10 Two-Sample Inference

Section 10.3 Exercises

CLARIFYING THE CONCEPTS

Question 10.113

1. ${\hat{p}}_{pooled}$ must always lie between which two quantities? (p. 608)

10.3.1

${\hat{p}}_{1}$ and ${\hat{p}}_{2}$

Question 10.114

2. Does it make sense to use ${\hat{p}}_{pooled}$ when calculating confidence intervals for $p_{1} - p_{2}$ ? Why or why not? (p. 612)

Question 10.115

3. What does $Z_{data}$ measure? What do extreme values of $Z_{data}$ indicate? (p. 608)

10.3.3

$Z_{data}$ measures the standardized distance between sample proportions.

Extreme values of $Z_{data}$ indicate evidence against the null hypothesis.

Question 10.116

4. What might we suggest if the $p$ -value is very close to the level of significance $α$ ? (p. 612)

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PRACTICING THE TECHNIQUES

CHECK IT OUT!

To do	Check out	Topic
Exercises 5–8	Example 15	$Z$ test for $p_{1} − p_{2}$ : critical-value method
Exercises 9–12	Example 16	$Z$ test for $p_{1} − p_{2}$ : $p$ -value method
Exercises 13–18	Example 17	$Z$ confidence interval for $p_{1} − p_{2}$
Exercises 19–22	Example 18	Equivalence between confidence intervals and $Z$ tests for $p_{1} − p_{2}$

The summary statistics in Exercises 5–7 and 9–11 were taken from random samples that were drawn independently. Let $n_{1}$ and $n_{2}$ denote the size of samples 1 and 2, respectively. Let $x_{1}$ and $x_{2}$ denote the number of successes in samples 1 and 2, respectively.

For Exercises 5–7, perform the indicated hypothesis test using the critical-value method. Answer (a)–(d) for each exercise.

State the hypotheses and find the critical value $Z_{crit}$ and the rejection rule.
Calculate ${\hat{p}}_{pooled}$ .
Calculate $Z_{data}$ .
Compare $Z_{data}$ with $Z_{crit}$ . State and interpret your conclusion.

Question 10.117

5. Test, at level of significance $α = 0.10$ , whether $p_{1} \neq p_{2}$ .

Sample 1	$n_{1} = 100$	$x_{1} = 80$
Sample 2	$n_{2} = 40$	$x_{2} = 30$

10.3.5

(a) $H_{0} : p_{1} = p_{2} vs . H_{a} : p_{1} \neq p_{2}$ ; $Z_{crit} = 1.645$ . Reject $H_{0}$ if $Z_{data} \leq - 1.645$ or $Z_{data} \geq 1.645$ . (b) 0.7857 (c) 0.65 (d) Since $Z_{data} \geq - 1.645$ and $Z_{data} \leq - 1.645$ , we do not reject $H_{0}$ . There is insufficient evidence that the population proportion from Population 1 is different from the population proportion from Population 2.

Question 10.118

6. Test, at level of significance $α = 0.05$ , whether $p_{1} < p_{2}$ .

Sample 1	$n_{1} = 10$	$x_{1} = 5$
Sample 2	$n_{2} = 12$	$x_{2} = 5$

Question 10.119

7. Test, at level of significance $α = 0.01$ , whether $p_{1} > p_{2}$ .

Sample 1	$n_{1} = 200$	$x_{1} = 60$
Sample 2	$n_{2} = 250$	$x_{2} = 40$

10.3.7

(a) $H_{0} : p_{1} = p_{2} vs . H_{a} : p_{1} > p_{2}$ . $Z_{crit} = 2.33$ . Reject $H_{0}$ if $Z_{crit} \geq 2.33$ (b) ${\hat{p}}_{pooled} = 100 / 450 \approx 0.2222$ . (c) $Z_{data} = 3.550$ . (d) Since $Z_{data} = 3.550$ is ≥ 2.33, we reject $H_{0}$ . There is evidence at the $α = 0.01$ level of significance that the population proportion of Population 1 is greater than the population proportion of Population 2.

Question 10.120

8. Refer to the data from Exercise 7. Test, at level of significance $α = 0.01$ , whether $p_{1} \neq p_{2}$ .

For Exercises 9–11, perform the indicated hypothesis test using the $p$ -value method. Answer (a)–(e) for each exercise.

State the hypotheses and the rejection rule.
Calculate ${\hat{p}}_{pooled}$ .
Calculate $Z_{data}$ .
Calculate the $p$ -value.
Compare the $p$ -value with $α$ . State and interpret your conclusion.

