Question 1 1.   Assume that the data has a normal distribution and the number of observations is greater than fifty. Find the critical z value used to test a null hypothesis. = 0.09 for a right-tailed test. Answer ±1.96 1.34 ±1.34 1.96 5 points   Question 2 1.   Find the value of the test statistic z using z = The claim is that the proportion of drowning deaths of children attributable to beaches is more than 0.25, and the sample statistics include n = 681 drowning deaths of children with 30% of them attributable to beaches. Answer 3.01 2.85 -2.85 -3.01 5 points   Question 3 1.   Use the given information to find the P-value. Also, use a 0.05 significance level and state the conclusion about the null hypothesis (reject the null hypothesis or fail to reject the null hypothesis). The test statistic in a left-tailed test is z = -1.83. Answer 0.0672; reject the null hypothesis 0.0336; reject the null hypothesis 0.9664; fail to reject the null hypothesis 0.0672; fail to reject the null hypothesis 5 points   Question 4 1.   Use the given information to find the P-value. Also, use a 0.05 significance level and state the conclusion about the null hypothesis (reject the null hypothesis or fail to reject the null hypothesis). With H1: p < 3/5, the test statistic is z = -1.68. Answer 0.093; fail to reject the null hypothesis 0.0465; fail to reject the null hypothesis 0.0465; reject the null hypothesis 0.9535; fail to reject the null hypothesis 5 points   Question 5 1.   Formulate the indicated conclusion in nontechnical terms. Be sure to address the original claim. The owner of a football team claims that the average attendance at games is over 694, and he is therefore justified in moving the team to a city with a larger stadium. Assuming that a hypothesis test of the claim has been conducted and that the conclusion is failure to reject the null hypothesis, state the conclusion in nontechnical terms. Answer There is not sufficient evidence to support the claim that the mean attendance is less than 694. There is sufficient evidence to support the claim that the mean attendance is greater than 694. There is sufficient evidence to support the claim that the mean attendance is less than 694. There is not sufficient evidence to support the claim that the mean attendance is greater than 694. 5 points   Question 6 1.   Assume that a hypothesis test of the given claim will be conducted. Identify the type I or type II error for the test. A consumer advocacy group claims that the mean mileage for the Carter Motor Company's new sedan is less than 32 miles per gallon. Identify the type I error for the test. Answer Fail to reject the claim that the mean is equal to 32 miles per gallon when it is actually greater than 32 miles per gallon. Reject the claim that the mean is equal to 32 miles per gallon when it is actually less than 32 miles per gallon. Reject the claim that the mean is equal to 32 miles per ...

1. Question 1
1.
Assume that the data has a normal distribution and the number
of observations is greater than fifty. Find the critical z value
used to test a null hypothesis. = 0.09 for a right-tailed test.
Answer
±1.96
1.34
±1.34
1.96
5 points
Question 2
1.
Find the value of the test statistic z using z = The claim is that
the proportion of drowning deaths of children attributable to
beaches is more than 0.25, and the sample statistics include n =
681 drowning deaths of children with 30% of them attributable
to beaches.
Answer
3.01
2.85

2. -2.85
-3.01
5 points
Question 3
1.
Use the given information to find the P-value. Also, use a 0.05
significance level and state the conclusion about the null
hypothesis (reject the null hypothesis or fail to reject the null
hypothesis). The test statistic in a left-tailed test is z = -1.83.
Answer
0.0672; reject the null hypothesis
0.0336; reject the null hypothesis
0.9664; fail to reject the null hypothesis
0.0672; fail to reject the null hypothesis
5 points
Question 4
1.
Use the given information to find the P-value. Also, use a 0.05
significance level and state the conclusion about the null
hypothesis (reject the null hypothesis or fail to reject the null
hypothesis). With H1: p < 3/5, the test statistic is z = -1.68.
Answer
0.093; fail to reject the null hypothesis

