1.

The following dataset represents the repair costs (in dollars) for a random

sample of 30 dishwashers. 41.82 52.81 57.80 68.16 73.48 78.88 88.13 88.79

90.07 90.35 91.68 91.72 93.01 95.21 95.34 96.50

100.05 101.32 103.59 104.19 105.62 111.32 117.14 118.42

118.77 119.01 120.70 140.52 141.84 147.06

(a) Find the point estimate of the population mean.

(b) Find the maximum error of estimate for a 95% level of confidence.

(c) Construct a 95% confidence intervalfor the population mean and

interpret the results.

5.

In a surveyof 2000 adults from the U.S. age 65 and over, 1320 received a flu

shot. ( (a) Find a point estimate for the population proportion p of those receiving

flu shots.

(b) Construct a 90% confidence interval for the population proportion.

(c) Find the minimum sample size needed to estimate the population

proportion at the 99% confidence level in order to ensure that the

estimate is accurate within 4% of the population proportion.

6.

Refer to the data set in Exercise 1. Assume the population of dishwasher

repair costs is normally distributed.

(a) Construct a 95% confidence interval for the population variance.

(b) Construct a 95% confidence interval for the population standard

deviation.

32-

Interpreting a Decision In Exercises 29-34, consider each claim. If a

hypothesis test is performed, how should you interpret a decision that

a) rejects the null hypothesis?

(b) fails to reject the null hypothesis?

32. Gas Mileage ◆ An automotive manufacturer claims the standard deviation

for the gas mileage of its models is 3.9 miles per gallon.

37. Writing Hypotheses: Refrigerator Manufacturer ◆ A refrigerator manufacturer

claims that the mean life of its refrigerators is about 15 years.You

are asked to test this claim. How would you write the null hypothesis if

(a) you represent the manufacturer and want to support the claim?

(b) you represent a consumer group and want to reject the claim?

Tea Drinkers ◆ A tea drinker’s society estimates that the mean consumption

of tea by a person in the U.S. is more than 7 gallons per year. In a

sample of 100 people, you find that the mean consumption of tea is 7.8

gallons per year with a standard deviationof 2.67 gallons.At can

you support the society’s claim?

35. Quitting Smoking ◆ The number of years it took a random sampleof 32

former smokers to quit permanently is listed. At = 0.05, test the claim

that the mean time it takes smokers to quit smoking permanently is 15

years.

15.7 13.2 22.6 13.0 10.7 18.1 14.7 7.0 17.3 7.5 21.8

12.3 19.8 13.8 16.0 15.5 13.1 20.7 15.5 9.8 11.9 16.9

7.0 19.3 13.2 14.6 20.9 15.4 13.3 11.6 10.9 21.6

42. Weight Loss ◆ A weight loss program claims that program participants

have a mean weight loss of at least 10 pounds after one month. You work

for a medical association and are asked to test this claim. A random sample

of 30 program participants and their weight losses (in pounds) after one

month is listed below.At do you have enough evidence to reject

the program’s claim?

43. Electric Usage ◆ You believe the mean annual kilowatt usage of U.S.

residential customers is less than 10,000.You do some research and find that

a random sample of 30 residential customers has a mean kilowatt usage of

9900 with a standard deviation of 280. You conduct a statistical experiment

where and At explain why you

cannot reject (Adapted from Edison Electric Institute)

44. Using Different Values of _ and n ◆ In Exercise 43, you believe that

is not valid.Which of the following allows you to reject ?

(a) Use the same values but increase from 0.01 to 0.02.

(b) Use the same values but increase from 0.01 to 0.03.

(c) Use the same values but increase n from 30 to 50.

(d) Use the same values but increase n from 30 to 100.

Extending the Basics

45. Writing ◆ Explain the difference between the classical z-testfor and the

z-test for m using a P-value.

_

a

a

H0

H0

H0 .

H0: m Ú 10,000 Ha: m 6 10,000. a = 0.01,

a = 0.03,

a = 0.06,

348 CHAPTER 7 Hypothesis Testingwith One Sample

Weight Loss (in pounds)

after One Month

5 7 7 Key: 5∣ 7 = 5.7

6 6 7

7 0 1 9

8 2 2 7 9

9 0 3 5 6 8

10 2 5 6 6

11 1 2 5 7 8

12 0 7 8

13 8

14

15 0

WK3

23.

