ch10-solns-all_skuce_2e

8/12/2019 ch10-solns-all_skuce_2e

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Instructors Solutions Manual - Chapter 10

Chapter 10 Solutions

Develop Your Skills 10.11. Call the defects on the night shift population 1, and the defects on the day shift

population 2.

H0: 1- 2= 0H1: 1- 2> 0= 0.05

1x = 35.4, 2x = 27.8, s1= 15.3, s2= 7.9, n1= 45, n2= 50

We are told that the population distributions of errors are normal.

2.992

50

9.7

45

3.15

0)8.274.35(

)(

22

2

2

2

1

2

1

2121

n

s

n

s

xxt

Degrees of freedom: minimum of (n1 1) and (n2 1), so minimum (44, 49) = 44.

Closest row in the table is for 45 degrees of freedom.p-value < 0.005

Reject H0. There is sufficient evidence to infer that the number of defects is higheron the night shift than on the day shift, on average.

Using the Excel template, we find a more exact p-value of 0.00196.

MakingDecisionsAbouttheDifferencein

PopulationMeanswithTwoIndependent

Samples

Dothesampledataappeartobenormally

distributed? yes

Sample1StandardDeviation 15.3


Sample1Mean 35.4

Sample2Mean 27.8

Sample1Size 45

Sample2Size 50

HypotheticalDifferenceinPopulationMeans 0

tScore 2.99245

OneTailedpValue 0.00196

TwoTailedpValue 0.00393

Copyright 2011 Pearson Canada Inc. 234


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2. Call the population of purchases by females population 1, and the purchases by

males population 2.

H0: 1- 2= 0

H1: 1- 2> 0= 0.025

We start by creating histograms of the sample data to check for normality.

0

1

2

3

4

5

6

7

Fre

quency

ValueofPurchase

DrugstorePurchasesbyFemales

0

1

2

3

4

5

6

NumberofPurchases

ValueofPurchase

DrugstorePurchasesbyMales

Neither histogram is perfectly normal, with the purchases by females in particular

showing some skewness to the right. Note that these histograms were not designedfor comparison (classes are different for each), but to assess normality. We will

proceed, but with some caution.



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We have the data in Excel, so can use the Data Analysis tool for the calculations.

tTest:TwoSampleAssumingUnequalVariances

PurchasesbyMales

PurchasesbyFemales

Mean 32.42933333 27.82428571

Variance 85.32606381 23.95908791

Observations 15 14

HypothesizedMeanDifference 0

df 22

tStat 1.692875747

P(T


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3. Use the Excel template to construct the confidence interval. Of course, this can also

be done manually with the formula 2

2

2

1

2

1

21n

s

n

sscoretxx .

Confidence

Interval

Estimate

for

the

DifferenceinPopulationMeans

Dothesampledataappeartobe

normallydistributed? yes



Sample1Mean $32.43

Sample2Mean $27.82

Sample1Size 15

Sample2Size 14

ConfidenceLevel

(decimal

form) 0.95

UpperConfidenceLimit 10.2621

LowerConfidenceLimit 1.052

With 95% confidence, we estimate that the interval (-1.05, 10.26) contains the trueaverage difference in the purchase of females, compared to males, at this drugstore.

We expect this interval to contain zero, since we failed to reject the hypothesis that

the difference was zero in Exercise 2.



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4. Call the population of daily sales last year population 1, and the daily sales this year

population 2.

H0: 1- 2= 0

H1: 1- 2< 0= 0.03

We start by creating histograms of the sample data to check for normality.

0

2

4

6

8

10

12

Frequency

DailySales

SampleofHotDogVendor'sDaily

Sales,LastYear

0

2

4

6

8

10

12

14

Frequency

DailySales

SampleofHotdogVendorDailySales,

ThisYear

While neither histogram is perfectly normal, they are approximately normal, and

sample sizes are fairly large, at 30 and 35.



