7.12: Chapter 9 Solution (Practice + Homework)
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 51832
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left#1\right}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)38. a righttailed test
40. a lefttailed test
42. This is a lefttailed test.
44. This is a twotailed test.
45.
 H_{0}: μ = 34; H_{a}: μ ≠ 34
 H_{0}: p ≤ 0.60; H_{a}: p > 0.60
 H_{0}: μ ≥ 100,000; H_{a}: μ < 100,000
 H_{0}: p = 0.29; H_{a}: p ≠ 0.29
 H_{0}: p = 0.05; H_{a}: p < 0.05
 H_{0}: μ ≤ 10; H_{a}: μ > 10
 H_{0}: p = 0.50; H_{a}: p ≠ 0.50
 H_{0}: μ = 6; H_{a}: μ ≠ 6
 H_{0}: p ≥ 0.11; H_{a}: p < 0.11
 H_{0}: μ ≤ 20,000; H_{a}: μ > 20,000
47.c
 Type I error: We conclude that the mean is not 34 years, when it really is 34 years. Type II error: We conclude that the mean is 34 years, when in fact it really is not 34 years.
 Type I error: We conclude that more than 60% of Americans vote in presidential elections, when the actual percentage is at most 60%.Type II error: We conclude that at most 60% of Americans vote in presidential elections when, in fact, more than 60% do.
 Type I error: We conclude that the mean starting salary is less than $100,000, when it really is at least $100,000. Type II error: We conclude that the mean starting salary is at least $100,000 when, in fact, it is less than $100,000.
 Type I error: We conclude that the proportion of high school seniors who get drunk each month is not 29%, when it really is 29%. Type II error: We conclude that the proportion of high school seniors who get drunk each month is 29% when, in fact, it is not 29%.
 Type I error: We conclude that fewer than 5% of adults ride the bus to work in Los Angeles, when the percentage that do is really 5% or more. Type II error: We conclude that 5% or more adults ride the bus to work in Los Angeles when, in fact, fewer that 5% do.
 Type I error: We conclude that the mean number of cars a person owns in his or her lifetime is more than 10, when in reality it is not more than 10. Type II error: We conclude that the mean number of cars a person owns in his or her lifetime is not more than 10 when, in fact, it is more than 10.
 Type I error: We conclude that the proportion of Americans who prefer to live away from cities is not about half, though the actual proportion is about half. Type II error: We conclude that the proportion of Americans who prefer to live away from cities is half when, in fact, it is not half.
 Type I error: We conclude that the duration of paid vacations each year for Europeans is not six weeks, when in fact it is six weeks. Type II error: We conclude that the duration of paid vacations each year for Europeans is six weeks when, in fact, it is not.
 Type I error: We conclude that the proportion is less than 11%, when it is really at least 11%. Type II error: We conclude that the proportion of women who develop breast cancer is at least 11%, when in fact it is less than 11%.
 Type I error: We conclude that the average tuition cost at private universities is more than $20,000, though in reality it is at most $20,000. Type II error: We conclude that the average tuition cost at private universities is at most $20,000
51. b
55. d
56.
 H_{0}: μ ≥ 50,000
 H_{a}: μ < 50,000
 Let X¯¯¯¯ = the average lifespan of a brand of tires.
 normal distribution
 z = 2.315
 pvalue = 0.0103
 Check student’s solution.

 alpha: 0.05
 Decision: Reject the null hypothesis.
 Reason for decision: The pvalue is less than 0.05.
 Conclusion: There is sufficient evidence to conclude that the mean lifespan of the tires is less than 50,000 miles.
 (43,537, 49,463)
58.
 H_{0}: μ = $1.00
 H_{a}: μ ≠ $1.00
 Let X¯¯¯¯ = the average cost of a daily newspaper.
 normal distribution
 z = –0.866
 pvalue = 0.3865
 Check student’s solution.

 Alpha: 0.01
 Decision: Do not reject the null hypothesis.
 Reason for decision: The pvalue is greater than 0.01.
 Conclusion: There is sufficient evidence to support the claim that the mean cost of daily papers is $1. The mean cost could be $1.
 ($0.84,
60.
 H_{0}: μ = 10
 H_{a}: μ ≠ 10
 Let X¯¯¯¯ the mean number of sick days an employee takes per year.
 Student’s tdistribution
 t = –1.12
 pvalue = 0.300
 Check student’s solution.

 Alpha: 0.05
 Decision: Do not reject the null hypothesis.
 Reason for decision: The pvalue is greater than 0.05.
 Conclusion: At the 5% significance level, there is insufficient evidence to conclude that the mean number of sick days is not ten.
 (4.9443, 11.806)
62.
 H_{0}: p ≥ 0.6
 H_{a}: p < 0.6
 Let P′ = the proportion of students who feel more enriched as a result of taking Elementary Statistics.
 normal for a single proportion
 1.12
 pvalue = 0.1308
 Check student’s solution.

