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Instructors Solutions Manual - Chapter 12

Chapter 12 Solutions

Develop Your Skills 12.11. We are looking for evidence of a decrease in the proportion of on-time flights after

the merger. Call the population of flights before the merger population 1, and the

population of flights after the merger population 2.

H0: p1 p2= 0

H1: p1 p2> 0

= 0.04

85.0100

85p1 , n1= 100, 78.0

100

78p 2 , n2= 100

Sampling is done without replacement, but it is likely that the airline handles many

thousands of flights, so we can still use the binomial distribution as the appropriateunderlying model.

Check for normality of the sampling distribution:

=100(0.85) = 85 > 1011pn

=100(1-0.85) = 100(0.15) = 15 > 1011qn

=100(0.78) = 78 > 1022pn

=100(1-0.78) = 100(0.22) = 22 > 1022 qn

Since the null hypothesis is that there is no difference in the proportions, we can pool

the sample data to estimate .p

815.0100100

7885p

We calculate the z-score as:

2747.1054913568.0

07.0

100

1

100

1)815.01)(815.0(

0)78.085.0(

n

1

n

1qp

0)pp(z

21

21

p-value = P(z 1.27) = 1 0.8980 = 0.102

Since p-value > , fail to reject H0. There is insufficient evidence to infer that theproportion of on-time flights decreased after the merger.

Copyright 2011 Pearson Canada Inc. 321


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2. Call the data on use of social network profiles by online Canadians in 2009 sample 1

from population 1, and the data on use of social network profiles by onlineCanadians 18 months previously sample 2 from population 2.

H0: p1 p2= 0.10

H1: p1 p2> 0.10

= 0.05

5607.0824

462

1 p , n1= 824, 39.02 p , n2= 800

Sampling is done without replacement, but there are millions of online Canadians, sowe can still use the binomial distribution as the appropriate underlying model.


= 462 > 1011pn = 824 - 462 = 362 > 1011qn

=800(0.39) = 312 > 1022pn

=800(1-0.39) = 800(0.61) = 488 > 1022 qn

Since the null hypothesis is that there is a 10% difference in the proportions, we

cannot pool the sample data to estimate .p


89.2

800

)61.0)(39.0(

824

)4393.0)(5607.0(

10.0)39.05607.0(

)()(

2

22

1

11

21

21 21

21

21

n

qp

n

qp

pp

s

ppz

pp

pp

pp

p-value = P(z 2.89) = 1 0.9981 = 0.0019

Since p-value < , reject H0. There is sufficient evidence to infer that the proportion

of online Canadians with a social network profile is more than 10% higher in 2009than it was 18 months previous.



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3. Call the data on perceptions of female bank employees sample 1 from population 1,

and the data on perceptions of male bank employees sample 2 from population 2.We want to know if the proportion of male employees who felt that female

employees had as much opportunity for advancement as male employees is more

than 10% higher than the proportion of female employees who thought so. So, we

are wondering if p2is more than 10% higher than p1, that is, if p2 p1> 0.10.Rewriting this in standard format, we ask the equivalent question: is p1p2< -0.10?

H0: p1 p2= -0.10H1: p1 p2< -0.10

= 0.05

, n1= 240, , n2= 350825.01p 943.0 2 p

Sampling is done without replacement, but Canadian banks are large employers (in

2006, the Royal Bank employed about 69,000 people, for instance), so we can stilluse the binomial distribution as the appropriate underlying model.


=240(0.825) = 198 > 1011pn

=240(1-0.825) = 42 > 1011qn

=350(0.943) = 330.05 > 1022pn

=350(1-0.943) =19.95 > 1022 qn

The null hypothesis is that there is 10% difference in the proportions, so we cannot

pool the sample data to estimate .p


1 2 1 2

1 1 2 2

1 2

( ) (0.825 0.943) ( 0.10)0.655

(0.825)(0.175) (0.943)(0.057)

240 350

p pp p

zp q p q

n n

p-value = P(z -0.66) = 0.2546

Since p-value > , fail to reject H0. There is not enough evidence to infer that theproportion of male employees who felt that female employees had as much

opportunity for advancement as male employees was more than 10% higher than the

proportion of female employees who thought so.



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4. Call the data on customers told about the extended warranty by cashiers sample 1

from population 1. Call the data on customers exposed to the display at the checkoutsample 2 from population 2.

