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) 9.88 $\pm$ 1.32.
D) 4.94 $\pm$ 1.32.

Question

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) 9.88 $\pm$ 1.32.
D) 4.94 $\pm$ 1.32.

Quizplus · Accepted Answer

The formula for the margin of error for a confidence interval is:
Margin of Error = Z* (Standard Deviation / sqrt(n))
where Z* is the critical value for the desired level of confidence, n is the sample size, and the standard deviation is either known or estimated from the sample.
Since we are comparing two sample sizes, we can use the formula to find the new margin of error:
Margin of Error 1 = Z* (Standard Deviation / sqrt(50))
Margin of Error 2 = Z* (Standard Deviation / sqrt(200))
We know that the interval estimate for the sample size of 50 is 19.76 ± 1.32. This means that the margin of error is 1.32. Therefore, we can solve for Z*:
1.32 = Z* (Standard Deviation / sqrt(50))
Z* = 1.32 * sqrt(50) / Standard Deviation
Now, we can use this value of Z* to solve for the new margin of error when n = 200:
Margin of Error 2 = Z* (Standard Deviation / sqrt(200))
Margin of Error 2 = [1.32 * sqrt(50) / Standard Deviation] * [Standard Deviation / sqrt(200)]
Margin of Error 2 = 1.32 * sqrt(50/200)
Margin of Error 2 = 1.32 * 0.5
Margin of Error 2 = 0.66
Therefore, the interval estimate for a sample size of 200 would be: 19.76 ± 0.66. So, the best choice is B).

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