ch03-solns-all_skuce_2e

8/13/2019 ch03-solns-all_skuce_2e

1/14

Instructors Solutions Manual - Chapter 3

Chapter 3 Solutions

Develop Your Skills 3.1

1. y = 2 + 4 + 6 + 8 = 20

2.

y

2

= 2

2

+ 4

2

+ 6

2

+ 8

2

= 120(y)2= (2 + 4 + 6 + 8) = 20 2= 400The answers are different because of the different order of operations

3. 54

20

n

y

44

16

n

x

4. 031)1(3)47()45()43()41()4x(

031)1(3)58()56()54()52()5y(

5. Consider the data set: 34, 67, 2, 31, 89, 35. For this data set, calculate:

a. 25835893126734x

b. 1575635893126734x 2222222

c. 436

258

n

x

d.

535.304.9325

4662

5

1109415756

5

6

25815756

1n

n

xx

22

2


6. The mean age is 41.2, the median age is 35.5, and the mode of the ages is 30. In thiscase, because the data set is severely skewed to the right (as we saw when we createdthe histogram of ages in Develop Your Skills 2.2, Exercise 6), the median is thebetter measure of central tendency.

7. The mean income is $47,868.10, and the median income is $44,925. This data set isskewed to the right (as we saw when we created the histogram of incomes inDevelop Your Skills 2.2, Exercise 7). As a result the unusually high incomes havepulled the mean to the right of the median. The median is the better measure ofcentral tendency.

Copyright 2011 Pearson Canada Inc. 51


2/14


8. From the stem and leaf display we constructed in Develop Your Skills 2.2, Exercise8, we can see that this data set is slightly skewed to the right, but not much. This isreflected in the calculations of mean and median (when in doubt, calculate both!).The mean of this data set is 26.12, and the median is 26. Either would be acceptableas a measure of central tendency, but the mean is preferred, because its calculation

depends on the value of every single data point in the data set.

9. Because the mean and the median are almost equal, we expect the distribution to besymmetric.

10. Because the quarterly operating profits of the oil and gas sector are highly skewed tothe right, the median of $1.816 billion is the appropriate measure of central tendency.Although the distribution of operating profits for the manufacturing sector is not asskewed, so we might have considered using the mean as a measure of centraltendency, we must use the median so that we are comparing the same measure forboth data sets. The median quarterly operating profit for the manufacturing sector is

$8.909 billion. Generally, the quarterly operating profits are much higher for themanufacturing sector than for the oil and gas sector.


11. Since the age data are skewed to the right, the IQR is the best measure of variability.Using Excel calculations, we find:

Q1= 31Q3= 42IQR = 11

The Empirical Rule could not be applied here, as the data are not symmetric and bell-shaped.

12. Since the data are skewed to the right, the IQR is the best measure of variability.Using Excel calculations we find:

Q1= $40,350Q3= $55,400IQR = $15,050

13. Since this data set is fairly symmetric with no obvious outliers, the standarddeviation is the preferred measure of variability.

429.719333.5524

64.132424

36.170561838124

25

65318381

1nn

xx

s

22

2



3/14


14. Because the distribution is reasonably symmetric and bell-shaped, the Empirical Rulecan be applied. You mustcreate a histogram to check this. Shown below is onepossible histogram for the data set.

0

1

2

3

4

5

6

7

8

9

10

Nu

mbero

fDay

s

NumberofCustomers

DailyCustomerCounts,DowntownAutomotive

15. The mean number of daily customers at Downtown Automotive is 26.12 (calculatedfor Develop Your Skills 3.2, Exercise 8). The standard deviation is 7.43 (the answerto Exercise 13 above).

The Empirical Rule says that about 95% of the data points will lie within 2 standard

deviations of the mean. s2x 26.12 + 2(7.43) = 40.98 s2x 26.12 - 2(7.43) = 11.26

If this sample is representative of the population, then 95% of the daily customercounts will be between 11.26 and 40.98. Since the data set is (more or less)symmetric, this means about 2% of the data will lie below 11.26, and about 2%will lie above 40.98. About 97.5% of the time, the maximum number of customersDoug would need to plan for is 41.


16. The scatter diagram showed some (not much) evidence of a positive relationshipbetween household income and monthly spending on restaurant meals. Since therelationship appears to be linear, the Pearson r is the appropriate measure ofassociation. Excel calculates it as 0.42. This is positive and less than 0.5, as wewould expect, since the relationship is not very strong.



4/14


17. The only choice is b (-0.88). Choices a and c are incorrect, because they are positiveand the relationship is clearly negative. Choice d is not correct, because the negativerelationship is obviously fairly strong.