H0: p1 p2= 0

H1: p1 p20

= 0.10

We presume the store has many thousands of customers, so although we are

sampling without replacement, we can still use the binomial distribution as the

appropriate underlying model.

The data are available in Excel, so we will use Excel to do this problem. First the

raw data must be organized. The Excel output from the Histogram tool is shownbelow.

Cashier Display

Bin Frequency Bin Frequency

0 122 0 145

1 28 1 55

We can then use the Excel template to proceed. The output is shown below.

MakingDecisionsAboutTwo

PopulationProportions

Sample1Size 150

Sample2Size 200

Sample1Proportion 0.18666667


n1p1hat 28

n1q1hat 122

n2p2hat 55

n2q2hat 145

Arenpandnq>=10? yes

HypothesizedDifference

in

Population

Proportions,p1p2(decimalform) 0

zScore 1.92275784

OneTailedpValue 0.02725523

TwoTailedpValue 0.05451047



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This is a two-tailed test, so the appropriate p-value is 0.0545. Since this is less than, reject H0. There is sufficient evidence to infer there is a difference in theproportion of customers who buy the extended warranty when exposed to promotion

by a display or informed by the cashier.

5. We will use Excel to calculate this confidence interval (it could also be done byhand, based on the information acquired in Exercise 4 above). The Excel template is

shown below.

ConfidenceIntervalEstimatefor

theDifferenceinPopulation

Proportions

ConfidenceLevel(decimalform) 0.9



Sample1Size

150

Sample2Size 200

n1p1hat 28

n1q1hat 122

n2p2hat 55

n2q2hat 145

Arenpand nq>=10? yes

UpperConfidenceLimit 0.0146

LowerConfidence

Limit

0.1621

With 90% confidence, we estimate that the interval (-0.162, -0.015) contains the true

difference in the proportion of customers who buy the extended warranty, when told

about it by the cashier, compared with being exposed to a prominent display at the

checkout. Another way to say this: a greater proportion of those exposed to thedisplay bought the extended warranty. We estimate the difference to be contained in

the interval (1.5%, 16.2%).

This confidence interval corresponds to the hypothesis test in the preceding exercise.

Since we rejected the hypothesis of no difference, we would not expect theconfidence interval to contain zero (and it does not).



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Develop Your Skills 12.26. First, summarize the data, and calculate expected values. See the table below.

Past % Observed Expected

Pay Now 0.26 23 19.5Pay In Six Months 0.37 32 27.75

Pay In One Year 0.37 20 27.75

75 75

Expected values are calculated as follows. In the past, 26% of customers paidimmediately. Out of the 75 customers surveyed, we would expect 26% 75 = 19.5

customers to pay immediately. The other expected values are calculated in a similar

fashion.

H0: The distribution of customers according to method of payment is the same nowas it was in the past.

H1: The distribution of customers according to method of payment is different now,compared to the past.

= 0.05 (given)

All expected values are more than 5, so we can proceed.

444.375.27

)75.2720(

75.27

)75.2732(

5.19

)5.1923(

e

)eo(X

222

i

2

ii2

Degrees of freedom = k 1 = 2.

Using the tables, we see that p-value > 0.100.Using CHITEST, we see that p-value = 0.1787.

Fail to reject H0. There is insufficient evidence to infer that there has been a changein customers preferences for the different payment plans.

The graph below summarizes the changes.

0%

5%

10%

15%

20%

25%

30%35%

40%

45%

PayNow PayinSixMonths PayinOneYear

CustomerPreferencesforPaymentPlansPastCustomerPreferences CurrentCustomerPreferences



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7. H0: The distribution of customers brand preferences is as claimed by the previous

manager.H1: The distribution of customers brand preferences is different from what was

claimed by the previous manager.= 0.05 (given)

Expand the table of claimed brand preferences to show expected and observed

values, as shown below.

Labatt

Blue

Labatt

Blue

Light

Molson

Canadian

Kokanee Rickards

Honey

Brown

Claimed

Preference

33% 8% 25% 19% 15%

Expected

(for 75Customers) 24.75 6.00 18.75 14.25 11.25

Observed 29 6 21 16 13

All expected values are more than 5, so we can proceed.

487.1

25.11

)25.1113(

25.14

)25.1416(

75.18

)75.1821(

00.6

)00.66(

75.24

)75.2429()( 222222

2

i

ii

e

eo

X


Using the tables, we see that p-value > 0.100.