Microwave Repair Costs ◆ A microwave oven repairer says that the

mean repair cost for damaged microwave ovens is less than $100.You work

for the repairer and want to test this claim.You find that a random sample

of five microwave ovens has a mean repair cost of $75 and a standard deviation

of $12.50. At ∝ = 0.01 do you have enough evidence to support the

repairer’s claim?

37)

Deciding on a Distribution In Exercises 37 and 38, decide whether you

should use a normal samplingdistribution or a t-sampling distribution to perform

the hypothesis test. Justify your decision.Then use the distribution to test the claim.

Write a short paragraph about the results of the test and what you can conclude

about the claim.

37. Gas Mileage ◆ A car company says that the mean gas mileage for its

luxury sedan is at least 21 miles per gallon (mpg).You believe the claim is

incorrect and find that a random sample of 5 cars has a mean gas mileage

of 19 mpg and a standard deviation of 4 mpg. Assume the gas mileage of all

of the company’s luxury sedans is normally distributed. At ∝ = 0.05,test

the company’s claim.

10 and 12

Testing Claims In Exercises 10 and 12 , (a) write the claim mathematically and

identify H0 and Ha (b) find the critical values and identify the rejection regions,

(c) find the standardized test statistic, and (d) decide whether to reject or fail to

reject the null hypothesis. Then interpret the decision in the context of the

original claim

10.

Do You Eat Breakfast? ◆ A medical researcher estimates that no more

than 55% of U.S. adults eat breakfast every day. In a random sample of 250

U.S. adults, 56.4% say that they eat breakfast every day. At ∝ = 0.01is

there enough evidence to reject the researcher’s claim?

12.

Just Say No to GMO ◆ An environmentalist claims that more than 50%

of British consumers want supermarkets to stop selling genetically modified

foods.You want to test this claim.You find that in a random sample of 100

British consumers, 53% say that they want supermarkets to stop selling

genetically modified foods. At ∝ = 0.10 can you support the environmentalist’s

claim?

Testing Claims In Exercises 17-26, (a) write the claim mathematically and

identify and (b) find the critical value(s) and identify the rejection region(s),

(c) use the test to find the standardized test statistic, and (d) decide whether to

reject or fail to reject the null hypothesis.Then interpret the decision in the context

of the original claim. Assume the populations are normally distributed.

17.

Life of Appliances ◆ A large appliance company estimates that the

variance of the life of its appliances is 3.You work for a consumer advocacy

group and are asked to test this claim.You find that a random sample of the

lives of 27 of the company’s appliances has a variance of 2.8. At ∝= 0.05,

do you have enough evidence to reject the company’s claim?

26.

Salaries ◆ An employment information service says that the standard

deviation of the annual salaries for public relations managers is at least

$14,500. The annual salaries for 18 randomly chosen public relations managers

are listed. At ∝= 0.10 can you reject the claim?

37,517 50,217 29,177 51,744 69,422 60,770

50,549 50,263 62,939 62,372 65,014 49,164

34,811 55,413 51,310 80,433 34,185 31,805

WK 4

For Exercises 1-8, refer to the data in the following table.The table lists the personal

income and outlays (both in trillions of dollars) for Americans for 11 recent years.

Personal Personal

income, x outlays, y

4.5 3.7

4.9 4.0

5.0 4.1

5.3 4.3

5.6 4.6

5.9 4.8

6.2 5.1

6.5 5.4

7.0 5.7

7.4 6.1

7.8 6.5

1. Construct a scatter plotfor the data. Do the data appear to have a positive

linear correlation, a negative linear correlation, or no linear correlation?

Explain.

2. Calculate the correlation coefficient r.What can you conclude?

3. Test the level of significance of the correlation coefficient Use

4. Find the equation of the regression line for the data. Include the regression

line in the scatter plot.

5. Use the regression line to predict the personal outlays when the personal

income is 5.3 trillion dollars.

6. Find the coefficient of determinationand interpret the results.

7. Find the standard errorof estimate and interpret the results.

8. Construct a 95% prediction interval for personal outlays when personal

income is 6.4 trillion dollars. Interpret the results.

9. The equation used to predict sunflower yield (in pounds) is

y = 1257 – 1.34×1 + 1.41×2

where x1is the number of acres planted (in thousands) and x2 is the number

of acres harvested (in thousands). Use the regression equation to predict the

values for the given values of the independent variables listed below.Then

determine which variable has a greater influence on the value of y

(a) x1 = 2103, x2 = 2037 (b) x1 = 3387, x2 = 3009

(c) x1 = 2185, x2 = 1980 (d) x1 = 3485, x2 = 3404

Wk 5

1 and 3

Graphical Analysis In Exercises 1-3, use the graph below to answer the

question.

