In a test of the effectiveness of garlic for lowering cholesterol, 4343 subjects
ID: 3385650 • Letter: I
Question
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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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
Explanation / Answer
a)
The best point estimate is the sample mean, 4.7. [ANSWER]
****************
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]
**********************
c)
OPTION A: The confidence interval limits do not contain 0, suggesting that the garlic treatment did affect the LDL cholesterol levels. [ANSWER, A]
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