Margin of Error

In a survey of $1000$ people, $60\%$ say they would vote for Candidate A and $40\%$ for someone else. How can you calculate a margin of error on the $60\%$ estimate?

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#The estimate is a sample proportion

Each reply is a Bernoulli draw, so the sample proportion $\hat{p} = 0.60$ estimates the true share $p$ . The spread of that estimate around $p$ is its standard error,

\SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = \sqrt{\frac{0.6 \cdot 0.4}{1000}} = \sqrt{0.00024} \approx 0.0155. \tag{1}

#The 95 percent margin

A $95\%$ confidence interval extends $1.96$ standard errors on each side, and that half-width is the margin of error,

\text{margin} = 1.96 \cdot \SE \approx 1.96 \cdot 0.0155 \approx 0.030. \tag{2}

The poll pins Candidate A at 60 percent, but with only a thousand replies the true share could sit anywhere in the 57 to 63 band with 95 percent confidence. The whiskers are the margin of error, plus or minus three points.

#Result

The margin of error is about $\pm 3$ percentage points, so the $95\%$ interval is roughly $60\% \pm 3\%$ , that is $[57\%, 63\%]$ . Quadrupling the sample would halve this, since the margin shrinks like $1/\sqrt{n}$ .