Question

n a test of the effectiveness of garlic for lowering​ cholesterol, 48 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 4.8 and a standard deviation of 17.6. Complete parts​ (a) and​ (b) below.

a)What is the best point estimate of the population mean net change in LDL cholesterol after the garlic​ treatment?

b)Construct a 95​% 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? What is the confidence interval estimate of the population mean μ​?

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

Answer 1

a) The best estimate for the population mean is the sample mean=4.8.

b)

The formula for estimation is:

μ = M ± Z(s_M)

where:

M = sample mean
Z = Z statistic determined by confidence level
s_M = standard error = √(s²/n)

M = 4.8
Z = 1.96
s_M = √(17.6²/48) = 2.54

μ = M ± Z(s_M)
μ = 4.8 ± 1.96*2.54
μ = 4.8 ± 4.979

95% CI [-0.179, 9.779].

c) we can be 95% confident that the population mean (μ) falls between -0.179 and 9.779.

The Z critical value is calculated using Z table shown below as