Question

In a test of the effectiveness of garlic for lowering​ cholesterol, 50 subjects were treated with garlic in a processed tablet form. Cholesterol levels were measured before and after the treatment. The changes in their levels of LDL cholesterol​ (in mg/dL) have a mean of 4.3 and a standard deviation of 19.6

Complete parts​ (a) and​ (b) below.

a. What is the best point estimate of the population mean net change in LDL cholesterol after the garlic​ treatment?

The best point estimate is

nothing

​mg/dL.

​(Type an integer or a​ decimal.)

b. Construct a

99​%

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

muμ​?

nothing

​mg/dLless than<muμless than<nothing

​mg/dL

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

Answer 1

Solution :

Given that,

a) Point estimate = sample mean = $\bar x$ =4.3

Population standard deviation = $\sigma$ = 19.6

Sample size = n =50

At 99% confidence level

$\alpha$ = 1-0.99% =1-0.99 =0.01

$\alpha$/2 =0.01/ 2= 0.005

Z$\alpha$/2 = Z_{0.005 = 2.576}

Z$\alpha$/2 = 2.576

Margin of error = E = Z$\alpha$/2 * ( $\sigma$ /$\sqrt$n)

= 2.576 * (19.6 /  $\sqrt$50 )

=7.14

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

$\bar x$ - E < $\mu$ < $\bar x$ + E

4.3 - 7.14 <  $\mu$ < 4.3+ 7.14

-2.84 <  $\mu$ < 11.44

(-2.84 ,11.44 )mg/dL