1.

Use the graph to describe the total variation about a regression line in words

and in symbols.

3.

Use the graph to describe the unexplained variation about a regression line

in words and in symbols.

11)

11. Retail Space and Sales ◆ The following table represents the total square

footage (in billions) of retailing space at shopping centers and their sales (in

billions of U.S. dollars) for 11 years.The equation of the regression line is

Total square footage, x 1.6 2.3 3.0 3.4 3.9 4.6

Sales, y 123.2 211.5 385.5 475.1 641.1 716.9

Total square footage, x 4.7 4.8 4.9 5.0 5.1

Sales, y 768.2 806.6 851.3 893.8 933.9

13)

Earnings of Men and Women ? The following table represents medianweekly earnings (in U.S. dollars) of full-time male and female workers for five years. The equation of the regression line is

(Source: U.S. Bureau of Labor Statistics)

Median weekly earnings of male workers,x 312 419 485 538 557

Median weekly earnings of female workers,y 201 290 348 406 418

15)

Campaign Money ? The money raised and spent (both in millions of U.S. dollars) by all congressional campaigns for eight recent years are shown in the table. The data can be modeled by the regression equation = 1.020x – 25.854. (Source: Federal Election Commission)

Money raised,x 354.7 397.2 472.0 477.6

Money spent,y 342.4 374.1 450.9 459.0

Money raised,x 471.7 659.3 740.5 790.5

Money spent,y 446.3 680.2 725.2 765.3

16)

9. Fund Assets ? The following table represents the total assets (in billions of U.S. dollars) of equity funds and bond and income funds for nine years. The equation of the regression line is = 0.689x + 68.861. (Source: Investment Company Institute)

Equity funds,x 35.9 41.2 77.0 116.9 180.7

Bond and income funds,y 13.1 14 36.6 134.8 273.1

Equity funds,x 249.0 411.6 749.0 1269.0

Bond and income funds,y 304.8 441.4 761.1 798.3

Old Vehicles In Exercises 25-31, use

the information given at the right.

25. Scatter Plot ◆ Construct a scatter

plot of the data. Show and on the

graph.

26. Regression Line ◆ Find and graph

the regression line.

27. Deviation ◆ Calculate the explained

deviation, the unexplained deviation,

and the total deviation for each data

point.

28. Variation ◆ Find the (a) explained variation, (b) unexplained variation,

and (c) total variation.

29. Coefficient of Determination ◆ Find the coefficient of determination.

What can you conclude?

30. Error of Estimate ◆ Find the standard error of estimate and interpret

the results.

31. Prediction Interval ◆ Construct a 95% prediction interval for the median

age of trucks in use when the median age of cars in use is 7.3.

32. Correlation Coefficient and Slope ◆ Recall that the formula for the

correlation coefficient r is and the formula for the slope m of a regression line is

Given a set of data, why must the slope m of the data’s regression line

always have the same sign as the data’s correlation coefficient r?

m =

ng xy – 1g x21g y2

ng x2 – 1g x22 .

r =

ng xy – 1g x21g y2

2ng x2 – 1g x222ng y2 – 1g y22

se

y x

c = 0.90.

c = 0.95.

Trucks, y

4.9

5.4

6.0

6.9

6.5

7.7

8.1

5.9

5.8

6.3

7.6

6.5

7.6

7.8

Cars, x

Keeping cars longer.

The median age of vehicles on U.S.

roads for seven different years:

Median age in years

Please see the attachment.

1. The following data set represents the repair costs (in dollars) for a random

sample of 30 dishwashers. 41.82 52.81 57.80 68.16 73.48 78.88 88.13 88.79

90.07 90.35 91.68 91.72 93.01 95.21 95.34 96.50 100.05 101.32 103.59 104.19 105.62 111.32 117.14 118.42 118.77 119.01 120.70 140.52 141.84 147.06

Answers

(a) Find the point estimate of the population mean.

Point estimate of the population mean = sample mean

=

= 98.11

(b) Find the maximum error of estimate for a 95% level of confidence.

Maximum error of estimate, E = , where = 2.045229611, s = 24.72242414, n = 30

Therefore, E = = 9.231504748

Details

Confidence Interval Estimate for the Mean

Data

Sample Standard Deviation 24.72242414

Sample Mean 98.11

Sample Size 30

Confidence Level 95%

Intermediate Calculations

Standard Error of the Mean 4.513676459

Degrees of Freedom 29

t Value 2.045229611

Margin of Error 9.231504748

(c) Construct a 95% confidence interval for the population mean and

interpret the results.

