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
(beforeminus
after)
in their levels of LDL cholesterol (in mg/dL) have a mean of
3.5
and a standard deviation of
16.5
.
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?
df = n - 1 = 43 - 1 = 42
t critical value at 0.01 significance level with 42 df = 2.698
99% confidence interval for is
- t * S / sqrt(n) < < + t * S / sqrt(n)
3.5 - 2.698 * 16.5 / sqrt(43) < < 3.5 + 2.698 * 16.5 / sqrt(43)
-3.289 < < 10.289
99% CI is (-3.289 , 10.289 )
Since 0 contained in confidence interval, test is not effective.
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