Question

justin read in the newspaper that 50% of voters in his city were voting
o\ on a local initiative measure. the poll claimed a margin of error of ±4%. justin wanted to know how many voters were polled and wrote an equation to solve. what answer should he have gotten?

Explanation:

Step1: Recall the formula for margin of error

The formula for margin of error \(E\) for a proportion is \(E = z\sqrt{\frac{p(1 - p)}{n}}\). For a \(95\%\) confidence level (common in polls), \(z\approx1.96\). When \(p = 0.5\) (since \(p(1 - p)\) is maximized at \(p = 0.5\)), the formula for margin of error simplifies to \(E=\frac{z}{\sqrt{4n}}\) (because \(\sqrt{p(1 - p)}=\sqrt{0.5\times(1 - 0.5)}=\sqrt{0.25}=0.5\)). Given \(E = 0.04\) and \(z = 1.96\).

Step2: Solve for \(n\)

From \(E=\frac{z}{\sqrt{4n}}\), we can rewrite it as \(\sqrt{n}=\frac{z}{2E}\). Substitute \(z = 1.96\) and \(E=0.04\). Then \(\sqrt{n}=\frac{1.96}{2\times0.04}=\frac{1.96}{0.08}=24.5\). Square both sides: \(n=(24.5)^2 = 600.25\approx601\) (since we can't poll a fraction of a person and we round up to ensure the margin of error is met).

Answer:

\(601\)