In a test of the effectiveness of garlic for lowering choles

In a test of the effectiveness of garlic for lowering cholesterol,

4343

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

4.74.7

and a standard deviation of

17.117.1.

Complete parts (a) and (b) below.Click here to view a t distribution table.

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Click here to view page 1 of the standard normal distribution table.

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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

9090%

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 than<muless 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 containdo not contain

0, suggesting that the garlic treatment

diddid

affect the LDL cholesterol levels.

B.

The confidence interval limits

containcontain

0, suggesting that the garlic treatment

diddid

affect the LDL cholesterol levels.

C.

The confidence interval limits

do not containdo not contain

0, suggesting that the garlic treatment

did notdid not

affect the LDL cholesterol levels.

D.

The confidence interval limits

containcontain

0, suggesting that the garlic treatment

did notdid not

affect the LDL cholesterol levels.

Show Work

Solution

a)

The best point estimate is the sample mean, 4.7. [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 =    4.7          
z(alpha/2) = critical z for the confidence interval =    1.644853627          
s = sample standard deviation =    17.1          
n = sample size =    43          
              
Thus,              
Margin of Error E =    4.289326833          
Lower bound =    0.410673167          
Upper bound =    8.989326833          
              
Thus, the confidence interval is              
              
(   0.410673167   ,   8.989326833   ) [ANSWER]

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c)

OPTION A: The confidence interval limits do not contain 0, suggesting that the garlic treatment did affect the LDL cholesterol levels. [ANSWER, A]