Question

In a study to assess the effectiveness of garlic for lowering cholesterol, 50 adults were treated with garlic tablets. Cholesterol levels were measured before and after treatment. The changes in their LDL cholesterol (in mg/dL) have a mean of 8 and a standard deviation of 4. (a) Construct a 95% interval estimate of the mean change in LDL cholesterol after the garlic tablet treatment. (Round the lower bound and upper bound of the confidence interval to two decimal places) Include description of how confidence interval was constructed. (b) Describe the results of the study in everyday language.

Answer 1

Let's write the given information:

n = sample size = 50

mean = = 4

sample standard deviation = s = 4

c = confidence level = 0.95

so level of significance = = 1 - c = 1 - 0.95 = 0.05

Formula of confidence interval for population mean is as follow:

+/- E

Where E is the margin of error.

Let's use excel to find E

E = "=CONFIDENCE.T(0.05,4,50)" = 1.1368

Lower limit = - E = 8 - 1.1368 = 6.8632

Upper limit = + E = 8 + 1.1368 = 9.1368

So 95% confidence interval for population mean is ( 6.86 , 9.14)

b) If we collect the same size of sample from the same population, many times then 95 of the times the confidence interval include the true population mean.

So here we are 95% confidence that the population mean change in LDL cholesterol after the garlic tablet treatment lies between 6.86 and 9.14