18. The Spearman rank correlation coefficient must be used here, since the data are

ranked. The Spearman r (calculated with Excel) is 0.61. This indicates a positiverelationship between the recruiters ranking and the supervisors ranking, but therelationship is not particularly strong.

19. These are quantitative data, and the graph created for Develop Your Skills 2.5,Exercise 24 shows a linear relationship. The Pearson r is the correct measure ofassociation. Excel calculates it at -0.67 (note that you mustcheck for linearity of therelationship before you calculate the Pearson r). There is a negative relationshipbetween the two variables. The greater the number of hours of paid employmentduring the semester, the lower the semester average mark.

20. Exhibit 3.44b is the graph that corresponds to the negative correlation coefficient of-0.90. This is obvious, since it is the only graph of the three showing a negativerelationship. Exhibits 3.44a and c share the same correlation coefficient of 0.73.This is interesting because the correlation probably looks stronger in Exhibit 3.44a.However, notice that these two graphs depict exactly the same data, but with the x-and y-axes reversed. Realize that you cannot reliably eyeball the strength of arelationship. The correlation coefficient allows us to make much more precisecomparisons.

Chapter Review Exercises

1. The mean mark is quite a bit higher than the median mark. This suggests that thedistribution of marks is skewed to the right. It is likely that there are a few unusuallyhigh marks in the distribution.

2. The mean weekly sales for both businesses are similar, although the mean sales at thehaircutting salon are a bit lower than at the day spa. However, the mean sales at thehaircutting salon are much less variable than at the day spa. This would result in agreater number of weeks with higher sales for the haircutting salon, and as a result, itwould be a better purchase (all other things being equal).

3. The mean age is 26.05, and the median age is 20.5. This is as expected. Because thedistribution of ages is skewed to the right, the mean is greater than the median. Thereare several modes in the data set: 8, 9, 12, 20. Clearly, the three lower modes are notgood indications of central tendency in this data set. The standard deviation is 17.1.Calculation of the interquartile range (manual method) is as follows: The location ofQ1is 5.25, and its value is 12. The location of Q3is 15.75, and its value is 38, so theIQR is 26.

4. Because the Pearson r is higher for Don's data set, the correlation between test marksand calories consumed (for Don) will be higher than the correlation between test



5/14


marks and hours spent studying (for Jane). However, there is no obvious reason whyeating more calories would result in higher test marks. There is a logical connectionbetween hours spent studying and test marks, so this cause and effect relationship isstronger.

5. First, remember that with sample data, we cannot absolutelyproveanything. As well,although the correlation coefficient is low, this does not mean that there is norelationship between incomes and purchases. As discussed in the answer to DevelopYour Skills Exercise 23 in Chapter 2, the lack of relationship between an individualpurchase and annual income does not preclude the existence of a relationshipbetween annual purchases and annual income.

6. Both histograms showed some right-skewness, particularly the purchases by females.However, the mean and median purchases for both groups are similar. The meanpurchase by females is $30.86, and the median purchase is $29.50. The meanpurchase for males is $28.90, and the median purchase is $28.38. Because the means

and medians are so close, we will use the mean as the measure of central tendency.On average, the purchases of males are slightly higher than the purchases of females.

Because we used the mean for the measure of central tendency, we will use thestandard deviation as the measure of variability. The standard deviation for purchasesby females is $10.97, while the standard deviation for purchases by males is $5.39.As we saw in the histograms we created in Chapter 2 (Chapter Review Exercise 6),there is less variability in purchases by males than purchases by females.

7. Because a histogram of the data is symmetric and bell-shaped, we can apply theEmpirical Rule. If the sample is representative of the population, then we can expectthat about 68% of the data lie within one standard deviation of the mean, that is,between 170 cm and 184.4 cm, with 32% divided between the two tails of thedistribution. This means that about 16% of young men aged 18-24 would be shorterthan 170 cm. Almost all of the heights would be within three standard deviations ofthe mean, that is, between 155.6 cm and 198.8 cm. Therefore, there would not bemany young men aged 18-24 who were taller than 199 cm.

8. This is a small data set, and so it is not possible to create a histogram to assess theshape of each locations sales distribution. However, if you order the two data sets,in both cases, there are more observations on the high end of the range thanelsewhere, suggesting some skewness. Therefore the interquartile range is probablythe best measure of variability.

If you do the calculations by hand, the results are as follows.Both data sets have 7 data points.

