Margin of error for a 400-person poll

Question

A poll of n = 400 people finds \hat{p} = 50\% support. Use z = 1.96 for 95\% confidence. What is the 95\% margin of error?

Accepted Answer

The margin of error is the z-multiplier times the standard error of the proportion: 	ext{ME} = z \cdot \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}. At \hat{p} = 0.50 the outcome spread \hat{p}(1-\hat{p}) = 0.25 is at its maximum, so this is the widest-margin case. The standard error is \sqrt{\frac{0.50 	imes 0.50}{400}} = \sqrt{\frac{0.25}{400}} = \sqrt{0.000625} = 0.025. Multiply by z = 1.96: 	ext{ME} = 1.96 	imes 0.025 = 0.049, about \pm 4.9 percentage points. A smaller poll of 400 carries a noticeably wider margin than a poll of 1000. Takeaway. Margin of error is z\sqrt{\hat{p}(1-\hat{p})/n}: …

Margin of error for a 400-person poll

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