In a test of the effectiveness of garlic for lowering cholesterol, 43 subjects w

Question

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 2.7 and a standard deviation of 16.3. 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?

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?

C. What does the confidence interval suggest about the effectiveness of the treatment? Do they contain 0? Did it affect the cholesterol levels?

Explanation / Answer

PART A.
given that,
best point estimate = sample mean, x =2.7

PART B.
standard deviation, s =16.3
sample size, n =43
i.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 16.3/ sqrt ( 43) )
= 2.486
ii.
margin of error = t /2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, = 0.1
from standard normal table, two tailed value of |t /2| with n-1 = 42 d.f is 1.682
margin of error = 1.682 * 2.486
= 4.181
iii.
ci = x ± margin of error
confidence interval = [ 2.7 ± 4.181 ]
= [ -1.481 , 6.881 ]
-----------------------------------------------------------------------------------------------
direct method
given that,
sample mean, x =2.7
standard deviation, s =16.3
sample size, n =43
level of significance, = 0.1
from standard normal table, two tailed value of |t /2| with n-1 = 42 d.f is 1.682
we use 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
confidence interval = [ 2.7 ± t a/2 ( 16.3/ sqrt ( 43) ]
= [ 2.7-(1.682 * 2.486) , 2.7+(1.682 * 2.486) ]
= [ -1.481 , 6.881 ]

ho: no difference
ha: is a difference
ince 0 is in the ci, we fail to reject ho. no diffrence b/w before and after cholesterol levels

PART C.
the confidence interval limits contain 0., suggesting that the garlic treatment did not affect the cholesterol levels.

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