Question 10.121

9. Test, at level of significance $α = 0.05$ , whether $p_{1} > p_{2}$ .

Sample 1	$n_{1} = 400$	$x_{1} = 250$
Sample 2	$n_{2} = 400$	$x_{2} = 200$

10.3.9

(a) $H_{0} : p_{1} = p_{2} vs . H_{a} : p_{1} > p_{2}$ . Reject $H_{0}$ if $p – value$ $p – value \leq 0.05$ . (b) ${\hat{p}}_{pooled} = 450 / 800 = 0.5625$ . (c) $Z_{data} = 3.563$ . (d) $p – value = 0.0002$ (e) Since $p – value = 0.0002$ is $\leq 0.05$ , we reject $H_{0}$ . There is evidence at the $α = 0.05$ level of significance that the population proportion of Population 1 is greater than the population proportion of Population 2.

Question 10.122

10. Test, at level of significance $α = 0.05$ , whether $p_{1} < p_{2}$ .

Sample 1	$n_{1} = 1000$	$x_{1} = 490$
Sample 2	$n_{2} = 1000$	$x_{2} = 620$

Question 10.123

11. Test, at level of significance $α = 0.10$ , whether $p_{1} \neq p_{2}$ .

Sample 1	$n_{1} = 527$	$x_{1} = 412$
Sample 2	$n_{2} = 613$	$x_{2} = 498$

10.3.11

(a) $H_{0} : p_{1} = p_{2} vs . H_{a} : p_{1} \neq p_{2}$ . Reject $H_{0}$ if $p – value \leq 0.10$ . (b) ${\hat{p}}_{pooled} = 910 / 1140 \approx 0.7982$ . (c) $Z_{data} \approx - 1.284$ . (d) $p – value \approx 0 .1991$ (e) Since $p – value \approx 0.1991$ is not ≤ 0.10, we do not reject $H_{0}$ . There is insufficient evidence at the $α = 0.10$ level of significance that the population proportion of Population 1 is different from the population proportion of Population 2.

Question 10.124

12. Refer to the data from Exercise 11. Test, at level of significance $α = 0.10$ , whether $p_{1} < p_{2}$ .

For Exercises 13–18, refer to the indicated data to answer (a)–(d).

We are interested in constructing a 95% confidence interval for $p_{1} - p_{2}$ . Is it appropriate to do so? Why or why not? If not appropriate, then do not perform (b)–(d).
Provide the point estimate of the difference in population proportions $p_{1} - p_{2}$ .
Calculate the margin of error for a confidence level of 95%. What does this number mean?
Construct and interpret a 95% confidence interval for $p_{1} - p_{2}$ .

Question 10.125

13. Data from Exercise 5

10.3.13

(a) $x_{1} = 80 \geq 5$ , $n_{1} = x_{1} = 20 \geq 5$ , $x_{2} = 30 \geq 5$ , and $n_{2} - x_{2} = 10 \geq 5$ , so it is appropriate. (b) 0.05 (c) 0.1554. The point estimate ${\hat{p}}_{1} - {\hat{p}}_{2}$ will lie within $E = 0.1554$ of the difference in population proportions $p_{1} - p_{2}$ 95% of the time. (d) (–0.1054, 0.2054). We are 95% confident that the difference in population proportions lies between −0.1054 and 0.2054.

Question 10.126

14. Data from Exercise 6

Question 10.127

15. Data from Exercise 7

10.3.15

(a) $x_{1} = 60 \geq 5$ , $n_{1} = x_{1} = 140 \geq 5$ , $x_{2} = 40 \geq 5$ , and $n_{2} - x_{2} = 210 \geq 5$ , so it is appropriate. (b) 0.14 (c) 0.078. The point estimate ${\hat{p}}_{1} - {\hat{p}}_{2}$ will lie within $E = 0.078$ of the difference in population proportions $p_{1} - p_{2}$ 95% of the time. (d) (0.062, 0.218). We are 95% confident that the difference in population proportions lies between 0.062 and 0.218.

Question 10.128

16. Data from Exercise 9

Question 10.129

17. Data from Exercise 10

10.3.17

(a) $x_{1} = 490 \geq 5$ , $n_{1} - x_{1} = 510 \geq 5$ , $x_{2} = 620 \geq 5$ , and $n_{2} - x_{2} = 380 \geq 5$ , so it is appropriate. (b) −0.13 (c) 0.0432. The point estimate ${\hat{p}}_{1} - {\hat{p}}_{2}$ will lie within $E = 0.0432$ of the difference in population proportions $p_{1} - p_{2}$ 95% of the time. (d) (–0.1732, −0.0868). We are 95% confident that the difference in population proportions lies between −0.1732 and −0.0868.

