Margin of error for a 1000-person poll

Question

A survey of $1000$ voters finds $60\%$ backing a particular candidate. How would you work out the poll's margin of error?

Short answer

About $\pm3$ percentage points. The standard error of a sample proportion is $\sqrt{\hat p(1-\hat p)/n}=\sqrt{0.6\cdot0.4/1000}\approx1.55\%$ , and a 95% interval multiplies by $1.96$ , giving roughly $\pm3.0\%$ . Pollsters often quote the conservative $1/\sqrt n\approx3.2\%$ (the $\hat p=0.5$ case); this ignores non-sampling biases.

Adapted from Crack, Heard on the Street, Q4.29, p.43; question and answer rewritten by SelfMadeQuant.com.

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Lessons behind this question

Maximum Likelihood Estimation: Score and Log-Likelihood
The Hypothesis Testing Framework
Pivotal Quantities and Exact Confidence Intervals

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