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In a test of the effectiveness of garlic for lowering cholesterol, 44 subjects w

ID: 3183668 • Letter: I

Question

In a test of the effectiveness 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

16.7.

Complete parts (a) and (b) below.

What is the best point estimate of the population mean net change in LDL cholesterol after the garlic treatment?

The best point estimate is

3.2

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?

negative 0.79

mg/dLless than

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

containcontain

0, suggesting that the garlic treatment

did notdid not

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

diddid

affect the LDL cholesterol levels.

D.

The confidence interval limits

do not containdo not contain

0, suggesting that the garlic treatment

did notdid not

affect the LDL cholesterol levels.

Explanation / Answer

a.
best point estimate = Mean(x)=3.6

b.
Confidence Interval
CI = x ± t a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Standard deviation( sd )=16.7
Sample Size(n)=44
Confidence Interval = [ 3.6 ± t a/2 ( 16.7/ Sqrt ( 44) ) ]
= [ 3.6 - 1.681 * (2.518) , 3.6 + 1.681 * (2.518) ]
= [ -0.632,7.832 ]

c.
when reducing the confidence level, interval willl be narrower

d.
The confidence interval limits
contain 0, suggesting that the garlic treatment
did affect the LDL cholesterol levels.

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