Question 10.130

18. Data from Exercise 11

For Exercises 19–22, a $100 (1 - α) %$ $Z$ confidence interval for $p_{1} - p_{2}$ is given. Use the confidence interval to test, using level of significance $α$ , whether $p_{1} - p_{2}$ differs from each of the indicated hypothesized values.

Question 10.131

19. A 95% $Z$ confidence interval for $p_{1} - p_{2}$ is (0.5, 0.6). Hypothesized values are

0
0.1
0.57

10.3.19

(a) $H_{0} : p_{1} - p_{2} = 0 vs . H_{a} : p_{1} - p_{2} \neq 0$ . The hypothesized value of 0 lies outside the interval (0.5, 0.6), so we reject $H_{0}$ at the $α = 0.05$ level of significance. (b) $H_{0} : p_{1} - p_{2} = 0.1 vs . H_{a} : p_{1} - p_{2} \neq 0.1$ . The hypothesized value of 0.1 lies outside the interval (0.5, 0.6), so we reject $H_{0}$ at the $α = 0.05$ level of significance. (c) $H_{0} : p_{1} - p_{2} = 0.57 vs . H_{a} : p_{1} - p_{2} \neq 0.57$ . The hypothesized value of 0.57 lies inside the interval (0.5, 0.6), so we do not reject $H_{0}$ at the $α = 0.05$ level of significance.

Question 10.132

20. A 99% $Z$ confidence interval for $p_{1} - p_{2}$ is (0.01, 0.99). Hypothesized values are

0.2
0
0.999

Question 10.133

21. A 90% $Z$ confidence interval for $p_{1} - p_{2}$ is (0.1, 0.11). Hypothesized values are

0.151
0.115
0.105

10.3.21

(a) $H_{0} : p_{1} - p_{2} = 0.151 vs . H_{a} : p_{1} - p_{2} \neq 0.151$ . The hypothesized value of 0.151 lies outside of the interval (0.1, 0.11), so we reject $H_{0}$ at the $α = 0.10$ level of significance. (b) $H_{0} : p_{1} - p_{2} = 0.115 vs . H_{a} : p_{1} - p_{2} \neq 0.115$ . The hypothesized value of 0.115 lies outside of the interval (0.1, 0.11), so we reject $H_{0}$ at the $α = 0.10$ level of significance. (c) $H_{0} : p_{1} - p_{2} = 0.105 vs . H_{a} : p_{1} - p_{2} \neq 0.105$ . The hypothesized value of 0.105 lies inside of the interval (0.1, 0.11), so we do not reject $H_{0}$ at the $α = 0.10$ level of significance.

Question 10.134

22. A 95% $Z$ confidence interval for $p_{1} - p_{2}$ is (0.43, 0.57). Hypothesized values are

0.41
0.51
0.61

APPLYING THE CONCEPTS

Question 10.135

23. Technological Change. Do attitudes about technological change differ between the young and their elders? The Pew report discussed earlier reported that 85 of 144 respondents 18 to 29 years old agreed that technology will lead to a better future, whereas 166 of 297 respondents 65 years old or older agreed.

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Is it appropriate to perform the $Z$ test for the difference in population proportions? Why or why not?
Clearly state the meaning of $p_{1}$ and $p_{2}$ .
Test whether the proportions agreeing that technology will lead to a better future differ between the younger group and the older group, using level of significance $α = 0.05$ .

10.3.23

(a) $x_{1} = 85 \geq 5$ , $n_{1} - x_{1} = 144 - 85 = 59 \geq 5$ , $x_{2} = 166 \geq 5$ , and $n_{2} - x_{2} = 297 - 166 = 131 \geq 5$ . Therefore it is appropriate to perform the $Z$ test for the difference in population proportions. (b) $p_{1}$ refers to the population proportion of people 18–29 years old who agree that technology will lead to a better future and $p_{2}$ refers to the population proportion of people 65 years old or older who agree that technology will lead to a better future. (c) $H_{0} : p_{1} - p_{2} vs . H_{a} : p_{1} \neq p_{2}$ . Reject $H_{0}$ if the $p – value$ $p – value \leq α = 0.05$ . $Z_{data} \approx 0.62$ ; $p – value = 0.5352$ (TI-84: 0.5329195849). The $p – value = 0.5352$ is not $\leq α = 0.05$ , so we do not reject $H_{0}$ . There is insufficient evidence at level of significance $α = 0.05$ that the population proportions agreeing that technology will lead to a better future differ between the younger group and the older group.

