Question

In a test of the effectiveness of garlic for lowering cholesterol, 45 subjects were treated with garlic in a processed tablet form. Cholesterol levels were measured before and after the treatment. The changes (before - after) in their levels of LDL cholesterol (in mg/dL) have a mean of 5.3 and a standard deviation of 19.7. 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? Click here to view at distribution table. Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table. What is the confidence interval estimate of the population mean u? mg/dL <u= mg/dL (Round to two decimal places as needed.)

Answer 1

Solution :

Let X be the changes (before -after) in levels of LDL cholesterol (in mg/dL). A random sample of 45 subject were taken and the changes in their levels were observed. the sample mean was 5.3 and standard deviation was 19.7.

The (1-α)*100% confidence interval is given by,

$(\bar X - t_{\alpha/2,n-1}*\frac{s}{\sqrt{n}} , \bar X + t_{\alpha/2,n-1}*\frac{s}{\sqrt{n}})$

Where, $\bar X$ is the sample mean, s is the sample standard deviation, n is the sample size and is the critical value. This can be obtained from the t table with α = 0.01 and df = n - 1 = 44. This gives, the critical value is, 2.690.

ta/2.n-1

And hence the 99% confidence interval for population mean µ is,

$(5.3 - 2.690*\frac{19.7}{\sqrt{45}} , 5.3 + 2.690*\frac{19.7}{\sqrt{45}})$

this is the required confidence interval. This can be interpreted as, there is 99% confidence that the population mean µ lies within -2.60 and 13.20.