3. 0.0465; fail to reject the null hypothesis
0.0465; reject the null hypothesis
0.9535; fail to reject the null hypothesis
5 points
Question 5
1.
Formulate the indicated conclusion in nontechnical terms. Be
sure to address the original claim. The owner of a football team
claims that the average attendance at games is over 694, and he
is therefore justified in moving the team to a city with a larger
stadium. Assuming that a hypothesis test of the claim has been
conducted and that the conclusion is failure to reject the null
hypothesis, state the conclusion in nontechnical terms.
Answer
There is not sufficient evidence to support the claim that the
mean attendance is less than 694.
There is sufficient evidence to support the claim that the mean
attendance is greater than 694.
There is sufficient evidence to support the claim that the mean
attendance is less than 694.
There is not sufficient evidence to support the claim that the
mean attendance is greater than 694.

4. 5 points
Question 6
1.
Assume that a hypothesis test of the given claim will be
conducted. Identify the type I or type II error for the test. A
consumer advocacy group claims that the mean mileage for the
Carter Motor Company's new sedan is less than 32 miles per
gallon. Identify the type I error for the test.
Answer
Fail to reject the claim that the mean is equal to 32 miles per
gallon when it is actually greater than 32 miles per gallon.
Reject the claim that the mean is equal to 32 miles per gallon
when it is actually less than 32 miles per gallon.
Reject the claim that the mean is equal to 32 miles per gallon
when it is actually 32 miles per gallon.
Fail to reject the claim that the mean is equal to 32 miles per
gallon when it is actually less than 32 miles per gallon.
5 points
Question 7
1.
Find the P-value for the indicated hypothesis test. In a sample
of 88 children selected randomly from one town, it is found that
8 of them suffer from asthma. Find the P-value for a test of the
claim that the proportion of all children in the town who suffer
from asthma is equal to 11%.
Answer

5. 0.2843
-0.2843
0.2157
0.5686
5 points
Question 8
1.
Find the P-value for the indicated hypothesis test. An article in
a journal reports that 34% of American fathers take no
responsibility for child care. A researcher claims that the figure
is higher for fathers in the town of Littleton. A random sample
of 225 fathers from Littleton, yielded 97 who did not help with
child care. Find the P-value for a test of the researcher's claim.
Answer
0.0019
0.0015
0.0038
0.0529
5 points
Question 9
1.
Find the critical value or values of based on the given

6. information. H1: > 3.5 n = 14 = 0.05
Answer
22.362
5.892
24.736
23.685
5 points
Question 10
1.
Find the critical value or values of based on the given
information. H1: > 26.1 n = 9 = 0.01
Answer
1.646
21.666
20.090
2.088
5 points
Question 11
1.
Find the number of successes x suggested by the given

7. statement. A computer manufacturer randomly selects 2850 of
its computers for quality assurance and finds that 1.79% of
these computers are found to be defective.
Answer
51
56
54
49
5 points
Question 12
1.
Assume that you plan to use a significance level of alpha = 0.05
to test the claim that p1 = p2, Use the given sample sizes
andnumbers of successes to find the pooled estimate Round
your answer to the nearest thousandth. n1 = 570;n2 = 1992 x1 =
143;x2 = 550
Answer
0.541
0.270
0.520

8. 0.216
5 points
Question 13
1.
Assume that you plan to use a significance level of alpha = 0.05
to test the claim that p1 = p2. Use the given sample sizes and
numbers of successes to find the z test statistic for the
hypothesis test. A report on the nightly news broadcast stated
that 10 out of 108 households with pet dogs were burglarized
and 20 out of 208 without pet dogs were burglarized.
Answer
z = -0.041
z = -0.102
z = 0.000
z = -0.173
5 points
Question 14
1.
Solve the problem. The table shows the number of smokers in a
random sample of 500 adults aged 20-24 and the number of
smokers in a random sample of 450 adults aged 25-29. Assume
that you plan to use a significance level of alpha = 0.10 to test
the claim that Find the critical value(s) for this hypothesis test.
Do the data provide sufficient evidence that the proportion of
smokers in the 20-24 age group is different from the proportion
of smokers in the 25-29 age group?
Answer