Using CHITEST, we see that p-value = 0.8289

.Fail to reject H0. There is insufficient evidence to infer that the distribution of

customers brand preferences is different from what the previous manager claimed.



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8. H0: The die is fair (the probability of occurrence of each side is 1/6).

H1: The die is not fair.= 0.025 (given)

Expand the table of observations from repeated tosses of a die to show expected and

observed values, as shown below.

1 Spot 2 Spots 3 Spots 4 Spots 5 Spots 6 Spots

Observed 18 24 17 25 16 25

Expected(Out Of 125) 20.83333 20.83333 20.83333 20.83333 20.83333 20.83333

All expected values are more than 5.

Using the formula as before, we calculate X2= 4.36.


From the table, we see p-value > 0.100.

Using CHITEST, we see that p-value = 0.4988.

Fail to reject H0. There is insufficient evidence to infer that the die is not fair.Johns troubles are of his own making. We have no way to know if Mary will ever

forgive him.

9. H0: The distribution of customer destination preferences at the travel agency is the

same as in the past.

H1: The distribution of customer destination preferences at the travel agency is

different from the past.= 0.04 (given)

Summarize the data from the random sample and calculate expected values.

Canada U.S. Caribbean Europe Asia Australia

/NewZealand

Other

Past Preferences 28% 32% 22% 12% 2% 3% 1%

Expected

(for a sample of54) 15.12 17.28 11.88 6.48 1.08 1.62 0.54

Observed 22 14 8 8 0 2 0



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In this case, there are three expected values < 5 (these are highlighted in the table).

It seems logical to combine these categories, and then proceed. The new table ofexpected and observed values is shown below.

Canada U.S. Caribbean Europe OtherPast Preferences 28% 32% 22% 12% 6%

Expected

(for a sample of 54)15.12 17.28 11.88 6.48 3.24

Observed 22 14 8 8 2

However, even this change still leaves us with an expected value < 5. We must

combine categories again. It is less satisfying to combine Europe with Asia,

Australia/New Zealand and Other. However, all of these destinations represent

destinations at a significant distance from the North American continent, so there is

some sense to combining them

The final table of expected and observed values is shown below.

Canada U.S. Caribbean Europe, Asia,

Australia/New

Zealand, Other

Past Preferences 28% 32% 22% 18%

Expected(for a sample of 54)

15.12 17.28 11.88 9.72

Observed 22 14 8 10

Now that all expected values are 5, we can proceed.Using the formula as before, we calculate X

2= 5.028.

Using the tables, with 3 degrees of freedom, we see p-value > 0.100.

Using CHITEST, we see that p-value = 0.1697.Fail to reject H0. There is insufficient evidence to infer that there has been a change

in customer destination preferences at this travel agency.

10. H0: The distribution of responses to the survey at the local branch is the same as the

national benchmarksH1: The distribution of responses to the survey at the local branch is not the same as

the national benchmarks

= 0.025; X2= 20.859; from the tables (4 degree of freedom), p-value < .005Reject H0. There is sufficient evidence to infer that the distribution of responses to

the survey at the local branch differs from the national benchmarks. The graph belowgives some indication of where the differences lie.



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0%

10%

20%

30%

40%

50%

60%

StronglyAgree

Agree Neit herAgreenorDisagree

Disagree StronglyDisagree

ResponsetoaSurveyofFinancialServicesCompanyCustomers:"Thestaffatmylocalbranchcanprovide

me

with

good

advice

on

my

financial

affairs"

NationalBenchmark

ObservedPercentage

Develop Your Skills 12.3

11. H0: There is no relationship between the views on the proposed health benefit

changes and the type of job held in the organization

H1: There is a relationship between the views on the proposed health benefitchanges and the type of job held in the organization

= 0.01

The calculations of expected values for a contingency table can be done manually,

but are somewhat tedious. We will use Excels Non-Parametric Tool for Chi-Squared Expected Value Calculations. The Excel output is shown below.

Chi-Squared Expected Values Calculations

Chi-squared test statistic 16.44338

# of expected values


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The p-value is 0.002478, which is < = 0.01. Reject H0. There is sufficient

evidence to infer that there is a relationship between the views on the proposedhealth benefits changes and the type of job held in the organization.

12. H0: There is no relationship between hair colour and tendency to use sunscreen.

H1: There is a relationship between hair colour and tendency to use sunscreen.= 0.05

The output of the Excel tool for Chi-Squared Expected Value Calculations is shownbelow.




