Question

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

43 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 3.5 and a standard deviation of 16.3. 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

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.01 /2, 43- 1 ) = 2.698
3.5 ± t(0.01/2, 43 -1) * 16.3/√(43)
Lower Limit = 3.5 - t(0.01/2, 43 -1) 16.3/√(43)
Lower Limit = -3.21
Upper Limit = 3.5 + t(0.01/2, 43 -1) 16.3/√(43)
Upper Limit = 10.21
99% Confidence interval is ( -3.21 , 10.21 )

Since 0 contained in confidence interval, the test for garlic in reducing LDL​cholesterol is

not effective.