In a test of the effectiveness of garlic for lowering cholesterol, 50 subjects w

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 in their levels of LDL cholesterol (in mg/dL) have a mean of 3.2 and a standard deviation of 15.4. Complete parts (a) and (b) below. Click here to view a t 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. 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 3.2 mg/dL Type an integer or a decimal.) b. Construct a 99% confidence interval estimate of he mean net change in L L cno estero after he garlic reatment cholesterol? What is the confidence interval estimate of the population mean ?? hat does the con dence interval suggest a out ee ectiveness o an n in D Round to two decimal places as needed.)

Explanation / Answer

b. Solution : Given that mean x = 3.2 , standard deviation s = 15.4 and n = 50

=> df = n - 1 = 49

For 99% confidence interval , t = 2.680

=> The 99% confidence interval for population mean is x - t*s/sqrt(n) < ? < x + t*s/sqrt(n)

=> 3.2 - 2.680*15.4/sqrt(50) < ? < 3.2 + 2.680*15.4/sqrt(50)

=> -2.64 < ? < 9.04

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

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