ch08-solns-all_skuce_2e

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Instructors Solutions Manual - Chapter 8

Chapter 8 Solutions

Develop Your Skills 8.1

1. First, check conditions.Sampling is done without replacement. We have no way to know if 300 smokers is

5% of the total number of construction workers. We proceed by noting this, andthat the estimate will not be correct if there are not at least 6,000 workers in total.

np = 56

n q = 244

Both are 10, so we can proceed.

)2446.0,1287.0(

)02249609.0(576.21866667.0

300

300

244

300

56

576.2300

56

n

qp)scorezcritical(p

A 99% confidence interval estimate for the proportion of smokers is (0.1287,0.2446).

The Excel template confirms this, with slightly more accurate results.

ConfidenceIntervalEstimateforthePopulationProportionConfidenceLevel(decimalform) 0.99

SampleProportion 0.18667

SampleSizen 300

nphat 56

nqhat 244

Arenphat andnqhat>=10? yes

UpperConfidenceLimit 0.24461

LowerConfidenceLimit 0.12872

Copyright 2011 Pearson Canada Inc. 161


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2. First, check conditions.Sampling is done without replacement. We have no way to know if 200 employeesis 5% of the total number of staff. We proceed by noting this, and that the estimatewill not be correct if there are not at least 4,000 workers in total.

np = 142n q = 58


)7729.0,6471.0()032085822.0(96.171.0

200

200

58

200

142

96.1200

142

n

qp)scorezcritical(p

A 95% confidence interval estimate for the proportion of employees who havechildren of daycare age is (0.6471, 0.7729).




SampleSizen 200

nphat 142

nqhat 58

Arenphat andnqhat>=10? yes





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3. First, check conditions.Sampling is done without replacement. However, the sample size of 1000 iscertainly no more than 5% of all Canadians.

np = 1000(0.48) = 480

n q = 1000(1-0.48) = 520 Both are 10, so we can proceed.

)5060.0,4540.0(

)015798734.0(645.14.0

1000

52.0)48.0(645.148.0

)(

n

qpscorezcriticalp

A 90% confidence interval estimate for the proportion of all Canadians who do notfeel knowledgeable about such television features is (0.4540, 0.5060).




SampleSizen 1

nphat 480

nqhat 520

Arenphatandnqhat>=10? yes



000



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4. First, check conditions.Sampling is done without replacement. The east end of St. John's, Newfoundland,would contain thousands of households, so the sample of 50 is no more than 5% ofthe population.

np = 37n q = 13


)8843.0,5957.0(

)062032.0(326.274.0

50

)26.0)(74.0(326.274.0

)(

n

qpscorezcriticalp

A 98% confidence interval estimate for the proportion of households in the east endof St. John's with high-speed internet access is (0.5957, 0.8843).


ConfidenceIntervalEstimateforthePopulationProportionConfidence

Level

(decimal

form) 0.98


SampleSizen 5

nphat 37

nqhat 13




0



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5. First, record the information given.n = 1,202

= 469/1202 = 0.390183p

confidence level = 19/20 = 0.95The result is accurate to with 2.9 percentage points. This means the half-width of the

confidence interval is 0.029.The 95% confidence interval will be(0. 390183 0.029, 0. 390183 + 0.029)(0.361, 0.419)The polling company has 95% confidence that the interval (36.1%, 41.9%) containsthe percentage of British Columbia residents who think that retailers should providebiodegradable plastic bags to consumers at no charge.


6. We are told the sample data appear approximately normal, so we will assume the

population data are normal.n = 25x = $112.36s = $32.45For 95% confidence, we need to identify t.025for 24 degrees of freedom.t.025= 2.064, from the tables

)76.125,96.98(

39536.1336.112

25

45.32)064.2(36.112

)(

n

sscoretcriticalx

A 95% confidence interval estimate for the average grocery bill of all householdswho shop at this store is ($98.96, $125.76).



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The Excel template confirms this.

