Question

In a test of the effectiveness of garlic for lowering cholesterol, 49 subjects were treated with garlic in a processed tablet form Cholesterol lovels were measured before and after the treatment. The changes (before - after) in their levels of LDL cholesterol (in mg/dL) have a mean of 3.3 and a standard deviation of 15.3 . 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 effectivenoss of garlic in reducing LOL cholesterol? Click here to view a t distribution table. Click here to vew page 1 of the standard normal distribution table.

Answer 1

To construct a 90% confidence interval estimate of the mean net change in LDL cholesterol after garlic treatment, use the formula CI = sample_mean ± (critical_value * standard_error). The confidence interval suggests that the garlic treatment is somewhat effective in reducing LDL cholesterol.

Step-by-step explanation:

To construct a 90% confidence interval estimate of the mean net change in LDL cholesterol after garlic treatment, we will use the formula:

CI = sample_mean ± (critical_value * standard_error)

Given that the sample mean is 3.3, the standard deviation is 15.3, and the sample size is 49, we can calculate the standard error as 15.3 / sqrt(49) = 2.18.

Using a t-distribution table or calculator, we can find the critical value for a 90% confidence interval with 48 degrees of freedom to be approximately 1.677.

Plugging in these values, the confidence interval estimate is 3.3 ± (1.677 * 2.18), which simplifies to (0.19, 6.41).

This means that we are 90% confident that the true mean net change in LDL cholesterol after the garlic treatment falls between 0.19 and 6.41 mg/dL.

The confidence interval suggests that the garlic treatment is somewhat effective in reducing LDL cholesterol, as the lower bound of the interval is greater than zero. However, further studies and analysis would be needed to determine the extent of the effectiveness.