10 Two-Sample Inference

Section 10.2 Exercises

CLARIFYING THE CONCEPTS

Question 10.52

1. What are the conditions that permit us to perform Welch's two-sample $t$ test? (p. 592)

10.2.1

The two populations are normally distributed. The sample sizes are large (at least 30).

Question 10.53

2. If a $100 (1 - α) %$ confidence interval for $μ_{1} - μ_{2}$ contains 0, then with level of significance $α$ , what is our conclusion regarding the hypothesis that there is no difference in the population means? (p. 596)

PRACTICING THE TECHNIQUES

CHECK IT OUT!

To do	Check out	Topic
Exercises 3–6	Example 7	$t$ test for $μ_{1} - μ_{2}$ : critical-value method
Exercises 7–10	Example 8	$t$ test for $μ_{1} - μ_{2}$ : $p$ -value method
Exercises 11–16	Example 9	$t$ confidence interval for $μ_{1} - μ_{2}$
Exercises 17–20	Example 10	Equivalence between confidence intervals and $t$ tests for $μ_{1} - μ_{2}$
Exercises 21–22	Example 11	Pooled variance $t$ test for $μ_{1} - μ_{2}$
Exercises 23–24	Example 12	Pooled variance $t$ confidence interval for $μ_{1} - μ_{2}$
Exercises 25–26	Example 13	$Z$ test for $μ_{1} - μ_{2}$
Exercises 27–28	Example 14	$Z$ confidence interval for $μ_{1} - μ_{2}$

For Exercises 3–6, perform the indicated Welch's hypothesis test using the critical-value method. The summary statistics were taken from random samples that were drawn independently. For each exercise, follow these steps:

State the hypotheses.
Find the critical value $t_{crit}$ and the rejection rule for this test.
Calculate $t_{data}$ .
Compare $t_{data}$ with $t_{crit}$ . State and interpret your conclusion.

Question 10.54

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

Sample 1	$n_{1} = 36$	${\bar{x}}_{1} = 10$	$s_{1} = 2$
Sample 2	$n_{2} = 36$	${\bar{x}}_{2} = 8$	$s_{2} = 2$

10.2.3

(a) $H_{0} : μ_{1} = μ_{2}$ vs. $H_{a} : μ_{1} \neq μ_{2}$ (b) $t_{crit} = 1.690$ . Reject $H_{0}$ if $t_{data} \leq - 1.690$ or $t_{data} \geq - 1.690$ . (c) $t_{data} = - 4.243$ . (d) Since $t_{data} = - 1.690$ , we reject $H_{0}$ . There is evidence that the population mean for Population 1 is different from the population mean for Population 2.

Question 10.55

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

Sample 1	$n_{1} = 64$	${\bar{x}}_{1} = 20$	$s_{1} = 3$
Sample 2	$n_{2} = 64$	${\bar{x}}_{2} = 18$	$s_{2} = 2$

Question 10.56

5. Test, at level of significance $α = 0.01$ , whether $μ_{1} < μ_{2}$ .

Sample 1	$n_{1} = 100$	${\bar{x}}_{1} = 70$	$s_{1} = 10$
Sample 2	$n_{2} = 50$	${\bar{x}}_{2} = 80$	$s_{2} = 12$

10.2.5

(a) $H_{0} : μ_{1} = μ_{2}$ vs. $H_{a} : μ_{1} < μ_{2}$ (b) $t_{crit} = - 2.423$ . Reject $H_{0}$ if $t_{data} \leq - 2.423$ . (c) $t_{data} = - 5.077$ . (d) Since $t_{data} = - 5.077$ is $\leq - 2.423$ , we reject $H_{0}$ . There is evidence at the $α = 0.01$ level of significance that the population mean of Population 1 is less than the population mean of Population 2.

Question 10.57

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

Sample 1	$n_{1} = 60$	${\bar{x}}_{1} = 100$	$s_{1} = 20$
Sample 2	$n_{2} = 40$	${\bar{x}}_{2} = 90$	$s_{2} = 10$

For Exercises 7–10, perform the indicated Welch's hypothesis test using the $p$ -value method. The summary statistics were taken from random samples that were drawn independently. For each exercise follow these steps:

State the hypotheses and the rejection rule.
Calculate $t_{data}$ .
Find the $p$ -value.
Compare the $p$ -value with level of significance $α$ . State and interpret your conclusion.

Question 10.58

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

Sample 1	$n_{1} = 64$	${\bar{x}}_{1} = 0$	$s_{1} = 3$
Sample 2	$n_{2} = 49$	${\bar{x}}_{2} = 1$	$s_{2} = 1$

10.2.7

(a) $H_{0} : μ_{1} = μ_{2} vs . H_{a} : μ_{1} \neq μ_{2}$ Reject $H_{0}$ if the $p – value$ ≤ 0.10.

