Question

In a test of the effectiveness of garlic for lowering​ cholesterol, 4545 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 5.15.1 and a standard deviation of 19.119.1. Construct a 9090​% 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

Answer:

$5.1 - 2.015 \frac{19.1}{\sqrt{45}}= -0.637$

$5.1 + 2.015 \frac{19.1}{\sqrt{45}}= 10.837$

And for this case since the confidence interval contains the value 0 we don't have significant evidence that we have a net change in the levels

Step-by-step explanation:

For this case we have the following info given:

$n=45$ represent the sample size

$\bar X= 5.1$ represent the sample mean

$s= 19.1$ represent the sample deviation

We can calculate the confidence interval for the mean with the following formula:

$\bar X \pm t_{\alpha/2} \frac{s}{\sqrt{n}}$

The confidence level is 0.90 then the significance level would be $\alpha=0.1$ and the degrees of freedom are given by:

$df= n-1 = 45-1=44$

And the critical value for this case would be:

$t_{\alpha/2}=2.015$

And replacing we got:

$5.1 - 2.015 \frac{19.1}{\sqrt{45}}= -0.637$

$5.1 + 2.015 \frac{19.1}{\sqrt{45}}= 10.837$

And for this case since the confidence interval contains the value 0 we don't have significant evidence that we have a net change or efectiveness in the levels