his Question: 4 pts 15 of 17 (15 complete) This Quiz: 33 pts poss In a test of t
ID: 3063095 • Letter: H
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his Question: 4 pts 15 of 17 (15 complete) This Quiz: 33 pts poss In a test of the effectiveness of garlic for lowering cholesterol, 49 subjeds were treated with garlic in a processed tablet form. Cholesterol levels were measured before and ate the treatment. The changes in their levels of LDL cholesterol (in mo/dl) have a mean of 4 5 and a standard dewation of 15.8 Complete parts (a) and (b) below Click here to view atdistribution table Clack here to view page 1 of the standard normal distribubon table a What is the best point estimate of the population mean net change in LDL cholesterol after the garlic treatment? The best point estimate is mg/dL. (Type an intéger or a decimal.) b. Construd a 90% confidence interval estimate of the mean net change in LDL cholesterol after te gartic treatment what does the confidence interval suggest about the effectveness of garlic in reducing LDL cholesterol? What is the confidence interval estimate of the population mean ? (Round to two decimal places as needed) What does the confidence interval suggest about the effectiveness of the treatment? A. The confidence interval limits contain 0, suggesting that the garlic treatment did not amea me LDL cholesterol levels OB. The confidence interval limits contain 0, suggesting that the garlic treatment did affect the LDL cholesterol levels c. The confidence interval limits do not contain 0, suggesting that the garlic treatment did not affect the LDL cholesterol levels. D. The conndence interval limits do not contain 0, suggesting that the garlic treatment did area te LDL cholesterol levelsExplanation / Answer
a. best point estimate is sample mean = 4.5
b. Degree of freedom of t distribution = 48
critical t for 90% = 1.68
Option selected will be D
Thus, S = std dev. /square root(n) = 15.8 / (7) = 2.257
confidence interval = 4.5 - 1.68*2.257 to 4.5 + 1.68*2.257 = 0.7 to 8.29
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