Help ME PLEASE do not understand.

Question

In a test of the effectiveness of garlic for lowering​ cholesterol, 45 subjects were treated with garlic in a processed tablet form. Cholesterol levels were measured before and after the treatment. The changes ​(before−​after) in their levels of LDL cholesterol​ (in mg/dL) have a mean of 3.8 and a standard deviation of 18.6. 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?

Answer 1

Hi Britany,

This is a paired test The results a re measured before and after for each individual.

The mean of the differences d=3.8

Stand. dev. sd=18.6

n=45 subjects in the test

SE=sd/√n=2.77

t₄₄=1.301.. a 90% confidence level α=.1

H₀=0, the garlic makes no difference in lowering cholesterol this is the Null hypothesis

The confidence interval is

Lower limit =3.8-1.3*2.77=.20

Upper limit =3.8+1.3*2.77=7.40

Notice that the confidence interval does not contain d=0

JIm