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Instructors Solutions Manual - Chapter 6

Chapter 6 Solutions

Develop Your Skills 6.11. Claim about the population: salaries of graduates from the Business programs earn

$40,000 a year.

Sample result: sample mean is $39,368 for a random sample of 35 graduatesIf the colleges claim is true, sample means of one-year-after-graduation salarieswould be normally distributed, with a mean of $40,000 and a standard deviation of$554.

The sample mean is lower than expected. We need to calculate P( x $39,368)

1271.0

)14.1(554

4000039368

)39368(

zP

zP

xP

It would not be highly unlikely to get a sample mean salary as low as $39,368, if thecolleges claim about salaries is true. Since this is not an unexpected result, we do nothave reason to doubt the colleges claim.

2. Claim about the population: at least 90% of customers would recommend the centreto friendsSample result: sample proportion is 87%, for 300 randomly-selected customersIf the centres claim is true, sample proportions would be normally distributed, with amean of 0.90 and a standard deviation of 0.01732.

The sample proportion is lower than expected. We need to calculate P(SR 0.87)

0418.0)73.1(

01732.090.087.0

)87.0(

zP

zP

pP

The probability of getting a sample proportion as low as 95% is just a little over 4%

(which is less than our cut-off of 5%). This is an unusual sample result, and not onewe would expect if the actual proportion of satisfied customers is 90%. Therefore, thesample provides evidence to doubt the centres ad about the proportion of satisfiedcustomers.

Copyright 2011 Pearson Canada Inc. 108


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There are two things to note here. First, 87% does not seem to be so far from 90%, soyou might have been surprised at the result of the probability calculation. This shouldserve as a caution. You cannot make a decision about how far away a sample resultis from what is expected, until you actually do a probability calculation.

The other point is that although the proportion of satisfied customers may not be 90%,from the sample evidence, we suspect it will be around 87% or so. This suggests thatalthough the centres claim may not be specifically correct, it still has a high

percentage of satisfied customers.

3. Desired characteristic about the population: no more than 25% of employees wouldenrol in education programsSample result: sample proportion is 26%, for 500 randomly-selected customersIf the population percentage is actually 25%, sample proportions would be normallydistributed, with a mean of 0.25 and a standard deviation of 0.021651.

The sample proportion is higher than expected. We need to calculate P(SR 0.26)

3228.06772.01

)46.0(021651.0

25.026.0

)26.0(

zP

zP

pP

The probability of getting a sample proportion as high as 26%, when the population

proportion is actually 25%, is over 32%. It would not be unusual to get such a sample proportion, if the actual population proportion is only 25%. The sample does not provide enough evidence to conclude that the actual percentage of employees whowould enrol in such programs is more than 25%. On the basis of the sample results,the company should conclude that it can afford the programs and extend the benefit.



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4. Claim about the population: average mark on the mid-term statistics exam forstudents in the Business program is 67%.Sample result: a random sample of 30 of the Business students taking statistics has anaverage mark of 62%.If the teachers claim is true, the sample means would be normally distributed, with a

mean of 67%, and a standard deviation of 3.2%

The sample mean is lower than expected. We need to calculate P( x 62).

0594.0)56.1(

2.36762

)62(

zP

zP

xP

The probability of getting a sample mean as low as 62%, if the true population mean is

actually 67%, is 0.0594. Such a sample result is not unusual. The sample does not provide enough evidence to suggest that the teachers claim overestimates the trueaverage mark for the mid-term stats exam. Notice that the sample result is almost unusual enough for us to doubt the teacher's claim. However, we have established arule for deciding when a sample result is unusual, and for long-run consistency in ourdecisions, we should follow the rule.

5. Claim about the population: average commuting time is 32 minutes.Sample result: a random sample of 20 commuters has an average commuting time of40 minutes.If the true average commuting time is 32 minutes, the sample means would be

normally distributed, with a mean of 32 minutes, and a standard deviation of 5minutes.

The sample mean is higher than expected. We need to calculate P( x 40).

0548.0

9452.01)6.1z(P5

3240zP

)40x(P

The probability of getting a sample average commuting time as high as 40 minutes, ifthe actual population average commuting time is 32 minutes, is 0.0548. This resultgives us pause. Such a sample result is not that usualit will happen with a

probability of only 5.48%. However, the cut-off we have decided to use for decidingwhat is unusual is a probability of 5% or less. This sample result does not meet that



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test. So, in this case, the sample does not give us enough evidence to conclude that theaverage commuting time has increased from 32 minutes.

You may not be entirely comfortable with this decision. We will discuss this pointfurther, when we talk about p-values (in Chapter 7). For now, stick to the rule, and

later, you will find out more about how these decisions are made.

