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 38 and 4.598136 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.)

Potential Answers:

(A) 90% CI: (34.91932, 41.08068), an average weight loss of 35 pounds is expected to see.

(B) 90% CI: (36.91082, 39.08918), an average weight loss of 35 pounds is expected to see.

(C) 90% CI: (36.91082, 39.08918), an average weight loss of 35 pounds is not expected to see.

(D) 90% CI: (34.91932, 41.08068), an average weight loss of 35 pounds is not expected to see

(E) 90% CI: (35.69965, 41.08068), an average weight loss of 35 pounds is not expected to see.

(F) 90% CI: (35.69965, 41.08068), an average weight loss of 35 pounds is expected to see.

Answer 1

Solution:

Given:

Sample size = n =8

Sample mean =

Sample standard deviation = s = 4.598136

Confidence level = c = 90%

We have to 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.

where

t_c is t critical value for c = 90%  confidence level

Thus two tail area = 1 - c = 1 - 0.90 = 0.10

df = n - 1 = 8 - 1 = 7

Look in  t table for df = 7 and two tail area = 0.10 and find t critical value

t_c = 1.895

thus

thus

thus correct answer is:

(A) 90% CI: (34.91932, 41.08068), an average weight loss of 35 pounds is expected to see.

Since 35 is within the limits of 90% confidence interval.