This Question: 1 pt 14 of 25 (4 complete) This Test: 25 pts possible In a test o
ID: 3326850 • Letter: T
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This Question: 1 pt 14 of 25 (4 complete) This Test: 25 pts possible In a test of the effectiveness of garlic for lowering cholesterol, 47 subjects were treated with garlic in a processed tablet form. Cholesterol levels were measured before and after the treatment. The changes in their levels of LDL cholesterol (in mg/dL) have a mean of 4.5 and a standard deviation of 17.1. Complete parts (a) and (b) below. Click here to view a t distribution table, Click here to 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 integer or a decimal.) b. Construct a 99% 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? 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? 0 A. The confidence interval limits contain 0, suggesting that the garlic treatment did not affect the LDL cholesterol levels. O B. The confidence interval limits do not contain 0, suggesting that the garlic treatment did not affect the LDL cholesterol levels. ° C. The confidence interval limits do not contain o, suggesting that the garlic treatment did affect the LDL cholesterol levels 0 D. The confidence interval limits contain 0, suggesting that the garlic treatment did affect the LDL cholesterol levels.Explanation / Answer
a.
Best point estimate of the population mean net change in LDL cholestrol
= sample mean = 4.5
b.
Standard error of the mean, SE = SD / sqrt(n) = 17.1 / sqrt(47) = 2.494291
As, we do not know the population standard deviation, we use t distribution to construct the confidence interval.
Degree of freedom = n - 1 = 47 - 1 = 46
t value for 99% confidence interval and degree of freedom = 46 is 2.69
99% confidence interval estimate of the population mean
= (4.5 - 2.69 * 2.494291, 4.5 + 2.69 * 2.494291)
= (-2.21 , 11.21)
The answer is
-2.21 < < 11.21
As, the confidence interval contains 0, there is significant evidence at 99% confidence interval, that the mean net change in LDL cholestrol is 0. Hence, the correct option is,
A. The confidence interval limits contain 0, suggesting that the garlic treatment did not affect the LDL cholestrol levels.
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