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 ​(beforeminus​after) in their levels of LDL cholesterol​ (in mg/dL) have a mean of 5.3 and a standard deviation of 16.3. 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

Answer 1

Confidence interval for Population mean is given as below:

Confidence interval = Xbar ± t*S/sqrt(n)

From given data, we have

Xbar = 5.3

S = 16.3

n = 44

df = n – 1 = 43

Confidence level = 90%

Critical t value = 1.6811

(by using t-table)

Confidence interval = Xbar ± t*S/sqrt(n)

Confidence interval = 5.3 ± 1.6811*16.3/sqrt(44)

Confidence interval = 5.3 ± 4.1309

Lower limit = 5.3 - 4.1309 = 1.17

Upper limit = 5.3 + 4.1309 = 9.43

Confidence interval = (1.17, 9.43)