Develop Your Skills 6.26. Claim about the population: average weight in cereal boxes is 645 grams.

Sample result: a random sample of 10 cereal boxes has an average weight of 648grams.We are told the cereal box weights are normally distributed, with = 5 grams.If the true average weight is 645 grams, the sample means would be normally

distributed, with a mean of 645, and a standard deviation of10

5 grams.

0287.09713.01

)90.1z(P10

5645648

zP

)648x(P

If the cereal line were properly adjusted, the probability of getting a sample mean ashigh as 648 grams is 0.0287. Yet we did get this unlikely result. We have evidencethat the cereal line is not working properly, and it should be adjusted.

7. Claim about the population: average salary of business program graduates one yearafter graduation is at least $40,000 a yearSample result: a random sample of 20 salaries of business program graduates one yearafter graduation has an average $38,000.We are told the salaries are normally distributed, with = $3,300.If the true average salary is $40,000, the sample means would be normally distributed,

with a mean of $40,000, and a standard deviation of20

3300.

0034.0)71.2(

203300

4000038000

)38000(

zP

zP

xP



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If the true average salary were $40,000, it would be almost impossible to get a samplemean as low as $38,000, under these conditions. Such a sample mean providesevidence that the average salary of graduates of the business program one year aftergraduation is less than $40,000.

8. Claim about the population: average tire life is 25,000 kilometres.Sample result: a random sample of 20 tires has an average life of 24,000 kilometres.We are told nothing about the population of tires. Sample data are not provided, so wecannot assess if the sample appears to be normally distributed. A sample size of 20 isnot large enough to ensure normality of the sampling distribution unless the populationis fairly normal. Without more information, we cannot proceed.

9. Claim about the population: it takes 1.5 working days on average to approve loanrequests.Sample result: a random sample of 64 loan requests has an average of 1.8 working

days.We are told the population are normally distributed, with = 2.0.If the true average time for a loan to be approved is 1.5 working days, the samplemeans would be normally distributed, with a mean of 1.5, and a standard deviation of

64

2.

1151.08849.01

)2.1z(P64

25.18.1

zP

)8.1x(P

If the claim about the average time to approve the loan requests is true, the probabilityof getting this sample result would be 0.1151. This is not unusual, and so there is notenough evidence to conclude that the bank understates the average amount of time toapprove loan requests.



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10. Claim about the population: mean weight of packages is 36.7 kgSample result: a random sample of 64 packages has an average weight of 32.1 kgWe are told the population are normally distributed, with = 14.2 kg.If the true average weight of packages is 36.7 kg, the sample means would be

normally distributed, with a mean of 36.7 kg, and a standard deviation of64

2.14.

0048.0)59.2z(P

642.14

7.361.32zP

)1.32x(P

The probability of getting a sample mean weight as low as 32.1 kg, if the average package weight is actually 36.7 kg, is only 0.0048. We would not expect to get thissample result, but we did. The sample result provides evidence that the average

package weight may have decreased.

Develop Your Skills 6.311. This is not really a random sample. It excludes anyone who eats the cereal but does

not visit the website set up for the survey. As well, the sample is likely to be biased.People who did not find a free ticket in their cereal box are probably more likely toanswer your survey. This sample data set cannot be reliably used to decide about the

proportion of cereal boxes with a free ticket.

12. Claim about the population: p = 0.97 (proportion of graduates to find jobs in theirfields within a year of graduation, from a particular college)Sample result: a random sample of 200 students reveals 5% who do not have a job intheir field, so = 1 0.05 = 0.95 p

Sampling is done without replacement. Do we know that 200 graduates representnot more than 5% of graduating class? We do not, and so should proceed withcaution. The binomial distribution may not be the appropriate underlying modelhere. We proceed by noting our assumption: that the sample of 200 graduates is notmore than 5% of the graduating class.

Check conditions:np = 200(0.97) = 194nq = 200(0.03) = 6nq is < 10 so normal approximation is not appropriate (the sampling distribution of

should not be used). p



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In the sample 95% (200) = 190 graduates got jobs in their field within a year ofgraduation.

Using Excel, we calculate P(x


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0038.0)67.2z(P

150)60.0)(40.0(

40.0293333.0zP

)293333.0 p(P

It would be very unusual to get a sample result such as this one, if in fact 40% ofretired people ate out at least once a week. The sample results suggest that fewer than40% of retired people eat out at least once a week in your city.

However, before deciding whether or not to focus on retired people, it would beimportant to know if this group tends to eat out more or less than other groups of

people. While the percentage of those who eat out more than once a week isapparently lower in your city than in the survey, it still might be higher than for othergroups, and might still be a good target market.

