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    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
    (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 interval for the population mean and
    interpret the results.

    5.
    In a survey of 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 deviation of 2.67 gallons.At can
    you support the society's claim?

    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

    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-test for 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 Testing with 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 sampling distribution 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 plot for 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 determination and interpret the results.
    7. Find the standard error of 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 - 1.34x1 + 1.41x2

    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 median weekly 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

    © BrainMass Inc. brainmass.com April 4, 2020, 12:29 am ad1c9bdddf
    https://brainmass.com/statistics/descriptive-statistics/statistics-data-set-557904

    Solution Preview

    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 ...

    Solution Summary

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

    $2.19

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