ConfidenceIntervalEstimateforthePopulationMeanDothesampledataappeartobe

normallydistributed? yes

ConfidenceLevel(decimalform) 0.95

SampleMean 112.36

SampleStandardDeviations 32.45

SampleSizen 2



5

7. Because the sample data set is clearly non-normal and highly skewed to the right, thenecessary conditions are not met, and we cannot construct a confidence interval.

8. We are told to assume the daycare costs are normally distributed.n = 50x = $460s = $65For 98% confidence, we need to identify t.010for 49 degrees of freedom.t.010= 2.403 for 50 degrees of freedom, the closest we could get from the tables.

)09.482,91.437(

08930874.22460

50

65)403.2(460

)(

nsscoretcriticalx

A 98% confidence interval estimate for the average monthly daycare costs forHalifax households is ($437.94, $482.09).



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The Excel template confirms this. The numbers are slightly different because theExcel template is more accurate.




SampleMean 460

SampleStandardDeviations 6

SampleSizen 5



5

0

9. First, check the sample data for normality. One possible histogram is shown below.

0

1

2

3

4

5

6

NumberofMarks

Mark(%)

RandomSampleofMarksona

StatisticsTest

Since the histogram is fairly normal, we proceed.



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A completed Excel template is shown below.

ConfidenceIntervalEstimateforthePopulationMeanDo

the

sample

data

appear

to

be



SampleMean 59.8


SampleSizen 2



0

A 95% confidence interval estimate for the average grade on the statistics test is(52.5, 67.1).

10. We are told to assume the salary data are approximately are normally distributed.n = 50x = $56,387s = $5,435For 90% confidence, we need to identify t.050for 49 degrees of freedom.t.050= 1.676 for 50 degrees of freedom, the closest we could get from the tables.

)22.57675,78.55098(

215619.128856387

50

5435)676.1(56387

)(

n

sscoretcriticalx

A 90% confidence interval estimate for the average salary of university graduateswith a bachelors degree in science, and 10 years of working experience, in theToronto area is ($55,098.78, $57,675.22).



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The Excel template confirms this. The numbers are slightly different because theExcel template is more accurate.




SampleMean 56387

SampleStandardDeviations 5435

SampleSizen 5



0


11. confidence level is 99%, so z-score = 2.576HW = 0.05

= 0.1867p

Substitute these values into the formula:

03.403

)31.2654)(15184311.0(

05.0576.2)1867.01)(1867.0(

2

2

n

HW

scorezqpn

A sample size of 404 is required to estimate the proportion of smokers on staff towithin 5%, with 99% confidence.



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12. Since no estimate of the proportion is available, we will use = 0.50.p

For 95% confidence, the z-score is 1.96.HW = 0.05

16.384

)64.1536)(25.0(

05.0

96.1)5.0)(5.0(

2

2

n

HW

scorez

qpn

A sample size of 385 is necessary to estimate the proportion of employees who havechildren of daycare age to within 5%, with 95% confidence.For 98% confidence, the z-score is 2.326.

03.541

)1104.2164)(25.0(

05.0

326.2)5.0)(5.0(

2

2

n

HW

scorezqpn

The sample size would have to increase to 542 to estimate the proportion ofemployees who have children of daycare age to within 5%, with 98% confidence.

13. For 95% confidence, the z-score is 1.96.HW = $10s = $32.45

5.40

10

)45.32)(96.1(

))((

2

2

HW

sscorezn

A sample size of 41 is necessary to estimate the average grocery bill of householdswho shop at this supermarket to within $10, with 95% confidence.



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14. For 95% confidence, the z-score is 1.96.HW = 5s = 15.54

1.37

5

)54.15)(96.1(

))((

2

2

HW

sscorezn

A sample size of 38 is necessary to estimate the average mark on the stats test towithin 5 marks, with 95% confidence.

15. We will use an estimate of of 0.5, since no other information is given. For 96%

confidence, the z-score is 2.05.

p

HW = 0.04

64.656

)5625.2626)(25.0(

04.0

05.2)5.01)(5.0(

2

2

n

HW

scorezqpn

A sample size of 657 would have to be taken to estimate the percentage of Canadian

internet users who visit social networking sites, to within 4%, with 96% confidence.