Question 10.136

24. Medicare Recipients. The Centers for Medicare and Medicaid Services reported in 2004 that 3305 of the 50,350 Medicare recipients living in Alaska were age 85 or over, and 73,289 of the 754,642 Medicare recipients living in Arizona were age 85 or over.

Find a point estimate of the difference in population proportions.
Clearly state the difference in meaning between $p_{1}$ and ${\hat{p}}_{1}$ .
Test whether the population proportions differ, using level of significance $α = 0.05$ .

Question 10.137

25. Women's ownership of Businesses. The U.S. Census Bureau tracks trends in women's ownership of businesses. A random sample of 100 Ohio businesses showed 34 that were woman-owned. A sample of 200 New Jersey businesses showed 64 that were woman-owned. Test whether the population proportions of female-owned businesses in Ohio is greater than that of New Jersey, using level of significance $α = 0.10$ .

10.3.25

$x_{1} = 34 \geq 5$ , $(n_{1} - x_{1}) = 66 \geq 5$ , $x_{2} = 64 \geq 5$ , and $(n_{2} - x_{2}) = 136 \geq 5$ . Therefore it is appropriate to perform the $Z$ test for the difference in population proportions. $p_{1}$ is the population proportion of Ohio businesses that are owned by women and $p_{2}$ is the population proportion of New Jersey businesses that are owned by women. Critical-value method: $H_{0} : p_{1} = p_{2} vs . H_{a} : p_{1} > p_{2} . Z_{crit} = 1.28$ . Reject $H_{0}$ if $Z_{data} \geq 1.28$ . ${\hat{p}}_{pooled} = 98 / 300 \approx 0.3267$ . $Z_{data} = 0.348$ . Since $Z_{data} = 0.348$ is not $\geq 1.28$ , we do not reject $H_{0}$ . There is insufficient evidence at the $α = 0.10$ level of significance that the population proportion of Ohio businesses that are owned by women is greater than the population proportion of New Jersey businesses that are owned by women. $p – value$ method: $H_{0} : p_{1} = p_{2} vs . H_{a} : p_{1} > p_{2}$ . Reject $H_{0}$ if the $p – value \leq 0.10$ . ${\hat{p}}_{pooled} = 98 / 300 \approx 0.3267$ . $Z_{data} = 0.348$ . $p – value = 0.3639$ . Since the $p – value = 0.3639$ is not $\leq 0.10$ , we do not reject $H_{0}$ . There is insufficient evidence at the $α = 0.10$ level of significance that the population proportion of Ohio businesses that are owned by women is greater than the population proportion of New Jersey businesses that are owned by women.

Question 10.138

26. Fetal Cells and Breast Cancer. A number of fetal stem cells may cross the placenta from the fetus to the mother during pregnancy and remain in the mother's tissue for decades. A recent study shows that the presence of fetal cells in the mother may offer some protection against the onset of breast cancer.¹⁶ Of the 54 women in the study with breast cancer, 14 had fetal cells. Of the 45 women without breast cancer, 25 had fetal cells. Test whether the population proportions of women with fetal cells is lower among women with breast cancer compared with women without breast cancer, using level of significance $α = 0.01$ .

Question 10.139

27. Technological Change. Refer to Exercise 23 to answer the following questions:

Construct and interpret a 95% confidence interval for the difference in population proportions.
Use the confidence interval from (a) to test, using level of significance $α = 0.05$ , whether the population proportions differ.
Does your conclusion from (b) agree with your conclusion from Exercise 23 (c)?

10.3.27

(a) TI-84 (–0.0668, 0.1295). We are 95% confident that the difference in the population proportions of people 18–29 years old and people age 65 years or older who agree that technology will lead to a better future lies between 20.0668 and 0.1295. (b) $H_{0} : p_{1} = p_{2} vs . H_{a} : p_{1} \neq p_{2}$ lies in the interval so we do not reject $H_{0}$ . (c) Yes.

Question 10.140

28. Medicare Recipients. Refer to Exercise 24 to answer the following questions:

Construct and interpret a 95% confidence interval for the difference in population proportions.
Use the confidence interval from (a) to test, using level of significance $α = 0.05$ , whether the population proportions differ.
Does your conclusion from (b) agree with your conclusion from Exercise 24 (c)?