(b) $t_{data} = - 2.492$ . (c) $p – value = 0.0162$ . (d) Since the $p – value$ is ≤ 0.10, we reject $H_{0}$ . There is evidence at the $α = 0.10$ level of significance that the population mean of Population 1 is different from the population mean of Population 2.

Question 10.59

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

Sample 1	$n_{1} = 255$	${\bar{x}}_{1} = 103$	$s_{1} = 17$
Sample 2	$n_{2} = 400$	${\bar{x}}_{2} = 95$	$s_{2} = 11$

Question 10.60

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

Sample 1	$n_{1} = 100$	${\bar{x}}_{1} = 50$	$s_{1} = 10$
Sample 2	$n_{2} = 100$	${\bar{x}}_{2} = 75$	$s_{2} = 15$

10.2.9

(a) $H_{0} : μ_{1} = μ_{2} vs . H_{a} : μ_{1} < μ_{2}$ . Reject $H_{0}$ if the $p – value$ ≤ 0.05.

(b) $t_{data} = - 13.868$ . (c) $p – value \approx 0$ . (d) Since the $p – value \approx 0$ is ≤ 0.05, we reject $H_{0}$ . There is evidence at the $α = 0.05$ level of significance that the population mean of Population 1 is less than the population mean of Population 2.

Question 10.61

10. Test, at level of significance $α = 0.01$ , whether $μ_{1} \neq μ_{2}$ .

Sample 1	$n_{1} = 30$	${\bar{x}}_{1} = 210$	$s_{1} = 5$
Sample 2	$n_{2} = 30$	${\bar{x}}_{2} = 25$	$s_{2} = 2$

For Exercises 11–16, do the following for the designated data:

Provide the point estimate of $μ_{1} - μ_{2}$ .
Calculate the margin of error for the confidence level indicated.
Construct and interpret a $t$ confidence interval for $μ_{1} - μ_{2}$ with the confidence level indicated.

Question 10.62

11. Data in Exercise 3, confidence level = 90%

10.2.11

(a) ${\bar{x}}_{1} - {\bar{x}}_{2} = 2$ (b) $E = 0.797$ . We can estimate the difference in the population means of Population 1 and Population 2 to within 0.797 with 90% confidence. (c) (1.203, 2.797). We are 90% confident that the difference in the population means of Population 1 and Population 2 lies between 1.203 and 2.797.

Question 10.63

12. Data in Exercise 4, confidence level = 95%

Question 10.64

13. Data in Exercise 5, confidence level = 99%

10.2.13

(a) ${\bar{x}}_{1} - {\bar{x}}_{2} = - 10$ (b) $E = 5.326$ . We can estimate the difference in the population means of Population 1 and Population 2 to within 5.326 with 99% confidence. (c) (−15.326, −4.674). We are 99% confident that the difference in the population means of Population 1 and Population 2 lies between 215.326 and 24.674.

Question 10.65

14. Data in Exercise 6, confidence level = 95%

Question 10.66

15. Data in Exercise 7, confidence level = 95%

10.2.15

(a) ${\bar{x}}_{1} - {\bar{x}}_{2} = - 1$ (b) $E = 0.811$ . We can estimate the difference in the population means of Population 1 and Population 2 to within 0.811 with 95% confidence. (c) (–1.811, −0.189). We are 95% confident that the difference in the population means of Population 1 and Population 2 lies between 2 1.811 and 20.189.

Question 10.67

16. Data in Exercise 8, confidence level = 90%

For Exercises 17–20 a $100 (1 - α) %$ $t$ confidence interval for $μ_{1} - μ_{2}$ is given. Use the confidence interval to test, at level of significance $α$ , whether $μ_{1} - μ_{2}$ differs from each of the designated hypothesized values.

Page 603

Question 10.68

17. A 95% $t$ confidence interval for $μ_{1} - μ_{2}$ is (10, 15). Hypothesized values are

10.2.17

(a) $H_{0} : μ_{1} - μ_{2} = 0 vs . H_{a} : μ_{1} - μ_{2} \neq 0$ lies outside of the interval (10, 15), so we reject $H_{0}$ at the $α = 0.05$ level of significance. (b) $H_{0} : μ_{1} - μ_{2} = 12 vs . H_{a} : μ_{1} - μ_{2} \neq 12$ . $μ_{0} = 12$ lies inside of the interval (10, 15), so we do not reject $H_{0}$ at the $α = 0.05$ level of significance. (c) $H_{0} : μ_{1} - μ_{2} = 16 vs . H_{a} : μ_{1} - μ_{2} \neq 16$ . $μ_{0} = 16$ lies outside of the interval (10, 15), so we reject $H_{0}$ at the $α = 0.05$ level of significance.