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Again, using Excel, we obtain the following results.


LastYear'sDailySales ThisYear'sDailySalesMean 298.2857143 356.0666667Variance 3368.915966 593.9954023

Observations 35 30


df 47

tStat 5.36356487

P(T


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Develop Your Skills 10.26. H0: there is no difference in the locations of the populations of ratings for trainees #1

and #2

H1: there is a difference in the locations of the populations of ratings for trainees #1

and #2= 0.025

Since these are ranked data, we must use the Wilcoxon Rank Sum Test. First, we

have to see if the distributions are similar in shape and spread. With so few datapoints, this can be difficult to see. Below are two quick dot plots (you could also do

a histogram with Excel) that reveal similarities in shape and spread.

Ratings for Trainee #1

*

* * *

* * * * *

1 2 3 4 5

Ratings for Trainee #2

*

*

* *

* * * *

1 2 3 4 5 6

The table below illustrates the ranking process.

Performance Ratings

Trainee #1 ordered

ratings

ranks Trainee #2 ordered

ratings

ranks

1 1 1 4 2 2.5

5 2 2.5 5 4 8

3 3 4.5 4 4 8

2 3 4.5 5 5 13.5

3 4 8 6 5 13.5

4 4 8 2 5 13.5

4 4 8 5 5 13.5

5 5 13.5 5 6 17

4 5 13.5

rank sum 63.5 rank sum 89.5



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We need only calculate the rank sum for the smallest sample, which is the ratings forTrainee #2, but both are shown here. Note that the tables are set up to for W1to be

calculated from the smallestsample, so W1= 89.5 here (even though these are the

ratings for Trainee #2). So, n1= 8, and n2= 9.

Because sample sizes are below 10, we must use the tables to estimate the p-value.

We see from the table that

0.046 < P(W 89.5) < 0.057Since this is a two-tailed test,

0.046 2 < p-value < 0.057 20.092 < p-value < 0.114

Fail to reject H0. There is insufficient evidence to infer there is a difference in thelocations of the ratings of Trainee #1 and Trainee #2. Note the implications of this

result. If you simply looked at the ratings, you probably would have concluded that

the ratings for Trainee #2 were higher. However, they are not significantly higher,and the difference could just be a result of sampling variability. This means that

Trainee #2 should not be promoted over Trainee #2 on the basis of these ratings.

Some other criteria will have to be used to decide which trainee to promote.



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7. H0: there is no difference in the locations of the populations of distances travelled by

the current best-selling golf ball and the new golf ballH1: the population of distances travelled by the current best-selling golf ball is to the

left of the population of distances travelled by the new golf ball

= 0.05

First, we must check the histograms for normality.

0

1

2

3

4

5

Frequency

Metres

DistancesTravelledbyCurrentBestSelling

GolfBall

0

1

2

3

4

Frequency

Metres

DistancesTravelledbyNewGolfBall

Both histograms are non-normal, but they are similar in shape and spread, so weproceed with the Wilcoxon Rank Sum Test.



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We will use Excel to analyze these data.

WilcoxonRankSum TestCalculations

sample1size 12

sample

2size

15

W1 221

W2 157

We will use the template for the Wilcoxon Rank Sum Test for independent samples.

MakingDecisionsAboutTwoPopulation

Locations,TwoIndependentSamplesofNon

NormalQuantitativeData orRankedData

(WRST)

Sample1Size 12

Sample2Size 15

Arebothsamplesizesatleast10? yes

Arethesamplehistogramssimilarinshape

andspread? yes

W1 221

W2 157

zScore(basedonW1) 2.58613519


TwoTailed

p

Value 0.00970589

Again, for consistency, we have selected Sample 1 as the smallest sample, so that we

assign W1as the rank sum of the smallest sample, which contains the distances

travelled by the new golf ball. We see that n1= 12, and n2= 15. Since both are 10,

we can use the normal approximation to the sampling distribution of W1.