Question 10.141

29. Women's ownership of Businesses. Refer to Exercise 25 to answer the following questions:

Construct and interpret a 90% confidence interval for the difference in population proportions.
Use the confidence interval from (a) to test, using level of significance $α = 0.10$ , whether the population proportions differ.
Explain whether or not we could use the confidence interval from part (b) to perform the hypothesis test in Exercise 25. Why or why not?

10.3.29

(a) (–0.0745, 0.1145). TI-83/84: (–0.0749, 0.1150). We are 90% confident that the difference of the population proportion of Ohio businesses that are owned by women and the population proportion of New Jersey businesses that are owned by women lies between −0.0745(–0.0749) and 0.1145(0.1150).

(b) $H_{0} : p_{1} = p_{2} vs . H_{a} : p_{1} \neq p_{2} .$ Our hypothesized value of 0 lies inside the interval in (a), so we do not reject $H_{0}$ . There is insufficient evidence that the population proportion of Ohio businesses that are owned by women differs from the population proportion of New Jersey businesses that are owned by women. (c) No, it is a one-sided test and confidence intervals can only be used to perform two-sided tests.

Question 10.142

30. Fetal Cells and Breast Cancer. Refer to Exercise 26 to answer the following questions:

Construct and interpret a 99% confidence interval for the difference in population proportions.
Use the confidence interval from (a) to test, using level of significance $α = 0.01$ , whether the population proportions differ.
Explain whether or not we could use the confidence interval from (b) to perform the hypothesis test in Exercise 26. Why or why not?

Question 10.143

31. Evidence for alternative Medical Therapies? A company called QT, Inc., sells “ionized” bracelets, called Q-Ray bracelets, that it claims help to ease pain through balancing the body's flow of “electromagnetic energy.” The Mayo Clinic decided to conduct a statistical experiment to determine whether the claims for the Q-Ray bracelets were justified.¹⁷ At the end of 4 weeks, of the 305 subjects who wore the “ionized” bracelet, 236 (77.4%) reported improvement in their maximum pain index (where the pain was the worst). Of the 305 subjects who wore the placebo bracelet (a bracelet identical in every respect to the “ionized” bracelet except that there was no active ingredient—presumably, here, “ionization”), 234 (76.7%) reported improvement in their maximum pain index. Using level of significance $α = 0.05$ , test whether the population proportions reporting improvement differ between wearers of the ionized bracelet and wearers of the placebo bracelet.

10.3.31

$H_{0} : p_{1} = p_{2} vs . H_{a} : p_{1} \neq p_{2}$ . Reject $H_{0}$ if the $p -value \leq 0.05$ . ${\hat{p}}_{pooled} = 0.7705$ . $Z_{data} = 0.19$ . $p – value = 0.8336$ . Since the $p – value$ is not $\leq 0.05$ , we do not reject $H_{0}$ . There is insufficient evidence that the proportion of the people who wore the ionized bracelets who reported improvement in their maximum pain index is different from the proportion of the people who wore the placebo bracelets who reported improvement in their maximum pain index.

BRINGING IT ALL TOGETHER

Males listening to the Radio. Use the following information for Exercises 32–40: The Arbitron Corporation tracks trends in radio listening. In their publication Radio Today, Arbitron reported that 92% of 18- to 24-year-old males listen to the radio each week, whereas 87% of males 65 years and older listen to the radio each week. Suppose each sample size was 1000.

Question 10.144

32. Is it appropriate to perform $Z$ inference for the difference in population proportions? Why or why not?

Question 10.145

33. Clearly describe what $p_{1}$ means and what $p_{2}$ means.

10.3.33

$p_{1} =$ the population proportion of 18- to 24-year-old males who listen to the radio each week and $p_{2} =$ the population proportion of males age 65 or older who listen to the radio each week.

Question 10.146

34. Explain what the difference is between $p_{1}$ and ${\hat{p}}_{1}$ .

Question 10.147

35. Calculate the margin of error for a 95% confidence interval for $p_{1} - p_{2}$ . Explain what this number means.

10.3.35

0.02678. The point estimate of the difference in the population proportion of 18- to 24-year-old males who listen to the radio each week and the population proportion of males 65 years and older who listen to the radio each week will lie within $E = 0.02678$ of the difference in population proportions $p_{1} - p_{2}$ 95% of the time.

