Sample size for a plus-or-minus 2-point poll

Question

You want a 95\% margin of error of at most \pm 2 percentage points. Plan for the worst case \hat{p} = 0.5. 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}}. Use the worst-case spread \hat{p} = 0.5, so \hat{p}(1-\hat{p}) = 0.25. Set the margin to 0.02 and solve for n: n = \left(\frac{z}{2 	imes 	ext{ME}}ight)^2 = \left(\frac{1.96}{2 	imes 0.02}ight)^2 = \left(\frac{1.96}{0.04}ight)^2 = (49)^2 = 2401. Here it comes out exactly to 2401. Compare this to the \pm 3-point poll, which needed only about 1068: tightening the margin from 3 points to 2 points more than doubles the sample. Takeaway. Margin scales with 1/\sqrt{n}, so n = (z/(2\,	ext{ME}))…

Sample size for a plus-or-minus 2-point poll

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