In a test of the effectiveness of garlic for lowering choles

In a test of the effectiveness of garlic for lowering cholesterol, 43 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 3.1 and a standard deviation of 19.3.

Complete parts (a) and (b) below.

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 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 ?

__ mg/dL<<__mg/dL (round to two decimal places as needed)

Solution

a)

it is the sample mean, 3.1 mg/dL. [ANSWER]

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b)

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 =    0.05          
X = sample mean =    3.1          
z(alpha/2) = critical z for the confidence interval =    1.644853627          
s = sample standard deviation =    19.3          
n = sample size =    43          
              
Thus,              
Margin of Error E =    4.841170052          
Lower bound =    -1.741170052          
Upper bound =    7.941170052          
              
Thus, the confidence interval is              
              
(   -1.741170052   ,   7.941170052   ) [ANSWER]