˆ  z

pˆ  qˆ (.44)(.56) = .44 ± 2.33 = .44 ± .07 = .37 < p < .51 n 275

n = 12

x = 28.4

2.995,11 = 2.60320

s2 = 44.9

99% C.I.

df = 12 – 1 = 11

2.005,11 = 26.7569

(12  1)(44.9) (12  1)(44.9) < 2 < 2.60320 26.7569

18.46 < 2 < 189.73

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

x = 4.37

n=7

s = 1.24

2.975,6 = 1.23734

s2 = 1.5376

95% C.I. df = 7 – 1 = 6

2.025,6 = 14.4494

(7  1)(1.5376) (7  1)(1.5376) < 2 < 1.23734 14.4494

0.64 < 2 < 7.46 c)

n = 20

x = 105

s = 32

2.95,19 = 10.11701

s2 = 1,024

90% C.I.

df = 20 – 1 = 19

2.05,19 = 30.1435

(20  1)(1,024) (20  1)(1,024) < 2 < 30.1435 10.11701

645.45 < 2 < 1923.10 d)

n = 17

s2 = 18.56

2.90,16 = 9.31224

80% C.I.

df = 17 – 1 = 16

2.10,16 = 23.5418

(17  1)(18.56) (17  1)(18.56) < 2 < 9.31224 23.5418

12.61 < 2 < 31.89

8.36

n = 16

s2 = 37.1833

2.99,15 = 5.22936

df = 16 – 1 = 15

98% C.I.

2.01,15 = 30.5780

(16  1)(37.1833) (16  1)(37.1833) < 2 < 30.5780 5.22936

18.24 < 2 < 106.66

8.37

n = 20

s = 4.3

s2 = 18.49

98% C.I.

df = 20 – 1 = 19

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

2.99,19 = 7.63270

2.01,19 = 36.1908

(20  1)(18.49) (20  1)(18.49) < 2 < 36.1908 7.63270

9.71 < 2 < 46.03 Point Estimate = s2 = 18.49

8.38

s2 = 3.067

n = 15

2.995,14 = 4.07466

df = 15 – 1 = 14

99% C.I. 2.005,14 = 31.3194

(15  1)(3.067) (15  1)(3.067) < 2 < 31.3194 4.07466

1.37 < 2 < 10.54

8.39

s2 = 26,798,241.76

n = 14

95% C.I.

df = 14 – 1 = 13

Point Estimate = s2 = 26,798,241.76 2.975,13 = 5.00874

2.025,13 = 24.7356

(14  1)(26,798,241.76) (14  1)(26,798,241.76) < 2 < 24.7356 5.00874

14,084,038.51 < 2 < 69,553,848.45

8.40

 = 36 n=

E=5

95% Confidence

z.025 = 1.96

z 2 2 (1.96) 2 (36) 2  = 199.15 E2 52

Sample 200

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

 = 4.13 n=

E=1

99% Confidence

z 2 2 (2.575) 2 (4.13) 2  E2 12

z.005 = 2.575

= 113.1

Sample 114

E = 10

Range = 500 – 80 = 420

1/4 Range = (.25)(420) = 105 90% Confidence

z.05 = 1.645

z 2 2 (1.645) 2 (105) 2  n = = 298.3 E2 10 2

Sample 299 d)

Range = 108 – 50 = 58

E=3

1/4 Range = (.25)(58) = 14.5 88% Confidence n =

z.06 = 1.555

z 2 2 (1.555) 2 (14.5) 2  E2 32

= 56.5

Sample 57

8.41

E = .02 n =

p = .40

96% Confidence

z.02 = 2.05

z 2 p  q (2.05) 2 (.40)(.60)  = 2521.5 E2 (.02) 2

Sample 2522 b)

E = .04

p = .50

95% Confidence

z.025 = 1.96

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

z 2 p  q (1.96) 2 (.50)(.50) = 600.25  E2 (.04) 2

n =

Sample 601 c)

E = .05 n =

p = .55

90% Confidence

z.05 = 1.645

z 2 p  q (1.645) 2 (.55)(.45) = 267.9  E2 (.05) 2

Sample 268 d)

E =.01 n =

p = .50

99% Confidence

z.005 = 2.575

z 2 p  q (2.575) 2 (.50)(.50)  = 16,576.6 E2 (.01) 2

Sample 16,577

8.42

E = $200

 = $1,000

99% Confidence

z.005 = 2.575

z 2 2 (2.575) 2 (1,000) 2  n = = 165.77 E2 200 2

Sample 166

8.43

E = $2.00 n =

 = $12.50

90% Confidence

z.05 = 1.645

z 2 2 (1.645) 2 (12.50) 2  = 105.7 E2 2.00 2

Sample 106

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

8.44

E = $100

Range = $2,500 - $600 = $1,900

  1/4 Range = (.25)($1,900) = $475 90% Confidence

z.05 = 1.645

z 2 2 (1.645) 2 (475) 2 n = = 61.05  E2 100 2

Sample 62

8.45

p = .20

q = .80

90% Confidence, n =

E = .02 z.05 = 1.645

z 2 p  q (1.645) 2 (.20)(.80)  = 1,082.41 E2 (.02) 2

Sample 1,083

8.46

p = .50

q = .50

95% Confidence, n =

E = .05 z.025 = 1.96

z 2 p  q (1.96) 2 (.50)(.50)  = 384.16 E2 (.05) 2

Sample 385

8.47 E = .10

p = .50

95% Confidence, n =

q = .50 z.025 = 1.96

z 2 p  q (1.96) 2 (.50)(.50)  = 96.04 E2 (.10) 2

Sample 97

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

8.48 x = 45.6

 = 7.75

80% confidence xz

 n

n = 35

z.10 = 1.28

 45.6  1.28

7.75 35

= 45.6 + 1.68

43.92 <  < 47.28

94% confidence xz

 n

z.03 = 1.88

 45.6  1.88

7.75 35

= 45.6 + 2.46

43.14 <  < 48.06

98% confidence xz

 n

z.01 = 2.33

 45.6  2.33

7.75 35

= 45.6 + 3.05

42.55 <  < 48.65

x = 45.6 (point estimate)

8.49

x = 12.03 (point estimate) For 90% confidence: xt

s n

 12.03  1.833

s = .4373

/2 = .05 (.4373) 10

n = 10

df = 9

t.05,9= 1.833

= 12.03 + .25

11.78 <  < 12.28 For 95% confidence:

/2 = .025

t.025,9 = 2.262

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

xt

s n

 12.03  2.262

(.4373) 10

= 12.03 + .31

11.72 <  < 12.34

/2 = .005

For 99% confidence: xt

s n

 12.03  3.25

(.4373) 10

t.005,9 = 3.25

= 12.03 + .45

11.58 <  < 12.48

8.50

n = 715 pˆ 

x = 329

95% confidence

z.025 = 1.96

329 = .46 715

pˆ  z

pˆ  qˆ (.46)(.54)  .46  1.96 = .46 + .0365 n 715

.4235 < p < .4965 b)

n = 284

pˆ  z

p̂ = .71

90% confidence

z.05 = 1.645

pˆ  qˆ (.71)(.29)  .71  1.645 = .71 + .0443 n 284

.6657 < p < .7543

n = 1,250 pˆ  z

p̂ = .48

95% confidence

z.025 = 1.96

pˆ  qˆ (.48)(.52)  .48  1.96 = .48 + .0277 n 1,250

.4523 < p < .5077

n = 457

x = 270

98% confidence

z.01 = 2.33

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

pˆ 

270 = .591 457

pˆ  z

pˆ  qˆ (.591)(.409)  .591  2.33 = .591 + .0536 n 457

.5374 < p < .6446

8.51

n = 10

s2 = 54.7667

s = 7.40045

90% confidence,

/2 = .05

2.95,9 = 3.32512

df = 10 – 1 = 9

1 - /2 = .95

2.05,9 = 16.9190

(10  1)(54.7667) (10  1)(54.7667) < 2 < 16.9190 3.32512

29.133 < 2 < 148.235

95% confidence,

/2 = .025

2.975,9 = 2.70039

1 - /2 = .975

2.025,9 = 19.0228

(10  1)(54.7667) (10  1)(54.7667) < 2 < 19.0228 2.70039

25.911 < 2 < 182.529

8.52

 = 44

E=3

95% confidence

z.025 = 1.96

z 2 2 (1.96) 2 (44) 2  n = = 826.4 E2 32

Sample 827

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Range = 88 – 20 = 68

E=2

use  = 1/4(range) = (.25)(68) = 17 90% confidence

z.05 = 1.645

z 2 2 (1.645) 2 (17) 2  n = = 195.5 E2 22

Sample 196

E = .04

p = .50

98% confidence

q = .50 z.01 = 2.33

z 2 p  q (2.33) 2 (.50)(.50)  n = = 848.3 E2 (.04) 2

Sample 849

E = .03

p = .70

95% confidence n =

q = .30 z.025 = 1.96

z 2 p  q (1.96) 2 (.70)(.30)  = 896.4 E2 (.03) 2

Sample 897

x = 10.765

8.53 n = 17

s = 2.223

99% confidence

/2 = .005

2.223

xt

 10.765  2.921

df = 17 – 1 = 16

t.005,16 = 2.921 = 10.765 + 1.575

9.19 < µ < 12.34

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

8.54

p = .40

E = .03

90% Confidence

z.05 = 1.645

z 2 p  q (1.645) 2 (.40)(.60) n = = 721.61  E2 (.03) 2

Sample 722

8.55

s2 = 4.941

n = 17

99% C.I.

2.995,16 = 5.14216

df = 17 – 1 = 16

2.005,16 = 34.2671

(17  1)(4.941) (17  1)(4.941) < 2 < 34.2671 5.14216

2.307 < 2 < 15.374

98% Confidence xz

 = 48

x = 213

8.56 n = 45

 n

 213  2.33

z.01 = 2.33 48 45

= 213 ± 16.67

196.33 < µ < 229.67

x = 37.256

8.57 n = 39

 = 3.891

90% confidence

z.05 = 1.645



3.891

xz

 37.256  1.645

= 37.256 ± 1.025

36.231 < µ < 38.281

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

8.58

 = 6

E=1

98% Confidence

z.01 = 2.33

z 2 2 (2.33) 2 (6) 2  = 195.44 E2 12

n =

Sample 196

8.59

n = 1,255 pˆ 

x = 714

95% Confidence

z.025 = 1.96

714 = .569 (point estimate) 1,255 pˆ  qˆ (.569)(.431) = .569 ± .027  .569  1.96 n 1,255

pˆ  z

.542 < p < .596

8.60

n = 41

x  128

s = 21

98% C.I.

df = 41 – 1 = 40

t.01,40 = 2.423 Point Estimate = $128 xt

s n

 128  2.423

21 41

= 128 + 7.947

120.053 <  < 135.947 Interval Width = 135.947 – 120.053 = 15.894 We are 98% confident that the population mean error is between 120.053 and 135.947 minutes. Or 98% of similarly constructed confidence intervals will contain the population mean.

8.61

n = 60

x = 6.717

98% Confidence

 = 3.06

N = 300

z.01 = 2.33

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition



N n 3.06 300  60 =  6.717  2.33 n N 1 60 300  1 6.717 ± 0.825 xz

5.892 < µ < 7.542

8.62 E = $20

Range = $600 – $30 = $570

1/4 Range = (.25)($570) = $142.50 95% Confidence

z.025 = 1.96

z 2 2 (1.96) 2 (142.50) 2  n = = 195.02 E2 20 2

Sample 196

8.63

n = 245 pˆ 

x = 189

90% Confidence

z.05= 1.645

x 189 = .77  n 245

pˆ  z

pˆ  qˆ (.77)(.23)  .77  1.645 = .77 ± .044 n 245

.726 < p < .814

8.64

n = 90 pˆ 

x = 30

95% Confidence

z.025 = 1.96

x 30 = .33  n 90

pˆ  z

pˆ  qˆ (.33)(.67)  .33  1.96 = .33 ± .097 n 90

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

.233 < p < .427 8.65

s2 = 228

x = 43.7

n = 12

df = 12 – 1 = 11

95% C.I.

t.025,11 = 2.201 s

xt

228

 43.7  2.201

= 43.7 + 9.59

34.11 <  < 53.29 2.99,11 = 3.05350

2.01,11 = 24.7250

(12  1)(228) (12  1)(228) < 2 < 24.7250 3.05350

101.44 < 2 < 821.35

8.66

x = 4.82

n = 27 95% CI: s

xt

s = 0.37

df = 26

t.025,26 = 2.056 0.37

 4.82  2.056

= 4.82 + .1464

4.6736 < µ < 4.9664 Since 4.50 is not in the interval, we are 95% confident that µ does not equal to 4.50.

8.67

 = 12

x = 2.48

n = 77

95% Confidence xz

 n

 2.48  1.96

z.025 = 1.96 12 77

= 2.48 ± 2.68

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

-0.20 < µ < 5.16 The point estimate is 2.48 The interval is inconclusive. It says that we are 95% confident that the average arrival time is somewhere between .20 of a minute (12 seconds) early and 5.16 minutes late. Since zero is in the interval, there is a possibility that, on average, the flights are on time.

8.68

n = 560

p̂ =.33

99% Confidence

z.005= 2.575

pˆ  qˆ (.33)(.67)  .33  2.575 = .33 ± .05 n 560

pˆ  z

.28 < p < .38

8.69

p = .50 n=

E = .05

98% Confidence

z.01 = 2.33

z 2 p  q (2.33) 2 (.50)(.50)  = 542.89 E2 (.05) 2

Sample 543

8.70

x = 2.10

n = 27

s = 0.86

98% confidence

/2 = .01

0.86

xt

 2.10  2.479

df = 27 – 1 = 26

t.01,26 = 2.479

= 2.10 ± 0.41

1.69 < µ < 2.51

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

8.71

df = 23 – 1 = 22

n = 23

2.95,22 = 12.33801

s = .014419

90% C.I.

2.05,22 = 33.9245

(23  1)(.014419) 2 (23  1)(.014419) 2 2 <  < 33.9245 12.33801

.00013 < 2 < .00037

8.72

n = 39 xz

 n

x = 1.294

 = 0.205

 1.294  2.575

0.205 39

99% Confidence

z.005 = 2.575

= 1.294 ± .085

1.209 < µ < 1.379

8.73

n = 1,000

pˆ  z

p̂ = .23

80% CI

z.10 = 1.28

pˆ  qˆ (.23)(.77) = .23 + 1.28 = .23 + .017 = .213 < p < .247 1,000 n

8.74

The sample mean fill for the 510 cans is 354.373 ml with a standard deviation of 1.5857 ml. The 99% confidence interval for the population fill is 353.837 ml to 354.908 ml which does not include 355 ml. We are 99% confident that the population mean is not 355 ml, indicating that the machine may be underfilling the cans.

8.75

The point estimate for the average length of burn of the new bulb is 2198.217 hours. Eighty-four bulbs were included in this study. A 90% confidence interval can be constructed from the information given. The margin of error of the confidence interval is + 27.76691. Combining this with the point estimate yields the 90% confidence interval of 2198.217 + 27.76691 = 2170.450 < µ < 2225.984.

8.76

The point estimate for the average age of a first time buyer is 27.63 years. The sample of 21 buyers produces a standard deviation of 6.54 years. We are 99%

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

confident that the actual population mean age of a first-time home buyer is between 24.0222 years and 31.2378 years.

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SOLUTIONS TO PROBLEMS IN

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

CHAPTER 9: STATISTICAL INFERENCE: HYPOTHESIS TESTING FOR SINGLE POPULATIONS

9.1 a) Two-tailed test b) One-tailed test c) One-tailed test d) Two-tailed test 9.2 a) If the null hypothesis is true, then the analyst would have made the correct decision. If the null hypothesis is false then the analyst would have committed a Type II error. b) If the null hypothesis is true, then the analyst would have made the correct decision. If the null hypothesis is false then the analyst would have committed a Type II error. c) If the null hypothesis is true, then the analyst would have committed a Type I error. If the null hypothesis is false then, the analyst would have made the correct decision. d) If the null hypothesis is true, then the analyst would have committed a Type I error. If the null hypothesis is false then, the analyst would have made the correct decision.

9.3 a) Ho: µ = 25 Ha: µ  25

x = 28.1

n = 57

 = 8.46

For two-tailed test, /2 = .005 z=

x

 n



 = .01

zc = z0.005 = ± 2.575

28.1  25 = 2.77 8.46 57

observed z = 2.77 > z0.005 = 2.575 Reject the null hypothesis b) from Table A.5, inside area between z = 0 and z = 2.77 is .4972 p-value = .5000 – .4972 = .0028 Since the p-value of .0028 is less than /2 = .005, the decision is to reject the null hypothesis

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

c) critical mean values: zc =

± 2.575 =

xc  

 n x c  25 8.46 57

x c = 25 ± 2.885 x c = 27.885 (upper value) x c = 22.115 (lower value)

9.4 Ho: µ = 7.48 Ha: µ < 7.48

x = 6.91

n = 24

For one-tailed test,  = .01 z =

x

 n



 = 1.21

 =.01

zc = z0.01 = – 2.33

6.91  7.48 = –2.31 1.21 24

observed z = – 2.31 > z0.01 = – 2.33 Fail to reject the null hypothesis

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

9.5 a) Ho: µ = 1,200 Ha: µ > 1,200

x = 1,215

n = 113

 = 100

For one-tailed test,  = .10 z =

x





1,215  1,200 100

 = .10

zc = z0.01 = 1.28 = 1.59

113

observed z = 1.59 > z0.01 = 1.28 Reject the null hypothesis b) from Table A.5, inside area between z = 0 and z = 1.59 is .4441 p-value = .5000 – .4441 = .0559 Since the p-value of .0559 is less than  = .10, the decision is to reject the null hypothesis

c) Critical mean value: zc =

1.28 =

xc  

 n x c  1,200 100 113

x c = 1,200 + 12.04 = 1,212.04 Since the observed x = 1,215 is greater than the critical mean value x c = 1212.04, the decision is to reject the null hypothesis.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

9.6 Ho: µ = 82 Ha: µ < 82

x = 78.125

n = 32

For one-tailed test, z=

x





 = 9.184

 = .01

z.01 = – 2.33

78.125  82 = – 2.39 9.184

Since observed z = – 2.39 < z.01 = – 2.33 Reject the null hypothesis Statistically, we can conclude that urban air soot in Hamilton is significantly lower than what the study reported. From a business and community point-of-view, assuming that the sample result is representative of how the air actually is now, is a reduction of suspended particles from 82 to 78.125 really an important reduction in air pollution (is it substantive?). Certainly it marks an important first step and perhaps a significant start. Whether or not it would really make a difference in the quality of life for people in the city of Hamilton remains to be seen. Most likely, politicians and city chamber of commerce folks would jump on such results as indications of improvement in city conditions.

9.7 H0:  = $424.20 Ha:   $424.20 n = 54

 = $33.90

Two-tailed test, /2 = .025

z.025 = + 1.96

x = $432.69

z =

x

 n



 = .05

432.69  424.20 = 1.84 33.90 54

Since the observed z = 1.84 < z.025 = 1.96 and z = 1.84 > z0.025 = -1.96, the decision is to fail to reject the null hypothesis. There is not sufficient evidence to claim that the mean weekly earnings of a production worker have changed.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

9.8

H0:  = $62,600 Ha:  < $62,600

x = $58,974

One-tailed test,  = .01 z =

x





 = $7,810

n = 18

 = .01

z.01 = – 2.33

58,974  62,600 = –1.97 7,810

Since the observed z = –1.97 > z.01 = –2.33, the decision is to fail to reject the null hypothesis. At   0.01 , there is not sufficient evidence to support the researcher‟s theory that the annual average earnings of a mobile phone user are lower than $62,600 per year.

9.9 H0:  = 500 Ha:   500

x = 506.11

n = 42

Two-tailed test, /2 = .05 z =

x

 n

N n N 1



N = 650

 = 28.03

 = .10

z.05 = + 1.645

506.11  500 28.03 650  42 650  1 42

= 1.46

Since the observed z = 1.46 < z.05 = 1.645 and , the decision is to fail to reject the null hypothesis. At   0.10 , there is not sufficient evidence to claim that an average tensile strength is different from 500 MPA.

9.10 Ho: µ = 18.2 Ha: µ < 18.2

x = 15.6

n = 32

 = 2.3

 = .10

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

For one-tailed test,  = .10, z =

x





15.6  18.2 2.3

z.10 = –1.28

= – 6.39

Since the observed z = – 6.39 < z.10 = –1.28, the decision is to reject the null hypothesis. At   0.10 , the sample data support the claim that the average number of orders is down.

9.11

Ho: µ = 41 Ha: µ > 41

= 8.95

 = .01

For one-tailed test,  = .01,

z.01 = 2.33

x = 43.4

n = 97

z =

x





43.4  41 = 2.64 8.95 97

Since the observed z = 2.64 > z.01 = 2.33, the decision is to reject the null hypothesis. From Table A.5, inside area between z = 0 and z = 2.64 is .4959 p-value = .5000 – .4959 = .0041 Since the p-value of .0041 is less than  = .01, the decision is to reject the null hypothesis.

At 1% level of significance, the sample data support the claim that the mean age of Canadian business women has increased.

9.12 Ho: µ = 492 Ha: µ > 492

 = .05

n = 40

40 people were sampled

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Sample mean: x = 529.44 This is a one-tailed test. Since the p-value = .016 is less than  = .05, we reject the null hypothesis. . At 5% level of significance, the sample data support the claim that the average water usage per person is greater than 492 L.

x = 16.45

9.13 n = 20

s = 3.59

df = 20 – 1 = 19

 = .05

Ho: µ = 16 Ha: µ  16 For two-tailed test, /2 = .025, t =

critical t.025,19 = ±2.093

x   16.45  16 = 0.56  s 3.59 n

Observed t = 0.56 < t.025,19 = 2.093 and t =0.56 > t.025,19 = -2.093 The decision is to fail to reject the null hypothesis.

9.14

s2 = 25.68

x = 58.42

n = 51

df = 51 – 1 = 50

=

.01 Ho: µ = 60 Ha: µ < 60 For one-tailed test,  = .01 t =

x   58.42  60  s 25.68 n

critical t.01,50 = –2.403

= –2.23

Observed t = –2.23 > t.01,7 = –2.403 The decision is to fail to reject the null hypothesis.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

x = 1,236.36

9.15 n = 11 = .05

s = 103.81

Ho: µ = 1,160 Ha: µ > 1,160 or one-tailed test,  = .05 t =



df = 11 - 1 = 10

critical t.05,10 = 1.812

x   1,236.36  1,160 = 2.44  s 103.81 n

Observed t = 2.44 > t.05,10 = 1.812 The decision is to reject the null hypothesis.

9.16

x = 8.37

n = 20

s = .1895

df = 20 – 1 = 19



.01 Ho: µ = 8.3 Ha: µ  8.3 For two-tailed test, /2 = .005 t =

critical t.005,19 = ±2.861

x   8.37  8.3 = 1.65  s .1895 n

Observed t = 1.65 < t.005,19 = 2.861 and t = 1.65 > t.005,19 = -2.861 The decision is to fail to reject the null hypothesis. At   0.01 , there is not sufficient evidence to warrant rejection of the claim that the parts average 8.3 kg.

9.17

n = 12

x = 1.85083

s = .02353

df = 12 – 1 = 11

=

.10 H0: µ = 1.84 Ha: µ  1.84 For a two-tailed test, /2 = .05

critical t.05,11 = 1.796

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

t =

x   1.85083  1.84 = 1.59  s .02353 n

Since t.05,11 = -1.796 < t = 1.59 < t.05,11 = 1.796, The decision is to fail to reject the null hypothesis. At   0.10 , there is not sufficient evidence to warrant rejection of the claim that the holes being punched have an average diameter of 1.84 cm.

9.18

x = 1.3784

n = 25

s = .0293

df = 25 – 1 = 24

=

.01 Ho: µ = $1.37 Ha: µ > $1.37 For one-tailed test, = .01

Critical t.01,24 = 2.492

t = ( x - μ)/(s/√ n) = (1.3784-1.37)/(0.0293/√ 25) = 1.43

Observed t = 1.43 < t.01,24 = 2.492 The decision is to fail to reject the null hypothesis. At 1% level of significance, there is not sufficient evidence to support the claim that the average price of gas is higher than $1.07.

9.19

x = $340.89

n = 49

s = $13.89

df = 49 – 1 = 48



= .05 H0:  = $347.46 Ha:   $347.46 Two-tailed test, /2 = .025 t =

t.025,48 ≈ t.025,50 = + 2.009

x   340.89  347.46 = –3.31  s 13.89 n

The observed t = – 3.31 < t.025,48 ≈ – 2.009, Solutions Manual 1-297 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

The decision is to reject the null hypothesis. At 5% level of significance, the sample data support the claim that the average price per square metre for warehouses in Canada has changed.

9.20

x = 3.72

n = 61

df = 61 – 1 = 60

s = 0.65

=

.01 H0:  = 3.51 Ha:  > 3.51 One-tailed test,  = .01 t =

t.01,60 = 2.390

x   3.72  3.51 = 2.52  s 0.65 n

The observed t = 2.52 > t.01,60 = 2.390, The decision is to reject the null hypothesis. At 1% level of significance, the sample data support the claim that the mean housing rating for the city of Calgary by businesses has significantly increased.

x = 1,031.32

9.21 n = 22  = .05

s = 240.37

df = 22 – 1 = 21

H0:  = 1,135 Ha:   1,135 Two-tailed test, /2= .025 t =

t.025,21 = +2.080

x   1,031.32  1,135 = –2.02  s 240.37 n

The observed t = –2.02 > t.025,21 = –2.080 and t = –2.02 < t.025,21 = 2.080, The decision is to fail to reject the null hypothesis. At   0.05 , the sample data support the claim that the figure of $1,135 is accurate.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

x = 42.167

9.22 n = 12 = .01

s = 9.124

df = 12 – 1 = 11



H0:  = 46 Ha:  < 46 One-tailed test, = .01 t =

t.01,11 = –2.718

x   42.167  46 = –1.46  s 9.124 n

The observed t = –1.46 > t.01,11 = –2.718, The decision is to fail to reject the null hypothesis. At   0.01 , there is not sufficient evidence to support the city leaders‟ claim that rates of particulate matter in Kabul have decreased.

9.23

n = 26

x = 19.534 minutes

s = 16.813 = 4.100 minutes

H0:  = 19 Ha:   19 Two-tailed test,

critical t value = + 2.06

Observed t value = 0.66.

Since the observed t = 0.66 < critical t value =

2.06, the decision is to fail to reject the null hypothesis. There is not sufficient evidence to support the claim that the average commuting time for people in the researcher‟s city is different from 19.0 minutes. She would not conclude that her city is any different from the ones in the national survey.

9.24

Ho: p = .45 Ha: p > .45 n = 310

p̂ = .465

For one-tailed test,  = .05

 = .05 z.05 = 1.645

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

z =

pˆ  p pq n

.465  .45



(.45)(.55) 310

= 0.53

observed z = 0.53 < z.05 = 1.645 The decision is to fail to reject the null hypothesis.

9.25

Ho: p = 0.63 Ha: p < 0.63 n = 100

x = 55

For one-tailed test,  = .01 z =

pˆ  p pq n



.55  .63 (.63)(.37) 100

pˆ 

x 55 = .55  n 100

z.01 = –2.33 = –1.66

observed z = –1.66 > zc = –2.33 The decision is to fail to reject the null hypothesis.

9.26

Ho: p = .29 Ha: p  .29 n = 740

x = 207

pˆ 

x 207 = .28  n 740

For two-tailed test, /2 = .025 z =

pˆ  p pq n



.28  .29 (.29)(.71) 740

 = .05

z.025 = ±1.96

= –0.60

observed z = –0.60 > zc = –1.96 and z = –0.60 < zc = 1.96 The decision is to fail to reject the null hypothesis.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

p-Value Method:

from Table A.5, inside area between z = 0 and z = –0.60 is .2257 p-value = Area in tail = .5000 – .2257 = .2743 Since the p-value = .2743 > /2 = .025, the decision is to fail to reject the null hypothesis. Solving for critical values: zc =

pˆ c  p pq n pˆ c  .29 (.29)(.71) 740

±1.96 =

p̂c = .29 ± .033 .257 and .323 are the critical values Since p̂ = .0.28 is not outside critical values in tails, the decision is to fail to reject the null hypothesis.

9.27

Ho: p = .48 Ha: p  .48 n = 380 x = 164 For two-tailed test, pˆ 

z =

 = .01

/2 = .005

z.005 = +2.575

x 164 = .4316  n 380

pˆ  p pq n



.4316  .48 (.48)(.52) 380

= –1.89

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Since the observed z = –1.89 > z.005= –2.575 and z = –1.89 < z.005=2.575, the decision is to fail to reject the null hypothesis. There is not enough evidence to declare that the proportion is any different than .48.

9.28

Ho: p = .79 Ha: p < .79 n = 415 x = 303 For one-tailed test,  = .01 pˆ 

z =

z.01 = –2.33

x 303 = .7301  n 415

pˆ  p pq n



.7301  .79

= –3.00

(.79)(.21) 415

Since the observed z = –3.00 is less than z.01= –2.33, the decision is to reject the null hypothesis. At 1% level of significance, the sample data support the researcher‟s claim that the proportion of accounting firms offering employees flexible scheduling is significantly lower than 79%.

9.29

Ho: p = .31 Ha: p  .31 n = 600 x = 200 For two-tailed test, pˆ 

z =

 = .10

/2 = .05

z.05 = +1.645

x 200 = .3333  n 600

pˆ  p pq n



.3333  .31 (.31)(.69) 600

= 1.23

Since the observed z = 1.23 < z.05= 1.645 and z = 1.23 > z.05= -1.645, the decision is to fail to reject the null hypothesis. There is not enough evidence to declare that the proportion is any different than .31. Solutions Manual 1-302 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Ho: p = .24 Ha: p < .24 n = 600 x = 158 For one-tailed test,  = .05 pˆ 

z =

z.05 = –1.645

x 158 = .2633  n 600

pˆ  p pq n



.2633  .24

= 1.34

(.24)(.76) 600

Since the observed z = 1.34 is greater than z.05= –1.645, the decision is to fail to reject the null hypothesis. There is not enough evidence to declare that the proportion is less than .24. 9.30

Ho: p = .18 Ha: p > .18 n = 376

For one-tailed test, z =

 = .01

p̂ = .22

pˆ  p pq n



 = .01

.22  .18 (.18)(.82) 376

z.01 = 2.33

= 2.02

Since the observed z = 2.02 is less than z.01= 2.33, the decision is to fail to reject the null hypothesis. There is not enough evidence to declare that the proportion is greater than .18.

9.31

Ho: p = .32 Ha: p < .32 n = 118

x = 22

For one-tailed test,

pˆ 

x 22 = .1864  n 118

 = .05

z.05 = –1.645

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

z =

pˆ  p pq n



.1864  .32 (.32)(.68) 118

= –3.11

Observed z = –3.11 < z.05 –1.645 Since the observed z = –3.11 is less than z.05= –1.645, the decision is to reject the null hypothesis. At   0.05 , the sample data support the claim that the proportion of late deliveries was reduced significantly.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

9.32

Ho: p = .47 Ha: p  .47 n = 67

x = 40

For a two-tailed test, pˆ 

 = .05 /2 = .025

z.025 = +1.96

x 40 = .597  n 67

z =

pˆ  p pq n



.597  .47 (.47)(.53) 67

= 2.08

Since the observed z = 2.08 is greater than z.025= 1.96, the decision is to reject the null hypothesis. At   0.05 , there is sufficient evidence to reject the findings of Robert Half International. The sample data support the claim that the proportion of CFOs who get their money news from newspapers is different from 47%.

9.33 a) H0: 2 = 20 Ha: 2 > 20

 = .05

For one-tailed test, 2 =

n = 15

df = 15 – 1 = 14

s2 = 32

2.05,14 = 23.6848

(15  1)(32) = 22.4 20

Since 2 = 22.4 < 2.05,14 = 23.6848, the decision is to fail to reject the null hypothesis. b) H0: 2 = 8.5 17

 = .10

/2 = .05

n = 22

df = n – 1 = 21

s2 =

Ha: 2  8.5 For two-tailed test, 2 =

21–.05,21 = 2.95,21 = 11.59132 and 2.05,21 = 32.6706

(22  1)(17) = 42 8.5

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Since 2 = 42 > 2.05,21 = 32.6706, the decision is to reject the null hypothesis. c) H0: 2 = 45 Ha: 2 < 45

 = .01

For one-tailed test, 2 =

n=8

df = n – 1 = 7

s 2 = 4.12

2 1– .01,7 = 2 .99,7 = 1.23903

(8  1)(4.12) = 0.64 45

Since 2 = 0.64 < 2.99,7 = 1.23903, the decision is to reject the null hypothesis. d) H0: 2 = 5 1.2

 = .05

/2 = .025

n = 11

df = 11 – 1 = 10

s2 =

Ha: 2  5 For two-tailed test,

2.025,10 = 20.4832

2.975,10 = 3.24696

(11  1)(1.2) = 2.4 5 Since 2 = 2.4 < 2.975,10 = 3.24696, the decision is to reject the null

2 = hypothesis.

9.34 30.0833

H0: 2 = 14

 = .05

/2 = .025

n = 12

df = 12 – 1 = 11

s2 =

Ha: 2  14 For two-tailed test, 2 =

2.025,11 = 21.9200

2.975,11 = 3.81574

(12  1)(30.0833) = 23.64 14

Since 2 = 23.64 > 2.025,11 = 21.9200, the decision is to reject the null hypothesis. At   0.05 , the sample data support the claim that the variance of a given process has changed.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

9.35 .00144667

H0: 2 = .001

 = .01

n = 16

s2 =

df = 16 – 1 = 15

Ha: 2 > .001 2.01,15 = 30.5780

For one-tailed test, 2 =

(16  1)(.00144667) = 21.7 .001

Since 2 = 21.7 < 2.01,15 = 30.5780, the decision is to fail to reject the null hypothesis. At   0.01 , there is not sufficient evidence to support the claim that the variance of the bearings diameters is more than 0.001 cm2. 9.36 = 12

H0: 2 = 199,996,164

 = .10

Ha: 2  199,996,164

s2 = 832,089,743.6

For two-tailed test,

2.05,12 = 21.0261

2 =

/2 = .05

n = 13

df =13 – 1

2.95,12 = 5.22603

(13  1)(832,089,743.6) = 49.93 199,996,164

Since 2 = 49.93 > 2.05,12 = 21.0261, the decision is to reject the null hypothesis. At   0.10 , the sample data support the claim that the variance for weekly deposits has changed.

9.37 .1156

H0: 2 = .04

 = .01

n=7

df = 7 – 1 = 6

s = .34

s2 =

Ha: 2 > .04 For one –tailed test, 2 =

2.01,6 = 16.8119

(7  1)(.1156) = 17.34 .04

Since 2 = 17.34 > 2.01,6 = 16.8119, the decision is to reject the null hypothesis. At   0.01 , the sample data support the claim that the variance of the wiring is Solutions Manual 1-307 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

more than 0.04cm2 . Thus, it is too great to meet specifications.

9.38

H0: µ = 100 Ha: µ < 100 n = 48 a) i.

 = .10 zc =

 = 14

µ1 = 99

z.10 = –1.28

xc  

 n x c  100 14

–1.28 =

x c = 97.4 z1 =

x c  1



97.4  99 = –0.79 14 48

from Table A.5, area for z1 = –0.79 is .2852

 = .2852 + .5000 = .7852 ii.

 = .05 zc =

z.05 = –1.645

xc  

–1.645 =

 n x c  100 14 48

x c = 96.68

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

x c  1

z1 =



96.68  99 = –1.15 14 48

from Table A.5, area for z1 = –1.15 is .3749

 = .3749 + .5000 = .8749 iii.

 = .01 zc =

z.01 = –2.33

xc  

 n x c  100 14

–2.33 =

x c = 95.29 z1 =

x c  1



95.29  99 = –1.84 14 48

from Table A.5, area for z1 = –1.84 is .4671

 = .4671 + .5000 = .9671

9.39

As  gets smaller (other variables remaining constant), gets larger. Decreasing the probability of committing a Type I error increases the probability of committing a Type II error if other variables are held constant.

 = .05

µa = 98.5 zc =

µ = 100

n = 48

 = 14

zc = –1.645

xc  

 n

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

–1.645 =

x c  100 14 48

x c = 96.68 za =

xc  a



96.68  98.5 = –0.90 14 48

from Table A.5, area for za = –0.90 is .3159

 = .3159 + .5000 = .8159 zc = –1.645

ii. µa = 98

x c = 96.68 za =

xc  a



96.68  98 = –0.65 14 48

from Table A.5, area for za = –0.65 is .2422

 = .2422 + .5000 = .7422 z.05 = –1.645

iii. µa = 97

x c = 96.68 za =

xc  a



96.68  97 = –0.16 14 48

from Table A.5, area for za = –0.16 is .0636

 = .0636 + .5000 = .5636 iv. µa = 96

z.05 = –1.645

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

x c = 96.68 za =

xc  a



96.68  96 = 0.34 14 48

from Table A.5, area for za = 0.34 is .1331

 = .5000 – .1331 = .3669 b) As the alternative value gets farther from the null hypothesized value, the probability of committing a Type II error reduces (all other variables being held constant). 9.40

Ho: µ = 50 Ha: µ  50 µa = 53

n = 35

 =7

Since this is two-tailed test, /2 = .005 zc =

±2.575 =

 = .01 z.005 = ±2.575

xc  

 n x c  50 7 35

x c = 50 ± 3.05 46.95 and 53.05 za =

xc  a



53.05  53 = 0.04 7 35

from Table A.5 for za = 0.04, area = .0160 Other end: Solutions Manual 1-311 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

za =

xc  a



46.95  53 = –5.11 7 35

Area associated with za = –5.11 is .5000

 = .5000 + .0160 = .5160

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

9.41

a) Ho: p = .65 Ha: p < .65

 = .05

n = 360 zc =

pa = .60

z.05 = –1.645

pˆ c  p pq n pˆ c  .65 (.65)(.35) 360

–1.645 =

p̂c = .65 – .041 = .609 za =

pˆ c  p a pa  qa n

.609  .60 = 0.35 (.60)(.40) 360

from Table A.5, area for za = 0.35 is .1368

 = .5000 – .1368 = .3632 b)

z.05 = –1.645

pa = .55 za =

pˆ c  p a pa  qa n

p̂ c = .609

.609  .55 = 2.25 (.55)(.45) 360

from Table A.5, area for z = 2.25 is .4878

 = .5000 – .4878 = .0122 c)

z.05 = –1.645

pa = .50 za =

pˆ c  p a pa  qa n

p̂ c = .609

.609  .50 = 4.14 (.50)(.50) 360

from Table A.5, the area for za = 4.14 is .5000

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

9.42

 = 8.7

x = 45.1

n = 58

H0: µ = 44 Ha: µ  44 For two-tailed test, z =

/2 = .025

 = .05

z.025 = ± 1.96

45.1  44 = 0.96 8.7 58

Since zc = -1.96 < z = 0.96 < zc = 1.96, the decision is to fail to reject the null

hypothesis. At   0.05 , there is not sufficient evidence to declare that

female shareholders in Vancouver are significantly different in age from female shareholders in the U.S. + 1.96 =

x c  44 8.7 58

± 2.239 = x c – 44

x c = 46.239 and 41.761 For 45 years (µa = 45): za =

46.239  45 = 1.08 8.7 58

from Table A.5, the area for za = 1.08 is .3599 za =

41.761  45 = – 2.84 8.7 58

from Table A.5, the area for za = – 2.84 is .4977

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

 = .4977 + .3599 = .8576 Power = 1 –  = 1 – .8576 = .1424

For 46 years (µa = 46): za =

46.239  46 = 0.21 8.7 58

From Table A.5, the area for za = 0.21 is .0832 za =

41.761  46 = – 3.71 8.7 58

From Table A.5, the area for za = – 3.71 is .4999

 = .4999 + .0832 = .5831 Power = 1 –  = 1 – .5831 = .4169

For 47 years (µa = 47):

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

za =

46.239  47 = – 0.67 8.7 58

From Table A.5, the area for za = –0.67 is .2486 za =

41.761  47 = – 4.59 8.7 58

From Table A.5, the area for za = – 4.59 is .49998

 = .49998 – .2486 = .2514 Power = 1 –  = 1 – .2514 = .7486

For 48 years (µa = 48): za =

46.239  48 = – 1.54 8.7 58

From Table A.5, the area for za = – 1.54 is .4382 za =

41.761  48 = – 5.46 8.7 58

From Table A.5, the area for za = – 5.46 is .5000

 = .5000 – .4382 = .0618 Power = 1 –  = 1 – .0618 = .9382

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

9.43

Eight-Step approach: 1) Ho: µ = 36 Ha: µ  36 2) z =

x

 n

3)  = .01 4) two-tailed test, /2 = .005, z.005 = + 2.575 If the observed value of z is greater than 2.575 or less than -2.575, the decision will be to reject the null hypothesis.

x = 38.4,

5) n = 63, 6) z =

x

 n

 = 5.93

38.4  36 = 3.21 5.93 63

7) Since the observed value of z = 3.21 is greater than z.005 = 2.575, the decision is to reject the null hypothesis. 8) At   0.01 , the sample data support the claim that the population mean is different from 36.

9.44 Eight-Step approach: 1) Ho: µ = 7.82 Ha: µ < 7.82 2) The test statistic is t =

x s n

3)  = .05

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

4) one-tailed test, df = n – 1 = 16, t.05,16 = –1.746. If the observed value of t is less than –1.746, then the decision will be to reject the null hypothesis.

x = 7.01

5) n = 17 6) t =

s = 1.69

7.01  7.82 x = = –1.98 1.69 s 17 n

7) Since the observed t = –1.98 is less than the table value of t = –1.746, the decision is to reject the null hypothesis. 8) At   0.05 , the sample data support the claim that the population mean is less than 7.82.

9.45

Eight-Step approach: a. 1) Ho: p = .28 Ha: p > .28 2) z =

pˆ  p pq n

3)  = .10 4) This is a one-tailed test, z.10 = 1.28. If the observed value of z is greater than 1.28, the decision will be to reject the null hypothesis. 5) n = 783 pˆ 

6) z =

x = 230

230 = .2937 783

.2937  .28 = 0.85 (.28)(.72) 783

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

7) Since z = 0.85 is less than z.10 = 1.28, the decision is to fail to reject the null hypothesis. 8) At   0.10 , there is not enough evidence to declare that p > .28.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

b. 1) Ho: p = .61 Ha: p  .61 2) z =

pˆ  p pq n

3)  = .05 4) This is a two-tailed test, z.025 = + 1.96. If the observed value of z is greater than 1.96 or less than –1.96, then the decision will be to reject the null hypothesis. 5) n = 401 6) z =

p̂ = .56

.56  .61 = –2.05 (.61)(.39) 401

7) Since z = –2.05 is less than z.025 = –1.96, the decision is to reject the null hypothesis. 8) At   0.05 , the sample data support the claim that the population proportion is different from .61.

9.46

Eight-Step approach: 1) H0: 2 = 15.4 Ha: 2 > 15.4 2)  = 2

(n  1) s 2

2 3)  = .01

4) n = 18, df = 17, one-tailed test, 2.01,17 = 33.4087 2 If the observed value of  is greater than 33.4087, then the decision will be to reject the null hypothesis.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

5) s2 = 29.6 (n  1) s 2

6)  = 2



(17)(29.6) = 32.675 15.4

7) Since the observed 2 = 32.675 is less than 33.4087, the decision is to fail to reject the null hypothesis. 8) At   0.01 , there is not sufficient evidence to support the claim that the population variance is more than 15.4.

9.47

H0: µ = 130 Ha: µ > 130 n = 75

 = 12

 = .01

z.01 = 2.33

µa = 135

Solving for x c: zc =

2.33 =

xc  

 n x c  130 12 75

x c = 133.23 za =

133.23  135 = –1.28 12 75

from table A.5, area for za = –1.28 is .3997

 = .5000 – .3997 = .1003 b)

H0: p = .44 Ha: p < .44

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

 = .05

n = 1,095

pa = .42

z.05 = –1.645

pˆ c  p

zc =

pq n

–1.645 =

pˆ c  .44 (.44)(.56) 1095

p̂c = .4153 .4153  .42 = –0.32 (.42)(.58) 1095

za =

from table A.5, area for za = –0.32 is .1255  = .5000 + .1255 = .6255

9.48

H0: p = .32 Ha: p > .32

 = .01

n = 80

For one-tailed test, z =

pˆ  p pq n



p̂ = .39

z.01 = 2.33

.39  .32 (.32)(.68) 80

= 1.34

Since the observed z = 1.34 < z.01 = 2.33, the decision is to fail to reject the null hypothesis. At   0.01 , there is not sufficient evidence to support the researcher‟s statement that a proportion of Canadian households trying to cut long distance phone bills by switching companies is significantly higher than 32%.

9.49

Ho: µ = 1.49 Ha: µ  1.49

x = 1.63

n = 64

2 = 0.59

 = .05

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

For two-tailed test, /2 = .025 z =

x

 n

1.63  1.49 0.59

zc = ±1.96

= 1.46

Since the observed z = 1.46 < zc = 1.96 and z = 1.46 > zc = –1.96, the decision is to fail to reject the null hypothesis. On the basis of the random sample data, we conclude that the Zero Population Growth figure for urban U.S. residents is also true for Canadians.

9.50

n = 210

x = 93

 = .10

pˆ 

x 93 = .443  n 210

Ho: p = .57 Ha: p < .57 For one-tailed test,  = .10 z =

pˆ  p pq n



.443  .57 (.57)(.43) 210

zc = –1.28 = –3.72

Since the observed z = –3.72 < zc = –1.28, the decision is to reject the null hypothesis. At   0.10 , the sample data support the claim that the proportion of executives with retirement fund investments in the bond market is significantly lower than 57%.

9.51

H0: 2 = 0.41 Ha: 2 > 0.41

n = 12

s = 0.1523

s2 = 0.0232

 = .05

df = 12 – 1 = 11

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

2.05,11 = 19.6752

For one-tailed test, 2 =

(12  1) 0.0232 = 0.62 0.41

Since 2 = 0.62 < 2.05,11 = 19.6752, then we fail to reject the null hypothesis. At   0.05 , there is not sufficient evidence to warrant rejection of the engineers‟ claim that the variance of stripes is not more than 0.41 m.

9.52

H0: µ = 8.4

 = .01 x = 5.6

n=7

df = 7 – 1 = 6

= 1.3 Ha: µ  8.4 For two-tailed test, t =

/2 = .005 t.005,6 = + 3.707

5.6  8.4 = –5.70 1.3 7

Since the observed t = – 5.70 < t.005,6 = –3.707, the decision is to reject the null hypothesis. At   0.01 , there is sufficient evidence to warrant rejection of the computer manufacturer‟s claim that its line of minicomputers has, on average, 8.4 days of downtime per year.

x = $26,650

9.53 a)

n = 100

Ho: µ = $25,000 Ha: µ > $25,000

 = .05

For one-tailed test,  = .05 z =

x

 n

 = $12,000

z.05 = 1.645

26,650  25,000 = 1.38 12,000 100

Since the observed z = 1.38 < z.05 = 1.645, the decision is to fail to reject the null hypothesis.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

At   0.05 , there is not sufficient evidence to warrant rejection of the life insurance salesperson‟s claim that the average worker in the city of Winnipeg has no more than $25,000 of personal life insurance. b)

µa = $30,000

zc = 1.645

Solving for x c: zc =

xc  

 n

1.645 =

( x c  25,000) 12,000 100

x c = 25,000 + 1,974 = 26,974 za =

26,974  30,000 = –2.52 12,000 100

from Table A.5, the area for za = –2.52 is .4941

 = .5000 – .4941 = .0059

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

9.54

H0: 2 = 4 Ha: 2 > 4

n=8

For one-tailed test,

s = 7.80

 = .10

df = 8 – 1 = 7

2.10,7 = 12.0170

(8  1)(7.80) 2  = = 106.47 4 2

Since observed 2 = 106.47 > 2.10,7 = 12.017, the decision is to reject the null hypothesis. At   0.10 , there is sufficient evidence to warrant rejection of the claim that the variance of the share price does not exceed $4.

9.55

H0: p = .46 Ha: p > .46 n = 125

x = 66

Using a one-tailed test, z =

pˆ  p pq n



.528  .46 (.46)(.54) 125

 = .05

pˆ 

x 66 = .528  n 125

z.05 = 1.645 = 1.53

Since the observed value of z = 1.53 < z.05 = 1.645, the decision is to fail to reject the null hypothesis. At   0.05 , there is not sufficient sample evidence to declare that the proportion of MBAs who expect stock options is significantly higher than 46%. Solving for p̂c : zc =

1.645 =

pˆ c  p pq n pˆ c  .46 (.46)(.54) 125

and therefore, p̂c = .533

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

pˆ c  p a

za =

pa  qa n



.533  .50

= 0.74

(.50)(.50) 125

from Table A.5, the area for za = 0.74 is .2704  = .5000 + .2704 = .7704 9.56

x = 175

n = 16

s = 14.28286

df = 16 – 1 = 15

 = .05

H0: µ = 185 Ha: µ < 185 For one-tailed test, t =

t.05,15 = – 1.753

175  185 x = = –2.80 14.28286 s 16 n

Since observed t = – 2.80 < t.05,15 = – 1.753, the decision is to reject the null hypothesis. At   0.05 , there is sufficient evidence to warrant rejection of the hospital administrator‟s claim that, on average, at least 185 beds are filled on any given day.

9.57

H0: p = .198 Ha: p > .198 n = 428

x = 90

For a one-tailed test, z =

 = .01

pˆ 

x 90 = .210  n 428

z.01 = 2.33

pˆ  p .210  .198 = 0.62  pq (.198)(.802) n 428

Since the observed z = 0.62 < z.01 = 2.33, the decision is to fail to reject the null hypothesis. At   0.01 , there is not sufficient sample evidence to support the market researcher‟s claim that Hewlett-Packard holds a higher than 19.8% share of the market in Ontario.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

The solution for finding the probability of making a Type II error is following. Given that pa =.22. Solving for p̂c :

pˆ c  p

zc =

pq n

2.33 =

. pˆ c  .198 (.198)(.802) 428

p̂c = .2429 za =

pˆ c  pa .2429  .22 = 1.14  p a  qa (.22)(.78) n 428

from Table A.5, the area for za = 1.14 is .3729

 = .5000 + .3729 = .8729

9.58

Ho: µ = $15 Ha: µ > $15

x = $19.34

n = 17

 = $4.52

For one-tailed test,  = .10 z =

x

 n

 = .10

zc = 1.28

19.34  15 = 3.96 4.52 17

Since the observed z = 3.96 > zc = 1.28, the decision is to reject the null hypothesis. At   0.10 , the sample data support the claim that, on average, a student living away from home spends more than $15 per month on laundry.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

9.59

H0: 2 = 103.2 Ha: 2 > 103.2 n = 22

df = 22 –1 = 21

s = 15.2

 = .05

2.05,21 = 32.6706

For one-tailed test,

(22  1)(15.2) 2  = = 47.01 103.2 2

Since the observed 2 = 47.01 > 2.05,21 = 32.6706, the decision is to reject the null hypothesis. At   0.05 , there is sufficient evidence to warrant rejection of the company‟s claim that the variance of the depth of the lines does not exceed 103.2 cm2.

