Could someone give me some help with this Please!!!!!

Question

I'M LOST ON THIS PROBLEM COULD SOMEONE PLEASE HELP ME !!!!!!!



I THINK I UNCOVERED ONE ANSWER, I'M STUCK ON THE OTHERS!!!!!



In a test of the effectiveness of garlic for lowering cholesterol, 49 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 2.7 and a standard deviation of 16.9. 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 16.9 mg/dL. (Type an integer or a decimal.)


b. 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?


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


? mg/dL ? < μ < ? mg/dL (Round to two decimal places as needed.)


What does the confidence interval suggest about the effectiveness of the treatment?


A. The confidence interval limits do not contain 0, suggesting that the garlic treatment did affect the LDL cholesterol levels.


B. The confidence interval limits contain 0, suggesting that the garlic treatment did not affect the LDL cholesterol levels.


C. The confidence interval limits do not contain 0, suggesting that the garlic treatment did not affect the LDL cholesterol

levels


D.The confidence interval limits contain 0, suggesting that the garlic treatment did affect the LDL cholesterol levels.

 

Answer 1

Hi Wendy,

 

First of all the best point estimate of the population net change is 2.7 mg/dL Point estimates are almost always incorrect. That is why we construct a confidence interval instead. A confidence interval is an interval z standard deviations on either side of the point estimate. 

 

                              µ =  x¯  ±   z (s/√n)   The z-score that corresponds to a 95% confidence interval is 1.96

Substituting the values into the formula results in 

 

                              μ = 2.7  ±  1.96 ( 16.9/√49)

 

                              μ = 2.7 ±  4.732

 

                                    12.168  ≤  μ  ≤  21.632     Rounding to two decimal places gives us 12.17≤μ≤21.63

 

Since the confidence interval does not contain zero, this implies that the garlic was effective.