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 :\(\boxed{\text{Final Answer: The confidence interval estimate of the population mean } \mu \text{ is } -3.13 \, \mathrm{mg} / \mathrm{dL}<\mu<9.73 \, \mathrm{mg} / \mathrm{dL}. \text{ The confidence interval limits contain 0, suggesting that the garlic treatment did not have a significant effect on LDL cholesterol levels.}}\)