1.
The random variable is the mean Internet speed in megabits per second.
3.
The random variable is the mean number of children an American family has.
5.
The random variable is the proportion of people picked at random in Times Square visiting the city.
7.
 H_{0}: p = 0.42
 H_{a}: p < 0.42
9.
 H_{0}: μ = 15
 H_{a}: μ ≠ 15
11.
Type I: The mean price of midsized cars is $32,000, but we conclude that it is not $32,000.
Type II: The mean price of midsized cars is not $32,000, but we conclude that it is $32,000.
13.
α = the probability that you think the bag cannot withstand –15 degrees F, when, in fact, it can.
β = the probability that you think the bag can withstand –15 degrees F, when, in fact, it cannot.
15.
Type I: The procedure will go well, but the doctors think it will not.
Type II: The procedure will not go well, but the doctors think it will.
17.
0.019
19.
0.998
21.
A normal distribution or a Student’s tdistribution
23.
Use a Student’s tdistribution
25.
a normal distribution for a single population mean
27.
It must be approximately normally distributed.
29.
They must both be greater than five.
31.
binomial distribution
33.
The outcome of winning is very unlikely.
35.
H_{0}: μ > = 73
H_{a}: μ < 73
The pvalue is almost zero, which means there is sufficient data to conclude that the mean height of high school students who play basketball on the school team is less than 73 inches at the 5 percent level. The data do support the claim.
37.
The shaded region shows a low pvalue.
39.
Do not reject H_{0}.
41.
means
43.
the mean time spent on homework for 26 students
45.
 3
 1.5
 1.8
 26
47.
$\overline{X}~N\left(2.5,\frac{1.5}{\sqrt{26}}\right)$
49.
This is a lefttailed test.
51.
This is a twotailed test.
53.
Figure 9.23
55.
a righttailed test
57.
a lefttailed test
59.
This is a lefttailed test.
61.
This is a twotailed test.
62.
 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
64.
c
66.
 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 percent of Americans vote in presidential elections, when the actual percentage is at most 60 percent.Type II error: We conclude that at most 60 percent of Americans vote in presidential elections when, in fact, more than 60 percent 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 take physical education daily is not 29%, when it really is 29%. Type II error: We conclude that the proportion of high school seniors who take physical education daily is 29% when, in fact, it is not 29%.
 Type I error: We conclude that fewer than 5 percent of adults ride the bus to work in Los Angeles, when the percentage that do is really 29%. Type II error: We conclude that 29%. or more adults ride the bus to work in Los Angeles when, in fact, fewer that 29% 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 percent, when it is really at least 11 percent. Type II error: We conclude that the proportion of women who develop breast cancer is at least 11 percent, when in fact it is less than 11 percent.
 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 when, in fact, it is more than $20,000.
68.
b
70.
d
72.
d
74.
 H_{0}: μ ≥ 50,000
 H_{a}: μ < 50,000
 Let $\overline{X}$ = 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)
75.
 H_{0}: μ ≥ 35.5
 H_{a}: μ < 35.5
 Let $\overline{x}$ = the average mpg for the sample of cars and trucks in the fleet
 normal distribution
 z = 0.648
 pvalue = 0.2578
 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 sufficient evidence to support the claim that the manufacturer’s fleet meets the fuel economy standards in the 2016 policy.
 (31.88 mpg, 37.32 mpg)
76.
 H_{0}: μ = $1.00
 H_{a}: μ ≠ $1.00
 Let $$\overline{x}$$ = 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, $1.06)
78.
 H_{0}: μ = 10
 H_{a}: μ ≠ 10
 Let $\overline{X}$ = 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 percent significance level, there is insufficient evidence to conclude that the mean number of sick days is not 10.
 (4.9443, 11.806)
80.
 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)
82.
 H_{0}: μ = 4
 H_{a}: μ ≠ 4
 Let $\overline{X}$ 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)
84.
 H_{0}: p ≥ 0.13
 H_{a}: p < 0.13
 Let P′ = the proportion of Americans who have the disease
 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 been diagnosed with the disease is less than 13 percent.
 (0, 0.0623).
The plus4s confidence interval is (0.0022, 0.0978)
86.
 H_{0}: μ ≥ 129
 H_{a}: μ < 129
 Let $\overline{X}$ = the average time in seconds that Terri finishes Lap 4.
 Student's tdistribution
 t = 1.209
 0.8792
 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 Terri’s mean lap time is less than 129 seconds.
 (128.63, 130.37)
88.
 H_{0}: p = 0.60
 H_{a}: p < 0.60
 Let P′ = the proportion of family members who shed tears at a reunion.
 normal for a single proportion
 –1.71
 0.0438
 Check student’s solution.
