Question

In a test of the effectiveness of garlic for lowering​ cholesterol, 44

subjects were treated with garlic in a processed tablet form. Cholesterol levels were measured before and after the treatment. The changes (before−​after)

in their levels of LDL cholesterol​ (in mg/dL) have a mean of 3.2 and a standard deviation of 16.7. Construct a 95​%

confidence interval estimate of the mean net change in LDL cholesterol after the garlic treatment. What does the confidence interval suggest about the effectiveness of garlic in reducing LDL​ cholesterol?

What is the confidence interval estimate of the population mean μ​?

​(Round to two decimal places as​ needed.)

Answer 1

Solution :

Given that mean x-bar = 3.2 , standard deviation s = 16.7 , n = 44

=> df = n - 1 = 43

=> for 95% confidence interval, t = 2.017

=> A 95% confidence interval of the population mean is

=> x-bar +/- t*s/sqrt(n)

=> 3.2 +/- 2.017*16.7/sqrt(44)

=> (-1.8780 , 8.2780)

=> (-1.88 , 8.28) (rounded)

=> The confidence interval limits contain 0, suggesting that the garlic treatment did not affect the LDL cholesterol levels