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13. H0: There is no relationship between household income and the section of the paper

read most closely.H1: There is a relationship between household income and the section of the paper

read most closely.= 0.25

The output of the Excel tool for Chi-Squared Expected Value Calculations is shown

below.


Chi-squared teststatistic 51.92698



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All expected values are 5, so we can proceed. The p-value is 0.14647 > = 0.05.

Fail to reject H0. There is insufficient evidence to infer that there are differences inthe proportions of students whose first language is English, French, or something

else for these four schools.

15. H0: The proportions of students drawn from inside or outside the local area are thesame for the Business, Technology and Nursing programs at a college.

H1: The proportions of students drawn from inside or outside the local area are

different for the Business, Technology and Nursing programs at a college.= 0.025


below.



# of expected values = 0.025.

Fail to reject H0. There is not enough evidence to infer that the proportions of

students drawn from inside or outside the local area are different for the Business,Technology and Nursing programs at a college.

Chapter Review Questions1. We can pool data when the null hypothesis is that there is NO difference in the

population proportions. If that is the case (as we assume), then we can pool thesample data, because both samples provide estimates of the sameproportion of

successes. We cannot pool the data when the null hypothesis is that the population

proportions differ by 5%, because the sample data are providing estimates of twodifferentproportions of success.

2. We can't pool the sample data when we are constructing a confidence intervalestimate of p1 p2, because we are not assuming there is no difference in the

population proportions. We do not have a null hypothesis in mind when we areestimating the difference in proportions.

3. H0: p1 p2= - 0.10

H1: p1 p2< -0.10

This may not be immediately obvious. Remember, the subscript 1 corresponds to last

year's results, and the subscript 2 corresponds to this year's results. If the proportion



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of people who pass this year is more than 10% higher, then when we subtract p 1-p2,

we will get a negative number, and it will be to the left of -0.10 on the number line.

4. The Chi-square goodness-of-fit test measures only how closely the observed

frequencies match the expected frequencies. The test does not take into account

whether the differences are positive or negative (differences are squared in thecalculation of the test statistic). Larger differences result in larger values of the Chi-

square test statistics. Unusually large values (in the right tail of the distribution)

signal that there are significant differences in the distributions.

5. Repeated tests on the same data set lead to higher chances of Type I error, and are

therefore not reliable. A Chi-square test allows us to compare all three proportionssimultaneously.

6. Call the data on members who were taking fitness classes sample 1 from population1. Call the data on members who were working with a personal trainer sample 2 from

population 2.

H0: p1 p2= 0H1: p1 p20

= 0.05

63333333.060

38p1 , n1= 60, 75.0

80

60p 2 , n2= 80

Sampling is done without replacement, but presumably the fitness club has hundreds

of members. This is an assumption that we should note before we proceed to use thebinomial distribution as the underlying model.


=38 > 1011pn

=60 - 38 = 22 > 1011qn

=60 > 1022pn

=80 - 60 = 20 > 1022 qn



70.08060

6038p



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49.1

80

1

60

1

)70.01)(70.0(

0)75.060

38(

n

1

n

1

qp

0)pp(z

21

21

(Note that p1is left in fractional form to preserve accuracy for calculations with a

P(z - 1.49) = 2 0.0681 = 0.1362

Since p-value > , fail to reject H0. There is insufficient evidence to infer that there

ith

. Since the Chi-square test is equivalent to the test of proportions, we expect to get the

Still working out in Quit working out in

calculator.)

p-value = 2

is a difference in the proportion of new members still working out regularly sixmonths after joining the club, when comparing those who attend fitness classes w

those who work out with a personal trainer.

7same answer. First, set up the appropriate contingency table for the data, as shown

below.

first six months first six months

Taking fitness classes 38 22

Working with a personal trainer 60 20

The setup of the problem is the same, with the same null and alternative hypotheses.

red Expected Values Calculations


below.

Chi-Squa




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8. Call the data on deliveries by the private courier sample 1 from population 1. Call

H0: p1 p2= 0.05

= 0.025

, n1= 100, , n2= 75

Sampling is done without replacement, but presumably both the private courier and


0

0

Since the null hypothesis is that there is a difference in the proportions, we cannot


the data on deliveries by Canada Post sample 2 from population 2.