16. A 95% confidence interval estimate for the average grocery bill of all householdswho shop at this store is ($98.96, $125.76). Since this does not contain $95, we canreject the owners claim that the average household grocery bill is $95, with a 5%level of significance.

17. A 99% confidence interval estimate for the proportion of smokers is (0.1287,0.2446). Since this contains 20%, we do not have sufficient evidence to reject thenurses claim that 20% of the staff are smokers. A 1% level of significance applies.

18. A 99% confidence interval estimate for the average weekly spending of members ofthe college community on morning coffees is ($3.24, $5.76). Since this does notcontain $6, there is sufficient evidence to reject the managers claim that the averageweekly spending of members of the college community on morning coffees is $6,with a 1% level of significance.



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19. A 95% confidence interval estimate for the average grade on the statistics test is(52.5, 67.1). Since this interval does not contain 50, there is sufficient evidence toreject your belief that the average mark on the test was 50, with a 5% level ofsignificance.

Chapter Review ExercisesThe solutions are shown as calculations done by hand with tables and a calculator, whensummary data are provided. The use of Excel tools and templates is shown when anExcel data set is provided.

1. n = 400= 0.26p

First, check conditions.Sampling is done without replacement. The population is all Canadian groceryshoppers, so the sample of 400 is definitely 5% of the population.

np = 400(0.26) = 104

n q = 400(0.74) = 296


a.

)2961.0,2239.0(

)021932.0(645.126.0

400

)74.0)(26.0(645.126.0

)(

n

qpscorezcriticalp

A 90% confidence interval estimate for the proportion of all Canadian groceryshoppers who are trying to make healthier choices is (0.2239, 0.2961).

b.

)3050.0,2150.0(

)021932.0(05.226.0

)(

n

qpscorezcriticalp




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c.

)3165.0,2035.0(

)021932.0(576.226.0

)(

n

qpscorezcriticalp


d. As the desired level of confidence increases, the confidence interval gets wider.


= 0.26p


2.821

03.0

96.1)74)(.26.0(

2

2

n

HW

scorezqpn

A sample size of 822 is required to estimate the proportion of Canadian groceryshoppers who are trying to make healthier food choices, to within 3%, with 95%confidence.

If there was no sample information available about the proportion of Canadian

grocery shoppers, we would have to use = 0.5.p

1.1067

03.0

96.1)5)(.5.0(

2

2

n

HW

scorezqpn

The required sample size would be 1068, which is larger, and would be more

expensive than a sample 822. Having some estimate of the proportion will likelyreduce the costs of sampling.



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3. There is no information about whether the sample data are normally distributed. Wewill proceed by assuming that they are, with the caution that our results are notreliable if in fact they are not.n = 212x = 1576

a. s = 521For 95% confidence, we need to identify t.025for 520 degrees of freedom.t.025= 1.972 for 200 degrees of freedom, the closest we could get from the tables.

)1647,1505(

)7824.35(972.11576

212

521)972.1(1576

)(

n

sscoretcriticalx

A 95% confidence interval estimate for the average number of web pages visited permonth by Canadian internet users is (1505, 1647).

b. s = 321For 95% confidence, we need to identify t.025for 520 degrees of freedom.t.025= 1.972 for 200 degrees of freedom, the closest we could get from the tables.

)16197,1533(

)0464.22(972.11576

212

321)972.1(1576

)(

n

sscoretcriticalx


c. s = 201For 95% confidence, we need to identify t.025for 520 degrees of freedom.t.025= 1.972 for 200 degrees of freedom, the closest we could get from the tables.

)1603,1549(

)8047.13(972.11576

212

201)972.1(1576

)(

n

sscoretcriticalx




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d. As the variability in the data decreases, the confidence intervals become narrower.They do not have to be so wide, because the distributions are not so wide.

4

a. For 95% confidence, the z-score is 1.96.HW = 100s = 521

3.104

100

)521)(96.1(

))((

2

2

HW

sscorezn

A sample size of 105 is necessary to estimate the average number of web pages

visited per month by Canadian internet users to within 100 pages, with 95%confidence.

b. For 95% confidence, the z-score is 1.96.HW = 50s = 521

1.417

50

)521)(96.1(

))((

2

2

HW

sscorezn

A sample size of 418 is necessary to estimate the average number of web pagesvisited per month by Canadian internet users to within 50 pages, with 95%confidence.



