\[ Z = \frac{\hat{p}-p}{\sqrt{\displaystyle \frac{p\ (1-p)}{n}}} \] は標準正規分布 $\mathcal{N}(0, 1^2)$ に従うとみなせる。

\[ \begin{align*} \Pr\ \left \{ \frac{\left |\ \hat{p}-p\ \right |}{\sqrt{\displaystyle \frac{p\ (1-p)}{n}}} \leqq z_{\alpha/2}\right \} &= \Pr\ \left \{ \left |\ \hat{p}-p\ \right | \leqq z_{\alpha/2}\ \sqrt{\displaystyle \frac{p\ (1-p)}{n}} \right \} \\ &= 1-\alpha \end{align*} \]

\[ \begin{align*} \Pr\ \left \{ \left |\ \hat{p}-p\ \right | \leqq z_{\alpha/2}\ \sqrt{\displaystyle \frac{p\ (1-p)}{n}} \right \} &= \Pr\ \left \{ \hat{p}-z_{\alpha/2}\ \sqrt{\displaystyle \frac{\hat{p}\ (1-\hat{p})}{n}} \leqq p \leqq \hat{p}+z_{\alpha/2}\ \sqrt{\displaystyle \frac{\hat{p}\ (1-\hat{p})}{n}} \right \} \\ &= 1-\alpha \end{align*} \]

\[ \left [ \ \hat{p}-z_{\alpha/2}\ \sqrt{\displaystyle \frac{\hat{p}\ (1-\hat{p})}{n}}\ , \ \ \ \hat{p}+z_{\alpha/2}\ \sqrt{\displaystyle \frac{\hat{p}\ (1-\hat{p})}{n}} \ \right ] \] となる。