Question

In a test of the effectiveness of garlic for lowering​ cholesterol, 8181 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 0.40.4 and a standard deviation of 16.216.2. Use a 0.010.01 significance level to test the claim that with garlic​ treatment, the mean change in LDL cholesterol is greater than 00. 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. What are the null and alternative​ hypotheses?

Answer 1

Answer:

With garlic​ treatment, the mean change in LDL cholesterol is not greater than 0.

Step-by-step explanation:

The dependent t-test (also known as the paired t-test or paired samples t-test) compares the two means associated groups to conclude if there is a statistically significant difference amid these two means.

In this case a paired t-test is used to determine the effectiveness of garlic for lowering​ cholesterol.

A random sample of 81 subjects were treated with raw garlic. Cholesterol levels were measured before and after the treatment.

The hypothesis for the test can be defined as follows:

H₀: With garlic​ treatment, the mean change in LDL cholesterol is not greater than 0, i.e. d ≤ 0.

Hₐ: With garlic​ treatment, the mean change in LDL cholesterol is greater than 0, i.e. d > 0.

The information provided is:

$\bar d=0.40\\SD_{d}=16.2\\\alpha =0.01$

Compute the test statistic value as follows:

$t=\frac{\bar d}{SD_{d}/\sqrt{n}}\\\\=\frac{0.40}{16.2/\sqrt{81}}\\\\=0.22$

The test statistic value is 0.22.

Decision rule:

If the p-value of the test is less than the significance level then the null hypothesis will be rejected and vice-versa.

Compute the p-value of the test as follows:

$p-value=P(t_{n-1}>0.22)\\=P(t_{80}>0.22)\\=0.4132$

*Use a t-table.

The p-value of the test is 0.4132.

p-value= 0.4132 > α = 0.01

The null hypothesis was failed to be rejected.

Thus, it can be concluded that with garlic​ treatment, the mean change in LDL cholesterol is not greater than 0.