9.60 =8

H0: µ = 2.5

x = 3.4

s = 0.6

 = .01

n=9

df = 9 – 1

Ha: µ > 2.5 For one-tailed test, t =

t.01,8 = 2.896

3.4  2.5 x = = 4.50 0.6 s 9 n

Since the observed t = 4.50 > t.01,8 = 2.896, the decision is to reject the null hypothesis. At   0.01 , the sample data support the claim that certain industrial emissions exceed 2.5 parts per million. So, the company is exceeding the safe limit.

9.61

Ho:  = 23.58 Ha:   23.58 n = 95

x = 22.83

 = 5.11

 = .05

Since this is a two-tailed test and using /2 = .025: z.025 = + 1.96

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

z =

x

 n

22.83  23.58 = –1.43 5.11 95

Since the observed z = –1.43 > z.025 = –1.96 and z = –1.43 < z.025 = 1.96, the decision is to fail to reject the null hypothesis. At   0.05 , there is not sufficient evidence to warrant rejection of the claim that the average cost per square metre for office rental space in Toronto‟s Bay Street business district is $23.58. b)

zc =

+ 1.96 =

xc  

 n xc  23.58 5.11 95

x c = 23.58 + 1.03 x c = 22.55, 24.61 for a = 22.30 za =

xc   a



za =

xc   a



22.55  22.30 = 0.48 5.11 95 24.61  22.30 = 4.41 5.11 95

from Table A.5, the areas for za = 0.48 and za = 4.41 are .1844 and .5000

 = .5000 – .1844 = .3156 The upper tail has no effect on .

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

9.62

x = 49.333

n = 12

s2 = 166.788

H0: 2 = 2.5 Ha: 2  2.5

 = .05

df = 11 For two-tailed test,

/2 = .025

2.025,11 = 21.9200 2..975,11 = 3.81574 If the observed 2 is greater than 21.9200 or less than 3.81574, the decision is to reject the null hypothesis. 2 =

(n  1) s 2



11(166.788) = 733.867 2.5

Since the observed 2 = 733.867 is greater than 2.025,11 = 21.92, the decision is to reject the null hypothesis. At a 5% level of significance, the sample data support the claim that the population variance for such water usage is significantly different from 2.5 L2. 9.63 H0: µ = 23 1 = 15 Ha: µ < 23

x = 18.5

s = 6.91

 = .10

n = 16

df = 16 –

t.10,15 = –1.341 t =

x 18.5  23 = = –2.60 s 6.91 n

Since the observed t = –2.60 < t.10,15 = –1.341, the decision is to reject the null hypothesis. At   0.10 , the sample data support the claim that the daily downtime has been significantly reduced and is less than 23 minutes.

9.64 The sample size is 22, the sample mean x is 1.80031. The population standard deviation σ = 0.39281. The hypotheses are:

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Ho:  = 2.0 Ha:  < 2.0 The observed z = –2.38. The p-value is .0087. The results are statistically significant at  = .05. The decision is to reject the null hypothesis. At   0.05 , the sample data support the claim that the average Canadian generates less than 2.0 kg of solid waste per day.

9.65

The hypotheses are: H0: p = .25 Ha: p  .25 This is a two-tailed test. The financial management association conducts the survey by randomly contacting 384 CFOs (n = 384). 79 CFOs choose Lexus as the favorite

luxury car (x = 79). The sample proportion, pˆ 

x 79   .205729. n 384

Since the p-value = .045 <  = .05, the decision is to reject the null hypothesis. At   0.05 , there is sufficient evidence to warrant rejection of the claim that the Lexus is the favorite luxury car for 25% of CFOs. 9.66 H0:  = 2747 Ha:  < 2747 This is a one-tailed test. Sixty-seven households were included in this study. The sample average amount spent on home-improvement projects was 2349.00. Since the observed z = –2.09 < z.05 = –1.645, the decision is to reject the null

null

hypothesis at  = .05. This is underscored by the p-value of .018 which is less than  = .05. However, the p-value of .018 also indicates that we would not reject the hypothesis at  = .01.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

At   0.05 , the sample data support the company‟s claim that the average household in Manitoba spend less than $2,747 on home improvement. However, at   0.01 , there is not sufficient sample evidence to support the claim that the average is lower in Manitoba.

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Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

SOLUTIONS TO PROBLEMS IN CHAPTER 10: STATISTICAL INFERENCES ABOUT TWO POPULATIONS

10.1

Sample 1 x1 = 51.3 σ12 = 52 n1 = 31 a)

Ho: Ha:

Sample 2 x 2 = 53.2 σ22 = 60 n2 = 32

µ1 - µ2 = 0 µ1 - µ2 < 0

For one-tail test,  = .10 z =

z.10 = -1.28

( x 1  x 2 )  ( 1   2 )

 12 n1



 22



(51.3  53.2)  (0)

52 60  31 32

= -1.01

Since the observed z = -1.01 > zc = -1.28, the decision is to fail to reject the null hypothesis.

Critical value method: zc =

( x 1  x 2 ) c  ( 1   2 )

 12 n1

-1.28 =



 22 n2

( x1  x 2 ) c  (0) 52 60  31 32

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

( x 1 - x 2)c = -2.41 c)

The area for z = -1.01 using Table A.5 is .3438. The p-value is .5000 - .3438 = .1562

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

10.2

Sample 1 n1 = 32 x 1 = 70.4 1 = 5.76

Sample 2 n2 = 31 x 2 = 68.7 2 = 6.1

For a 90% C.I.,

z.05 = 1.645

( x1  x 2 )  z

 12 n1



 22 n2

5.762 6.12  32 31

(70.4 – 68.7) + 1.645 1.7 ± 2.46 -.76 < µ1 - µ2 < 4.16

10.3

Sample 1

Sample 2

x 1 = 88.23 12 = 22.74 n1 = 30

x 2 = 81.2 22 = 26.65 n2 = 30

Ho: µ1 - µ2 = 0 Ha: µ1 - µ2  0 For two-tail test, use /2 = .01 z =

( x 1  x 2 )  ( 1   2 )

1



2



z.01 = + 2.33

(88.23  81.2)  (0) 22.74 26.65  30 30

= 5.48

Since the observed z = 5.48 > z.01 = 2.33, the decision is to reject the null hypothesis.

b) ( x1  x 2 )  z

 12 n1



 22 n2

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

(88.23 – 81.2) + 2.33

22.74 26.65  30 30

7.03 + 2.99 4.04 <  < 10.02 This supports the decision made in a) to reject the null hypothesis because zero is not in the interval.

10.4

Computers/electronics

Food/Beverage

x 1 = 1.96 12 = 1.0188 n1 = 50 Ho: Ha:

x 2 = 3.02 22 = .9180 n2 = 50

µ1 - µ2 = 0 µ1 - µ2  0

For two-tail test, /2 = .005 z =

( x 1  x 2 )  ( 1   2 )

 12 n1



 22 n2



z.005 = ±2.575

(1.96  3.02)  (0) 1.0188 0.9180  50 50

= -5.39

Since the observed z = -5.39 < zc = -2.575, the decision is to reject the null hypothesis. At  = .01, there is a significant difference between people in these two industries.

10.5

A n1 = 40 x 1 = 5.3 12 = 1.99 For a 95% C.I.,

B n2 = 37 x 2 = 6.5 22 = 2.36 z.025 = 1.96

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

( x1  x 2 )  z

 12 n1

(5.3 – 6.5) + 1.96



 22 n2

1.99 2.36  40 37 -1.86 <  < -.54

-1.2 ± .66

The results indicate that we are 95% confident that, on average, Plumber B does between 0.54 and 1.86 more jobs per day than Plumber A. Since zero does not lie in this interval, we are confident that there is a difference between Plumber A and Plumber B.

10.6

Managers n1 = 35 x 1 = 1.84 1 = .38

Specialty n2 = 41 x 2 = 1.99 2 = .51

for a 98% C.I., ( x1  x 2 )  z

 12 n1

z.01 = 2.33 

(1.84 - 1.99) ± 2.33

 22 n2

.382 .512  35 41

-.15 ± .2384 -.3884 < µ1 - µ2 < .0884 Point Estimate = -.15 Hypothesis Test: 1) Ho: µ1 - µ2 = 0 Ha: µ1 - µ2  0

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

2) z =

( x 1  x 2 )  ( 1   2 )

 12



 22 n2

3)  = .02 4) For a two-tailed test, z.01 = + 2.33. If the observed z value is greater than 2.33 or less than -2.33, then the decision will be to reject the null hypothesis. 5) Data given above 6) z =

(1.84  1.99)  (0) (.38) 2 (.51) 2  35 41

= -1.47

7) Since z = -1.47 > z.01 = -2.33, the decision is to fail to reject the null hypothesis. 8) There is no significant difference in the hourly rates of the two groups.

10.7

2004

2014

x 1 = 190 1 = 18.50 n1 = 51

x 2 = 198 2 = 15.60 n2 = 47

 = .01

H0: 1 - 2 = 0 Ha: 1 - 2 < 0 For a one-tailed test, z =

z.01 = -2.33

( x 1  x 2 )  ( 1   2 )

1



2



(190  198)  (0) (18.50) 2 (15.60) 2  51 47

= -2.32

Since the observed z = -2.32 > z.01 = -2.33, the decision is to fail to reject the null hypothesis. There is not sufficient evidence to support the company‟s belief that the per diem average expense in Windsor has gone up significantly.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

10.8

Vancouver

Montreal

n1 = 31 x 1 = 3.88 12 = .12

n2 = 31 x 2 = 5.59 22 = .06

For a 99% C.I.,

z.005 = 2.575

 12

( x1  x 2 )  z



 22 n2

(3.88 – 5.59) ± 2.575

.12 .06  31 31 −1.91 <  < −1.51

− 1.71 ± .20

Between $ 1.51 and $ 1.91 difference with Montreal being more expensive.

10.9

Manan x 1 = 5.8 1 = 1.7 n1 = 36 Ho: Ha:

Prairie x 2 = 5.0 2 = 1.4 n2 = 45

µ1 - µ2 = 0 µ1 - µ2  0

For two-tail test, /2 = .025

z.025 = ±1.96

( x 1  x 2 )  ( 1   2 )

(5.8  5.0)  (0)

z =

1



2



(1.7) 2 (1.4) 2  36 45

= 2.27

Since the observed z = 2.27 > zc = 1.96, the decision is to reject the null hypothesis. At  = .05, there is a significant difference in mean number of suggestions a month per employee between the Manan Corporation and the Prairie Corporation. Solutions Manual 1-342 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

10.10

A x 1 = 8.05 1 = 1.36 n1 = 50 Ho: Ha:

B x 2 = 7.26 2 = 1.06 n2 = 38

µ1 - µ2 = 0 µ1 - µ2 > 0

For one-tail test,  = .10 z =

( x 1  x 2 )  ( 1   2 )

1



2

z.10 = 1.28



(8.05  7.26)  (0) (1.36) 2 (1.06) 2  50 38

= 3.06

Since the observed z = 3.06 > zc = 1.28, the decision is to reject the null hypothesis. For  = .10, the sample data support the claim that operation A takes significantly longer to perform than operation B.

 = .01

10.11 Ho: µ1 - µ2 = 0 Ha: µ1 - µ2 < 0

df = 8 + 11 - 2 = 17

Sample 1 n1 = 8 x 1 = 24.56 s12 = 12.4

Sample 2 n2 = 11 x 2 = 26.42 s22 = 15.8

For one-tail test,  = .01

Critical t.01,17 = -2.567

t =

( x1  x 2 )  ( 1   2 ) s1 (n1  1)  s 2 (n2  1) 1 1  n1  n2  2 n1 n2 2

(24.56  26.42)  (0) = 12.4(7)  15.8(10) 1 1  8  11  2 8 11

-1.05

Since the observed t = -1.05 > t.01,19 = -2.567, the decision is to fail to reject the null hypothesis.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

10.12 a)

 =.10

Ho: µ1 - µ2 = 0 Ha: µ1 - µ2  0

df = 20 + 20 - 2 = 38

Sample 1 n1 = 20 x 1 = 118 s1 = 23.9

Sample 2 n2 = 20 x 2 = 113 s2 = 21.6

For two-tail test, /2 = .05 Critical t.05,38 = 1.684 (used df=40) t =

t =

( x1  x 2 )  ( 1   2 ) s1 (n1  1)  s 2 (n2  1) 1 1  n1  n2  2 n1 n2 2

(118  113)  (0) (23.9) (19)  (21.6) 2 (19) 1 1  20  20  2 20 20 2

= 0.69

Since the observed t = 0.69 < t.05,38 = 1.684, the decision is to fail to reject the null hypothesis.

s (n  1)  s 2 (n2  1) 1 1 ( x1  x 2 )  t 1 1  n1  n2  2 n1 n2 2

(118 – 113) + 1.684

(23.9) 2 (19)  (21.6) 2 (19) 1 1  20  20  2 20 20

5 + 12.130 -7.13 < 1 - 2 < 17.13

10.13

Ho: µ1 - µ2 = 0 Ha: µ1 - µ2 > 0

 = .05

Sample 1 n1 = 10 x 1 = 45.38 s1 = 2.357

Sample 2 n2 = 10 x 2 = 40.49 s2 = 2.355

df = n1 + n2 - 2 = 10 + 10 - 2 = 18

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

For one-tail test,  = .05 t =

t =

Critical t.05,18 = 1.734

( x1  x 2 )  ( 1   2 ) s1 (n1  1)  s 2 (n2  1) 1 1  n1  n2  2 n1 n2 2

(45.38  40.49)  (0)

= 4.64

(2.357) 2 (9)  (2.355) 2 (9) 1 1  10  10  2 10 10

Since the observed t = 4.64 > t.05,18 = 1.734, the decision is to reject the null hypothesis. The sample data support the belief that population 1 has a greater mean than population 2.

10.14

 =.01

Ho: µ1 - µ2 = 0 Ha: µ1 - µ2  0

df = 18 + 18 - 2 = 34

Sample 1 n1 = 18 x 1 = 5.333 s12 = 12

Sample 2 n2 = 18 x 2 = 9.444 s22 = 2.026

For two-tail test, /2 = .005

Critical t.005,34 = ±2.75 (used

df=30) t =

t =

( x 1  x 2 )  ( 1   2 ) s1 (n1  1)  s 2 (n2  1) 1 1  n1  n2  2 n1 n2 2

(5.333  9.444)  (0) = -4.66 12(17)  (2.026)17 1 1  18  18  2 18 18

Since the observed t = -4.66 < t.005,34 = -2.75, we reject the null hypothesis. For  =.01, the average values for populations 1 and 2 are different.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

For 98% confidence, t.01, 30 = 2.457 s1 (n1  1)  s 2 (n2  1) 1 1  n1  n2  2 n1 n2 2

( x1  x 2 )  t

(5.333 – 9.444) + 2.457

(12)(17)  (2.026)(17) 1 1  18  18  2 18 18

-4.111 + 2.169 -6.280 < 1 - 2 < -1.942

10.15

Montreal n1 = 21 x1 = 328,000 s1 = 14,900

Halifax n2 = 26 x 2 = 331,000 s2 = 13,700

df = 21 + 26 – 2 = 45

90% level of confidence, /2 = .05 1.684  1.676 t .05,45 = = 1.680 (used the average of df = 40 and 2 df=50) s (n  1)  s 2 (n2  1) 1 1 ( x1  x 2 )  t 1 1  n1  n2  2 n1 n2 2

(328,000 – 331,000) ± 1.680 (√(14,9002(20)+13,7002(25))/45) (√(1/21)+(1/26))

= -3,000±7,021.81 -10,021.81 < 1 - 2 < 4,021.81 Since zero does fall into the 90% confidence interval, we can state at  = 0.10 that there is no difference in the mean prices.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

10.16

 = .10 df = 12 + 12 - 2 = 22

Ho: µ1 - µ2 = 0 Ha: µ1 - µ2  0 Co-op n1 = 12 x 1 = $15.645 s1 = $1.093

Interns n2 = 12 x 2 = $15.439 s2 = $0.958

For two-tail test, /2 = .05 Critical t.05,22 = ± 1.717 t =

t =

( x 1  x 2 )  ( 1   2 )

s1 (n1  1)  s 2 (n2  1) 1 1  n1  n2  2 n1 n2 2

(15.645  15.439)  (0) (1.093) 2 (11)  (0.958) 2 (11) 1 1  12  12  2 12 12

= 0.49

Since the observed t = 0.49 < t.05,22 = 1.717, the decision is to fail to reject

the null hypothesis. We can state at  = 0.10 that there is no

difference in the mean hourly wages of university co-ops and interns.

90% Confidence Interval:

s1 (n1  1)  s 2 (n2  1) 1 1  n1  n2  2 n1 n2 2

( x1  x 2 )  t

t.05,22 = ± 1.717

(15.645 – 15.439) + 1.717

(1.093) 2 (11)  (0.958) 2 (11) 1 1  = 12  12  2 12 12

0.206 + 0.7204 -0.5144 < 1 - 2 < 0.9264

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

10.17 Let Toronto be group 1 1) Ho: µ1 - µ2 = 0 Ha: µ1 - µ2 > 0 2) t =

( x 1  x 2 )  ( 1   2 ) s1 (n1  1)  s 2 (n2  1) 1 1  n1  n2  2 n1 n2 2

3)  = .05 4) For a one-tailed test and df = 8 + 9 - 2 = 15, t.05,15 = 1.753. If the observed value of t is greater than 1.753, the decision is to reject the null hypothesis. 5) Toronto n1 = 8 x 1 = 47 s1 = 3 6) t =

Montreal n2 = 9 x 2 = 44 s2 = 3

(47  44)  (0) (3) 2 7  (3) 2 8 1 1  15 8 9

= 2.06

7) Since t = 2.06 > t.05,15 = 1.753, the decision is to reject the null hypothesis. 8) The rental rates in Toronto are significantly higher than in Montreal.

10.18

nE = 22 x E = 112 sE = 11

nQ = 20 x Q = 122 sQ = 12

df = nE + nQ - 2 = 22 + 20 - 2 = 40 For a 98% Confidence Interval, /2 = .01 and s E (nE  1)  sQ (nQ  1) 2

(x E  xQ )  t

nE  nQ  2

1 1  nE nQ

t.01,40 = 2.423

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

(112 – 122) + 2.423

(11) 2 (21)  (12) 2 (19) 1 1  22  20  2 22 20

-10 ± 8.60 -$18.60 < µE - µQ < -$1.40 10.19

Point Estimate = -$10 Ho: µ1 - µ2 = 0 Ha: µ1 - µ2  0 df = n1 + n2 - 2 = 11 + 11 - 2 = 20 Perth n1 = 11 x 1 = $67,381.82 s1 = $2,067.28

Mexico City n2 = 11 x 2 = $63,390.91 s2 = $1,526.08

For a two-tail test, /2 = .005 t =

t =

Critical t.005,20 = ±2.845

( x 1  x 2 )  ( 1   2 ) s1 (n1  1)  s 2 (n2  1) 1 1  n1  n2  2 n1 n2 2

(67,381.82  63,390.91)  (0) (2,067.28) 2 (10)  (1,526.08) 2 (10) 1 1  11  11  2 11 11

= 5.15

Since the observed t = 5.15 > t.005,20 = 2.845, the decision is to reject the null

hypothesis. Thus, at  = 0.01, the executive can state that there is a difference in the average annual cost of supporting her family of four in Perth and Mexico City.

95% Confidence Interval:

s1 (n1  1)  s 2 (n2  1) 1 1  n1  n2  2 n1 n2 2

( x1  x 2 )  t

t.025,20 = 2.086 2

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

(67,381.82 – 63,390.91) + 2.086 (2,067.28) (10)  (1,526.08) 2 (10) 1 1  = 11  11  2 11 11 3,990.91 + 1,616.12 2

2,374.79 < 1 - 2 < 5,607.03

10.20

Ho: µ1 - µ2 = 0 Ha: µ1 - µ2 > 0 df = n1 + n2 - 2 = 9 + 10 - 2 = 17 Men n1 = 9 x 1 = $110.92 s1 = $28.79

Women n2 = 10 x 2 = $75.48 s2 = $30.51

This is a one-tail test,  = .01 t =

t =

Critical t.01,17 = 2.567

( x 1  x 2 )  ( 1   2 ) s1 (n1  1)  s 2 (n2  1) 1 1  n1  n2  2 n1 n2 2

(110.92  75.48)  (0) (28.79) 2 (8)  (30.51) 2 (9) 1 1  9  10  2 9 10

= 2.60

Since the observed t = 2.60 > t.01,17 = 2.567, the decision is to reject the null

hypothesis. At  = .01, the sample data support the claim that, on average, men actually spend significantly more than women on Valentine‟s Day.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

10.21 Ho: D = 0 Ha: D > 0 Sample 1 38 27 30 41 36 38 33 35 44

Sample 2 22 28 21 38 38 26 19 31 35

d 16 -1 9 3 -2 12 14 4 9

d = 7.11

n=9

sd = 6.45

 = .01

df = n - 1 = 9 - 1 = 8 For one-tail test and  = .01, t =

the critical t.01,8 = 2.896

d  D 7.11  0 = 3.31  sd 6.45 n

Since the observed t = 3.31 > t.01,8 = 2.896, the decision is to reject the null hypothesis.

10.22 Ho: Ha: Before 107 99 110 113 96 98 100 102 107 109 104

D=0 D0 After 102 98 100 108 89 101 99 102 105 110 102

d 5 1 10 5 7 -3 1 0 2 -1 2

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

99 101

96 100

3 1

d = 2.5385 n = 13 df = n - 1 = 13 - 1 = 12

sd = 3.4789

For a two-tail test and /2 = .025

 = .05

Critical t.025,12 = ±2.179

d  D 2.5385  0 = 2.63  sd 3.4789

t =

Since the observed t = 2.63 > t.025,12 = 2.179, the decision is to reject the null hypothesis.

d = 40.56

10.23 n = 22

sd = 26.58

For a 98% Level of Confidence, /2 = .01, and df = n - 1 = 22 - 1 = 21 t.01,21 = 2.518 d t

sd n

40.56 ± (2.518)

26.58 22

40.56 ± 14.27 26.29 < D < 54.83

10.24

Before 32 28 35 32 26 25

After 40 25 36 32 29 31

d -8 3 -1 0 -3 -6

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

37 16 35

39 30 31

-2 -14 4

d = -3 n=9 df = n - 1 = 9 - 1 = 8

sd = 5.6347

For 90% level of confidence and /2 = .05, d t

 = .025 t.05,8 = 1.86

sd n

-3 + (1.86)

5.6347 = -3 ± 3.49 9

-6.49 < D < 0.49

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

10.25 City

Cost

Resale

Calgary Edmonton Fredericton Halifax London, ON Montreal Regina Toronto Vancouver Victoria Winnipeg

20,427 27,255 22,115 23,256 21,887 24,255 19,852 23,624 25,885 28,999 20,836

25,163 24,625 12,600 24,588 19,267 20,150 22,500 16,667 26,875 35,333 16,292

-4,736 2,630 9,515 -1,332 2,620 4,105 -2,648 6,957 - 990 -6,334 4,544

d = 1,302.82

sd = 4,938.22

 = .01 d t

/2 = .005

sd n

n = 11,

df = 10

t.005,10= 3.169

= 1,302.82 + 3.169

4,938.22 = 1,302.82 + 4,718.42 11

-3,415.60 < D < 6,021.24

10.26 Ho: Ha:

D=0 D<0

Before 2 4 1 3 4 2 2 3 1 n=9

After 4 5 3 3 3 5 6 4 5

d = -1.778

d -2 -1 -2 0 1 -3 -4 -1 -4 sd = 1.716

 = .05

df = n - 1 = 9

-1=8

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

For a one-tail test and  = .05, the critical t.05,8 = -1.86

t =

d  D  1.778  0 = -3.11  sd 1.716 9

Since the observed t = -3.11 < t.05,8 = -1.86, the decision is to reject the null hypothesis. At  = .05, we can state that significantly more site visits were made after the seminar.

10.27

Before 6.59 5.95 7.50 6.26 7.76 6.46 5.56 5.95 5.82 5.66 6.10

After 5.09 5.82 5.56 5.56 6.21 6.08 4.91 6.21 5.17 5.25 5.77

n = 11

d = 0.7255

d 1.50 0.13 1.94 0.70 1.55 0.38 0.65 -0.26 0.65 0.41 0.33 sd = 0.6684

df = n - 1 = 11 - 1 = 10

For a 98% level of confidence and /2=.01, t.01,10 = 2.764 d t

sd n

0.7255 ± (2.764)

0.6684 11

= 0.7255 ± 0.5570

0.1685 < D < 1.2825

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Ha: D > 0 n = 27 df = 27 – 1 = 26 Since  = .01, the critical t.01,26 = 2.479 t =

d = 3.71

sd = 5

d  D 3.71  0 = 3.86  sd 5 27

Since the observed t = 3.86 > t.01,26 = 2.479, the decision is to reject the null hypothesis. At  = .01, the sample date support the claim that the audit delays after the merger were significantly lower than before the merger.

d = 75

10.29 n = 21

sd = 30

df = 21 - 1 = 20

For a 90% confidence level, /2=.05 and t.05,20 = 1.725 d t

sd n

75 + 1.725

30 = 75 ± 11.29 21

63.71 < D < 86.29

10.30 Ho: Ha:

D=0 D0

d = 1.32

n = 15

sd = 2.08

 = .01

df = 15 - 1

= 14 For a two-tail test, /2 = .005 and the critical t.005,14 = + 2.977 t =

d  D 1.32  0 = 2.46  sd 2.08 n

Since the observed t = 2.46 < t.005,14 = 2.977 and t = 2.46 > t.005,14 = -2.977, we fail Solutions Manual 1-357 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

to reject the null hypothesis. At  = .01, there is no sufficient sample evidence to support the idea that there is a significant difference in fuel efficiency between regular unleaded and premium unleaded gas. 10.31 a)

Sample 1 n1 = 368 x1 = 175

Sample 2 n2 = 405 x2 = 182

pˆ 1 

x1 175  = .476 n1 368

p

x1  x 2 175  182 357   = .462 n1  n2 368  405 773

pˆ 2 

x 2 182  = .449 n2 405

Ho: p1 - p2 = 0 Ha: p1 - p2  0 For two-tail, /2 = .025 and z.025 = ±1.96

z

( pˆ 1  pˆ 2 )  ( p1  p 2 ) 1 1 p  q    n1 n2 



(.476  .449)  (0) 1   1 (.462)(.538)    368 405 

= 0.75

Since the observed z = 0.75 < zc = 1.96, the decision is to fail to reject the null hypothesis.

Sample 1 p̂ 1 = .38 n1 = 649

Sample 2 p̂ 2 = .25 n2 = 558

p

n1 pˆ 1  n2 pˆ 2 649(.38)  558(.25)  = .32 n1  n2 649  558

Ho: Ha:

p1 - p2 = 0 p1 - p2 > 0

For a one-tail test and  = .10,

zc = z.10 = 1.28

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

z

10.32

( pˆ 1  pˆ 2 )  ( p1  p 2 ) 1 1 p  q    n1 n2 



(.38  .25)  (0)

= 4.83

1   1 (.32)(.68)    649 558 

Since the observed z = 4.83 > zc = 1.28, the decision is to reject the null hypothesis. a) n1 = 85 n2 = 90 p̂ 1 = .75 p̂ 2 = .67 For a 90% Confidence Level, z.05 = 1.645

( pˆ 1  pˆ 2 )  z

pˆ 1 qˆ1 pˆ 2 qˆ 2  n1 n2

(.75 - .67) ± 1.645

(.75)(.25) (.67)(.33)  = .08 ± .11 85 90

-.03 < p1 - p2 < .19

n1 = 1,100

n2 = 1,300

p̂ 1 = .19

p̂ 2 = .17

For a 95% Confidence Level, /2 = .025 and z.025 = 1.96

( pˆ 1  pˆ 2 )  z

pˆ 1 qˆ1 pˆ 2 qˆ 2  n1 n2

(.19 - .17) + 1.96

(.19)(.81) (.17)(.83) = .02 ± .03  1,100 1,300

-.01 < p1 - p2 < .05

n1 = 430

pˆ 1 

n2 = 399

x1 275  = .64 n1 430

x1 = 275

pˆ 2 

x2 = 275

x 2 275  = .69 n2 399

For an 85% Confidence Level, /2 = .075 and

z.075 = 1.44

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

pˆ 1 qˆ1 pˆ 2 qˆ 2  n1 n2

( pˆ 1  pˆ 2 )  z

(.64)(.36) (.69)(.31)  = -.05 ± .047 430 399

(.64 - .69) + 1.44

-.097 < p1 - p2 < -.003 n1 = 1,500 n2 = 1,500 pˆ 1 

x1 = 1,050

x2 = 1,100

x1 1,050 x 1,100 = .70 pˆ 2  2  = .733  n1 1,500 n2 1,500

For an 80% Confidence Level, /2 = .10 and z.10 = 1.28

pˆ 1 qˆ1 pˆ 2 qˆ 2  n1 n2

( pˆ 1  pˆ 2 )  z

(.70 - .733) ± 1.28

(.70)(.30) (.733)(.267) = -.033 ± .021  1,500 1,500

-.054 < p1 - p2 < -.012

10.33 H0: pm - pw = 0 Ha: pm - pw < 0 pˆ m  0.59 pˆ w  0.70

374

For a one-tailed test and  = .05,

p

z

481

z.05 = -1.645

nm pˆ m  n w pˆ w 374(.59)  481(.70)  = .652 nm  nw 374  481

( pˆ 1  pˆ 2 )  ( p1  p 2 ) 1 1 p  q    n1 n2 



(.59  .70)  (0) 1   1 (.652)(.348)    374 481 

= -3.35

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Since the observed z = -3.35 < z.05 = -1.645, the decision is to reject the null hypothesis. At 5% level of significance, the sample data support the claim that a significantly higher proportion of women than men believe that weight is an extremely/very important factor in purchasing a laptop computer.

10.34 n1 = 210

p̂1 = .24

n2 = 176

p̂2 = .35

For a 90% Confidence Level, /2 = .05 and

( pˆ 1  pˆ 2 )  z

z.05 = + 1.645

pˆ 1 qˆ1 pˆ 2 qˆ 2  n1 n2

(.24 - .35) + 1.645

(.24)(.76) (.35)(.65)  = -.11 + .0765 210 176

-.1865 < p1 – p2 < -.0335

10.35 Computer Firms p̂ 1 = .48 n1 = 56

Banks p̂ 2 = .56

n2 = 89

p

n1 pˆ 1  n2 pˆ 2 56(.48)  89(.56)  = .529 n1  n2 56  89

Ho: Ha:

p1 - p2 = 0 p1 - p2  0

For two-tail test, /2 = .10 and zc = ±1.28

z

( pˆ 1  pˆ 2 )  ( p1  p2 ) 1 1 p  q    n1 n2 



(.48  .56)  (0) 1   1 (.529)(.471)    56 89 

= -0.94

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Since the observed z = -0.94 > zc = -1.28 and z = -0.94 < zc = 1.28 , the decision is to fail to reject the null hypothesis. At a level of significance of 0.20, there is not sufficient sample evidence to support the claim that there is a significant difference in the responses.

10.36

A n1 = 35 x1 = 5

pˆ 1 

B n2 = 35 x2 = 7

x1 5  = .14 n1 35

pˆ 2 

x2 7  = .20 n2 35

For a 98% Confidence Level, /2 = .01 and z.01 = 2.33

( pˆ 1  pˆ 2 )  z

pˆ 1 qˆ1 pˆ 2 qˆ 2  n1 n2

(.14 - .20) ± 2.33

(.14)(.86) (.20)(.80)  35 35

= -.06 ± .21

-.27 < p1 - p2 < .15

10.37 H0: p1 – p2 = 0 Ha: p1 – p2  0

 = .10

p̂ 2 = .06

n1 = 780

For a two-tailed test, /2 = .05 and

z.05 = + 1.645

p

z

p̂1 = .09

n2 = 915

n1 pˆ 1  n2 pˆ 2 780(.09)  915(.06)  = .0738 n1  n2 780  915 ( pˆ 1  pˆ 2 )  ( p1  p2 ) 1 1 p  q    n1 n2 



(.09  .06)  (0) 1   1 (.0738)(.9262)    780 915 

= 2.35

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Since the observed z = 2.35 > z.05 = 1.645, the decision is to reject the null hypothesis.

At  = .10, the sample data support the claim that there is a significant difference between small businesses and all businesses on the issue.

10.38 n1 = 850

p̂1 = .60

n2 = 910

p̂ 2 = .52

For a 95% Confidence Level, /2 = .025 and z.025 = + 1.96

( pˆ 1  pˆ 2 )  z

pˆ 1 qˆ1 pˆ 2 qˆ 2  n1 n2

(.60 - .52) + 1.96

(.60)(.40) (.52)(.48)  = .08 + .046 850 910

.034 < p1 – p2 < .126

10.39 H0: 12 = 22 Ha: 12 < 22

 = .01

dfnum = 10 - 1 = 9 Since F.99, 9,11  F.01,11, 9 

s12 = 562 s22 = 1,013

dfdenom = 12 - 1 = 11

1 F.01,11, 9

F.01,10, 9  F.01,12, 9

n1 = 10 n2 = 12



and from Table A.7

5.26  5.11  5.185 2

1  0.193 5.185 2 s1 562  F = = .5548 2 1,013 s2 F.99, 9,11 

Since the observed F = .5548 >F.99,9,11 = .193, the decision is to fail to reject the null hypothesis.

10.40 H0: 12 = 22

 = .05

n1 = 5

s1 = 4.68

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Ha: 12  22

n2 = 19

dfnum = 5 - 1 = 4

dfdenom = 19 - 1 = 18

The critical table F values are:

F =

 2

s2 = 2.78

F.025,4,18 = 3.61

F..975,18,4 = .277

(4.68) 2 = 2.83 (2.78) 2

Since the observed F = 2.83 < F.025,4,18 = 3.61 and F = 2.83 > F..975,18, 4 = .277 , the decision is to fail to reject the null hypothesis. 10.41

City 1

City 2

1.338 1.326 1.322 1.303 1.295 1.322 1.318 1.318 1.318 1.278

1.299 1.334 1.322 1.310 1.287 1.349 1.322 1.342 1.314 1.318

n1 = 10

df1 = 9

n2 = 10

df2 = 9

s12 = 0.0002971

s22 = 0.0003553

H0: 12 = 22 Ha: 12  22

 = .01 /2 = .005

Upper tail critical F value = F.005,9,9 = 6.54 Lower tail critical F value = F.995,9,9 = 0.153 F =



.0002971 = 0.8362 .0003553

Since the observed F = 0.8362 is greater than the lower tail critical value of 0.153 and less than the upper tail critical value of 6.54, the decision is to fail to reject the null hypothesis. At  = .01, there is not sufficient Solutions Manual 1-364 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

evidence to support the claim that there is a significant difference in variances of the prices of unleaded regular gas between two cities.

10.42 Let Thunder Bay = group 1 and Moncton = group 2 1) H0: 12 = 22 Ha: 12  22 2 s1 2) F = 2 s2 3)  = .01 4) df1 = 12

df2 = 10 This is a two-tailed test

The critical table F values are:

F.005,12,10 = 5.66

F.995,10,12 = .177

If the observed value is greater than 5.66 or less than .177, the decision will be to reject the null hypothesis. 5) s12 = 393.4 6) F =

s22 = 702.7

393.4 = 0.56 702.7

7) Since F = 0.56 is greater than .177 and less than 5.66, the decision is to fail to reject the null hypothesis. 8) There is no significant difference in the variances of number of days between Thunder Bay and Moncton.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

10.43 Let Nova Scotia households = group 1 and Alberta households = group 2. H0: 12 = 22 Ha: 12 > 22

 = .05

dfnum = 12 - 1 = 11

dfdenom = 15 - 1 = 14

n1 = 12 n2 = 15

s1 = 7.52 s2 = 6.08

The critical table F value is F.05,11,14 F  F.05,12,14 2.60  2.53  .05,10,14   2.565 2 2 F=

s1 s2

 2

(7.52) 2 = 1.53 (6.08) 2

Since the observed F = 1.53 < F.05,11,14 = 2.565, the decision is to fail to reject the null hypothesis. At  = .05, there is not sufficient sample evidence to conclude that the variance of annual hotdog wiener purchases for Nova Scotia households is greater than the variance of annual hotdog wiener purchases for Alberta households.

10.44 Let public service workers = group 1 and private-sector workers = group 2. H0: 12 = 22 Ha: 12  22

 = .01

dfnum = 15 - 1 = 14

n1 = 15 n2 = 15

s12 = 91.5 s22 = 67.3

dfdenom = 15 - 1 = 14

The critical table F values are: F.005,14,14  F.005,15,14 = 4.25 and F.995,15,14  1/ 4.25 = .235 F =



91.5 = 1.36 67.3

Since the observed F = 1.36 < F.005,14,14 = 4.25 and > F.995,15,14 = ..235 , the decision is to fail to reject the null hypothesis. At  = .01, there is not sufficient sample evidence to support the claim that there is a difference in the variation of ages of men in the public service and men in the private-sector. Solutions Manual 1-366 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

10.45 Ho: Ha:

µ1 - µ2 = 0 µ1 - µ2  0

For  = .10 and a two-tailed test, /2 = .05 and z.05 = + 1.645 Sample 1 x 1 = 138.4 1 = 6.71 n1 = 48 z =

Sample 2 x 2 = 142.5 2 = 8.92 n2 = 39

( x 1  x 2 )  ( 1   2 )

 12



 22 n2



(138.4  142.5)  (0) (6.71) 2 (8.92)  48 39

= -2.38

Since the observed value of z = -2.38 is less than the critical value of z = 1.645, the decision is to reject the null hypothesis. There is a significant difference in the means of the two populations.

10.46 Sample 1 x 1 = 34.9 12 = 2.97 n1 = 34

Sample 2 x 2 = 27.6 22 = 3.50 n2 = 31

For 98% CI, z.01 = 2.33 ( x1  x 2 )  z

 12 n1



(34.9 – 27.6) + 2.33

 22 n2

2.97 3.50  34 31

= 7.3 + 1.04

6.26 < 1 - 2 < 8.34

10.47 Ho: µ1 - µ2 = 0 Ha: µ1 - µ2 > 0 Solutions Manual 1-367 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Sample 1 x 1= 2.06 s12 = .176 n1 = 12

Sample 2 x 2 = 1.93 s22 = .143 n2 = 15

 = .05

This is a one-tailed test with df = 12 + 15 - 2 = 25. The critical value is t.05,25 = 1.708. If the observed value is greater than 1.708, the decision will be to reject the null hypothesis. t =

t =

( x 1  x 2 )  ( 1   2 ) s1 (n1  1)  s 2 (n2  1) 1 1  n1  n2  2 n1 n2 2

(2.06  1.93)  (0) (.176)(11)  (.143)(14) 1 1  25 12 15

= 0.85

Since the observed value of t = 0.85 is less than the critical value of t = 1.708, the decision is to fail to reject the null hypothesis. The mean for population one is not significantly greater than the mean for population two.

10.48

Sample 1 x 1 = 74.6 s12 = 10.5 n1 = 18

Sample 2 x 2 = 70.9 s22 = 11.4 n2 = 19

For 95% confidence, /2 = .025. Using df = 18 + 19 - 2 = 35 and Table A.6, t t 2.042  2.021 t.025,35  .025,30 .025, 40   2.03 2 2 s (n  1)  s 2 (n2  1) 1 1 ( x1  x 2 )  t 1 1  n1  n2  2 n1 n2 2

(74.6 – 70.9) + 2.03

(10.5)(17)  (11.4)(18) 1 1  18  19  2 18 19

3.7 + 2.21 1.49 < 1 - 2 < 5.91 Solutions Manual 1-368 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

 = .01

10.49 Ho: D = 0 Ha: D < 0 n = 21

d = -1.16

df = 20

sd = 1.01

The critical t.01,20 = -2.528. If the observed t is less than -2.528, then the decision will be to reject the null hypothesis. t =

d  D  1.16  0 = -5.26  sd 1.01 21

Since the observed value of t = -5.26 is less than the critical t value of 2.528, the decision is to reject the null hypothesis. The population difference is less than zero.

10.50

Respondent 1 2 3 4 5 6 7 8 9

Before 47 33 38 50 39 27 35 46 41

After 63 35 36 56 44 29 32 54 47

d -16 -2 2 -6 -5 -2 3 -8 -6

d = -4.44 sd = 5.703 df = 8 For a 99% Confidence Level, /2 = .005 and t.005,8 = 3.355 d t

sd n

= -4.44 + 3.355

5.703 = -4.44 + 6.38 9

-10.82 < D < 1.94

10.51 Ho: p1 - p2 = 0 Ha: p1 - p2  0

 = .05

/2 = .025

z.025 = + 1.96

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

If the observed value of z is greater than 1.96 or less than -1.96, then the decision will be to reject the null hypothesis. Sample 1 x1 = 345 n1 = 783

Sample 2 x2 = 421 n2 = 896

p

x1  x 2 345  421  = .4562 n1  n2 783  896

pˆ 1 

x1 345  = .4406 n1 783

z

( pˆ 1  pˆ 2 )  ( p1  p2 ) 1 1 p  q    n1 n2 

pˆ 2 



x 2 421  = .4699 n2 896

(.4406  .4699)  (0) 1   1 (.4562)(.5438)    783 896 

= -1.20

Since the observed value of z = -1.20 is greater than -1.96 and less than 1.96, the decision is to fail to reject the null hypothesis. There is no significant difference in two population proportions.

10.52 Sample 1 n1 = 409 p̂ 1 = .71

Sample 2 n2 = 378 p̂ 2 = .67

For a 99% Confidence Level, /2 = .005 and

( pˆ 1  pˆ 2 )  z

z.005 = 2.575

pˆ 1 qˆ1 pˆ 2 qˆ 2  n1 n2

(.71 - .67) + 2.575

(.71)(.29) (.67)(.33)  = .04 ± .085 409 378

-.045 < p1 - p2 < .125 10.53 H0: 12 = 22 Ha: 12  22 dfnum = 8 - 1 = 7

 = .05

n1 = 8 n2 = 10

s12 = 46 s22 = 37

dfdenom = 10 - 1 = 9

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

The critical F values are:

F.025,7,9 = 4.20

F.975,9,7 = .238

If the observed value of F is greater than 4.20 or less than .238, then the decision will be to reject the null hypothesis. F =



46 = 1.24 37

Since the observed F = 1.24 is less than F.025,7,9 =4.20 and greater than F.975,9,7 = .238, the decision is to fail to reject the null hypothesis. There is no significant difference in the variances of the two populations.

10.54

Term x t = $75,000 st = $22,000 nt = 27

Whole Life x w = $45,000 sw = $15,500 nw = 29

df = 27 + 29 - 2 = 54 For a 95% Confidence Level, /2 = .025 and t.025,54  t.025,50 = 2.009 s t (nt  1)  s w (nw  1) 1 1  nt  nw  2 nt nw 2

(xt  xw )  t

(75,000 – 45,000) + 2.009

(22,000) 2 (26)  (15,500) 2 (28) 1 1  27  29  2 27 29

30,000 ± 10,160.11 19,839.89 < µ1 - µ2 < 40,160.11 10.55 Morning Afternoon 43 41 51 49 37 44 24 32 47 46 44 42 50 47 55 51 46 49 n=9

d = -0.444

For a 90% Confidence Level:

d 2 2 -7 -8 1 2 3 4 -3 sd = 4.447

df = 9 - 1 = 8

/2 = .05 and t.05,8 = 1.86

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

d t

-0.444 + (1.86)

4.447 = -0.444 ± 2.757 9

-3.201 < D < 2.313

10.56 Marketing (group 1) n1 = 400 x1 = 220 Ho: p1 - p2 = 0 Ha: p1 - p2 > 0

Accountants (group 2) n2 = 450 x2 = 216

 = .01

The critical table z value is:

z.01 = 2.33

220 = .55 400

216 = .48 450

p̂1 = p

z

p̂2 =

x1  x 2 220  216  = .513 n1  n2 400  450 ( pˆ 1  pˆ 2 )  ( p1  p2 ) 1 1 p  q    n1 n2 



(.55  .48)  (0) 1   1 (.513)(.487)    400 450 

= 2.04

Since the observed z = 2.04 is less than z.01 = 2.33, the decision is to fail to reject the null hypothesis. There is not sufficient sample evidence to support the claim that a greater proportion of marketing managers keep track of obligations in their head than do accountants.

10.57

Accounting Clerks (group 1)

Data Entry Operators (group

n1 = 16 x 1 = 36,400 s1 = 2,200

n2 = 14 x 2 = 35,800 s2 = 2,050

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Ha: 12  22

 = .05

and 

dfnum = 16 – 1 = 15

dfdenom = 14 – 1 = 13

The critical F values are:

F.025,15,13 = 3.05

F.975,13,15 = 0.33

F = = s12/s22 = 4,840,000/4,202,500 = 1.15 Since the observed F = 1.15 is less than F.025,15,13 = 3.05 and greater than F.975,13,15 = 0.33, the decision is to fail to reject the null hypothesis. At  = .05, there is not sufficient sample evidence to warrant rejection of the claim that the population variance of salaries are the same for accounting clerks as it is for data entry operators.

10.58

Men

Women

n1 = 60 x 1 = 631 1 = 100

n2 = 41 x 2 = 848 2 = 100

For a 95% Confidence Level, /2 = .025 and z.025 = 1.96 ( x1  x 2 )  z

 12 n1



 22 n2

1002 1002  (631 – 848) + 1.96 = -217 ± 39.7 60 41

10.59 Ho: Ha:

-256.7 < µ1 - µ2 < -177.3 µ1 - µ2 = 0  = .01 µ1 - µ2  0 df = 20 + 24 - 2 = 42 Guelph (group 1)

Sudbury (group 2)

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

n1 = 20 x1 = 7.97 s1 = 1.5

n2 = 24 x 2 = 6.7 s2 = 1.2

For two-tail test, /2 = .005 and the critical t.005,42  ±2.704 (used df=40) t =

t =

( x1  x 2 )  ( 1   2 ) s1 (n1  1)  s2 (n2  1) 1 1  n1  n2  2 n1 n2 2

(7.97  6.7)  (0) (1.5) (19)  (1.2) (23) 1 1  42 20 24 2

= 3.12

Since the observed t = 3.12 > t.005,42  2.704, the decision is to reject the null

hypothesis. At  = .01, the sample data support the claim that there is a difference in the mean mass of turkeys sold in Guelph and Sudbury.

10.60

With Fertilizer x 1 = 97.5 1 = 24.9 n1 = 35 Ho: Ha:

Without Fertilizer x 2 = 58.7 2 = 18.8 n2 = 35

µ1 - µ2 = 0 µ1 - µ2 > 0

For one-tail test,  = .01 and z.01 = 2.33 z =

( x 1  x 2 )  ( 1   2 )

1



2



(97.5  58.7)  (0) (24.9) 2 (18.8) 2  35 35

= 7.36

Since the observed z = 7.36 > z.01 = 2.33, the decision is to reject the null hypothesis. At  = .01, the sample data support the theory that the population of trees with the fertilizer grew significantly larger than the nonfertilized trees.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

10.61

Specialty n1 = 350 p̂ 1 = .75

Discount n2 = 500 p̂ 2 = .52

 = 0.10, /2 = .05 and z.05 = ±1.645 p

n1 pˆ 1  n2 pˆ 2 350(.75)  500(.52)   .615 n1  n2 350  500

Ho: p1 - p2 = 0 Ha: p1 - p2  0

z

( pˆ 1  pˆ 2 )  ( p1  p 2 ) 1 1 p  q    n1 n2 



(.75  .52)  (0) 1   1 (.615)(.385)    350 500 

 6.78

Since the observed z = 6.78 > zc = 1.645, the decision is to reject the null hypothesis. For  = 0.10, there is a significant difference between the proportion of shoppers at specialty stores and the proportion of shoppers at discount stores who say that quality of merchandise is a determining factor in their perception of a store‟s image.

10.62 Let 8 A.M. to 4 P.M. data = group 1 and 4 P.M. to Midnight data = group 2. H0: 12 = 22 Ha: 12  22 dfnum = 7

F =

n1 = 8 n2 = 7 2 s1 = 72,909 s2 = 129,569 2

dfdenom = 6

The critical F values are:

 = .01



F.005,7,6 = 10.79

F.995,6,7 = .09

72,909 = .56 129,569

Since F = .56 < F.005,7,6 = 10.79 and F = .56 > F.995,6,7 = .09, the decision is to fail to reject the null hypothesis. There is no difference in the variances of productivity for the shifts.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

10.63

Name Brand 54 55 59 53 54 61 51 53

Store Brand 49 50 52 51 50 56 47 49

d = 4.5

n=8

d 5 5 7 2 4 5 4 4

sd = 1.414

For a 90% Confidence Level, /2 = .05 and d t

df = 8 - 1 = 7 t.05,7 = 1.895

sd n

4.5 + 1.895

1.414 = 4.5 ± .947 8

3.553 < D < 5.447

10.64 Ho: Ha:

 = .01

µ1 - µ2 = 0 µ1 - µ2 < 0

df = 23 + 19 - 2 = 40

Alberta n1 = 23 x 1 = 20.917 s12 = 3.104

Saskatchewan n2 = 19 x 2 = 22.074 s22 = 1.422

For one-tail test,  = .01 and the critical t.01,40 = -2.423 t =

( x 1  x 2 )  ( 1   2 ) s1 (n1  1)  s 2 (n2  1) 1 1  n1  n2  2 n1 n2 2

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

t =

(20.917  22.074)  (0) (3.104)(22)  (1.422)(18) 1 1  40 23 19

= -2.44

Since the observed t = -2.44 < t.01,40 = -2.423, the decision is to reject the null hypothesis. At  = .01, the sample data support the claim that the average temperature of a house in Saskatchewan is significantly higher.

10.65

Wednesday 71 56 75 68 74

Friday 53 47 52 55 58

d = 15.8

n=5

sd = 5.263

d 18 9 23 13 16 df = 5 - 1 = 4

 = .05

Ho: D = 0 Ha: D > 0

For one-tail test,  = .05 and the critical t.05,4 = 2.132 t =

d  D 15.8  0 = 6.71  sd 5.263 n

Since the observed t = 6.71 > t.05,4 = 2.132, the decision is to reject the null

hypothesis. At  = .05, the sample data support the claim that productivity

drops on Friday.

10.66 Ho: p1 - p2 = 0 Ha: p1 - p2  0 Machine 1 x1 = 38 n1 = 191

 = .05 Machine 2 x2 = 21 n2 = 202

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

pˆ 1 

p

x1 38  = .199 n1 191

pˆ 2 

x2 21  = .104 n2 202

n1 pˆ 1  n2 pˆ 2 (.199)(191)  (.104)(202)  = .15 n1  n2 191  202

For two-tail, /2 = .025 and the critical z values are:

z

( pˆ 1  pˆ 2 )  ( p1  p2 ) 1 1 p  q    n1 n2 

(.199  .104)  (0)



1   1 (.15)(.85)    191 202 

z.025 = ±1.96

= 2.64

Since the observed z = 2.64 > zc = 1.96, the decision is to reject the null hypothesis. At  = .05, there is a significant difference in the proportion of sheets drilled with defective holes between machine 1 and machine 2.