If the distances travelled by the new golf ball are longer, we would expect W1to behigh.

p-value = P(W1> 221) = 0.005

The p-value < = 0.05. Reject H0. There is sufficient evidence to suggest that thepopulation of distances travelled by the current best-selling ball is to the left of the

population of distances travelled by the new golf ball. Note this is equivalent to

saying the distances travelled by the new golf ball are to the right of the distances

travelled by the current best-selling ball.



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8. H0: there is no difference in the locations of the flight delays before takeoff before

and after the redesign of the airportH1: the location of the population of flight delays before takeoff before the redesign

of the airport is to the right of the population of flight delays before takeoff after

the redesign of the airport

= 0.05

These are quantitative data, so we must check histograms.

0

2

46

8

10

12

14

Freq

uency

Minutes

FlightDelaysBeforeTakeoff,Before

AirportRedesign

0

2

4

6

8

10

12

14

16

Frequency

Minutes

FlightDelaysBeforeTakeoff,After

AirportRedesign

The histogram for the flight delays after the airport redesign appears non-normal.

The data sets are not that similar in shape and spread. We will use the WilcoxonRank Sum Test, but we will be cautious about drawing conclusions about location

only.



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Since the data are available in Excel, we can use the Wilcoxon Rank Sum Test

Calculations tool to get the rank sums, and then use the template to calculate the p-value. The results are shown below.

Wilcoxon Rank Sum Test Calculations

sample 1 size 35 sample 2 size 35

W1 1352.5

W2 1132.5




(WRST)

Sample1Size 35

Sample2Size 35



andspread? no

W1 1352.5

W2 1132.5




This is a one-tailed test, so p-value = 0.098 > = 0.05. There is insufficientevidence to infer that the population of flight delays before takeoff before the airport

redesign is to the left of the population of flight delays before takeoff after the airport

redesign.



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9. H0: there is no difference in the locations of the weight losses for young women aged

18-25 who take the diet pill, compared with those who do not take the diet pillH1: the location of the population of weight losses for the young women aged 18-25

who take the diet pill is to the right of the population of weight losses of the

young women who do not take the diet pill

= 0.04

We are told the distributions of weight-loss are non-normal, and that both are skewed

to the right, so there is similarity in shape of the distributions. No indication is givenof the spread of the data, so we will assume similar spreads, noting that our

conclusions may not be valid if this is not the case.

We are given the rank sums, and can proceed manually, or use the Excel template.

The completed Excel template is shown below. Of course, you could also do this

calculation with the formulas.




(WRST)

Sample1Size 25

Sample2Size 25



andspread? yes

W1 700W2 575




If the weight losses with the diet pill are higher, we would expect W 1to be high. Thep-value for a one-tailed test is 0.113, which is > = 0.04.

Fail to reject H0. There is insufficient evidence to infer that the population of weightlosses of the young women aged 18-25 who took the diet pill is to the right of thepopulation of weight losses for those who did not take the diet pill.



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10. H0: there is no difference in the locations of the populations of food ratings by

weeknight and weekend diners at a restaurantH1: there is a difference in the locations of the populations of food ratings by

weeknight and weekend diners at a restaurant

= 0.05

These are ranked data, so we must examine the distributions for similarity in shape

and spread. The dot plots below (created simply in Excel) illustrate.

Dot Plot for Ratings of Weeknight Diners

*

* *

* *

* * * *

1

2

3

4

5

Dot Plot for Ratings of Weekend Diners

*

* *

* * *

* * *

1 2 3 4 5

There is some similarity in shape, as both dot plots are skewed to the right. However,

there is much less variability in the ratings of the weekend diners. We will proceed

with the Wilcoxon Rank Sum Test, but we must be cautious about makingconclusions about location.



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The assignment of ranks is illustrated in the following table.

