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?

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:

represent the sample size

represent the sample mean

represent the sample deviation

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

The confidence level is 0.90 then the significance level would be and the degrees of freedom are given by:

And the critical value for this case would be:

And replacing we got:

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

4

step-by-step explanation:

15%

step-by-step explanation:

use the table and count the number of guesses that has two correct answers.

the following are two correct answers

10010, 01100, 11000

a total of 3 numbers having 2 correct guesses out of 20 questions.

probability of getting 2 correct guesses = the number of questions getting 2 correct answers / total number of questions.

= 3/20

to convert it into percent multiply by 100, we get

= 3/20 * 100

= 15%

the answer is 15%

you.