Statistical Help

Question

In a test of the effectiveness of garlic for lowering cholesterol, 44 subjects were treated with garlic in a processed tablet form.  Cholesterol levels were mearsured before and after the treatment.  The changes in their levels of LDL cholesterol (in mg/dl) have a mean of 4.6 and a standard deviation of 18.9.

 

A) What is the best point estimate of the population mean net change in LDL Cholesterol after the garlic treatment?

 

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.

 

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

Answer 1

To answer Part A):

Let x=mean net change in LDL Cholesterol after garlic treatment. Note what I've defined as the variable 'x'

The best estimate of a population mean is the sample mean, which we're told is 4.6.

I think it's always a good idea to write down what we know:

To answer Part (B):

Let x=mean net change in LDL Cholesterol after garlic treatment. We're told the µ_x=4.6, and the σ_x=18.9.

We also know the sample size(i.e. N=44).Since we're told to find the 99% CI, from the tables for the Z-score, we see the correct score need to use for a 99% confidence Interval is=2.576. Now we have everything we need to calculate the 99% Confidence Interval, and the formula for it is: μ ± [ (Z*σ) / √n ]. I've told you all the numbers, now you can just plug them in the formula & you'll have the answer to Part B)

To answer Part (C): I should can't do all the work for you, so here's a hint to help you: What do you know about how the width of a CI relates to the significance of any treatment?

That should be sufficient help for you to understand how to get to the answers you need for this problem.