Sample size for a proportion with no prior guess

Question

You want to estimate a population proportion to within a margin of error of E = 0.03, with 95% confidence. You have no prior estimate of the rate. What is the minimum sample size?

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Setup. The margin of error for a proportion is E = z^{}\sqrt{p(1-p)/n}. Solving for n: n = \frac{(z^{})^{2}\,p(1-p)}{E^{2}}. With no prior guess, use the worst case p = 0.5, which maximizes p(1-p) = 0.25. Here z^{} = 1.96. Compute. n = \frac{(1.96)^{2}(0.25)}{(0.03)^{2}} = \frac{3.8416 	imes 0.25}{0.0009} = \frac{0.9604}{0.0009} = 1067.1. Round up: n = 1068. Interpret. Using p = 0.5 guarantees the margin is met no matter what the true rate turns out to be. It is the safe, conservative choice. Takeaway. With no prior information, plug p = 0.5 into the sample size formula. This gives the large…

Sample size for a proportion with no prior guess

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