 Alpha: 0.05
 Decision: Reject the null hypothesis.
 Reason for decision: pvalue < alpha
 Conclusion: At the 5 percent significance level, there is sufficient evidence to conclude that the proportion of family members who shed tears at a reunion is less than 0.60. However, the test is weak because the pvalue and alpha are quite close, so other tests should be done.
 We are 95 percent confident that between 38.29 percent and 61.71 percent of family members will shed tears at a family reunion. (0.3829, 0.6171). The plus4s confidence interval (see chapter 8) is (0.3861, 0.6139)
Note that here the largesample 1 – PropZTest provides the approximate pvalue of 0.0438. Whenever a pvalue based on a normal approximation is close to the level of significance, the exact pvalue based on binomial probabilities should be calculated whenever possible. This is beyond the scope of this course.
89.
 H_{0}: μ ≥ 22
 H_{a}: μ < 22
 Let $\overline{X}$ = the mean number of bubbles per blow.
 Student's tdistribution
 –2.667
 pvalue = 0.00486
 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 number of bubbles per blow is less than 22.
 (18.501, 21.499)
91.
 H_{0}: μ ≤ 1
 H_{a}: μ > 1
 Let $\overline{X}$ = the mean cost in dollars of macaroni and cheese in a certain town.
 Student's tdistribution
 t = 0.340
 pvalue = 0.36756
 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: The mean cost could be $1, or less. At the 5 percent significance level, there is insufficient evidence to conclude that the mean price of a box of macaroni and cheese is more than $1.
 (0.8291, 1.241)
93.
 H_{0}: p = 0.01
 H_{a}: p > 0.01
 Let P′ = the proportion of errors generated
 Normal for a single proportion
 2.13
 0.0165
 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 percent significance level, there is sufficient evidence to conclude that the proportion of errors generated is more than 0.01.
 Confidence interval: (0, 0.094).
The plus4s confidence interval is (0.004, 0.144).
95.
 H_{0}: p = 0.50
 H_{a}: p < 0.50
 Let P′ = the proportion of friends that has a pierced ear.
 normal for a single proportion
 –1.70
 pvalue = 0.0448
 Check student’s solution.
 Alpha: 0.05
 Decision: Reject the null hypothesis.
 Reason for decision: The pvalue is less than 0.05. (However, they are very close.)
 Conclusion: There is sufficient evidence to support the claim that less than 50 percent of his friends have pierced ears.
 Confidence interval: (0.245, 0.515): The plus4s confidence interval is (0.259, 0.519).
97.
 H_{0}: p = 0.40
 H_{a}: p < 0.40
 Let P′ = the proportion of schoolmates who fear public speaking.
 normal for a single proportion
 –1.01
 pvalue = 0.1563
 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 support the claim that less than 40 percent of students at the school fear public speaking.
 Confidence interval: (0.3241, 0.4240): The plus4s confidence interval is (0.3257, 0.4250).
99.
 H_{0}: p = 0.14
 H_{a}: p < 0.14
 Let P′ = the proportion of nursing home residents that have the disease.
 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 percent significance level, there is insufficient evidence to conclude that the proportion of nursing home residents that have the disease is less than 0.14.
 Confidence interval: (0.0502, 0.2070): The plus4s confidence interval (see chapter 8) is (0.0676, 0.2297).
101.
 H_{0}: μ = 69,110
 H_{a}: μ > 69,110
 Let $\overline{X}$ = 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 percent significance level, there is sufficient evidence to conclude that the mean salary of California registered nurses exceeds $69,110.
 ($68,757, $73,485)
103.
 H_{0}: p ≥ 0.14, H_{a}: p < 0.14
 pvalue < 0.0002
 Alpha: 0.05
 Reject the null hypothesis.
 At the 5 percent significance level, there is sufficient evidence to conclude that the proportion of Harleys stolen is significantly less than their share of all motorcycles. (conclusion a)
105.
c
107.
 H_{0}: p = 0.488 H_{a}: p ≠ 0.488
 pvalue = 0.0114
 alpha = 0.05
 Reject the null hypothesis.
 At the 5 percent level of significance, there is enough evidence to conclude that 48.8 percent of families own stocks.
 The survey does not appear to be accurate.
109.
 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 percent 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.
111.
 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 percent 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 percent because the pvalue is almost 0.
113.
 H_{0}: µ ≤ 5.8H_{a}: µ > 5.8
 pvalue = 0.9987
 alpha = 0.05
 Do not reject the null hypothesis.
 At the 5 percent 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.
115.
 H_{0}: µ ≥ 150H_{a}: µ < 150
 pvalue = 0.0622
 alpha = 0.01
 Do not reject the null hypothesis.
 At the 1 percent 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.