Question 10.69

18. A 99% $t$ confidence interval for $μ_{1} - μ_{2}$ is (0, 100). Hypothesized values are

Question 10.70

19. A 90% $t$ confidence interval for $μ_{1} - μ_{2}$ is (−10, 10). Hypothesized values are

−10.1
−9.9
0

10.2.19

(a) $H_{0} : μ_{1} - μ_{2} = - 10 .1 vs . H_{a} : μ_{1} - μ_{2} \neq - 10.1$ . $μ_{0} = - 10.1$ lies outside of the interval (–10, 10), so we reject $H_{0}$ at the $α = 0.10$ level of significance. (b) $H_{0} : μ_{1} - μ_{2} = - 9.9 vs . H_{a} : μ_{1} - μ_{2} \neq - 9.9$ . $μ_{0} = - 9.9$ lies inside of the interval (–10, 10), so we do not reject $H_{0}$ at the $α = 0.10$ level of significance. (c) $H_{0} : μ_{1} - μ_{2} = 0 vs . H_{a} : μ_{1} - μ_{2} \neq 0$ . $μ_{0} = 0$ lies inside of the interval (–10, 10), so we do not reject $H_{0}$ at the $α = 0.10$ level of significance.

Question 10.71

20. A 95% $t$ confidence interval for $μ_{1} - μ_{2}$ is (−25, −15). Hypothesized values are

−16
−26
0

For Exercises 21–22, perform the indicated hypothesis test using the pooled variance method. The summary statistics were taken from random samples that were drawn independently. Assume $σ_{1}^{2} = σ_{2}^{2}$ .

Question 10.72

21. Test, at level of significance $α = 0.10$ , whether $μ_{1}$ > $μ_{2}$ .

Sample 1	$n_{1} = 36$	${\bar{x}}_{1} = 54$	$s_{1} = 10$
Sample 2	$n_{2} = 36$	${\bar{x}}_{2} = 52$	$s_{2} = 11$

10.2.21

$H_{0} : μ_{1} = μ_{2} vs . H_{a} : μ_{1} > μ_{2}$ . $t_{crit} = 1.294$ . Reject $H_{0}$ if $t_{data} \geq 1.294$ . $S_{pooled}^{2} = 110.5$ . $t_{data} \approx 0.807$ . Since $t_{data} \approx 0.807$ is not ≥ 1.294, we do not reject $H_{0}$ . There is insufficient evidence at the $α = 0.10$ level of significance that the population mean of Population 1 is greater than the population mean of Population 2.

Question 10.73

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

Sample 1	$n_{1} = 250$	${\bar{x}}_{1} = 3.0$	$s_{1} = 0.25$
Sample 2	$n_{2} = 150$	${\bar{x}}_{2} = 3.2$	$s_{2} = 0.30$

For Exercises 23 and 24, construct a 95% confidence interval for $μ_{1} - μ_{2}$ for the indicated data using the pooled variance method.

Question 10.74

23. The data in Exercise 21

10.2.23

(–2.940, 6.940). We are 95% confident that the difference in the population means of Population 1 and Population 2 lies between −2.940 and 6.940.

Question 10.75

24. The data in Exercise 22

For Exercises 25 and 26, perform the indicated hypothesis test using the $Z$ test. The summary statistics were taken from random samples that were drawn independently. Assume that $σ_{1}$ and $σ_{2}$ are known.

Question 10.76

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

Sample 1	$n_{1} = 49$	${\bar{x}}_{1} = 100$	$σ_{1} = 1$
Sample 2	$n_{2} = 36$	${\bar{x}}_{2} = 99$	$σ_{2} = 2$

10.2.25

$H_{0} : μ_{1} = μ_{2} vs . H_{a} : μ_{1} > μ_{2}$ . $Z_{crit} = 1.645$ . Reject $H_{0}$ if $Z_{data} \geq 1.645$ . $Z_{data} \approx 2.757$ . Since $Z \approx 2.757$ is $\geq 1.645$ , we reject $H_{0}$ . There is evidence at the $α = 0.05$ level of significance that the population mean of Population 1 is greater than the population mean of Population 2.

Question 10.77

26. Test, at level of significance $α = 0.10$ , whether $μ_{1} < μ_{2}$ .

Sample 1	$n_{1} = 64$	${\bar{x}}_{1} = 72$	$σ_{1} = 3$
Sample 2	$n_{2} = 100$	${\bar{x}}_{2} = 76$	$σ_{2} = 5$

For Exercises 27 and 28, construct a 95% $Z$ confidence interval for $μ_{1} - μ_{2}$ for the indicated data.

