Probability and Statistics for Engineering and the Sciences
9th Edition
ISBN: 9781305251809
Author: Jay L. Devore
Publisher: Cengage Learning
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Chapter 9.2, Problem 27E
Anorexia Nervosa (AN) is a psychiatric condition leading to substantial weight loss among women who are fearful of becoming fat. The article “Adipose Tissue Distribution After Weight Restoration and Weight Maintenance in Women with Anorexia Nervosa” (Amer. J. of Clinical Nutr., 2009: 1132–1137) used whole-body magnetic resonance imagery to determine various tissue characteristics for both an AN sample of individuals who had undergone acute weight restoration and maintained their weight for a year and a comparable (at the outset of the study) control sample. Here is summary data on intermuscular adipose tissue (IAT; kg)
| Condition | Sample |
Sample SD | |
| AN | 16 | .52 | .26 |
| Control | 8 | .35 | .15 |
Assume that both samples were selected from
a. Calculate an estimate for true average IAT under the described AN protocol, and do so in a way that conveys information about the reliability and precision of the estimation.
b. Calculate an estimate for the difference between true average AN IAT and true average control IAT, and do so in a way that conveys information about the reliability and precision of the estimation. What does your estimate suggest about true average AN IAT relative to true average control IAT?
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The article “Treadmill Exercise and Resistance Training in Patients With Peripheral Arterial Disease With and Without Intermittent Claudication. A Randomized Controlled Trial” (M. McDermott, P. Ades, et al., Journal of the American Medical Association, 2009:165–174) reported the results of a study to determine whether treadmill exercise could improve the walking ability of patients suffering from claudication, which is pain caused by insufficient blood flow to the muscles of the legs. A sample of 48 patients walked on a treadmill for six minutes every day. After six months, the mean distance walked in six minutes was 348 meters, with a standard deviation of 80 m. For a control group of 46 patients who did not walk on a treadmill, the mean distance was 309 m with a standard deviation of 89 m. Can you conclude that the mean distance walked for patients using a treadmill is greater than the mean for the controls? Use the α = 0.05 level of significance.
The article “Treadmill Exercise and Resistance Training in Patients With Peripheral Arterial Disease With and Without Intermittent Claudication. A Randomized Controlled Trial” (M. McDermott, P. Ades, et al., Journal of the American Medical Association, 2009:165–174) reported the results of a study to determine whether treadmill exercise could improve the walking ability of patients suffering from claudication, which is pain caused by insufficient blood flow to the muscles of the legs. A sample of 48 patients walked on a treadmill for six minutes every day. After six months, the mean distance walked in six minutes was 348 m, with a standard deviation of 80 m. For a control group of 46 patients who did not walk on a treadmill, the mean distance was 309 m with a standard deviation of 89 m. Find a 95% confidence interval for the difference in mean distance walked between the two groups of patients.
Chapter 9 Solutions
Probability and Statistics for Engineering and the Sciences
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