A medical researcher wishes to investigate the effectiveness of exercise versus

Question

A medical researcher wishes to investigate the effectiveness of exercise versus diet to toting weight. Two groups of 25 overweight adults we used, with a subject in each group matched to a similar subject. In the other group on the basis of a number of physiological variables. One group is placed on a regular program of vigorous exercise, but with no restriction on diet, and the other group on a strict diet, but with no requirement to exercise. The weight losses after 20 weeks are determined for each subject, and the difference between matched pairs of subjects (weight loss of subject in exercise group - weight loss of matched subject in diet group) is computed. The mean of these differences in weight toss is found to be -2 pounds with standard deviation s = 6 pounds. Is this evidence of a difference to mean weight loss for the two methods? To test this, consider the population of differences (weight loss an overweight adult would experience after 20 weeks on the exercise program) - (weight loss the same adult would experience after 20 weeks on the strict diet). Let mu be the mean of true population of differences and assume their distribution is approximately Normal. We test the null hypothesis that mu = 0 versus the alternative hypothesis mu 0 using the matched pairs t-test. The P-value for this test is: greater than 0.10. between 0.10 and 0.05. between 0.05 and 0.01. less than 0.01. If we accept the null hypothesis when, in fact, it is false, we have: committed a Type I error. committed a Type II error. a probability of being correct that is equal to the P-value. a probability of being correct that is equal to 1 - P-value. The scores of a certain population on the Wechsier Intelligence Scale for Children (WISC) are thought to be Normally distributed with mean mu and standard deviation sigma = 10. A simple random sample of 25 children from this population is taken and each is given the WISC. The mean of the 25 scores is r = 104.32. Based on these data, a 95% confidence interval for mu is: 104.32 plusminus 0.78. 104.32 plusminus 3.29. 104.32 plusminus 3.92. 104.32 plusminus 19.60

Explanation / Answer

4) b is correct

5) c is correct

6) a is correct . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. / . . . .. . .. . . . . . . .. .. ............ . . . . .. . . . . . . . . .. .

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