10 Two-Sample Inference

Chapter 10 Review Exercises

section 10.1

Question 10.195

1. Assume that a sample of differences for the matched pairs in the table follows a normal distribution, and perform Steps (a) and (b).

Subject	1	2	3	4	5	6	7	8
Sample 1	100.7	110.2	105.3	107.1	95.6	109.9	112.3	94.7
Sample 2	104.4	112.5	105.9	111.4	99.8	109.9	115.7	97.7

Calculate ${\bar{x}}_{d}$ and $s_{d}$ .
Construct a 95% confidence interval for $μ_{d}$ .

10.99.1

(a) ${\bar{x}}_{d} = - 2.6875$ , $s_{d} = 1.6146$ (b) (–4.0376, –1.3374)

Question 10.196

2. For the data in Exercise 1, test whether $μ_{d} < 0$ , using the critical-value method and level of significance $α = 0.05$ .

Question 10.197

3. For the data in Exercise 1, test whether $μ_{d}, 0$ , using the $p$ -value method and level of significance $α = 0.05$ .

10.99.3

$H_{0} : μ_{d} = 0 vs . H_{a} : μ_{d} < 0$ . Reject $H_{0}$ if $p -value < 0.05. t_{data} = - 4.708$ . $p – value = 0.0010939869$ . Since the $p – value \leq 0 .05$ , we reject $H_{0}$ . There is evidence that the population mean of the differences is less than 0.

section 10.2

Refer to the following summary statistics for two independent samples for Exercises 4–7.

Sample 1	$n_{1} = 36$	${\bar{x}}_{1} = 14.4$	$s_{1} = 0.01$
Sample 2	$n_{2} = 81$	${\bar{x}}_{2} = 14.3$	$s_{2} = 0.02$

Question 10.198

4. We are interested in constructing a 95% confidence interval for $μ_{1} - μ_{2}$ . Explain why it is appropriate to do so.

Question 10.199

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

10.99.5

0.1

Question 10.200

6. Calculate the margin of error for a confidence level of 95%.

Question 10.201

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

10.99.7

(0.094, 0.106). We are 95% confident that the interval captures the difference in population means.

Question 10.202

8. A random sample of 49 young persons with college degrees had a mean salary of $30,000 with a standard deviation of $5000. An independent random sample of 36 young persons without college degrees had a mean salary of $25,000 and a standard deviation of $4000.

Test, at level of significance $α = 0.10$ , whether the population mean salary of college graduates $μ_{1}$ is greater than the population mean salary of those without a degree $μ_{2}$ .
Construct and interpret a 90% confidence interval for $μ_{1} - μ_{2}$ .

section 10.3

Question 10.203

9. The Centers for Disease Control, in their Pregnancy Risk Assessment Monitoring System, reported that 641 of 823 new mothers living in Florida and 658 of 824 new mothers living in North Carolina took their babies in for a checkup within one week of delivery.²²

Find a 99% confidence interval for the difference in population proportions.
Perform a hypothesis test of whether the population proportions differ, at level of significance $α = 0.01$ , using the $p$ -value method.

10.99.9

(a) (–0.0715, 0.0322) (b) Since the $p – value = 0.3320$ is not ≤ 0.01, we do not reject $H_{0}$ . There is insufficient evidence that the population proportion of new mothers in Florida who took their babies in for a checkup within one week of delivery is different from the proportion of new mothers in North Carolina who took their babies in for a checkup within one week of delivery.

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section 10.4

For Exercises 10 and 11, assume that the populations are normally distributed. Perform the indicated hypothesis tests using the critical-value method.

Question 10.204

10. $H_{0} : σ_{1} = σ_{2} vs . H_{a} : σ_{1} > σ_{2}, α = 0.01, n_{1} = 55, n_{2} = 107, s_{1} = 20.2, s_{2} = 5.6$

Question 10.205

11. $H_{0} : σ_{1} = σ_{2} vs . H_{a} : σ_{1} < σ_{2}, α = 0.05, n_{1} = 125, n_{2} = 27, s_{1} = 5.7, s_{2} = 10.2$

10.99.11

$H_{0} : σ_{1} = σ_{2} vs . H_{a} : σ_{1} < σ_{2}$ . $F_{crit} = F_{1 - α, n_{1} - 1, n_{2} - 1} = F_{0.95, 124, 26} = 0.6173$ . Reject $H_{0}$ if $F_{data} \leq 0.6173$ . $F_{data} = 0.3123$ . Since $F_{data} = 0.3123$ is ≤ 0.6173, we reject $H_{0}$ . There is evidence at the $α = 0.05$ level of significance that the population standard deviation of Population 1 is less than the population standard deviation of Population 2.

For Exercises 12 and 13, assume that the populations are normally distributed. Perform the indicated hypothesis tests using the $p$ -value method for each exercise.

Question 10.206

12. $H_{0} : σ_{1} = σ_{2} vs . H_{a} : σ_{1} > σ_{2}, α = 0.05, n_{1} = 1050, n_{2} = 250, s_{1} = 10.4, s_{2} = 4.6$

Question 10.207

13. $H_{0} : σ_{1} = σ_{2} vs . H_{a} : σ_{1} < σ_{2}, α = 0.01, n_{1} = 150, n_{2} = 80, s_{1} = 25.1, s_{2} = 35.8$

10.99.13

$H_{0} : σ_{1} = σ_{2} vs . H_{a} : σ_{1} < σ_{2}$ . Reject $H_{0}$ if the $p – value \leq 0 .01$ . $F_{data} = 0.4916$ . $p – value = 0.0001$ . Since the $p – value = 0.0001$ is ≤ 0.01, we reject $H_{0}$ . There is evidence at the $α = 0.01$ level of significance that the population standard deviation of Population 1 is less than the population standard deviation of Population 2.