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.