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c. For 95% confidence, the z-score is 1.96.HW = 10s = 521

7.10427

10

)521)(96.1(

))((

2

2

HW

sscorezn

A sample size of 10,428 is necessary to estimate the average number of web pagesvisited per month by Canadian internet users to within 10 pages, with 95%confidence.

d. As the desired level of accuracy increases, the sample size required increases.

5. We have no estimate for , so we will use = 0.5.p p confidence level is 90%, so z-score = 1.645

HW = 0.02Substitute these values into the formula:

3.1691

)0625.6765)(25.0(

02.0

645.1)5.01)(5.0(n

HW

scorezqpn

2

2

A sample size of 1692 is required to estimate the proportion of new graduates of aBusiness program who are willing to relocate to find a job. Since your collegegraduates only about 350 students from the Business program, this indicates that theentire population should be surveyed for the desired level of accuracy.



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6. Sampling is done without replacement, but there are millions of adults in the greaterToronto are, so we can be sure that the sample of 316 is 5% of the total population.

np = 316(0.40) = 126.4

n q = 316(1-0.40) = 189.6


)4453.0,3547.0(

)027558913.0(645.14.0

316

)6.0)(4.0(645.140.0

)(

n

qpscorezcriticalp

A 90% confidence interval estimate for the proportion of adults in the greater

Toronto are who would keep their jobs if they won $10-million in the lottery is(0.3547, 0.4453).

7. We are told the sample data appear normally distributed, so we will assume thepopulation data are as well.n = 30x = 54.2s = 3.2For 99% confidence, we need to identify t.005for 29 degrees of freedom.t.005= 2.756.

)81.55,59.52(

610158.12.54

30

2.3)756.2(2.54

)(

nsscoretcriticalx

A 99% confidence interval estimate for average hours of work per week for theseemployees is (52.59, 55.81 hours).



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8. For 95% confidence, the z-score is 1.96.HW = 1There is no estimate of s. However, the range is (10 1) = 9, so we will use 9/4 asan estimate of s.

s 9/4 = 2.25

4.19

1

)25.2)(96.1(

HW

)s)(scorez(n

2

2

A sample size of 20 is necessary to estimate the average number of hours, per week,that students in statistics classes spend (outside class) working on statistics, to within1 hour, with 95% confidence.

The results of any such survey should be interpreted with care, because it would beimpossible to get the data independent of the students themselves. It is not unlikelythat students would inflate their estimates of the amount of work they do, particularlyif they wanted to impress their professor favourably.


= 0.10p


36.864

)9604)(09.0(

02.0

96.1)1.01)(1.0(

2

2

n

HW

scorezqpn

A sample size of 865 is required to estimate the percentage of the adult population inCanada who would consider buying a hybrid vehicle for their next purchase, towithin 2%, with 95% confidence.



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no estimate of p is given, so we will use = 0.5p


11.1067

)444444.4268)(25.0(

03.0

96.1)5.01)(5.0(

2

2

n

HW

scorez

qpn

A sample size of 1068 is required to estimate the percentage of households that makeconsistent efforts to separate recyclable materials from their garbage, to within 3percentage points, with 95% confidence.

11. Sampling is done without replacement. The sample size is 2450, which is quitelarge. If the total population is all workers, though, there would be millions, and sothe sample would be 5% of the population.

np = 2450(0.43) = 1053.5

n q = 2450(1 0.43) = 1396.5

Both are 10, so we can proceed.For a 99% confidence level, the z-score is 2.576.

)4558.0,4042.0(

)010002041.0(576.243.0

2450

)57.0)(43.0(

576.243.0

)(

n

qpscorezcriticalp

A 99% confidence interval estimate for the proportion of workers who phone in sickwhen they are not ill is (0.4042, 0.4558). It is hard to assess the reliability of theseresults. Would people tell the truth when they were asked such a question? Thereare a number of reasons why they might not.



