math statistics

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

4.1

and a standard deviation of

16.6.

Construct a

99​%

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?

Answer 1

Making the standard assumptions of normality, independence, randomness, etc a CI around a mean with a given x¯ (x-bar) of 4.1 mg/dl , a standard deviation s of 16.6, and a sample size n of 44 and a 99% confidence interval is calculated using the following formula:

(x¯ - t_α/2 * s / √n , x¯ + t_α/2 * s / √n)

A Gosset's (Student's) t distribution is used as the data are from samples and the question is for the CI of a sample mean.

Looking up the t statistic for 43 degrees of freedom and significance level (α/2) of .005 we get 2.695102

The rest is just using the formula.

Good luck.