Question

In a test of the effectiveness of garlic for lowering​ cholesterol, 42 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 4.9 and a standard deviation of 16.9. 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 a t distribution table.

Click here to view page 1 of the standard normal distribution table.

Click here to view page 2 of the standard normal distribution table.

What is the confidence interval estimate of the population mean μ​?

?? ​mg/dL  < μ < ?? ​mg/dL

​(Round to two decimal places as​ needed.)

Answer 1

Solution :

Given that,

= 4.9

= 16.9

n = 42

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 * ( 16.9/ 42)

= 5.11

At 99% confidence interval estimate of the population mean is,

- E < < + E

4.9 - 5.11 < < 4.9 + 5.11

< <

(, )