Question

n a test of the effectiveness of garlic for lowering​ cholesterol, 36 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 0.1 and a standard deviation of 1.67 Use a 0.05 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. H0​: muμ=00 ​mg/dL H1​: muμ>00 ​mg/dL

.Determine the test statistic.   .36

Determine the​ P-value. ???

Answer 1

SOLUTION:

From given data,

In a test of the effectiveness of garlic for lowering​ cholesterol, 36 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 0.1 and a standard deviation of 1.67 Use a 0.05 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?

Mean = = 0.1

Standard deviation = S = 1.67

Sample size = n = 36

Test hypotheses:

The null hypothesis

: μ = 0  mg/dL

The Alternative hypothesis

: μ > 0  mg/dL

Determine the test statistic.

Standard error = SE = S/sqrt(n) = 1.67/sqrt(36) = 0.2783

test statistic = t = (-μ) / SE = (0.1-0) / 0.2783 = 0.1/ 0.2783 = 0.359 0.36

Determine the​ P-value.

Degree of freedom = n-1 = 36-1 = 35

P-value = 0.3605

Final conclusion:

significance level = = 0.05

Where,

P-value = 0.3605 > significance level = = 0.05

Then There is not sufficient evidence to support the claim.