Question

A study was conducted to determine the effectiveness of a diet program in reducing weight of obeys people. Eight randomly selected patients were put on this diet and the amount of weight losses (in pounds) were recorded. Suppose the sample mean and the sample standard deviation of weight losses were 36 and 5.291503 pounds, respectively. Construct a 90% confidence interval for the average weight loss and based on the confidence interval constructed determine if you would expect to see an average weight loss of 35 pounds. (Assume that samples were taken independently from a normal distribution.)

Group of answer choices

90% CI: (32.45478, 39.54522), an average weight loss of 35 pounds is expected to see.

90% CI: (32.45478, 39.54522), an average weight loss of 35 pounds is not expected to see.

90% CI: (34.74658, 37.25342), an average weight loss of 35 pounds is expected to see.

90% CI: (34.74658, 37.25342), an average weight loss of 35 pounds is not expected to see.

90% CI: (33.35278, 38.64722), an average weight loss of 35 pounds is expected to see.

90% CI: (33.35278, 38.64722), an average weight loss of 35 pounds is not expected to see.

Answer #1

For n - 1 = 7 degrees of freedom, we have from the t
distribution tables here:

P( t_{7} < 1.895) = 0.95

Therefore, due to symmetry, we get here:

P( - 1.895 < t_{7} < 1.895) = 0.9

Therefore the confidence interval here is obtained as:

As 35 lies in the above interval, therefore we do expect to see
it. **Therefore a is the correct answer here.**

90% CI: (32.45478, 39.54522), an average weight loss of 35 pounds is expected to see.

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