10.67 Construction n1 = 338 x1 = 297

pˆ 1 

Telephone Repair n2 = 281 x2 = 192

x1 297  = .879 n1 338

pˆ 2 

x 2 192  = .683 n2 281

For a 90% Confidence Level, /2 = .05 and z.05 = 1.645

( pˆ 1  pˆ 2 )  z

pˆ 1 qˆ1 pˆ 2 qˆ 2  n1 n2

(.879 - .683) + 1.645

(.879)(.121) (.683)(.317)  = .196 ± .054 338 281

.142 < p1 - p2 < .250

10.68

Aerospace n1 = 33

Automobile n2 = 35

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

x 1 = 12.4 1 = 2.9

x 2 = 4.6 2 = 1.8

For a 99% Confidence Level, /2 = .005 and z.005 = 2.575 ( x1  x 2 )  z

 12 n1



 22 n2

(2.9) 2 (1.8) 2  (12.4 – 4.6) + 2.575 33 35

= 7.8 ± 1.52

6.28 < µ1 - µ2 < 9.32

10.69

Discount x 1 = $47.20 1 = $12.45 n1 = 60

Specialty x 2 = $27.40 2 = $9.82 n2 = 40

 = .01

Ho: µ1 - µ2 = 0 Ha: µ1 - µ2  0

For two-tail test, /2 = .005 and zc = ±2.575 z =

( x 1  x 2 )  ( 1   2 )

 12 n1



 22 n2



(47.20  27.40)  (0) (12.45) 2 (9.82) 2  60 40

= 8.86

Since the observed z = 8.86 > zc = 2.575, the decision is to reject the null hypothesis. At  = .01, there is a significant difference in the average size of purchases at these stores.

10.70

Before 12 7 10 16 8

After 8 3 8 9 5

d 4 4 2 7 3

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

d = 4.0

n=5

sd = 1.8708

df = 5 - 1 = 4

 = .01

Ho: D = 0 Ha: D > 0

For one-tail test,  = .01 and the critical t.01,4 = 3.747 t =

d  D 4.0  0 = 4.78  sd 1.8708 5

Since the observed t = 4.78 > t.01,4 = 3.747, the decision is to reject the null hypothesis. At  = .01, the sample data support the quality-control consulting group claim.

 = .01

10.71 Ho: µ1 - µ2 = 0 Ha: µ1 - µ2  0

df = 10 + 6 - 2 = 14

A n1 = 10 x 1 = 18.3 s12 = 17.122

B___ n2 = 6 x 2 = 9.667 s22 = 7.467

For two-tail test, /2 = .005 and the critical t.005,14 = ±2.977 t =

t =

( x1  x 2 )  ( 1   2 ) s1 (n1  1)  s2 (n2  1) 1 1  n1  n2  2 n1 n2 2

(18.3  9.667)  (0) = 4.52 (17.122)(9)  (7.467)(5) 1 1  14 10 6

Since the observed t = 4.52 > t.005,14 = 2.977, the decision is to reject the null hypothesis. At  = .01, there is a significant difference in the average number of pages dedicated to advertising.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

10.72 A t test was used to determine if , on average, Hong Kong has significantly different rental rates than Mumbai. Let group 1 be Hong Kong and group 2 be Mumbai. Ho: Ha:

µ1 - µ2 = 0 µ1 - µ2  0

x 1 = 130.4 x 2 = 128.4 n1 = 19 n2 = 23 s12 = 166.41 s22 = 193.21 98% C.I. and /2 = .01 t = 0.48 with a p-value of .634 which is greater than  = .02. There is not enough evidence in these data to declare that there is a difference in the average rental rates of the two cities.

10.73 H0: D = 0 Ha: D  0 This is a related measure before and after study. Fourteen people were involved in the study. Before the treatment, the sample mean was 4.357 and after the treatment, the mean was 5.214. The higher number after the treatment indicates that subjects were more likely to “blow the whistle” after having been through the treatment. The observed t value was –3.12 which was more extreme than two-tailed table t value of + 2.16 and as a result, the researcher rejects the null hypothesis. This is underscored by a p-value of .0081 which is less than  = .05. The study concludes that there is a significantly higher likelihood of “blowing the whistle” after the treatment. 10.74 Ho: p1 - p2 = 0 Ha: p1 - p2  0 The point estimates from the sample data indicate that in the Alberta city the market share is .310782 and in the Ontario city the market share is .270130. The point estimate for the difference in the two proportions of market share is .0406524. Since the 99% confidence interval ranges from -.0393623 to +.120667 and zero is in the interval, any hypothesis testing decision based on this interval would result in failure to reject the null hypothesis. Alpha is .01 with a two-tailed test. This is underscored by an observed z value of 1.31 which has an associated p-value of .191 which, of course, is not significant for any of the usual values of . Solutions Manual 1-381 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

10.75 A test of differences of the variances of the populations of the two machines is being computed. The hypotheses are: H0: 12 = 22 Ha: 12 > 22 Twenty-six pipes were measured for sample one and twenty-eight pipes were measured for sample two. The observed F = 2.0575 is significant at  = .05 for a one-tailed test since the associated p-value is .0348. The variance of pipe lengths for machine 1 is significantly greater than the variance of pipe lengths for machine 2.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Legal Notice Copyright

SOLUTIONS TO PROBLEMS IN CHAPTER 11: ANALYSIS OF VARIANCE AND DESIGN OF EXPERIMENTS

11.1

a) Time Period, Market Condition, Day of the Week, Season of the Year b) Time Period - 4 P.M. to 5 P.M. and 5 P.M. to 6 P.M. Market Condition - Bull Market and Bear Market

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Day of the Week - Monday, Tuesday, Wednesday, Thursday, Friday Season of the Year – Summer, Winter, Fall, Spring c) Volume, Value of the Dow Jones Average, Earnings of Investment Houses

11.2

a) Type of 737, Age of the plane, Number of Landings per Week of the plane, City that the plane is based b) Type of 737 - Type I, Type II, Type III Age of plane - 0-2 years, 3-5 years, 6-10 years, over 10 years Number of Flights per Week - 0-5, 6-10, over 10 City - Toronto, Montreal, Calgary, Vancouver c) Average annual maintenance costs, Number of annual hours spent on maintenance

11.3

a) Type of Card, Age of User, Economic Class of Cardholder, Geographic

Region b) Type of Card - Mastercard, Visa, Discover, American Express Age of User - 21-25 , 26-32, 33-40, 41-50, over 50 years old Economic Class - Lower, Middle, Upper Geographic Region – Atlantic Provinces, Central Canada, the Prairies, the West Coast, the North

11.4

11.5

c) Average number of card usages per person per month, Average balance due on the card, Average per expenditure per person, Number of cards possessed per person Average dollar expenditure per day/night, Age of adult registering the family, Number of days stay (consecutive)

Source Treatment Error Total

 = .05

df 2 14 16

SS MS F__ 22.20 11.10 11.07 14.03 1.00______ 36.24 Critical F.05,2,14 = 3.74

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Since the observed F = 11.07 > F.05,2,14 = 3.74, the decision is to reject the null hypothesis.

11.6

Source Treatment Error Total

df 4 18 22

SS MS F__ 93.77 23.44 15.82 26.67 1.48______ 120.43

 = .01

Critical F.01,4,18 = 4.58

Since the observed F = 15.82 > F.01,4,18 = 4.58, the decision is to reject the null hypothesis.

11.7

Source Treatment Error Total

df 3 12 15

SS MS F_ 544.25 181.4 13.00 167.5 14.0______ 711.75

 = .01

Critical F.01,3,12 = 5.95

Since the observed F = 13.00 > F.01,3,12 = 5.95, the decision is to reject the null hypothesis.

11.8

Source Treatment Error Total

df 1 12 13

SS 64.29 43.43 107.71

MS F__ 64.29 17.76 3.62______

 = .05

Critical F.05,1,12 = 4.75

Since the observed F = 17.76 > F.05,1,12 = 4.75, the decision is to reject the null hypothesis. Observed t value using t test: 1 n1 = 7 x 1 = 29

2 n2 = 7 x 2 = 24.71429

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

s1 2 = 3 t =

s22 = 4.238095

(29  24.71429)  (0) 3(6)  (4.238095)(6) 1 1  772 7 7

Also, t =

= 4.21

F  17.76 = 4.21

Since the observed t = 4.21 > t.025,12 = 2.179, the decision is to reject the null hypothesis. Hence, the results are the same.

11.9

Given C = 5 and N = 55. Source SS Treatment 583.39 Error 972.18 Total 1,555.57

df MS F__ 4 145.8475 7.50 50 19.4436______ 54

11.10 Given C = 3 and N = 17. Source Treatment Error Total

SS 29.64 68.42 98.06

df 2 14 16

MS F _ 14.820 3.03 4.887___ __

F.05,2,14 = 3.74 Since the observed F = 3.03 < F.05,2,14 = 3.74, the decision is to fail to reject the null hypothesis

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

11.11 Source Treatment Error Total

 = .01

df 3 15 18

SS .007076 .003503 .010579

MS F__ .002359 10.10 .000234________

Critical F.01,3,15 = 5.42

Since the observed F = 10.10 > F.01,3,15 = 5.42, the decision is to reject the null hypothesis. At  = .01, there is a significant difference in the average fill for these machines.

11.12 We restated data: Nova Scotia 4.05 4.15 4.00 4.10 4.15

Ontario 5.10 4.95 4.90 4.80 4.95

British Columbia 4.55 4.35 4.50 4.65 4.60

SS 1.807 0.117 1.924

MS F__ 0.9035 92.67 0.00975_________

Source Treatment Error Total

df 2 12 14

 = .01

Critical F.01,2,12 = 6.93

Since the observed F = 92.67 > F.01,2,12 = 6.93, the decision is to reject the null hypothesis. For 1% level of significance, there is a significant difference in the mean starting salaries for new accounting graduates from these three provinces.

11.13 Source Treatment Error Total

 = .05

df 2 15 17

SS 29.61 18.89 48.50

MS F___ 14.80 11.76 1.26________

Critical F.05,2,15 = 3.68

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Since the observed F = 11.76 > F.05,2,15 = 3.68, the decision is to reject the null

hypothesis. At  = .05, there is a significant difference in the evaluations according to manager level. 11.14 Source df SS MS F__ Treatment 3 456,630 152,210 11.03 Error 16 220,770 13,798_______ Total 19 677,400

 = .05

Critical F.05,3,16 = 3.24

Since the observed F = 11.03 > F.05,3,16 = 3.24, the decision is to reject the null

hypothesis. At  = .05, there is a significant difference in monthly

transportation costs for families living in these cities.

11.15 There are 4 treatment levels. The sample sizes are 18, 15, 21, and 11. The F value is 2.95 with a p-value of .04. Fcrit = 2.7555 is approximately equal to F.05, 3, 60 . So,   0.05 . Since the observed F = 2.95 > Fcrit = 2.7555, the decision is to reject the null hypothesis. There is an overall significant difference at alpha of .05. The means are 226.73, 238.79, 232.58, and 239.82.

11.16 The independent variable for this study was plant with five classification levels (the five plants). There were a total of 43 workers who participated in the study. The dependent variable was number of hours worked per week. An observed F value of 3.10 was obtained with an associated pvalue of .026595. With an alpha of .05, there was a significant overall difference in the average number of hours worked per week by plant. A cursory glance at the plant averages revealed that workers at plant 3 averaged 61.47 hours per week (highest number) while workers at plant 4 averaged 49.20 (lowest number).

11.17 Given n3 = 8

C=6 n6 = 7

MSE = .3352

x 3 = 15.85

 = .05

N = 46

x 6 = 17.21

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Since q.05,6,40 = 4.23,

HSD = 4.23

.3352  1 1     = 0.896 2 8 7

x 3  x 6  15.85  17.21 = 1.36

Since 1.36 > 0.896, there is a significant difference between the means of groups 3 and 6. 11.18 C = 4 n n1  n2  n3  n4  6 N = 4  6  24 N – C = 24 – 4 = 20  = .05 MSE = 2.389 Since q.05,4,20 = 3.96, HSD = qα, C, N-C

MSE 2.389 = (3.96) = 2.50 n 6

11.19 From problem 11.5,

C=3

MSE = 1.002381

 = .05

17 N – C = 14

n1 = 6

n2 = 5

x1=2

x 2 = 4.6

Since q.05,3,14 = 3.70, HSD = 3.70

1.002381  1 1     = 1.586 2 6 5

x1  x 2  2  4.6 = 2.6

Since 2.6 > 1.586, there is a significant difference between the means of groups 1 and 2.

11.20 From problem 11.6,

MSE = 1.481481

C=5

N = 23

N – C

= 18

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

n2 = 5

n5 = 6

x 2 = 10

 = .01

x 5 = 13

Since q.01,5,18 = 5.38, HSD = 5.38

1.481481  1 1     = 2.80 2 5 6

x 2  x 5  10  13 = 3

Since 3 > 2.80, there is a significant difference in the means of groups 2 and 5.

11.21 From problem 11.7,

N = 16

n=4

C=4

N – C = 12

MSE = 13.95833 Since q.01,4,12 = 5.50,

HSD = qα, C, N-C

MSE 13.95833 = 5.50 = 4 n

10.27 x 1 = 115.25

x 2 = 125.25

x 3 = 131.5

x 4 = 122.5

Since only x1  x 3  10.27 , x 1 and x 3 are the only pair that are significantly different using the Tukey’s HSD test.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

11.22 From problem 11.8, N – C = 14 – 2 = 12

n=7

C=2

MSE = 3.619048

x 1 = 29 and

x 2 = 24.71429

For  = .05,

q.05,2,12 = 3.08

HSD = qα, C, N-C

MSE 3.619048 = 3.08 = 2.215 7 n

N = 14

Since x 1  x 2 = 4.28571 > HSD = 2.215, there is a significant difference in means.

11.23 From problem 11.11,

C=4

MSE = .000234

n2 = 6

n3 = 5

 = .01

19 N – C = 15

n1 = 4

n4 = 4

x 1 = 4.03, x 2 = 4.001667, x 3 = 3.974, x 4 = 4.005

Since q.01,4,15 = 5.25, HSD1,2 = 5.25

.000234  1 1     = .0367 2 4 6

HSD1,3 = 5.25

.000234  1 1     = .0381 2  4 5

HSD1,4 = 5.25

.000234 = .0402 4

HSD2,3 = 5.25

.000234  1 1     = .0344 2 6 5

HSD2,4 = 5.25

.000234  1 1     = .0367 2 6 4

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

.000234  1 1     = .0381 2 5 4

HSD3,4 = 5.25

Only for pair, machine 1 and machine 3, the difference between the means is significant because x 1  x 3 = .056 > 0.0381. This is the only pair of machines, 1 and 3, produces significantly different mean fills.

11.24 Let Nova Scotia = group 1, Ontario = group 2, British Columbia = group 3. From problem 11.12,

C=3

n=5

N – C = 12

N = 15

MSE = 0.00975 For  = .01: q.01,3,12 = 5.04 MSE 0.00975 HSD = qα, C, N-C = 5.04 5 n From problem 11.12,

x 1 = 4.09

= 0.22256

x 2 = 4.94

x 3 = 4.53

x 1  x 2 = 0.85 x 1  x 3 = 0.44 x 2  x 3 = 0.41

Using Tukey's HSD, all three pairwise comparisons are significantly Different in mean starting salary figures.

11.25 From problem 11.13, C = 3 MSE = 1.259365 x 1 = 7.6

n1 = 5 x 2 = 8.8571

N = 18 n2 = 7

N – C = 15 n3 = 6

x 3 = 5.8333

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

HSD1,2 = 3.67

1.259365  1 1     = 1.705 2 5 7

HSD1,3 = 3.67

1.259365  1 1     = 1.763 2 5 6

HSD2,3 = 3.67

1.259365  1 1     = 1.620 2 7 6

x 1  x 3 = 1.767 (is significant) x 2  x 3 = 3.024 (is significant)

11.26 Let Edmonton=group 1, Toronto=group 2, Vancouver=group 3, Halifax=group 4. From problem 11.14,

n=5

MSE = 13,798.13

x 1 = 591

C=4

N – C = 16

N = 20

x 2 = 350

x 3 = 776

x4=

563 For  = .05,

q.05,4,16 = 4.05.

HSD = qα, C, N-C

MSE 13,798.13 = 4.05 = 212.76 5 n

x 1  x 2 = 241

x 1  x 3 = 185

x 1  x 4 = 28

x 2  x 3 = 426

x 2  x 4 = 213

x 3  x 4 = 213

Using Tukey's HSD = 212.76, Edmonton and Toronto (1 and 2), Toronto and Vancouver (2 and 3), Toronto and Halifax (2 and 4), Vancouver and Halifax (3 and 4) have significantly different mean costs. 11.27  = .05. There were five plants and ten pairwise comparisons. The Minitab output reveals that the only significant pairwise difference is between plant 2 and plant 3 where the reported confidence interval (0.180 to 22.460) contains the same sign throughout indicating that 0 is not in the Solutions Manual 1-393 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

interval. Since the confidence interval does not contain zero, then we are 95% confident that there is a significant difference in the pair of means. The lower and upper values for all other confidence intervals have different signs. It indicates that zero is included in the interval, and there is no difference in the means for these pairs.

11.28 H0: µ1 = µ2 = µ3 = µ4 Ha: At least one treatment mean is different from the others Source Treatment Blocks Error Total

df 3 4 12 19

SS 62.95 257.50 45.30 365.75

MS F__ 20.9833 5.56 64.3750 17.05 3.7750______

 = .05

Critical F.05,3,12 = 3.49 for treatments

For treatments, the observed F = 5.56 > F.05,3,12 = 3.49, the decision is to reject the null hypothesis.

11.29 H0: µ1 = µ2 = µ3 Ha: At least one treatment mean is different from the others Source df Treatment 2 Blocks 3 Error 6 Total 11

 = .01

SS MS F_ .001717 .000858 1.48 .076867 .025622 44.13 .003483 .000581_______ .082067 Critical F.01,2,6 = 10.92 for treatments

For treatments, the observed F = 1.48 < F.01,2,6 = 10.92 and the decision is to

fail to reject the null hypothesis. At  = .01, there is no significant

difference in the treatment levels.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Source Treatment Blocks Error Total

df 5 9 45 59

SS 2,477.53 3,180.48 11,661.38 17,319.39

MS F__ 495.506 1.91 353.387 1.36 259.142______

 = .05

Critical F.05,5,45  F.05,5,40 2.45 for treatments

For treatments, the observed F = 1.91 < Fcr = 2.45 and decision is to fail to reject the null hypothesis.

11.31 Given C = 4 and n = 7. Source Treatment Blocks Error Total

df 3 6 18 27

SS 199.48 265.24 306.59 771.31

MS F__ 66.493 3.90 44.207 2.60 17.033______

 = .01

Critical F.01,3,18 = 5.09 for treatments

For treatments, the observed F = 3.90 < F.01,3,18 = 5.09 and the decision is to fail to reject the null hypothesis.

11.32 H0: µ1 = µ2 = µ3 = µ4 Ha: At least one treatment mean is different from the others Source Treatment Blocks Error Total

df 3 9 27 39

SS 2302.5 5402.5 1322.5 9027.5

MS F__ 767.5000 15.67 600.2778 12.26 48.9815____ __

 = .05

Critical F.05,3,27 = 2.96 for treatments

For treatments, the observed F = 15.67 > F.05,3,27 = 2.96 and the decision is to

reject the null hypothesis. At  = .05, there is a significant difference in

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

safety ratings of the four motels.

11.33 H0: µ1 = µ2 = µ3 Ha: At least one treatment mean is different from the others Source Treatment Blocks Error Total

df 2 4 8 14

SS 64.5333 137.6000 16.8000 218.9333

MS F 32.2667 15.37 34.4000 16.38 2.1000_ _____

 = .01

Critical F.01,2,8 = 8.65 for treatments

For treatments, the observed F = 15.37 > F.01,2,8 = 8.65 and the decision is to

reject the null hypothesis. At  = .01, there is a significant difference in

mean responses according to political affiliation.

11.34 This is a randomized block design with 3 treatments (machines) and 5 block levels (operators). The F for treatments is 6.72 with a p-value of .019. There is a significant difference in machines at  = .05. The F for blocking effects is 0.22 with a p-value of .807. There are no significant blocking effects. The blocking effects reduced the power of the treatment effects since the blocking effects were not significant.

11.35 The p value for Phone Type, .00018, indicates that there is an overall significant difference in treatment means at alpha .001. The lengths of calls differ according to type of telephone used. The p-value for managers, .00028, indicates that there is an overall difference in block means at alpha .001. The lengths of calls differ according to Manager. The significant blocking effects have improved the power of the F test for treatments.

11.36 This is a two-way factorial design with two independent variables and one dependent variable. It is 2x4 in that there are two row treatment levels and Solutions Manual 1-396 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

four column treatment levels. Since there are three measurements per cell, interaction can be analyzed. dfrow treatment = 1

dfcolumn treatment = 3

dfinteraction = 3

dferror = 16

dftotal

= 23

11.37 This is a two-way factorial design with two independent variables and one dependent variable. It is 4x3 in that there are four row treatment levels and three column treatment levels. Since there are two measurements per cell, interaction can be analyzed. dfrow treatment = 3

dfcolumn treatment = 2

dfinteraction = 6

dferror = 12

dftotal

= 23

11.38 Source Row Column Interaction Error Total

df 3 4 12 60 79

SS 126.98 37.49 380.82 733.65 1278.94

MS F__ 42.327 3.46 9.373 0.77 31.735 2.60 12.228______

 = .05 Critical F.05,3,60 = 2.76 for rows. For rows, the observed F = 3.46 > F.05,3,60 = 2.76 and the decision is to reject the null hypothesis. Critical F.05,4,60 = 2.53 for columns. For columns, the observed F = 0.77 < F.05,4,60 = 2.53 and the decision is to fail to reject the null hypothesis. Critical F.05,12,60 = 1.92 for interaction. For interaction, the observed F = 2.60 > F.05,12,60 = 1.92 and the decision is to reject the null hypothesis. Since there is significant interaction, the researcher should exercise extreme caution in analyzing the "significant" row effects.

11.39 Source

F__

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Row Column Interaction Error Total

1 3 3 16 23

1.047 3.844 0.773 6.968 12.632

1.047 2.40 1.281 2.94 0.258 0.59 0.436______

 = .05 Critical F.05,1,16 = 4.49 for rows. For rows, the observed F = 2.40 < F.05,1,16 = 4.49 and decision is to fail to reject the null hypothesis. Critical F.05,3,16 = 3.24 for columns. For columns, the observed F = 2.94 < F.05,3,16 = 3.24 and the decision is to fail to reject the null hypothesis. Critical F.05,3,16 = 3.24 for interaction. For interaction, the observed F = 0.59 < F.05,3,16 = 3.24 and the decision is to fail to reject the null hypothesis.

11.40 Source Row Column Interaction Error Total

df 1 2 2 6 11

SS 60.750 14.000 2.000 9.500 86.250

MS F___ 60.750 38.37 7.000 4.42 1.000 0.63 1.583________

 = .01 Critical F.01,1,6 = 13.75 for rows. For rows, the observed F = 38.37 > F.01,1,6 = 13.75 and the decision is to reject the null hypothesis. Critical F.01,2,6 = 10.92 for columns. For columns, the observed F = 4.42 < F.01,2,6 = 10.92 and the decision is to fail to reject the null hypothesis. Critical F.01,2,6 = 10.92 for interaction. For interaction, the observed F = 0.63 < F.01,2,6 = 10.92 and the decision is to fail to reject the null hypothesis. Therefore, neither significant interaction nor column effects are present. However,

a significant difference in row effects is evident at  = .01.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

11.41 Source Treatment 1 Treatment 2 Interaction Error Total

df 1 3 3 24 31

SS 1.24031 5.09844 0.12094 0.46750 6.92719

MS F__ 1.24031 63.67 1.69948 87.25 0.04031 2.07 0.01948______

 = .05 Critical F.05,3,24 = 3.01 for interaction. For interaction, the observed F = 2.07 < F.05,3,24 = 3.01 and the decision is to fail to reject the null hypothesis. Because no significant interaction effects are present, it is possible to examine the main effects. Critical F.05,1,24 = 4.26 for treatment 1. For treatment 1, the observed F = 63.67 > F.05,1,24 = 4.26 and the decision is to reject the null hypothesis. Critical F.05,3,24 = 3.01 for treatment 2. For treatment 2, the observed F = 87.25 > F.05,3,24 = 3.01 and the decision is to reject the null hypothesis.

11.42 Source df Age 3 No. Children 2 Interaction 6 Error 12 Total 23

SS 42.4583 49.0833 4.9167 11.5000 107.9583

MS F__ 14.1528 14.77 24.5417 25.61 0.8194 0.86 0.9583______

 = .05 Critical F.05,3,12 = 3.49 for Age. For Age, the observed F = 14.77 > F.05,3,12 = 3.49 and the decision is to reject the null hypothesis. Critical F.05,2,12 = 3.89 for No. Children. For No. Children, the observed F = 25.61 > F.05,2,12 = 3.89 and the decision is to reject the null hypothesis. Critical F.05,6,12 = 3.00 for interaction. For interaction, the observed F = 0.86 < F.05,6,12 = 3.00 and fail to reject the null hypothesis. Solutions Manual 1-399 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

11.43 Source Location Competitors Interaction Error Total

df 2 3 6 24 35

SS 1736.22 1078.33 503.33 607.33 3925.22

MS F__ 868.11 34.31 359.44 14.20 83.89 3.32 25.31_______

 = .05 Critical F.05,2,24 = 3.40 for rows. For rows, the observed F = 34.31 > F.05,2,24 = 3.40 and the decision is to reject the null hypothesis. Critical F.05,3,24 = 3.01 for columns. For columns, the observed F = 14.20 > F.05,3,24 = 3.01 and decision is to reject the null hypothesis. Critical F.05,6,24 = 2.51 for interaction. For interaction, the observed F = 3.32 > F.05,6,24 = 2.51 and the decision is to reject the null hypothesis. Note: There is a significant interaction in this study. This may confound the interpretation of the main effects, Location and Number of Competitors.

11.44 This two-way design has 3 row treatments and 5 column treatments. There are 45 total observations with 3 in each cell. FR =

MSR 46.156 = 13.23  MSE 3.489

p-value = .000 and the decision is to reject the null hypothesis for rows. FC =

MSC 249.70 = 71.57  MSE 3.489

p-value = .000 and the decision is to reject the null hypothesis for columns. FI =

MSI 55.267 = 15.84  MSE 3.489

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

p-value = .000 and the decision is to reject the null hypothesis for interaction. Because there is a significant interaction, the analysis of main effects is confounded. The graph of means displays the crossing patterns of the line segments indicating the presence of interaction.

11.45 The null hypotheses are that there are no interaction effects, that there are no significant differences in the means of the valve openings by machine, and that there are no significant differences in the means of the valve openings by shift. Since the p-value for interaction effects is .876, there are no significant interaction effects and that is good since significant interaction effects would confound that study. The p-value for columns (shifts) is .008 indicating that column effects are significant at alpha of .01. There is a significant difference in the mean valve opening according to shift. No multiple comparisons are given in the output. However, an examination of the shift means indicates that the mean valve opening on shift one was the largest at 6.47 followed by shift three with 6.3 and shift two with 6.25. The p-value for rows (machines) is .9368 and that is not significant.

11.46 This two-way factorial design has 3 rows and 3 columns with three observations per cell 18  3  3  (n  1)  n  1  2  n  3 . The observed F value for rows is 0.190 = 0.148/0.778, for columns is 1.190 = 0.926/0.778, and for interaction is 1.405 = 1.093/0.778. Using an alpha of .05, the critical F value for rows and columns (same df) is F.05,2,18, = 3.55. Neither the observed F value for rows nor the observed F value for columns is significant. The critical F value for interaction is F.05,4,18 = 2.93. There is no significant interaction.

11.47 H0: µ1 = µ2 = µ3 = µ4 Ha: At least one mean is different from the others Source Treatment Error

df 3 12

SS 66.69 30.25

MS F__ 22.23 8.82 2.52______

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Total

96.94

 = .05

Critical F.05,3,12 = 3.49

Since the treatment F = 8.82 > F.05,3,12 = 3.49, the decision is to reject the null hypothesis. There is a significant difference in treatment levels. For Tukey's HSD: MSE = 2.52

n=4

N = 16

C=4

N – C = 12

q.05,4,12 = 4.20 HSD = q.05,4,12

x 1 = 12

MSE 2.52 = (4.20) = 3.33 n 4

x 2 = 7.75

x 3 = 13.25

x 4 = 11.25

Using HSD of 3.33, there are significant pairwise differences between means 1 and 2, means 2 and 3, and means 2 and 4.

11.48

11.49 Given C = 6 and N = 42. Then, dfC = 5, dfE = 36, and dfT = 41.

11.50 Source Treatment Error

df 2 22

SS 150.91 102.53

MS F__ 75.46 16.19 4.66________

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Total

253.44

 = .01

Critical F.01,2,22 = 5.72

Since the observed F = 16.19 > F.01,2,22 = 5.72, the decision is to reject the null hypothesis. There is a significant difference in treatment levels.

x 1 = 9.200

x 2 = 14.250

x 3 = 8.714286

n1 = 10

n2 = 8

n3 = 7

MSE = 4.66  = .01

C=3 N = 25 N – C = 22 q.01,3,22  (q.01,3,20 + q.01,3,24 ) /2 (4.64 +4.54)/2 = 4.59

HSD1,2 = 4.59

4.66  1 1     = 3.32 2  10 8 

HSD1,3 = 4.59

4.66  1 1     = 3.45 2  10 7 

HSD2,3 = 4.59

4.66  1 1     = 3.63 2 8 7

Since x 1  x 2 = 5.05 > 3.32 and

x 2  x 3 = 5.5357 > 3.63, there is a

significant pairwise difference between means 1and 2, means 2 and 3 at  = .01.

11.51 This design is a repeated-measures type random block design. There is one treatment variable with three levels. There is one blocking variable with six people in it (six levels). The degrees of freedom treatment are two. The degrees of freedom block are five. The error degrees of freedom are ten. The total degrees of freedom are seventeen. There is one dependent variable.

11.52

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

 = .05

Critical F.05,3,27 = 2.96 for treatments

Since the observed F = 5.58 > F.05,3,27 = 2.96 for treatments, the decision is to

reject the null hypothesis. At  = .05, there is a significance in treatment

effects.

11.53 Source Treatment Blocks Error Total

 = .05

df 3 5 15 23

SS 240.125 548.708 38.125 826.958

MS F__ 80.042 31.49 109.742 43.2 2.542_ _____

Critical F.05,3,15 = 3.29 for treatments

Since for treatments the observed F = 31.49 > F.05,3,15 = 3.29, the decision is to reject the null hypothesis. There is a significant difference in the treatment effects. For Tukey's HSD: Ignoring the blocking effects, the sum of squares blocking and sum of squares error are combined together for a new SSerror = 548.708 + 38.125 = 586.833. Combining the degrees of freedom error and blocking yields a new dferror = 20. Using these new figures, we compute a new mean square error, MSE = (586.833/20) = 29.34165. n=6

C=4

N = 24

N – C = 20

q.05,4,20 = 3.96

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

HSD  q , C , N C

MSE = (3.96) n

29.34165 = 8.757 6

x 1 = 16.667

x 2 = 12.333

x 3 = 12.333

x 4 = 19.833

None of the pairs of means are significantly different using Tukey's HSD = 8.757. This may be due in part to the fact that we compared means by folding the blocking effects back into error and the blocking effects were highly significant.

11.54 Source Treatment 1 Treatment 2 Interaction Error Total

df 4 1 4 30 39

SS MS F__ 29.13 7.2825 1.98 12.67 12.6700 3.44 73.49 18.3725 4.99 110.38 3.6793______ 225.67

 = .05 Critical F.05,4,30 = 2.69 for treatment 1. For treatment 1, the observed F = 1.98 < F.05,4,30 = 2.69 and the decision is to fail to reject the null hypothesis. Critical F.05,1,30 = 4.17 for treatment 2. For treatment 2 observed F = 3.44 < F.05,1,30 = 4.17 and the decision is to fail to reject the null hypothesis. Critical F.05,4,30 = 2.69 for interaction. For interaction, the observed F = 4.99 > F.05,4,30 = 2.69 and the decision is to reject the null hypothesis. Since there are significant interaction effects, examination of the main effects should not be done in the usual manner. However, in this case, there are no significant treatment effects anyway.

11.55

Source Treatment 2 Treatment 1 Interaction Error

df 3 2 6 24

SS 257.889 1.056 17.611 54.000

MS F___ 85.963 38.21 0.528 0.23 2.935 1.30 2.250________

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Total

330.556

 = .01 Critical F.01,3,24 = 4.72 for treatment 2. For the treatment 2 effects, the observed F = 38.21 > F.01,3,24 = 4.72 and the decision is to reject the null hypothesis. Critical F.01,2,24 = 5.61 for Treatment 1. For the treatment 1 effects, the observed F = 0.23 < F.01,2,24 = 5.61 and the decision is to fail to reject the null hypothesis. Critical F.01,6,24 = 3.67 for interaction. For the interaction effects, the observed F = 1.30 < F.01,6,24 = 3.67 and the decision is to fail to reject the null hypothesis.

11.56 Source Age Column Interaction Error Total

df 2 3 6 24 35

SS 49.3889 1.2222 1.2778 15.3333 67.2222

MS F___ 24.6944 38.65 0.4074 0.64 0.2130 0.33 0.6389_______

 = .05 Critical F.05,2,24 = 3.40 for Age. For the age effects, the observed F = 38.65 > F.05,2,24 = 3.40 and the decision is to reject the null hypothesis. Critical F.05,3,24 = 3.01 for Region (Column). For the region effects, the observed F = 0.64 < F.05,3,24 = 3.01 and the decision is to fail to reject the null hypothesis. Critical F.05,6,24 = 2.51 for interaction. For interaction effects, the observed F = 0.33 < F.05,6,24 = 2.51 and the decision is to fail to reject the null hypothesis. There are no significant interaction effects. Only the Age effects are significant. Compute Tukey's HSD for Age. Let 21-35 years old = group 1, Solutions Manual 1-406 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

36-50 years old = group 2, and over 50 years old = group 3. x 1 = 2.667

x 2 = 4.917

n = 12

N = 36

C=4

x 3 = 2.250

N – C = 32

From the table above: MSE = 0.6389 q.05,4,32  q.05,4,30 = 3.85 HSD = q

MSE = (3.85) n

0.6389 = 0.8884 12

Using HSD, there are significant pairwise differences between means of groups1 (age: 21-35) and 2 (age: 36-50) and between means of groups 2 and 3 (age: over 50). Shown below is a graph of the interaction using the cell means by Age.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

11.57 Source Treatment Error Total

 = .05

df 3 20 23

SS 90,477,679 81,761,905 172,239,584

MS F__ 30,159,226 7.38 4,088,095_______

Critical F.05,3,20 = 3.10

For treatment the observed value F = 7.38 > F.05,3,20 = 3.10 and the decision is to

reject the null hypothesis. At  = .05, there is a significant difference in

the mean maximum distance of these four brands.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

11.58

Source Treatment Blocks Error Total

df 2 5 10 17

SS 6.35034 0.46836 0.30079 7.11949

MS F_ _ 3.17517 105.56 0.09367 3.11 0.03008_______

 = .01

Critical F.01,2,10 = 7.56 for treatments

Since the treatment observed F = 105.56 > F.01,2,10 = 7.56, the decision is to

reject the null hypothesis. At  = .01, there is a significant difference in

yields among the planting methods.

11.59 Source Treatment Error Total

 = .05

df 2 18 20

SS 9.555 185.1337 194.6885

MS F__ 4.777 0.46 10.285_______

Critical F.05,2,18 = 3.55

Since the treatment (job type) F = 0.46 < F.05,2,18 = 3.55, the decision is to fail to

reject the null hypothesis. At  = .05, there is not sufficient evidence to

support the claim that there is a significant difference in mean hourly wages for these three jobs. Since there are no significant treatment effects, it would make no sense to compute Tukey-Kramer values and do pairwise comparisons.

11.60 Source Years Size Interaction Error Total

df 2 3 6 36 47

SS 4.875 17.083 2.292 17.000 41.250

MS F___ 2.437 5.16 5.694 12.06 0.382 0.81 0.472_______

 = .05

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Critical F.05,2,36  F.05,2,40 = 3.23for Years. For Years, the observed F = 5.16 > Fcr = 3.23 and the decision is to reject the null hypothesis. Critical F.05,3,36  F.05,3,40 = 2.84 for Size. For Size, the observed F = 12.06 > Fcr = = 2.84 and the decision is to reject the null hypothesis. Critical F.05,6,36  F.05,6,40 = 2.34 for interaction. For interaction, the observed F = = 0.81 < Fcr = 2.34 and the decision is to fail to reject the null hypothesis. There are no significant interaction effects. There are significant row and column

effects at  = .05. Thus, there are significant differences in the mean

responses categorized by size of company as well as by years CEOs have been with their company.

11.61 Source Treatment Blocks Error Total

 = .05

df 4 7 28 39

SS 53.400 17.100 27.400 97.900

MS F___ 13.350 13.64 2.443 2.50 0.979________

Critical F.05,4,28 = 2.71 for treatments

For treatments, the observed F = 13.64 > F.05,4,28 = 2.71 and the decision is to reject the null hypothesis.

11.62 This is a one-way ANOVA with four treatment levels. There are 36 observations in the study. The p-value of .045 indicates that there is a significant overall difference in the means at  = .05. An examination of the mean analysis shows that the sample sizes are different with sizes of 8, 7, 11, and 10, respectively. No multiple comparison technique was used here to conduct pairwise comparisons. However, a study of sample means shows that the two most extreme means are from levels one and four. These two means would be the most likely candidates for multiple comparison tests. Note that the confidence intervals for means one and Solutions Manual 1-410 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

four (shown in the graphical output) are seemingly non-overlapping indicating a potentially significant difference.

11.63 Excel reports that this is a two-factor design without replication indicating that this is a randomized block design. Neither the row nor the column pvalues are less than .05 indicating that there are no significant treatment or blocking effects in this study. Also displayed in the output to underscore this conclusion are the observed and critical F values for both treatments and blocking. In both cases, the observed value is less than the critical value.

11.64 This is a two-way ANOVA with 5 rows and 2 columns. There are 2 observations per cell. For rows, FR = 0.98 with a p-value of .461 which is not significant. For columns, FC = 2.67 with a p-value of .134 which is not significant. For interaction, FI = 4.65 with a p-value of .022 which is significant at  = .05. Thus, there are significant interaction effects and the row and column effects are confounded. An examination of the interaction plot reveals that most of the lines cross verifying the finding of significant interaction.

11.65 This is a two-way ANOVA with 4 rows and 3 columns. There are 3 observations per cell. FR = 4.30 with a p-value of .0146 is significant at  = .05. The null hypothesis is rejected for rows. Hence, a significant difference in row effects is evident at  = .05. FC = 0.53 with a p-value of .594 is not significant. We fail to reject the null hypothesis for columns. FI = 0.99 with a p-value of .453 for interaction is not significant. We fail to reject the null hypothesis for interaction effects. Therefore, neither significant column effects nor significant interaction effects are present.

11.66 This was a randomized block design with 5 treatment levels and 5 blocking levels. For both treatment and blocking effects, the critical value is F.05,4,16 = 3.01. The observed F value for treatment effects is MSC / MSE = 35.98 / 7.36 = 4.89 which is greater than the critical value. The null hypothesis for treatments is rejected, and we conclude that there is a significant difference in treatment means. No multiple comparisons have been Solutions Manual 1-411 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

computed in the output. The observed F value for blocking effects is MSR / MSE = 10.36 /7.36 = 1.41 which is less than the critical value. There are no significant blocking effects.

11.67 This one-way ANOVA has 4 treatment levels and 24 observations. The F = 3.51 yields a p-value of .034 indicating significance at  = .05. Since the sample sizes are equal, Tukey‟s HSD is used to make multiple comparisons. The computer output shows that means 1 and 3 are the only pairs that are significantly different (same signs in confidence interval). Observe on the graph that the confidence intervals for means 1 and 3 barely overlap.

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Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

CHAPTER 12: CORRELATION AND SIMPLE REGRESSION ANALYSIS

12.1

x = 80 y2 = 815

r =

x2 = 1,148 xy = 624

 xy 

y = 69 n=7

 x y n

 ( x )   ( y ) 2  2 2  x    y   n   n  

(80)(69) 7 = 2 2     (80) (69) 1,148   815   7  7   624 

r =

 164.571 = (233.714)(134.857)

r =  164.571 = –0.927 177.533

12.2

x = 1,087 y2 = 878,686

r =

x2 = 322,345 xy= 507,509

 xy 

y = 2,032 n=5

 x y n

 ( x )   ( y )  2 2  x    y   n   n   2

(1,087)(2,032) 5 = 2 2  (1,087)   (2,032)  322,345   878,686   5  5   507,509 

r =

65,752.2 65,752.2 = = .975 67,449.5 (86,031.2)(52,881.2)

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

12.3

Air Canada (x) 25.12 27.22 31.68 33.36 32.79 36.21

WestJet (y) 17.10 17.53 20.65 20.58 19.54 18.52

x = 186.38 x2 = 5,874.803

r =

 xy 

y = 113.92xy = 3,558.785 y2 = 2,174.472

 x y n

 ( x )   ( y )  2 2  x    y   n   n   2

x = 112,968,343 y = 92,274 xy = 521,218,059,167

12.4

r =

 xy 

x2 = 1,187,929,640,160,610 y2 = 606,130,090 n = 23

 x y n

 ( x )   ( y ) 2  2 2  x    y   n   n  

(112,968,343)(92,274) 23 = 2 2  (112,968,343)   (92,274)  1,187,929,640,160,610   606,130,090   23 23    521,218,059,167 

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

x = 160,067,978 y = 116,792 xy = 1,009,792,014,950

x2 = 2,469,099,947,194,680 y2 = 753,828,796 n = 25

= 0.478 12.5

Correlation between Year 1 and Year 2: x = 17.09 y = 15.12 xy = 48.97

r =

x2 = 58.7911 y2 = 41.7054 n=8

 xy 

 x y n

 ( x )   ( y ) 2  2 2  x    y   n   n  

(17.09)(15.12) 8 = 2 2     (17.09) (15.12) 58.7911   41.7054   8 8    48.97 

r =

16.6699 = 16.6699 = .975 17.1038 (22.28259)(13.1286)

Correlation between Year 2 and Year 3: x = 15.12 x2 = 41.7054 y = 15.86 y2 = 42.0396 xy = 41.5934 n=8

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

r =

 xy 

 x y n

 ( x )   ( y ) 2  2 2  x    y   n   n  

(15.12)(15.86) 8 = 2  (15.12)   (15.86) 2  41.7054   42.0396   8 8    41.5934 

r =

11.618 = 11.618 = .985 11.795 (13.1286)(10.59715)

Correlation between Year 1 and Year 3: x = 17.09 x2 = 58.7911 y = 15.86 y2 = 42.0396 xy = 48.5827 n=8

r =

 xy 

 x y n

 ( x )   ( y )  2 2  x    y   n   n   2

(17.09)(15.86) 8 2  (17.09)   (15.86) 2  58 . 7911  42 . 0396     8 8    48.5827 

r =

14.702

= 14.702 = .957 15.367 (22.2826)(10.5972)

The Years 2 and 3 are the most correlated with r = .985.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

12.6

x 12 21 28 8 20

y 17 15 22 19 24

x = 89 x2 = 1,833

y = 97 y2 = 1,935

 xy   n b1 =  ( x) SS x   x

SS xy

b0 =

xy = 1,767 n=5

(89)(97) 5 = 0.162379 (89) 2 1,833  5

1,767 

 y  b  x  97  0.162379 89 = 16.50965 n

ŷ = 16.50965 + 0.162379 x

12.7

x_ 140 119 103 91 65 29 24

_y_ 25 29 46 70 88 112 128

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

x = 571 2 x = 58,293

y= 498 y2 = 45,154

 xy   n SS b1 =  ( x) SS x   x

b0 =

xy = 30,099 n=7 (571)( 498) 7 (571) 2 58,293  7

30,099 

= –0.89824

 y  b  x  498  (0.89824) 571 = 144.4136 n

ŷ = 144.414 – 0.898 x

12.8

(Advertising) x 12.5 3.7 21.6 60.0 37.6 6.1 16.8 41.2

(Sales) y 148 55 338 994 541 89 126 379

x = 199.5 x2 = 7,667.15

y y2

 xy   n SS b1 =  ( x) SS x   x

= 2,670 = 1,587,328

xy = 107,610.4 n=8

(199.5)( 2,670) 8 (199.5) 2 7,667.15  8

107,610.4 

= 15.23977

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

b0 =

 y  b  x  2,670  15.23977 199.5 = -46.29176 1

ŷ = -46.292 + 15.240 x

12.9

(Prime) x 16 6 8 4 7 x = 41 x2 = 421

r =

 xy 

(Bond) y 5 12 9 15 7 y = 48 y2 = 524  x y

xy = 333 n=5

 ( x )   ( y )  2 2  x    y   n   n   2

(41)(48) 5 2  (41)   (48) 2  421  5  524  5     333 

r =

 60.6

= – .828

(84.8)(63.2) Since r = – 0.828, the bond rate can be predicted by the prime interest rate.

 xy   n b1 =  ( x) SS x   x

SS xy

b0 =

( 41)( 48) 5 = –0.71462 ( 41) 2 421  5

333 

 y  b  x  48  (0.71462) 41 = 15.4599 n

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

ŷ = 15.460 – 0.715 x 12.10 Firm Births (x) 58.1 55.4 57 58.5 57.4 58

Bankruptcies (y) 34.3 35 38.5 40.1 35.5 37.9

x = 344.4

y = 221.3

y2 = 8188.41

xy = 12,708.08  x y

r =

 xy 

x2 = 19,774.78

 ( x )   ( y )  2 2  x    y   n   n   2

n=6

(344.4)(221.3) 6 2  (344.4)   (221.3) 2  19 , 774 . 78  8188 . 41   6   6   12,708.08 

r =

5.46 = 0.428 (6.22)(26.1283)

r = 0.428 represents a relatively moderate positive linear relationship between the number of firm births and the number of business bankruptcies. .

 xy   n SS b1 =  ( x) SS x   x

(344.4)( 221.3) 6 = (344.4) 2 19,774.78  6

12,708.08 

b1 = 0.87781 b0 =

 y  b  x  221.3  (0.878) 344.4 = –13.503 n

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

ŷ = –13.503 + 0.878 x

The slope of the line, b1 = 0.878, means that for every 10,000 increase of firm births, the number of business bankruptcies is predicted to increase by 878. 12.11 Number of Farms (1000s) (x) 480.88 430.50 366.11 338.55 318.36 293.09 280.04 276.55 246.92 229.37 205.73 193.49

Average Size (ha) (y) 145 164 188 202 207 231 242 246 273 295 315 332

x = 3,659.59

y = 2,840

y2 = 710,362

xy = 811,067.61

x2 = 1,200,729 n = 12

ŷ = 434.87– 0.650 x

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Steel (x) 90.6 88.8 89.7 79.7 84.3 88.8 91.3 95.2 95.5 98.5

12.12

r =

Orders (y) 2.74 2.87 2.93 2.87 2.98 3.09 3.36 3.61 3.75 3.95

x = 902.4

y = 32.15

x2 = 81,705.14

y2 = 104.9815

xy = 2917.906

n = 10

 xy 

 x y n

 ( x )   ( y )  2 2  x    y   n   n   2

(902.4)(32.15) 10 2  (902.4)   (32.15) 2  81 , 705 . 14  104 . 9815   10   10   2917.906 

r =

16.69 = 0.794 (272.564)(1.61925)

r = 0.794 represents a relatively strong positive linear relationship between the raw steel production and the annual new orders. .

 xy   n SS b1 =  ( x) SS x   x

(902.4)(32.15) 10 = (902.4) 2 81,705.14  10

2917.906 

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

b0 =

 y  b  x  32.15  (0.061233) 902.4 = –2.3107 1

ŷ = –2.3107 + 0.06123 x

12.13

x 15 8 19 12 5

y 47 36 56 44 21

ŷ = 13.625 + 2.303 x

Residuals: x 15 8 19 12 5

y 47 36 56 44 21

ŷ 48.1694 32.0489 57.3811 41.2606 25.1401

Residuals (y- ŷ ) -1.1694 3.9511 -1.3811 2.7394 -4.1401

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

12.14

x 12 21 28 8 20

y 17 15 22 19 24

Predicted ( ŷ ) 18.4582 19.9196 21.0563 17.8087 19.7572

Residuals (y- ŷ ) -1.4582 -4.9196 0.9437 1.1913 4.2428

ŷ = 16.510 + 0.162 x

12.15

x 140 119 103 91 65 29 24

y 25 29 46 70 88 112 128

Predicted ( ŷ ) Residuals (y- ŷ ) 18.6597 6.3403 37.5229 -8.5229 51.8948 -5.8948 62.6737 7.3263 86.0280 1.9720 118.3648 -6.3648 122.8561 5.1439

ŷ = 144.414 – 0.898 x

12.16

x 12.5 3.7 21.6 60.0 37.6 6.1 16.8 41.2

y 148 55 338 994 541 89 126 379

Predicted ( ŷ ) 144.2053 10.0953 282.8873 868.0945 526.7236 46.6708 209.7364 581.5868

Residuals (y- ŷ ) 3.7947 44.9047 55.1127 125.9055 14.2764 42.3292 -83.7364 -202.5868

ŷ = – 46.292 + 15.240x

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

12.17

x 16 6 8 4 7

y 5 12 9 15 7

Predicted ( ŷ ) 4.0259 11.1722 9.7429 12.6014 10.4575

Residuals (y- ŷ ) 0.9741 0.8278 -0.7429 2.3986 -3.4575

ŷ = 15.460 – 0.715 x

12.18

_ x_ 58.1 55.4 57.0 58.5 57.4 58.0

_ y_ 34.3 35.0 38.5 40.1 35.5 37.9

Predicted ( ŷ ) 37.4978 35.1277 36.5322 37.8489 36.8833 37.4100

Residuals (y- ŷ ) -3.1978 -0.1277 1.9678 2.2511 -1.3833 0.4900

ŷ = –13.503 + 0.878 x

The residual for x = 58.1 is relatively large, but the residual for x = 55.4 is quite small.

12.19

_x_ _ y_ 5 47 7 38 11 32 12 24 19 22 25 10

Predicted ( ŷ ) 42.2756 38.9836 32.3996 30.7537 19.2317 9.3558

Residuals (y- ŷ ) 4.7244 -0.9836 -0.3996 -6.7537 2.7683 0.6442

ŷ = 50.506 - 1.646 x

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

No apparent violation of assumptions 12.20

Distance (x) 1,245 425 1,346 973 255 865 1,080 296

Cost y 2.64 2.31 2.45 2.52 2.19 2.55 2.40 2.37

( ŷ ) 2.5376 2.3322

(y- ŷ ) .1024 -.0222

2.5629 2.4694 2.2896 2.4424 2.4962 2.2998

-.1128 .0506 -.0996 .1076 -.0962 .0702

ŷ = 2.2257 – 0.00025 x

No apparent violation of assumptions

12.21

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Error terms appear to be nonindependent

12.22

There appears to be nonlinear regression

12.23

The Residuals vs. Fits graphic is strongly indicative of a violation of the homoscedasticity assumption of regression. Because the residuals are very close together for small values of x, there is little variability in the residuals at the left end of the graph. On the other hand, for larger values of x, the graph flares out indicating a much greater variability at the upper end. Thus, there is a lack of homogeneity of error across the values of the independent variable.

12.24 SSE = y2 – b0y – b1xy = 1,935 – (16.50965)(97) – 0.162379(1,767) = 46.6403 Solutions Manual 1-427 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

se 

SSE  n2

46.6403 = 3.94 3

Approximately 68% of the residuals should fall within ±1se. 3 out of 5 or 60% of the actual residuals fell within ± 1se.

