Sample size to shrink the margin fivefold

Question

A 95% confidence interval for a mean has a margin of error of $10$ at a sample size of $n = 25$ . You want to reduce the margin to $2$ , keeping the spread and confidence level the same.

What sample size is required?

Accepted Answer

Setup. The margin scales as 1/\sqrt{n}. Reducing the margin by a factor of 10/2 = 5 requires multiplying n by 5^{2} = 25. Compute. n{	ext{new}} = 25 	imes 25 = 625. Check. At n = 625 the margin is 10 	imes \sqrt{25/625} = 10 	imes 	frac{1}{5} = 2, as required. Interpret. Shrinking the margin fivefold demands twenty-five times as much data, a steep but predictable cost. Takeaway. The cost of precision grows with the square of the improvement factor: five times tighter means twenty-five times the sample.

Sample size to shrink the margin fivefold

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