Rating by Weeknight

Diners

Ordered

RatingsRank

Ratings by

Weekend

Diners

Ordered

RatingsRank

4 1 4.5 1 1 4.5

5 1 4.5 3 1 4.5

1 1 4.5 2 1 4.5

2 1 4.5 1 1 4.5

1 2 11 1 2 11

2 2 11 1 2 11

2 2 11 3 3 15

1 4 17 2 3 15

1 5 18 3 3 15

86 85

If there was a difference in the food ratings by weeknight and weekend diners at the

restaurant, we would expect W1and W2to be different. They are very similar here.

p-value = 2 P(W1> 86)

Since both samples are of size 9, we must use the tables to approximate the p-value.

The closest value in the table to 86 is 104, so we can be sure P(W1> 86) > 0.057.

This means the p-value > 2 0.057 = 0.114.

Fail to reject H0. There is not enough evidence to conclude there is a difference in the

food ratings by weekend and weeknight diners at the restaurant.

Chapter Review ExercisesThroughout these exercises, it is often possible to do the calculations manually, or withExcel. Manual calculations are sometimes illustrated, and when they are not, the results

should be close to the Excel output.

1. It is preferable to use the t-test, if the necessary conditions are met, because it is

harder to reject the null hypothesis with the Wilcoxon Rank Sum Test. The t-testuses all of the information available from the sample data, while the WRST uses the

ranks, not the actual values. Any time we can use the actual values to make a

decision, we should.



16/41


2. If population 1 was to the right of population 2, we would expect the values in

sample 1 to be higher than the values in sample 2. As a result, the rank sum forsample 1 should be larger than for sample 2. However, when there are 10

observations in each sample (so 20 values have to be ranked), the ranks have to add

up to 210. This means that the rank sum for sample 2 has to equal 210 78 = 132.

This tells us that the values in sample 1 are generally smaller and to the left of thevalues in sample 2. Therefore, there is no evidence that population 1 is to the right of

population 2. The p-value here would be 1 P(W178) = 1 0.022 = 0.978. Be sure

that you think about what the rank sums are telling you. This sample result would behighly unexpected, but if you didn't think about it, you might slip and draw exactly

the wrong conclusion!

3. When samples are different sizes, they will tend to have different rank sums, even if

they come from equivalent populations. The smaller sample will have a smaller rank

sum, simply because there are fewer data points. So, when comparing rank sums, wehave to take this into consideration. The table is based on the rank sum being

calculated from the smallest of the two samples. The conclusions could be wrong ifyou mistakenly calculate W from the larger sample.

4. The unequal-variances version is preferred because:

i.The unequal-variances version of the t-test will lead to the right decision, even if thevariances are in fact equal (with very few exceptions).

ii.It can be hard to determine if variances are in fact equal, especially with smallsample sizes. Really, you should do another sample to test for equal variances.

Remember, the more times you skate across the same frozen lake, the more likelyyou are to observe a rare eventfalling in!and the greater the chance of a Type

I error.iii.If you mistakenly assume that variances are equal when they are not, results will be

unreliable, particularly when sample sizes are unequal (and especially when the

smaller sample has the larger variance).

5. The Excel template is preferred because the t-score will be more accurate than the

one used for the manual calculation.

6. Call the times managers spent on email in the past population 1, and the times

managers spend on emails after the new procedures have been implemented

population 2.

H0: 1- 2= 0

H1: 1- 2> 0= 0.05

1x = 49.2, 2x = 39.6, s1= 22.3, s2= 10.6, n1= 27, n2= 25

We are told that the population distributions of errors are normal.



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006.2

256.10

273.22

0)6.392.49(

)(

22

2

2

2

1

2

1

2121

n

s

n

s

xxt

Degrees of freedom: minimum of (n1 1) and (n2 1), so minimum (26, 24) =24.

Using the table for 24 degrees of freedom, we see

0.025 < p-value < 0.05Reject H0. There is sufficient evidence to infer that the average time spent by

managers on email was lower after the new procedures were implemented.