12.25 SSE = y2 – b0y – b1xy = 45,154 – 144.4139(498) – (– .89824)(30,099) = SSE = 272.121

6 out of 7 = 85.7% fall within + 1se 7 out of 7 = 100% fall within + 2se The numbers do not compare well with the empirical rule. From the empirical rule, 68% are within +1se,while 95% are within +2se . 12.26 SSE = y2 – b0y – b1xy = 1,587,328 – (–46.29176)(2,670) – 15.23977(107,610.4) = SSE = 70,969.254

se 

SSE 70,969.254  = 108.8 n2 6

Six out of eight (75%) of the sales estimates are within $108.8 million.

12.27 SSE = y2 – b0y – b1xy = 524 – 15.4599(48) – (–0.71462)(333) = 19.89326

se 

SSE 19.89326  = 2.575 n2 3

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Four out of five (80%) of the estimates are within 2.575 of the actual rate for bonds. This amount of error is probably not acceptable to financial analysts.

12.28

_ x_

_ y_

Predicted ( ŷ )

58.1 55.4 57.0 58.5 57.4 58.0

34.3 35.0 38.5 40.1 35.5 37.9

37.4978 35.1277 36.5322 37.8489 36.8833 37.4100

Residuals (y– ŷ )

( y  yˆ ) 2

-3.1978 10.2259 -0.1277 0.0163 1.9678 3.8722 2.2511 5.0675 -1.3833 1.9135 0.4900 0.2401 2  ( y  yˆ ) = 21.3355

SSE =  ( y  yˆ ) 2 = 21.3355

se 

SSE  n2

21.3355 = 2.3095 4

The standard error of the estimate indicates that the regression model is with + 2.3095(1,000) bankruptcies about 68% of the time. In this particular problem, 5/6 or 83.3% of the residuals are within this standard error of the estimate.

12.29

(y– ŷ ) 4.7244 -0.9836 -0.3996 -6.7537 2.7683 0.6442

(y– ŷ )2 22.3200 .9675 .1597 45.6125 7.6635 .4150 2 (y- ŷ ) = 77.1382

SSE =  ( y  yˆ ) 2 = 77.1382

se 

SSE 77.1382  = 4.391 n2 4

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

12.30

(y- ŷ )2 .0105 .0005 .0127 .0026 .0099 .0116 .0093 .0049 2 (y– ŷ ) = .0620

(y- ŷ ) .1024 -.0222 -.1128 .0506 -.0996 .1076 -.0962 .0702

SSE =  ( y  yˆ ) 2 =

.0620

SSE .0620  = .102 n2 6 The model produces estimates that are ±.102 or within about 10 cents 68% of the time. However, the range of milk costs is only 45 cents for this data. se 

12.31

Volume (x) 728.6 497.9 439.1 377.9 375.5 363.8 276.3

Sales (y) 10.5 48.1 64.8 20.1 11.4 123.8 89.0

n=7 x = 3059.1 x2 = 1,464,071.97

y = 367.7 y2 = 30,404.31

xy =

141,558.6 b1 = –.1504

b0 = 118.257

ŷ = 118.257 – .1504x

SSE = y2 – b0y – b1xy = 30,404.31 – (118.257)(367.7) – (–0.1504)(141,558.6) = 8211.6245

se 

SSE 8211.6245  = 40.53 n2 5

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

This is a relatively large standard error of the estimate given the sales values (ranging from 10.5 to 123.8).

12.32 r2 = 1 

SSE 46.6403 = .123  1 2 ( y ) (97) 2 2 1,935  y  n 5

This is a low value of r2. Only 12.3% of the variation of y is accounted for by x but 87.7% is unaccounted for.

12.33

This is a high value of r2. 97.2% of the variation of y is accounted for by x but 2.8% is unaccounted for.

12.34 r2 = 1 

SSE 70,969.254 = .898  1 2 ( y ) (2,670) 2 2 1,587,328  y  n 8

This value of r2 is relatively high. 89.8% of the variation of y is accounted for by x but 10.2% is unaccounted for.

12.35 r2 = 1 

SSE 19.89326  1 2 ( y ) (48) 2 2 524  y   5 n

= .685

This value of r2 is a modest value. 68.5% of the variation of y is accounted for by x but 31.5% is unaccounted for.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

12.36 r2 = 1 

SSE 21.3355 = .183  1 2 2 ( y ) ( 221 . 3 )  8,188.41   y2  n 6

This value is a low value of r2. Only 18.3% of the variability of y is accounted for by the x values and 81.7% are unaccounted for.

12.37 Long-term Interest Rates (y) 5.29 4.81 4.58 4.07 4.21 4.27 3.61 3.23 3.24 2.78 1.87 2.26 2.23

GDP Growth (x) 2.8 1.9 3.1 3.2 2.6 2.0 1.2 -2.7 3.4 3.0 1.9 2.0 2.4 x = 26.8

y =

y2 = 179.8869

xy = 99.05

 xy   n b1 =  ( x) SS x   x

n = 13

SS xy

x2 = 84.72

46.45

= =

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

ŷ = 3.3428 + 0.1117 x or Long-term Interest Rates = 3.3428 + 0.1117 (Real GDP Growth)

=13.5410 Standard Error of the Estimate

Real GDP Growth is not a very good predictor of the Long-term Interest Rates because only 2.6% of the variation in Long-term Interest Rates can be explained by the variation in Real GDP Growth.

12.38 sb =

x  2

( x) n



3.94 (89) 2 1,833  5

= .2498

b1 = 0.162 Ho: 1 = 0 Ha: 1  0

 = .05

This is a two-tail test, /2 = .025

df = n – 2 = 5 – 2 = 3

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

t =

b1   1 0.162  0  = 0.65 sb .2498

Since the observed t = 0.65 < t.025,3 = 3.182 and t = 0.65 > t.025,3 = –3.182, the decision is to fail to reject the null hypothesis.

12.39 sb =

x  2

( x )

7.38



58,293 

(571) 7

= .06812

b1 = – 0.898 Ho: 1 = 0 Ha: 1  0

 = .01

Two-tail test, /2 = .005

df = n – 2 = 7 – 2 = 5

t.005,5 = ±4.032 t=

b1  1  0.898  0  = –13.17 sb .0682

Since the observed t = –13.17 < t.005,5 = – 4.032, the decision is to reject the null hypothesis.

12.40 sb =

x  2

( x) 2 n



108.8 (199.5) 2 7,667.15  8

= 2.097

b1 = 15.240 Ho: 1 = 0 Ha: 1  0

 = .10

For a two-tail test, /2 = .05

df = n – 2 = 8 – 2 = 6

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

t =

b1  1 15.240  0  = 7.27 sb 2.097

Since the observed t = 7.27 > t.05,6 = 1.943, the decision is to reject the null hypothesis.

12.41 sb =

x  2

( x ) 2 n



2.575 (41) 2 421  5

= .2796

b1 = – 0.715 Ho: 1= 0 Ha: 1  0

 = .05

For a two-tail test, /2 = .025

df = n – 2 = 5 – 2 = 3

t.025,3 = ±3.182 t =

b1  1  0.715  0  = – 2.56 sb .2796

Since the observed t = –2.56 > t.025,3 = –3.182 and t = –2.56 < t.025,3 = 3.182, the decision is to fail to reject the null hypothesis.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

12.42 sb =

x  2

( x )



2.3095 (344.4) 2 19,774.78  6

= 0.9260

b1 = 0.878 Ho: 1 = 0 Ha: 1  0

 = .05

For a two-tail test, /2 = .025

df = n – 2 = 6 – 2 = 4

t.025,4 = ±2.776 t =

b1  1 0.878  0  = 0.948 sb .9260

Since the observed t = 0.948 < t.025,4 = 2.776 and t = 0.948 > t.025,4 = – 2.776, the decision is to fail to reject the null hypothesis.

12.43 F = 8.26 with a p-value of .021. The overall model is significant at  = .05 but

not at  = .01. For simple regression, t =

F = 2.874

If  = .05, then tcr = t.025,8 = ±2.306 and t =2.874 >2.306. For  = .01 tcr = t.005,8 = ±3.355 and t =.2.874 does not fall into rejection region. Thus, the slope is significant at  = .05 but not at  = .01.

12.44 x0 = 25 95% confidence df = n – 2 = 5 – 2 = 3

x

/2 = .025 t.025,3 = ±3.182

 x  89 = 17.8 n

x = 89

x2 = 1,833

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

se = 3.94 ŷ = 16.50965 + 0.162379(25) = 20.57

ŷ ± t /2,n-2 se

1  n

( x0  x) 2 ( x ) 2 2 x  n

1 (25  17.8) 2 20.57 ± 3.182(3.94)  (89) 2 5 1,833  5

= 20.57 ± 3.182(3.94)(.63903)

= 20.57 ± 8.01 12.56 < E(y25) < 28.58 12.45 x0 = 100 For 90% confidence, /2 = .05 df = n – 2 = 7 – 2 = 5 t.05,5 = ±2.015

x

 x  571 n

= 81.57143

x= 571

x2 = 58,293

se = 7.38

ŷ = 144.414 – .898(100) = 54.614

ŷ ± t /2,n-2 se 1 

1  n

( x0  x) 2 = ( x ) 2 2 x  n

54.614 ± 2.015(7.38) 1 

1 (100  81.57143) 2 =  (571) 2 7 58,293  7

54.614 ± 2.015(7.38)(1.08252) = 54.614 ± 16.098 38.516 < y < 70.712

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

For x0 = 130,

ŷ = 144.414 – 0.898(130) = 27.674

y ± t /2,n-2 se 1 

1  n

( x0  x) 2 ( x ) 2 2 x   n

27.674 ± 2.015(7.38) 1 

1 (130  81.57143) 2 =  (571) 2 7 58,293  7

27.674 ± 2.015(7.38)(1.1589) =

27.674 ± 17.234

10.440 < y < 44.908 The confidence interval of y for x0 = 130 is wider than the confidence interval of y for x0 = 100, because x0 = 100 is nearer to the value of x = 81.57 than is x0 = 130.

12.46 x0 = 20 For 98% confidence, /2 = .01 df = n – 2 = 8 – 2 = 6 t.01,6 = + 3.143

x

 x  199.5 = 24.9375 n

x = 199.5

x2 = 7,667.15

se = 108.8

ŷ = – 46.29 + 15.24(20) = 258.51

ŷ ± t /2,n-2 se

1  n

( x0  x) 2 ( x ) 2 2 x   n

258.51 ± (3.143)(108.8)

1  8

(20  24.9375) 2 (199.5) 2 7,667.15  8

258.51 ± (3.143)(108.8)(0.36614) = 258.51 ± 125.20

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

133.31 < E(y20) < 383.71 For single y value: ŷ ± t /2,n-2 se 1 

1  n

( x0  x) 2 ( x ) 2 2 x   n

258.51 ± (3.143)(108.8) 1 

1  8

(20  24.9375) 2 (199.5) 2 7,667.15  8

258.51 ± (3.143)(108.8)(1.06492) = 258.51 ± 364.16 – 105.65 < y < 622.67 The prediction interval for the single value of y is wider than the confidence interval for the average value of y because the average is more towards the middle and individual values of y can vary more than values of the average.

12.47 x0 = 10 For 99% confidence df = n – 2 = 5 – 2 = 3 t.005,3 = + 5.841

x

/2 = .005

 x  41 = 8.20 n

x = 41

x2 = 421

se = 2.575

ŷ = 15.46 – 0.715(10) = 8.31

ŷ ± t /2,n-2 se

1  n

( x0  x) 2 ( x ) 2 2 x  n

8.31 ± 5.841(2.575)

1 (10  8.2) 2 =  (41) 2 5 421  5

8.31 ± 5.841(2.575)(.488065) = 8.31 ± 7.34 Solutions Manual 1-439 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

0.97 < E(y10) < 15.65 If the prime interest rate is 10%, we are 99% confident that the average bond rate is between 0.97% and 15.65%.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

12.48 Year (x) 2015 2016 2017 2018 2019

Export – Balance of payments (unadjusted) Wheat ($millions) (y) 646.4 546.4 570.0 592.4 831.2

Σx =10,085 Σy =3,186.4 Σxy =6,427,384 Σ =20,341,455 Σ =2,083,117.1 n = 5

ŷ = -83,189.24 + 41.56 x Export – Balance of Payment = -83,189.24 + 41.56 (Year)

ŷ (2021) = -83,189.24 + 41.56 (2021) = 803.52

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

12.49 Time Period

Recoded Time Period (x)

Long-Term Interest Rate (y)

Nov-18 Dec-18 Jan-19 Feb-19 Mar-19 Apr-19 May-19

1 2 3 4 5 6 7

2.41 2.06 1.95 1.91 1.71 1.72 1.67

x = 28 x2 = 140

y = 13.43 y = 26.1737 2

xy = 50.58 n=7

ŷ = 2.367 -0.11214 x or Long-term Interest Rate = 2.367 -0.11214 (Recoded Time Period)

Since June 2019 can be coded as 8, we forecast the long-term interest rate for June 2013 as: ŷ (8) = 2.367 -0.11214(8) = 1.47

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

12.50 Time Period 2003 2004 2005 2006 2007 2008 2009 2010 2011 2012 2013 2014 2015 2016 2017 2018 x = 136 x2 = 1,496

Recoded Time Period (x) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

New Housing Prices Index (y) 66.5 69.9 74.1 82.0 87.1 87.5 86.7 88.4 90.7 92.7 93.9 95.5 97.0 100.0 103.3 103.3

y = 1,418.6 y2 = 127,605.2

xy = 12,816.7 n = 16

ŷ = 69.6975 + 2.23118 x or New Housing prices Index = 69.6975 + 2.23118(Recoded Time Period)

Since 2019 can be coded as 17, we forecast new housing prices index for 2019 as: ŷ (17) = 69.6975 + 2.23118 (17) = 107.6 12.51

x = 36 y = 44

x2 = 256 y2 = 300

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

xy = 188

n =7

 xy 

 x y n

 ( x )   ( y ) 2  2 2  x    y   n   n  

(36)( 44) 7 2  (36)   (44) 2  256  300     7  7   188 

r =

12.52

 38.2857

=  38.2857 = –.940 40.7441 (70.85714)( 23.42857)

x 5 7 3 16 12 9

y 8 9 11 27 15 13

x = 52 y = 83 xy = 865

x2 = 564 y2 = 1,389 n=6

ŷ = 2.6941 + 1.2853 x

ŷ (Predicted Values) 9.1206 11.6912 6.5500 23.2588 18.1177 14.2618

b1 = 1.2853 b0 = 2.6941

(y- ŷ ) residuals -1.1206 -2.6912 4.4500 3.7412 -3.1177 -1.2618

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

(y- ŷ )2 1.2557 7.2426 19.8025 13.9966 9.7201 1.5921 SSE = 53.6096

se 

SSE 53.6096  = 3.661 n2 4 SSE 53.6096  1 2 ( y ) (83) 2 2 1 , 389  y  n 6

r2 = 1 

Ho: 1 = 0  = .01 Ha: 1  0 Two-tailed test, /2 = .005

= .777

df = n – 2 = 6 – 2 = 4

t.005,4 = ±4.604

sb =

x  2

t =

( x ) 2 n



3.661 (52) 2 564  6

= .34389

b1   1 1.2853  0  = 3.74 sb .34389

Since the observed t = 3.74 < t.005,4 = 4.604 and t = 3.74 > t.005,4 = – 4.604 , the decision is to fail to reject the null hypothesis. f)

12.53

The r2 = 77.7% is modest. There appears to be some prediction with this model. The slope of the regression line is not significantly different from zero using  = .01. However, for  = .05, the null hypothesis of a zero slope is rejected. The standard error of the estimate, se = 3.661 is not particularly small given the range of values for y (27 – 8 = 19).

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

53 47 41 50 58 62 45 60

5 5 7 4 10 12 3 11

x = 416 y = 57 xy = 3,106

x2 = 22,032 y2 = 489 n=8

ŷ = –11.335 + 0.355 x

ŷ (Predicted Values) 7.48 5.35 3.22 6.415 9.255 10.675 4.64 9.965

b1 = 0.355 b0 = –11.335

(y- ŷ ) residuals -2.48 -0.35 3.78 -2.415 0.745 1.325 -1.64 1.035

(y- ŷ )2 6.1504 0.1225 14.2884 5.8322 0.5550 1.7556 2.6896 1.0712 SSE = 32.4649

SSE 32.4649  = 2.3261 n2 6

se =

r2 = 1 

SSE 32.4649  1 2 ( y ) (57) 2 2 489  y  n 8

= .608

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Ho: 1 = 0 Ha: 1  0

 = .05

Two-tailed test, /2 = .025

df = n – 2 = 8 – 2 = 6

t.025,6 = ±2.447

sb =

x  2

t =

( x ) 2 n



2.3261 (416) 2 22,032  8

= 0.116305

b1  1 0.355  0  = 3.05 sb .116305

Since the observed t = 3.05 > t.025,6 = 2.447, the decision is to reject the null hypothesis. The population slope is different from zero. g)

This model produces only a modest r2 = .608. Almost 40% of the variance of y is unaccounted for by x. The range of y values is 12

–3=9 and the standard error of the estimate is 2.33. Given this small range, the se is not small. 12.54 x = 1,263 y = 417 xy = 88,288 b0 = 25.42778

x2 = 268,295 y2 = 29,135 n=6 b1 = 0.209369

SSE = y2 - b0y - b1xy = 29,135 – (25.42778)(417) – (0.209369)(88,288) = 46.845468 r2 = 1 

SSE 46.845468 = .695  1 2 153 . 5 ( y )   y2  n

Coefficient of determination = r2 = .695

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

12.55 a) x0 = 60 x = 524 y = 215 xy = 15,125

x2 = 36,224 y2 = 6,411 n=8

se = 3.201 95% Confidence Interval df = n – 2 = 8 – 2 = 6

b1 = .5481 b0 = –9.026

/2 = .025

t.025,6 = ±2.447 ŷ = –9.026 + 0.5481(60) = 23.86

x

 x  524 = 65.5 n

ŷ ± t /2,n-2 se

1  n

( x0  x) 2 ( x ) 2 2 x  n 1  8

23.86 + 2.447(3.201)

(60  65.5) 2 (524) 2 36,224  8

23.86 + 2.447(3.201)(.375372) = 23.86 + 2.94 20.92 < E(y60) < 26.8 b) x0 = 70 ŷ = –9.026 + 0.5481(70) = 29.341

ŷ + t/2,n-2 se 1 

1  n

( x0  x) 2 ( x ) 2 2 x  n

1 29.341 + 2.447(3.201) 1   8

(70  65.5) 2 (524) 2 36,224  8

29.341 + 2.447(3.201)(1.06567) = 29.341 + 8.347 Solutions Manual 1-448 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

20.994 < y < 37.688 c) The confidence interval for (b) is much wider because part (b) is for a single value of y which produces a much greater possible variation. In actuality, x0 = 70 in part (b) is slightly closer to the mean (x) than x0 = 60. However, the width of the single interval is much greater than that of the average or expected y value in part (a).

12.56

Year 1 2 3 4 5

Cost 56 54 49 46 45

x = 15 x2 = 55

b1 =

b0 =

SS xy SS xx

y = 250 y2 = 12,594



 xy 

 x y

x  2

n ( x) 2

xy = 720 n=5 (15)(250) 5 = –3 (15) 2 55  5

720 

 y  b  x  250  (3) 15 = 59 n

ŷ = 59 – 3 x ŷ (7) = 59 – 3(7) = 38 ($ millions)

12.57 y = 267 x = 21 xy = 1,256 b0 = 9.234375

y2 = 15,971 x2 = 101 n =5 b1 = 10.515625

SSE = y2 – b0y – b1xy = 15,971 – (9.234375)(267) – (10.515625)(1,256) = 297.7969

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

r2 = 1 

SSE 297.7969 = .826  1 2 1 , 713 . 2 ( y )   y2  n

If a regression model would have been developed to predict number of cars sold by the number of sales people, the model would have had an r2 of 82.6%. This result means that 82.6% of the variation in the number of cars sold is explained by variation in the number of salespeople.

12.58 = y.

Let Percentage Purchases to be Made from Catalogues = x and Average Spending x = 468 y = 8,731 xy = 379,186

b1 =

SS xy SS xx



x2 = 20,404 y2 = 7,423,101 n = 11

 xy 

 x y n ( x ) 2

= 15.67122

x  n  y  b  x  126.98819 b0 = n

ŷ = 126.98819 + 15.67122 x or Average Spending = 126.98819 + 15.67122 (Percentage Purchases to be Made from Catalogues) b0 = 126.98819

b1 = 15.67122

SSE = y2 - b0y - b1xy = 7,423,101 – (126.98819)( 8,731) – (15.67122)( 379,186) = 372,059.8862 r2 = 1 

SSE 372,059.8862  1  0.2454 2 493,068.1818 ( y ) 2 y  n

Coefficient of determination = r2 = 0.2454 Solutions Manual 1-450 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

This regression model accounts for only 24.54% of the variation in the average amounts of planned spending on gifts. It also means that 75.46% of the variability of the spending is unexplained by the regression model.

12.59

Sales(y)

Number of Units(x)

37.48 15.0 13.17 13.9 10.80 25.9 10.03 7.2 9.79 6.4 9.29 5.8 5.90 5.6 5.51 7.5 4.42 4.1 3.64 4.5 y = 110.03 y2 = 2075.545 x = 95.9 2 x = 1,340.13 xy = 1,322.604 n = 10 b1 = 0.63603

b0 = 4.90352

ŷ = 4.90352 + 0.63603 x

SSE = y2 – b0y – b1xy = 2,075.545 – (4.90352)(110.03) – (0.63603)(1,322.604) = 694.7944

Thus, 19.7% of the variation of the restaurant chain’s sales is explained by variation in the number of of units, and 80.3% of the variation in restaurant chain’s sales is due to other factors.  x y  xy  n r = = 2 2     ( x ) ( y )   y 2    2  x   n   n   Solutions Manual 1-451 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

r = = 0.4438 represents a low positive correlation between, the restaurant chain‟s sales and its number of units. 12.60 Time Period

Labor Force (y)

2009 2010 2011 2012 2013 2014 2015 2016 2017 2018

Recoded Time Period (x) 1 2 3 4 5 6 7 8 9 10

y = 611.9 x2 = 385

y2 = 37,600 xy = 3,477.21

x = 55 n = 10

55.86 56.81 57.88 58.96 60.05 61.16 62.28 63.39 67.14 68.37

ŷ = 53.7393 + 1.3547 x ŷ (2021) = ŷ (13) = 53.7393 + 1.3547 (13) = 71.35

12.61

x 2.4 3.9 1.1 2.8 2.4 1.6 2.8 2.2 2.2 0.1

y 6.7 7.0 7.8 7.9 7.2 6.9 6.1 6.0 6.1 8.4

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

r =

 xy 

 x y n

 ( x )   ( y )  2 2  x    y   n   n   2

(21.5)(70.1) 10  (21.5) 2   (70.1) 2  55 . 87  497 . 57     10   10   146.94 

r =

 3.775

= –.489

(9.645)(6.169)

The r value of –.489 denotes a moderate negative correlation.

12.62

Let Nondurables = x and Durables = y. x = 3,991,629 y = 1,900,874 2 x = 1,001,164,850,933 y2 = 232,554,188,906 xy = 480,204,660,282 n = 16

r =

 xy 

 x y n

 ( x )   ( y ) 2  2 2  x    y   n   n   2

r = 0.9977 The r value of 0.9977 denotes a strong positive correlation. b1 = 1.118725 b0 = -160,291.4164 ŷ = –160,291.4164+ 1.118725 x or Durables = –160,291+ 1.118725 (Non-Durables) ŷ (200,000) = –160,291.4164+ 1.118725(200,000) = 63,453.66 Thus, the predicted amount for durables shipment is $63,453.66 million.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

ŷ + t/2,n-2se

1  n

( x0  x) 2 ( x ) 2 2 x  n

 = .05, t.025,14 = + 2.145 x0 = 200,000, n = 16 x = 249,477

SSE = y2 – b0y – b1xy = 232,554,188,906 – (-160,291.4164)(1,900,874) – (1.118725)(480,204,660,282) = 30,834,608.02

Confidence Interval =

61,157.22 to 65,750.10 H0: 1 = 0 Ha: 1  0 Solutions Manual 1-454 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

 = .05

df = 14

Table t.025,14 = + 2.145

Since the observed t = 55.11 > t.025,14 = 2.145, the decision is to reject the null hypothesis that the population slope is zero. The linear regression model adds something to the explanation of the variation of the shipments of durables that the average value of y model does not.

12.63 x = 10.797 y = 516.8 xy = 1091.2081

x2 = 20.673549 y2 = 61,899.06 n=7

b1 = 73.15543 b0 = –39.00846

ŷ = –39.00846+ 73.15543 x or

Cost = –39.00846+ 73.15543 Mass SSE = y2 – b0 y – b1 xy SSE = 61,899.06 – (–39.00846)(516.8) – (73.15543)( 1091.2081) = 2,230.83435

se 

SSE  n2

r2 =

1

2,230.83435 = 21.12266 5

SSE 2,230.83435  1 2 ( y ) (516.8) 2 2 61 , 899 . 06  y  n 7

= 1 – .09395 =

.90605 12.64 Let trade balance with the rest of the world = x and trade balance with the U.S. = y . x = 32,401 y = 47,958 x2 = 109,124,859 y2 = 230,544,076

n = 10

xy = 154,518,237

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

ŷ

= 5,476.67 – 0.21014 x

r2 for this model is 0.334 . There is less predictability in this model. Test for slope: t = -2.00 with a p-value of 0.07995. The null hypothesis that the population slope is zero is not rejected at  = .05 . Time-Series Trend Line: (x = recoded years of 1 through 10, y = trade balance with the rest of the world) x = 55 x2 = 385

y = 32,401 y2 = 109,124,859

xy = 193,754 n = 10

ŷ = 46555.666667+ 1406.442424 x ŷ = 2,203.53+ 188.467 x ŷ (2008) = ŷ (11) = 46,555.666667 + 1406.442424 (11) = 62,026.5 ($ millions) ŷ (2020) = ŷ (11) = 2,203.53 + 188.467 (11) = 4,276.67 ($ millions)

12.65 x = 1034.7 x2 = 217,604.49

y = 18,840 y2 = 62,486,228

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

xy = 2,356,711.6

b1 =

SS xy



SS xx

n=7

 xy 

 x y

x  2

n ( x) 2

(1034.7)(18,840) 7 (1034.7) 2 217,604.49  7

2,356,711.6 

= –

6.620826 b0 =

 y  b  x  18,840  (6.620826) 1034.7 = 3670.081237 1

ŷ = 3670.081237– 6.620826 x

SSE = y2 – b0y – b1xy = 62,486,228 – (3670.081237)(18,840) – (–6.620826)(2,356,711.6) SSE = 8,945,274.9307

se 

SSE 8,945,274.9307  = 1337.5556 n2 5

r2 = 1 

SSE 8,945,274.9307  1 2 ( y ) (18,840) 2 2 62,486,228  y  n 7

= 1 – .7594 =

.2406 The r value of –0.4905 denotes a moderate negative correlation. H0: 1 = 0 Ha: 1 0

 = .05

t.025,5 = + 2.571

 x  (1034.7) x   217,604.49  = 64,661.048571 2

SSxx = 

t =

b1  0  6.620826  0 = – 1.26  se 1337.5556 SS xx

64,661.048571

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Since the observed t = – 1.26 > t.025,5 = – 2.571 and t = – 1.26 < t.025,5 = 2.571, the decision is to fail to reject the null hypothesis.

12.66 Let Water Use = y and Temperature = x x = 196 y = 3,880 xy = 112,006

x2 = 5,834 y2 = 2,188,612 n=8

b1 = 16.42054 b0 = 82.69671

ŷ = 82.69671 + 16.42054 x or

Water Use = 82.69671 + 16.42054 Temperature For x = 38o C: ŷ = 82.69671 + 16.42054 (38) = 707 (millions of litres) SSE = y2 – b0 y – b1 xy SSE = 2,188,612– (82.69671)( 3,880) – (16.42054)( 112,006) = 28,549.76196

se 

SSE  n2

r2 = 1 

28,549.76196 = 68.980386 6

SSE 28,549.76196  1 2 ( y ) (3,880) 2 2 2 , 188 , 612  y   8 n

1 – .093053 =

.9069 About 91% of the variation in the water usage can be explained by the variation in the temperatures, but 9% of the variation in the water usage is due to other factors. Testing the slope: Ho: 1 = 0 Ha: 1  0

 = .01

Since this is a two-tailed test, /2 = .005 df = n – 2 = 8 – 2 = 6 t.005,6 = ±3.707

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

sb =

x  2

t =

( x )



b1  1 16.42054  0  sb 2.147266

68.980386 (196) 2 5,834  8

= 2.147266

= 7.65

Since the observed t = 7.65 > t.005,6 = 3.707, the decision is to reject the null hypothesis.

12.67 The F value for overall significance of the model is 7.12 with an associated p-value of .0205 which is significant at  = .05. It is not significant at alpha of .01. The t statistic for testing to determine if the slope is significantly different from zero is – 2.67 with a p value of 0.0205. Both statistics state that there are significant regression effects in the model at  = .05, but not at alpha of 0.01. The coefficient of determination is .372 with an adjusted r2 of .32. This represents the modest predictability. The standard error of the estimate is 982.219 which in units of 1,000 labourers means that about 68% of the predictions are within 982,219 of the actual figures. The regression model is: Number of Union Members = 22,348.97 – 0.0524 Labour Force. For a labour force of 100,000 (thousand, actually 100 million), substitute x = 100,000 and get a predicted value of 17,108.97 (thousand) which is actually 17,108,970 union members.

12.68 The Residual Diagnostics from output indicate a relatively healthy set of residuals. The Histogram indicates that the error terms are generally normally distributed. This is somewhat confirmed by the semi straight line Normal Plot of Residuals. However, the Residuals vs. Fitted Values graph indicates that there may be some heteroscedasticity with greater error variance for small x values.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

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SOLUTIONS TO PROBLEMS IN CHAPTER 13: MULTIPLE REGRESSION ANALYSIS

13.1

The regression model is:

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

ŷ = 25.0287 – 0.0497 x1 + 1.9282 x2 Predicted value of y for x1 = 200 and x2 = 7 is:

ŷ = 25.0287 – 0.0497(200) + 1.9282(7) = 28.586

13.2

The regression model is:

ŷ = 118.5595 – 0.0794 x1 – 0.8843 x2 + 0.3769 x3 Predicted value of y for x1 = 33, x2 = 29, and x3 = 13 is:

ŷ = 118.5595 – 0.0794(33) – 0.8843(29) + 0.3769(13) = 95.1943

13.3

The regression model is:

ŷ = 121.62 – 0.174 x1 + 6.02 x2 + 0.00026 x3 + 0.0041 x4 There are four independent variables. If x2, x3, and x4 are held constant, the predicted y will decrease by 0.174 for every unit increase in x1. Predicted y will increase by 6.02 for every unit increase in x2 as x1, x3, and x4 are held constant. Predicted y will increase by 0.00026 for every unit increase in x3 holding x1, x2, and x4 constant. If x4 is increased by one unit, the predicted y will increase by 0.0041 if x1, x2, and x3 are held constant.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

13.4

The regression model is:

ŷ = 31,409.5 + 0.08425 x1 + 289.62 x2 – 0.0947 x3 For every unit increase in x1, the predicted y increases by 0.08425 if x2 and x3 are held constant. The predicted y will increase by 289.62 for every unit increase in x2 if x1 and x3 are held constant. The predicted y will decrease by 0.0947 for every unit increase in x3 if x1 and x2 are held constant.

13.5

The regression model is: Per Capita Personal Consumption = –7,655.99 + 116.66 Paper Consumption – 265.09 Fish Consumption + 45.63 Gasoline Consumption. For every kg per person increase in paper consumption, the predicted per capita personal consumption increases by $116.66 if fish and gasoline consumptions are held constant. For every kg per person increase in fish consumption, the predicted per capita personal consumption decreases by $265.09 if paper and gasoline consumptions are held constant. For every litre per person increase in gasoline consumption, the predicted per capita personal consumption increases by $45.63 if paper and fish consumptions are held constant.

13.6

The regression model is: Insider Ownership = 17.6772 – 0.0594Debt Ratio – 0.1184 Dividend Payout For every unit of increase in debt ratio there is a predicted decrease of 0.0594 in insider ownership if dividend payout is held constant. If dividend payout is increased by one unit, then there is a predicted drop of insider ownership by 0.1184 with debt ratio held constant.

13.7

There are 9 predictors in this model. The F test for overall significance of the model is 1.99 with a p value of .0825. This model is not significant at  = .05. Only one of the t values is statistically significant. Predictor x1 has a t of 2.73 which has an associated probability (P-value) of .011 and this is significant at  = .05.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

13.8

This model contains three predictors. The F test is significant at  = .05 but not at  = .01. The t values indicate that only one of the three predictors is significant. Predictor x1 yields a t value of 3.41 with an associated probability (P-value) of .005. The recommendation is to rerun the model using only x1 and then search for other variables besides x2 and x3 to include in future models.

13.9

The regression model is: Per Capita Personal Consumption = –7,655.99 + 116.66 Paper Consumption – 265.09 Fish Consumption + 45.63 Gasoline Consumption. This model yields an F = 14.32 with p-value = .0023. Thus, there is overall significance at  = .01. One of the three predictors is significant. Gasoline Consumption has a t = 2.66 with p-value of .033 which is statistically significant at  = .05. The p-values of the t statistics for the other two predictors are insignificant indicating that a model with just Gasoline Consumption as a single predictor might be nearly as strong.

13.10 The regression model is: Insider Ownership = 17.6772 – 0.0594Debt Ratio – 0.1184 Dividend Payout The overall value of F is only 0.02 with p-value of .982. This model is not significant. Neither of the t values are significant (tDebt = – 0.189 with a p-value of 0.855 and tDividend = – 0.113 with a p-value of .913). 13.11 The regression model is: ŷ = 3.98077 + 0.07322 x1 – 0.03232 x2 – 0.00389 x3

The overall F for this model is 100.47 with p-value of .000 000 03. This model is significant at  = .000 000 1. Only one of the predictors, x1, has a significant t value (t = 3.50, p-value of .005). The other independent variables have non-significant t values (x2: t = –1.55, p-value of .150 and x3: t = –1.01, p-value of .332). Since x2 and x3 are non-significant predictors, the researcher should consider using a simple regression model with only x1 as a predictor. The R2 would drop but the model would be much more parsimonious.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

13.12 The regression equation for the model using both x1 and x2 is:

ŷ = 243.4408 – 16.6079 x1 – 0.0732 x2

The overall F = 156.89 with a p-value of .000. x1 is a significant predictor of y as indicated by t = – 16.10 and a p-value of .000. For x2, t = – 0.39 with a p-value of .702. x2 is not a significant predictor of y when included with x1. Since x2 is not a significant predictor, the researcher might want to rerun the model using just x1 as a predictor. The regression model using only x1 as a predictor is:

ŷ = 235.1429 – 16.7678 x1 There is very little change in the coefficient of x1 from model one (2 predictors) to this model. The overall F = 335.47 with a p-value of .000 is highly significant. By using the one-predictor model, we get virtually the same predictability as model with the two predictors and it is more parsimonious.

13.13 There are 3 predictors in this model and 15 observations. The regression equation is:

ŷ = 657.053 + 5.7103 x1 – 0.4169 x2 –3.4715 x3 Since F = 8.96 with a p-value of .0027, the model is significant overall at  = .005, but not at  = .001. x1 is significant at  = .01 (t = 3.19, p-value of .0087). x2 is not a significant predictor (t = – 1.29, p-value of .2222). x3 is significant at  = .05 (t = – 2.41, p-value of .0349).

13.14 The standard error of the estimate is 3.503. R2 is .408 and the adjusted R2 is only .203. This indicates that there are a lot of insignificant predictors in the model. That is underscored by the fact that eight of the nine predictors have non-significant t values. 13.15 S = 9.722, R2 = .515 but the adjusted R2 is only .404. The difference in the two is due to the fact that two of the three predictors in the model are nonSolutions Manual 1-465 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

significant. The model fits the data only modestly. The R2 indicates that 51.5% of the variation of y is accounted for by this model and 48.5% is unaccounted for by the model. 13.16 The standard error of the estimate of 14,599.85 indicates that approximately 68% of the error values (residuals) for Per Capita Personal Consumption would be within +$ 14,599.85.. The entire range of Per Capita Personal Consumption for the data is slightly less than 110,000. Relative to this range, the standard error of the estimate is modest. R2 = .85989 and the adjusted value of R2 is .799848 indicating that there are potentially some non-significant variables in the model. An examination of the t statistics reveals that two of the three predictors are not significant. The model has relatively good predictability. 13.17 S = 6.544. R2 = .0045. R2 (adj.) = – .244. This model has no predictability. 13.18 The value of S = se = 0.2331, R2 = .965, and adjusted R2 = .955. The standard error of the estimate of 0.2331 indicates that approximately 68% of the error values (residuals) would be within 0.2331. The entire range of y for the data is 7 – 2.9 = 4.1. Relative to this range, the standard error of the estimate is small. Since 96.5% of the variation in y can be explained by regression model, this is a very strong regression model. However, since x2 and x3 are not significant predictors, the researcher should consider using a simple regression model with only x1 as a predictor. The R2 would drop but the model would be much more parsimonious. 13.19 For the regression model using both x1 and x2, S = se = 6.333, R2 = .963 and adjusted R2 = .957. Overall, this is a very strong model. For the regression model using only x1 as a predictor, the standard error of the estimate is 6.124, R2 = .963 and the adjusted R2 = .960. The value of R2 is the same as it was with the two predictors. However, the adjusted R2 is slightly higher with the one-predictor model because the non-significant variable has been removed. In conclusion, by using the one predictor model, we get virtually the same predictability as with the two predictor model and it is more parsimonious. 13.20 R2 = .710, adjusted R2 = .630, S = se = 109.43. Since 71% of the variation in y can be explained by the regression model, the model is moderately strong. A comparison of R2 with the adjusted R2 shows that the adjusted

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

R2 reduces the overall proportion of variation of the dependent variable accounted for by the independent variables by a factor of 0.08, or 8%.

13.21 The Histogram indicates that there may be some problem with the error terms being normally distributed as does the Normal Probability Plot of the Residuals in which the plotted points are not completely lined up on the line. The Residuals vs. Fits plot reveals that there may be some lack of homoscedasticity of error variance.

13.22 There are four predictors. The equation of the regression model is: ŷ = –55.93 + 0.01049 x1 – 0.1072 x2 + 0.57922 x3 – 0.8695 x4 The test for overall significance yields an F = 55.52 with a p-value of .000 which is significant at  = .001. Three of the t tests for regression coefficients are significant at  = .01 including the coefficients for x2, x3, and x4. The R2 value of 80.2% indicates strong predictability for the model. The value of the adjusted R2 (78.7%) is close to R2 and S = se is 9.025.

13.23 There are two predictors in this model. The equation of the regression model is: ŷ = 203.3937 + 1.1151 x1 – 2.2115 x2 The F test for overall significance yields a value of 24.55 with an associated p-value of .0000013 which is significant at  = .00001. Both variables yield t values that are significant at a 5% level of significance. x2 is significant at  = .001. The R2 is a rather modest 66.3% and the standard error of the estimate is 51.761. 13.24 The regression equation for new model (based on 24 observations) is:

ŷ = 155.73 - 0.0046 x1 - 2.5213 x2 There are huge changes in the coefficient of x1, x2, and the regression constant as compared with the old model based on 23 observations. However, at  = 0.05 the regression model is not significant overall because F = 1.80 with a p-value of 0.1906. As indicated by t = -0.25 with p-value of .81, x1 is not a significant predictor of y at  = .05. For x2, t = -1.87 with p-value of .07. Thus, x2 is also not a significant predictor of y at  = .05 . For the old model, both predictors, x1 and x2, have the significant t values, but the levels of significance have been changed.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

R2 = .146 and the adjusted value of R2 is 0.065. Thus, the new model does not have a good predictability with smaller coefficient of determination. For the previous regression model the standard error of the estimate is about $12,000, and range of Market Price is $77,000. For the new regression model, se = $73,760 and range of Market Price = $387,000. 13.25 Given p = number of independent variables = 2 (see Figure 13.5) and n = sample size = 23. Then, the expected value of R2 is p 2 2 E R2     0.091 . n  1 23  1 22

 

13.26 Given p = number of independent variables = 10 and n = sample size = 23 (see Figure 13.5). Then, the expected value of R2 is p 10 10 E R2     0.455 . n  1 23  1 22

 

13.27 The regression model is: ŷ = 137.268 + 0.002515 x1 + 29.2061 x2 The F test for overall significance yields a value of 10.89 with an associated p-value of .0052 which is significant at  = .01. For x1, t = 0.01 with p = .99 and for x2, t = 4.47 with p = .002. The t values show x1 has virtually no predictability and x2 is a significant predictor of y. S = se = 9.401, R2 = .731, adjusted R2 = .664. This model has good predictability. The gap between R2 and adjusted R2 indicates that there may be a non-significant predictor in the model.

13.28 The regression model is:

ŷ = 362.3054 – 4.74552 x1 – 13.8997 x2 + 1.874297 x3 The F test for overall significance yields a value of 16.05 with an associated p-value of .000955 which is significant at  = .001. For x1, t = – 4.35 with p = .002; for x2, t = – 0.73 with p = .483, for x3, t = 1.96 with p = .086. Thus, only one of the three predictors, x1, is a significant predictor in this model. S = se = 37.07, R2 = .858, adjusted R2 = .804. This model has very good predictability (R2 = .858). The gap between R2 and adjusted R2 underscores the fact that there are two non-significant predictors in this model. 13.29 Solutions Manual 1-468 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

The regression model is: Gold = – 51.5749 + 0.0696 Copper + 18.7835 Silver + 3.5378 Aluminum The overall F for this model is 12.19 with a p-value of .002 which is significant at  = .01. The t test for Silver is significant at  = .01 ( t = 4.94, p = .0011). The t test for Aluminum yields a t = 3.03 with a p-value of .016 which is significant at  = .05. The t test for Copper is insignificant with a p-value of .939. The value of R2 was 82.1% compared to an adjusted R2 of 75.3%. The gap between the two indicates the presence of some insignificant predictors (Copper). The standard error of the estimate is 53.44.

13.30 The regression equation is: Total Goods = – 10.1928 + 0.2525Durable Goods + + 0.2875Semi-durable Goods + 0.5604Non-durable Goods The F test for overall significance yields a value of 7,090.68 with an associated p-value of .000 which is highly significant. All variables are significant (Durable Goods t = 7.14, p-value of .000; Semi-durable Goods t = 4.70, p-value of .0002; and Non-durable Goods t = 83.47, p-value of .000). R2 = 0.9992 and adjusted R2 = 0.9990 The value of adjusted R2 is very close to R2. The high value of R2 indicates that the model has a very strong predictability.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

13.31

The regression equation is: Exchange rate(C$ per US$) = 2.9154 – 0.0166 Price Index – 0.0020 change % + 0.0009 (PPI-Mfg) – 0.0003 (CPI) The F test for overall significance yields a value of 242.39 with an associated p-value of .000 which is highly significant. Only one of the predictors, Price Index, has a significant t value (t = – 8.91, p-value of .000 02). The other independent variables have non-significant t values (change %: t = – 0.365, p-value of .725 ; PPI-Mfg: t = 0.252, p-value of .807; CPI: t = – 0.113, p-value of .913 ). Since x2 and x3 are nonsignificant predictors, the researcher should consider using a simple regression model with only x1 as a predictor. The R2 would drop but the model would be much more parsimonious. This model has very good predictability (R2 = .992). The gap between R2 and adjusted R2 = 0.988 underscores the fact that there are three nonsignificant predictors in this model.

13.32 The regression model was: New Car Dealers = – 6854.01+ 3.957893Used Vehicles and Parts + + 0.248643 Total Excluding Used Vehicles and Parts – – 1.43775 Gas Stations F = 348.6134 with p = .000. S = se = 1838.944, R2 = 0.987722, and adjusted R2 = 0.984889. The high value of adjusted R2 indicates that the model has a very strong predictability. The t test for Used Vehicles and Parts is significant (t = 5.37, p = 0.000128). The t test for Total Excluding Used Vehicles and Parts is significant at  = .05 (t = 2.85, p = 0.013687). The t test for Gas Stations yields a t = – 4.69 with a p-value of 0.000422.

13.33 The regression equation is: ŷ = 87.890 – 0.256 x1 – 2.714 x2 + 0.071 x3

F = 47.571 with a p-value of .000 significant at  = .001. S = se = 0.8503, R2 = .9407, adjusted R2 = .9209. All three predictors produced significant t tests with two of them (x2 and x3) significant at .01 and the other, x1 significant at  = .05. This is Solutions Manual 1-470 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

a very strong model.

13.34 Two of the diagnostic charts indicate that there may be a problem with the error terms being normally distributed. The histogram indicates that the error term distribution might be skewed to the right and the normal probability plot is somewhat nonlinear. In addition, the residuals vs. fits chart indicates a potential heteroscedasticity problem with residuals for middle values of x producing more variability that those for lower and higher values of x.

13.35 The regression equation for new model (based on 10 observations) is:

ŷ = 72.9796 + 0.0037 x1 – 0.2594 x2 The regression model is not significant overall because F = 1.46 with a pvalue of .294. The p-values for x1 (0.365) and x2 (0.179) indicate that in case of 10 observations, Total number of square feet and Age of house are not significant predictors of Market Price . R2 is 0.295. The small value of R2 is underscored by the fact that all predictors have non-significant t values. Since p = number of independent variables = 2 and n = sample size = 10. Then, the expected value of R2 is p 2 2 E R2     0.222 . n  1 10  1 9

 

Thus, we can expect to find R2 close to the true value of 0.295 purely by chance.

13.36 Given p = number of independent variables = 6 and n = sample size = 10. Then, the expected value of R2 is p 6 6 E R2     0.667 . n  1 10  1 9

 

The gap between the real R2 =0.5 and the expected value of R2 = 0.667 is not large. To be confident we have to work with the larger sample size. If we want to keep 6 independent variables, then the sample size should be at least 60.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

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Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

CHAPTER 14: BUILDING MULTIPLE REGRESSION MODELS

14.1

Simple Regression Model: ŷ = – 147.27 + 27.128 x

F = 229.67 with p = .000, se = 27.27, R2 = .97, adjusted R2 = .966, and t = 15.15 (for x) with p = .000. This is a very strong simple regression model. Quadratic Model (Using both x and x2):

ŷ = – 22.01 + 3.385 x + 0.9373 x2 F = 578.76 with p = .000, se = 12.3, R2 = .995, adjusted R2 = .993, for x: t = 0.75 with p = .483, and for x2: t = 5.33 with p = .002. The quadratic model is also very strong with an even higher R2 value. However, in this model only the x2 term is a significant predictor.

14.2

The model is:

ŷ = b0b1x Using logs:

log ŷ = log b0 + x log b1

The regression model is solved for in the computer using the values of x and the values of log y. The resulting regression equation is: log ŷ = 0.5797 + 0.82096 x F = 68.83 with p = .000, se = 0.1261, R2 = .852, and adjusted R2 = .839, t = 8.30 (for x) with p = .000. This model has relatively strong predictability. An examination of the simple regression model, (x2 , y) and (x3 , y) models reveals that the exponential model gives a better fit.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

14.3

Simple regression model: ŷ = - 1,456.6 + 71.017 x

R2 = .928 and adjusted R2 = .910. t = 7.17 (for x) with p = .002. Quadratic regression model:

ŷ = 1,012 - 14.06 x + 0.6115 x2 R2 = .947 but adjusted R2 = .911. The t statistic for the x term is t = - 0.17 with p = .876. The t statistic for the x2 term is t = 1.03 with p = .377 Neither predictor is significant in the quadratic model. Also, the adjusted R2 for this model is virtually identical to the simple regression model. The quadratic model adds virtually no predictability that the simple regression model does not already have. The scatter plot of the data follows:

7000 6000

Ad Exp

5000 4000 3000 2000 1000 30

100

110

Eq & Sup Exp

14.4

The model is:

ŷ = b0b1x Using logs:

log ŷ = log b0 + x log b1

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

The regression model is solved for in the computer using the values of x and the values of log y where x is failures and y is liabilities. The resulting regression equation is: log liabilities = 3.1256 + 0.012846 failures F = 19.98 with p = .001, se = 0.2862, R2 = .666, and adjusted R2 = .633. This model has modest predictability. Let us try to use the model: yˆ  b0 x b1

Using log:

log y = log b0 + b1 log x

The regression model is solved for in the computer using the values of log x and the values of log y where x is failures and y is liabilities. The resulting regression equation is: log liabilities = -0.03806 + 2.24009 log failures F = 24.89 with p = .0005, se = 0.2653, R2 = .713, and adjusted R2 = .685. This model has better predictability than the first one.

14.5

The regression model is:

ŷ = - 28.61 - 2.68 x1 + 18.25 x2 - 0.2135 x12 - 1.533 x22 + 1.226 x1x2 F = 63.43 with p = .000 significant at  = .001 se = 4.669, R2 = .958, and adjusted R2 = .943 None of the t statistics for this model are significant. They are t(x1) = 0.25 with p = .805, t(x2) = 0.91 with p = .378, t(x12) = - 0.33 with .745, t(x22) = - 0.68 with .506, and t(x1x2) = 0.52 with p = .613. This model has a high R2 yet none of the predictors are individually significant. The same thing occurs when the interaction term is not in the model. None of the t statistics are significant. The R2 remains high at .957 indicating that the loss of the interaction term was insignificant.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

14.6

The F value shows very strong overall significance with a p-value of .00000073. This is reinforced by the high R2 of .910 and adjusted R2 of .878. An examination of the t values reveals that only one of the regression coefficients is significant at  = .05 and that is the interaction term with a p-value of .039. Thus, this model with both variables, the square of both variables, and the interaction term contains only one significant t statistic and that is for interaction. Without interaction, the R2 drops to .877 and adjusted R2 to .844. With the interaction term removed, both variable x2 and x22 are significant at  = .01.

14.7

The regression equation is: ŷ = 13.619 - 0.01201 x1 + 2.988 x2

The overall F = 8.43 is significant at  = .01 (p = .009). se = 1.245, R2 = .652, adjusted R2 = .575 The t statistic for the x1 variable is only t = -0.14 with p = .893. However the t statistic for the dummy variable, x2 is t = 3.88 with p = .004. The indicator variable is the significant predictor in this regression model that has some predictability (adjusted R2 = .575).

14.8

The indicator variable has c = 4 categories as shown by the c - 1 = 3 categories of the predictors (x2, x3, x4). The regression equation is:

ŷ = 7.909 + 0.581 x1 + 1.458 x2 - 5.881 x3 - 4.108 x4 Overall F = 13.54, p = .000 significant at  = .001 se = 1.733, R2 = .806, and adjusted R2 = .747 For the predictors, t = 0.56 with p = .585 for the x1 variable (not significant), t = 1.32 with p = .208 for the first indicator variable (x2) and is non-significant, t = -5.32 with p = .000 for x3 the second indicator variable and this is significant at  = .001, t = -3.21 with p = .007 for the third indicator variable (x4) which is significant at  = .01. This model has Solutions Manual 1-476 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

14.9

strong predictability and the only significant predictor variables are the two dummy variables, x3 and x4. This regression model has relatively strong predictability as indicated by R2 = .795. Of the three predictor variables, only x1 and x2 have significant t statistics (using  = .05). x3 (a non-indicator variable) is not a significant predictor. x1, the indicator variable, plays a significant role in this model along with x2.

