Question 1101699
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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