Sample size when you expect a lopsided result

Question

You want a 95\% margin of error of at most \pm 2 percentage points. From prior data you expect support near \hat{p} = 0.20 (a lopsided, not fifty-fifty, result). Use z = 1.96. How many people must you survey?

Accepted Answer

The margin of error for a proportion is 	ext{ME} = z \cdot \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}. Here you have a credible expectation that \hat{p} \approx 0.20, so instead of the worst-case 0.25 you can use the smaller outcome spread \hat{p}(1-\hat{p}) = 0.20 	imes 0.80 = 0.16. A more lopsided rate is less noisy, so it needs fewer people. Set the margin to 0.02 and solve for n: n = \frac{z^2\,\hat{p}(1-\hat{p})}{	ext{ME}^2} = \frac{(1.96)^2 	imes 0.16}{(0.02)^2} = \frac{3.8416 	imes 0.16}{0.0004} = \frac{0.6147}{0.0004} \approx 1536.6. Round up: n = 1537. Compare to the worst-case (\hat{p…

Sample size when you expect a lopsided result

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