In developing an interval estimate for a population mean, a sample of 40 observations was used. The interval estimate was 17.25 2.42. Had the sample size been 160 instead of 40, the interval estimate would have been:
A) 17.259.68
B) 17.251.21
C) 69.009.68
D) 34.504.82
E) 69.001.21

Question

In developing an interval estimate for a population mean, a sample of 40 observations was used. The interval estimate was 17.25  2.42. Had the sample size been 160 instead of 40, the interval estimate would have been:
A) 17.259.68 
B) 17.251.21 
C) 69.009.68 
D) 34.504.82 
E) 69.001.21

Quizplus · Accepted Answer

Since the sample size is larger, the standard error of the mean would be smaller, resulting in a narrower interval estimate. Therefore, we can narrow down the options to either (A) or (B). To determine which one is correct, we can use the formula for the standard error of the mean:

SE = s / sqrt(n)

Where s is the sample standard deviation and n is the sample size. Since the sample size is 4 times larger, we can expect the standard error to be half its original value. Therefore, the standard error for a sample of size 160 is:

SE' = 2.42 / sqrt(160) = 0.191

To create the interval estimate, we add and subtract the margin of error from the sample mean, where the margin of error is the product of the t-distribution value for the desired level of confidence (we assume 95% here) and the standard error of the mean. Since we have twice the sample size, the t-distribution value will be smaller. Using the t-distribution table, we find that the value for the 95% confidence level with a sample size of 160 is:

t = 1.98

Therefore, the interval estimate for a sample of size 160 is:

17.25 ± 1.98(0.191) = 17.25 ± 0.378

Which simplifies to:

(16.872, 17.628)

Comparing this with the original interval estimate of (14.83, 19.67), we can see that only option (B) is within this range, so that is our answer.

In Developing an Interval Estimate for a Population Mean, a Sample