Sample size needed to halve the margin

Question

A 95% confidence interval for a mean has a margin of error of 6 at a sample size of n = 50. You want to cut the margin to 3, keeping the same spread and confidence level. What sample size is required?

Accepted Answer

Setup. Since the margin scales as 1/\sqrt{n}, halving the margin requires quadrupling n. Compute. To halve the margin, multiply n by 2^{2} = 4: n{	ext{new}} = 50 	imes 4 = 200. Check. At n = 200 the margin is 6 	imes \sqrt{50/200} = 6 	imes 	frac{1}{2} = 3, as required. Interpret. The square-root law is unforgiving: each halving of the margin costs a fourfold increase in data. Takeaway. To divide the margin of error by k, multiply the sample size by k^{2}.

Sample size needed to halve the margin

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