Question 10.78

27. The data in Exercise 25

10.2.27

(0.289, 1.711). We are 95% confident that the difference in the population means of Population 1 and Population 2 lies between 0.289 and 1.711.

Question 10.79

28. The data in Exercise 26

APPLYING THE CONCEPTS

For Exercises 29–49, assume normality and use Welch's $t$ test and $t$ interval unless otherwise indicated.

Question 10.80

29. PC Sales. A personal computer company launched an advertising campaign in the hopes of boosting sales. A random sample (sample 1) of 16 days before the advertising blitz showed mean sales of 120 computers per day with a standard deviation of 30. A random sample (sample 2) of 15 days after the advertisements appeared showed mean sales of 125 computers per day with a standard deviation of 35. If it is appropriate, test whether $μ_{1} < μ_{2}$ . If not, explain why not.

10.2.29

Since both sample sizes are less than 30 and the distribution of both populations is unknown, it is not appropriate to use Welch's $t$ test.

Question 10.81

30. Flight Delays. The U.S. Customs and Border Protection Agency keeps track of the flight delays at all U.S. international airports. The summary statistics for the waiting times at Los Angeles International Airport (Tom Bradley Terminal) on July 1, 2014 at 5 P.M. and at 11 P.M. are provided in the following table.

Waiting Times at 5 P.M.	$n_{1} = 1490$	${\bar{x}}_{1} = 27$ minutes	$s_{1} = 20$ minutes
Waiting Times at 11 P.M.	$n_{2} = 150$	${\bar{x}}_{2} = 11$ minutes	$s_{2} = 15$ minutes

Test whether the mean waiting times at 5 P.M.are longer than those at 11 P.M., using level of significance $α = 0.01$ .

Question 10.82

31. Income in California Counties. According to random samples taken in 2010 by the Bureau of Economic Analysis, the mean income for Sacramento County and Los Angeles County, California, was $31,987 and $33,179, respectively. Suppose the samples had the following sample statistics.

Sacramento County	$n_{1} = 36$	${\bar{x}}_{1} = $31,987$	$s_{1} = $5000$
los Angeles County	$n_{2} = 49$	${\bar{x}}_{2} = $33,179$	$s_{2} = $6000$

Provide the point estimate of the difference in population means $μ_{1} - μ_{2}$ .
Calculate the margin of error for a confidence level of 95%.
Construct and interpret a 95% confidence interval for $μ_{1} - μ_{2}$ .
Test, at level of significance $α = 0.05$ , whether $μ_{1} < μ_{2}$ .
Explain whether the confidence interval in (c) could have been used to perform the hypothesis test in (d). Why or why not?

10.2.31

(a) –$1192 (b) $2426.7954 (c) (–3618.7954, 1234.7954). We are 95% confident that the difference in population mean incomes $μ_{1} - μ_{2}$ lies between –$3618.7954 and $1234.7954. (d) $H_{0} : μ_{1} - μ_{2}$ versus $H_{a} : μ_{1} < μ_{2}$ . Reject $H_{0}$ if the $p – value \leq α = 0.05$ . $t_{data} \approx - 0.9971$ ; $p – value = 0.1627799835$ (TI-84: 0.1608312442). The $p – value = 0.1627799835$ 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 mean income of Sacramento County is less than that of Los Angeles County. (e) No. The hypothesis test in (d) is a one-tailed test and confidence intervals can only be used for two-tailed tests.

Question 10.83

32. Math Scores. The Institute of Educational Sciences published the results of the Trends in International Math and Science Study. In 2011, the sample mean mathematics scores for 8th-grade students from the United States and Hong Kong were 509 and 586, respectively. Suppose independent random samples are drawn from each population, and assume that the populations are normally distributed with the following summary statistics.

USA	$n_{1} = 10$	${\bar{x}}_{1} = 509$	$s_{1} = 80$
Hong Kong	$n_{2} = 12$	${\bar{x}}_{2} = 586$	$s_{2} = 70$

Provide the point estimate of the difference in population means $μ_{1} - μ_{2}$ .

Page 604
Calculate the margin of error for a confidence level of 90%.
Construct and interpret a 90% confidence interval for $μ_{1} - μ_{2}$ .
Test, at level of significance $α = 0.01$ , whether $μ_{1} < μ_{2}$ .
Provide two reasons why the confidence interval in (c) could not have been used to perform the hypothesis test in (d).

Question 10.84

33. Children per Classroom. According to LocalSchoolDirectory.com, the sample mean number of children per teacher in the towns of Cupertino, California, and Santa Rosa, California, are 20.9 and 19.3, respectively. Suppose random samples of classrooms are taken from each county, with the following sample statistics.