14.10 The regression model is:

ŷ = 41.225 + 1.081 x1 – 18.404 x2 F = 8.23 with p = .0017 which is significant at  = .01. se = 11.744, R2 = .388 and the adjusted R2 = .341. The t-statistic for x2 (the dummy variable) is –4.05 which has an associated p-value of .0004 and is significant at  = .001. The t-statistic of 0.80 for x1 is not significant (p-value = .4316). With x2 = 0, the regression model becomes ŷ = 41.225 + 1.081x1. With x2 = 1, the regression model becomes ŷ = 22.821 + 1.081x1. The presence of x2 causes the y intercept to drop by 18.404. The graph of each of these models (without the dummy variable and with the dummy variable equal to one) is shown below:

14.11 The regression equation is: Price = 3.4394 - 0.0195 Hours + 9.113 ProbSeat + 10.528 Downtown The overall F = 6.58 with p = .0099 which is significant at  = .01. se = 3.94, R2 = .664, and adjusted R2 = .563. The difference between R2 and Solutions Manual 1-477 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

adjusted R2 indicates that there are some non-significant predictors in the model. The t statistics, t = - 0.13 with p = .901 and t = 1.34 with p = .209, of Hours and Probability of Being Seated are non-significant at  = .05. The only significant predictor is the dummy variable, Downtown location or not, which has a t statistic of 3.95 with p = .003 which is significant at  = .01. The positive coefficient on this variable indicates that being in the Downtown adds to the price of a meal.

14.12 There will be six predictor variables in the regression analysis: three for occupation, two for industry, and one for marital status. The dependent variable is job satisfaction. In total, there will be seven variables in this analysis.

14.13 Stepwise Regression: Step 1:

After developing a simple regression model for each independent variable, we select the model with x2 with t = - 7.35 and R2 = .794. The model is ŷ = 36.15 - 0.146 x2.

Step 2:

x3 enters the model and x2 remains in the model. t for x2 is -4.60, t for x3 is 2.93. R2 = .876. The model is ŷ = 26.40 - 0.101 x2 + 0.116 x3.

Step 3:

The regression model is explored that contains x1 in addition to x2 and x3. The model does not produce any significant result. No new variable is added to the model produced in Step 2. Note that at every step of the procedure, the variable x1 appears to be non-significant.

14.14 Stepwise Regression: Step 1:

After developing a simple regression model for each independent variable, we select the model with x4 with t = - 4.20 and R2 = .525. The model is ŷ = 133.53 - 0.78 x4.

Step 2:

x2 enters the model and x4 remains in the model.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

t for x4 is - 3.22 and t for x2 is 2.15. R2 = .637 The model is ŷ = 91.01 - 0.60 x4 + 0.51 x2 Step 3:

The regression models are explored that contain x1 (or x3) in addition to x2 and x4. None of the two models produce significant results. No new variables are added to the model produced in Step 2. Note that at every step of the procedure, the variables x1 and x3 appear to be non-significant.

14.15 The output shows that the final model had four predictor variables, x3, x1, x2, and x6. The variables, x4 and x5 did not enter the stepwise analysis. The procedure took four steps. The final model was: y1 = 5.96 – 5.00 x3 + 3.22 x1 + 1.78 x2 + 1.56 x6 The R2 for this model is .5929, and se is 3.36. The t ratios are: x3 : t = 3.07; x1 :t = 2.05; x2: t = 2.02; and x6: t = 1.98.

14.16 The output indicates that the stepwise process only went two steps. Variable x3 entered at step one. However, at step 2, x3 dropped out of the analysis and x2 and x4 entered as the predictors. Note that at every step of the procedure, the variable x1 appears to be non-significant. The final regression model is: ŷ = 22.30 + 12.38 x2 + 0.0047 x4. R2 = .682 and se = 9.47. For x2: t = 2.64 and for x4: t = 2.01.

14.17 Stepwise Regression: Step 1:

After developing a simple regression model for each independent variable, we select the model for Durability with t = 3.32. For this model: R2 = .379 and se = 15.48. The regression equation is: Amount Spent = 17.093 + 7.135 Durability

Step 2:

The regression models are explored that contain Value

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

or Service in addition to Durability. The t value of the regression coefficient for Value (Service) is not significant. No new variable is added to the model produced in Step 1.

14.18 Stepwise Regression: Step 1:

After developing a simple regression model for each independent variable, we select the model for Natural Gas with t = 11.48. For this model: R2 = .9295 and se = 0.490. The regression equation is: Electricity = 1.748 + 0.994 Natural Gas

Step 2:

The regression models are explored that contain Fuel Oil or Gasoline in addition to Natural Gas. The t value of the regression coefficient for Fuel Oil (Gasoline) is not significant. No new variable is added to the model produced in Step 1.

14.19

y y

x1 -.653

-.653 -.891

.821

x2 -.891 .650

.650 -.615 1

-.688

x3 .821 -.615 -.688

There appears to be some correlation between all pairs of the predictor variables, x1, x2, and x3. All pairwise correlations between independent variables are in the .600 to .700 range.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

14.20

-.241

.621

.278 -.724

x1 -.241

-.359 -.161

.325

.621 -.359

.243

-.442

.278 -.161

.243

-.278

-.442 -.278

x4 -.724

.325

An examination of the intercorrelations of the predictor variables reveals that the highest pairwise correlation exists between variables x2 and x4 (-.442). Other correlations between independent variables are less than .400. Multicollinearity may not be a serious problem in this regression analysis.

14.21 The predictor intercorrelations are: Value

Durability

Service

.559

.533

Durability

.559

.364

Service

.533

.364

Value

An examination of the predictor intercorrelations reveals that Service and Durability have very little correlation, but Value and Durability have a correlation of .559 and Value and Service a correlation of .533. These correlations might suggest multicollinearity.

14.22 The intercorrelations of the predictor variables are: Natural Gas

Fuel Oil

Gasoline

Natural Gas

.570

.701

Fuel Oil Gasoline

.570 .701

1 .934

.934 1

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Each of these intercorrelations is not small. Of particular concern is the correlation between Fuel Oil and Gasoline which is .934. These two variables seem to be adding about the same predictability to the model. In the stepwise regression analysis only Natural Gas entered the procedure. Perhaps the overlapping information between Natural Gas and Fuel Oil and Gasoline was such that Fuel Oil and Gasoline did not have significant unique variance to add to the prediction.

14.23 The log of the odds ratio or logit equation is: ln ( S )  0.932546  0.0000323 Payroll Expenditures.

The G statistic is 11.175 which with one degree of freedom has a p-value of 0.001. Thus, there is overall significance in this model. The predictor, Payroll Expenditures, is significant at  = .01 because the associated p-value of 0.008 is less than  = .01. If the payroll expenditures are $80,000, then ln ( S )  0.932546  0.0000323 (80,000) 14.5 PR ln(S )  3.516546

OBLEMS

3.516546

S e  0.0297. From this, the probability that the hospital with the $80,000 payroll expenditure is a psychiatric hospital can be determined by S 0.0297  p   0.0288 or 2.88% . S  1 0.0297  1

14.24 The log of the odds ratio or logit equation is: ln ( S )  0.705987  0.0001257 Annual Food Spending .

The G statistic is 6.957 which with one degree of freedom has a p-value of 0.008. Thus, there is overall significance in this model at  = .01. The p-value associated with the predictor variable, Annual Food Spending, is 0.010. This indicates that Annual Food Spending is a significant predictor in the model. If the annual food spending by household is $12,000, then ln ( S )  0.705987  0.0001257 (12,000)

ln(S )  0.802413 S  e 0.802413  0.4482. Solutions Manual 1-482 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

From this, the probability that family with an average annual food expenditure of $12,000 is outside a metro area can be determined by S 0.4482  p   0.3095 or about 31% . S  1 0.4482  1 In this case n0  120, n1  80, and N  200. The Log Likelihood of the constant-only model is:  120 120  80  80  Log Likelihood " constant - only" model  ln       134.602 .  200   200   Using Log Likelihood = - 131.124 (see Minitab output) we can also check: G = 2{[Log Likelihood with variable] - [Log Likelihood without variable]} G = 2{[-131.124] – [-134.602]} = 6.957.

14.25 The log of the odds ratio or logit equation is: ln ( S )  3.07942  0.0544532 Number of Production Workers .

The G statistic is 97.492 which with one degree of freedom has a p-value of 0.000. Thus, there is overall significance in this model. The p-value associated with the predictor variable, Number of Production Workers, is 0.000. This indicates that Number of Production Workers is a significant predictor in the model at  = .001. If the number of production workers is 30, then ln ( S )  3.07942  0.0544532 30

ln ( S )  1.445824 S  e 1.445824  0.23555. From this, the probability that that a company with 30 production workers has a large value of industrial shipments can be determined by S 0.23555  p   0.1906 or 19.06% . S  1 0.23555  1

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

14.26 The log of the odds ratio or logit equation is:

ln ( S )  5.40701  0.0295051 Number of Production Workers   0.0074783 New Capital Expenditures . The G statistic is 139.132 and is greater than the G = 97.492 for the model with the single predictor in problem 14.25. In this case, the degrees of freedom is equal to 2. The G statistic (139.132) also has a p-value of 0.000 . As in the problem 14.25, there is overall significance in this model. The p-value associated with the first predictor variable, Number of Production Workers, is 0.002. This indicates that Number of Production Workers is a significant predictor in the model at  = .01. The p-value associated with the second predictor variable, New Capital Expenditures, is 0.000. This indicates that New Capital Expenditures is a significant predictor in the model at  = .001. Therefore, both variables are statistically significant in this model and the added variable, New Capital Expenditures, is more significant than Number of Production Workers. If the number of production workers is 80 and new capital expenditures are 1,000, then ln ( S )  5.40701  0.0295051(80)  0.00747831,000

ln ( S )  4.4317 S  e 4.4317  84.074 . From this, the probability that that a company with 80 production workers and 1,000 in new capital expenditures has a large value of industrial shipments can be determined by S 84.074  p   0.9882 or about 99% . S  1 84.074  1 This indicates that there is very high probability that this company is large in terms of values of industrial shipments. The log of the odds ratio or logit equation is: ln(S) = -1.1143 + 0.00485 Market Value The P-value associated with the predictor variable, Market Value is 0.000. This indicates that market value is a significant predictor. If the market Value is 300, then ln(S) = -1.1143 + 0.00485(300) = 0.34445

The probability that a company with market value of 300 has headquarters in North America is 58.4%. If the market Value is 10, then ln(S) = -1.1143 + 0.00485(10) = -1.0658

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

The probability that a company with market value of 10 has headquarters in North America is 25.6%.

14.27 The regression model is:

ŷ = 564.2 - 27.99 x1 - 6.155 x2 - 15.90 x3 F = 11.32 with p = .003, se = 42.88, R2 = .809, adjusted R2 = .738.Thus, overall there is statistical significance at  = .01, For x1, t = -0.92 with p = .384, for x2, t = -4.34 with p = .002, for x3, t = -0.71 with p = .497. Thus, only one of the three predictors, x2, is a significant predictor in this model. This model has very good predictability (R2 = .809). The gap between R2 and adjusted R2 underscores the fact that there are two non-significant predictors in this model. x1 and x3 are non-significant indicator variables.

14.28 Stepwise Regression: Step 1:

After developing a simple regression model for each independent variable (x1 , x12 , x2 , x22 , x1 x2) , we select the model for x1 because it has the largest absolute value of t = 11.55. For this model: R2 = .9112. The model appears in the form ŷ = 1540 + 48.2 x1.

Step 2:

The regression models with two predictors are explored that contain x12 (or x2 , x22 , x1 x2) in addition to x1. At this step, analysis of the t statistics shows the best model: ŷ = 1237 + 136.1 x1 - 5.887 x12. The R2 at this step is .9723, the t ratio for x1 is 7.89 with p = .000, and the t ratio for x12 is - 5.14 with p = .0002.

Step 3:

A search is made to determine which of the x2, x22, x1 x2 variables in conjunction with x1 and x12 produces the largest significant absolute t value in the model. None of the models produce significant results. No new variables are added to the model produced in Step 2.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

14.29 Stepwise Regression: Step 1:

After developing a simple regression model for each independent variable(x1 , Log x1) , we select the model for Log x1 because it has the largest absolute value of t = 17.36 ( p-value of 0.000). For this model: R2 = .9617. The model appears in the form: ŷ = - 13.20 + 11.64 Log x1.

Step 2:

The regression model with two predictors is explored that contains x1 in addition to Log x1. At this step, the t ratio for x1 is 0.90 with the p-value = 0.386. It indicates that the predictor x1 is non-significant. No new variable is added to the model produced in Step 1.

14.30 Let Oilseeds = x1 and Livestock = x2. Stepwise Regression: Step 1:

Using graphs and Tukey‟s ladder of transformations we develop a simple regression model for each independent variable (x1 , Log x1 , x2 , Log x2). We select the model for Log x1 because it has the largest absolute value of t = 5.37. For this model: R2 = .762. The model appears in the form ŷ = -9.562 + 19.71Log x1.

Step 2:

A search is made to determine which of the x1 or x2 or log

x2 variables in conjunction with Log x1 produces the largest significant absolute t value in the model. None of the models produce significant results. No new variables are added to the model produced in Step 2. Note that if we use a linear regression model, then the model appears to be ŷ = – 4.675 + 0.4732 x1 + 1.180 x2. The value of R2 is .9014 and adjusted R2 = .8767.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Step 1:

After developing a simple regression model for each independent variable (Copper , Silver, Aluminum), we select the model with Silver because it has the largest absolute value t statistic: tSilver = 3.32 ( p-value of 0.007). The predictor Silver is significant at  = .01. For this model: R2 = 0.5244. The regression equation is Gold = 233.4 + 17.74 Silver .

Step 2:

The regression models with two predictors are explored that contain Copper (or Aluminum) in addition to Silver. At this step, analysis of the t statistics shows the best model: Gold = – 50.07 + 18.86 Silver +3.587 Aluminum. 2 The R at this step is .8204, the t ratio for Silver is 5.43 with p = .0004, and the t ratio for Aluminum is 3.85 with p = .004.

Step 3:

A search is made to determine whether the variable Copper in conjunction with Silver and Aluminum produces the largest significant absolute t value in the model. The model does not produce significant result. No new variable is added to the model produced in Step 2.

14.32 The graphs (see figures below) are not shaped like the curves in Tukey‟s ladder of transformations. Each graph contains at least three visible outliers.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Step 1:

We develop a simple regression model for each independent variable (Number of Families, Population). The results are following: model with Number of Families: R2 = 0.0027, t-ratio = 0.146, p-value = 0.887

; model with Population: R2 = 0.0056, t-ratio = 0.212, p-value = 0.837 None of the independent variables produces a t value that is significant. Thus, the search procedure stops at Step 1 and recommends no model.

14.33 Let Beef = x1, Chicken = x2, Eggs = x3, Bread = x4, Coffee = x5, and Price Index = y. Stepwise Regression: Step 1:

Using graphs and Tukey‟s ladder of transformations we develop a simple regression model for each independent variable (x1, Log x2, x3 , x4 , x5 ). We select the model for x1 because it has the largest absolute value of t = 13.67. For this model: R2 = .8696. The model appears in the form ŷ = 93.62 + 0.2080 x1.

Step 2:

The regression models with two predictors are explored that contain Logx2 (or x3, x4, x5) in addition to x1. At this

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

step, analysis of the t statistics shows the best model: ŷ = 86.96 + 0.1427 x1 + 0.08561 x4. 2 The R at this step is .9033, the t ratio for x1 is 5.67 with p = .000, and the t ratio for x4 is 3.06 with p = .005 (it is significant at  = .01). Step 3:

A search is made to determine which of the remaining independent variables in conjunction with x1 and x4 produces the largest significant absolute t value in the model. None of the models produce significant results. No new variables are added to the model produced in Step 2.

14.34 Stepwise Regression: Step 1:

After developing a simple regression model for each independent variable (Soybeans , Wheat) , we select the model with Soybeans because it has the largest absolute value t statistic: tSoybeans = 5.43 ( p-value of 0.0006). The predictor Soybeans is significant at  = .001. For this model: R2 = 0.7868. The regression equation is Corn = - 2,962 + 5.445 Soybeans.

Step 2:

A search is made to determine whether the variable Wheat in conjunction with Soybeans produces the significant absolute t value in the model. The model does not produce significant result. No new variable is added to the model produced in Step 1.

14.35 Stepwise Regression: Step 1:

After developing a simple regression model for each independent variable (Familiarity, Satisfaction , Proximity), we select the model with Familiarity because it has the largest absolute value t statistic: t Familiarity = 6.71 ( p-value of 0.000). The predictor Familiarity is significant at  = .001.For this model: R2 = 0.6167. The regression equation is Number of Visits = 0.05488 + 1.0915 Familiarity.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Step 2:

A search is made to determine whether the variable Satisfaction or Proximity in conjunction with Familiarity produces the significant absolute t value in the model. None of the models produce significant results. No new variables are added to the model produced in Step 1.

14.36 The output shows that the stepwise regression procedure stopped at Step 2. At step 1, the model with x1 is selected. R2 = .7539 and t statistic for x1: t = 7.42 . The regression equation is ŷ = 152.2 – 50.6 x1. At step 2, x2 is entered into the model along with x1. The procedure stops here with a final model of: ŷ = 124.5 - 43.4 x1 + 1.36 x2. The t statistics are : t x1  6.13 and t x2  2.13 . The R2 for this model is .8059 indicating relatively strong predictability with two independent variables (x1 and x2) out of four independent variables. 14.37 The output shows that the stepwise regression procedure stopped at Step 3. At step 1, the model with x3 is selected. R2 = .8124 and t statistic for x3: t = 6.90 . The regression equation is ŷ = 74.81 + 0.099 x3. At step 2, x2 is entered into the model along with x3. The regression equation is ŷ = 82.18 + 0.067 x3 – 2.26 x2. The t statistics are t x3  3.65 and t x2  2.32 . The R2 for this model is .8782. At step 3, x1 is entered into the model along with x3 and x2. The procedure stops here with a final model of: ŷ = 87.89 + 0.071 x3 – 2.71 x2 – 0.256 x1. The t statistics are : t x3  5.22 , t x2  3.71 and t x1  3.08 . The R2 for this model is .9407 indicating very strong predictability.

14.38 The R2 for the full model is .321. After dropping out variable, x3, the R2 is still .321. Variable x3 added virtually no information to the model. This is underscored by the fact that the p-value for the t test of the slope for x3 is .878 indicating that there is no significance. The standard error of the estimate actually drops slightly after x3 is removed from the model.

14.39 The log of the odds ratio or logit equation is: Solutions Manual 1-490 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

ln ( S )  3.94828  1.36988 Number of kilometres .

The G statistic is 100.537 with p-value of 0.000. Thus, the model is significant overall. The degree of freedom is equal to 1. The p-value associated with the predictor variable, Number of kilometres, is 0.000. This indicates that Number of kilometres is a significant predictor in the model at  = .001. If a shopper drives 5 kilometres to get to the store, then ln ( S )  3.94828  1.36988 (5)

ln ( S )  2.90112 S  e 2.90112  18.1945 . From this, the probability that that a person would purchase something can be determined by S 18.1945  p   0.948 or about 95% . S  1 18.1945  1 This indicates that there is very high probability that the person who drives 5 kilometres would purchase something. For 4 kilometres, the probability drops to .822. For 3 kilometres, the probability drops to .540 (almost a coin toss). For 2 kilometres, the probability drops to .230. For 1 kilometer, the probability drops to .071.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Legal Notice Copyright

SOLUTIONS TO PROBLEMS IN CHAPTER 15: TIME-SERIES FORECASTING AND INDEX NUMBERS Solutions Manual 1-492 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

15.1

1 2 3 4 5 6 7 8 9 Total

2.30 1.60 -1.40 1.10 0.30 -0.90 -1.90 -2.10 0.70 -0.30

2.30 1.60 1.40 1.10 0.30 0.90 1.90 2.10 0.70 12.30

5.29 2.56 1.96 1.21 0.09 0.81 3.61 4.41 0.49 20.43

MAD =

MSE =

15.2

Period i

e

Number of Forecasts

e

MAD =

MSE =

12.30 = 1.367 9



20.43 = 2.270 9

Number of Forecasts

Period Value F 1 2 3 4 5 6 7 8 9 10 11



202 191 202 173 192 169 181 171 174 175 172 182 174 196 179 204 189 219 198 227 211 Total

-11 11 121 -19 19 361 -12 12 144 -3 3 9 3 3 9 8 8 64 17 17 289 15 15 225 21 21 441 16 16 256 35 125 1,919

e

Number of Forecasts

e



Number of Forecasts



125.00 = 12.5 10

1,919 = 191.9 10

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

15.3

Period Value F

19.4 16.6

2.8

7.84

3 4 5 6

24.0 22.0 2.0 26.8 24.8 2.0 29.2 25.9 3.3 35.5 28.6 6.9 Total 21.5

2.0 2.0 3.3 6.9 21.5

4.00 4.00 10.89 47.61 94.59

MAD =

MSE =

15.4

e

Number of Forecasts

e



21.5 = 3.583 6



94.59 = 15.765 6

Number of Forecasts

Year Hectares

Forecast

1 2 3 4 5 6 7 8 9

140,000 141,730 134,590 131,710 131,910 134,250 135,220 131,020 120,640

140,000 141,038 137,169 133,894 132,704 133,632 134,585 132,446

1,730 -6,448 -5,459 -1,984 1,546 1,588 -3,565 -11,806

1,730 6448 5459 1984 1546 1588 3565 11806

115,190

125,362

-10,172

10172

114,510

119,259

-4,749

4749

-39,319

49,047

2,992,900 41,576,704 29,800,681 3,936,256 2,390,116 2,521,744 12,709,225

139,381,636 103,469,584 22,553,001 Total 361,331,847

MA MSE =

e

Number of Forecasts



361,331,847 = 36,133,184.7 10

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

15.5

a. Time Period

Value

4 5 6 7 8 9 10

63 59 66 71 86 101 97

4-Month Moving Average

44.75 52.75 61.50 64.75 70.50 81.00

14.25 13.25 9.50 21.25 30.50 16.00 104.75

Total MAD =

e

Number of Forecasts



104.75 = 17.46 6

b. Time Period

Value

4 5 6 7 8 9 10

63 59 66 71 86 101 97

4-Month Weighted Moving Average

53.250 56.375 62.875 67.250 76.375 89.125

5.750 9.625 8.125 18.750 24.625 7.875 74.750

Total MAD =

e

Number of Forecasts



74.750 = 12.46 6

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

c. Comparing MAD in a. with MAD in b., we conclude that the fourmonth moving average produces greater error of forecast than the four-month weighted moving average.

15.6 Time Period

Value

1 2 3 4 5 6 7 8

211 228 236 241 242 227 217 203

Forecast (  0.1)

211 213 215 218 220 221 221 Total

e

MAD =

Number of Forecasts

Time Period

Value

1 2 3 4 5 6 7 8

211 228 236 241 242 227 217 203



e

Forecast (  0.8)

211 225 234 240 242 230 220

Number of Forecasts



23 26 24 7 4 18 102

102 = 17.00 6

Total MAD =

11 7 2 15 13 17 65

65 = 10.83 6

Comparing MAD for α = 0.1 with MAD for α = 0.8, we conclude that the Solutions Manual 1-496 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

exponential smoothing with α = 0.1 produces greater error of forecast than the exponential smoothing with α = 0.8.

15.7 Time Period

Value

Forecast (  0.3)

9.4

2 3 4 5 6 7 8 9

8.2 7.9 9.0 9.8 11.0 10.3 9.5 9.1

9.4 9.0 8.7 8.8 9.1 9.7 9.9 9.8

1.1 0.3 1.0 1.9 0.6 0.4 0.7 6.0

Total MAD =

e

Number of Forecasts



6 .0 = 0.86 7

Time Period

Value

Forecast (  0.7)

9.4

2 3 4 5 6 7 8 9

8.2 7.9 9.0 9.8 11.0 10.3 9.5 9.1

9.4 8.6 8.1 8.7 9.5 10.6 10.4 9.8

Total MAD =

e

Number of Forecasts



0.7 0.9 1.1 1.5 0.3 0.9 0.7 6.1

6 .1 = 0.87 7

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Time Period

Value

9.4

8.2

3 4 5 6 7 8 9

7.9 9.0 9.8 11.0 10.3 9.5 9.1

MAD =

e

3-Month Moving Average

8.5 8.4 8.9 9.9 10.4 10.3

0.5 1.4 2.1 0.4 0.9 1.2 6.5

Total i

Number of Forecasts



6 .5 = 1.08 6

Comparing MAD for α = 0.3, MAD for α = 0.7, and the MAD for a 3-month moving average we conclude that the 3-month moving average produces greater error of forecast than the exponential smoothing with α = 0.3 and with α = 0.7.

15.8 a.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Year 2005 2006 2007 2008 2009 2010 2011 2012 2013 2014 2015 2016 2017 2018

Median Total Income 82,900 87,200 91,560 95,290 95,090 97,850 101,890 105,580 109,090 112,860 116,680 118,620 124,020

Year 2005 2006

5-Year Moving Average 90,408 93,398 96,336 99,140 101,900 105,454 109,220 112,566 116254

| ei | 7,442 8,492 9,244 9,950 10,960 11,226 9,400 11,454 Total = 78,168

Median Total Income

5-Year Weighted Moving Average

82,900 87,200

| ei | -

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

2007 2008 2009 2010 2011 2012 2013 2014 2015 2016 2017 2018

91,560 95,290 95,090 97,850 101,890 105,580 109,090 112,860 116,680 118,620 124,020

93,209 95,486 98,555 101,937 105,256 108,887 112,655 115,631 119,565

4,641 6,404 7,025 7,153 7,604 7,793 5,965 8,389 Total = 54,975

c. Comparing MAD in a. with MAD in b., we conclude that the five-year moving average produces greater error of forecast than the five-year weighted moving average.

15.9 Year 1 2 3 4 5 6 7 8 9 10 11 12 13

Number of Issues 332 694 518 222 209 172 366 512 667 571 575 865 609

Forecast (  0.2) 332 404 427 386 351 315 325 363 423 453 477 555 Total

362 114 205 177 179 51 187 304 148 122 388 54 2,290

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Year 1 2 3 4 5 6 7 8 9 10 11 12 13

Number of Issues 332 694 518 222 209 172 366 512 667 571 575 865 609

Forecast (  0.9) 332 658 532 253 213 176 347 496 650 579 575 836 Total

365 140 310 44 41 190 165 171 79 4 290 227 2,023

Comparing MAD for α = 0.2 with MAD for α = 0.9, we conclude that the exponential smoothing with α = 0.2 produces greater error of forecast than the exponential smoothing with α = 0.9.

15.10 Let x = Year and y = Total Number of New Orders. Linear Trend Analysis: ŷ = 37,969.6 + 9899.0 x F = 1603 (p = .000), R2 = .988, adjusted R2 = .988, se = 6,861, tx = 40.04 (p = .000) Quadratic Regression Trend Model: ŷ = 35,767.3 + 10,473.5 x - 26.1 x2 Solutions Manual 1-502 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

F = 772.68 (p = .000), R2 = .988, adjusted R2 = .987 se = 6,988, tx = 9.91 (p = .000), t x 2 = -0.56 (p = .583) A comparison of the models shows that both models accounts for 98.8% of the variability in Total Number of New Orders, but the quadratic model has a gap between R2 and adjusted R2. It can be explained by the fact that the predictor x2 is not significant (see p-value). Thus, the simple linear regression trend model is superior; the x2 variable is not a significant addition to the model.

15.11 We recode years as 1 through 21 the time periods. Let x = Year and y = Consumer Price Index for Food. Linear Trend Analysis: ŷ = 84.86 +2.9517 x F = 2,312.23 (p-value = 0.000) R2 = 0.992 adjusted R2 = 0.991 1.703 tx = 48.085 (p = .000)

se =

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Quadratic Trend Analysis:

ŷ = 86.34 – 2.5645 x + 0.018 x2 F = 1,254.34 (p = .000) R2 = .993 adjusted R2 = .992 tx = 10.36 (p = .000), t x 2 = 1.61 (p = .125)

se = 1.636

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

A comparison of the scatter plots indicates a quadratic fit rather than a linear fit. The quadratic model produced R2 = 0.993 which is slightly greater than R2 = 0.992 for linear trend. It also confirms a better fit for the quadratic model. In addition, the standard error of the estimate drops from 1.703 to 1.636 with the quadratic model. The t values for both predictors in the quadratic model are significant. Thus, the quadratic regression model is superior; the x2 variable is a significant addition to the model.

15.12

Let x = Year and y = Part-Time Employment (%). We recode years as 1 through 10 the time periods. Linear Trend Analysis: ŷ = 19.13+0.002x F = 0.006 (p-value = 0.94) R2 = 0.0007 adjusted R2 = -0.124 se = 0.288 tx = 0.076 (p = .94)

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Quadratic Trend Analysis:

ŷ = 18.84 + 0.148 x – 0.013 x2 F = 0.57 (p = .589) R2 = .14 adjusted R2 = -.11 tx = 1.057 (p = .0.326), t x 2 = -1.067 (p = .321)

se = 0.286

Both regression models show quite poor predictability. 15.13

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Year Month 2017 August September October November December

Actual Values (T.C.S.I) 4.09 4.02 3.98 4.04 4.05

12 Month Moving Total

12 Month 2 Year Moving Total

T.C

(S.I)x100

97.13

4.047

99.33

97.3

4.054

96.69

97.41

4.059

98.31

97.68

4.070

98.77

97.95

4.081

100.21

98.34

4.098

101.04

98.81

4.117

100.07

99.14

4.131

103.13

99.2

4.133

97.26

99.2

4.133

98.95

99.33

4.139

101.48

99.45

4.144

100.39

48.48 2018 January

4.02 48.65

February

3.92 48.65

March

3.99 48.76

April

4.02 48.92

May

4.09 49.03

June

4.14 49.31

July

4.12 49.5

August

4.26 49.64

September

4.02 49.56

October

4.09 49.64

November

4.2 49.69

December

4.16 49.76

2019 January February March April May

4.3 4.11 4.13 3.94 4.17

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

June July

4.19 4.19

Month

12-Month 12-Month 2-Year Moving Moving Total Total T.C

15.14

T.C.I

Jan(Yr1) 1891

1952.50

2042.72

Feb

1986

1975.73

2049.87

Mar

1987

1973.78

2057.02

Apr

1987

1972.40

2064.17

May

2000

1976.87

2071.32

June

2082

1982.67

2078.46

July 94.49

1878

Ship.

S.I

23822 47689

1987.04

94.51 1970.62

2085.61

47852

1993.83 104.02 2011.83

2092.76

48109

2004.54 104.06 2008.47

2099.91

48392

2016.33 101.42 1969.76

2107.06

48699

2029.13

95.85 2024.57

2114.20

49126

2046.92

90.92 2002.80

2121.35

49621

2067.54

93.64 1998.97

2128.50

49989

2082.88 101.01 2093.12

2135.65

23867 Aug 96.13

2074 23985

Sept 95.65

2086

Oct 93.48

2045

24124

24268 Nov 95.76

1945 24431

Dec 94.41

1861 24695

Jan(Yr2) 1936 93.91 24926 Feb 98.01

2104

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

25063 Mar 98.56

2126

Apr 98.39

2131

May 99.11

2163

50308

2096.17 101.42 2111.85

2142.80

50730

2113.75 100.82 2115.35

2149.94

51132

2130.50 101.53 2137.99

2157.09

51510

2146.25 109.31 2234.07

2164.24

51973

2165.54

97.39 2213.01

2171.39

52346

2181.08 101.37 2144.73

2178.54

52568

2190.33 103.55 2183.71

2185.68

52852

2202.17 103.76 2200.93

2192.83

53246

2218.58

94.97 2193.19

2199.98

53635 2234.79 92.94 2235.26 26909 53976 2249.00 97.07 2254.00 27067 54380 2265.83 98.42 2218.46 27313 54882 2286.75 97.17 2207.21 27569 55355 2306.46 100.54 2301.97 27786 55779 2324.13 101.93 2341.60 27993 56186 2341.08 108.03 2408.34

2207.13

25245

25485

25647 June 2346 103.23 25863 July 2109 101.92 26110 Aug 98.45

2211 26236

Sept 99.91

2268 26332

Oct 2285 100.37 26520 Nov 99.69

2107 26726

Dec 2077 101.27 Jan(Yr3) 2183 101.79 Feb 2230 99.87 Mar 2222 99.04 Apr 2319 102.96 May 2369 104.40 June 2529 107.04

2214.28 2221.42 2228.57 2235.72 2242.87 2250.02

28193 July 2267 105.39

56539

2355.79

96.23 2378.80

2257.17

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Aug 2457 105.26

56936

2372.33 103.57 2383.35

2264.31

57504

2396.00 105.34 2430.19

2271.46

58075

2419.79 103.40 2409.94

2278.61

58426

2434.42

95.05 2408.66

2285.76

58573

2440.54

93.30 2450.50

2292.91

58685

2445.21

95.53 2411.98

2300.05

58815

2450.63 100.95 2461.20

2307.20

58806

2450.25 103.91 2529.06

2314.35

58793

2449.71 104.75 2547.15

2321.50

58920

2455.00 100.73 2444.40

2328.65

59018

2459.08 104.59 2449.29

2335.79

59099

2462.46

94.86 2451.21

2342.94

59141

2464.21 102.18 2442.53

2350.09

59106

2462.75

99.64 2362.80

2357.24

58933

2455.54 104.21 2464.84

2364.39

28590 Sept 2524 106.99 28914 Oct 2502 105.76 29161 Nov 2314 105.38 29265 Dec 2277 106.87 29308 Jan(Yr4) 2336 104.87 29377 Feb 2474 106.67 29438 Mar 2546 109.28 29368 Apr 2566 109.72 29425 May 2473 104.97 29495 June 2572 104.86 29523 July 2336 104.62 29576 Aug 103.93

2518 29565

Sept 100.24

2454 29541

Oct 2559 104.25 29392

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Nov 2384 104.64

58779

2449.13

97.34 2481.52

2371.53

58694

2445.58

94.25 2480.63

2378.68

58582

2440.92

97.87 2466.70

2385.83

58543

2439.29 100.97 2450.26

2392.98

58576

2440.67 103.33 2505.22

2400.13

58587

2441.13

99.01 2399.25

2407.27

58555

2439.79 101.16 2439.46

2414.42

58458

2435.75 102.31 2373.11

2421.57

58352

2431.33

94.76 2417.63

2428.72

58258

2427.42 103.44 2435.74

2435.87

57922

2413.42 103.34 2401.31

2443.01

57658

2402.42 105.31 2436.91

2450.16

57547

2397.79

99.30 2478.40

2457.31

57400

2391.67

92.45 2379.47

2464.46

57391

2391.29

99.40 2454.31

2471.61

29387 Dec 104.29

2305 29307

Jan(Yr5) 2389 103.39 29275 Feb 2463 102.39 29268 Mar 2522 104.38 29308 Apr 99.67

2417 29279

May 2468 101.04 29276 June 98.00

2492

July 99.54

2304

Aug 99.99

2511

29182

29170

29088 Sept 98.29

2494

Oct 99.46

2530

28834

28824 Nov 2381 100.86 28723 Dec 96.55

2211 28677

Jan(Yr6) 2377 99.30 28714

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Feb 95.56

2381

57408

2392.00

99.54 2368.68

2478.76

57346

2389.42

94.92 2252.91

2485.90

57335

2388.96 100.76 2389.32

2493.05

57362

2390.08

99.03 2339.63

2500.20

57424

2392.67 102.23 2329.30

2507.35

28694 Mar 90.63

2268 28652

Apr 95.84

2407

May 93.58

2367

28683

28679 June 92.90

2446

July Aug Sept Oct Nov Dec

2341 2491 2452 2561 2377 2277

28745

Seasonal Indexing: Month Year1 Year2 Year3 Year4 Jan 93.64 97.07 95.53 Feb 101.01 98.42 100.95 Mar 101.42 97.17 103.91 Apr 100.82 100.54 104.75 May 101.53 101.93 100.73 June 109.31 108.03 104.59 July 94.51 97.39 96.23 94.86 Aug 104.02 101.37 103.57 102.18 Sept 104.60 103.55 105.34 99.64 Oct 101.42 103.76 103.40 104.21 Nov 95.85 94.97 95.05 97.24 Dec 90.92 92.94 93.30 94.25 Total

Year5 Year6 Index 97.87 99.40 96.82 100.97 99.54 100.49 103.33 94.92 100.64 99.01 100.76 100.71 101.16 99.03 101.14 102.31 102.23 104.98 94.76 95.28 103.44 103.06 103.34 103.83 105.31 103.79 99.30 96.05 92.45 92.90 1199.69

Adjust each seasonal index by 1.0002584 Final Seasonal Indexes: Month Index Jan 96.85 Feb 100.52 Mar 100.67 Solutions Manual 1-512 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Apr May June July Aug Sept Oct Nov Dec

100.74 101.17 105.01 95.30 103.09 103.86 103.82 96.07 92.92

Ŷ = 2035.58 + 7.1481 X R2 = .682, se = 102.9 Note: Trend Line was determined after seasonal effects were removed (based on TCI column). 15.15 Regression Analysis: Let X = U.S. Rate and Y = Canadian Rate. The regression equation is Canadian Rate = 0.371 + 0.4643 U.S. Rate Predictor Coef t-ratio p Constant 0.371 0.308 0.763 U.S. Rate 0.4643 2.463 0.029 Regression Output for Trend Line:

se = 1.137

Year U.S. Canada 2003 5.8 2 2004 5.3 2.6

R-sq = 0.318

3.064 2.832

-1.064 -0.232

R-sq(adj) = 0.266

et2

et - et-1

(et - et-1)2

1.132 0.054

0.832

0.692

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

2005 2006 2007 2008 2009 2010 2011 2012 2013 2014 2015 2016 2017

3.3 4 3.9 5.2 6.3 6.8 7.4 9.1 6.6 7.6 7.8 6.9 6.9

1.5 2.6 2.2 3.4 4.7 4.4 4.3 4.8 5.3 3.6 4.2 1.3 1.8

1.903 2.228 2.182 2.785 3.296 3.528 3.807 4.596 3.435 3.900 3.993 3.575 3.575

-0.403 0.372 0.018 0.615 1.404 0.872 0.493 0.204 1.865 -0.300 0.207 -2.275 -1.775

0.163 0.138 0.000 0.378 1.971 0.760 0.243 0.042 3.476 0.090 0.043 5.175 3.150 16.814

-0.171 0.775 -0.354 0.596 0.789 -0.532 -0.379 -0.289 1.661 -2.164 0.507 -2.482 0.500

0.029 0.601 0.125 0.356 0.623 0.283 0.143 0.084 2.758 4.684 0.257 6.161 0.250 17.047

Using Table A.9 for n = 15, k = 1(1 independent variable), and  = .05, we find the critical values: dL = 1.08 and dU = 1.36. Since D = 0.014 is below dL =

1.08, we reject the null hypothesis. At  = .05, there is enough evidence

that significant autocorrelation is present in the model.

15.16

Regression Analysis: Let X = first differences in U.S. Rate and Y = firstdifferences in Canadian Rate. The regression equation is: First Diff. in Canadian Rate = – 0.03994 + 0.3202 First Diff. in U.S. Rate Predictor Coef t-ratio p Constant – 0.0394 – 0.125 0.902 First Diff 0.3202 1.191 0.257 se = 1.177

R-sq = 0.106

R-sq(adj) = 0.031

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

1-Diff. in U.S. Rate

1-Diff. in Canadian Rate

et2

e t - et-1

(e t - et-1)2

-0.5 -2 0.7 -0.1 1.3 1.1 0.5 0.6 1.7 -2.5 1 0.2 -0.9 0

0.6 -1.1 1.1 -0.4 1.2 1.3 -0.3 -0.1 0.5 0.5 -1.7 0.6 -2.9 0.5

-0.200 -0.680 0.185 -0.071 0.377 0.313 0.121 0.153 0.505 -0.840 0.281 0.025 -0.328 -0.039

0.800 -0.420 0.915 -0.329 0.823 0.987 -0.421 -0.253 -0.005 1.340 -1.981 0.575 -2.572 0.539 Total

0.639 0.177 0.838 0.108 0.678 0.975 0.177 0.064 0.000 1.795 3.923 0.331 6.617 0.291 16.613

-1.220 1.336 -1.244 1.152 0.164 -1.408 0.168 0.248 1.345 -3.321 2.556 -3.148 3.112

1.488 1.784 1.547 1.327 0.027 1.982 0.028 0.061 1.808 11.026 6.534 9.909 9.684 47.205

Using Table A.9 for n = 14 n  15 , k = 1(1 independent variable), and  = .05, Solutions Manual 1-515 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

we find the critical values: dL = 1.08 and dU = 1.36. Since D = 2.841 is above dU =

1.36, we fail to reject the null hypothesis. At  = .05, there is not enough

evidence to support the claim that significant autocorrelation is present in the model.

15.17 The simple regression forecasting model is:

CPI = -15.044 + 1.277 Industry Price Index R2 = 0.963

adjusted R2 = 0.960

se = 2.438

F = 462.048

.000

Year

Industry Product Price Index

et2

et - et-1

(et - et-1)2

CPI

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

1999 2000 2001 2002 2003 2004 2005 2006 2007 2008 2009 2010 2011 2012 2013 2014 2015 2016 2017 2018

86.3 90.0 90.9 90.9 89.8 92.7 94.2 96.4 97.9 102.1 98.5 100.0 107.0 108.1 108.6 111.3 110.3 110.1 113.6 117.9

92.9 95.4 97.8 100.0 102.8 104.7 107.0 109.1 111.5 114.1 114.4 116.5 119.9 121.7 122.8 125.2 126.6 128.4 130.4 133.4

95.173 99.899 101.048 101.048 99.643 103.347 105.263 108.072 109.988 115.352 110.754 112.670 121.610 123.015 123.653 127.102 125.824 125.569 130.039 135.531

-2.273 -4.499 -3.248 -1.048 3.157 1.353 1.737 1.028 1.512 -1.252 3.646 3.830 -1.710 -1.315 -0.853 -1.902 0.776 2.831 0.361 -2.131 Total

5.167 20.238 10.550 1.098 9.965 1.831 3.019 1.056 2.286 1.567 13.291 14.669 2.924 1.729 0.728 3.616 0.601 8.014 0.130 4.540 107.021

-2.225 1.251 2.200 4.205 -1.804 0.384 -0.710 0.484 -2.764 4.898 0.184 -5.540 0.395 0.461 -1.048 2.677 2.055 -2.470 -2.492

4.952 1.564 4.840 17.681 3.253 0.148 0.504 0.235 7.640 23.987 0.034 30.691 0.156 0.213 1.099 7.167 4.225 6.101 6.208 120.697

From Table A.9 the critical table values for k = 1, n = 20, and α = 0.05 are dL = 1.20 and dU = 1.41. Since the observed value of D = 1.13 is below dU, we reject the null hypothesis. There is significant autocorrelation.

15.18

The regression equation is: First Diff. in CPI = 1.7758 + 0.2139 First Diff. in Industry Price Index R2 = 0.546

adjusted R2 = 0.519

se = 0.481

F = 20.411 with p –

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

1-Diff. in Industry Price Index 3.7 0.9 0.0 -1.1 2.9 1.5 2.2 1.5 4.2 -3.6 1.5 7.0 1.1 0.5 2.7

1-Diff. in CPI 2.5 2.4 2.2 2.8 1.9 2.3 2.1 2.4 2.6 0.3 2.1 3.4 1.8 1.1 2.4

et2

et - et-1

(e t - et-1)2

2.567 1.968 1.776 1.541 2.396 2.097 2.246 2.097 2.674 1.006 2.097 3.273 2.011 1.883 2.353

-0.067 0.432 0.424 1.259 -0.496 0.203 -0.146 0.303 -0.074 -0.706 0.003 0.127 -0.211 -0.783 0.047

0.005 0.186 0.180 1.586 0.246 0.041 0.021 0.092 0.006 0.498 0.000 0.016 0.045 0.613 0.002

0.499 -0.007 0.835 -1.756 0.699 -0.350 0.450 -0.378 -0.632 0.709 0.124 -0.338 -0.572 0.829

0.249 0.000 0.698 3.082 0.489 0.122 0.202 0.143 0.399 0.503 0.015 0.114 0.327 0.688

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

-1.0 -0.2 3.5 4.3

1.4 1.8 2.0 3.0

1.562 1.733 2.524 2.696

-0.162 0.067 -0.524 0.304 Total

0.026 0.004 0.275 0.093 3.936

-0.209 0.229 -0.591 0.829

0.044 0.052 0.350 0.687 8.164

From Table A.9 the critical table values for k = 1, n = 19, and α = 0.05 are dL = 1.18 and dU = 1.40. Since the observed value of D = 2.074 is above dU, we fail to reject the null hypothesis. There is no significant autocorrelation.

15.19

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Year 1991 1992 1993 1994 1995 1996 1997 1998 1999 2000 2001 2002 2003 2004 2005 2006 2007 2008 2009 2010 2011 2012 2013 2014 2015 2016 2017

Crude Oil Production Yt 68.7 71.6 74.9 77.3 79.9 81.8 86.5 91.1 84.4 89.9 90.2 93.1 101.4 104.8 106.8 115.5 109.5 110.0 103.7 110.4 117.0 127.6 138.3 152.7 158.0 161.5 174.2

One Period Lagged Yt-1 (X1) 68.7 71.6 74.9 77.3 79.9 81.8 86.5 91.1 84.4 89.9 90.2 93.1 101.4 104.8 106.8 115.5 109.5 110.0 103.7 110.4 117.0 127.6 138.3 152.7 158.0 161.5

Two-Period Lagged Yt-2 (X2) -

68.7 71.6 74.9 77.3 79.9 81.8 86.5 91.1 84.4 89.9 90.2 93.1 101.4 104.8 106.8 115.5 109.5 110.0 103.7 110.4 117.0 127.6 138.3 152.7 158.0

The model with 1 lagged variable:

F = 786.956

p = .000 R2 = 0.970 adjusted R2 = 0.969

se = 5.0005

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

For the predictor variable Yt-1: t = 28.053 ( p = .000) The model with 2 lagged variables:

F = 336.65 p = .000 R2 = 0.968 adjusted R2 = 0.965 For the predictor variable Yt-1: t = 4.989 ( p = .000) For the predictor variable Yt-2: t = 0.029 ( p = .977)

se = 5.212

Both models have fairly strong predictability. However, the two-period lagged variable is not significant in the second model, indicating the presence of firstorder autocorrelation.

15.20 The model with 2 lagged variables:

F = 4.53 p = .019 R2 = 0.232 adjusted R2 = 0.181 For the predictor variable Yt-1: t = 2.68 ( p = .012) For the predictor variable Yt-2: t = -0.090 ( p = .93)

se = 0.59

The model has modest predictability. The gap between R2 and adjusted R2 indicates presence non-significant predictor. Analysis of p-values shows that the two-period lagged variable is not significant. The first-order autocorrelation is possible.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Year 1983 1984 1985 1986 1987 1988 1989 1990 1991 1992 1993 1994 1995 1996

Energy Supply 17.2 17.3 17.9 18.1 17.8 16.7 15.5 17.2 17.7 17.7 17.4 17.7 17.8 17.9

One-Period Lagged Yt-1 (X1) Two-Period Lagged Yt-2 (X2) 17.2 17.3 17.9 18.1 17.8 16.7 15.5 17.2 17.7 17.7 17.4 17.7 17.8

17.2 17.3 17.9 18.1 17.8 16.7 15.5 17.2 17.7 17.7 17.4 17.7

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

1997 1998 1999 2000 2001 2002 2003 2004 2005 2006 2007 2008 2009 2010 2011 2012 2013 2014 2015 2016 2017

17.6 17.3 17.6 17.6 16.6 17.6 16.2 16.3 16.7 15.9 16.6 16.8 17.2 16.7 17.4 17.6 18.2 17.5 17.6 17.4 17.3

17.9 17.6 17.3 17.6 17.6 16.6 17.6 16.2 16.3 16.7 15.9 16.6 16.8 17.2 16.7 17.4 17.6 18.2 17.5 17.6 17.4

17.8 17.9 17.6 17.3 17.6 17.6 16.6 17.6 16.2 16.3 16.7 15.9 16.6 16.8 17.2 16.7 17.4 17.6 18.2 17.5 17.6

a. Using formulae I i 

Xi 100 , we calculate, for example, the simple Xo

15.21

index number for year 1955: X 31.40 100  139.866  139.9 . I 1955  1955 100  X 1950 22.45 X b. Using formulae I i  i 100 , we calculate, for example, the simple Xo index number for year 1955: X 31.40 100  45.018  45.0 . I 1955  1955 100  X 1980 69.75 The table below displays all the index numbers for given data. Note that in a. 1950 is the base year, Xo = X1950 = 22.45; in b. 1980 is the base year, Xo = X1980 = 69.75.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

15.22

Year

Price

1950 1955 1960 1965 1970 1975 1980 1985 1990 1995 2000 2005 2010

22.45 31.40 32.33 36.50 44.90 61.24 69.75 73.44 80.05 84.61 87.28 89.56 93.22

a. Index (Xo = X1950) 100.0 139.9 144.0 162.6 200.0 272.8 310.7 327.1 356.6 376.9 388.8 398.9 415.2

b. Index (Xo = X1980) 32.2 45.0 46.4 52.3 64.4 87.8 100.0 105.3 114.8 121.3 125.1 128.4 133.6

Given that 2005 is the base year, Xo = X2005 = 721. X Using formulae I i  i 100 , we calculate, for example, the simple Xo

index number for year 2006:

The table below displays all the index numbers for given data.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Year 2005 2006 2007 2008 2009 2010 2011 2012 2013 2014 2015 2016

Triadic Patent Simple Index Families for (Xo = X2005) Canadian R&D 721 100.0 668 92.6 368 51.0 685 95.0 672 93.2 555 77.0 580 80.4 527 73.1 620 86.0 581 80.6 553 76.7 536 74.3

15.23 1995 3.37 4.86 4.22 7.44 19.89

Total

Year 2002 3.08 4.73 5.9 6.82 20.53

2013 4.77 5.52 5.72 8.80 24.81

Index2002 =

20.53 (100) = 103.2 19.89

Index2013 =

24.81 (100) = 124.7 19.89

15.24 2010 0.58 0.83 5.71 4.20 Total 11.32

2011 0.61 0.85 5.77 4.21 11.44

2012 0.65 0.94 6.28 4.35 12.22

2013 0.65 0.94 6.21 4.34 12.14

Year 2014 2015 0.57 0.82 0.85 0.90 6.15 6.02 4.33 4.25 11.90 11.99

2016 0.98 1.00 6.09 4.29 12.36

2017 1.37 1.08 6.15 4.28 12.88

2018 1.53 1.10 6.21 4.35 13.19

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

15.25 Item

Quantity 2005

2005

Price 2016 2017

2018

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

1 2 3 4

21 6 17 43

0.50 1.23 0.84 0.15

0.67 1.85 0.75 0.21

0.68 1.90 0.75 0.25

P2005Q2005 P2016Q2005 P2017Q2005

Totals

0.71 1.91 0.80 0.25 P2018Q2005

10.50 7.38 14.28 6.45

14.07 11.10 12.75 9.03

14.28 11.40 12.75 10.75

14.91 11.46 13.60 10.75

38.61

46.95

49.18

50.72

15.26 Item 1 2 3

2005 Price 22.50 10.90 1.85

2017 Price Quantity 27.80 13 13.10 5 2.25 41

P2005Q2017 P2005Q2018 292.50 270.00 54.50 87.20 75.85 81.40 Totals 422.85 438.60

2018 Price Quantity 28.11 12 13.25 8 2.35 44

P2017Q2017 P2018Q2018 361.40 337.32 65.50 106.00 92.25 103.40 519.15 546.72

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

15.27 a)

The linear model:

Yield = 9.956 - 0.1403 Month

F = 219.24 p = .000

R2 = .909

adjusted R2 = .905 se =

.3212 tMonth = -14.807 (p = .000) The quadratic model:

Yield = 10.44 - 0.2516 Month +

.004455 Month2 F = 176.21 p = .000

R2 = .944

adjusted R2 = .938

= .2582 In the quadratic model, both t ratios are significant, for Month: t = - 7.925, p = .000 and for Month2: t = 3.613, p = .002 Both models are very strong. The quadratic term adds some predictability but has a smaller t ratio than does the linear term.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

x 10.08 10.05 9.24 9.23 9.69 9.55 9.37 8.55 8.36 8.59 7.99 8.12 7.91 7.73 7.39 7.48 7.52 7.48 7.35 7.04 6.88 6.88 7.17 7.22

MAD = c)

│e │ 9.65 .04 9.55 .00 9.43 .06 9.46 .91 9.29 .93 8.96 .37 8.72 .73 8.37 .25 8.27 .36 8.15 .42 7.94 .55 7.79 .31 7.63 .11 7.53 .05 7.47 .12 7.46 .42 7.35 .47 7.19 .31 7.04 .13 6.99 .23  e = 6.77 F -

6.77 = .3385 20

 = .3 x F e 10.08 10.05 10.08 .03 9.24 10.07 .83 9.23 9.82 .59 9.69 9.64 .05 9.55 9.66 .11 9.37 9.63 .26 8.55 9.55 1.00 8.36 9.25 .89

 = .7 F 10.08 10.06 9.49 9.31 9.58 9.56 9.43 8.81

.03 .82 .26 .38 .03 .19 .88 .45

7.99 8.12

8.56 8.16

.57 .04

8.59 8.86 8.60

.87 .48

8.9

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

7.91 7.73 7.39 7.48 7.52 7.48 7.35 7.04 6.88 6.88 7.17 7.22

8.46 .55 8.13 .22 8.30 .57 7.98 .25 8.13 .74 7.81 .42 7.91 .43 7.52 .04 7.78 .26 7.49 .03 7.70 .22 7.51 .03 7.63 .28 7.49 .14 7.55 .51 7.39 .35 7.40 .52 7.15 .27 7.24 .36 6.96 .08 7.13 .04 6.90 .27 7.14 .08 7.09 .13  e = 10.06  e = 5.97

MAD=.3 =

10.06 = .4374 23

MAD=.7 =

5.97 = .2596 23

 = .7 produces better forecasts based on MAD. d).

