In a test of the effectiveness of garlic for lowering cholesterol, 47 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 2.9 and a standard deviation of 17.7. 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?
Click here to view a t distribution table.
Click here to view page 1 of the standard normal distribution table.
Click here to view page 2 of the standard normal distribution table.
What is the confidence interval estimate of the population mean $\mu$ ?
\[
\mathrm{mg} / \mathrm{dL}< \mu< \square \mathrm{mg} / \mathrm{dL}
\]
(Round to two decimal places as needed.)
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 affect the LDL cholesterol levels.
B. The confidence interval limits do not contain 0 , suggesting that the garlic treatment did not affect the LDL cholesterol levels.
C. The confidence interval limits contain 0 , suggesting that the garlic treatment did not affect the LDL cholesterol levels.
D. The confidence interval limits do not contain 0 , suggesting that the garlic treatment did affect the LDL cholesterol levels.
Answer
Since the confidence interval contains 0, this suggests that the garlic treatment did not have a significant effect on LDL cholesterol levels. Therefore, the correct interpretation is that the garlic treatment did not affect the LDL cholesterol levels.
Step 1 ：Given that the sample mean (\(\bar{x}\)) is 2.9, the sample standard deviation (\(s\)) is 17.7, and the sample size (\(n\)) is 47.
Step 2 ：We need to construct a 99% confidence interval for the mean change in LDL cholesterol. The formula for a confidence interval is \(\bar{x} \pm t \left( \frac{s}{\sqrt{n}} \right)\), where \(t\) is the t-score corresponding to our desired level of confidence.
Step 3 ：The degrees of freedom is \(n - 1\), which is 46 in this case. The t-score for a 99% confidence interval with 46 degrees of freedom can be found in a t-distribution table. The t-score is approximately 2.687.
Step 4 ：Substitute the given values into the formula, we get the margin of error as \(2.687 \times \frac{17.7}{\sqrt{47}}\), which is approximately 6.937.
Step 5 ：Subtract and add the margin of error from the sample mean to get the confidence interval. The lower limit is \(2.9 - 6.937\), which is -4.04. The upper limit is \(2.9 + 6.937\), which is 9.84.
Step 6 ：Thus, the confidence interval estimate of the population mean \(\mu\) is \(-4.04 \, \mathrm{mg} / \mathrm{dL}<\mu<9.84 \, \mathrm{mg} / \mathrm{dL}\).
Step 7 ：Since the confidence interval contains 0, this suggests that the garlic treatment did not have a significant effect on LDL cholesterol levels. Therefore, the correct interpretation is that the garlic treatment did not affect the LDL cholesterol levels.