Cupertino	$n_{1} = 36$	${\bar{x}}_{1} = 20.9$	$s_{1} = 5$
Santa Rosa	$n_{2} = 64$	${\bar{x}}_{2} = 19.3$	$s_{2} = 4$

Construct and interpret a 99% confidence interval for $μ_{1} - μ_{2}$ .
Use the confidence interval in (a) to test, at level of significance $α = 0.01$ , whether $μ_{1}$ differs from $μ_{2}$ .

10.2.33

(a) (–1.047, 4.247). We are 99% confident that the interval captures the difference in the population mean number of children per teacher in the towns of Cupertino, California, and Santa Rosa, California.

(b) $H_{0} : μ_{1} - μ_{2} = 0 vs . H_{a} : μ_{1} - μ_{2} \neq 0$ . $μ_{0} = 0$ lies inside of the interval (–1.047, 4.247), so we do not reject $H_{0}$ . There is insufficient evidence at the $α = 0.01$ level of significance that the population mean number of children per teacher in the town of Cupertino, California, differs from the population mean number of children per teacher in the town of Santa Rosa, California.

Question 10.85

propertytax

34. Property Taxes. Suppose you want to move to either a small town in Ohio (sample 1) or a small town in North Carolina (sample 2). You did some research on property taxes in each state and chose two random samples shown in the table. The data represent the property taxes in dollars for a residence assessed at $250,000. Test whether $μ_{1} \neq μ_{2}$ using level of significance $α = 0.05$ .

North Carolina		Ohio
164	206	298	270
147	129	270	315
207	176	165	177
138	120	400	245
143	154	268	180
201	123	289	292
		285	291
		225

Question 10.86

35. Engineering Starting Salaries. The National Association of Colleges (NAC) reports mean starting salaries for college graduates in various disciplines. NAC reported that the mean salary for 2014 engineering graduates was $62,719, whereas the mean salary for 2013 engineering graduates was $62,535. Assume that these statistics were drawn from independent samples of size 100, each with a sample standard deviation of $10,000. Test whether there was a significant change in the salary of engineering graduates from 2013 to 2014, using level of significance $α = 0.10$ .

10.2.35

$H_{0} : μ_{1} - μ_{2}$ versus $μ_{1} - μ_{2} \neq μ_{2}$ ; Reject $H_{0}$ if the $p – value \leq α = 0.10$ . $t_{data} \approx 0.1301$ . $p – value = 0.896751292$ (TI-84: 0.8966133443). The $p – value = 0.896751292$ is not $\leq α = 0.10$ , so we do not reject $H_{0}$ . There is insufficient evidence at level of significance $α = 0.10$ that the population mean salary of engineering graduates has changed from 2013 to 2014.

Question 10.87

36. Park Usage. Suppose that planners for the town of The Woodlands, Texas, were interested in assessing usage of their parks. Random samples were taken of the number of daily visitors to Windvale Park and Cranebrook Park, with the statistics as reported here.

Windvale Park	$n_{1} = 36$	${\bar{x}}_{1} = 110$	$s_{1} = 60$
Cranebrook Park	$n_{2} = 30$	${\bar{x}}_{2} = 150$	$s_{2} = 75$

Construct and interpret a 95% confidence interval for $μ_{1} - μ_{2}$ .
Test at $α = 0.05$ whether $μ_{1}$ is less than $μ_{2}$ .
Explain whether the confidence interval in (a) could have been used to perform the hypothesis test in (b). Why or why not?

Coaching for the SAT. Use this information for Exercises 37–39. The College Board reports that a pretest and post-test study was done to investigate whether coaching had a significant effect on SAT scores. The improvement from pretest to post-test was 29 points for the coached sample of students, with a standard deviation of 59 points. For the uncoached students, the pretest to post-test improvement was 21 points with a standard deviation of 52 points.

Question 10.88

37. Suppose we consider a sample of 100 students from each group. Perform a test, at level of significance $α = 0.05$ , to determine whether the population mean coached SAT pretest–post-test improvement is greater than that for the uncoached students.

10.2.37

$H_{0} : μ_{1} = μ_{2} vs . H_{a} : μ_{1} > μ_{2}$ . Reject $H_{0}$ if $p – value \leq 0.05$ . $t_{data} = 1.017$ . $p – value = 0.1558$ . Since the $p – value$ is not $\leq 0.05$ , we do not reject $H_{0}$ . There is insufficient evidence that the population coached SAT score improvement is greater than the population noncoached SAT score improvement. Critical-value method: $H_{0} : μ_{1} = μ_{2} vs . H_{a} : μ_{1} > μ_{2}$ . $t_{crit} = 1.662$ . Reject $H_{0}$ if $t_{data} \geq 1.662$ . $t_{data} = 1.017$ . Since $t_{data} = 1.017$ is not $\geq 1.662$ , we do not reject $H_{0}$ . There is insufficient evidence at the $α = 0.05$ level of significance that the population mean coached SAT improvement is greater than the population mean noncoached improvement.