MAD for b) .3385, c) .4374 and .2596. Exponential smoothing with  =

.7 produces the lowest error (.2596 from part c). e) .

T C S I 10.08 10.05

4-Month Moving Total

8-Month Moving Total

T.C

S . I__

76.81

9.60

96.25

75.92

9.49

97.26

75.55

9.44

102.65

75.00

9.38

101.81

72.99

9.12

102.74

70.70

8.84

96.72

68.36

8.55

97.78

66.55

8.32

103.25

65.67

8.21

97.32

38.60 9.24 38.21 9.23 37.71 9.69 37.84 9.55 37.16 9.37 35.83 8.55 34.87 8.36 33.49 8.59 33.06 7.99 32.61 Solutions Manual 1-531 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

8.12

64.36

8.05

100.87

62.90

7.86

100.64

61.66

7.71

100.26

60.63

7.58

97.49

59.99

7.50

99.73

59.70

7.46

100.80

59.22

7.40

101.08

58.14

7.27

101.10

56.90

7.11

99.02

56.12

7.02

98.01

56.12

7.02

98.01

31.75 7.91 31.15 7.73 30.51 7.39 30.12 7.48 29.87 7.52 29.83 7.48 29.39 7.35 28.75 7.04 28.15 6.88 27.97 6.88 28.15 7.17 7.22 1st Period 2nd Period 3rd Period 4th Period

102.65 97.78 100.64 100.80 98.01 101.81 103.25 100.26 101.08 98.01 96.25 102.74 97.32 97.49 101.10 97.26 96.72 100.87 99.73 99.02

The highs and lows of each period (underlined) are eliminated and the others are averaged resulting in: 1st 99.82 2nd 101.05 3rd 98.64 4th 98.67 total 398.18 Since the total is not 400, adjust each seasonal index by multiplying by 400 = 1.004571 resulting in the final seasonal indexes of: 398.18 Seasonal Indexes:

1st 100.28 2nd 101.51 3rd 99.09 Solutions Manual 1-532 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

4th 15.28

99.12

The table below displays all the index numbers for given data. Note that

2002 is the base year, Xo = X2006 = 2,073. Using formulae I i 

Xi 100 , we Xo

calculate, for example, the simple index number for year 2007:

15.29

Year 2006 2007 2008 2009 2010 2011 2012 2013 2014 2015 2016 2017 2018 2019 2020

Quantity 2,073 2,290 2,349 2,313 2,456 2,508 2,463 2,499 2,520 2,529 2,483 2,467 2,397 2,351 2,308

Item 1 2 3 4 5 6 Totals

2016 3.21 0.51 0.83 1.30 1.67 0.62 8.14

Index Number 100.0 110.5 113.3 111.6 118.5 121.0 118.8 120.5 121.6 122.0 119.8 119.0 115.6 113.4 111.3

2017 3.37 0.55 0.90 1.32 1.72 0.67 8.53

2018 3.80 0.68 0.91 1.33 1.90 0.70 9.32

2019 3.73 0.62 1.02 1.32 1.99 0.72 9.40

2020 3.65 0.59 1.06 1.30 1.98 0.71 9.29

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

15.30

2017 2018 2019 Item Price Quantity Price Quantity Price Quantity Quantity 1 2.75 12 2.98 9 3.10 9

2020_ Price 3.21

11 2

0.85

0.89

0.95

0.98

1.33

1.32

1.36

1.40

66 32 Compute Laspeyres price index: P2017Q2017 P2019Q2017 33.00 37.20 39.95 44.65 26.60 27.20 Totals 99.55 109.05

Compute Paasche price index: P2017Q2020 30.25 56.10 42.56 Totals 128.91

P2020Q2020 35.31 64.68 44.80 144.79

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

15.31 a) and b) α =0.2

3-year Year 1977 1978 1979 1980 1981 1982 1983 1984 1985 1986 1987 1988 1989 1990 1991 1992 1993 1994 1995 1996 1997 1998 1999 2000 2001 2002 2003 2004 2005 2006 2007 2008 2009

Emission s 406 409 423 428 411 393 385 402 403 394 406 437 453 429 423 435 435 450 461 476 493 498 508 533 526 533 557 554 559 544 569 551 525

Moving Average – – – 412.7 420.0 420.7 410.7 396.3 393.3 396.7 399.7 401.0 412.3 432.0 439.7 435.0 429.0 431.0 440.0 448.7 462.3 476.7 489.0 499.7 513.0 522.3 530.7 538.7 548.0 556.7 552.3 557.3 554.7

│ei│ – – – 15.3 9.0 27.7 25.7 5.7 9.7 2.7 6.3 36.0 40.7 3.0 16.7 0.0 6.0 19.0 21.0 27.3 30.7 21.3 19.0 33.3 13.0 10.7 26.3 15.3 11.0 12.7 16.7 6.3 29.7

F – 406.0 406.6 409.9 413.5 413.0 409.0 404.2 403.8 403.6 401.7 402.6 409.5 418.2 420.4 420.9 423.7 426.0 430.8 436.8 444.6 454.3 463.0 472.0 484.2 492.6 500.7 512.0 520.4 528.1 531.3 538.8 541.2

│ei│ – – 16.4 18.1 2.5 20.0 24.0 2.2 0.8 9.6 4.3 34.4 43.5 10.8 2.6 14.1 11.3 24.0 30.2 39.2 48.4 43.7 45.0 61.0 41.8 40.4 56.3 42.0 38.6 15.9 37.7 12.2 16.2

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

2010

537

548.3 Total

e

11.3 529.1

538.0

1.0 808.2

529.1  17.1 Number of Forecasts 31 808.2  ei   25.3 MAD=.2 = Number of Forecasts 32

MADmoving average =



The three-year moving average produced a smaller MAD (17.1) than did exponential smoothing with  = .2 (MAD = 25.3). Using MAD as the criterion, the three-year moving average was a better forecasting tool than the exponential smoothing with  = .2.

15.32

Year 2014

Month January February March April May June

Actual Value

12-Month Moving T  C  S  I  Total

12-Month Ratios of Actual 2-Year Centered Moving Average Moving Total T  C 

Values to Moving Averages

S  I   100

1,591 1,337 2,122 2,781 2,216 1,518 24,698

July

1,167

49,283

2,053

56.84

49,864

2,078

96.15

50,656

2,111

121.51

51,409

2,142

126.14

53,765

2,240

99.29

56,861

2,369

104.56

59,504

2,479

59.62

61,919

2,580

78.72

24,585 August

1,998 25,279

September

2,565 25,377

October

2,702 26,032

November

2,224 27,733

December

2,477 29,128

2015

January

1,478 30,376

February

2,031

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

31,543 March

2,220

63,025

2,626

84.54

63,256

2,636

130.35

65,056

2,711

144.49

66,974

2,791

104.37

67,621

2,818

85.70

68,689

2,862

110.59

69,624

2,901

86.32

70,483

2,937

101.94

69,821

2,909

128.29

67,979

2,832

101.94

69,483

2,895

59.24

71,251

2,969

96.40

72,218

3,009

77.23

74,473

3,103

135.06

75,494

3,146

79.47

75,464

3,144

79.13

76,869

3,203

135.62

77,464

3,228

93.06

77,132

3,214

113.01

77,103

3,213

128.26

31,482 April

3,436 31,774

May

3,917 33,282

June

2,913 33,692

July

2,415 33,929

August

3,165 34,760

September

2,504

October

2,994

34,864 35,619 November

3,732 34,202

December

2,887 33,777

2016

January

1,715 35,706

February

2,862 35,545

March

2,324 36,673

April

4,191 37,800

May

2,500

June

2,488

37,694 37,770 July

4,344 39,099

August

3,004 38,365

September

3,632 38,767

October

4,121 38,336

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

November

3,626

77,977

3,249

111.60

80,623

3,359

88.21

79,829

3,326

91.52

79,172

3,299

64.50

80,039

3,335

81.74

79,005

3,292

114.22

78,844

3,285

115.83

79,358

3,307

115.78

80,663

3,361

65.72

82,099

3,421

131.01

82,820

3,451

87.54

83,277

3,470

106.57

83,146

3,464

112.24

82,572

3,441

93.43

84,754

3,531

116.03

86,866

3,619

69.38

86,630

3,610

84.88

87,300

3,638

106.62

88,150

3,673

96.79

88,049

3,669

95.53

39,641 December

2,963 40,982

2017

January

3,044 38,847

February

2,128 40,325

March

2,726 39,714

April

3,760 39,291

May

3,805 39,553

June

3,829 39,805

July

2,209 40,858

August

4,482 41,241

September

3,021 41,579

October

3,698 41,698

November

3,888 41,448

December

3,215 41,124

2018

January

4,097 43,630

February

2,511 43,236

March

3,064 43,394

April

3,879 43,906

May

3,555 44,244

June

3,505 43,805

July

4,715

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

August September October November December

4,088 3,179 4,210 4,226 2,776

Seasonal Indexes Month January February March April May June July August September October November December

2014

56.84 96.15 121.51 126.14 99.29 104.56

2015 59.62 78.72 84.54 130.35 144.49 104.37 85.70 110.59 86.32 101.94 128.29 101.94

2016 59.24 96.40 77.23 135.06 79.47 79.13 135.62 93.06 113.01 128.26 111.60 88.21

2017 91.52 64.50 81.74 114.22 115.83 115.78 65.72 131.01 87.54 106.57 112.24 93.43

Final 2018 116.03 69.38 84.88 106.62 96.79 95.53

Index 75.57 74.05 83.14 122.29 106.31 99.95 75.71 103.37 100.28 116.36 111.92 97.69

Adjuste d Index 77.73 76.17 85.52 125.79 109.35 102.81 77.87 106.33 103.15 119.69 115.12 100.48

Total 1,166.64

1,200  1.028594939 1,166.64 The adjusted final seasonal indexes are shown in the table above.

Adjust the seasonal indexes by:

15.33 Year

Month

2014

January February March

Actual Values

T  C  S  I  1,591 1,337 2,122

Seasonal Indexes S 77.73 76.17 85.52

Deseasonalized Data

T  C  I  2,047 1,755 2,481

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

2015

2016

2017

April May June July August September October November December January February March April May June July August September October November December January February March April May June July August September October November December January February March April May June July August

2,781 2,216 1,518 1,167 1,998 2,565 2,702 2,224 2,477 1,478 2,031 2,220 3,436 3,917 2,913 2,415 3,165 2,504 2,994 3,732 2,887 1,715 2,862 2,324 4,191 2,500 2,488 4,344 3,004 3,632 4,121 3,626 2,963 3,044 2,128 2,726 3,760 3,805 3,829 2,209 4,482

125.79 109.35 102.81 77.87 106.33 103.15 119.69 115.12 100.48 77.73 76.17 85.52 125.79 109.35 102.81 77.87 106.33 103.15 119.69 115.12 100.48 77.73 76.17 85.52 125.79 109.35 102.81 77.87 106.33 103.15 119.69 115.12 100.48 77.73 76.17 85.52 125.79 109.35 102.81 77.87 106.33

2,211 2,027 1,477 1,499 1,879 2,487 2,257 1,932 2,465 1,901 2,666 2,596 2,732 3,582 2,833 3,101 2,977 2,428 2,501 3,242 2,873 2,206 3,757 2,717 3,332 2,286 2,420 5,579 2,825 3,521 3,443 3,150 2,949 3,916 2,794 3,188 2,989 3,480 3,724 2,837 4,215

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

2018

September October November December January February March April May June July August September October November December

3,021 3,698 3,888 3,215 4,097 2,511 3,064 3,879 3,555 3,505 4,715 4,088 3,179 4,210 4,226 2,776

103.15 119.69 115.12 100.48 77.73 76.17 85.52 125.79 109.35 102.81 77.87 106.33 103.15 119.69 115.12 100.48

2,929 3,090 3,377 3,200 5,271 3,297 3,583 3,084 3,251 3,409 6,055 3,845 3,082 3,517 3,671 2,763

15.34 Linear model: Monthly Sales = 2,049.8 + 31.536 Month F = 39.20 p = .000 R2 = 0.403 adjusted R2 = 0.393 For the predictor variable Month: t = 6.26 ( p = .000)

se = 675.7

Quadratic model: Monthly Sales = 1,699.5 + 65.441 Month - 0.55582 Month2 F = 21.79 p = .000 R2 = 0.433 adjusted R2 = 0.413 se = 664.2 For the predictor variable Month: t = 3.25 ( p = .002) For the predictor variable Month2: t = -1.74 ( p = .09) Both models have modest predictability. In the quadratic model the gap between R2 and adjusted R2 indicates presence non-significant predictor. Analysis of pvalues shows that the variable Month2 contributes very little to the model. Therefore, the linear model is a better choice.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

15.35 2011

2012

2013 Item

Price Quantity

Price

Margarine (500 g)

Price Quantity Quantity 1.26 21

1.32

1.39

Shortening (500 g)

0.94

0.97

1.12

Milk (2 L)

1.43

1.56

1.62

Cola (2 litres)

1.05

1.02

1.25

Potato Chips (220 g) 2.81

2.86

2.99

22 4 65 11 28

Index2011 =

Total

7.49

7.73

P P

2011

100  7.49 100 = 100.0

P P

2012

P P

2013

7.49

2011

Index2012 =

100  7.73 100 = 103.2 7.49

2011

Index2013 =

100  8.37 100 = 111.7 7.49

2011

Totals

P2011Q2011

P2012Q2011

P2013Q2011

26.46 4.70 100.10 12.60 75.87 219.73

27.72 4.85 109.20 12.24 77.22 231.23

29.19 5.60 113.40 15.00 80.73 243.92

IndexLaspeyres 2012 =

IndexLaspeyres 2013 = P2011Q2012

8.37

P P

2012

Q2011

2011

Q2011

P P

P2011Q2013

2013

Q2011

2011

Q2011

100  =

23123 . (100) = 105.2 219.73

100  =

243.92 (100) = 111.0 219.73

P2012Q2012

P2013Q2013

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Total

28.98 2.82 97.24 13.65 81.49 224.18

27.72 3.76 92.95 11.55 78.68 214.66

IndexPaasche 2012 =

IndexPaasche 2013 =

15.36

30.36 2.91 106.08 13.26 82.94 235.55

P P

2012

Q2012

2011

Q2012

P P

2013

Q2013

2011

Q2013

30.58 4.48 105.30 13.75 83.72 237.83

100 

23555 . (100) = 105.1 224.18

100 

237.83 (100) = 110.8 214.66

Regression Analysis: Let X = Number of Business Establishments and Y = SelfEmployment Rate. The regression equation is Yˆ  9.5382  0.2716 X Predictor Coef t-ratio p Constant 9.5382 20.59 .000 X - 0.2716 -3.575 .002 F = 12.78 (p = .002)

se = 0.265

R-sq = 0.402

R-sq(adj) =

0.371 The self-employment rate for a year in which there are 7.0 (million) business establishments: Yˆ 7.0  9.5382  0.2716 7.0  7.637  7.6% Number of SelfBusiness Employmen Establishments t (millions) Rate (%) Ŷ et et2 4.54317 8.1 8.304 -0.204 0.042 4.58651 8.0 8.293 -0.293 0.086 4.63396 8.1 8.280 -0.180 0.032 5.30679 8.2 8.097 0.103 0.011 5.51772 8.2 8.040 0.160 0.026 5.70149 8.0 7.990 0.010 0.000 5.80697 7.9 7.961 -0.061 0.004

et - et-1

(et - et-1)2

-0.089 0.113 0.283 0.057 -0.150 -0.071

0.008 0.013 0.080 0.003 0.023 0.005

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

5.93706 6.01637 6.10692 6.17556 6.20086 6.3193 6.40123 6.50907 6.61272 6.73848 6.89487 6.94182 7.00844 7.07005

8.0 8.2 8.1 8.0 8.1 7.8 8.0 8.1 7.9 7.8 7.7 7.5 7.2 6.9

7.926 7.904 7.880 7.861 7.854 7.822 7.800 7.770 7.742 7.708 7.666 7.653 7.635 7.618

0.074 0.296 0.220 0.139 0.246 -0.022 0.200 0.330 0.158 0.092 0.034 -0.153 -0.435 -0.718 Total

0.005 0.088 0.048 0.019 0.061 0.000 0.040 0.109 0.025 0.008 0.001 0.023 0.189 0.516 1.333

0.135 0.222 -0.076 -0.081 0.107 -0.268 0.222 0.130 -0.172 -0.066 -0.058 -0.187 -0.282 -0.283

0.018 0.049 0.006 0.007 0.011 0.072 0.049 0.017 0.030 0.004 0.003 0.035 0.080 0.080 0.593

 e  e   0.593  0.445 D 1.333 e 2

t 1

Using Table A.9 for n = 21, k = 1(1 independent variable), and  = .05, we find the critical values: dL = 1.22 and dU = 1.42. Since D = 0.445 is below dL =

1.22, we reject the null hypothesis. At  = .05, there is enough evidence

that significant autocorrelation is present in the model.

15.37

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Year

Month

4-Month Moving Average, F

128.9

128.91 128.8 128.64

0.0784 0.16 0.5929

0.0961 0.16 0.7396

August

128.7

September

128.8

October

129.1 128.6 128.4 129.5

128.88 128.80 128.73

November December 2017 January

et2 in case b. 4-Month Weighted Moving Average

CPI

2016 July

4-Month Weighted Moving Average, F

et2 in case a.: 4-Month Moving Average,

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

February March April May June July August September October November December 2018 January February March April May June July August September October November December 2019 January February March April May June

129.7 129.9 130.4 130.5 130.4 130.4 130.5 130.8 130.9 131.3 130.8 131.7 132.5 132.9 133.3 133.4 133.6 134.3 134.2 133.7 134.1 133.5 133.4 133.6 134.5 135.4 136 136.6 136.3

128.90 129.05 129.38 129.88 130.13 130.30 130.43 130.45 130.53 130.65 130.88 130.95 131.18 131.58 131.98 132.60 133.03 133.30 133.65 133.88 133.95 134.08 133.88 133.68 133.65 133.75 134.23 134.88 135.63

128.95 129.27 129.61 130.02 130.27 130.38 130.42 130.45 130.59 130.74 131 130.97 131.27 131.8 132.33 132.86 133.18 133.41 133.81 134.03 133.96 134.02 133.79 133.6 133.57 133.91 134.57 135.28 135.97 Total

0.64 0.7225 1.0404 0.3844 0.0729 0.01 0.0049 0.1225 0.1369 0.4225 0.0064 0.5625 1.7424 1.7424 1.7424 0.64 0.3249 1 0.3025 0.0324 0.0225 0.3364 0.2304 0.0064 0.7225 2.7225 3.1329 2.9584 0.4489 23.0651

0.5625 0.3969 0.6241 0.2304 0.0169 0.0004 0.0064 0.1225 0.0961 0.3136 0.04 0.5329 1.5129 1.21 0.9409 0.2916 0.1764 0.7921 0.1521 0.1089 0.0196 0.2704 0.1521 0 0.8649 2.2201 2.0449 1.7424 0.1089 16.5466

The weighted moving average does a better job of forecasting the data using MSE as the criterion.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

15.38 The regression model with one-month lag is: Cotton Prices = - 61.24 + 1.1035 LAG1 F = 130.46 (p = .000), R2 = .839, adjusted R2 = .833, se = 17.57, t = 11.42 (p = .000). The regression model with four-month lag is: Cotton Prices = 303.9 + 0.4316 LAG4 F = 1.24 (p = .278), R2 = .053, adjusted R2 = .010, se = 44.22, t = 1.11 (p = .278). The model for the four-month lag does not have overall significance and has an adjusted R2 of 1%. This model has virtually no predictability. Since the pvalue for predictor is .278, the variable LAG4 is not significant. The model for the one-month lag has relatively strong predictability with adjusted R2 of 83.3%. Analysis of p-value for LAG1 indicates that the variable is significant predictor at  = .001. In addition, the F value is significant at  = .001 and the standard error of the estimate is less than 44.22 as large as the standard error for the four-month lag model.

15.39

Actual Value Time Period 1st quarter (year 1) 2nd quarter

T  C  S  I 

4-Quarter Moving Total

Ratios of Actual Centered Moving Average

Values to Moving Averages

T  C 

S  I   100

425.044

53.131

94.43

421.546

52.693

100.38

423.402

52.925

98.09

4-Quarter 2-Year Moving Total

54.019 56.495 213.574

3rd quarter

50.169 211.470

4th quarter

52.891 210.076

1st quarter (year 2)

51.915 213.326

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

2nd quarter

55.101

430.997

53.875

102.28

440.49

55.061

97.02

453.025

56.628

101.07

467.366

58.421

97.68

480.418

60.052

104.06

492.176

61.522

98.13

503.728

62.966

100.58

512.503

64.063

97.91

518.498

64.812

105.51

524.332

65.542

96.51

526.685

65.836

100.93

526.305

65.788

99.48

526.72

65.840

103.30

521.415

65.177

97.04

511.263

63.908

104.64

501.685

62.711

95.22

491.099

61.387

103.59

217.671 3rd quarter

53.419 222.819

4th quarter

57.236 230.206

1st quarter (year 3)

57.063 237.160

2nd quarter

62.488 243.258

3rd quarter

60.373 248.918

4th quarter

63.334 254.810

1st quarter (year 4)

62.723 257.693

2nd quarter

68.38 260.805

3rd quarter

63.256 263.527

4th quarter

66.446 263.158

1st quarter (year 5)

65.445 263.147

2nd quarter

68.011 263.573

3rd quarter

63.245 257.842

4th quarter

66.872 253.421

1st quarter (year 6)

59.714 248.264

2nd quarter

63.59 242.835

3rd quarter 4th quarter

58.088 61.443

Seasonal Indexes Quarte Year 1 r 1

Year 2

Year 3

Year 4

Year 5

Year 6

98.09

97.68

97.91

99.48

95.22

Final Index 97.89

Adjusted Index 98.07

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

2 3 4

94.43 100.38

102.28 97.02 101.07

104.06 98.13 100.58

105.51 96.51 100.93

103.30 97.04 104.64

103.59

103.65 96.86 100.86 Total 399.86

103.84 97.04 101.05

400  1.00185343 399.86 The adjusted final seasonal indexes are shown in the table above.

Adjust the seasonal indexes by:

15.40 Actual Values

Year 1

Quarter 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4

Seasonal Indexes, T  C  S  I  S 54.019 56.495 50.169 52.891 51.915 55.101 53.419 57.236 57.063 62.488 60.373 63.334 62.723 68.380 63.256 66.446 65.445 68.011 63.245 66.872 59.714 63.590 58.088 61.443

98.07 103.84 97.04 101.05 98.07 103.84 97.04 101.05 98.07 103.84 97.04 101.05 98.07 103.84 97.04 101.05 98.07 103.84 97.04 101.05 98.07 103.84 97.04 101.05

Deseasonalized Data

T  C  I 

55.082 54.406 51.699 52.341 52.937 53.063 55.048 56.641 58.186 60.177 62.215 62.676 63.957 65.851 65.185 65.756 66.733 65.496 65.174 66.177 60.889 61.238 59.860 60.805

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

15.41 Let x = Time Period (from the 1st quarter (1st of year 1) to the 24th quarter (4th of year 6) and y = Industrial machinery and equipment shipments. Linear Model: ŷ = 53.410 + 0.53249 x F = 27.65 p = .000 R2 = 0.557 adjusted R2 = 0.537 For the predictor variable x: t = 5.26 ( p = .000)

se = 3.43

Quadratic Model: ŷ = 47.687 + 1.8533 x – 0.052834 x2 F = 34.37 p = .000 R2 = 0.766 adjusted R2 = 0.744 se = 2.55 For the predictor variable x: t = 5.90 ( p = .000) For the predictor variable x 2: t = -4.33 ( p = .0003) In the quadratic regression model, both the linear and squared terms have significant t statistics at alpha .001 indicating that both are contributing. In addition, the R2 for the quadratic model is considerably higher than the R2 for the linear model. Also, se is smaller for the quadratic model. All of these indicate that the quadratic model is a stronger model.

15.42

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Investments

Year 2007 2008 2009 2010 2011 2012 2013 2014 2015 2016 2017 2018

One-Period Lagged Yt-1 (X)

116,809 61,520 22,733 28,399 39,667 43,118 69,371 59,008 43,853 35,992 24,826 42,231

116,809 61,520 22,733 28,399 39,667 43,118 69,371 59,008 43,853 35,992 24,826

The autoregression model is: The low value of R2 = 0.22 and the very high value of se = 14,146.5 indicate that this regression model has poor predictability. Since F = 2.543 with p = 0.145, the model is not significant overall. There is no evidence to the presence of first-order autocorrelation.

15.43 The following regression equation was obtained: Foreign Inflows = 19,622.89 +0.508373 Foreign Outflows. Note that Y = Foreign Inflows and X = Foreign Outflows.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Year 2005 2006 2007 2008 2009 2010 2011 2012 2013 2014 2015 2016 2017 2018

Inflows Outflows Ŷ et et2 US$ US$ millions millions 25,693 27,540 33623.47 -7930.47 62892429.86 60,298 46,215 43117.33 17180.67 295175265.55 116,809 64,621 52474.44 64334.56 4138935342.96 61,520 79,236 59904.31 1615.691 2610458.10 22,733 39,660 39784.95 -17052 290769054.39 28,399 34,721 37274.1 -8875.1 78767381.67 39,667 52,144 46131.48 -6464.48 41789452.00 43,118 55,875 48028.21 -4910.21 24110207.66 69,371 57,364 48785.18 20585.82 423775921.48 59,008 60,273 50264.04 8743.962 76456877.25 43,853 67,467 53921.27 -10068.3 101370075.37 35,992 69,948 55182.54 -19190.5 368276953.73 24,826 79,802 60192.05 -35366 1250757331.56 42,231 49,593 44834.62 -2603.62 6778823.84 Total 7,162,465,575.43

et - et-1

(et - et-1) 2

25111.14 47153.89 -62718.87 -18667.64 8176.85 2410.62 1554.26 25496.03 -11841.86 -18812.23 -9122.27 -16175.50 32762.43

630569362.83 2223489574.54 3933656241.17 348480889.32 66860919.48 5811102.32 2415728.91 650047702.84 140229556.47 353900112.52 83215857.71 261646942.01 1073376837.18 9,773,700,827.30

From Table A.9 the critical table values for k = 1, n = 14 (use n=15), and α = 0.01 are dL = 0.81 and dU = 1.07. Since the observed value of D = 1.365 is above dU, we

fail to reject the null hypothesis. At α = 0.01, there is not sufficient sample evidence that significant autocorrelation is present in the model.

15.44

 = .1

 = .5

 = .8

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Year

Value

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

6.04 5.92 5.57 5.40 5.17 5.00 4.91 4.73 4.55 4.34 4.67 5.01 4.86 4.72 4.60 4.48 4.86 5.15

.12 .41 .38 .42 .38 .28 .32 .34 .38 .14 .41 .05 .12 .18 .21 .27 .42 4.83

6.04 5.94 5.64 5.45 5.23 5.05 4.94 4.77 4.59 4.39 4.61 4.93 4.87 4.75 4.63 4.51 4.79

MADα=0.1 =

MAD α=0.5 =

MAD α=0.8 =

with  =

6.04 6.03 5.98 5.92 5.85 5.77 5.68 5.59 5.49 5.38 5.31 5.28 5.24 5.19 5.13 5.07 5.05 Total

.12 .46 .58 .75 .85 .86 .95 1.04 1.15 .71 .30 .42 .52 .59 .65 .21 .10 10.26

e

Number of Forecasts

e

Number of Forecasts

e

Number of Forecasts

6.04 5.98 5.78 5.59 5.38 5.19 5.05 4.89 4.72 4.53 4.60 4.81 4.84 4.78 4.69 4.59 4.73

.12 .37 .24 .28 .23 .14 .21 .22 .25 .28 .40 .07 .15 .15 .15 .35 .36 3.97

10.26 = .60 17

4.83 = .28 17

3.97 = .23 17

The smallest mean absolute deviation error is produced using  = .8. Thus, to predict the figure for year 19 we use exponential smoothing .8.Then, the forecast for year 19 is: F(19) = (.8)(5.15) + (.2)(4.79) = 5.078 or $5.08 .

15.45 The model is: Consumer Bankruptcy

Business Bankruptcy = 75,532.43621 – 0.01574

Since R2 = .28 and the adjusted R2 = .23, the model has very Solutions Manual 1-554 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

modest predictability. Since F = 5.44 with p = 0.0351, the model is significant overall at α = .05, but not significant at α = .01. et2 1,791,849.96 73,758,896.89 49,710,960.36 1,243,225.00 163,131,647.29 216,466,483.84 9,177,264.36 6,755,320.81 387,381.76 95,009,857.29 86,281,805.44 189,051.04 118,274,325.16 96,196,864.00 18,298,717.29 65,946.24 936,739,596.73

et -1,338.6 -8,588.3 -7,050.6 1,115.0 12,772.3 14,712.8 -3,029.4 -2,599.1 622.4 9,747.3 9,288.8 -434.8 -10,875.4 -9,808.0 -4,277.7 -256.8 Total

et – et-1

(et – et-1)2

-7,249.7 1,537.7 8,165.6 11,657.3 1,940.5 -17,742.2 430.3 3,221.5 9,124.9 -458.5 -9,723.6 -10,440.6 1,067.4 5,530.3 4,020.9

52,558,150.09 2,364,521.29 66,677,023.36 135,892,643.29 3,765,540.25 314,785,660.84 185,158.09 10,378,062.25 83,263,800.01 210,222.25 94,548,396.96 109,006,128.36 1,139,342.76 30,584,218.09 16,167,636.81 921,526,504.70

 (e  e )  921,526,504.70 = 0.98 D = 936,739,596.73 e t 1

From Table A.9 for k = 1, n = 16,  = .05: dL = 1.10 and dU = 1.37;  = .01: dL = 0.84 and dU = 1.09. Since D = 0.98 < dL = 1.10, the decision is to reject the null hypothesis and

conclude that there is significant autocorrelation at  = .05 . In case of  = .01, we have 0.84 < D = 0.98 < 1.09. Then, the Durbin-

Watson test is inconclusive.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Legal Notice Copyright

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

SOLUTIONS TO THE PROBLEMS IN CHAPTER 16: ANALYSIS OF CATEGORICAL DATA

16.1

f0 53 37 32 28 18 15

( fo  fe )2 fe

fe 68 42 33 22 10 8

3.309 0.595 0.030 1.636 6.400 6.125

Ho: The observed distribution is the same as the expected distribution. Ha: The observed distribution is not the same as the expected distribution. Observed  2  

( f0  fe )2 = 18.095 fe

df = k - 1 = 6 - 1 = 5,  = .05 2.05,5 = 11.0705 Since the observed 2 = 18.095 > 2.05,5 = 11.0705, the decision is to reject the null hypothesis. The observed frequencies are not distributed the same as the expected frequencies.

16.2

f0 19 17 14

fe 18 18 18

( fo  fe )2 fe

0.056 0.056 0.889

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

18 19 21 18 18 fo = 144

18 18 18 18 18 fe = 144

0.000 0.056 0.500 0.000 0.000 1.557

Ho: The observed frequencies are uniformly distributed. Ha: The observed frequencies are not uniformly distributed. The total for the expected frequencies must equal to the total for the observed frequencies (144). If the frequencies are uniformly distributed, the same number is expected for each category, f e 

144  18 . 8

df = k – 1 = 8 – 1 = 7,  = .01 2.01,7 = 18.4753 Observed  2  

( f0  fe )2 = 1.557 fe

Since the observed 2 = 1.557 < 2.01,7 = 18.4753, the decision is to fail to reject the null hypothesis There is no reason to conclude that the frequencies are not uniformly distributed.

16.3

Number 0 1 2 >3

f0 28 17 11 _5

(Number)(f0) 0 17 22 15 61

Ho: The frequency distribution is Poisson. Ha: The frequency distribution is not Poisson.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

=

54 = 0.9 61

Number 0 1 2 >3

Expected Probability .4066 .3659 .1647 .0628

Expected Frequency 24.803 22.320 10.047 3.831

The probability for x > 3 is determined by summing the probabilities for the values of x = 3, 4, 5, … 7 (see Table A.3). Since fe for > 3 is less than 5, combine categories 2 and >3:

Number 0 1 >2

fo 28 17 16 61

fe 24.803 22.320 13.878 61.001

( fo  fe )2 fe

0.412 1.268 0.324 2.004

df = k - 2 = 3 - 2 = 1,  = .05 2.05,1 = 3.8415 Observed  2  

( f0  fe )2 = 2.004 fe

Since the observed 2 = 2.004 < 2.05,1 = 3.8415, the decision is to fail to reject the null hypothesis. There is insufficient evidence to reject the distribution as Poisson distributed. The conclusion is that at  = .05 the distribution is Poisson distributed.

16.4 Ho: Men entrepreneurs feel the same way Ha: Men entrepreneurs do not feel the same way.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

16.4

Definition Happiness Sales/Profit Helping Others Achievement/ Challenge

fo 42 95 27 63 227

Exp.Prop. fe .39 227(.39)= 88.53 .12 227(.12)= 27.24 .18 40.86 .31

70.37

( fo  fe )2 fe

24.46 168.55 4.70 0.77 198.48

Ho: The observed frequencies are distributed the same as the expected frequencies. Ha: The observed frequencies are not distributed the same as the expected frequencies. Observed 2 = 198.48 df = k – 1 = 4 – 1 = 3,

 = .05

2.05,3 = 7.8147 Since the observed 2 = 198.48 > 2.05,3 = 7.8147, the decision is to reject the null hypothesis. At  = .05, the observed frequencies for men are not distributed the same as the expected frequencies which are based on the responses of women.

16.5

Age 10-14 15-19

fo 22 50

Prop. from survey .09 .23

fe (.09)(212)=19.08 (.23)(212)=48.76

( fo  fe )2 fe

0.45 0.03

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

20-24 25-29 30-34 > 35

43 29 19 49 212

.22 .14 .10 .22

29.68

46.64 0.02 21.20 46.64 1.13

0.28 0.23 0.12

Ho: The distribution of observed frequencies is the same as the distribution of expected frequencies. Ha: The distribution of observed frequencies is not the same as the distribution of expected frequencies.

 = .01, df = k - 1 = 6 - 1 = 5 2.01,5 = 15.0863 The observed 2 = 1.13. Since the observed 2 = 1.13 < 2.01,5 = 15.0863, the decision is to fail to reject the null hypothesis. At  = .01, there is not enough evidence to declare that the distribution of observed frequencies is different from the distribution of expected frequencies. Thus, we do not need to reject the national survey distribution for local recordedmusic shoppers.

16.6

Number 0 1 2 3 4 5 6 or more

f 18 28 47 21 16 11 9

(f) (number) 0 28 94 63 64 55 54

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

f = 150 f(number) = 358  f  number  358 = 2.4 = 150 f Ho: The observed frequencies are Poisson distributed. Ha: The observed frequencies are not Poisson distributed. The probability for x > 6 is determined by summing the probabilities for the values of x = 6, 7, 8, … 12 (see Table A.3).

Number 0 1 2 3 4 5 6 or more

Probability .0907 .2177 .2613 .2090 .1254 .0602 .0358

18 28 47 21 16 11 9

13.61 32.66 39.20 31.35 18.81 9.03 5.37

fe (.0907)(150) = 13.61 (.2177)(150) = 32.66 39.20 31.35 18.81 9.03 5.37

( f0  fe )2 f0 1.42 0.66 1.55 3.42 0.42 0.43 2.45 10.35

The observed 2 = 10.35

 = .01, df = k – 2 = 7 – 2 = 5, 2.01,5 = 15.0863 Since the observed 2 = 10.35 < 2.01,5 = 15.0863, the decision is to fail to reject the null hypothesis. At  = .01, there is not enough evidence to reject the claim that the observed frequencies are Poisson distributed. Solutions Manual 1-562 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

16.7 Category of Car Manufacturers

Exp. Prop.

Luxury car

0.40

Sport car

0.12

Economy car

157

0.48

280(0.40) = 112.00 280(0.12) = 33.60 280(0.48) = 134.40

280

(f o  f e )2 fe

2.29 1.30 3.80 7.39

H0: The observed distribution is the same as the expected distribution. Ha: The observed distribution is not the same as the expected distribution. Observed 2 = 7.39 df = k – 1 = 3 – 1 = 2,

 = .01

2.01,2 = 9.2104 Since the observed 2 = 7.39 < 2.01,2 = 9.2104, the decision is to fail to reject the

null hypothesis. At  = .01, there is not sufficient evidence to declare

that there is a significant difference between the distribution of car manufacturers in the survey and the distribution released by the researcher.

16.8 Category Containers & Packaging Nondurable Goods

(f o  f e )2 fe

Exp. Prop.

0.303

300(0.303) = 90.90

0.26

0.213

300(0.213) = 63.90

1.6

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Durable Goods Yard Trimmings /Other Food Scraps

0.196

58.80

2.13

0.149

44.70

0.31

0.139

41.70

3.87

300 8.17 H0: The observed frequencies are distributed the same as the expected frequencies. Ha: The observed frequencies are not distributed the same as the expected frequencies. Observed 2 = 8.17 df = k – 1 = 5 – 1 = 4,

 = .10

2.10,4 = 7.7794 Since the observed 2 = 8.17 > 2.10,4 = 7.7794, the decision is to reject the null hypothesis. At  = .10, there is a significant difference between the distribution of municipal solid waste in this Midwestern city and the distribution released by the EPA.

16.9 Variable One

Variable Two 203 326 68 110 271 436

529 178 707

Ho: Variable One is independent of Variable Two. Ha: Variable One is not independent of Variable Two.

e11 =

(529)(271) = 202.77 707

e12 =

(529)(436) = 326.23 707

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

e21 =

(271)(178) = 68.23 707

e22 =

Variable Two Variable (202.77) (326.23) One 203 326 (68.23) (109.77) 68 110

2

271 (203  202.77) 2 + 202.77

(436)(178) = 109.77 707

529 178

436 707 (326  326.23) 2 + 326.23

(68  68.23) 2 68.23

(110  109.77) 2 = 109.77

.00 + .00 + .00 + .00 = 0.00

 = .01, df = (c-1)(r-1) = (2-1)(2-1) = 1 2.01,1 = 6.6349 Since the observed 2 = 0.00 < 2.01,1 = 6.6349, the decision is to fail to reject

the null hypothesis. At  = .01, there is not sufficient evidence to support

the statement that Variable 1 is not independent of Variable 2.

16.10

Variable One

24 93

Variable Two 13 47 59 187

58 244

142 583

117

302

725

234

Ho: Variable One is independent of Variable Two. Ha: Variable One is not independent of Variable Two.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

e11 =

(142)(117) = 22.92 725

e12 =

(142)(72) = 14.10 725

e13 =

(142)(234) = 45.83 725

e14 =

(142)(302) = 59.15 725

e21 =

(583)(117) = 94.08 725

e22 =

(583)(72) = 57.90 725

e23 =

(583)(234) = 188.17 725

e24 =

(583)(302) = 242.85 725

Variable Two Variable (22.92) (14.10) (45.83) (59.15) One 24 13 47 58 142 (94.08) (57.90) (188.17) (242.85) 583 93 59 187 244 117 72 234 302 725 2 2 2 (24  22.92) (13  14.10) (47  45.83) (58  59.15) 2 2  = + + + + 22.92 14.10 45.83 59.15 (93  94.08) 2 94.08

(59  57.90) 2 57.90

(188  188.17) 2 188.17

(244  242.85) 2 = 242.85

.05 + .09 + .03 + .02 + .01 + .02 + .01 + .01 = 0.24

 = .01, df = (c-1)(r-1) = (4-1)(2-1) = 3, 2.01,3 = 11.3449 Since the observed 2 = 0.24 < 2.01,3 = 11.3449, the decision is to fail to

reject the null hypothesis. At  = .01, there is not sufficient

evidence to support the statement that Variable 1 is not independent of Variable 2.

16.11

Number

Social Class Lower Middle Upper 7 18 6

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

of Children

1 2 or 3 >3

9 34 47 97

38 97 31 184

23 58 30 117

70 189 108 398

Ho: Social Class is independent of Number of Children. Ha: Social Class is not independent of Number of Children. e11 =

(31)(97) = 7.56 398

e31 =

(189)(97) = 46.06 398

e12 =

(31)(184) = 14.33 398

e32 =

(189)(184) = 87.38 398

e13 =

(31)(117) = 9.11 398

e33 =

(189)(117) = 55.56 398

e21 =

(70)(97) = 17.06 398

e41 =

(108)(97) = 26.32 398

(70)(184) = 32.36 398 (70)(117) e23 = = 20.58 398

e22 =

0 Number of Children

1 2 or 3 >3

2 =

(108)(184) = 49.93 398 (108)(117) e43 = = 31.75 398

e42 =

Social Class Lower Middle Upper (7.56) (14.33) (9.11) 7 18 6 (17.06) (32.36) (20.58) 9 38 23 (46.06) (87.38) (55.56) 34 97 58 (26.32) (49.93) (31.75) 47 31 30 97 184 117

31 70 189 108 398

(7  7.56) 2 (18  14.33) 2 (6  9.11) 2 (9  17.06) 2 + + + + 14.33 17.06 7.56 9.11 (38  32.36) 2 (23  20.58) 2 (34  46.06) 2 (97  87.38) 2 + + + + 46.06 32.36 20.58 87.38

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

(58  55.56) 2 (47  26.32) 2 (31  49.93) 2 (30  31.75) 2 + + + = 26.32 55.56 49.93 31.75 .04 + .94 + 1.06 + 3.81 + .98 + .29 + 3.16 + 1.06 + .11 + 16.25 + 7.18 + .10 = 34.98

 = .05, df = (c-1)(r-1) = (3-1)(4-1) = 6 2.05,6 = 12.5916 Since the observed 2 = 34.98 > 2.05,6 = 12.5916, the decision is to reject the

null hypothesis. At  = .05, Social Class is not independent of Number of Children.

16.12

Region

O/Q A/P BC

Type of Music Preferred Rock R&B Country Classic 140 32 5 18 134 41 52 8 154 27 8 13 428 100 65 39

195 235 202 632

Ho: Type of music preferred is independent of region. Ha: Type of music preferred is not independent of region. e11 =

(195)(428) = 132.06 632

e23 =

(235)(65) = 24.17 632

e12 =

(195)(100) = 30.85 632

e24 =

(235)(39) = 14.50 632

e13 =

(195)(65) = 20.06 632

e31 =

(202)(428) = 136.80 632

e14 =

(195)(39) = 12.03 632

e32 =

(202)(100) = 31.96 632

e21 =

(235)(428) = 159.15 632

e33 =

(202)(65) = 20.78 632

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

e22 =

(235)(100) = 37.18 632

O/Q Region A/P BC

2 =

Type of Music Preferred Rock R&B Country (132.06) (30.85) (20.06) 140 32 5 (159.15) (37.18) (24.17) 134 41 52 (136.80) (31.96) (20.78) 154 27 8 428 100 65

e34 =

(202)(39) = 12.47 632

Classic (12.03) 18 (14.50) 8 (12.47) 13 39

195 235 202 632

(141  132.06) 2 (32  30.85) 2 (5  20.06) 2 (18  12.03) 2 + + + 132.06 20.06 12.03 30.85

+ (134  159.15) 2 (41  37.18) 2 (52  24.17) 2 (8  14.50) 2 + + + + 159.15 24.17 37.18 14.50 (154  136.80) 2 (27  31.96) 2 (8  20.78) 2 (13  12.47) 2 + + + 31.96 20.78 12.47 136.80

= .48 + .04 + 11.31 + 2.96 + 3.97 + .39 + 32.04 + 2.91 + 2.16 + .77 + 7.86 + .02 = 64.91

 = .01, df = (c-1)(r-1) = (4-1)(3-1) = 6 2.01,6 = 16.8119 Since the observed 2 = 64.91 > 2.01,6 = 16.8119, the decision is to reject the null hypothesis. At  = .01, music preference is not independent of geographic location.

16.13 Transportation Mode Air Train Truck Industry Publishing 32 12 41

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Comp.Hard.

5 37

6 18

24 65

35 120

H0: Type of Industry is independent of Transportation Mode. Ha: Type of Industry is not independent of Transportation Mode. e11 =

(85)(37) = 26.21 120

e21 =

(35)(37) = 10.79 120

e12 =

(85)(18) = 12.75 120

e22 =

(35)(18) = 5.25 120

e13 =

(85)(65) = 46.04 120

e23 =

(35)(65) = 18.96 120

Transportation Mode Air Train Truck Industry Publishing (26.21) (12.75) (46.04) 32 12 41 Comp.Hard. (10.79) (5.25) (18.96) 5 6 24 37 18 65

85 35 120

(32  26.21) 2 (12  12.75) 2 (41  46.04) 2  = + + + 12.75 26.21 46.04 (5  10.79) 2 (6  5.25) 2 (24  18.96) 2 + + = 10.79 5.25 18.96 1.28 + .04 + .55 + 3.11 + .11 + 1.34 = 6.43 2

 = .05, df = (c-1)(r-1) = (3-1)(2-1) = 2 2.05,2 = 5.9915 Since the observed 2 = 6.43 > 2.05,2 = 5.9915, the decision is to reject the null hypothesis. At  = .05, Type of Industry is not independent of Transportation Mode.

16.14 Number of Bedrooms <2 3 >4 Solutions Manual 1-570 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Number of Stories

1 2

116 90 206

101 325 426

57 160 217

274 575 849

H0: Number of bedrooms is independent of Number of Stories. Ha: Number of bedrooms is not independent of Number of Stories. e11 =

(274)(206) = 66.48 849

e21 =

(575)(206) = 139.52 849

e12 =

(274)(426) = 137.48 849

e22 =

(575)(426) = 288.52 849

e13 =

(274)(217) = 70.03 849

e23 =

(575)(217) = 146.97 849

2 =

(101  137.48) 2 (57  70.03) 2 (116  66.48) 2 + + + 66.48 137.48 70.03

(90  139.52) 2 + 139.52 (325  288.52) 2 (160  146.97) 2 + = 288.52 146.97

2 = 36.89 + 9.68 + 2.42 + 17.58 + 4.61 + 1.16 = 72.34

 = .10

df = (c-1)(r-1) = (3-1)(2-1) = 2

2.10,2 = 4.6052 Since the observed 2 = 72.34 > 2.10,2 = 4.6052, the decision is to reject the null hypothesis. At  = .10, Number of bedrooms is not independent of Number of stories . 16.15

Type of Store

Shoppers‟ Residence U.S. Department 24 Discount 20 Hardware 11 Shoe 32 87

Canada 17 15 19 28 79

41 35 30 60 166

Ho: Shoppers‟ residence is independent of type of border city retailer. Ha: Shoppers‟ residence is not independent of type of border city retailer. Solutions Manual 1-571 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

e11 =

(41)(87) = 21.49 166

e31 =

(30)(87) = 15.72 166

e12 =

(41)(79) = 19.51 166

e32 =

(30)(79) = 14.28 166

e21 =

(35)(87) = 18.34 166

e41 =

(60)(87) = 31.45 166

e22 =

(35)(79) = 16.66 166

e42 =

(60)(79) = 28.55 166

Type of Store

2 =

Shoppers‟ Residence U.S, Canada Dept. (21.49) (19.51) 24 17 Disc. (18.34) (16.66) 20 15 Hard. (15.72) (14.28) 11 19 Shoe (31.45) (28.55) 32 28 87 79

41 35 30 60 166

(24  21.49) 2 (17  19.51) 2 (20  18.34) 2 (15  16.66) 2 + + + 21.49 19.51 16.66 18.34

+ (11  15.72) 2 (19  14.28) 2 (32  31.45) 2 (28  28.55) 2 + + + 15.72 14.28 28.55 31.45

= .29 + .32 + .15 + .17 + 1.42 + 1.56 + .01 + .01 = 3.93

 = .05, df = (c-1)(r-1) = (2-1)(4-1) = 3 2.05,3 = 7.8147 Since the observed 2 = 3.93 < 2.05,3 = 7.8147, the decision is to fail to reject the null hypothesis. At  = .05, there is not sufficient evidence that the shoppers‟ residence is not independent of type of border city retailer.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

16.16  = .01, k = 7, df = 6 H0: The observed distribution is the same as the expected distribution Ha: The observed distribution is not the same as the expected distribution ( f0  fe )2 Use:    fe 2

critical 2.01,6 = 16.8119 fo 214 235 279 281 264 254 211

2  

_ fe 206 232 268 284 268 232 206

(f0-fe)2 64 9 121 9 16 484 25

( fo  fe )2 fe

0.311 0.039 0.451 0.032 0.060 2.086 0.121 3.100

( f0  fe )2 = 3.100 fe

Since the observed value of 2 = 3.1 < 2.01,6 = 16.8119, the decision is to fail to

reject the null hypothesis. At  = .01, there is not sufficient evidence

that the observed distribution is different from the expected distribution.

