Question

In a test of the effectiveness of garlic for lowering​ cholesterol, 50 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 4.2 and a standard deviation of 15.5. Construct a 90​% 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 a t distribution table.LOADING... Click here to view page 1 of the standard normal distribution table.LOADING... Click here to view page 2 of the standard normal distribution table.LOADING... What is the confidence interval estimate of the population mean mu​? nothing ​mg/dLless thanmuless than nothing ​mg/dL ​(Round to two decimal places as​ needed.) What does the confidence interval suggest about the effectiveness of the​ treatment

Answer 1

As we are not given here that the distribution is normal, but the sample size is greater than 30, therefore we can use the Central limit theorem here to get the distribution of sample mean as a normal distribution.

For n - 1 = 49 degrees of freedom, we get from t distribution tables here:
P( t49 < 1.677 ) = 0.95

Therefore, due to symmetry, we have here:
P( -1.677 < t49 < 1.677) = 0.9

Therefore the confidence interval here is obtained as:

This is the required 90% confidence interval for the population mean here.

As the whole confidence interval lies above 0, therefore we have sufficient evidence here that there is a significant mean net change in LDL cholesterol after the garlic treatment at the 90% confidence that is 10% level of significance.