In a test of the effectiveness of garlic for lowering cholesterol, 48 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 3.3 and a standard deviation of 16.6 . 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 affect the LDL cholesterol levels.
Answer
\(\boxed{\text{Final Answer: The confidence interval estimate of the population mean } \mu \text{ is } -3.13 \, \mathrm{mg} / \mathrm{dL}<\mu<9.73 \, \mathrm{mg} / \mathrm{dL}. \text{ The confidence interval limits contain 0, suggesting that the garlic treatment did not have a significant effect on LDL cholesterol levels.}}\)
Step 1 ：We are given that the sample mean (\(\bar{x}\)) is 3.3, the sample standard deviation (s) is 16.6, and the sample size (n) is 48. We want to construct a 99% confidence interval for the mean net change in LDL cholesterol.
Step 2 ：The formula for a confidence interval is \(\bar{x} \pm t \left( \frac{s}{\sqrt{n}} \right)\), where t is the t-score for the desired level of confidence.
Step 3 ：We need to find the t-score for a 99% confidence level with 47 degrees of freedom (since n - 1 = 48 - 1 = 47). Using a t-distribution table or a calculator, we find that the t-score is approximately 2.685.
Step 4 ：Substituting the given values into the formula, we get \(3.3 \pm 2.685 \left( \frac{16.6}{\sqrt{48}} \right)\).
Step 5 ：Calculating the margin of error, we get approximately 6.43.
Step 6 ：Subtracting and adding this margin of error from the sample mean, we get the confidence interval as \(-3.13 < \mu < 9.73\).
Step 7 ：\(\boxed{\text{Final Answer: The confidence interval estimate of the population mean } \mu \text{ is } -3.13 \, \mathrm{mg} / \mathrm{dL}<\mu<9.73 \, \mathrm{mg} / \mathrm{dL}. \text{ The confidence interval limits contain 0, suggesting that the garlic treatment did not have a significant effect on LDL cholesterol levels.}}\)