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12. We are told the sample data appear approximately normal, so we will assume thepopulation data are normal.n = 40x = $68.52s = $14.89

For 98% confidence, we need to identify t.010for 39 degrees of freedom. There is norow in the t-table for 39 degrees of freedom. We will use the row for 40 degrees offreedom, and we see that t.010= 2.423.

)22.74$,82.62($

704506985.552.68

40

89.14)423.2(52.68

n

s)scoretcritical(x

A 98% confidence interval estimate for of the average amount spent (per person) inthis restaurant by diners with business expense accounts is ($62.82, $74.22).

13. We are told the sample data appear approximately normal, so we will assume thepopulation data are normal.n = 40x = $543.21s = $47.89For 95% confidence, we need to identify t.025for 39 degrees of freedom. Since thereis no row in the table for 39 degrees of freedom, we will use the row for 40 degreesof freedom. We see that t.025= 2.021.

)51.558$,91.527($

30316127.1521.543

40

89.47)021.2(21.543

n

s)scoretcritical(x

A 95% confidence interval estimate for the average monthly rent for students at thiscollege is ($527.91, $558.51). Because this interval does not contain $500, there is

sufficient evidence to reject the claim that the average monthly rent is $500, with 5%significance.



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= 0.35p


6.1367

)417778.6011)(2275.0(

03.0

326.2)35.01)(35.0(n

HW

scorez

qpn

2

2

A sample size of 1368 is required estimate the proportion of this colleges studentswho live at home with their parents.

15. First we have to organize the data. We can use Excels Histogram tool to organize

the data, and then produce the following table.

Customer Survey for an Ice Cream StoreWhich flavour would you like to try, if any?

Response Frequency Relative Frequency

Pecan and Fudge 37 24.7%

Apple Pie 16 10.7%

Banana Caramel Ripple 29 19.3%

Ginger and Honey 32 21.3%

Would Not Try Any Of These Flavours 36 24.0%

Total 150

From this we see = 0.24.p

Sampling is done without replacement. We do not know the total number ofcustomers at the ice cream store. As long as the sample of 150 is not more than 5%of this total number, we can proceed.



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SampleSize

n 1

nphat 29

nqhat 121

Arenphat andnqhat>=10?



50

A 95% confidence interval estimate for the percentage of customers who would not

try any of the new flavours is (0.1301, 0.2565).



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16. A histogram of the sample data appears approximately normal.

0

2

4

6

8

10

12

NumerofTubes

mL

QuantityofToothpaste inaRandom

Sample

of

30

Tubes




SampleMean 129.5


SampleSizen 3



165

0

A 95% confidence interval for the average amount of toothpaste in the tubes is(128.64, 130.37).



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17. The data appear to be normally distributed.

0

2

4

6

8

10

12

14

16

18

NumberofDays

NumberofCustomers

No.ofCustomersRentingaCarinthe8a.m.

10a.m.PeriodataCarRentalAgency




Sample

Mean

21.48SampleStandardDeviations 3.56422

SampleSizen 5



0

A 99% confidence interval estimate for the average number of customers renting acar in the 8 a.m.-10 a.m. period at this car rental agency is (20.13, 22.83). Of course,it is not possible for there to be 20.13 customers. This number arises because the

underlying data are not continuous (they are counts). However, it is not unusual tocreate a confidence interval estimate for such data sets. Realistically, the confidenceinterval would be (20, 23).



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18. The sample data are approximately normally distributed, although somewhat skewedto the right. In this case, with a fairly large sample size of 60, we will rely on therobustness of the t-distribution and proceed.

0

2

4

6

8

10

12

14

16

18

20

NumberofCars

AnnualMaintenanceCosts

AnnualCar

Maintenance

Costs,

Third

Year

of

LifeofEntrylevelCompact

A completed Excel template for the data set is shown below.




SampleMean 138.556


SampleSizen 6


LowerConfidence

Limit 116.990

0

A 98% confidence interval estimate for the annual maintenance costs of this entry-level compact in the 3rdyear of its life is ($116.99, $160.12).



