In a test of the effectiveness of garlic for lowering cholesterol, 43 subjects w

Question

In a test of the effectiveness of garlic for lowering cholesterol, 43 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 3.9 and standard deviation of 18.3 Complete parts(a) and (b) below. b. Construct a 95% confidence interval estimate of 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 does the confidence interval suggest about the effective4nsess of the treatment? A. The confidence interval limits do into contain 0, suggesting that the4 garlic treatment did affect the LDL cholesterol levels. B. The confidence interval limits contain 0 suggesting that the garlic treatment did affect the LDL cholesterol levels C. the confidence interval limits contain 0, suggesting that the garlic treatment did affect the LDL cholesterol levels D. the confidence interval limits contain 0, suggesting that the garlic treatment did affect the LDL cholesterol levels.

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
Mean(x)=3.9
Standard deviation( sd )=18.3
Sample Size(n)=43
Confidence Interval = [ 3.9 ± t a/2 ( 18.3/ Sqrt ( 43) ) ]
= [ 3.9 - 2.018 * (2.791) , 3.9 + 2.018 * (2.791) ]
= [ -1.732,9.532 ]
Interpretations:
1) We are 95% sure that the interval [-1.732 , 9.532 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population mean  

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