In a test of the effectiveness of garlic for lowering cholesterol, 42 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 mg dL) have a mean of 4.6 and a standard deviation of 16.7 . Construct a 90%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 is the confidence interval estimate of the population mean u? (Round to two decimal places as needed.)

Step 1 ：Given that the sample mean (\(\bar{x}\)) is 4.6, the sample standard deviation (s) is 16.7, the sample size (n) is 42, and the z-score corresponding to the desired 90% confidence level is 1.645.

Step 2 ：The formula for the confidence interval for a population mean is \(\bar{x} \pm z \frac{s}{\sqrt{n}}\).

Step 3 ：Substitute the given values into the formula to calculate the confidence interval: mean = 4.6, std_dev = 16.7, n = 42, z = 1.645.

Step 4 ：Calculate the margin of error: margin_of_error = 4.2389444784836074.

Step 5 ：Calculate the lower bound of the confidence interval: lower_bound = 0.3610555215163922.

Step 6 ：Calculate the upper bound of the confidence interval: upper_bound = 8.838944478483608.

Step 7 ：The confidence interval for the mean net change in LDL cholesterol after the garlic treatment is approximately (0.36, 8.84). This means that we are 90% confident that the true mean net change in LDL cholesterol after the garlic treatment is between 0.36 and 8.84.

Step 8 ：Since this interval includes positive values, it suggests that the garlic treatment may not be effective in reducing LDL cholesterol. However, since it also includes values close to zero, it's also possible that the garlic treatment has no effect on LDL cholesterol levels. More data or a larger sample size would be needed to make a more definitive conclusion.

Step 9 ：Final Answer: The 90% confidence interval estimate of the population mean is approximately \(\boxed{(0.36, 8.84)}\).