H1: p1 p2> 0.05

89.01 p 80.0p2

Canada Post make a very large number of deliveries, so we can use the binomialdistribution as the appropriate underlying model.

11pn = 100(0.89) = 89 > 10

= 100(1-0.89) = 11 111qn

2 =75(0.80) = 60 > 102pn

=75(1 - 0.80) = 15 > 122 qn

pool the sample data.

72.0

75

)20.0)(80.0(

100

)11.0)(89.0(

05.0)80.089.0(

)(

2

22

1

11

21 21

n

qp

n

qp

ppz

pp

p-value = P(z 0.72) = 1 0.7642 = 0.2358

Since p-value > , fail to reject H0. There is insufficient evidence to infer that the

uldon-time or early percentage for the private courier is more than 5% higher thanCanada Posts. Using this criterion for the decision, the mail order company sho

not use the private courier service.



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9. Call the data on students who were called by program faculty sample 1 from

population 1. Call the data on students who were only sent a package in the mailsample 2 from population 2.

H0: p1 p2= 0

H1: p1 p2> 0

= 0.025

841726618.0278

234

1 p , n1= 278, 76821192.0302

232

2 p , n2= 302

Sampling is done without replacement. We have no information on collegeenrolment. Sample sizes are fairly large. For these samples to be at most 5% of the

relevant populations, the college would have to have about 5560 students who were

called by faculty in total, and about 6040 who were sent acceptance packages. This

means a fairly large potential first-year enrolment. This is an assumption that weshould note before we proceed to use the binomial distribution as the underlying

model.


= 234 > 1011pn

= 278 - 234 = 44 > 1011qn

= 232 > 1022pn

= 302 - 232 = 70 > 1022 qn

Since the null hypothesis is that there is no difference in the proportions, we can poolthe sample data to estimate .p

80344828.0302278

232234

p


23.2

3021

2781

5804661

580466

0302

232

278

234

11

0)(

21

21

nnqp

ppz

(Note that some proportions are left in fractional form to preserve accuracy for

calculations with a calculator.)

p-value = P(z 2.23) = 1 - 0.9871 = 0.0129



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Since p-value < , reject H0. There is sufficient evidence to infer that the proportionof prospective students who send acceptances is higher when they get calls from

program faculty (compared with receiving a package in the mail).

10. This confidence interval could be calculated manually. The output of the Exceltemplate is shown below (manual calculations should be very close).



Proportions

ConfidenceLevel(decimalform) 0.95



Sample1Size 278

Sample2Size 302

n1p1hat 234

n1q1hat 44

n2p2hat 232

n2q2hat 70

Arenpandnq>=10? yes


LowerConfidenceLimit 0.00944

With 95% confidence, we estimate that the proportion of students who send

acceptances when called by program faculty is 0.9% to 13.8% higher than the

proportion who send acceptances when they receive only a package in the mail.

11. Call the data on managers who have been sent to conflict resolution training sample

1 from population 1. Call the data on non-managerial employees who have been sentto conflict resolution training sample 2 from population 2.

H0: p1 p2= 0H1: p1 p20

= 0.025

72.050

36p1 , n1= 50, 50.0

76

38p 2 , n2= 76



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Sampling is done without replacement. We have no information on the total number

of employees who have been sent to conflict resolution training. We are told that the

company is large. For these samples to be at most 5% of the relevant populations,

the company would have had to send 1000 managerial employees to the training, and1520 non-managerial employees. This is an assumption that we should note before

we proceed to use the binomial distribution as the underlying model.


= 36 > 1011pn

= 50 - 36 = 14 > 1011qn

= 38 > 1022pn

= 76 - 38 = 38 > 1022 qn

Since the null hypothesis is that there is no difference in the proportions, we can poolthe sample data to estimate .p

587301587.07650

3836p


45.2

76

1

50

1

126

741

126

74

0)50.072.0(

n

1

n

1qp

0)pp(z

21

21



p-value = 2 P(z 2.45) = 2 (1 - 0.9929) = 2 0.0071 = 0.0142

Since p-value < , reject H0. There is sufficient evidence to infer there is a

difference in the proportions of managers and non-managers who thought thatconflict resolution training was a waste of time.

12. In the sample, 72% of managers and 50% of non-managers thought the training wasa waste of time. There is no way to know why, and this is something that might beworthy of further research. Was the training perceived as a waste of time because

the employees felt they did not benefit? If they did not benefit, was this because they

learned nothing new, or they thought the training was poorly done? Was it a waste

of time only because they felt they had more important tasks to complete?