Question 10.148

36. Construct and interpret a 95% confidence interval for $p_{1} - p_{2}$ .

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Question 10.149

37. Use the confidence interval from Exercise 36 to test, using level of significance $α = 0.05$ , whether $p_{1} - p_{2}$ differs from the following:

0
0.01
0.05

10.3.37

(a) $H_{0} : p_{1} - p_{2} = 0 vs . H_{a} : p_{1} - p_{2} \neq 0$ . The hypothesized value of 0 does not lie in the interval from Exercise 36, so we reject $H_{0}$ . There is evidence that the difference in the population proportion of 18- to 24-year-old males who listen to the radio each week and the population proportion of males 65 years and older who listen to the radio each week differs from 0. (b) $H_{0} : p_{1} - p_{2} = 0.01 vs . H_{a} : p_{1} - p_{2} \neq 0.01$ . The hypothesized value of 0.01 does not lie in the interval from Exercise 36, so we reject $H_{0}$ . There is evidence that the difference in the population proportion of 18- to 24-year-old males who listen to the radio each week and the population proportion of males 65 years and older who listen to the radio each week differs from 0.01. (c) $H_{0} : p_{1} - p_{2} = 0.05 vs . H_{a} : p_{1} - p_{2} \neq 0.05$ . The hypothesized value of 0.05 lies in the interval from Exercise 36, so we do not reject $H_{0}$ . There is insufficient evidence that the difference in the population proportion of 18- to 24-year-old males who listen to the radio each week and the population proportion of males 65 years and older who listen to the radio each week differs from 0.05.

Question 10.150

38. Explain whether we could use the confidence interval from Exercise 36 to test whether the proportion of 18- to 24-year-old males who listen to the radio each week is greater than the proportion of males 65 years and older who do so. Why or why not?

Question 10.151

39. Test, using level of significance $α = 0.05$ , whether the proportion of 18- to 24-year-old males who listen to the radio each week is greater than the proportion of males 65 years and older who do so.

10.3.39

Critical-value method: $H_{0} : p_{1} = p_{2} vs . H_{a} : p_{1} > p_{2}$ . $Z_{crit} = 1.645$ . Reject $H_{0}$ if $Z_{data} \geq 1.645$ . ${\hat{p}}_{pooled} = 1790 / 2000 = 0.895$ . $Z_{data} = 3.647$ . Since $Z_{data} = 3.647$ is $\geq 1.645$ , we reject $H_{0}$ . There is evidence at the $α = 0.05$ level of significance that the population proportion of 18- to 24-year-old males who listen to the radio each week is greater than the population proportion of males 65 years and older who listen to the radio each week. $p – value$ method: $H_{0} : p_{1} = p_{2} vs . H_{a} : p_{1} > p_{2}$ . Reject $H_{0}$ if the $p – value$ $p -value0 \leq 0.05$ . ${\hat{p}}_{pooled} = 1790 / 2000 = 0.895$ . $Z_{data} = 3.647$ . $p – value = 0.00013$ . Since the $p – value 0.00013$ is $\leq 0.05$ , we reject $H_{0}$ . There is evidence at the $α = 0.05$ level of significance that the population proportion of 18- to 24-year-old males who listen to the radio each week is greater than the population proportion of males 65 years and older who listen to the radio each week.

Question 10.152

40. What if, instead of 1000, each sample size was 100? How would this change affect each of the following measures?

Margin of error in Exercise 35.
$p$ -value in Exercise 39.
Conclusion of the hypothesis test in Exercise 39.

WORKING WITH LARGE DATA SETS

Case Study: Bank loans.

Open the Chapter 10 Case Study data sets, BankLoans_approved, and BankLoans_Denied. Here, we will examine whether a difference exists in the proportion of applicants with credit scores of 700 or higher between those approved for and those denied a loan. Use technology to do the following:

Question 10.153

bankloan_approved

41. Obtain independent random samples of size 100, one each from the BankLoans_approved data set and the BankLoans_Denied data set. bankloans_denied

10.3.41

Answers will vary.

Question 10.154

42. For each sample, calculate the proportion of applicants who have credit scores of 700 or higher.

Question 10.155

43. Perform and interpret a $Z$ test for the difference in the proportion of applicants who have credit scores of 700 or higher, using level of significance $α = 0.05$ .

10.3.43

$H_{0} : p_{1} = p_{2} vs . H_{a} : p_{1} \neq p_{2}$

Rest of answer will vary.

Question 10.156

44. Construct and interpret a 95% $Z$ confidence interval for the difference in proportion of applicants who have credit scores of 700 or higher.