Question

In a test of the effectiveness of garlic for lowering​ cholesterol, 50 subjects were treated with garlic in a processed tablet form. Cholesterol levels were measured before and after the treatment. The changes ​(beforeminus​after) in their levels of LDL cholesterol​ (in mg/dL) have a mean of 3.2 and a standard deviation of 18.5. 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? Click here to view a t distribution table.LOADING... Click here to view page 1 of the standard normal distribution table.LOADING... Click here to view page 2 of the standard normal distribution table.LOADING... What is the confidence interval estimate of the population mean mu​? nothing ​mg/dLless thanmuless than nothing ​mg/dL

Answer 1

For n - 1 = 49 degree of freedom, we have from t distribution tables here:

P( t₄₉ < 2.010) = 0.975

Therefore, due to symmetry, we have here:
P( - 2.010 < t₄₉ < 2.010) = 0.95

Therefore the confidence interval here is computed as:

This is the required 95% confidence interval for the population mean here.