Standard error of a large-sample vote share

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In a poll of n = 2500 voters, a fraction \hat{p} = 0.40 favored a ballot measure. What is the standard error of the sample vote share?

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The vote share is a proportion, so its standard error is 	ext{SE} = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}. With \hat{p} = 0.40 and n = 2500: 	ext{SE} = \sqrt{\frac{0.40 	imes 0.60}{2500}} = \sqrt{\frac{0.24}{2500}} = \sqrt{0.000096} \approx 0.0098. So the 40\% share is measured to just under 1 percentage point. The large sample of 2500 is what buys that tight precision, since the standard error falls with \sqrt{n}. Takeaway. A proportion's standard error is \sqrt{\hat{p}(1-\hat{p})/n}: with 2500 voters even a middling 40\% share is pinned to about \pm 1 point, thanks to the \sqrt{n} in the d…

Standard error of a large-sample vote share

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