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 $\mathrm{mg} / \mathrm{dL}$ ) have a mean of 5.6 and a standard deviation of 19.4. 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?
What does the confidence interval suggest about the effectiveness of the treatment?
A. The confidence interval limits contain 0 , suggesting that the garlic treatment did not affect the LDL cholesterol levels.
B. The confidence interval limits contain 0 , suggesting that the garlic treatment did affect the LDL cholesterol levels.
C. The confidence interval limits do not contain 0 , suggesting that the garlic treatment did affect the LDL cholesterol levels.
D. The confidence interval limits do not contain 0 , suggesting that the garlic treatment did not affect the LDL cholesterol levels.
Answer
Therefore, the answer to the question is \(\boxed{\text{A. The confidence interval limits contain 0 , suggesting that the garlic treatment did not affect the LDL cholesterol levels.}}\)
Step 1 ：We are given that the sample mean (\(\bar{x}\)) is 5.6, the sample standard deviation (s) is 19.4, and the sample size (n) is 43.
Step 2 ：We need to construct a 99% confidence interval for the mean net change in LDL cholesterol. The formula for a confidence interval is given by: \(\bar{x} \pm z \frac{s}{\sqrt{n}}\)
Step 3 ：We need to find the z-score for a 99% confidence level. This can be found using a standard normal distribution table, or calculated using a statistical software. The z-score for a 99% confidence level is approximately 2.576.
Step 4 ：Substituting the given values into the formula, we get the confidence interval as \(5.6 \pm 2.576 \frac{19.4}{\sqrt{43}}\)
Step 5 ：Calculating the above expression, we get the confidence interval as approximately (-2.02, 13.22).
Step 6 ：This interval contains 0, which suggests that the garlic treatment did not have a significant effect on LDL cholesterol levels.
Step 7 ：Therefore, the answer to the question is \(\boxed{\text{A. The confidence interval limits contain 0 , suggesting that the garlic treatment did not affect the LDL cholesterol levels.}}\)