Question 10.89

38. Refer to Exercise 37.

Find a point estimate of the difference in population means.
Find a 99% confidence interval for the difference in population means.
Determine whether the population means differ, at level of significance $α = 0.01$ .

Question 10.90

39. What if the sample sizes for each group were some number greater than $n = 100$ ?

How would this affect the width of the confidence interval in Exercise 38 (b)? Is this good? Explain.
Would this change have any effect on our conclusion in the hypothesis test in Exercise 38 (c)? Explain why or why not.

10.2.39

(a) Since the width of the confidence interval is $2 \cdot t_{a / 2} \cdot \sqrt{\frac{s_{1}^{2}}{n_{1}} + \frac{s_{2}^{2}}{n_{2}}}$ , an increase in the sample sizes will result in a decrease in the width of the confidence interval. This is good because smaller confidence intervals give a more precise estimate. (b) It depends on how large the new sample sizes are.

Question 10.91

40. Nursing Support Services. A statistical study found that when nurses made home visits to pregnant teenagers to provide support services, discourage smoking, and otherwise provide care, the sample mean birth weight of the babies was higher for this treatment group (3285 grams) than for the control group (2922 grams) when the visits began before mid-gestation.¹⁰ There were 21 patients in the treatment group and 11 in the control group. Suppose the birth weights for the babies in both groups follow a normal distribution. Assume that the population standard deviation in each sample is 500 grams.

Construct and interpret a 95% $Z$ confidence interval for $μ_{1} - μ_{2}$ .
Test, at level of significance $α = 0.05$ , whether the population birth weight differs between the two groups. Use the $Z$ test.
Assess the strength of evidence against the null hypothesis.

Page 605

Question 10.92

41. Nursing Support Services. Refer to Exercise 40. What if the birth weights of the babies in each group are the same certain amount greater? Explain how this would affect the following:

${\bar{x}}_{1} - {\bar{x}}_{2}$
$t_{data}$
$p$ -value
conclusion

10.2.41

(a) Remain the same (b) Remain the same (c) Remain the same

(d) Remain the same

Zooplankton and Phytoplankton. Refer to the table below for Exercises 42 and 43. Meta-analysis refers to the statistical analysis of a set of similar research studies. In a meta-analysis, each data value represents an effect size calculated from the results of a particular study. The table contains effect sizes calculated in a meta-analysis for zooplankton and phytoplankton.¹¹ Not surprisingly, the paper found that zooplankton biomass was reduced by the introduction of zooplanktivorous fish (that is, fish that eat zooplankton), but that phytoplankton biomass increased, because there were now fewer zooplankton, and the zooplankton eat the phytoplankton. Such an effect is called a trophic cascade. See if you can replicate the scientists' results in the following exercises.

Zooplankton		Phytoplankton
−2.37	−3.00	10.61	3.04
−0.64	−0.68	2.97	0.65
−2.05	−1.39	1.58	2.55
−1.54	−0.64	2.55	1.05
−6.60	−3.88	5.67	2.11
0.26		1.57

Question 10.93

plankton

42. Use technology to construct a comparison dot plot or a comparison histogram of the zooplankton and the phytoplankton effect sizes. Is there evidence of a difference in mean effect size for the two groups?

Question 10.94

plankton

43. Test whether the population means effect sizes differ, at level of significance $α = 0.01$ .

10.2.43

$H_{0} : μ_{1} = μ_{2} vs . H_{a} : μ_{1} \neq μ_{2}$ . Reject $H_{0}$ if the $p – value \leq α = 0.01$ . $Z_{data} \approx - 5.0267$ . $p – value = 0.0002583121504$ (TI-84: 0.000092880601). The $p – value = 0.0002583121504$ is $\leq α = 0.01$ , so we reject $H_{0}$ . There is evidence at level of significance $α = 0.01$ that the population mean effect sizes of zooplankton differs from that of the phytoplankton.

Studying Time for Biologists and Psychologists. Use the following information for Exercises 44–47: The National Survey of Student Engagement reported in 2012 that the mean number of hours spent studying per week for biology majors was 16.7, whereas the mean for psychology majors was 13.9. Assume that these are sample means from sample sizes of 100, and that the sample standard deviations were each 5 hours.

Question 10.95

44. Calculate the margin of error for a 95% confidence interval for the difference in population mean studying times (Biology – Psychology).