9. z = ± 1.645; yes
z = ± 1.28; yes
z = ± 1.96; no
z = 1.28; no
5 points
Question 15
1.
Assume that you plan to use a significance level of alpha = 0.05
to test the claim that p1 = p2. Use the given sample sizes and
numbers of successes to find the P-value for the hypothesis test.
n1 = 50;n2 = 75 x1 = 20;x2 = 15
Answer
0.0032
0.0146
0.1201
0.0001
5 points
Question 16
1.
Construct the indicated confidence interval for the difference
between population proportions p1 - p2. Assume that the

10. samples are independent and that they have been randomly
selected. x1 = 61, n1 = 105 and x2 = 82, n2 = 120; Construct a
98% confidence interval for the difference between population
proportions p1 - p2.
Answer
0.456 < p1 - p2 < 0.707
0.432 < p1 - p2 < 0.730
-0.228 < p1 - p2 < 0.707
-0.252 < p1 - p2 < 0.047
5 points
Question 17
1.
Construct the indicated confidence interval for the difference
between the two population means. Assume that the two
samples are independent simple random samples selected from
normally distributed populations. Do not assume that the
population standard deviations are equal. A researcher was
interested in comparing the resting pulse rates of people who
exercise regularly and the pulse rates of people who do not
exercise regularly. She obtained independent simple random
samples of 16 people who do not exercise regularly and 12
people who do exercise regularly. The resting pulse rates (in
beats per minute) were recorded and the summary statistics are
as follows. Construct a 95% confidence interval for mu1 - mu2,
the difference between the mean pulse rate of people who do not
exercise regularly and the mean pulse rate of people who
exercise regularly.
Answer

11. -3.22 beats/min < mu1 - mu2 < 11.62 beats/min
-3.55 beats/min < mu1 - mu2 < 11.95 beats/min
-3.74 beats/min < mu1 - mu2 < 12.14 beats/min
-4.12 beats/min < mu1 - mu2 < 14.72 beats/min
5 points
Question 18
1.
State what the given confidence interval suggests about the two
population means. A researcher was interested in comparing the
heights of women in two different countries. Independent
simple random samples of 9 women from country A and 9
women from country B yielded the following heights (in
inches). The following 90% confidence interval was obtained
for mu1 - mu2, the difference between the mean height of
women in country A and the mean height of women in country
B.-4.34 in. < mu1 - mu2 < -0.03 in What does the confidence
interval suggest about the population means?
Answer
The confidence interval includes only negative values which
suggests that the mean height of women from country A is
greater than the mean height of women from country B.
The confidence interval includes only negative values which
suggests that the two population means might be equal. There
doesn't appear to be a significant difference between the mean

12. height of women from country A and the mean height of women
from country B.
The confidence interval includes only negative values which
suggests that the mean height of women from country A is
smaller than the mean height of women from country B.
The confidence interval includes 0 which suggests that the two
population means might be equal. There doesn't appear to be a
significant difference between the mean height of women from
country A and the mean height of women from country B.
5 points
Question 19
1.
Construct the indicated confidence interval for the difference
between the two population means. Assume that the two
samples are independent simple random samples selected from
normally distributed populations. Also assume that the
population standard deviations are equal (sigma1 = sigma2), so
that the standard error of the difference between means is
obtained by pooling the sample variances. A paint manufacturer
wanted to compare the drying times of two different types of
paint. Independent simple random samples of 11 cans of type A
and 9 cans of type B were selected and applied to similar
surfaces. The drying times, in hours, were recorded. The
summary statistics are as follows. Construct a 99% confidence
interval for mu1 - mu2, the difference between the mean drying
time for paint type A and the mean drying time for paint type B.
Answer
-2.73 hrs < mu1 - mu2 < 7.73 hrs

13. -0.64 hrs < mu1 - mu2 < 5.64 hours
-1.50 hrs < mu1 - mu2 < 6.50 hrs
-2.01 hrs < mu1 - mu2 < 7.01 hrs
5 points
Question 20
1.
The two data sets are dependent. Find to the nearest tenth.
Answer
44.1
20.3
33.9
203.4