10.4 Margin of Error and Sample Size Calculation for Proportion

A [latex](1 – \alpha) \times 100\%[/latex] confidence interval for the population proportion [latex]p[/latex] is [latex]\hat{p} \pm z_{\alpha / 2} \sqrt{\frac{\hat{p} (1 - \hat{p})}{n}}[/latex]. The margin of error is [latex]E = z_{\alpha /2 } \sqrt{\frac{\hat{p} (1 - \hat{p})}{n}}[/latex], which is half of the length of the interval; solving for [latex]n[/latex] yields [latex]n = \hat{p} (1 - \hat{p}) \left( \frac{z_{\alpha / 2}}{E} \right)^2[/latex]. Consequently, we are [latex](1 – \alpha) \times 100\%[/latex] confident that the margin of error is at most [latex]E[/latex] if the sample size [latex]n \geq \hat{p} (1 - \hat{p}) \left( \frac{z_{\alpha / 2}}{E} \right)^2[/latex]. However, this formula cannot be used because the sample proportion [latex]\hat{p} = \frac{x}{n}[/latex] is unknown until the sample is obtained. One solution to this problem is to use the maximum value of [latex]\hat{p} (1 - \hat{p})[/latex], which is 0.25 when [latex]\hat{p} = 0.5[/latex]. This leads to the conservative bound on the sample size.

[latex]n = 0.5(1 - 0.5) \left( \frac{z_{\alpha / 2}}{E} \right)^2 = 0.25 \left( \frac{z_{\alpha / 2}}{E} \right)^2[/latex]

rounded up to the nearest integer. However, if we have some extra information about the value of [latex]\hat{p}[/latex], we can use that information to obtain the guess [latex]\hat{p} = p_g[/latex]. This alternative approach leads to the sample size

[latex]n = p_g (1 - p_g) \left( \frac{z_{\alpha / 2}}{E} \right)^2[/latex]

rounded up to the nearest integer.

[latex]\hat{p}[/latex] [latex]\hat{p} (1 - \hat{p})[/latex] [latex]0[/latex] [latex]0 \times (1 - 0) = 0[/latex] [latex]0.2[/latex] [latex]0.2 \times (1 - 0.2) = 0.16[/latex] [latex]0.5[/latex] [latex]0.5 \times (1 - 0.5) = 0.25[/latex] [latex]0.8[/latex] [latex]0.8 \times (1-0.8) = 0.16[/latex] [latex]1[/latex] [latex]1 \times (1 - 1) = 0[/latex] Table 10.2: Relationship Between [latex]\hat p[/latex] and [latex]\hat p (1-\hat p)[/latex].