In a test of the effectivness of garlic for lowering cholest

In a test of the effectivness 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 in their levels of LDL cholesterol (in mg/dl) have a mean of 3.6 and a standard deviation of 17.5. Complete the following.

A. what is tge best point estimate of the population mean net change in LDL cholesterol after the garlic treatment.

B. Construct a 95% interval estimate of the mean and net change in LDL cholesterol aftr 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?

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 affect the LD cholesterol level.

B. the cconfidence interval limit do not contain 0, suggesting that the garlic treatment did not affect the LDL cholesterol level.

C. the confidence interval limits contain 0, suggesting that the grlic treatment did affect the LDL levels.

D. The confidence interval limits do not contain 0, suggesting that the garlic treatment did affect the LDL cholesterol levels

Solution

A. what is tge best point estimate of the population mean net change in LDL cholesterol after the garlic treatment.

It is the sample mean difference, 3.6 [ANSWER]

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B.

What is the confidence interval estimate of the population mean?

Note that              
Margin of Error E = z(alpha/2) * s / sqrt(n)              
Lower Bound = X - z(alpha/2) * s / sqrt(n)              
Upper Bound = X + z(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.025          
X = sample mean =    3.6          
z(alpha/2) = critical z for the confidence interval =    1.959963985          
s = sample standard deviation =    17.5          
n = sample size =    44          
              
Thus,              
Margin of Error E =    5.170824543          
Lower bound =    -1.570824543          
Upper bound =    8.770824543          
              
Thus, the confidence interval is              
              
(   -1.570824543   ,   8.770824543   ) [ANSWER]

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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 affect the LD cholesterol level. [ANSWER]