In developing an interval estimate for a population mean,a sample of 50 observations was used.The interval estimate was 19.76 $\pm$ 1.32.Had the sample size been 200 instead of 50,the interval estimate would have been:
A) 19.76$\pm$ .33
B) 19.76 $\pm$ .66
C) 19.76 $\pm$ 5.28
D) None of these choices.

Question

Quizplus · Accepted Answer

The standard error of the mean (SE) is calculated as SE = s/sqrt(n), where s is the sample standard deviation and n is the sample size. Since SE decreases as n increases, a larger sample size of 200 will result in a smaller SE. Using the formula for the confidence interval for the mean, CI = Xbar ± z*SE, where Xbar is the sample mean and z* is the z-score based on the desired level of confidence, we can see that a smaller SE will result in a narrower interval estimate. Therefore, the interval estimate with a sample size of 200 will have a smaller margin of error, which can be calculated as z*SE. Since the confidence level and sample mean are the same in both cases, the margin of error is the only difference. The difference between the margin of error for the two sample sizes is given by: ΔME = z*(s/sqrt(200) - s/sqrt(50)) = z*s*(sqrt(2) - 1)/10. Plugging in the values from the given information, we get ΔME = 1.32*(5/10) = 0.66. Therefore, the interval estimate with a sample size of 200 will have a margin of error that is 0.66 smaller than the interval estimate with a sample size of 50. Since the interval estimate is centered on the same sample mean of 19.76, the new interval estimate with a sample size of 200 will be 19.76 ± 0.66, or (19.10, 20.42). Therefore, the best choice is B.

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