16.17

Variable 1

12 8 7 27

Variable 2 23 17 11 51

21 20 18 59

56 45 36 137

Ho: Variable One is independent of Variable Two. Ha: Variable One is not independent of Variable Two. e11 = 11.04 e12 = 20.85 e13 = 24.12 e21 = 8.87

e22 = 16.75

e23 = 19.38

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

e31 = 7.09

e32 = 13.40

e33 = 15.50

(12  11.04) 2 (23  20.85) 2 (21  24.12) 2 (8  8.87) 2  = + + + + 20.85 11.04 24.12 8.87 (17  16.75) 2 (20  19.38) 2 (7  7.09) 2 (11  13.40) 2 + + + + 13.40 16.75 19.38 7.09 (18  15.50) 2 = 15.50 .084 + .222 + .404 + .085 + .004 + .020 + .001 + .430 + .403 = 1.652 df = (c-1)(r-1) = (2)(2) = 4  = .05 2

2.05,4 = 9.4877 Since the observed value of 2 = 1.652 < 2.05,4 = 9.4877, the decision is to fail

to reject the null hypothesis. At  = .05, there is not sufficient evidence

that Variable One is not independent of Variable Two.

16.18 Location ON/QU West Customer Industrial 230 115 Type Retail 185 143 415 258 Ho:

Atlantic 68 89 157

413 417 830

Manufacturer‟s geographic location is independent of type of

customer. Ha: Manufacturer‟s geographic location is not independent of type of customer. e11 =

(413)(415) = 206.5 830

e21 =

(417)(415) = 208.5 830

e12 =

(413)(258) = 128.38 830

e22 =

(417)(258) = 129.62 830

e13 =

(413)(157) = 78.12 830

e23 =

(417)(157) = 78.88 830

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

ON/QU West Atlantic Customer Industrial (206.5) (128.38) (78.12) Type 230 115 68 Retail (208.5) (129.62) (78.88) 185 143 89 415 258 157

413 417 830

(230  206.5) 2 (115  128.38) 2 (68  78.12) 2  = + + + 128.38 78.12 206.5 (185  208.5) 2 (143  129.62) 2 (89  78.88) 2 + + = 208.5 129.62 78.88 2.67 + 1.39 + 1.31 + 2.65 + 1.38 + 1.30 = 10.70 2

 = .10 and df = (c - 1)(r - 1) = (3 - 1)(2 - 1) = 2 2.10,2 = 4.6052 Since the observed 2 = 10.70 > 2.10,2 = 4.6052, the decision is to reject the

null hypothesis. At  = .10, the sample data support the claim that manufacturer‟s geographic location is not independent of type of customer.

16.19 Cookie Type fo Chocolate Chip 189 Peanut Butter 168 Cheese Cracker 155 Lemon Flavoured 161 Chocolate Mint 216 Vanilla Filled 165 fo = 1,054 Ho: Ha:

Cookie Sales is uniformly distributed across kind of cookie. Cookie Sales is not uniformly distributed across kind of cookie.

If cookie sales are uniformly distributed, then fe =

f

 no.t y p e s 0

1,0 5 4 = 6

175.67

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

fo 189 168 155 161 216 165

fe__ 175.67 175.67 175.67 175.67 175.67 175.67

( fo  fe )2 fe

1.01 0.33 2.43 1.23 9.26 0.65 14.91

The observed 2 = 14.91  = .05 df = k - 1 = 6 - 1 = 5 2.05,5 = 11.0705 Since the observed 2 = 14.91 > 2.05,5 = 11.0705, the decision is to reject the

null hypothesis. At  = .05, the sample data support the claim that sales for these six kinds of cookies are not uniformly distributed.

16.20

Bought Car

Y N

Gender M F 207 65 811 984 1,018 1,049

272 1,795 2,067

Ho: Gender is independent of being a major decision maker in purchasing car last year. Ha: Gender is not independent of being a major decision maker in purchasing car last year. (272)(1,018) = 133.96 2,067 (1,795)(1,018) e21 = = 884.04 2,067

e11 =

(27)(1,049) = 138.04 2,067 (1,795)(1,049) e22 = = 2,067

e12 =

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

M F Bought Y (133.96) (138.04) Car 207 65 272 N (884.04) (910.96) 811 984 1,795 1,018 1,049 2,067 2 2 (207  133.96) (65  138.04) (811  884.04) 2 2 = + + + 133.96 138.04 884.04 (984  910.96) 2 = 39.82 + 38.65 + 6.03 + 5.86 = 90.36 910.96

 = .05

df = (c-1)(r-1) = (2-1)(2-1) = 1

2.05,1 = 3.8415 Since the observed 2 = 90.36 > 2.05,1 = 3.8415, the decision is to reject the

null hypothesis. At  = .05, the sample data support the claim that gender

is not independent of being a major decision maker in purchasing a car last year.

16.21

Arrivals 0 1 2 3 4 5 >6

 =

fo 26 40 57 32 17 12 8 fo = 192

(fo)(Arrivals) 0 40 114 96 68 60 48 (fo)(arrivals) = 426

 ( f )(arrivals)  426 = 2.2 192 f o

Ho: The observed frequencies are Poisson distributed. Ha: The observed frequencies are not Poisson distributed. Arrivals 0 1

Probability .1108 .2438

fe___ (.1108)(192) = 21.27 (.2438)(192) = 46.81

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

2 3 4 5 >6

.2681 .1966 .1082 .0476 .0249

fo 26 40 57 32 17 12 8

51.48 37.75 20.77 9.14 4.78

fe__ 21.27 46.81 51.48 37.75 20.77 9.14 4.78

( fo  fe )2 fe

1.05 0.99 0.59 0.88 0.68 0.89 2.17 7.25

Observed 2 = 7.25  = .05 df = k - 2 = 7 - 2 = 5 2.05,5 = 11.0705 Since the observed 2 = 7.25 < 2.05,5 = 11.0705, the decision is to fail to reject

the null hypothesis. At  = .05, there is not enough evidence to reject the

claim that the observed frequency of arrivals is Poisson distributed.

16.22 Ho: The distribution of observed frequencies is the same as the distribution of expected frequencies. Ha: The distribution of observed frequencies is not the same as the distribution of expected frequencies. Soft Drink Classic Coke Pepsi Diet Coke Mt. Dew Diet Pepsi

fo 314 219 212 121 98

proportions fe___ .179 (.179)(1726) = 308.95 .115 (.115)(1726) = 198.49 .097 167.42 .063 108.74 .061 105.29

( fo  fe )2 fe

0.08 2.12 11.87 1.38 0.50

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Sprite 93 Dr. Pepper 88 Others 581 fo = 1,726

.057 .056 .372

98.32 96.66 642.07

0.29 0.78 5.81 22.83

Observed 2 = 22.83  = .05 df = k - 1 = 8 - 1 = 7 2  .05,7 = 14.0671 Since the observed 2 = 22.83 > 2.05,6 = 14.0671, the decision is to reject the

null hypothesis. At  = .05, the observed frequencies are not distributed

the same as the expected frequencies from the national poll. 16.23 Professional Position

Years

0-3 4-8 >8

Ho: worked in

Systems Manager Programmer Operator Analyst 6 37 11 13 28 16 23 24 47 10 12 19 81 63 46 56

67 91 88 246

Type of job held in the computer industry is independent of years

the industry. Ha: Type of job held in the computer industry is not independent of years worked in the industry. e11 = e12 = e13 = e14 = e21 = e22 =

(67)(81) = 22.06 246 (67)(63) = 17.16 246 (67)(46) = 12.53 246 (67)(56) = 15.25 246 (91)(81) = 29.96 246 (91)(63) = 23.30 246

(91)(46) = 17.02 246 (91)(56) e24 = = 20.72 246 (88)(81) e31 = = 28.98 246 (88)(63) e32 = = 22.54 246 (88)(46) e33 = = 16.46 246 (88)(56) e34 = = 20.03 246

e23 =

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Professional Position

0-3 Years 4-8 >8

2 =

Systems Manager Programmer Operator Analyst (22.06) (17.16) (12.53) (15.25) 6 37 11 13 (29.96) (23.30) (17.02) (20.72) 28 16 23 24 (28.98) (22.54) (16.46) (20.03) 47 10 12 19 81 63 46 56

67 91 88 246

(6  22.06) 2 (37  17.16) 2 (11  12.53) 2 (13  15.25) 2 + + + + 22.06 12.53 15.25 17.16 (28  29.96) 2 (16  23.30) 2 (23  17.02) 2 (24  20.72) 2 + + + + 29.96 20.72 23.30 17.02 (47  28.98) 2 (10  22.54) 2 (12  16.46) 2 (19  20.03) 2 + + + = 28.98 22.54 16.46 20.03 11.69 + 22.94 + .19 + .33 + .13 + 2.29 + 2.1 + .52 + 11.2 + 6.98 + 1.21 + .05 = 59.63

 = .01

df = (c-1)(r-1) = (4-1)(3-1) = 6  .01,6 = 16.8119 2

Since the observed 2 = 59.63 > 2.01,6 = 16.8119, the decision is to reject the

null hypothesis. At  = .01, the sample data support the claim that type of professional job held in the computer industry is not independent of

number of years of worked in the industry.

16.24 Category

Exp. Prop.

120

0.43

114

0.37

Technology

0.05

315(0.05) = 15.75

3.34

Others

0.15

315(0.15) = 47.25

2.45

More work, more business Insufficient number of workers

(f o  f e )2 fe

315(0.43) = 135.45 315(0.37) = 116.55

1.76 0.06

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

315

7.61

H0: The observed distribution is the same as the expected distribution. Ha: The observed distribution is not the same as the expected distribution. Observed 2 = 7.61 df = k – 1 = 4 – 1 = 3,  = .10 2.10,3 = 6.2514 Since the observed 2 = 7.61 > 2.10,3 = 6.2514, the decision is to reject the

null hypothesis. At  = .10, there is a significant difference between

the distribution of responses of U.S. workers and Canadian workers.

16.25

Number of Children

0 1 2 >3

Type of College or University Community Large Small College University College 25 178 31 49 141 12 31 54 8 22 14 6 127 387 57

234 202 93 42 571

Ho: Number of Children is independent of Type of College or University. Ha: Number of Children is not independent of Type of College or University. e11 = e12 = e13 = e21 = e22 =

(234)(127) = 52.05 571 (234)(387) = 158.60 571 (234)(57) = 23.36 571 (202)(127) = 44.93 571 (202)(387) = 136.91 571

(93)(127) = 20.68 571 (193)(387) e32 = = 63.03 571 (93)(57) e33 = = 9.28 571

e31 =

(42)(127) = 9.34 571 (42)(387) e42 = = 28.47 571 e41 =

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

e23 =

(202)(57) = 20.16 571

Number of Children

e43 =

(42)(57) = 4.19 571

Type of College or University Community Large Small College University College (52.05) (158.60) (23.36) 25 178 31 (44.93) (136.91) (20.16) 49 141 12 (20.68) (63.03) (9.28) 31 54 8 (9.34) (28.47) (4.19) 22 14 6 127 387 57

0 1 2 >3

234 202 93 42 571

(25  52.05) 2 (178  158.6) 2 (31  23.36) 2 (49  44.93) 2  = + + + + 52.05 158.6 44.93 23.36 2

(141  136.91) 2 (12  20.16) 2 (31  20.68) 2 (54  63.03) 2 + + + + 136.91 20.16 20.68 63.03 (8  9.28) 2 (22  9.34) 2 (14  28.47) 2 (6  4.19) 2 + + + = 9.34 9.28 28.47 4.19

14.06 + 2.37 + 2.50 + 0.37 + 0.12 + 3.30 + 5.15 + 1.29 + 0.18 + 17.16 + 7.35 + 0.78 = 54.63

 = .05,

df= (c - 1)(r - 1) = (3 - 1)(4 - 1) = 6

2.05,6 = 12.5916 Since the observed 2 = 54.63 > 2.05,6 = 12.5916, the decision is to reject the

null hypothesis. At  = .05, number of children is not independent of type

of College or University.

16.26 The observed chi-square is 30.18 with a p-value of .0000043. The chisquare Solutions Manual 1-582 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

goodness-of-fit test indicates that there is a significant difference between the observed frequencies and the expected frequencies. The distribution of responses to the question is not the same for adults between 21 and 30 years of age as they are for others. Marketing and sales people might reorient their 21 to 30 year old efforts away from home improvement and pay more attention to leisure travel/vacation, clothing, and home entertainment.

16.27 Ho: Gender is independent of color preference for cars. Ha: Gender is not independent of color preference for cars. The observed chi-square value for this test of independence is 5.366. The associated p-value of .252 indicates failure to reject the null hypothesis. There is not enough evidence here to say that color choice is dependent of gender. Automobile marketing people do not have to worry about which colors especially appeal to men or to women because car color is independent of gender. In addition, design and production people can determine car color quotas based on other variables.

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Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

SOLUTIONS TO PROBLEMS IN CHAPTER 17: NONPARAMETRIC STATISTICS

17.1

Ho: The observations in the sample are randomly generated. Ha: The observations in the sample are not randomly generated. This is a small sample runs test since n1, n2 < 20

 = .05, The lower tail critical value is 6 and the upper tail critical value is 16 n1 = 10

n2 = 10

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

R = 11 Since R = 11 is between the two critical values (6 and 16), the decision is to fail to reject the null hypothesis. At  = .05, there is not sufficient evidence to support the claim that the observations in the sample are not randomly generated.

17.2

Ho: The observations in the sample are randomly generated. Ha: The observations in the sample are not randomly generated.

 = .05, /2 = .025, z.025= + 1.96 n1 = 26

R 

R 

n2 = 21

n = 47

2n1n2 2(26)( 21) 1   1 = 24.234 n1  n2 26  21

2n1n2 (2n1n2  n1  n2 )  (n1  n2 ) 2 (n1  n2  1)

2(26)(21)2(26)(21)  26  21 = 3.351 (26  21) 2 (26  21  1)

R=9

z

R  R

R



9  24.234 = -4.55 3.351

Since the observed value of z = -4.55 < z.025 = -1.96, the decision is to reject the

17.3

null hypothesis. At  = .05, the sample data support the claim that the observations in the sample are not generated randomly.

n1 = 8

n2 = 52

 = .05

The observed number of runs is 11. The mean or expected number of runs is 14.8667. The p-value from the printout is .0264. Since p-value = .0264 < 0.05,

the decision is to reject the null hypothesis. At  = .05, the sample data support the claim that the observations in the sample were not generated randomly.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

17.4 runs

The observed number of runs is 18. The mean or expected number of is 14.333. The p-value for this test is .1452, and it is greater than  = .05

. Thus, the test is not significant at alpha of .05. The decision is to fail to reject

the null hypothesis. . At  = .05, there is not enough evidence to declare

that the observations in the sample are not randomly generated.

17.5

Ho: The observations in the sample are randomly generated. Ha: The observations in the sample are not randomly generated. Since n1, n2 > 20, use large sample runs test

 = .05 Since this is a two-tailed test, /2 = .025, z.025 = + 1.96. If the observed value of z is greater than 1.96 or less than -1.96, the decision is to reject the null hypothesis. R = 27 n1 = 40 n2 = 24

R 

2n1n2 2(40)( 24) 1   1 = 31 n1  n2 64

R 

2n1n2 (2n1n2  n1  n2 )  (n1  n2 )2 (n1  n2  1)

z

R  R

R



2(40)(24)2(40)(24)  40  24 = 3.716 (64) 2 (63)

27  31 = -1.08 3.716

Since the observed z of -1.08 is greater than the critical lower tail z value of -1.96 and less than 1.96 , the decision is to fail to reject the null hypothesis. At  = .05, there is not enough evidence to declare that the observations in the sample are not randomly generated.

17.6

Ho: The observations in the sample are randomly generated. Ha: The observations in the sample are not randomly generated. n1 = 5

n2 = 8

n = 13

 = .05

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Since this is a two-tailed test, /2 = .025 From Table A.11, the lower critical value is 3 From Table A.11, the upper critical value is 11 R=4 Since R = 4 is greater than the lower critical value of 3 and less than the upper critical value of 11, the decision is to fail to reject the null hypothesis. At  = .05, there is not enough evidence to declare that the observations in the sample are not randomly generated.

17.7

Ho: Group 1 is identical to Group 2 Ha: Group 1 is not identical to Group 2 Use the small sample Mann-Whitney U test since both n1, n2 < 10,  = .05. Since this is a two-tailed test, /2 = .025. The p-value is obtained using Table A.13. Value 11 13 13 14 15 17 18 18 21 21 22 23 23 24 26 29

Rank 1 2.5 2.5 4 5 6 7.5 7.5 9.5 9.5 11 12.5 12.5 14 15 16

Group 1 1 2 2 1 1 1 2 1 2 2 2 2 2 1 1

n1 = 8 and n2 = 8 W1 = 1 + 2.5 + 5 + 6 + 7.5 + 9.5 + 15 + 16 = 62.5 U  n1  n2 

n1 (n1  1) (8)(9)  W1  (8)(8)   62.5 = 37.5 2 2

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

U   n1  n2  U = 64 – 37.5 = 26.5 We use the small U which is 26.5. From Table A.13, the p-value for U = 27 is .3227(2) = .6454 Since this p-value is greater than /2 = .025, the decision is to fail to reject the

null hypothesis. At  = .05, there is not sufficient evidence to support the

claim that there is a significant difference between values of group 1and group2. 17.8

Ho: The populations 2 and 1 are the same. Ha: The population 2 is shifted to the left of the population 1. Value Rank Group 203 1 2 208 2 2 209 3 2 211 4 2 214 5 2 216 6 1 217 7 1 218 8 2 219 9 2 222 10 1 223 11 2 224 12 1 227 13 2 229 14 2 230 15.5 2 230 15.5 2 231 17 1 236 18 2 240 19 1 241 20 1 248 21 1 255 22 1 256 23 1 283 24 1 n1 = 11 n2 = 13 W1 = 6 + 7 + 10 + 12 + 17 + 19 + 20 + 21 + 22 + 23 + 24 = 181 n (n  1) (11)(12) U 1  U  n1  n2  1 1  W1  (11)(13)   181 = 28 2 2 n n (11)(13) = 71.5  1 2  2 2

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

n1  n2 (n1  n2  1) (11)(13)(25)  = 17.26 12 12 U   28  71.5 = -2.52 z   17.26

 

 = .01,

z.01 = –2.33

Since z = – 2.52 < z.01 = –2.33, the decision is to reject the null hypothesis. At  = .01, the sample data support the claim that the values of population 1 are significantly larger than the values of population 2.

17.9

Let Group 1 = people 65 to 74 years old and Group 2 = people 75 and

older. H0: The populations of number of contacts with physicals by people 65 to 74 years old and by people 75 or older are the same. Ha: The population of number of contacts with physicals by people 75 or older is shifted to the right of the population of number of contacts by people 65 to 74 years old. Contacts 6 8 9 9 10 11 11 12 12 13 13 13 14 15 16 17 W1 = 39

Rank 1 2 3.5 3.5 5 6.5 6.5 8.5 8.5 11 11 11 13 14 15 16 n1 = 7

Group 1 1 1 2 2 1 1 1 2 1 2 2 2 2 2 2 n2 = 9

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

U1  n1  n2 

n1 (n1  1) (7)(8)  W1  (7)(9)   39 = 52 2 2

U 2  n1  n2  U1 = (7)(9) – 52 = 11 U = 11. From Table A.13, the p-value for one-tailed test = .0156. Since this pvalue is At  =

greater than  = .01, the decision is to fail to reject the null hypothesis. .01, there is not sufficient sample evidence to support the claim that the

number of contracts with physicians by people 75 and older is greater than the number by people 65 to 74 years old.

17.10 Ho: Urban and rural households spend the same amounts. Ha: Urban and rural households spend different amounts. Expenditure 1,950 2,050 2,075 2,110 2,175 2,200 2,480 2,490 2,540 2,585 2,630 2,655 2,685 2,710 2,750 2,770 2,790 2,800 2,850 2,850 2,975 2,995 2,995

Rank 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19.5 19.5 21 22.5 22.5

Group U R R U U U U R U R U U R U U R R R U U R R R

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

3,100

Group 1 = Urban households and n1 = 12 Group 2 = Rural households and n2 = 12 W1 = 1 + 4 + 5 + 6 + 7 + 9 + 11 + 12 + 14 + 15 + 19.5 + 19.5 = 123



n1  n2 (12)(12) = 72  2 2



n1  n2 (n1  n2  1)  12

(12)(12)(25) = 17.32 12

n1 (n1  1) (12)(13)  W1  (12)(12)   123 = 99 2 2 U 2  n1  n2  U1 = (12)(12) – 99 = 45 U 1  n1  n2 

z

U 





 = .05

45  72 = – 1.56 17.32

/2 = .025

z.025 = +1.96 Since the observed z = – 1.56 > z.025 = –1.96 and less than 1.96, the decision is to

fail to reject the null hypothesis. At  = .05, there is not sufficient

sample evidence to support the claim that there is a significant difference between urban and rural households in the amounts spent for food at home.

17.11

Ho: The populations of female incomes and male incomes are the same. Ha: The population of female incomes is shifted to the left of the

population of male incomes. Earnings $28,900 31,400 36,600 40,000 40,500

Rank 1 2 3 4 5

Gender F F F F F

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

41,200 42,300 42,500 44,500 45,000 47,500 47,800 47,800 48,000 50,100 51,000 51,500 51,500 53,850 55,000 57,800 61,100 63,900

6 7 8 9 10 11 12.5 12.5 14 15 16 17.5 17.5 19 20 21 22 23

F F F F M F F M F M M M M M M M M M

Let Group 1 = incomes of male stock market investors and Group 2 = incomes of female investors. n1 = 11 n2 = 12 W1 = 10 + 12.5 + 15 + 16 + 17.5 + 17.5 + 19 + 20 + 21 + 22 + 23 = 193.5



n1  n2 (11)(12) = 66 and    2 2

n1  n2 (n1  n2  1)  12

(11)(12)(24) 12

= 16.25 n1 (n1  1) (11)(12)  W1  (11)(12)   193.5 = 4.5 2 2

U  n1  n2  z

U 



 = .01,



4.5  66 = -3.78 16.25

z.01 = -2.33

Since the observed z = -3.78 < z.01 = -2.33, the decision is to reject the null

hypothesis. At  = .01, the male investor earns significantly more than the

female investor.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

17.12 H0: There is no difference in the price of a single-family home in Calgary and Halifax. Ha: There is a difference in the price of a single-family home in Calgary and Halifax.

Price 274,127 274,514 275,238 276,333 276,419 277,359 277,741 277,867 279,114 280,031 282,012 282,136 283,947 283,968 284,500 372,405 374,157 375,062 375,940 376,981 377,016 378,057 379,638 380,102 380,479 381,408 381,730 381,861 n1 = 13 n2 = 15

Rank 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28

City H H H H H H H H H H H H H H H C C C C C C C C C C C C C

Group 1 = price of a single-family home in Calgary Group 2 = price of a single-family home in Halifax

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

W1 = 16 + 17 + 18 + 19 + 20 + 21 + 22 + 23 + 24 + 25 + 26 + 27 + 28 = 286

U  n1  n2 

n1 (n1  1) (13)(14)  W1  (13)(15)   286 = 0 2 2



n1  n2 (13)(15) = 97.5  2 2



n1  n2 (n1  n2  1)  12

z

U 





(13)(15)(29) = 21.708 12

0  97.5 = – 4.49 21.708

For  = .05 and a two-tailed test, /2 = .025 and z.025 = + 1.96. Since the observed z = – 4.49 < z.025 = – 1.96, we reject the null hypothesis. At  = .05 , there is enough evidence to declare that there is a significant difference in the price of a single family home in Calgary and Halifax.

17.13 Ho: Md = 0 Ha: Md  0 1 212 234 219 199 194 206 234 225 220 218 234 212

2 179 184 213 167 189 200 212 221 223 217 208 215

d 33 50 6 32 5 6 22 4 -3 1 26 -3

Rank +15 +16 +7.5 +13.5 +6 +7.5 +11 +5 - 3.5 +1 +12 -3.5

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

219 196 178 213

187 198 189 201

32 -2 -11 12

+13.5 -2 -9 +10

n = 16 > 15 T = min (T+ , T-) = T- = 3.5 + 3.5 + 2 + 9 = 18



(n)(n  1) (16)(17) = 68  4 4

n(n  1)(2n  1) 16(17)(33)  = 19.34 24 24

  z

T 



 = .10



18  68 = -2.59 19.34

/2 = .05

z.05 = ±1.645

Since the observed z = -2.59 < z.05 = -1.645, the decision is to reject the null

hypothesis. At  = .10, the sample data support the claim that there is a significant difference between the two groups.

17.14 Ho: Md = 0 Ha: Md  0 Before 49 41 47 39 53 51 51 49 38 54 46 50 44 49 45

After 43 29 30 38 40 43 46 40 42 50 47 47 39 49 47

d 6 12 17 1 13 8 5 9 -4 4 -1 3 5 0 -2

Rank +9 +12 +14 +1.5 +13 +10 +7.5 +11 - 5.5 +5.5 - 1.5 +4 + 7.5 OMIT -3

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

n = 15 but after dropping the zero difference, n = 14

 = .05(2-sided) and from Table A.14, Tcr = 21 T+ = 9 + 12+ 14 + 1.5 + 13 + 10 + 7.5 + 11 + 5.5 + 4 + 7.5 = 95 T- = 5.5 + 1.5 + 3 = 10 T = min(T+,T-) = min(95, 10) = 10 Since the observed value of T = 10 < Tcr = 21, the decision is to reject the null hypothesis. At  = .05, there is a significant difference in before and after measurements.

17.15 Ho: Md = 0 Ha: Md < 0 Before 10,500 8,870 12,300 10,510 5,570 9,150 11,980 6,740 7,340 13,400 12,200 10,570 9,880 12,100 9,000 11,800 10,500

After 12,600 10,660 11,890 14,630 8,580 10,115 14,320 6,900 8,890 16,540 11,300 13,330 9,990 14,050 9,500 12,450 13,450

d -2,100 -1,790 410 -4,120 -3,010 -965 -2,370 -160 -1,550 -3,140 900 -2,760 -110 -1,950 -500 -650 -2,950

Rank -11 -9 +3 -17 -15 -7 -12 -2 -8 -16 +6 -13 -1 -10 -4 -5 -14

Since n = 17, use the large sample test T+ = 3 + 6 = 9 T =9

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition



(n)(n  1) (17)(18) = 76.5  4 4

n(n  1)(2n  1) 17(18)(35)  = 21.12 24 24

  z

T 



 = .05



9  76.5 = -3.20 21.12

z.05 = -1.645

Since the observed z = -3.20 < z.05 = -1.645, the decision is to reject the null

hypothesis. At  = .05, the sample data support the claim that sales

increased significantly after the advertising campaign.

17.16 Ho:Md = 0 Ha:Md < 0 Manual 426 387 410 506 411 398 427 449 407 438 418 482 512 402

Scanner 473 446 421 510 465 409 414 459 502 439 456 499 517 437

d -47 -59 -11 -4 -54 -11 13 -10 -95 -1 -38 -17 -5 -35

Rank -11 -13 -5.5 -2 -12 -5.5 +7 -4 -14 -1 -10 -8 -3 -9

n = 14 T+ = (+7) = 7 T- = (11 + 13 + 5.5 + 2 + 12 + 5.5 + 4 + 14 + 1 + 10 + 8 + 3 + 9) = 98 T = min(T+,T-) = min(7, 98) = 7 from Table A.14 with  = .05, n = 14, T.05,14 = 26 Solutions Manual 1-597 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Since the observed T = 7 < T.05,14 = 26, the decision is to reject the null hypothesis. At  = .05, the sample data support the claim that Md < 0. The optical scanners register significantly more items than manual entry systems do.

17.17 Ho: Md = 0 Ha: Md < 0 1997 49 27 39 75 59 67 22 61 58 60 72 62 49 48 19 32 60 80 55 68

2021 54 38 38 80 53 68 43 67 73 55 58 57 63 49 39 34 66 90 57 58

d -5 -11 1 -5 6 -1 -21 -6 -15 5 14 5 -14 -1 -20 -2 -6 -10 -2 10

Rank -7.5 -15 +2 -7.5 +11 -2 -20 -11 -18 +7.5 +16.5 + 7.5 -16.5 -2 -19 -4.5 -11 -13.5 -4.5 +13.5

n = 20 T+ = 2 + 11 + 7.5 + 16.5 + 7.5 + 13.5 = 58 T = 58



 

(n)(n  1) (20)(21) = 105  4 4

n(n  1)(2n  1)  24

20(21)(41) = 26.79 24

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

z

T 





For  = .10,

58  105 = -1.75 26.79

z.10 = -1.28

Since the observed z = -1.75 < z.10 = -1.28, the decision is to reject the null

hypothesis. At  = .10, the sample data support the claim that the scores

from 2021 are significantly higher than the scores from 1997.

17.18 Ho: Md = 0 Ha: Md > 0 April 2020 63.1 67.1 65.5 68.0 66.6 65.7 69.2 67.0 65.2 60.7 63.4 59.2 62.9 69.4 67.3 66.8

April 2021 57.1 66.4 61.8 65.3 63.5 66.4 64.9 65.2 65.1 62.2 60.3 57.4 58.2 65.3 67.2 64.1

d 5.7 0.7 3.7 2.7 3.1 -0.7 4.3 1.8 0.1 -1.5 3.1 1.8 4.7 4.1 0.1 2.7

Rank +16 +3.5 +12 +8.5 +10.5 -3.5 +14 +6.5 +1.5 -5 +10.5 +6.5 +15 +13 +1.5 +8.5

n = 16 T- = 8.5 T = 8.5



 

(n)(n  1) (16)(17) = 68  4 4

n(n  1)(2n  1) 16(17)(33)  = 19.339 24 24

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

z

T 





For  = .05,

8.5  68 = -3.08 19.339

z.05 = 1.645

Since the observed z =  3.08 > z.05 = 1.645, the decision is to reject the null

hypothesis. At  = .05, the sample data support the claim that people were

less optimistic in 2020 than in 2021.

17.19 Ho: The 5 populations are identical Ha: At least one of the 5 populations is different 1 157 188 175 174 201 203

2 165 197 204 214 183

3 219 257 243 231 217 203

4 286 243 259 250 279

5___ 197 215 235 217 240 233 213

4 29 23.5 27 25 28

1 1 6 4 3 9 10.5

2 2 7.5 12 14 5

BY RANKS 3 18 26 23.5 19 16.5 10.5

Tj 33.5

40.5

113.5

___ 132.5

5__ 7.5 15 21 16.5 22 20 13 115

6 Tj

n

 j

(33.5) 2 (40.5) 2 (113.5) 2 (132.5) 2 (115) 2     = 8,062.67 6 5 6 5 7

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Tj 12 12 K  3(n  1)  (8,062.67)  3(30) = 21.21  n(n  1) nj 29(30)

 = .01

df = c - 1 = 5 - 1 = 4

2.01,4 = 13.2767 Since the observed K = 21.21 > 2.01,4 = 13.2767, the decision is to reject the null

hypothesis. At  = .01, the sample data support the claim that groups 1

through 5 come from different populations.

17.20 Ho: The 3 populations are identical Ha: At least one of the 3 populations is different Group 1 19 21 29 22 37 42

Group 2 30 38 35 24 29

Group 3 39 32 41 44 30 27 33

Group 2 8.5 14 12 4 6.5 _ 45

Group 3 15 10 16 18 8.5 5 11_ 83.5

By Ranks Group 1 1 2 6.5 3 13 17 __ Tj 42.5 nj

n

 j

(42.5) 2 (45) 2 (83.5) 2 = 1,702.08   6 5 7

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

n = 18 2

Tj 12 12 K  3(n  1)  (1,702.08)  3(19) = 2.72  n(n  1) nj 18(19)

 = .05,

df = c - 1 = 3 - 1 = 2

2.05,2 = 5.9915 Since the observed K = 2.72 < 2.05,2 = 5.9915, the decision is to fail to reject

the null hypothesis. At  = .05, there is not sufficient sample evidence to

support the claim that there is a significant difference in the groups.

17.21 Ho: The 4 populations are identical Ha: At least one of the 4 populations is different Region 1 $1,200 450 110 800 375 200

Region 2 $225 950 100 350 275

Region 3 $ 675 500 1,100 310 660

Region 4 $1,075 1,050 750 180 330 680 425

By Ranks Region 1 23 12 2 18 10 4

Region 2 5 19 1 9 6

Region 3 15 13 22 7 14

Region 4 21 20 17 3 8 16 11 96

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

n

 j

(69) 2 (40) 2 (71) 2 (96) 2 = 3,438.27    6 5 5 7

n = 23 2

Tj 12 12 K  3(n  1)  (3,438.27)  3(24) = 2.75  n(n  1) nj 23(24)

 = .05

df = c - 1 = 4 - 1 = 3

2.05,3 = 7.8147 Since the observed K = 2.75 < 2.05,3 = 7.8147, the decision is to fail to reject

the null hypothesis. At  = .05, there is not sufficient sample evidence to

support the claim that there is a significant difference in the amount of customers‟ initial deposits when they open savings accounts according to geographic region of Canada.

17.22 Ho: The 3 populations are identical Ha: At least one of the 3 populations is different Small Town $21,800 22,500 21,750 22,200 21,600

City $22,300 21,900 21,900 22,650 21,800

Suburb $22,000 22,600 22,800 22,050 21,250 22,550

City 11 6.5 6.5 15 4.5 ____ 43.5

Suburb 8 14 16 9 1 13 61

By Ranks Small Town 4.5 12 3 10 2 ____ Tj 31.5

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

(31.5) 2 (43.5) 2 (61) 2  n  5  5  6 = 1,197.07 j Tj

n = 16 2

Tj 12 12 K  3(n  1)  (1,197.07)  3(17) = 1.81  n(n  1) nj 16(17)

 = .05

df = c - 1 = 3 - 1 = 2

2.05,2 = 5.9915 Since the observed K = 1.81 < 2.05,2 = 5.9915, the decision is to fail to reject

the null hypothesis. At  = .05, there is not sufficient sample evidence to

support the claim that there is a significant difference between prices according to the area in which the dealership is located.

17.23 Ho: The 4 populations are identical Ha: At least one of the 4 populations is different Amusement Parks

Lake Area

City

National

0 1 1 0 2 1 0

3 2 3 5 4 4 3 5 2

2 2 3 2 3 2 3 3 1 3

2 4 3 4 3 5 4 4

Park

By Ranks Amusement Parks

Lake Area

City

20.5

11.5

National

Park 11.5

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

5.5

11.5

5.5

20.5

11.5

28.5

20.5

5.5 2

28.5 20.5

11.5 20.5

20.5

____

11.5 ____

5.5 20.5

207.5

28.5 20.5 28.5 20.5 33

28.5 28.5

____ 154.0

199.5 10

(34) 2 (207.5) 2 (154) 2 (199.5) 2  n  7  9  10  8 = 12,295.80 j n = 34 2 Tj 12 12 K  3(n  1)  (12,295.80)  3(35) = 18.99  n(n  1) nj 34(35) Tj

 = .05

df = c - 1 = 4 - 1 = 3  .05,3 = 7.8147 2

Since the observed K = 18.99 > 2.05,3 = 7.8147, the decision is to reject the

null hypothesis. At  = .05, the sample data support the claim that there is

a significant difference in the duration of stay by type of vacation destination.

17.24 Ho: The 3 populations are identical Ha: At least one of the 3 populations is different Day Shift 52 57

Swing Shift 45 48

Graveyard Shift 41 46

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

53 56 55 50 51

44 51 48 54 49 43

39 49 42 35 52

Day Shift 16.5 22 18 21 20 13 14.5 ____ 125 7

Swing Shift 7 9.5 6 14.5 9.5 19 11.5 _5_ 82 8

Graveyard Shift 3 8 2 11.5 4 1 16.5 ___ 46 7

By Ranks

Tj nj 2

(125) 2 (82) 2 (46) 2  n  7  8  7 = 3,374.93 j Tj

n = 22 2

Tj 12 12 K  3(n  1)  (3,374.93)  3(23) = 11.04  n(n  1) nj 22(23)

 = .05

df = c - 1 = 3 - 1 = 2

2.05,2 = 5.9915 Since the observed K = 11.04 > 2.05,2 = 5.9915, the decision is to reject the

null hypothesis. At  = .05, the sample data support the claim that there is

a significant difference in the number of hours of sleep per week for workers on these shifts.

17.25 Ho: The treatment populations are equal

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Ha: At least one of the treatment populations yields larger values than at least one other treatment population. Use the Friedman test with  = .05 c = 5, b = 5, df = c - 1 = 4, 2.05,4 = 9.4877 If the observed value of 2 > 9.4877, then the decision will be to reject the null hypothesis. Shown below are the data ranked by blocks: 1 2 3 5 1 2 2 2 3 2.5 4 1 5 1

2.5

11.5

Rj2

132.25

144

324

625

8.5

72.25 Rj2 = 1,297.5 12 12 2 r2  R j  3b(c  1)  (1,297.5)  3(5)(6) = 13.8  bc (c  1) (5)(5)(6) Since the observed value of r2 = 13.8 > 2.05,4 = 9.4877, the decision is to

reject the null hypothesis. At  = .05, the sample data support the claim that at least one treatment population yields larger values than at least one other treatment population.

17.26 Ho: The treatment populations are equal Solutions Manual 1-607 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Ha: At least one of the treatment populations yields larger values than at least one other treatment population. Use the Friedman test with  = .05 c = 6, b = 9, df = c - 1 = 5, 2.05,5 = 11.0705 If the observed value of 2 > 11.0705, then the decision will be to reject the null hypothesis. Shown below are the data ranked by blocks: 1 1 3 1 1 3 1 1 3 1

2 3 5 3 3 1 3 2 1 2

3 2 1 2 4 2 2 4 2 3

4 6 6 6 6 4 6 6 6 6

5 5 4 5 5 6 5 5 5 5

6 4 2 4 2 5 4 3 4 4

Rj2 225 Rj2 = 6,991

529

484

2,704

2,025

1,024

1 2 3 4 5 6 7 8 9 Rj

r 2 

12 12 2 R j  3b(c  1)  (6,991)  3(9)(7) = 32.94  bc (c  1) (9)(6)(7)

Since the observed value of r2 = 32.94 >2.05,5 = 11.0705, we reject the null

hypothesis. At  = .05, the sample data support the claim that at least one treatment population yields larger values than at least one other treatment population.

17.27 Ho: The treatment populations are equal Ha: At least one of the treatment populations yields larger values than at least one other treatment population. Use the Friedman test with  = .01 c = 4, b = 6, df = c - 1 = 3, 2.01,3 = 11.3449 Solutions Manual 1-608 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

If the observed value of 2 > 11.3449, then the decision will be to reject the null hypothesis. Shown below are the data ranked by blocks:

1 2 3 4 5

1 1 2 1 1 1

2 4 3 4 3 3

3 3 4 3 4 4

4 2 1 2 2 2

Rj2

400

484

100

Rj2 = 1,048

r2 

12 12 2 R j  3b(c  1)  (1,048)  3(6)(5) = 14.8  bc (c  1) (6)(4)(5)

Since the observed value of r2 = 14.8 > 2.01,3 = 11.3449, the decision is to

reject the null hypothesis. At  = .01, the sample data support the claim that at least one treatment population yields larger values than at least one other treatment population. Thus, there is a significant difference in recovery times for the four different medical treatments.

17.28 Ho: The treatment populations are equal Ha: At least one of the treatment populations yields larger values than at least one other treatment population. Use the Friedman test with  = .05 c = 3, b = 10, df = c - 1 = 2, 2.05,2 = 5.9915 If the observed value of 2 > 5.9915, then the decision will be to reject the null Solutions Manual 1-609 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

hypothesis. Shown below are the data ranked by blocks: Worker 1 2 3 4 5 6 7 8 9 10

5-day 3 3

4-day 2 2

3.5 day 1 1

3 2 3 3 3 3 3

2 3 2 1 2 2 1

1 1 1 2 1 1 2

Rj2

841

324

169

Rj2 = 1,334

r 2 

12 12 2 R j  3b(c  1)  (1,334)  3(10)(4) = 13.4  bc (c  1) (10)(3)(4)

Since the observed value of r2 = 13.4 > 2.05,2 = 5.9915, the decision is to

reject the null hypothesis. At  = .05, the sample data support the claim that at least one treatment population yields larger values than at least one other treatment population. Thus, there is a difference in productivity by workweek configuration. 17.29 c = 4 treatments

b = 5 blocks

S = r2 = 2.04 with a p-value of .564. Since the p-value of .564 >  = .10, .05, or .01, the decision is to fail to reject the null hypothesis. There is not sufficient sample evidence to support the claim that there is a significant difference in treatments.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

17.30 The experimental design is a random block design that has been analyzed using a Friedman test. There are five treatment levels and seven blocks. Thus, the degrees of freedom are four. The observed value of S = 13.71 is the equivalent of r2. The p-value is .009 indicating that this test is significant at alpha .01. The null hypothesis is rejected. That is, at least one population yields larger values than at least one other population. An examination of estimated medians shows that treatment 1 has the lowest value and treatment 3 has the highest value

17.31

x 23 41 37 29 25 17 33 41 40 28 19

y 201 259 234 240 231 209 229 246 248 227 200

x Ranked 3 10.5 8 6 4 1 7 10.5 9 5 2

y Ranked 2 11 7 8 6 3 5 9 10 4 1

d d2 1 1 -.5 0.25 1 1 -2 4 -2 4 -2 4 2 4 1.5 2.25 -1 1 1 1 1 1 2 d = 23.5

n = 11 rs  1 

6 d 2

n(n  1) 2

 1

6(23.5) = .893 11(120)

There is a quite strong positive correlation between the two variables.

17.32

x 4 5 8 11 10 7 3 1

y 6 8 7 10 9 5 2 3

d -2 -3 1 1 1 2 1 -2

d2 4 9 1 1 1 4 1 4

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

2 9 6

1 11 4

1 -2 2

1 4 4 2 d = 34

n = 11 rs  1 

6 d 2

n(n  1) 2

 1

6(34) = .845 11(120)

There is a quite strong positive correlation between the two variables.

17.33

x 99 67 82 46 80 57 49 91

y 108 139 117 168 124 162 145 102

x Ranked 8 4 6 1 5 3 2 7

y Ranked 2 5 3 8 4 7 6 1

d d2 6 36 -1 1 3 9 -7 49 1 1 -4 16 -4 16 6 36 2 d = 164

n =8 rs  1 

6 d 2

n(n  1) 2

 1

6(164) = -.95 8(63)

There is a strong negative correlation between the two variables.

17.34 x 0.92 0.96 0.91 0.89 0.91 0.88

y 9.3 9.0 8.5 8.0 8.3 8.4

Ranked x 8 9 6.5 5 6.5 4

Ranked y 9 8 7 3 5 6

d -1 1 -0.5 2 1.5 -2

d2 1 1 0.25 4 2.25 4

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

0.84 0.81 0.83

8.1 7.9 7.2 n =9

3 1 2

4 2 1

6 d 2

-1 -1 1

1 1 1 2 d = 15.5

6(15.5) = .871 n(n  1) 9(80) There is a quite strong positive correlation between prime interest rates and rs  1 

 1

the value of the dollar.

17.35 Bank Credit Year Card (x) 1 2.51 2 2.86 3 2.33 4 2.54 5 2.54 6 2.18 7 3.34 8 2.86 9 2.74 10 2.54 11 3.18 12 3.53 13 3.51 14 3.11

Home Equity Loan (y) 2.07 1.95 1.66 1.77 1.51 1.47 1.75 1.73 1.48 1.51 1.25 1.44 1.38 1.3 n = 14

6 d 2

Ranked x 3 8.5 2 5 5 1 12 8.5 7 5 11 14 13 10

Ranked y 14 13 9 12 7.5 5 11 10 6 7.5 1 4 3 2

d -11 -4.5 -7 -7 -2.5 -4 1 -1.5 1 -2.5 10 10 10 8  d2 =

d2 121 20.25 49 49 6.25 16 1 2.25 1 6.25 100 100 100 64 636

6(636) = -.398 n(n  1) 14(14 2  1) There is a modest negative correlation between overdue payments for rs  1 

 1

bank credit cards and home equity loans.

17.36 Total Pig Iron Year (x) 1 43,952,000

Raw Steel (y) 81,606,000

Ranked x 1

Ranked y 1

d 0

d2 0

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

2 3 4 5 6 7 8 9 10 11 12

48,410,000 55,745,000 55,873,000 54,750,000 48,637,000 52,224,000 53,082,000 54,426,000 56,097,000 54,485,000 54,679,000 n = 12

89,151,000 99,924,000 97,943,000 98,906,000 87,896,000 92,949,000 97,877,000 100,579,000 104,930,000 105,309,478 108,561,182 6 d 2

2 10 11 9 3 4 5 6 12 7 8

3 8 6 7 2 4 5 9 10 11 12

-1 2 5 2 1 0 0 -3 2 -4 -4 d2 =

1 4 25 4 1 0 0 9 4 16 16 80

6(80) = 0.72 n(n  1) 12(144  1) There is a quite strong positive correlation between production of pig iron rs  1 

 1

and raw steel.

17.37 Number of Companies Year 1 2 3 4 5 6 7 8 9 10 11

Number of Equity Issues on AMEX (y) Ranked x Ranked y 1,063 1 11 1,055 2 10 943 3 7 1,005 4 9 981 5 8 936 6 6 896 8 5 893 10 4 862 11 3 769 9 2 765 7 1

on NYSE (x) 1,774 1,885 2,088 2,361 2,570 2,675 2,907 3,047 3,114 3,025 2,862 n = 11

6 d 2

d -10 -8 -4 -5 -3 0 3 6 8 7 6 d2 =

d2 100 64 16 25 9 0 9 36 64 49 36 408

6(408) = -0.855 n(n  1) 11(112  1) There is a strong negative correlation between the number of companies rs  1 

 1

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

on the NYSE and the number of equity issues on the American Stock Exchange. 17.38  = .05 H0: The observations in the sample are randomly generated Ha: The observations in the sample are not randomly generated n1 = 13 n2 = 21 R = 10 Since this is a two-tailed test, we use /2 = .025. The critical value is: z.025 = + 1.96 2n1n2 2(13)( 21) R  1   1 = 17.06 n1  n2 13  21

R 

z

2n1n2 (2n1n2  n1  n2 )  (n1  n2 ) 2 (n1  n2  1)

R  R

R



2(13)(21)2(13)(21)  13  21 = 2.707 (13  21) 2 (13  21  1)

10  17.06 = -2.61 2.707

Since the observed z = - 2.61 < z.025 = - 1.96, the decision is to reject the null

hypothesis. At  = .05, the sample data support the claim that the observations in the sample are not randomly generated.

17.39 Ho: The populations 1 and 2 are identical. Ha: The population 1 and 2 are not identical Sample 1 573 532 544 565 540 548 536 523

Sample 2 547 566 551 538 557 560 557 547

 = .01 Since n1 = 8, n2 = 8 < 10, use the small sample Mann-Whitney U test.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Value 523 532

Rank 1 2

538 540 544 547 547 548 551 557 557 560 565 566 573

4 5 6 7.5 7.5 9 10 11.5 11.5 13 14 15 16

Group 1 1 536 2 1 1 2 2 1 2 2 2 2 1 2 1

W1 = 1 + 2 + 3 + 5 + 6 + 9 + 14 + 16 = 56 n (n  1) (8)(9) U1  n1  n2  1 1  W1  (8)(8)   56 = 44 2 2 U 2  n1  n2  U1 = 8(8) - 44 = 20 Take the smaller value of U1, U2 : U = 20 From Table A.13, the p-value (1-tailed) is .1172, for 2-tailed, the p-value is .2344. Since the p-value is greater than  = .01, the decision is to fail to reject the null hypothesis. At  = .01, there is not sufficient sample evidence to support the claim that the two populations are not identical.

17.40  = .05, n = 9 H0: Md = 0 Ha: Md  0 Group 1

Group 2

Rank

5.6 1.3 4.7 3.8 2.4 5.5 5.1 4.6

6.4 1.5 4.6 4.3 2.1 6.0 5.2 4.5

-0.8 -0.2 0.1 -0.5 0.3 -0.5 -0.1 0.1

-8.5 -4 +2 -6.5 +5 -6.5 -2 +2

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

3.7 4.5 -0.8 -8.5 Since n = 9, from Table A.14 (2-sided), Tcr = 6 T+ = 2 + 5 + 2 = 9 T- = 8.5 + 4 + 6.5 + 6.5 + 2 + 8.5 = 36 T = min(T+, T-) = 9 Since the observed value of T = 9 > Tcr = 6, the decision is to fail to reject the

null hypothesis. At  = .05, there is not sufficient sample evidence to support the claim that there is a difference between the two groups.

17.41 nj = 7, c = 4, Group 1

Group 2

Group 3

Group 4 1 4 5 6 9 6 7 By Ranks:

Group 1 9.5 25 17.5 23 27.5 13.5 23 139

Group 2 3.5 27.5 9.5 17.5 26 20.5 17.5 122

Group 3 2 13.5 13.5 6 23 17.5 6 81.5

Group 4 1 3.5 6 9.5 20.5 9.5 13.5 63.5

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

n

= 2,760.14 + 2,126.29 + 948.89 + 576.04 = 6,411.36 j

n = 28

K

Tj 12 12  3(n  1)  (6,411.36)  3(29) = 7.75  n(n  1) nj 28(29)

 = .01

df = c - 1 = 4 - 1 = 3

The critical value of chi-square is: 2.01, 3 = 11.3449. Since K = 7.75 < 23,.01 = 11.3449, the decision is to fail to reject the null hypothesis. At  = .01, there is not sufficient sample evidence to support the claim that the four groups come from different populations. 17.42  = .05, b = 7, c = 4, df = 3 2.05,3 = 7.8147 H0: The treatment populations are equal Ha: At least one treatment population yields larger values than at least one other treatment population Blocks 1 2 3 4 5 6 7

Group 1 16 8 19 24 13 19 21

Group 2 14 6 17 26 10 11 16

Group 3 15 5 13 25 9 18 14

Group 4 17 9 18 21 11 13 15

By Ranks:

Blocks 1 2 3

Group 1 3 3 4

Group 2 1 2 2

Group 3 2 1 1

Group 4 4 4 3

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

4 5 6 7 Rj

2 4 4 4 24

4 2 1 3 15

3 1 3 1 12

1 3 2 2 19

Rj2

576

225

144

361

Rj2 = 576 + 225 + 144 + 361 = 1,306

r 2 

12 12 2 R j  3b(c  1)  (1,306)  3(7)(5) = 6.94  bc (c  1) (7)(4)(5)

Since r2 = 6.94 < 2.05,3 = 7.8147, we fail to reject the null hypothesis. At  = .05, there is not sufficient sample evidence to support the claim that the treatment groups come from different populations.