 Alpha: 0.05
 Decision: Do not reject the null hypothesis.
 Reason for decision: The pvalue is greater than 0.05.
 Conclusion: There is insufficient evidence to conclude that less than 60 percent of her students feel more enriched.
 Confidence Interval: (0.409, 0.654)
The “plus4s” confidence interval is (0.411, 0.648)
64.
 H_{0}: μ = 4
 H_{a}: μ ≠ 4
 Let X¯¯¯¯ the average I.Q. of a set of brown trout.
 twotailed Student's ttest
 t = 1.95
 pvalue = 0.076
 Check student’s solution.

 Alpha: 0.05
 Decision: Reject the null hypothesis.
 Reason for decision: The pvalue is greater than 0.05
 Conclusion: There is insufficient evidence to conclude that the average IQ of brown trout is not four.
 (3.8865,5.9468)
66.
 H_{0}: p ≥ 0.13
 H_{a}: p < 0.13
 Let P′ = the proportion of Americans who have seen or sensed angels
 normal for a single proportion
 –2.688
 pvalue = 0.0036
 Check student’s solution.

 alpha: 0.05
 Decision: Reject the null hypothesis.
 Reason for decision: The pvalue is less than 0.05.
 Conclusion: There is sufficient evidence to conclude that the percentage of Americans who have seen or sensed an angel is less than 13%.
 (0, 0.0623).
The“plus4s” confidence interval is (0.0022, 0.0978)
69.
 H_{0}: p = 0.14
 H_{a}: p < 0.14
 Let P′ = the proportion of NYC residents that smoke.
 normal for a single proportion
 –0.2756
 pvalue = 0.3914
 Check student’s solution.

 alpha: 0.05
 Decision: Do not reject the null hypothesis.
 Reason for decision: The pvalue is greater than 0.05.
 At the 5% significance level, there is insufficient evidence to conclude that the proportion of NYC residents who smoke is less than 0.14.
 Confidence Interval: (0.0502, 0.2070): The “plus4s” confidence interval (see chapter 8) is (0.0676, 0.2297).
71.
 H_{0}: μ = 69,110
 H_{a}: μ > 69,110
 Let X¯¯¯¯ = the mean salary in dollars for California registered nurses.
 Student's tdistribution
 t = 1.719
 pvalue: 0.0466
 Check student’s solution.

 Alpha: 0.05
 Decision: Reject the null hypothesis.
 Reason for decision: The pvalue is less than 0.05.
 Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the mean salary of California registered nurses exceeds $69,110.
 ($68,757, $73,485)
73. c
75. c
77.
 H_{0}: p = 0.488 H_{a}: p ≠ 0.488
 pvalue = 0.0114
 alpha = 0.05
 Reject the null hypothesis.
 At the 5% level of significance, there is enough evidence to conclude that 48.8% of families own stocks.
 The survey does not appear to be accurate.
79.
 H_{0}: p = 0.517 H_{a}: p ≠ 0.517
 pvalue = 0.9203.
 alpha = 0.05.
 Do not reject the null hypothesis.
 At the 5% significance level, there is not enough evidence to conclude that the proportion of homes in Kentucky that are heated by natural gas is 0.517.
 However, we cannot generalize this result to the entire nation. First, the sample’s population is only the state of Kentucky. Second, it is reasonable to assume that homes in the extreme north and south will have extreme high usage and low usage, respectively. We would need to expand our sample base to include these possibilities if we wanted to generalize this claim to the entire nation.
81.
 H_{0}: µ ≥ 11.52 H_{a}: µ < 11.52
 pvalue = 0.000002 which is almost 0.
 alpha = 0.05.
 Reject the null hypothesis.
 At the 5% significance level, there is enough evidence to conclude that the mean amount of summer rain in the northeaster US is less than 11.52 inches, on average.
 We would make the same conclusion if alpha was 1% because the pvalue is almost 0.
83.
 H_{0}: µ ≤ 5.8 H_{a}: µ > 5.8
 pvalue = 0.9987
 alpha = 0.05
 Do not reject the null hypothesis.
 At the 5% level of significance, there is not enough evidence to conclude that a woman visits her doctor, on average, more than 5.8 times a year.
85.
 \(H_0: \mu \geq 150 H_a: \mu < 150\)
 \(p\)value = 0.0622
 alpha = 0.01
 Do not reject the null hypothesis.
 At the 1% significance level, there is not enough evidence to conclude that freshmen students study less than 2.5 hours per day, on average.
 The student academic group’s claim appears to be correct.