95% confidence interval for the population mean = (Sample mean ± E)

= (98.11 ± 9.231504748)

= (88.88. 107.34)

Thus with 95% confidence we can claim that population mean is within (88.88. 107.34).

Details

Confidence Interval Estimate for the Mean

Data

Sample Standard Deviation 24.72242414

Sample Mean 98.11

Sample Size 30

Confidence Level 95%

Intermediate Calculations

Standard Error of the Mean 4.513676459

Degrees of Freedom 29

t Value 2.045229611

Margin of Error 9.231504748

Confidence Interval

Interval Lower Limit 88.88

Interval Upper Limit 107.34

5. In a survey of 2000 adults from the U.S. age 65 and over, 1320 received a flu

shot.

Answers

(a) Find a point estimate for the population proportion p of those receiving flu shots.

Point estimate for the population proportion p of those receiving flu shots

= Sample proportion, p

= 1320/2000

= 0.66

(b) Construct a 90% confidence interval for the population proportion.

90% Confidence Interval for proportion is given by

where p = 0.66, = 1.644853627, n = 2000

= (0.6426, 0.6774)

Thus with 95% confidence we can claim that the proportion of those receiving flu shots is within (0.6426, 0.6774).

Details

Confidence Interval Estimate for the Proportion

Data

Sample Size 2000

Number of Successes 1320

Confidence Level 90%

Intermediate Calculations

Sample Proportion 0.66

Z Value -1.644853627

Standard Error of the Proportion 0.010592450

Interval Half Width 0.017423030

Confidence Interval

Interval Lower Limit 0.6426

Interval Upper Limit 0.6774

(c) Find the minimum sample size needed to estimate the population

proportion at the 99% confidence level in order to ensure that the

estimate is accurate within 4% of the population proportion.

The sample size is given by where p = 0.66

Given that E = 4% = 0.04, = 2.575829304

Therefore, sample size,

That is, n > 930.5442

Hence the minimum sample size required is n = 931

Details

Sample Size Determination

Data

Estimate of True Proportion 0.66

Sampling Error 0.04

Confidence Level 99%

Intermediate Calculations

Z Value -2.575829304

Calculated Sample Size 930.5442

Result

Sample Size Needed 931

6. Refer to the data set in Exercise 1. Assume the population of dishwasher

repair costs is normally distributed.

41.82 52.81 57.80 68.16 73.48 78.88 88.13 88.79

90.07 90.35 91.68 91.72 93.01 95.21 95.34 96.50

100.05 101.32 103.59 104.19 105.62 111.32 117.14 118.42

118.77 119.01 120.70 140.52 141.84 147.06

Answers

(a) Construct a 95% confidence interval for the population variance.

95% confidence interval for the population variance is given by,

, where n = 32, s = 24.72242414, = 48.23188958, = 17.53873879

That is,

= (392.83, 1080.30)

Details

Confidence Interval Estimate for the Population Variance

Data

Sample Size 32

Sample Standard Deviation 24.72242414

Confidence Level 95%

Intermediate Calculations

Degrees of Freedom 31

Sum of Squares 18947.14592

Single Tail Area 0.025

Lower Chi-Square Value 17.53873879

Upper Chi-Square Value 48.23188958

Results

Interval Lower Limit for Variance 392.8344106

Interval Upper Limit for Variance 1080.302645

(b) Construct a 95% confidence interval for the population standard

deviation.

95% confidence interval for the population standard deviation is given by,

, where n = 32, s = 24.72242414, = 48.23188958, = 17.53873879

That is,

= (19.82, 32.87)

Details

Confidence Interval Estimate for the Population Variance

Data

Sample Size 32

Sample Standard Deviation 24.72242414

Confidence Level 95%

Intermediate Calculations

Degrees of Freedom 31

Sum of Squares 18947.14592

Single Tail Area 0.025

Lower Chi-Square Value 17.53873879

Upper Chi-Square Value 48.23188958

Results

Interval Lower Limit for Variance 392.8344106

Interval Upper Limit for Variance 1080.302645

Interval Lower Limit for Standard Deviation 19.82005072

Interval Upper Limit for Standard Deviation 32.86795773

32. Interpreting a Decision In Exercises 29-34, consider each claim. If a

hypothesis test is performed, how should you interpret a decision that

a) Rejects the null hypothesis?

b) Fails to reject the null hypothesis?