7. With 90% confidence, we estimate that the interval (1.52 minutes, 17.68 minutes)

contains the true reduction in the average amount of time managers spend on email

after the new procedures. The completed Excel template is shown below. Of course,

this could also be done manually, using the formula 2

2

2

1

2

1

21n

s

n

sscoretxx .

ConfidenceIntervalEstimateforthe


Dothe

sample

data

appear

to

be




Sample1Mean 49.2

Sample2Mean 39.6

Sample1Size 27

Sample2Size 25

ConfidenceLevel(decimalform) 0.9


LowerConfidence

Limit 1.52438



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8. Call the hours spent doing unpaid work around the home by men in 2000 population

1, and the hours spent doing such work in 2009 population 2. Since it does notappear that the samemen were involved in the surveys, we will treat these as

independent samples.

H0: 1- 2= 0

H1: 1- 2< 0= 0.025

1x = 2.2, 2x = 2.6, s1= 0.6, s2= 1.3, n1= 55, n2= 55

We are told that the samples appear normally distributed, and so will assume the

population distributions are.

We can proceed to do the calculations manually, or with the Excel template. The

completed Excel template is shown below.



Samples


distributed? yes



Sample1Mean 2.2

Sample2Mean 2.6

Sample1Size

55

Sample2Size 55


tScore 2.0719



This is a one-tailed test, so the p-value is 0.021 < = 0.025.

Reject H0. There is sufficient evidence to infer that the average number of hours

men spend doing unpaid work around the home has increased in 2009, comparedwith 2000.



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9. This question can be done manually with the formula, or with the Excel template.

The completed template is shown below.

ConfidenceInterval

Estimate

for

the






Sample1Mean 2.2

Sample2Mean 2.6

Sample1Size 55

Sample2Size 55

ConfidenceLevel

(decimal

form) 0.95



We have 95% confidence that the interval (-0.78 hours, -0.02 hours) contains thechange in the average amount of time men spend doing unpaid work around the

house in 2000, compared with 2009. This means that (0.02 hours, 0.78 hours)

contains the increase in the average amount of time spend doing unpaid work around

the house in 2009 compared with 2000.

10. H0: there is no difference in the locations of the population of ratings of the

appearance of the grocery storeH1: the location of the population of ratings of the appearance of the grocery store

six months ago is different from the location of the population of current ratings

of the grocery store

= 0.05

These are ranked data, so we must examine the distributions for similarity in shapeand spread. The diagrams below illustrate.

Ratings for Grocery Store Appearance Six Months Ago

* *

* *

* * *

* * * * *

1 2 3 4 5



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Current Ratings for Grocery Store Appearance

* *

*

** * *

* * * *

1 2 3 4 5

The ratings appear to be similar in shape and spread. The assignment of ranks is

illustrated below.

Appearance Ratings SixMonths Ago

OrderedRatings

RanksCurrent Appearance

RatingsOrdered

RatingsRanks

5 1 1.5 1 1 1.5

4 2 3 5 3 5.5

5 3 5.5 4 3 5.5

4 3 5.5 5 4 11.5

2 4 11.5 4 4 11.5

1 4 11.5 4 4 11.5

4 4 11.5 5 4 11.5

4 4 11.5 3 5 19.53 5 19.5 5 5 19.5

3 5 19.5 4 5 19.5

5 5 19.5 3 5 19.5

5 5 19.5 W1 136.5

W2 139.5

As usual, we focus on the rank sum of the smallest sample, which contains the

current ratings for grocery store appearance.

n1= 11, n2= 12, W1= 136.5

Since both sample sizes are larger than 10, we can use the normal approximation tothe sampling distribution of W1.



