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 3.6 and a standard deviation of 19.6. 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? Click here to view at distribution table Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distnbution table What is the confidence interval estimate of the population mean u? mgidl<< mg/dL (Round to two decimal places as needed.)

Answer 1

Solution :

Given that,

Point estimate = sample mean = $\bar x$ = 3.6

sample standard deviation = s = 19.6

sample size = n = 45

Degrees of freedom = df = n - 1 = 45-1 = 44

At 95% confidence level

$\alpha$ = 1-0.95% =1-0.95 =0.05

$\alpha$/2 =0.05/ 2= 0.025

t$\alpha$/2,df = t0.025,44 = 3.50

t$\alpha$ /2,df = 2.02

Margin of error = E = t$\alpha$/2,df * (s /$\sqrt$n)

=2.02* (19.6 / $\sqrt$ 45)

Margin of error = E =5.888

The 95% confidence interval estimate of the population mean is,

$\bar x$ - E < $\mu$ <  $\bar x$+ E

3.6 - 5.888 < $\mu$ < 3.6 + 5.888

-2.29 mg/dL < $\mu$ < 9.49 mg/dL

(-2.29,9.49)