Estimating Sample Sizes for Proportions

A confidence interval for a population proportion $p$ has margin of error

$$E = z^{*} \sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}},$$

where $z^{*}$ is the critical value for the chosen confidence level and $\hat{p}$ is the sample proportion.

When planning a study, we fix the desired margin of error $E$ and solve for the required sample size $n$:

$$n = \left(\dfrac{z^{*}}{E}\right)^{2} \hat{p}(1-\hat{p}).$$

Because $n$ must be a whole number and the margin of error must be at most $E$, we always round up to the next integer.

The most common critical values are:

• $90\%$ confidence: $z^{*} = 1.645$
• $95\%$ confidence: $z^{*} = 1.96$
• $99\%$ confidence: $z^{*} = 2.576$

For example, suppose past data suggest $\hat{p} = 0.4$ and we want a margin of error of $0.05$ at $95\%$ confidence. Then

$$n = \left(\dfrac{1.96}{0.05}\right)^{2} \cdot 0.4 \cdot 0.6 = 1536.64 \cdot 0.24 = 368.79,$$

so we round up to $n = 369$.