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24807681.16

12)11211)(12(11

12

)1nn(nn 2121

W1

132

2

)11211(11

2

)1nn(n 211

W1

28.0

24807681.16

1325.136

1

11

W

WW

z

This is a two-tailed test. If the ratings are different, then W1would be high. The p-

value will be 2 P(W1136.5) = 2 P(z 0.28) = 2 (1 0.6103) = 0.7794.Fail to reject H0. There is insufficient evidence to infer that there is a difference

between the locations of the populations of grocery store ratings for appearance sixmonths ago and currently.



22/41


Of course, you could also use the Excel template to do these calculations. It is shown

below.


Locations,Two

Independent

Samples

of

Non


(WRST)

Sample1Size 11

Sample2Size 12



andspread? yes

W1 136.5

W2 139.5

zScore

(based

on

W1) 0.27695585



11. First, realize these are matched-pairs data. Prices are for the same book each year.(Remember to think about whether you have independent or matched-pairs samples,

because the techniques for each are different.)

Next check to see if the differences are normally distributed. One possible histogramof differences is shown below.

0

1

2

3

4

5

6

7

8

9

Fre

quency

(BookPriceLastYear)(BookPriceThisYear)

BookPriceComparison



23/41


The histogram is skewed to the right, but somewhat normal in shape.

The results of the Data Analysis tool for the t-test are shown below.

t

Test:

Paired

Two

Sample

for

Means

BookPriceLastYear BookPriceThisYear

Mean 14.54 12.426

Variance 30.37956842 30.895162

Observations 20 20

PearsonCorrelation 0.879011986


df 19

tStat 3.471780487

P(T


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12. We are provided with summary data, and so can proceed either manually or with the

Excel template. We are told that the sample data are normally distributed, so the t-test of the difference in means is appropriate.

We will refer to the population of the number of exercises required to master the

topic, according to professors, as population 1. The population of the number ofexercises required, according to the students experience, as population 2. We are

asked if the professors have unrealistic expectation of the number of exercises that

students need to master the topic. We interpret this to mean unrealistically high.

In this case, the alternative hypothesis will be that 1- 2 > 0.

H0: 1- 2= 0

H1: 1- 2> 0= 0.01

1x = 19.2, 2x = 12.3, s1= 5.2, s2= 3.6, n1= 15, n2= 20

We are told the sample data appear normally distributed.The completed Excel template is shown below.



Samples


distributed? yes


Sample2Standard

Deviation 3.6

Sample1Mean 19.2

Sample2Mean 12.3

Sample1Size 15

Sample2Size 20


tScore 4.40765



The one tailed p-value is 0.0001, which is less than 1%. Reject H0. There issufficient evidence to infer that professors have unrealistically high expectations of

the number of exercises that students need to do to master this topic.



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13. The completed Excel template is shown below.


Differencein

Population

Means





Sample1Mean 19.2

Sample2Mean 12.3

Sample1Size 15

Sample2Size 20


UpperConfidence

Limit 11.2948


We have 99% confidence that the interval (2.5, 11.3) contains the true

overestimation of the number of exercises required to master this topic, compared to

the actual experience of students. We would not particularly expect this interval tocontain zero, since the hypothesis test in Exercise 12 concluded that professors have

higher expectations about the number of exercises required to master a topic,

compared with students. The 99% confidence interval is wider than the interval thatdirectly corresponds to the hypothesis test in exercise 12 (the tail area there would be

1%; for a 99% confidence interval, there is only % in each tail). However, even

the wider interval does not contain zero.



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14. H0: A- B= 0

H1: A- B0

= 0.05

Ax = 862, Bx = 731, sA= 362, sB= 223, nA= 31, nB= 25




Samples


distributed? yes

Sample1StandardDeviation 362

Sample2Standard

Deviation 223

Sample1Mean 862

Sample2Mean 731

Sample1Size 31

Sample2Size 25


tScore 1.66151



The two-tailed p-value is 0.103 > . Fail to reject H0. There is insufficient evidence

to infer there is a difference in the number of pages produced by the two brands of

cartridges, under these conditions.