32. Gas Mileage

An automotive manufacturer claims the standard deviation for the gas mileage of its models is 3.9 miles per gallon.

Answers

a) The sample provides enough evidence to reject the claim that the standard deviation for the gas mileage of its models is 3.9 miles per gallon.

b) The sample does not provide enough evidence to reject the claim that the standard deviation for the gas mileage of its models is 3.9 miles per gallon.

37. Writing Hypotheses: Refrigerator Manufacturer ◆ A refrigerator manufacturer claims that the mean life of its refrigerators is about 15 years. You

are asked to test this claim. How would you write the null hypothesis if

Answers

(a) You represent the manufacturer and want to support the claim?

The null hypothesis tested is

H0: The mean life of refrigerators = 15 years (µ = 15)

(b) You represent a consumer group and want to reject the claim?

The null hypothesis tested is

H0: The mean life of refrigerators ≤ 15 years (µ ≤ 15)

Tea Drinkers ◆ A tea drinker’s society estimates that the mean consumption

of tea by a person in the U.S. is more than 7 gallons per year. In a

sample of 100 people, you find that the mean consumption of tea is 7.8

gallons per year with a standard deviation of 2.67 gallons. At can

you support the society’s claim?

Answer

The null hypothesis tested is

H0: Mean consumption of tea by a person in the U.S. ≤ 7 gallons per year. (µ ≤ 7)

The alternative hypothesis is

H1: Mean consumption of tea by a person in the U.S. > 7 gallons per year. (µ > 7)

Significance level = 0.05

The test statistic used is , where =7.8, n = 100, = 2.67

Therefore, = 2.996254682

Rejection criteria: Reject the null hypothesis, if the calculated value of test statistic is greater than the critical value at the 0.05 significance level.

Upper critical value = 1.644853627

Conclusion: Reject the null hypothesis, since the calculated value of test statistic is greater than the critical value. The sample provides enough evidence to support the claim that the mean consumption of tea by a person in the U.S. is more than 7 gallons per year.

Details

Z Test of Hypothesis for the Mean

Data

Null Hypothesis = 7

Level of Significance 0.05

Population Standard Deviation 2.67

Sample Size 100

Sample Mean 7.8

Intermediate Calculations

Standard Error of the Mean 0.267000000

Z Test Statistic 2.996254682

Upper-Tail Test

Upper Critical Value 1.644853627

p-Value 0.001366590

Reject the null hypothesis

35. Quitting Smoking ◆ The number of years it took a random sample of 32 former smokers to quit permanently is listed. At = 0.05, test the claim that the mean time it takes smokers to quit smoking permanently is 15 years.

15.7 13.2 22.6 13.0 10.7 18.1 14.7 7.0 17.3 7.5 21.8 12.3 19.8 13.8 16.0 15.5 13.1 20.7 15.5 9.8 11.9 16.9 7.0 19.3 13.2 14.6 20.9 15.4 13.3 11.6 10.9 21.6

Answer

The null hypothesis tested is

H0: Mean time it takes smokers to quit smoking permanently =15 years (µ= 15)

The alternative hypothesis is

H1: Mean time it takes smokers to quit smoking permanently ≠15 years (µ≠ 15)

Significance level = 0.05

Test Statistic used is . Given that = 14.834375, n = 32, s = 4.287612267

Therefore, = -0.218517074

Decision rule: Reject the null hypothesis, if the absolute value of calculated test statistic is greater than the critical value of t with 31 d.f. at the significance level 0.05.

Critical values = ±2.039513438

Conclusion: Fails to reject the null hypothesis, since the absolute value of calculated test statistic is less than the critical value of t. The sample does not provide enough evidence to reject the claim that the mean time it takes smokers to quit smoking permanently is 15 years.

Details

t Test for Hypothesis of the Mean

Data

Null Hypothesis = 15

Level of Significance 0.05

Sample Size 32

Sample Mean 14.834375

Sample Standard Deviation 4.287612267

Intermediate Calculations

Standard Error of the Mean 0.757949927

Degrees of Freedom 31

t Test Statistic -0.218517074

Two-Tail Test

Lower Critical Value -2.039513438

Upper Critical Value 2.039513438

p-Value 0.828458472

Do not reject the null hypothesis

42. Weight Loss ◆ A weight loss program claims that program participants

have a mean weight …

The statistics set representations for repair costs are provided. The maximum errors of estimated are found. The levels of confidence are given.