Problem
In a test of the effectiveness of garlic for lowering cholesterol, 49 subjects were treated with raw garlic. Cholesterol levels were measured before and after the treatment. The changes (before minus after) in their levels of LDL cholesterol (in $\mathrm{mg} / \mathrm{dL}$ ) have a mean of 0.4 and a standard deviation of 2.08 . Use a 0.05 significance level to test the claim that with garlic treatment, the mean change in LDL cholesterol is greater than 0 . What do the results suggest about the effectiveness of the garlic treatment? Assume that a simple random sample has been selected. Identify the null and alternative hypotheses, test statistic, P-value, and state the final conclusion that addresses the original claim. Determine the test statistic. (Round to two decimal places as needed.)
Solution
Step 1 ：Identify the null and alternative hypotheses. The null hypothesis is that the mean change in LDL cholesterol is equal to 0, and the alternative hypothesis is that the mean change in LDL cholesterol is greater than 0.
Step 2 ：Calculate the test statistic using the formula \((\text{sample mean} - \text{hypothesized mean}) / (\text{standard deviation} / \sqrt{\text{sample size}})\).
Step 3 ：Substitute the given values into the formula: \((0.4 - 0) / (2.08 / \sqrt{49})\).
Step 4 ：Simplify the expression to get the test statistic: \(1.3461538461538463\).
Step 5 ：Round the test statistic to two decimal places: \(\boxed{1.35}\).
