In a test of the effectiveness of garlic for lowering cholesterol, 49 subjects w

Question

In a test of the effectiveness of garlic for lowering cholesterol, 49 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 5.4 and a standard deviation of 18.5. Complete parts (a) and (b) below. Click here to view a t distribution table. LOADING... Click here to view page 1 of the standard normal distribution table. LOADING... Click here to view page 2 of the standard normal distribution table. LOADING... 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 nothing mg/dL. (Type an integer or a decimal.) b. 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? What is the confidence interval estimate of the population mean mu? nothing mg/dLless thanmuless than nothing mg/dL (Round to two decimal places as needed.) What does the confidence interval suggest about the effectiveness of the treatment? A. The confidence interval limits do not contain 0, suggesting that the garlic treatment did not affect the LDL cholesterol levels. B. The confidence interval limits contain 0, suggesting that the garlic treatment did not affect the LDL cholesterol levels. C. The confidence interval limits do not contain 0, suggesting that the garlic treatment did affect the LDL cholesterol levels. D. The confidence interval limits contain 0, suggesting that the garlic treatment did affect the LDL cholesterol levels. Click to select your answer(s).

Explanation / Answer

Hi,

Please find the below of standard devitaion of the:

a) The best point estimate for the mean of the change in the levels is already given in the problem and would be equal to:

5.4 mg/dL ( as given in the problem itself )

b) The 99% confidence interval for the mean change could be computed as:

ar X pm t_{n-1}rac{s}{sqrt{n}}

Here the sample size n = 48. Now for n-1 = 47 degrees of freedom and 99% confidence interval we get from the t-distribution table that:

P(t_{47} < 2.685) = 0.995

Therefore the critical t-value to be used here is 2.685. Therefore now the confidence interval could be computed as:

ar X pm t_{n-1}rac{s}{sqrt{n}}

5.4 pm 2.685*rac{18.4}{sqrt{48}}

5.4 pm 7.1309

(-1.7309, 12.5309)

This is the required 99% confidence interval.

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