27/41




Differencein

Population

Means





Sample1Mean 862

Sample2Mean 731

Sample1Size 31

Sample2Size 25


UpperConfidence

Limit 263.135


We have 90% confidence that the interval (-1.1, 263.19) contains the difference in

the number of pages produced by the two brands of printer cartridge, under these

conditions.



28/41


16. We will refer to the population of wait times for ITM support as population 1, and

the population of wait times for Dull support as population 2.

H0: 1- 2= 0

H1: 1- 20

= 0.051x = 8.5, 2x = 6.5, s1= 2.6, s2= 1.9, n1= 34, n2= 36




Samples

Dothe

sample

data

appear

to

be

normally

distributed? yes



Sample1Mean 8.5

Sample2Mean 6.5

Sample1Size 34

Sample2Size 36


tScore 3.65697

OneTailed

p

Value 0.00027


The two-tailed p-value is 0.00054 < . Reject H0. There is sufficient evidence toinfer there is a difference in average wait times for support between the ITM and

Dull computers.



29/41




Differencein

Population

Means





Sample1Mean 8.5

Sample2Mean 6.5

Sample1Size 34

Sample2Size 36


UpperConfidence

Limit 3.09397


We have 95% confidence that the interval (0.91 minutes, 3.09 minutes) contains the

true extra average wait time for support for the ITM computers, compared with the

Dull computers.

This confidence interval corresponds directly to the two-tailed hypothesis test in

Exercise 16. Since the null hypothesis of no difference was rejected there, we wouldnot expect this confidence interval to contain zero (and it does not).



30/41


31/41


The data are available on Excel, so it seems reasonable to use Excel to do the t-test.

t-Test: Two-Sample Assuming Unequal Variances

MinutesSpent with

Each Client

Last January

MinutesSpent with

Each Client

This January

Mean 50.63333333 59

Variance 281.7574713 597.8823529

Observations 30 35

Hypothesized Mean Difference 0

df 60

t Stat -1.62607466

P(T


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20. Call the amount of time spent by sales reps in a two week period with the old

software population 1, and the amount of time spent by sales reps with the newsoftware population 2.

H0: 1- 2= 0

H1: 1- 2> 0= 0.04

First we must examine the sample data. Histograms indicate the data are

approximately normal, although the sample data for the old software are skewed tothe right. Also, sample sizes, at 30 and 35, are fairly large.

0

2

4

6

8

10

12

Frequency

MinutesinaTwoWeekPeriod

MinutesSpentbySalesRepsOn

Computer,OldSoftware

0

2

4

6

8

10

12

F

requency

MinutesinaTwoWeekPeriod

MinutesSpentbySalesRepsOn

Computer,NewSoftware



34/41


The Excel output for the t-test is shown below.


OldSoftware

NewSoftware

Mean 799.8 608

Variance 172199.5 68901.88

Observations 30 35


df 48

tStat 2.184543

P(T


35/41





Dothe

sample

data

appear

to

be




Sample1Mean 799.8

Sample2Mean 608

Sample1Size 30

Sample2Size 35



LowerConfidence

Limit 6.34049

At a 96% confidence level, it is estimated that the interval (6.3 minutes, 377.3

minutes) contains the reduction in the amount of time that sales reps would spendover a two-week period, if the new software was adopted.



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22. The data are ranked, so the Wilcoxon Rank Sum Test will be used to make the

comparisons.

H0: there is no difference in the locations of the populations of ratings of high-speed

Internet service for the cable TV company and the telephone company

H1: there is a difference in the locations of the populations of ratings of high-speedInternet service for the cable TV company and the telephone company

= 0.025

Before we use the Wilcoxon Rank Sum Test, we must examine the data to see we can

reasonably assume that the populations are similar in shape and spread.

Two possible bar graphs of the data are shown below.

