Question

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 0.7 and a standard deviation of 18.2. Use a 0.10 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.

Answer 1

Answer: Given dat a nr 36, Xa= 0.7 Sd = 18.2 20.10 Test the claim that with garlic treatment the mean change ท LDU cholesterol is greater than o. The nwl and alternative hypothesis is Ho ed=0 VS Heldzo is 0 - The test statistic t = Xd - eld 0.7-O sd 18.2 V36 0.23 t = 0.23 . The p-value is

P-value = P [t>t>] (Hypothesis is right tailed) p[t> 0.23 ] ("idfant=36-1 df=35) 0.40971 ~ from t-table) 0.4097 0.4097 P-value p-value The Decision Rule is: p-value Ho. ke reject Here, 0.4097 > We failed to reject Ho. conclusion: failed to reject Ho, there is insuffi- cent evident to claim that with garlic treatment the mean change LDL cholesterol is greater than o. the garlic treatment - The result son show that is not effectiveness.