Sample size for a poll's margin of error

Question

You want to estimate a fill rate (the fraction of your orders that get filled) to a 95\% margin of error of \pm 1\%. How many orders must you sample, and how does the standard error drive the answer?

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For an estimated proportion \hat p from n independent observations, the standard error is \operatorname{SE}(\hat p) = \sqrt{\frac{p(1-p)}{n}}. This is the spread of the estimate, not of a single outcome. A single order is filled or not (a variance of p(1-p) per order), but the SE of the estimated rate shrinks like 1/\sqrt{n} as you gather more orders. The margin of error at 95\% confidence is 1.96 	imes \operatorname{SE}. To be conservative, use p = 0.5, which maximizes p(1-p) = 0.25. Set the margin equal to 1\% = 0.01: 1.96 \sqrt{\frac{0.25}{n}} = 0.01 \implies \sqrt{\frac{0.25}{n}} = \frac…

Sample size for a poll's margin of error

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