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

​(beforeminus−​after)

in their levels of LDL cholesterol​ (in mg/dL) have a mean of

5.2

and a standard deviation of

15.8.

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

Solution :

Given that,

Point estimate = sample mean = = 5.2

Population standard deviation =    = 15.8

Sample size = n =45

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

Margin of error = E = Z/2 * ( /n)

= 1.96 * (15.8 /  45 )

= 4.6164
At 95% confidence interval estimate of the population mean
is,

- E < < + E

5.2 - 4.6164 <   < 5.2 + 4.6164

0.5836<   < 5.1364

( 0.5836, 5.1364 )