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The initial research was not really that helpful. It would have been more appropriate

to ask the employees what they learned at the training, and whether they were likelyto put what they learned into practice.

However, generally, the sample results raise a question about whether the training is

accomplishing its intended goals. Before continuing to spend money on training, thedecision to implement the training should be revisited, with further data collection a

possibility.

13. We can set this up as a Chi-square test, with the information organized as in the tablebelow.

Manufacturer#1

Manufacturer#2

Manufacturer#3

Defective Components 36 30 38

Non-DefectiveComponents 89 95 87

Total 125 125 125

H0: The proportions of defective items are the same for all three manufacturers.

H1: The proportions of defective items are different among the three manufacturers.= 0.05

The output from the Excel tool for Chi-Squared Expected Value Calculations is

shown below.


Chi-squared test statistic 1.383764# of expected values = 0.05.

Fail to reject H0. There is not enough evidence to infer that the proportions of

defective items are different among the three manufacturers.



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14. Refer to the two plants as Plant 1 and Plant 2.

H0: p1 p2= 0

H1: p1 p20

= 0.05

15333333.0150

23p1 , n1= 150, 184.0

125

23p 2 , n2= 125

Sampling is done without replacement. We have no information on the total number

of employees at the two plants. As long as Plant 1 has 3000 employees, and Plant 2

has 2500 employees, we can still use the binomial distribution as the appropriateunderlying model. We note this assumption and proceed.


= 23 > 1011pn = 150 - 23 = 127 > 1011qn

= 23 > 1022pn

= 125 - 23 = 102 > 1022 qn



71672727272.0275

46

125150

2323p


68.0

125

1

150

1

275

461

275

46

0184.0150

23

n

1

n

1qp

0)pp(z

21

21



p-value =2 P(z -0.68) = 2 0.2483 = 0.4966

Using Excel, we calculate the exact p-value as 0.497. (See the output from the Exceltemplate for Making Decisions About Two Population Proportions, Qualitative Data,

shown below.)



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MakingDecisionsAboutTwo

PopulationProportions

Sample1Size 150

Sample2Size 125



n1p1hat 23

n1q1hat 127

n2p2hat 23

n2q2hat 102

Arenpand nq>=10? yes

HypothesizedDifferenceinPopulation

Proportions,p1

p2

(decimal

form) 0

zScore 0.67847976

OneTailedpValue 0.24873378

TwoTailedpValue 0.49746755

Since p-value > , fail to reject H0. There is insufficient evidence to infer there is adifference in the proportions of employees who had accidents at the two plants.

15. Exercise 14 could also be done as a Chi-square test. First, organize the data as

shown below.

Plant 1 Plant 2

Employees Who Had An Accident 23 23

Employees Who Had No Accident 127 102

Total 150 125

H0: The proportions of employees who had accidents are the same at the two plants.H1: The proportions of employees who had accidents are different at the two plants.

= 0.05



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The output from the Excel tool for Chi-Squared Expected Value Calculations is

shown below.


Chi-squared test statistic 0.460335# of expected values , fail to reject H0. There is insufficient evidence toinfer there is a difference in the proportions of employees who had accidents at the

two plants.

16. H0: There is no relationship between an individuals family status and his/her

willingness to accept a foreign posting

H1: There is a relationship between an individuals family status and his/her

willingness to accept a foreign posting= 0.05

The output of the Excel tool for Chi-squared Expected Value Calculations is shownbelow.


Chi-squared teststatistic 5.474924



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17. H0: The absences are equally distributed across the five working days of the week.

H1: The absences are not equally distributed across the five working days of theweek.

= 0.05

There are 48 absences in total, in the sample. If the absences are equally distributedacross the five working days of the week, then we would expect each of the five days

to have 48/5 = 9.6 absences.

625.126.9

)6.916(

6.9

)6.97(

6.9

)6.94(

6.9

)6.96(

6.9

)6.915()( 222222

2

i

ii

e

eoX

p-value = P(X2> 12.625)=0.013261 (using Excels CHITEST).

Using the table, for four degrees of freedom, we see 0.010 < P(X2> 12.625) < 0.025.

Reject H0. There is enough evidence to suggest that the absences are not equally

distributed across the five working days of the week.

