Question

7.2.14 Question Help $ In a test of the effectiveness of garlic for lowering cholesterol, 44 subjects were treated with garlic in a processed tablet form. Cholesterol levels were measured before and after the treatment. The changes (before-after) in their levels of LDL cholesterol (in mg/dL) have a mean of 5.7 and a standard deviation of 18.2. Construct a 90% confidence interval estimate of the mean net change in LDL cholesterol after the garlic treatment. What does the confidence interval suggest about the effectiveness of garlic in reducing LDL cholesterol? Click here to view at distribution table Click here to view.rage 1 of the standard normal distribution table. Click here to view.rage 2 of the standard normal distribution table What is the confidence interval estimate of the population mean? mg/dL <3<mgid (Round

Answer 1

Sample mean = x = 5.7 Sample size = n = 44 Sample S.D = s = 18.2 Standard Error = s/Vn = 18.2/V44 = 2.744 Confidence Level = 90 Significance Level = a = (100-90)% = 0.1 Degrees of freedom = n-1 = 44 -1 = 43 Critical value = t* = 1.681 [ using Excel =TINV(0.1,43)] Margin of Error = t*(s/Vn )= 1.681*2.744 = 4.613 Lower Limit = x - Margin of Error = 5.7 - 4.613 = 1.09 Upper Limit = x + Margin of Error = 5.7 + 4.613 = 10.31 Ans: 1.09 <u < 10.31