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 3.8 and a standard deviation of 19.1.Complete parts (a) and (b) below.

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

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?

What is the confidence interval estimate of the population mean mu?

mg/d ____ < _____ mg/dL

Explanation / Answer

Confidence Interval
CI = x ± t a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Point of estimate = Mean(x)=3.8
Standard deviation( sd )=19.1
Sample Size(n)=50
Confidence Interval = [ 3.8 ± t a/2 ( 19.1/ Sqrt ( 50) ) ]
= [ 3.8 - 1.677 * (2.701) , 3.8 + 1.677 * (2.701) ]
= [ -0.73,8.33 ]

Interpretations:
1) We are 90% sure that the interval [-0.73 , 8.33 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 90% of these intervals will contains the true population mean  

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