What is the margin of error on a 400-person poll?

Question

A survey of 400 randomly chosen users finds that 52\% prefer a new order type. What is the 95% margin of error on that 52\%?

Accepted Answer

The 95% confidence interval for a proportion is \hat{p} \pm 1.96\sqrt{\hat{p}(1-\hat{p})/n}. With \hat{p} = 0.52 and n = 400: \hat{p}(1 - \hat{p}) = 0.52 	imes 0.48 = 0.2496, 	ext{margin} = 1.96\sqrt{\frac{0.2496}{400}} = 1.96\sqrt{0.000624} = 1.96 	imes 0.02498 \approx 0.049. So the margin is about \pm 4.9 percentage points: the true preference is plausibly anywhere from about 47\% to 57\%. Because \hat{p} = 0.52 is close to the worst case of 0.5, this margin is essentially the maximum for n = 400; using the conservative p = 0.5 gives 1.96 	imes \sqrt{0.25/400} = 1.96 	imes 0.025 = 0.04…

What is the margin of error on a 400-person poll?

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