Chapter Review Exercises1. Your sketch should look something like the diagram below.

119.5 131.5 143.5 155.5 167.5 179.5 191.5 203.5 215.5



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b. The sampling distribution of the sample means (samples of size 25) will be normallydistributed, because the heights in the population are normally distributed. The mean

of the sample means will be 167.5 cm, and the standard error will be 4.225

12cm.

The sampling distribution of the sample means will therefore be much narrower thanthe population distribution of heights. It will have to be much taller as well, since thetotal area under the distribution must be 1 (it is a probability distribution).

119 .5 131 .5 143 .5 1 55.5 1 67.5 1 79.5 1 91.5 2 03.5 21 5.5

c. The sampling distribution of the sample means (samples of size 40) will be normallydistributed, because the heights in the population are normally distributed. The meanof the sample means will be 167.5 cm, and the standard error will be

8974.140

12cm. It will be narrower still.

119 .5 131 .5 143 .5 1 55.5 1 67.5 1 79.5 1 91.5 2 03.5 21 5.5



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3. p = 0.86 (claimed proportion of those taking the pill who get relief within 1 hour) p = 287/350 = 0.82n = 350 (fairly large)

Sampling is done without replacement. Presumably, there are thousands and

thousands of back pain sufferers who take this medication, so it is still appropriate touse the binomial distribution as the underlying model.

Check for normality:np = 350(0.86) = 301 > 10nq = 350(1-0.86) = 49 > 10

The binomial distribution could be approximated by a normal distribution, and so wecan use the sampling distribution of . p

0154.0

)16.2(350

)14.0)(86.0(86.082.0

)82.0(

zP

zP

pP

The probability of getting a sample result as extreme as the one we got, if the true proportion of sufferers who get relief within one hour is 86%, is 0.0154, which is less

than 5%. The sample result qualifies as an unexpected or unusual event. Since we gotthis sample results, we have enough evidence to suggest that fewer than 86% of back pain sufferers who take this pill get relief within one hour.

4. p = 0.90 (claimed proportion customers who are satisfied with the range of foodserved and prices)

= 438/500 = 0.876 p n = 500 (fairly large)

Sampling is done without replacement. The sample size is fairly large, at 500. Wehave no information about the total number of customers served by the cafeteria. As

long as this sample is no more than 5% of the total population of customers, it isappropriate to use the binomial distribution as the underlying model. We proceed bynoting that we are making this assumption, and that the conclusions are not valid ifthis assumption is not correct.

Check for normality:np = 500 (0.90) = 450 > 10nq = 500 (1 0.90) = 50 > 10



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6. p = 0.65 (% of visitors who had an enjoyable experience)= 282/400 = 0.705 p

n = 400 (fairly large)

Sampling is done without replacement. The sample size is 400, but we are told the

attraction gets 50,000 visitors a year. 400 is less than 5% of the population of 50,000.It is appropriate to use the binomial distribution as the underlying model.

Check for normality:np = 400(0.65) = 260 > 10nq = 400(1 0.65) = 140 > 10

The binomial distribution could be approximated by a normal distribution, so we canuse the sampling distribution of p .

0104.09896.01

)31.2(023848.0

65.0705.0

)705.0(

zP

zP

pP

There is only a very small probability of getting a sample result as extreme as this one,if the percentage of visitors who had an enjoyable experience is 65%. The sample

provides evidence that the percentage of visitors who had an enjoyable experience hasincreased. It seems reasonable to assume that the upgrades are the cause, but this isnot something that can be concluded directly from these data.

7. x = $756 = $132 = $700 (the average cost of textbooks per semester for a college student)

n = 75We are told that the population data are normally distributed, so the samplingdistribution will also be normal, with a mean of $500, and a standard error of

75132

x .



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0001.0

9999.01)67.3(

75132

700756

)756(

zP

zP

xP

There is almost no chance of getting a sample mean as high as $756 if the populationaverage textbook cost is $700. The sample provides evidence that the average cost oftextbooks per semester for college students has increased.

8. x = $8400 = $3700 = $7500 (the average total withdrawal at the ATM over the weekend)

n = 36We are told to assume the population data are normally distributed, so the samplingdistribution will also be normal, with a mean of $7500, and a standard error of

36

3700x .

0721.0

9279.01)46.1(

363700

75008400

)8400(

zP

zP

xP

If the true average total withdrawal is $7500, the probability of getting a sample meantotal withdrawal of $8400 is 0.0721. This sample result is not unexpected enough toconclude that the bank managers claim about the total average withdrawal over theweekend is too low.



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9. Start by analyzing the sample data set.x = 6188.3875 hourss = 234.6176

= 6200 hours (claimed average lifespan of electronic component)n = 40

Histogram of sample data is shown below.