18. H0: There is no relationship between gender and preferred movie type.

H1: There is a relationship between gender and preferred movie type.

= 0.04

This problem could be done manually, of course. The output of the Excel tool for

Chi-squared Expected Value Calculations is shown below.





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19. H0: The proportions of workers who travel to work via the different methods are the

same for the software firm and the accounting firm.H1: The proportions of workers who travel to work via the different methods at the

software firm are different from the proportions of workers who travel to work

via the different methods at the accounting firm.

= 0.05


below.


Chi-squared test statistic n/a



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Since the expected values are now all 5, we can proceed. The p-value is verysmall, at 0.00045. We have very convincing evidence that the proportions of

workers who travel to work via the different methods at the software firm are

different from the proportions of workers who travel to work via the different

methods at the accounting firm.

20. H0: The distribution of preferences for beer, wine and other alcoholic drinks is the

same for males and females.H1: The distribution of preferences for beer, wine and other alcoholic drinks is

different for males and females.

= 0.05


below.

ChiSquaredExpectedValuesCalculations

Chisquaredteststatistic 0

#ofexpectedvalues


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All of the expected values are 5, so we can proceed.

X2= 1.827

From the table (degrees of freedom = k 1 = 4), we see that p-value > 0.100.

Using CHITEST, we see that p-value = 0.7676.Fail to reject H0. There is insufficient evidence to infer that the proportions of mixed

nuts are not as specified.

22. H0: p = 0.50

H1: p > 0.50

= 0.025

= 190/374 = 0.50802139p

n = 374

Sampling is done without replacement. The company presumably produces a

significant quantity of mixed nuts, so the sample is presumably not more than 5% of

the population. This means the binomial distribution is still the appropriateunderlying model. In this case, we are presuming that one package of mixed nuts

constitutes a random sample.

np = 374(0.50) = 187

nq = 374(1 0.50) = 187

Both are 10, so the sampling distribution of will be approximately normal, with

a mean of 0.50, and a standard error of

p

025854384.0374

)50.0)(50.0(

n

pqp .

3783.0

6217.01

)31.0z(P

374

)50.0)(50.0(

50.0374

190

zP

)

374

190p(P

p-value = 0.3783 > = 0.025Fail to reject H0. There is insufficient evidence to infer that there are more than 50%

peanuts in the mixed nuts packages.



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23. H0: The proportions of types of nuts are the same for the two companies.

H1: The proportions of types of nuts are different for the two companies.

= 0.05


below.

Chi-Squared Expected ValuesCalculations


# of expected values 10AApn

=184 >10AA qn

=375 >10BBpn =363 > 10BB qn





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508093525.0738374

375190p

z-score = - 0.003

P(z -0.003) 50%Fail to reject H0. There is insufficient evidence to suggest that there is a difference in

the proportions of peanuts in the mixed-nuts packages of the two companies.

25. The Excel template output is shown below.



Proportions

ConfidenceLevel

(decimal

form) 0.9



Sample1Size 374

Sample2Size 738

n1p1hat 190

n1q1hat 184

n2p2hat 375

n2q2hat 363

Arenp

and

nq

>=10? yes


LowerConfidenceLimit 0.0523

With 90% confidence, we estimate the interval (-0.0523, 0.0521) contains the truedifference in the proportion of peanuts in the mixed-nuts packages of the two

companies.

This confidence interval is wider than the confidence interval that would correspondto the hypothesis test in the previous exercise. Since we failed to reject the

hypothesis of no difference in that test, we would expect that narrower confidence

interval to contain zero. The wider confidence interval for this exercise, then, wouldalso contain zero.

26. You cannot use the Chi-square test on the weights of the different-coloured candies

directly. The Chi-square test works with discrete qualitative data, and weights arecontinuous quantitative data. As well, it is important to use the correct counts for the

test. Don't be lazy!



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ghts into an approximate number of candies, you can proceed,although you are only approximating. So, for example, if you knew each candy

ppose the weight breakdown was as follows:

Red Yellow Green Black Orange Total

If you convert the wei

weighed 1.5 grams, you could convert the weights for each colour into a number of

candies.

Try it! Su

Weight 62 46 58 39 45 250

Use Excel to do a Chi-square test for this data set. Then divide each of the weightsby 1.5 grams to get the number of candies of each colour, and repeat. You will see

that you do not get the same Chi-square statistic or p-value for the two versions.

Only the second version, based on counts, is correct.

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