Question 1101699: In a test of the effectiveness of garlic for lowering cholesterol, 47
subjects were treated with garlic in a processed tablet form. Cholesterol levels were measured before and after the treatment. The changes (before-after) in their levels of LDL cholesterol (in mg/dL) have a mean of 3.5 and a standard deviation of 18.1. 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?
1. What is the confidence interval estimate of the population mean u?
___mg/dL
2. What does the confidence interval suggest about the effectiveness of the treatment?
*The confidence interval limits contain 0, suggesting that the garlic treatment did affect LDL cholesterol levels
*The confidence interval limits do not contain 0, suggesting that the garlic treatment did affect LDL cholesterol levels
*The Confidence levels contain 0, suggesting that the garlic treatment did not affect LDL cholesterol levels
*The Confidence levels do not contain 0, suggesting that the garlic treatment did not affect the LDL cholesterol levels.
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! sample size is 47, sample mean is 3.5 and sample standard deviation is 18.1
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standard error(SE) = 18.1 / square root(47) = 2.6402
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alpha(a) = 1 - (95/100) = 0.05
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critical probability(p*) = 1 - (a/2) = 0.975
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because the sample size > 30, we can express the critical value as a z-score by looking in the z-tables for p* and using the associated z-score.
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critical value(CV) for 0.975 is 1.96
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margin of error(ME) = CV * SE = 1.96 * 2.6402 = 5.1748 approximately 5.2
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1. (-1.7, 8.7) mg/dL
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*** Note Whenever an effect is significant, all values in the confidence interval will be on the same side of zero (either all positive or all negative).
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2. The confidence interval limits contain 0, suggesting that the garlic treatment did affect LDL cholesterol levels
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