0

2

4

6

8

10

12

14

16

18

N u m b e r o f C o m p o n e n t s

Lifespan in Hours

Lifespans of a Random Sample of 40 Electronic Components

The histogram appears to be fairly normal, with some skewness to the right. Thesample size is fairly large, at 40, so it seems likely the sampling distribution would benormally distributed. The sampling distribution would have a mean of 6200, and a

standard error of406176.234

x .

We are told to assume = s = 234.6176.

3783.0)31.0(

406176.234

62003875.6188

)3875.6188(

zP

zP

xP

The probability of getting a sample result as small as 6188.3875 hours, if the trueaverage lifespan of the components is 6200, is 0.3783. This sample result is notunusually small. The sample evidence does not give us reason to doubt the producersclaim that the average lifespan of the electronic components is 6200 hours.



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11. Start by analyzing the data set.x = 1756.48s = 599.773

= $2000 (claimed average daily sales)n = 29

The histogram is skewed to the right. The sample size, at 29, is reasonably large. Wewill proceed by assuming that with this sample size, the sampling distribution will beapproximately normal.

0144.0)19.2(

29773.599

200048.1756

)48.1756(

zP

zP

xP

The probability of getting average daily sales as low as $1,756.48, if the true averagedaily sales are $2,000, is quite small, at 1.44%. The fact that we got this unusualsample result gives us reason to doubt the former owner's claim that average dailysales at the shop were $2,000. However, it may be that the sales have changed underthe new owner, for a variety of reasons. The data cannot allow us to conclude that theformer owner misrepresented the daily sales figures.



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12.

es = 82 (where "success" is defined as a student living in themediate area)

of 10,000. Itis appropriate to use the binomial distribution as the underlying model.

nq = 400(1 0.25) = 300 > 10

e proximated by a normal distribution, so we canuse the sampling distribution of .

. P(x 82, n = 400, p = 0.25) = 0.0200

. = 82/400 = 0.205

c.

p = 0.25n=400number of successim

Sampling is done without replacement. The sample size is 400, but we are told thecollege has over 10,000 students. 400 is less than 5% of the population

Check for normality:np = 400(0.25) = 100 > 10

The binomial distribution could b ap p

a p b

0188.0

)078.2(021651.0

25.0205.0

)205.0(

zP

zP

pP

d.calculated in part a.

The probabilities are close in value, but not exactly the same.

e.as

provides evidence that the percentage of students from the local area has declined.

The probabilities calculated in parts a and c are different because the probability in part c is the normal approximation of the binomial probability

There is evidence that the percentage of college students from the immediatecatchment area is lower than in the past. The probability of getting a sample resultlow as 82 out of 400 is about 2%, presuming that 25% of students come from thelocal catchment area. This sample result is unusually low, and the fact that we got it



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13. p = 0.20n=300number of successes = 77 (where "success" is defined as a student with a laptop)

Sampling is done without replacement. The sample size is 300, and we have no

information about the total number of students at the college. We can proceed, first by noting that we are assuming there are at least 30020=6000 students at thecollege. We will use the binomial distribution as the underlying model.

Check for normality:np = 300(0.20) = 60 > 10nq = 300(1 0.20) = 240 > 10

The binomial distribution could be approximated by a normal distribution, so we canuse the sampling distribution of . p

0071.09929.01

)45.2(023094.0

20.02567.0

)2567.0(

zP

zP

pP

It would be almost impossible to find 77 laptops among 300 students, if the claimed percentage ownership was still 20%. Since we found this result, we have evidence tosuggest that the percentage of students with laptop computers is now more than 20%.

14. Use Excel's Histogram tool to create a frequency distribution of 0's and 1's. Theresults are as follows (the total was calculated with an Excel formula).

Bin Frequency 0 25

1 29

Total 54

n = 54 p = 29/54 = 0.537037 p=0.6

Sampling is done without replacement. The sample size is 54, and we have noinformation about the total number of visitors to the winery's retail shop. We can

proceed, first by noting that we are assuming there are at least 5420=1080 visitors tothe winery. We will use the binomial distribution as the underlying model.



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Check conditions:np = 54(0.6) = 32.4 > 10nq = 54(1-0.6) = 21.6 > 10

1725.0

)9444.0(066667.0

60.0537037.0

)537037.0(

zP

zP

pP

(using Excel; the answer is 0.1736 using the tables)

The probability of getting a sample proportion of female visitors as low as 53.7%, if theactual proportion of females is actually 60%, is over 17%. Although the proportion offemales in the sample is lower than expected, it is not unusually low. The sample does

not give us evidence that the proportion of female customers is lower than the owner believes.

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