Q1location is the 0.25(n+1) = 0.25(8) = 2nd

placeQ3location is the 0.75(8) = 6

thplace



6/14


Red Deer Vernon

Q1 109.55 112.30

Q3 122.48 122.01

IQR 12.93 9.71

The Red Deer locations sales are more variable than the Vernon locations salesover the period.

If you do the calculations with Excel, the numerical results are different but theconclusion is the same.

Red Deer Vernon

Q1 112.23 114.795

Q3 122.42 121.675

IQR 10.19 6.88

9. The mean price of the inkjet printer cartridges is $26.93. The median price is $25.95.

10. The mean weight of the honey in the jars is 497.3 grams. While this is below 500grams, it is not much below. Without knowing more about the variability of theweights of honey in the jars, we cannot make a conclusion about whether the jars arebeing consistently underfilled. When you master the techniques of Chapter 7, youwill be able to decide.



7/14


11. This data set is reasonably symmetric and bell-shaped. The mean weekly sales forthe sample of stores trying out the new marketing approach are $5101.07, with astandard deviation of $325.60. Applying the Empirical Rule, almost all of the saleswould be between $4124.28 and $6077.85.

0

1

2

3

4

5

Nu

mbero

fStores

WeeklySales

WeeklySalesforStoreswithNewMarketingApproach

12. First we must assess the shape of the distribution. The histogram below shows a

reasonably symmetric and bell-shaped data set.

0

2

4

6

8

10

12

Nu

mbero

fEmp

loy

ees

DaysOff

AnnualDaysOff(OtherThanVacation)foraRandomSampleofEmployees



8/14


The mean days off is 6.04, with a standard deviation of 1.29. The sample mean isbelow the average days off in the past, so there is a reason to hope that there has beenan improvement. However, we need to do a formal hypothesis test (covered inChapter 7) to determine whether there is sufficient evidence to conclude that averagedays off among all employees has actually decreased. As well, even if we conclude

that there has been a decrease in average days off, we cannot necessarily concludethat the wellness program is the cause. Other factors may account for the difference,such as a change in the workforce.

13. Since there are many tied values, it is a bit of a challenge to do the ranking process.The results are as follows.

Ratings by Customers Rank Ratings by Bosses Rank

2 4 3 7.5

2 4 4 9.5

3 8 2 4.54 10 1 1.5

3 8 2 4.5

3 8 2 4.5

2 4 4 9.5

2 4 1 1.5

1 1 3 7.5

2 4 2 4.5

The Spearman rank correlation coefficient is -0.574. This indicates a negativecorrelation between the ratings by customers and the ratings by bosses, that is, theratings tend to be higher by customers when the ratings by bosses are lower.However, the correlation coefficient indicates that this relationship is weak.

14. Since the data sets are fairly symmetric, the mean and the standard deviation are theappropriate measures of central tendency and variability. The results are shownbelow.

Location 1 Location 2

Mean 108.4 124.4

Standard Deviation 17.3 29.6

Mean daily pedestrian traffic is higher at Location 2, at 124.4, compared with 108.4at Location 1. The daily pedestrian traffic at Location 2 (standard deviation is 29.6)is also more variable than at Location 1 (standard deviation of 17.3).



9/14


15. Because the data are quantitative and appear to be linearly related, the Pearson r isthe appropriate measure of association. The Pearson r is 0.958, indicating a highcorrelation between the mark in Business Math and in Statistics.

16. Because the data are quantitative and appear to be linearly related, the Pearson r is

the appropriate measure of association. The Pearson r is 0.941, indicating a highcorrelation between annual advertising expenditure and annual sales.

17. The customer incomes are skewed to the right, with a few incomes much higher thanthe rest in the data set. Therefore, the median and the interquartile range are theappropriate measures. The median income of the drugstore customers is $44,925.The interquartile range is 15,050 (Excel) or 15,762.5 (by hand).



10/14


18. Before we can decide on appropriate numerical measures to compare the data sets,we must examine the shapes of the distributions, by creating histograms, as shownbelow. Note that these histograms are not appropriate for comparison of thedistributionsthey are just for deciding on the appropriate measures.

0

10

20

30

40

50

60

NumberofClients

RRSPHolding

Kate'sClients'RRSPHoldings

0

5

10

15

20

25

30

35

40

NumberofClie

nts

RRSPHolding

Wally'sClients'RRSPHoldings



11/14


Since both data sets are reasonably symmetric, we can use the mean and the standarddeviation to compare them. The results are shown in the table below.

Kates Clients

RRSP Holdings

Wallys Clients

RRSP HoldingsMean $111,021.86 $101,092.89

Standard Deviation $ 32,050.79 $ 40,192.47

The mean holdings of Kates clients RRSPs are higher, at $111,021.86, than themean holdings of Wallys clients RRSPs, at $101,092.89. The variability of theRRSP holdings of Kates clients is less than for Wallys clients (standard deviationof $32,050.79, compared with $40,192.47).

