Sample size for a five-point margin

Question

You want to estimate a population proportion to within a margin of error of E = 0.05, with 95% confidence, and you have no prior estimate of the rate. What is the minimum sample size?

Accepted Answer

Setup. Use n = (z^{})^{2}\,p(1-p)/E^{2} with the conservative p = 0.5 and z^{} = 1.96. Compute. n = \frac{(1.96)^{2}(0.25)}{(0.05)^{2}} = \frac{0.9604}{0.0025} = 384.16. Round up: n = 385. Interpret. Relaxing the margin from 3 points to 5 points cuts the required sample from over a thousand to under 400, because n scales with 1/E^{2}. Takeaway. A 5-point margin at 95% confidence needs roughly 385 respondents. Loosening the precision requirement is the cheapest way to shrink a required sample.

Sample size for a five-point margin

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