Question

In a test of the effectiveness of garlic for lowering cholesterol, 81 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 mg/dL) have a mean of 0.7 and a standard deviation of 24.8. 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 finaloclusion that addresses the original claim. А Ho: Omg/dl. H1: Omg/dL B. HO: 0 mg/dL Ht: >0 mg/dL 1 C. HO: Omgldl H10 mgdL D. HO: 0mg/. HI: 0mgdL Identify the test statistic? Pvalue Identify the conclusion (reject or fail to reject?

Answer 1

The null and alternate hypothesis are:

H0: $\mu=0$

Ha: $\mu>0$

Hence option B.

The test statistic is given by:

$Z=\frac{\overline{x}-\mu}{\frac{s}{\sqrt{n}}}=\frac{0.7-0}{\frac{24.8}{\sqrt{81}}}=0.254$

Since this is a right-tailed test, so the p-value is given by:

$P(Z>0.254)=0.3997$

Since p-value is greater than 0.05, so we do not have sufficient evidence to reject the null hypothesis H0.

Thus, we reject null hypothesis H0.