19. First, we must examine the shape of the distribution. A histogram shows a

reasonably symmetric data set (see below).

0

2

4

6

8

10

12

14

Num

berofCans

ContentsinMillilitres

ContentsofaSampleofSoupCans

The mean measurement is 540.4 mL, with a standard deviation of 5.17 mL. Themaximum measurement in the sample is 551 mL, and this does not give any causefor concern that the cans contain more than 556 mL. As well, if we apply the

Empirical Rule, we note that almost all of the measurements would be between 524.9mL and 555.9 mL, which again does not give any cause for concern that the canscontain more than 556 mL. A measurement of 530 mL is about two standarddeviations below the mean. The Empirical Rule says that about 95% of the data willlie within two standard deviations of the mean, with the remaining 5% split betweenthe two tails of the distribution. If this can be applied to the population data, thenabout 2 % of the cans would contain less than 530 mL.



12/14


20. Once again, we must check the distribution of the data set to see if the EmpiricalRule applies.

0

2

4

6

8

10

12

NumberofCans

ContentsinMillilitres

ContentsofaSampleofSoupCans

Since the distribution is approximately bell-shaped and symmetric, we can apply theEmpirical Rule.

The mean measurement is 543.63 mL, with a standard deviation of 6.44 mL. Thereis one can of soup in the sample that contains more than 556 mL. Applying theEmpirical Rule, we note that 95% of the soup cans would contain between 530.7 mLand 556.5 mL. This leaves 2% of the soup cans with more than 556 mL, and 2%

of the soup cans with less than 530 mL.

21. Once again, the answers will depend on the most up-to-date data available when youare answering this question.

(As a guide, the retail sector data that matches the data in the text for themanufacturing sector are discussed. Because of revisions, the more recent data setsmay not exactly match these data. For example, the retail series was significantlychanged between the time of the original download in May of 2009, and asubsequent download in November 2009.

It is a challenge to come up with a class width for comparison for the two sectors,since quarterly operating profits are much smaller for the retail sector than for themanufacturing sector. The compromise choice of $1 billion is really too narrow forthe manufacturing data, and not wide enough for the retail sector data. However,these histograms give a starting point for the analysis.)



13/14


0

5

10

15

20

25

30

35

40

45

NumberofQuarters

MillionsofDollars

QuarterlyOperatingProfitsCanadianRetailSector,I1988toIII2008

0

2

4

6

8

10

12

Num

berofQuarters

MillionsofDollars

QuarterlyOperatingProfitsCanadianManufacturing,I1988toIII2008

The histograms show that quarterly operating profits are smaller for the Canadianretail sector than for the manufacturing sector. For over half the period, quarterly

operating profits for the retail sector were < $2 billion. In over 60% of the quarters inthe period under study, the quarterly operating profits of the Canadian manufacturingsector were $8 billion or more.

The distributions of quarterly operating profits also differ. The distribution for theretail sector profits is skewed to the right, with profits above $3 billion in a fewquarters. The distribution for the manufacturing sector profits is skewed to the left,with a few quarters where operating profits were unusually low (below $5 billion).



14/14



Median quarterly operating profits for the manufacturing sector were $8.909 billion,much greater than the median quarterly operating profits for the retail sector, at$1.827 billion. Quarterly operating profits for the manufacturing sector were muchmore variable over the period, with an interquartile range of $4.648 billion,

compared with only $1.1 billion for the retail sector.

There appears to be some slight positive correlation between the quarterly operatingprofits of the two sectors, but it is not strong. The scatter diagram below illustrates.

$0

$2,000

$4,000

$6,000

$8,000

$10,000

$12,000

$14,000

$16,000

$0 $1,000 $2,000 $3,000 $4,000 $5,000Qu

arterly

OperatingPro

fitso

ftheM

anu

facturing

Sector

($M

illions)

QuarterlyOperatingProfitsoftheRetailSector($Millions)

QuarterlyOperatingProfitsforTwoCanadianSectors,I1988toIII2008

The Pearson r is 0.42, confirming the impression from the scatter diagram of a weakpositive relationship. When quarterly operating profits of the retail sector are higher,the quarterly operating profits of the manufacturing sector tend to be higher, but thecorrelation is weak.

(Your comparison, with more up-to-date data, should contain all of the elementsshown in this answer.)

ch03-solns-all_skuce_2e

Documents

Transcript of ch03-solns-all_skuce_2e