Question 10.96

45. Construct a 95% confidence interval for the difference in population mean studying times (Biology – Psychology).

10.2.45

(1.3950, 4.2050)

Question 10.97

46. Use the confidence interval to determine whether a significant difference exists between the population mean studying times.

Question 10.98

47. Test, using level of significance $α = 0.05$ , whether the population mean studying times are equal. Does your conclusion agree with that in Exercise 46?

10.2.47

$H_{0} : μ_{1} = μ_{2} vs . H_{a} : μ_{1} \neq μ_{2}$ . Reject $H_{0}$ if the $p – value \leq α = 0.05$ . $t_{data} \approx 3.9598$ . $p – value = 0.000141495588$ (TI-84: 0.0001045746); The $p – value = 0.000141495588$ is $\leq α = 0.05$ , so we reject $H_{0}$ . There is evidence at level of significance $α = 0.05$ that the population mean study time of biology majors differs from that of psychology majors. Yes.

Question 10.99

48. Does Multitasking Degrade Performance? The journal Computers and Education reported that multitasking during class may degrade student performance on assessments.¹² Students were randomly assigned into two groups: multitaskers and non-multitaskers. The multitaskers have their laptops open during class, doing things such as surfing Facebook and so forth. The following table provides the summary statistics for a class quiz for the two groups. Test whether the population mean score for the multitaskers is lower than that of the non-multitaskers, using level of significance $α = 0.05$ .

Multitaskers	$n_{1} = 20$	${\bar{x}}_{1} = 55$	$s_{1} = 11$
non-multitaskers	$n_{2} = 20$	${\bar{x}}_{2} = 66$	$s_{2} = 12$

Question 10.100

49. Does Multitasking Degrade your neighbor's Performance? Refer to the previous exercise. The same journal article reported that those sitting next to the multitaskers may have also suffered from degraded quiz performance, due to being distracted by their neighbor's laptop multitasking. The following table provides the summary statistics for a class quiz for the two groups: those sitting where they could see the multitasker's screen (multitasker's neighbors), and those who could not (multitasker's non-neighbors). Test whether the population mean score for the multitasker's neighbors is lower than that of the multitasker's non-neighbors, using level of significance $α = 0.05$ .

Multitasker's neighbors	$n_{1} = 19$	${\bar{x}}_{1} = 56$	$s_{1} = 12$
Multitasker's non-neighbors	$n_{2} = 19$	${\bar{x}}_{2} = 73$	$s_{2} = 12$

10.2.49

$H_{0} : μ_{1} = μ_{2} vs . H_{a} : μ_{1} < μ_{2}$ . Reject $H_{0}$ if the $p – value \leq α = 0.05$ . $t_{data} \approx - 4.3665$ s. $p – value = 0.0001860220303$ (TI-84: 0.00005106366) The $p – value = 0.0001860220303$ is $\leq α = 0.05$ , so we reject $H_{0}$ . There is evidence at level of significance $α = 0.05$ that the population mean score for the multitasker's neighbors is less than that of the multitasker's non-neighbors.

BRINGING IT ALL TOGETHER

Do Prior Student Evaluations Influence Students' Ratings of Professors? A study in 1950 reported that instructor reputation affected students' ratings of their instructors.¹³ Towler and Dipboye uncovered experimental evidence in support of this phenomenon.¹⁴ They randomly assigned to students one of two summaries of prior student evaluations: one for a “charismatic instructor” and the other for a “punitive instructor.” The “charismatic” summary included such phrases as “always lively and stimulating in class” and “always approachable and treated students as individuals.” The “punitive” summary included such phrases as “did not show an interest in students' progress” and “consistently seemed to grade students harder.” All subjects were then shown the same 20-minute lecture video given by the same instructor. They were asked to rate the instructor using three questions, and a summary rating score was calculated. The summary statistics are shown below. Assume that both populations are normally distributed and that the samples are drawn independently. Were students' ratings influenced by the prior student evaluations? Find out by answering Exercises 50–58.

Page 606

Reputation	Subjects	Sample mean rating	Sample standard deviation
Charismatic (sample 1)	$n_{1} = 25$	${\bar{x}}_{1} = 2.613$	$s_{1} = 0.533$
Punitive (sample 2)	$n_{2} = 24$	${\bar{x}}_{2} = 2.236$	$s_{2} = 0.543$

Question 10.101

50. Test whether the population mean student evaluation rating for the charismatic group is greater than that of the punitive group, using a $t$ test and level of significance $α = 0.05$ .