0

2

4

6

8

10

12

14

1 2 3 4 5

Frequency

1=VerySatisfied,5=VeryDissatisfied

Ratingsof

Internet

Service

Provided

by

theTelephoneCompany

0

2

4

6

8

10

12

14

1 2 3 4 5

Frequency

1=VerySatisfied,5=VeryDissatisfied

Ratingsof InternetServiceProvidedby

theCableTVCompany

The distributions appear similar in shape and spread.



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Since the data are available in an Excel file, it seems appropriate to do the

calculations in Excel. The output of the Wilcoxon Rank Sum Test Calculations isshown below.

Wilcoxon Rank Sum Test Calculations

sample 1 size 35

sample 2 size 35

W1 1349

W2 1136

The relevant template is shown below.


Locations,

Two

Independent

Samples

of

Non


(WRST)

Sample1Size 35

Sample2Size 35



andspread? yes

W1 1349

W2 1136

zScore

(based

on

W1) 1.25095882



The two-tailed p-value is 0.21095. Fail to reject H0. There is insufficient evidence

to infer there is a difference in the locations of the populations of ratings of Internet

service by the cable TV company and the telephone company.



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23.

a. The data are ranked, so we consider the Wilcoxon Rank Sum Test.

H0: there is no difference in the locations of the populations of ratings of the old

instructions and the new instructions for lawnmower assembly

H1: the location of the population of ratings of the old instructions is to the right of thepopulation of ratings of the new instructions for lawnmower assembly (a higher-

numbered rating means greater difficulty)

= 0.05

The requirement is that the distributions are similar in shape and spread. Two graphs

of the data are shown below.

0

5

10

15

1 2 3 4 5

Frequency

1=VeryEasytoReadandFollow,5=Very

Difficultto

Read

and

Follow

RatingsforOldLawnmower

AssemblyInstructions

0

5

10

15

20

1 2 3 4 5

Fre

quency

1=VeryEasytoReadandFollow,5=Very

DifficulttoReadandFollow

RatingsforNewLawnmower

AssemblyInstructions



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The distributions are similar in spread, but not in shape. Any conclusion we make

from the Wilcoxon Rank Sum Test will be weaker, as a result.

The ranking could of course be completed manually. The output from the WilcoxonRank Sum Test Calculations is shown below.

WilcoxonRankSumTestCalculations

sample1size 37

sample2size 42

W1 1697

W2 1463

Since both sample sizes are more than 10, we will use the Excel template to estimate

p-value.




(WRST)

Sample1Size 37

Sample

2

Size 42Arebothsamplesizesatleast10? yes


andspread? no

W1 1697

W2 1463




p-value = 0.016

Reject H0. There is sufficient evidence to infer that population distributions ofratings for the old and new instructions are different. We have seen that there is a

difference in shape, but given the marked differences in the frequencies of the "1"

and "5" ratings, we are probably safe to conclude that customers find the newinstructions easier to read and follow.



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b. These data are quantitative, and the samples are independent. We must check fornormality before proceeding. Two possible histograms for the sample data are

shown below.

0

2

4

6

8

10

12

14

Frequency

Minutes

LawnmowerAssemblyTimeswithOld

Instructions

0

2

4

6

8

10

12

14

16

F

requency

Minutes

LawnmowerAssemblyTimeswithNew

Instructions

Both distributions seem approximately normal. The distribution of times for the new

instructions is somewhat skewed to the right. However, samples sizes (37 and 42) are

fairly large, and we will proceed with the t-test.

H0: 1- 2= 0H1: 1- 2> 0

= 0.04



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The Excel output for this data set is shown below.


AssemblyTimes

(Minutes)WithOld

Instructions

AssemblyTimes

(Minutes)WithNew

Instructions

Mean 64.43243243 47.5952381

Variance 379.9189189 216.4907085

Observations 37 42


df 67

tStat 4.287366909

P(T

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