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19. For 98% confidence, the z-score is 2.326.HW = $10s = 69.861423

05.264

10

)861423.69)(326.2(

HW

)s)(scorez(n

2

2

A sample size of 265 is necessary to estimate the annual maintenance costs of thisentry-level compact in the 3rdyear of its life to within $10, with 98% confidence.

20. First we must analyze the data. We can use Excels Histogram tool to organize thedata, and then create the table shown below.

Age of A/R Frequency Relative Frequency

0-30 Days Old 47 0.47

31-60 Days Old 41 0.41

More Than 60 Days Old 12 0.12

Total 100

From the table we see = 0.47.p

We are sampling without replacement. The company is described as large. We

must assume that the sample of 100 is not more than 5% of the population, and notethat our analysis might not be correct otherwise.

np = 47

n q = 53


A completed Excel template is shown below.



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SampleSizen 1

nphat 47

nqhat 53




00

A 95% confidence interval estimate for the proportion of accounts receivable that are

0-30 days old is (0.3722, 0.5678).


= 0.47p


77.382

)64.1536)(2491.0(

05.0

96.1)47.01)(47.0(n

HW

scorezqpn

2

2

A sample size of 383 is required to estimate the proportion of accounts receivablethat are 0-30 days old, to within 5%, with 95% confidence.



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22. We have already examined this data set Chapter 7 Review Exercise 24.

a. Sampling is done without replacement. We are considering these students as arandom sample of all students entering Business programs in Canadian colleges, sothe sample is 5% of the population.

First, organize the data. "0" means "no", and "1" means "yes.

Bin Fre que ncy

0 371

1 139

Total 510

The completed Excel template is shown below.



SampleSizen 510

nphat 139

nqhat 371




We see that np and nq > 10, so conditions for normality of the sampling distribution

are met.

A 96% confidence interval estimate of the proportion of incoming students who havelaptops is (0.23, 0.31).



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b. confidence level is 96%, so z-score = 2.05HW = 0.02

= 0.27255p


03.2083

)25.10506)(198266.0(

02.0

05.2)27255.01)(27255.0(

2

2

n

HW

scorez

qpn

A sample size of 2084 is required estimate the proportion of incoming students whohave laptops to within 2%, with 96% confidence.

c. First, analyze the data.

WhereDoYouLive? Frequency RelativeFrequencyWithParentsorRelatives 164 32.16%

InResidence 130 25.49%

RentalHouseorApartment 113 22.16%

OwnHome 27 5.29%

RentRoom 76 14.90%

Total 510

Sampling is done without replacement. We are considering these students as arandom sample of all students entering Business programs in Canadian colleges, sothe sample is 5% of the population.



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ConfidenceIntervalEstimateforthePopulationProportionConfidence

Level

(decimal

form) 0.96


SampleSizen 510

nphat 76

nqhat 434

Arenphat and nqhat >=10? yes


LowerConfidence

Limit 0.11663

We see from the template that np and nq > 10.

A 96% confidence interval estimate of the proportion of incoming students who rentrooms is (0.1166, 0.1814).



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d. First, analyze the data and create a histogram to check for normality.

0

20

40

60

80

100

120

NumberofStudents

Mark

MarksofIncomingStudents

The histogram is fairly normal.

The completed Excel template for the confidence interval is shown below.




SampleMean 85.1922


SampleSizen 510.00



A 96% confidence interval estimate of the average mark of incoming students is(84.5, 85.9).



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e. First, analyze the data and create a histogram.

0

10

20

30

40

50

60

70

80

90

100

NumberofStudents

AmountofSavingsAvailable

IncomingStudents,AmountofSavingsAvailablefor

Education

The histogram is fairly normal, and the sample size is quite large.





SampleMean 6427.25


SampleSizen 510.00



A 96% confidence interval estimate of the amount of savings available for educationamong incoming students is ($6,133.02, $6,721.49).



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f. For 96% confidence, the z-score is 2.05.HW = $200s = 3227.1172

15.1094

200

)1172.3227)(05.2(

))((

2

2

HW

sscorezn

A sample size of 1,095 is necessary to estimate the amount of savings available foreducation to within $200, with 96% confidence.

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