Question

In a test of the effectiveness of garlic for lowering​ cholesterol, 49 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 2.7 and a standard deviation of 16.8. 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

a) For a test of the effectiveness of garlic for lowering​ cholesterol, n = 49 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 M = 2.7 and a standard deviation of s = 16.8.

Thus the confidence interval is calculated as:

μ = M ± t(sM)

since population standard deviation is not given hence t-distribution is applicable.

where:

M = sample mean df = n-1
t = t statistic determined by the confidence level and the degrees of freedom.
sM = standard error = √(s2/n)

M = 2.7 df= 49-1 = 48
t = 2.01 calculated using excel formula which is =T.INV.2T(0.05,48)
sM = √(16.82/49) = 2.4

μ = M ± t(sM)
μ = 2.7 ± 2.01*2.4
μ = 2.7 ± 4.826

Thus the confidence interval is computed as:

[-2.126, 7.526].

b) Based on the confidence interval we can see that it goes form -ve value to a positive value hence it is said that there is no significant effect of garlic for lowering cholesterol.