Question 10.102

51. Construct a 95% $t$ confidence interval for $μ_{1} - μ_{2}$ .

10.2.51

(0.0588, 0.6952) {TI-84: (0.0676, 0.6864)}

Question 10.103

52. Recall we explored some equivalences between confidence intervals and hypothesis tests. Could we have used the confidence interval in Exercise 51 to perform the hypothesis test in Exercise 50? Explain.

Question 10.104

53. Test whether the population mean student ratings differ, using a $t$ test and level of significance $α = 0.05$ .

10.2.53

$H_{0} : μ_{1} = μ_{2} vs . H_{a} : μ_{1} \neq μ_{2}$ . Reject $H_{0}$ if the $p – value \leq α = 0.05$ . $t_{data} \approx 2.4515$ . $p – value = 0.0222425869$ (TI-84: 0.0180119956) The $p – value = 0.0222425869$ is $\leq α = 0.05$ , so we reject $H_{0}$ . There is evidence at level of significance $α = 0.05$ that the population mean student evaluation rating for the charismatic group differs from that of the punitive group.

Question 10.105

54. Use the confidence interval in Exercise 51 to conduct $t$ tests, at level of significance $α = 0.05$ , for whether $μ_{1} - μ_{2}$ takes the following values:

Question 10.106

55. Do the following:

Find the pooled variance.
Test whether the population mean student ratings differ, using the pooled variance $t$ test and level of significance $α = 0.05$ .
Compare your results from (b) with Exercise 53.

10.2.55

(a) $s_{pooled}^{2} = 0.2893545319$ (b) $H_{0} : μ_{1} = μ_{2} vs . H_{a} : μ_{1} \neq μ_{2}$ . Reject $H_{0}$ if the $p – value$ $p – value \leq α = 0.05$ . $t_{data} \approx 2.4525$ . $p – value = 0.0179558037$ . The $p – value = 0.0179558037$ is $\leq α = 0.05$ , so we reject $H_{0}$ . There is evidence at level of significance $α = 0.05$ that the population mean student evaluation rating for the charismatic group differs from that of the punitive group. (c) The value of $t_{data}$ in (b) is a little bit larger than the value of $t_{data}$ in Problem 53, so the $p – value$ in (b) is a little smaller than the $p – value$ in Problem 53. The conclusions of both tests are to reject $H_{0}$ .

Question 10.107

56. Construct a 95% pooled variance $t$ confidence interval for $μ_{1} - μ_{2}$ . Compare your results with Exercise 51.

For Exercises 57 and 58, assume that the given standard deviations are population standard deviations.

Question 10.108

57. Use a $Z$ test to test whether the population mean student ratings differ, using level of significance $α = 0.05$ . Compare your results to Exercises 53 and 55.

10.2.57

$H_{0} : μ_{1} = μ_{2} vs . H_{a} : μ_{1} \neq μ_{2}$ . Reject $H_{0}$ if the $p – value$ $p – value \leq α = 0.05$ . $Z_{data} \approx 2.4515$ . $p – value = 0.0142254108$ . The $p – value = 0.0142254108$ is $\leq α = 0.05$ , so we reject $H_{0}$ . There is evidence at level of significance $α = 0.05$ that the population mean student evaluation rating for the charismatic group differs from that of the punitive group. The value of $Z_{data}$ in this problem equals the value of $t_{data}$ in Problem 53 but it is a little bit smaller than the value of $t_{data}$ in Problem 55 (b). The $p – value$ is smaller than the other $p – value$ s. All three hypothesis tests have the same conclusion of reject $H_{0}$ .

Question 10.109

58. Construct a 95% $Z$ confidence interval for $μ_{1} - μ_{2}$ . Compare your results with Exercises 51 and 56.

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 loan request amount between those approved for and those denied a loan. Use technology to do the following:

Question 10.110

bankloan_approved

bankloan_denied

59. Obtain independent random samples of size 100, one each from the bankloans_approved data set and the BankLoans_Denied data set.

Question 10.111

bankloan_approved

bankloan_denied

60. For each sample, find the summary statistics for the variable Request amount, that is, the sample size, sample mean, and sample standard deviation.

Question 10.112

BankLoan_Approved

bankloan_denied

61. Perform a $t$ test, using level of significance $α = 0.05$ , to determine whether the population mean request amount differs between those approved for a loan and those denied the loan.

North Carolina		Ohio
164	206	298	270
147	129	270	315
207	176	165	177
138	120	400	245
143	154	268	180
201	123	289	292
		285	291
		225

North Carolina		Ohio
164	206	298	270
147	129	270	315
207	176	165	177
138	120	400	245
143	154	268	180
201	123	289	292
		285	291
		225

North Carolina		Ohio
164	206	298	270
147	129	270	315
207	176	165	177
138	120	400	245
143	154	268	180
201	123	289	292
		285	291
		225