Question

In a test of the effectiveness of garlic for lowering​ cholesterol, 4747 subjects were treated with garlic in a processed tablet form. Cholesterol levels were measured before and after the treatment. The changes ​(beforeminus−​after) in their levels of LDL cholesterol​ (in mg/dL) have a mean of 5.3 and a standard deviation of 19.7. Construct a 95​% 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

Answer)

As the population s.d is unknown here we will use t distribution table to construct the interval

Given info

Mean = 5.3

S.d = 19.7

N = 47

Confidence level = 95%

Degrees of freedom is = n-1 = 46

For 46 dof and 95% confidence level

Critical value t is = 2.01

Margin of error (MOE) = t*s.d/√n = 2.01*19.7/√47 = 5.784

Interval is given by

(Mean - MOE, Mean + MOE)

[-0.484, 11.084].

We are 95% confident that the population mean difference falls between -0.484 and 11.084.

Null hypothesis Ho : u1 - u2 = 0

Alternate hypothesis Ha : u1-u2 not equal to 0

So as the interval contains the null hypothesised value 0 in it

We fail to reject the null hypothesis Ho

So do not have enough evidence to conclude that garlic is effective in reducing LDL