17.43

Ranks 1 101 129 133 147 156 179 183 190

n=8 rs  1 

2 87 89 84 79 70 64 67 71

6 d 2

n(n  1) 2

1 1 2 3 4 5 6 7 8

 1

2 7 8 6 5 3 1 2 4

d d2 -6 36 -6 36 -3 9 -1 1 2 4 5 25 5 25 4 16 2 d = 152

6(152) = -.81 8(63)

There is a quite strong negative correlation between the two variables.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

17.44 Ho: The 3 populations are identical Ha: At least one of the 3 populations is different 4 L per Fish 2.8 3.6 4.3 3.3 4.8 3.6 5.3

20 L per Fish 7.4 6.4 6.6 5.6 5.3 5.1 6.9

40 L per Fish 7.9 6.1 7.6 5.8 7.4 4.8

4 L per Fish 1 3.5 5 2 6.5 3.5 9.5

20 L per Fish 17.5 14 15 11 9.5 8 16

40 L per Fish 20 13 19 12 17.5 6.5

Tj 31

By Ranks

n

 j

(31) 2 (91) 2 (88) 2 = 2,610.95   7 7 6

n = 20 2

Tj 12 12 K  3(n  1)  (2,610.95)  3(21) = 11.60  n(n  1) nj 20(21)

 = .01

df = c - 1 = 3 - 1 = 2

2.01,2 = 9.2104

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Since the observed K = 11.60 > 2.01,2 = 9.2104, the decision is to reject the null

hypothesis. At  = .01, the sample data support the claim that there is a significant difference in fish growth by volume of allotted water.

17.45 Let group 1 = “S” and group 2 = “L”. N = 40 n1 = 24 n2 = 16  = .05 Use the large-sample runs test since n1 > 20. H0: The observations are randomly generated Ha: The observations are not randomly generated With a two-tailed test, /2 = .025, z.025 = + 1.96. If the observed z > 1.96 or < -1.96, the decision will be to reject the null hypothesis. R = 19

R 

2n1n2 2(24)(16) 1   1 = 20.2 n1  n2 24  16

R 

2n1n2 (2n1n2  n1  n2 )  (n1  n2 ) 2 (n1  n2  1)

z

R  R

R



2(24)(16)2(24)(16)  24  16 = 2.993 (40) 2 (39)

19  20.2 = -0.40 2.993

Since z = -0.40 > z.025 = -1.96 and z = -0.40 < z.025 =-1.96 , the decision is to fail to reject the null hypothesis. At  = .05, there is not sufficient sample evidence to support the claim that the sample does not appear to be random.

17.46 Use the Friedman test. Let  = .05 H0: The treatment populations are equal Ha: At least one treatment population yields larger values than at least one other treatment population c = 3 and b = 10 Solutions Manual 1-621 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Operator 1 2 3 4 5 6 7 8 9 10

Machine 1 231 233 229 232 235 234 236 230 228 237

Machine 2 229 232 233 235 228 237 233 229 230 238

Machine 3 234 231 230 231 232 231 230 227 229 234

Machine 1 2 3 1 2 3 2 3 3 1 2

Machine 2 1 2 3 3 1 3 2 2 3 3

Machine 3 3 1 2 1 2 1 1 1 2 1

22 484

23 529

15 225

By ranks: Operator 1 2 3 4 5 6 7 8 9 10 Rj Rj2 df = c - 1 = 2

2.05,2 = 5.9915.

If the observed

2r > 5.9915, the decision will be to reject the null

hypothesis. Rj2 = 484 + 529 + 225 = 1,238 12 12 2 r 2  R j  3b(c  1)  (1,238)  3(10)(4) = 3.8  bc (c  1) (10)(3)(4) Solutions Manual 1-622 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Since 2r = 3.8 < 2.05,2 = 5.9915, we fail to reject the null hypothesis. At  = .05, there is not sufficient sample evidence to support the claim that the machines are producing parts that are significantly different from each other in mass.

17.47 Ho: The populations of age of EMS workers and firefighters are the same. Ha: The population of age of firefighters is shifted to the left of the population of age of EMS workers . Age 21 23 24 25 27 27 27 28 28 28 29 30 30 30 32 33 33 36 36 37 39 41

Rank 1 2 3 4 6 6 6 9 9 9 11 13 13 13 15 16.5 16.5 18.5 18.5 20 21 22

Group 1 1 1 1 1 2 2 1 2 2 2 2 2 2 1 2 2 1 2 1 2 1

Let group 1 = age of Firefighters and group 2 = age of EMS workers. n1 = 10 n2 = 12 W1 = 1 + 2 + 3 + 4 + 6 + 9 + 15 + 18.5 + 20 + 22 = 100.5 n n (10)(12) = 60  1 2  2 2 n  n (n  n  1) (10)(12)(23)  1 2 1 2  = 15.17 12 12 Solutions Manual 1-623 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

n1 (n1  1) (10)(11)  W1  (10)(12)   100.5 = 74.5 2 2 U 2  n1  n2  U 1  (10)(12)  74.5  45.5 and U  U 2  45.5 U   45.5  60 with  = .05, z.05 = -1.645 z   0.96  15.17 U 1  n1  n2 

Since the observed z =- 0.96 > z.05 = -1.645, the decision is to fail to reject the

null hypothesis. At  = .05, there is not sufficient sample evidence to

support the claim that EMS workers are significantly older than firefighters.

17.48 Ho: Md = 0 Ha: Md  0 With 1,180 874 1,071 668 889 724 880 482 796 1,207 968 1,027 1,158 670 849 559 449 992 1,046 852

Without 1,209 902 862 503 974 675 821 567 602 1,097 962 1,045 896 708 642 327 483 978 973 841

d -29 -28 209 165 -85 49 59 -85 194 110 6 -18 262 -38 207 232 -34 14 73 11

Rank -6 -5 +18 +15 -12.5 +9 10 -12.5 +16 +14 +1 -4 +20 -8 +17 +19 -7 +3 +11 +2

n = 20 T- = 6 + 5 + 12.5 + 12.5 + 4 + 8 + 7 = 55 T = 55

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition



  z

(n)(n  1) (20)(21) = 105  4 4

n(n  1)(2n  1)  24 T 





20(21)(41) = 26.79 24

55  105 = -1.87 26.79

 = .01, /2 = .005

z.005 = ±2.575

Since the observed z = -1.87 > z.005 = -2.575 and z = -1.87 < z.005 = 2.575, the

decision is to fail to reject the null hypothesis. At  = .01, there is not

sufficient sample evidence to support the claim that there is a significant difference between the number of units sold when the dealership is listed in the white pages and the number sold when it is not listed.

17.49 H0: There is no difference between March and June Ha: There is a difference between March and June GMAT 350 430 460 470 490 500 510 520 530 530 540 550 550 560 570 570 590 600 610

Rank 1 2 3 4 5 6 7 8 9.5 9.5 11 12.5 12.5 14 15.5 15.5 17 18 19

Month J M J J M M M J M J M M J M M J J M J

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

630

Let group 1 = June and group 2 = March. n1 = 10

n2 = 10

W1 = 1 + 3 + 4 + 8 + 9.5 + 12.5 + 15.5 + 17 + 19 + 20 = 109.5 U1  n1  n2 

n1 (n1  1) (10)(11)  W1  (10)(10)   109.5 = 45.5 2 2

U 2  n1  n2  U1 = (10)(10) - 45.5 = 54.5 U = min (U1 , U2) = 45.5 From Table A.13, the p-value for U = 46 is .3980 and for 45 is .3697. For a

two-tailed test, double the p-value to at least .739. Using  = .10, the

decision is

to fail to reject the null hypothesis. At  = .10, the sample data support

the claim that there is no significant difference between the GMAT scores for test given in March and the scores for the test given in June.

17.50 Use the Friedman test. b = 6, c = 4, df = 3,  = .05 H0: The treatment populations are equal Ha: At least one treatment population yields larger values than at least on other treatment population The critical value is: 2.05,3 = 7.8147

Brand A 176 B 156 C 203 D 183 E 147 F 190

Location 2 58 111 62 98 89 117 73 118 46 101 83 113

4 120 117 105 113 114 115

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Brand A B C D E F

1 4 4 4 4 4 4

2 1 1 1 1 1 1

Location 3 2 2 3 3 2 2

4 3 3 2 2 3 3

Rj Rj2

24 576

6 36

14 196

16 256

Rj2 = 1,064

r2 

12 12 2 R j  3b(c  1)  (1,064)  3(6)(5) = 16.4  bc (c  1) (6)(4)(5)

Since r2 = 16.4 > 2.05,3 = 7.8147, the decision is to reject the null hypothesis.

At  = .05, the sample data support the claim that at least one treatment population yields larger values than at least one other treatment population. An examination of the data shows that location one produced the highest sales for all brands and location two produced the lowest sales of gum for all brands.

17.51 Ho: Md = 0 Ha: Md  0 Box 185 109 92 105 60 45 25 58 161 108 89 123 34 68

No Box 170 112 90 87 51 49 11 40 165 82 94 139 21 55

d 15 -3 2 18 9 -4 14 18 -4 26 -5 -16 13 13

Rank +11 -3 +2 +13.5 +7 -4.5 +10 +13.5 -4.5 +15.5 -6 -12 +8.5 +8.5

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

59 78

60 52

-1 26

-1 +15.5

n = 16 T- = 3 + 4.5 + 4.5 + 6 + 12 + 1 = 31 T = 31 (n)(n  1) (16)(17) = 68   4 4 n(n  1)(2n  1) 16(17)(33)    = 19.34 24 24 T   31 68 = -1.91 z   19.34  = .05, /2 = .025 z.025 = ±1.96 Since the observed z = -1.91 > z.025 = -1.96 and z = -1.91 < z.025 = 1.96, the decision is to fail to reject the null hypothesis. At  = .05, there is not sufficient sample evidence to support the claim that there is a significant difference in the number of units of perfume sold with and without the additional packaging.

17.52 Cups 25 41 16 0 11 28 34 18 5 n=9

Stress 80 85 35 45 30 50 65 40 20 6 d 2

Ranked Cups 6 9 4 1 3 7 8 5 2

Ranked Stress 8 9 3 5 2 6 7 4 1

d d2 -2 4 0 0 1 1 -4 16 1 1 1 1 1 1 1 1 1 1 2 d = 26

6(26) = .783 n(n  1) 9(80) There is a quite strong positive correlation between the number of cups of rs  1 

 1

coffee consumed on the job and perceived job stress. 17.53 n1 = 15, n2 = 15 Use the small sample Runs test Solutions Manual 1-628 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

 = .05, /.025 H0: The observations in the sample were randomly generated. Ha: The observations in the sample were not randomly generated From Table A.11, lower tail critical value = 10 From Table A.12, upper tail critical value = 22 R = 21 Since R = 21 between the two critical values, the decision is to fail to reject the

null hypothesis. At  = .05, there is not sufficient evidence to support the

claim that the observations were not randomly generated.

17.54 Ho: Md = 0 Ha: Md < 0 Before 430 485 520 360 440 500 425 470 515 430 450 495 540

After 465 475 535 410 425 505 450 480 520 430 460 500 530

d -35 10 -15 -50 15 -5 -25 -10 -5 0 -10 -5 10

Rank -11 +5.5 - 8.5 -12 +8.5 -2 -10 -5.5 -2 OMIT -5.5 -2 +5.5

n = 12 T+ = 5.5 + 8.5 + 5.5 = 19.5

T = 19.5

From Table A.14 (1-sided), using n = 12 and  = .01, the critical T is 10. Since T = 19.5 is greater than the critical T = 10, the decision is to fail to reject the null hypothesis. At  = .01, there is not sufficient sample evidence to support the claim that the scores after the statistics course are significantly higher than the scores before. Solutions Manual 1-629 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

17.55 Ho: The populations of scores of managers with ties and without ties are the same. Ha: The population of scores of managers without ties is shifted to the left of the population of scores of managers with ties. Rating 16 17 19 19 20 21 21 22 22 22 23 23 24 25 25 25 25 26 26 26 27 28

Rank 1 2 3.5 3.5 5 6.5 6.5 9 9 9 11.5 11.5 13 15.5 15.5 15.5 15.5 19 19 19 21 22

Group 2 2 2 2 2 2 1 1 1 2 1 2 2 1 1 1 2 1 1 2 1 1

Let group 1 = managers with ties and group 2 = managers without ties. n1 = 11

n2 = 11

W1 = 6.5 + 9 + 9 + 11.5 + 15.5 + 15.5 + 15.5 + 19 + 19 + 21 + 22 = 163.5 n (n  1) (11)(12) U 1  n1  n2  1 1  W1  (11)(11)   163.5 = 23.5 2 2 U 2  n1  n2  U 1  (11)(11)  23.5  97.5 U  23.5 n n (11)(11) = 60.5  1 2  2 2 n  n (n  n  1) (11)(11)(23)   1 2 1 2  = 15.23 12 12

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

z

U 





23.5  60.5 = -2.43 15.23

For  = .05,

z.05 = 1.645

Since the observed z = -2.43 < z.05 = -1.645, the decision is to reject the null

hypothesis. At  = .05, the sample data support the claim that the

managers with ties received significantly higher professionalism scores.

17.56 Ho: The populations of the amounts of sales by workers with automatic and manual dispensers are the same. Ha: The population of the amount of sales by workers with manual dispensers is shifted to the left of the population of the amount of sales by workers with automatic dispensers. Sales 92 105 106 110 114 117 118 118 125 126 128 129 137 143 144 152 153 168 n1 = 9

Rank 1 2 3 4 5 6 7.5 7.5 9 10 11 12 13 14 15 16 17 18

Type of Dispenser M M M A M M A M M M A M A A A A A A

n2 = 9

W1 = 4 + 7.5 + 11 + 13 + 14 + 15 + 16 + 17 + 18 = 115.5 n (n  1) (9)(10) U1  n1  n2  1 1  W1  (9)(9)   115.5 = 10.5 2 2 Solutions Manual 1-631 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

U 2  n1  n2  U1 = 81 – 10.5 = 70.5 U = U1 = 10.5  = .01 From Table A.13, for U = 10 the p-value = .0028 and for U = 11 the pvalue =

.0039. Thus, the p-value for U = 10.5 is definitely less than  = .01. The

decision

is to reject the null hypothesis. At  = .01, the sample data support the

claim that the workers with automatic dispensers are significantly more productive. 17.57 Ho: The 4 populations are identical Ha: At least one of the 4 populations is different 5o 216 215 218 216 219 214

15o 228 224 225 222 226 225

20o 219 220 221 223 224

30o 218 216 217 221 218 217

15o

20o

30o

23 18.5 20.5 16 22 20.5 120.5

11.5 13 14.5 17 18.5

4 2 9 4 11.5 1 31.5

74.5

9 4 6.5 14.5 9 6.5 49.5

By Ranks:

n

 j

(31.5) 2 (120.5) 2 (74.5) 2 (49.5)    6 6 5 6

= 4,103.84

n = 23

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Tj 12 12 K  3(n  1)  (4,103.84)  3(24) = 17.21  n(n  1) nj 23(24)

 = .01

df = c - 1 = 4 - 1 = 3

2.01,3 = 11.3449 Since the observed K = 17.21 > 2.01,3 = 11.3449, the decision is to reject the

null hypothesis. At  = .01, the sample data support the claim that there is a significant difference in the strength of the part for different

temperatures.

17.58 Sales 150,000 210,000 285,000 301,000 335,000 390,000 400,000 425,000 440,000 n=9

Kilometres 1,500 2,100 3,200 2,400 2,200 2,500 3,300 3,100 3,600 6 d 2

Ranks Sales 1 2 3 4 5 6 7 8 9

Ranks km 1 2 7 4 3 5 8 6 9

d 0 0 -4 0 2 1 -1 2 0

0 0 16 0 4 1 1 4 0 2 d = 26

6(26) = .783 n(n  1) 9(80) There is a quite strong positive correlation between the number of rs  1 

 1

kilometres driven by a salesperson and sales volume achieved.

17.59 Ho: The 3 populations are identical Ha: At least one of the 3 populations is different 3-day 9 11 17 10 22

Quality 27 38 25 40 31

Mgmt. Inv. 16 21 18 28 29

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

15 6

By Ranks: 3-day 2 4 7 3 12 5 1 Tj 34 nj 2 Tj

n

7 

19 35

20 31

Quality 14 20 13 21 17.5 9 19 113.5

Mgmt. Inv. 6 11 8 15 16 10 17.5 83.5

(34) 2 (113.5) 2 (83.5) 2 = 3,001.5   7 7 7

n = 21 2

Tj 12 12 K  3(n  1)  (3,001.5)  3(22) = 11.96  n(n  1) nj 21(22)

 = .10

df = c - 1 = 3 - 1 = 2

2.10,2 = 4.6052 Since the observed K = 11.96 > 2.10,2 = 4.6052, the decision is to reject the

null hypothesis. At  = .10, the sample data support the claim that there is

a significant difference between the three groups, as measured by the ratings.

17.60 Ho: Md = 0 Ha: Md < 0 Husbands 27 22 28

Wives 35 29 30

d -8 -7 -2

Rank -12 -11 -6.5

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

19 28 29 18 21 25 18 20 24 23 25 22 16 23 30

20 27 31 22 19 29 28 21 22 33 38 34 31 36 31

-1 1 -2 -4 2 +6.5 -4 -10 -1 2 -10 -13 -12 -15 -13 -1

-2.5 +2.5 -6.5 -9.5 -9.5 -13.5 -2.5 +6.5 -13.5 -16.5 -15 -18 -16.5 -2.5

n = 18 T+ = 2.5 + 6.5 + 6.5 = 15.5 T = 15.5 (n)(n  1) (18)(19) = 85.5   4 4

n(n  1)(2n  1) 18(19)(37)  = 22.96 24 24 T   15.5  85.5 = -3.05 z   22.96  = .01 z.01 = -2.33 Since the observed z = -3.05 < z.01 = -2.33, the decision is to reject the

 

null

hypothesis. At  = .01, the sample data support the claim that the wives‟

scores are significantly higher on the marketing measure than the husbands‟.

17.61 This problem uses a random block design which is analyzed by the Friedman nonparametric test. There are 4 treatments and 10 blocks. The value of the observed r2 (shown as S) is 12.16 (adjusted for ties) and has an associated

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

p-value of .007 that is significant at  = .01. At least one treatment population yields larger values than at least one other treatment population. Examining the treatment medians, treatment one has an estimated median of 20.125 and treatment two has a treatment median of 25.875. These two are the farthest apart.

17.62 This is a Runs test for randomness. n1 = 21, n2 = 29. Because of the size of the n‟s, this is a large sample Runs test. The observed number of runs is 28, R = 28. The mean or expected number of runs is 25.36. µR = 25.36 z

R = 3.41

28  25.36 = 0.77 3.41

The p-value for this statistic is .4387 for a two-tailed test , and it is greater than

 = .05. The decision is to fail to reject the null hypothesis at  = .05.

17.63 A large sample Mann-Whitney U test is being computed. There are 16 observations in each group. The null hypothesis is that the two populations are identical. The alternate hypothesis is that the two populations are not identical. The value of W is 191.5. The p-value for the test is .0066. The test is significant at  = .01. The decision is to reject the null hypothesis. At  = .01, the two populations are not identical. An examination of medians shows that the median for group two (46.5) is larger than the median for group one (37.0).

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

17.64 A Kruskal-Wallis test has been used to analyze the data. The null hypothesis is that the four populations are identical; and the alternate hypothesis is that at least one of the four populations is different. The H statistic (same as the K statistic) is 11.28 when adjusted for ties. The p-value for this H value is .010 which indicates that there is a significant difference in the four groups at  = .05 and marginally so for  = .01. An examination of the medians reveals that all group medians are the same (35) except for group 2 that has a median of 25.50. It is likely that it is group 2 that differs from the other groups.

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Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

SOLUTIONS TO PROBLEMS IN CHAPTER 18: STATISTICAL QUALITY CONTROL

18.2 Complaints Website down Too long a Wait Error message Link to incorrect page Incorrect date and time Could not locate reservation button Total

Number of Complain ts 420 184 85 37 10 8 744

% of Total 56.5 24.7 11.4 5.0 1.3 1.1 100

Cumulative % 56.5 81.2 92.6 97.6 98.9 100.0

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

18.4

x 1 = 27.00, x 2 = 24.29, x 3 = 25.29, x 4 = 27.71, x 5 = 25.86

R1 = 8, R2 = 8, R3 = 9, R4 = 7, R5 = 6 x = 26.03

R = 7.6

For x Chart:

Since n = 7, A2 = 0.419

Centerline:

x = 26.03

UCL:

x + A2 R = 26.03 + (0.419)(7.6) = 29.21

LCL:

x - A2 R = 26.03 – (0.419)(7.6) = 22.85

For R Chart:

Since n = 7, D3 = 0.076

D4 = 1.924

Centerline: R = 7.6 UCL: D4 R = (1.924)(7.6) = 14.62 LCL:

D3 R = (0.076)(7.6) = 0.58

x Chart:

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

UCL

LCL

R Chart:

All samples are within the control limits. 18.5

x 1 = 4.55, x 2 = 4.10, x 3 = 4.80, x 4 = 4.70,

x 5 = 4.30, x 6 = 4.73, x 7 = 4.38 R1 = 1.3, R2 = 1.0, R3 = 1.3, R4 = 0.2, R5 = 1.1, R6 = 0.8, R7 = 0.6 x = 4.51

R = 0.90

For x Chart:

Since n = 4, A2 = 0.729

Centerline:

x = 4.51

UCL:

x + A2 R = 4.51 + (0.729)(0.90) = 5.16

LCL:

x - A2 R = 4.51 – (0.729)(0.90) = 3.85

For R Chart: Since n = 4, D3 = 0 D4 = 2.282 Centerline: R = 0.90 UCL: D4 R = (2.282)(0.90) = 2.05 Solutions Manual 1-640 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

LCL:

D3 R = 0

x Chart:

R Chart:

18.6

p̂1 = .02, p̂2 = .07, p̂3 = .04, p̂4 = .03, p̂5 = .03 p̂6 = .05, p̂7 = .02, p̂8 = .00, p̂9 = .01, p̂10 = .06 p

 pˆ  0.33 = .033

k 10 Centerline: p = .033

(.033)(.967) = .033 + .054 = .087 100 (.033)(.967) LCL: .033 - 3 = .033 - .054 ~ .000 100 UCL: .033 + 3

p Chart:

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

The proportion for sample 8 is very close to the lower quality control limit.

18.7

p̂1 = .025, p̂2 = .000, p̂3 = .025, p̂4 = .075, p̂5 = .050, p̂6 = .125, p̂7 = .050 p

 pˆ  0.350 = .050

k Centerline:

7 p = .050

UCL: .05 + 3

(.05)(.95) = .05 + .1034 = .1534 40

LCL:

(.05)(.95) = .05 - .1034 ~ .000 40

.05 - 3

p Chart:

The proportion for sample 2 is very close to the lower quality control limit.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

18.8

22 c  c  c    ci = = 0.6286 c 1 2 3 i 35 Centerline: c = 0.6286

UCL:

c  3 c = 0.6286 + 3 0.6286 = 0.6286 + 2.3785 = 3.0071

LCL:

c  3 c = 0.6286 - 3 0.6286 = 0.6286 – 2.3785 ~ .000

c Chart: UCL

LCL

None of the points are beyond the control limits. However, four consecutive points are close to LCL.

18.9

43 c  c  c    ci = = 1.344 c 1 2 3 32 i

Centerline: c = 1.344 UCL:

c  3 c = 1.344 + 3 1.344 = 1.344 + 3.478 = 4.822

LCL:

c  3 c = 1.344 - 3 1.344 = 1.344 - 3.478 ~ 0.000

c Chart:

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

None of the points are beyond the control limits. However, some points are close to LCL.

18.10 a.) Eight consecutive points are decreasing. Two of three consecutive points are in the outer one-third (near LCL). Four out of five points are in the outer two-thirds (near LCL). b.) This is a relatively healthy control chart with no obvious rule violations. c.) One point is above the UCL. Two out of three consecutive points are in the outer one-third (both near LCL and near UCL). There are six consecutive increasing points.

18.11 While there are no points outside the limits, the first chart exhibits some problems. The chart ends with 9 consecutive points below the centerline. Of these 9 consecutive points, there are at least 4 out of 5 in the outer 2/3 of the lower region. The second control chart contains no points outside the control limits. However, near the end, there are 8 consecutive points above the centerline. The p chart contains no points outside the upper control limit. Three times, the chart contains two out of three points in the outer third. However, this occurs in the lower third where the proportion of noncompliance items approaches zero and is probably not a problem to be concerned about. Overall, this seems to display a process that is in control. One concern might be the wide swings in the proportions at samples 15, 16 and 22 and 23.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

18.12

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

18.13 Problem 1 6 8 4 10 3 5 9 2 7 Total

Frequency

% of Total

Cumulative %

673 564 402 379 202 108 73 54 29 12 2,496

27.0 22.6 16.1 15.2 8.1 4.3 2.9 2.2 1.2 0.5 100

27.0 49.6 65.7 80.9 89.0 93.3 96.2 98.4 99.6 100

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

18.14 Fishbone Diagram:

18.15

p̂1 = .06, p̂2 = .22, p̂3 = .14, p̂4 = .04, p̂5 = .10, p̂6 = .16, p̂7 = .00, p̂8 = .18, p̂9 = .02, p̂10 = .12 pˆ 1.04 p  = .104 k 10 Centerline: p = .104

(.104)(.896) = .104 + .130 = .234 50 (.104)(.896) LCL: .104 - 3 = .104 - .130 ~ .000 50 UCL: .104 + 3

p Chart:

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

18.16 x 1 = 1.89174, x 2 = 1.89380, x 3 = 1.88970, x 4 = 1.89000,

x 5 = 1.88984, x 6 = 1.89142, x 7 = 1.89000, x 8 = 1.89268, x 9 = 1.89112, x 10 = 1.89016, x 11 = 1.89096, x 12 = 1.89174 R1 = 0.0047, R2 = 0.0071, R3 = 0.0063, R4 = 0.0024, R5 = 0.004, R6 = 0.0039, R7 = 0.004, R8 = 0.0063, R9 = 0.0024, R10 = 0.0008, R11 = 0.0032, R12 = 0.0039 x = 1.89110

R = 0.004083

For x Chart:

Since n = 5 A2 = .577

Centerline:

x = 1.89110

UCL:

x + A2 R = 1.89110+ (0.577)( 0.004083) = 1.89110 + .00236 = 1.89346

LCL:

x - A2 R = 1.89110 - (0.577)( 0.004083) = 1.89110 - .00236 = 1.88874

For R Chart:

Since n = 5, D3 = 0

D4 = 2.115

Centerline: R = 0.004083 UCL: D4 R = (2.115)( 0.004083) = .00864 LCL: D3 R = (0)( 0.004083) = 0

x Chart:

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

The sample mean for sample 2 is above UCL. R Chart:

Two out of three consecutive values (ranges for samples 9, 10, 11) are in the outer one third. 77 c  c  c    ci 18.17 c  1 2 3 = = 2.139 i 36

Centerline: c = 2.139 The centreline is the average of the numbers of nonconformances for all sheets. UCL: c  3 c = 2.139 + 3 2.139 = 2.139 + 4.388 = 6.527 LCL:

c  3 c = 2.139 - 3 2.139 = 2.139 – 4.388 ~ .00000

c Chart:

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

18.18 x 1 = 1.2100, x 2 = 1.2050, x 3 = 1.1900, x 4 = 1.1725,

x 5 = 1.2075, x 6 = 1.2025, x 7 = 1.1950, x 8 = 1.1950, x 9 = 1.1850 R1 = .04, R2 = .02, R3 = .04, R4 = .04, R5 = .06, R6 = .02, R7 = .07, R8 = .07, R9 = .06, x = 1.1958

R = 0.0467

For x Chart:

Since n = 4, A2 = .729

Centerline:

x = 1.1958

UCL:

x + A2 R = 1.1958 + .729(.0467) = 1.1958 + .0340 = 1.2298

LCL:

x - A2 R = 1.1958 - .729(.0467) = 1.1958 - .0340 = 1.1618

For R Chart:

Since n = 4, D3 = 0

D4 = 2.282

Centerline: R = .0467 UCL: D4 R = (2.282)(.0467) = .1066 LCL: D3 R = (0)(.0467) = 0

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

R chart:

The process appears to be in control.

18.19 x 1 = 14.993, x 2 = 15.000, x 3 = 14.978, x 4 = 14.990,

x 5 = 15.013, x 6 = 15.000, x 7 = 15.017, x 8 = 14.997, R1 = .03, R2 = .07, R3 = .05, R4 = .05, R5 = .04, R6 = .05, R7 = .05, R8 = .06 x = 14.999

R = 0.05

For x Chart:

Since n = 6, A2 = .483

Centerline:

x = 14.999

UCL:

x + A2 R = 14.999 + .483(.05) = 14.999 + .024 = 15.023

LCL:

x - A2 R = 14.999 - .483(.05) = 14.999 - .024 = 14.975

For R Chart:

Since n = 6, D3 = 0

D4 = 2.004

Centerline: R = .05 UCL: D4 R = 2.004(.05) = .100 LCL: D3 R = 0(.05) = .000

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Two out of three consecutive points on the x chart are in the outer one third.

18.20

p̂1 = .12000, p̂2 = .04000, p̂3 = .00000, p̂4 = .02667, p̂5 = .09333, p̂6 = .18667, p̂7 = .14667, p̂8 = .10667, p̂9 = .06667, p̂10 = .05333, p̂11 = .0000, p̂12 = .09333 p k

pˆ

LCL:

.07778 - 3

0.93334 = .07778 Centerline: p = .07778 12 (.07778)(.92222) UCL: .07778 + 3 = .07778 + .09278 = .17056 75 

(.07778)(.92222) = .07778 - .09278 ~ .00000 75

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

p Chart: UCL

LCL

The proportion for sample 6 is above UCL. Then, there are also six consecutive decreasing values.

16 c  c  c    ci 18.21 c  1 2 3 = = 0.640 i 25

Centerline: c = 0.640 UCL:

c  3 c = 0.640 + 3 0.640 = 0.640 + 2.400 = 3.040

LCL:

c  3 c = 0.640 - 3 0.640 = 0.640 – 2.400 ~ .000

c Chart:

None of the points are beyond the control limits, but three consecutive points (bottle number = 14, 15, 16) are close to LCL. Solutions Manual 1-653 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

81 c  c  c    ci 18.22 c  1 2 3 = = 2.025 i 40

Centerline: c = 2.025 UCL:

c  3 c = 2.025 + 3 2.025 = 2.025 + 4.269 = 6.294

LCL:

c  3 c = 2.025 - 3 2.025 = 2.025 – 4.269 ~ .000

c Chart:

None of the points are beyond the control limits. However, two out of three consecutive values are in the outer one third.

18.23

p̂1 = .050, p̂2 = .000, p̂3 = .150, p̂4 = .075, p̂5 = .025, p̂6 = .025, p̂7 = .125, p̂8 = .000, p̂9 = .100, p̂10 = .075, p̂11 = .050, p̂12 = .050, p̂13 = .150, p̂14 = .025, p̂15 = .000 p

 pˆ  0.900 = .06 k

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Centerline:

p = .060

UCL: .060 + 3

(.060)(.940) = .060 + .113= .173 40

LCL:

(.060)(.940) = .060 - .113 ~ .000 40

.060 - 3

p Chart:

Twice two out of three consecutive values fall in the outer one third.

18.24 The following list provides summary of the control chart abnormalities that should be of concern to a statistical process controller:  Mean of the sample 17 is below LCL.  There are four consecutive points (not > 8) above the centerline. It happens twice.  There are no more than five consecutive points decreasing; three consecutive points increasing. However, we cannot say that a trend is present.

18.25 The following list provides summary of the control chart abnormalities that should be of concern to a statistical process controller:  Range of the sample 25 is above UCL.  Near the beginning of the chart there are nine consecutive sample ranges below the centerline. Solutions Manual 1-655 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition



Near the end of the chart there are nine consecutive sample ranges above the centerline. The controller might want to determine if there is some systematic reason why there is a string of ranges below the centerline and, perhaps more importantly, why there are a string of ranges above the centerline.

18.26 The following list provides summary of the control chart abnormalities that should be of concern to a statistical process controller:  Two of the sixty samples (about 3%) produce proportions that are above UCL.  In the left to the middle of the chart there are ten consecutive points below the centerline.  In the right to the middle of the chart there are fourteen consecutive points below the centerline.

18.27 The centerline of the c chart indicates that the process is averaging 0.7400 nonconformances per part. Twenty-five of the fifty sampled items have zero nonconformances. None of the samples exceed the upper control limit for nonconformances. However, the upper control limit is 3.321 nonconformances which, in and of itself, may be too many. Indeed, three of the fifty (6%) samples actually had three nonconformances. An additional six samples (12%) had two nonconformances. One matter of concern may be that there is a run of ten samples in which nine of the samples exceed the centerline (samples 12 through 21). The question raised by this phenomenon is whether or not there is a systematic flaw in the process that produces strings of nonconforming items.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Legal Notice Copyright

SOLUTIONS TO PROBLEMS IN CHAPTER 19: DECISION ANALYSIS

19.1

Max

Min

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

d1 250

175

-25

250

-25

d2 110

100

110

d3 390

140

-80

390

-80

a.) Max {250, 110, 390} = 390

decision: Select d3

b.) Max {-25, 70, -80} = 70

decision: Select d2

c.) For  = .3 d1: .3(250) + .7(-25) = 57.5 d2: .3(110) + .7(70) = 82 d3: .3(390) + .7(-80) = 61 Max {57.5, 82, 61} = 82

decision: Select d2

For  = .8 d1: .8(250) + .2(-25) = 195 d2: .8(110) + .2(70) = 102 d3: .8(390) + .2(-80) = 296 Max {195, 102, 296} = 296

decision: Select d3

Comparing the results for the two different values of alpha, with a more pessimist point-of-view ( = .3), the decision is to select d2 and the payoff is 82. Selecting by using a more optimistic point-of-view ( = .8) results in choosing d3 with a higher payoff of 296. d.) The opportunity loss table is: S1

Max

d1 140

140

d2 280

280

150

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

The minimax regret = min {140, 280, 150} = 140 Decision: Select d1 to minimize the regret.

19.2

Max

Min

120

110

120

100

d4 100

-30

-20

100

-30

-10

a.) Maximax = Max {120, 100, 60, 100, 80} = 120 Decision: Select d1 b.) Maximin = Max {50, 20, 20, -30, -10} = 50 Decision: Select d1 c.)  = .5 d1: .5(120) + .5(50) = 85 d2: .5(100) + .5(20) = 60 d3: .5(60) + .5(60) = 40 d4: .5(100) + .5(-30) = 35 d5: .5(80) + .5(-10) = 35 Max { 85, 60, 40, 35, 35 } = 85 Decision: Select d1 d.) Opportunity Loss Table:

Max

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

150

130

150

d5 100

100

Min {50, 65, 90, 150, 100} = 50 Decision: Select d1

19.3

Conservatives

Liberals

NDP

Max Min

-25

-10

Maximax : Max {60, 30, 40, 25} = 60 Decision: Select A Maximin : Max {-25, 10, -10, 5} = 10 Decision: Select B

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

19.4

Don't Produce Produce Few Produce Many

Not Receptive -50 -200 -600

Somewhat Very Receptive Max -50 300 100

-50 400 1,000

Min

-50 400 1,000

-50 -200 -600

Very Receptive 1,050 600 0

Max 1,050 600 550

a.) For Hurwicz criterion using  = .6: Don‟t Produce: .6(-50) + .4(-50) = -50 Produce Few: .6(400) + .4(-200) = 160 Produce Many: .6(1,000) + .4(-600) = 360 Max {-50, 160, 360}= 360 Decision: Select "Produce Many".

b.) Opportunity Loss Table:

Don't Produce Produce Few Produce Many

Not Receptive Somewhat 0 350 150 0 550 200

Minimax regret: Min {1,050, 600, 550} = 550 Decision: Select "Produce Many"

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

19.5, 19.6

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Decision: Select d2 .

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

19.7

Expected Monetary Payoff with Perfect Information = 75(.15) + 50(.25) + 20(.30) + 8(.10) + 6(.20) = 31.75 Expected Value of Perfect Information = 31.75 - 25.25 = 6.50

19.8 a.) & b.)

100 200 400

75 150 700

125 450 650

c.) Expected Monetary Payoff with Perfect Information = 150(.40) + 450(.35) + 700(.25) = 392.50 Expected Value of Perfect Information = 392.5 - 370 = 22.50 Solutions Manual 1-664 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

19.9

a State of Interest Rates Up (.30) Down (.65)

Nature will go Unchangeable (.05)

Decision

Lock in

-150

200

Alternativ e

Wait

175

-250

b. EMV for “Lock in” = (-150)(.30) + 200(.65) + 0(.05) = 85 EMV for “Wait” = 175(.30) + (-250)(.65) + 0(.05) = -110 Max {85 , -110} = 85 Decision:

Lock in a mortgage loan interest rate, do not wait until

closing.

c. Expected Monetary Payoff with Perfect Information = = 175(.30) + (200)(.65) + 0(.05) = 182.50 From b. Expected Monetary Payoff without Perfect Information = 85 Expected Value of Perfect Information = 182.50 – 85 = 97.50

19.10 EMV for “No Layoffs” = 100(.10) + (-300)(.40) + (-1,700)(.50) = -960 EMV for “Lay off 1,000” = (-100)(.10) + 100(.40) + (700)(.50) = -320 EMV for “Lay off 5,000” = (-200)(.10) + 300(.40) + 600(.50) = 400 Max {-960 , -320, 400} = 400 Decision: Based on maximum EMV = 400, Lay off 5,000.

Expected Monetary Payoff with Perfect Information = 100(.10) + 300(.40) + 600(.50) = 430 Solutions Manual 1-665 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Expected Value of Perfect Information = 430 - 400 = 30

19.11 a.) EMV = 200,000(.50) + (-50,000)(.50) = 75,000 b.) Risk Avoider because the EMV is more than the investment (75,000 > 50,000). c.) You would have to offer more than 75,000 which is the expected value.

19.12 a.)

S1(.30)

S2(.70)

EMV

350

-100

350(.30) + (-100)(.70) = 35.00

-200

325

(-200)(.30) + 325(.70) =

167.50 maximum {35.00, 167.50} = 167.50 Decision without sample information: Based on EMV = 167.50, select d2. b. & c.)

For Forecast F1: Prior

Conditional

Joint

Revised

P(S1) = .30

P(F1| S1) = .90

P(F1  S1) = .270

.270/.445

P(S2) = .70

P(F1| S2) = .25

P(F1  S2) = .175

.175/.445

= .607

= .393 P(F1) = .445 For Forecast F2:

Prior

Conditional

P(S1) = .30

P(F2| S1) = .10

Joint P(F2  S1) = .030

Revised .030/.555

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

P(S2) = .70

P(F2| S2) = .75

P(F2  S2) = .525

.525/.555

= .946 P(F2) = .555

EMV with Sample Information = 241.69

d.) EMV of Sample Information = 241.69 - 167.50 = 74.19

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

19.13 a. State of Nature Fuel Fuel Decrease Increase (0.60) (0.40) Small Cars

-225

425

Medium Cars

125

-150

Alternativ Large Cars e

350

-400

Decision

EMV (-225)(.60) + 425(.40) = 35 125(.60) + (-150)(.40) = 15 350(.60) + (-400)(.40) = 50

Maximum {35, 15, 50} = 50 Decision: Based on EMV = 50, purchase a fleet of large cars.

b. Let states of nature: S1 = Fuel Decrease and S2 = Fuel Increase; the forecasts: F1 = Fuel Decrease and F2 = Fuel Increase. For Forecast F1: Prior

Conditional

Joint

Revised

P(S1) = .60

P(F1| S1) = .75

P(F1  S1) = .45

.45/.51 =

P(S2) = .40

P(F1| S2) = .15

P(F1  S2) = .06

.06/.51 =

.8824

.1176 P(F1) = P(FDec.) = .51 For Forecast F2: Prior

Conditional

Joint

Revised

P(S1) = .60

P(F2| S1) = .25

P(F2  S1) = .15

.15/.49 =

P(S2) = .40

P(F2| S2) = .85

P(F2  S2) = .34

.34/.49 =

.3061

.6939 P(F2) = P(FInc.) = .49

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

EMV with Sample Information = 244.28 EMV of Sample Information = EVSI = 244.28 – 50.00 = 194.28 The agency should buy the forecast.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

19.14

a. State of Demand Decline Same Increase (0.20) (0.30) (0.50) Don't Plant Small Tree Large Tree

Decision Alternative

EMV

-40

20(.20) + 0(.30) + (-40)(.50) = -16

-90

175

(-90)(.20) + 10(.30) +175(.50) = 72.5

-600

-150

800

(-600) (.20) + (-150)(.30) + 800(.50) =

235

Max {-16, 72.5, 235} = 235 Decision: Based on Maximum EMV = 235, plant a large tree farm. b. Let states of demand: S1 = Decline, S2 = Same, and S3 = Increase; the forecasts: F1 = Decrease, F2 = Same, and F3 = Increase. For Forecast F1: Prior

Conditional

Joint

Revised S1

P(S1) = .20

P(F1| S1) = .70

P(F1  S1) = .140

.140/.156

P(S2) = .30

P(F1| S2) = .02

P(F1  S2) = .006

.006/.156

P(S3) = .50

P(F1| S3) = .02

P(F1  S3) = .010

.010/.156

= .8974

= .0385

= .0641 P(F1) = P(FDec.) = .156 For Forecast F2: Prior

Conditional

Joint

Revised S1

P(S1) = .20

P(F2| S1) = .25

P(F2  S1) = .050

.050/.375

P(S2) = .30

P(F2| S2) = .95

P(F2  S2) = .285

.285/.375

P(S3) = .50

P(F2| S3) = .08

P(F2  S3) = .040

.040/.375

= .1333

= .7600

= .1067 P(F2) = P(FSame) = .375 For Forecast F3: Solutions Manual 1-670 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Prior

Conditional

Joint

Revised S1

P(S1) = .20

P(F3| S1) = .05

P(F3  S1) = .010

.010/.469 =

P(S2) = .30

P(F3| S2) = .03

P(F3  S2) = .009

.009/.469

P(S3) = .50

P(F3| S3) = .90

P(F3  S3) = .450

.450/.469

.0213

= .0192

= .9595 P(F3) = P(FInc.) = .469

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

EMV with Sample Information = 360.41 EMV of Sample Information = EVSI = 360.41 – 235.00 = 125.41 19.15 a. The probability that there is no oil is 1 – 0.11 = 0.89

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

State of Nature Oil (0.11) No Oil (0.89) Decision

Drill 1,000,000

Alternative

Don't Drill

-100,000

EMV 1,000,000(.11) + (-100,000)(.89) = 21,000

0(.11) +0(.89) = 0

Max {21,000, 0} = 21,000 Decision: The decision is to Drill.

b. Let states of nature: S1 = Oil and S2 = No Oil; the forecasts: F1 = Oil is present and F2 = No Oil. Given P(F1| S1) = .20, P(F2| S1) = .80, P(F2| S2) = .90, and P(F1| S2) =.10 For Forecast F1: Prior

Conditional

Joint

Revised

P(S1) = .11

P(F1| S1) = .20

P(F1  S1) = .022

.022/.111

P(S2) = .89

P(F1| S2) = .10

P(F1  S2) = .089

.089/.111

= .1982

= .8018 P(F1) = P(FOil) = .111 For Forecast F2: Prior

Conditional

Joint

Revised S1

P(S1) = .11

P(F2| S1) = .80

P(F2  S1) = .088

.088/.889

P(S2) = .89

P(F2| S2) = .90

P(F2  S2) = .801

.801/.889

= .0990

= .9010 P(F2) = P(FNo Oil) = .889

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

EMV with Sample Information = 21,012.32

EMV of Sample Information = EVSI = 21,012.32 - 21,000 = 12.32

19.16

Max.

Min.

100

-75

200

-75

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

a.) Maximax:

Max {100, 200, 40, 75} = 200

Decision: Select d2 b.) Maximin:

Max {50, -75, 25, 10} = 50

Decision: Select d1 c.) Hurwicz with  = .6 d1: 100(.6) + 50(.4) = 80 d2: 200(.6) + (-75)(.4) = 90 d3: 40(.6) + 25(.4) = 34 d4: 75(.6) + 10(.4) = 49 Max {80, 90, 34, 49} = 90 Decision: Select d2 d.) Opportunity Loss Table: S1

Maximum

100

150

160

190

Min {100, 150, 160, 190} = 100 Decision: Select d1

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

19.17 a.

b.) EMV for d1: 400(.30) + 250(.25) + 300(.20) + 100(.25) = 267.5 EMV for d2: 300(.30) + (-100)(.25) + 600(.20) + 200(.25) = 235 Max {267.5 , 235} = 267.5 Decision: Select d1 c.) Expected Payoff of Perfect Information: 400(.30) + 250(.25) + 600(.20) + 200(.25) = 352.5 Value of Perfect Information = 352.5 - 267.5 = 85

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

19.18

S1(.40)

S2(.60)

EMV

200

150

200(.40) + 150(.60) = 170

-75

450

(-75)(.40) + 450(.60) = 240

175

125

175(.40) + 125(.60) = 145

Max {170, 240, 145} = 240 Decision: Based on Maximum EMV = 240, select d2. For Forecast F1 (predict S1): Prior

Conditional

Joint

Revised

P(S1) = .40

P(F1| S1) = .90

P(F1  S1) = .36

.36/.54

P(S2) = .60

P(F1| S2) = .30

P(F1  S2) = .18

.18/.54

= .667

= .333 P(F1) = P(Predict S1) = .54 For Forecast F2 (predict S2): Prior

Conditional

Joint

P(S1) = .40

P(F2| S1) = .10

P(F2  S1) = .04

.04/.46

P(S2) = .60

P(F2| S2) = .70

P(F2  S2) = .42

.42/.46

Revised

= .087

= .913 P(F2) = P(Predict S2) = .46

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

The Expected Value With Sample Information is 285.00. Value of Perfect Information = 285 - 240 = 45 Solutions Manual 1-678 Chapter 1 Copyright © 2020 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission of this page is strictly prohibited.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

19.19

Small

Moderate

Large

Min

Max

Small

200

250

300

200

300

Modest

100

300

600

100

600

Large

-300

400

2,000

-300

2,000

a.) Maximax: Max {300, 600, 2,000} = 2,000 Decision: Large Number Maximin: Max {200, 100, -300} = 200 Decision: Small Number b.) Opportunity Loss: Small

Moderate

Large

Max

Small

150

1,700

Modest

100

1,400

Large

500

Min {1,700, 1,400, 500} = 500 Decision: Large Number c.) Minimax regret criteria leads to the same decision as Maximax.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

19.20

Slow

Fast

Max

Min

Low

-700

-400

1,200

-700

Medium

-300

-100

550

-300

High

100

125

150

100

a. i. = .10: Low: 1,200(.1) + (-700)(.9) = -510 Medium: 550(.1) + (-300)(.9) = -215 High: 150(.1) + 100(.9) = 105 Max {-510, -215, 105} = 105 Decision: Price High ii.  = .50: Low: 1,200(.5) + (-700)(.5) = 250 Medium: 550(.5) + (-300)(.5) = 125 High: 150(.5) + 100(.5) = 125 Max {250, 125, 125} = 250 Decision: Price Low iii.  = .80: Low: 1,200(.8) + (-700)(.2) = 820 Medium: 550(.8) + (-300)(.2) = 380 High: 150(.8) + 100(.2) = 140 Max {820, 380, 140} = 820 Decision: Price Low b. Two of the three alpha values (.50 and .80) lead to a decision of pricing low. Alpha of .10 suggests pricing high as a strategy. For optimists (high alphas), pricing low is a better strategy; but for more pessimistic people, pricing high may be the best strategy.

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

19.21

Mild (.75)

Severe (.25)

EMV

Reg.

2,000

-2,500

Weekend

1,200

-200

1,200(.75) + (-200)(.25) = 850

Not Open

-300

100

(-300)(.75) + 100(.25) = -200

2,000(.75) + (-2,500)(.25) =

875

Max{875, 850, -200} = 875 Decision: Based on Max EMV = 875 , open parks during regular hours.

Expected Value with Perfect Information = 2,000(.75) + 100(.25) = 1,525 Value of Perfect Information = 1,525 - 875 = 650

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

19.22 Weaker (.35) Same (.25) Stronger (.40)

EMV

Don't Produce

-700

-200

150

(-700)(.35) + (-200)(.25) +150(.40)

Produce

1,800

400

-1,600

1,800(.35) + 400(.25) +(-1,600)(.40)

= -235

= 90

Max {-235, 90} = 90 Decision: Based on Max EMV = 90, select “Produce the Product”. Expected Payoff With Perfect Information = 1,800(.35) + 400(.25) + 150(.40) = 790 Value of Perfect Information = 790 - 90 = 700

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

19.23 a. Red Con. Inc. (.15) (.35) (.50) Automate -40,000 -15,000 60,000

EMV (-40,000)(.15) + (-15,000)(.35) + 60,000(.50) =

Do Not

5,000(.15) + 10,000(.35) + (-30,000)(.50) = -

18,750 5,000

10,000 -30,000

10,750

Max {18,750 , -10,750} = $18,750 Decision: Based on Max EMV = $18,750, purchase an automated checkout system. b. Let states of sales: S1 = Reduction, S2 = Constant, and S3 = Increase; the forecasts to predict: F1 = Reduction, F2 = Constant, and F3 = Increase. For Forecast F1: S1

Prior P(S1) = .15

Conditional P(F1| S1) = .60

Joint P(F1  S1) = .090

Revised .090/.150 =

P(S2) = .35

P(F1| S2) = .10

P(F1  S2) = .035

.035/.150 =

P(S3) = .50

P(F1| S3) = .05

P(F1  S3) = .025

.025/.150 =

.6000

.2333

.1667 P(F1) = P(FRed.) = .150 For Forecast F2: Prior S1 P(S1) = .15

Conditional P(F2| S1) = .30

Joint P(F2  S1) = .045

Revised .045/.450 =

P(S2) = .35

P(F2| S2) = .80

P(F2  S2) = .280

.280/.450 =

P(S3) = .50

P(F2| S3) = .25

P(F2  S3) = .125

.125/.450 =

.1000

.6222

.2778 P(F2) = P(FConst.) = .450 For Forecast F3: Prior S1 P(S1) = .15

Conditional P(F3| S1) = .10

Joint P(F3  S1) = .015

Revised .015/.400 =

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

P(S2) = .35

P(F3| S2) = .10

P(F3  S2) = .035

.035/.400 =

P(S3) = .50

P(F3| S3) = .70

P(F3  S3) = .350

.350/.400 =

.0875

.8750 P(F3) = P(FInc.) = .400 Expected Value With Sample Information = 21,425.55

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Value of Perfect Information = 21,425.55 - 18,750 = 2,675.55

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

19.24 a. Chosen (.20) Not Chosen (.80)

EMV

Build

12,000

-8,000

12,000(.20) + (-8,000)(.80) = -4,000

Don't

-1,000

2,000

(-1,000)(.20) + 2,000(.80) = 1,400

Max {-4000, 1400} = 1,400 Decision: Based on Max EMV = 1,400, choose "Don't Build Terminal" as a strategy.

b. Let states of nature: S1 = City Chosen and S2 = City Not Chosen; the forecasts to predict: F1 = City Chosen and F2 = City Not Chosen. For Forecast F1: Prior

Conditional

Joint

Revised

P(S1) = .20

P(F1| S1) = .45

P(F1  S1) = .09

.09/.41 =

P(S2) = .80

P(F1| S2) = .40

P(F1  S2) = .32

.32/.41 =

.2195

.7805 P(F1) = P(FChosen) = .41 For Forecast F2: Prior

Conditional

Joint

Revised

P(S1) = .20

P(F2| S1) = .55

P(F2  S1) = .11

.11/.59 =

P(S2) = .80

P(F2| S2) = .60

P(F2  S2) = .48

.48/.59 =

.1864

.8136 P(F2) = P(FNot Chosen) = .59

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

Expected Value With Sample Information = 1,400.09 Value of Perfect Information = 1,400.09 - 1,400 = .09

Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition

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Solution Manual for Black, Bayley, Castillo: Business Statistics, Third Canadian Edition