两独立样本率差异性检验¶
对于双侧检验，统计学假设如下：
\[
\begin{align}
H_0 &: p_1 - p_2 = 0 \\
H_1 &: p_1 - p_2 \neq 0
\end{align}
\]
对于左单侧检验，统计学假设如下：
\[
\begin{align}
H_0 &: p_1 - p_2 \geq 0 \\
H_1 &: p_1 - p_2 \lt 0
\end{align}
\]
对于右单侧检验，统计学假设如下：
\[
\begin{align}
H_0 &: p_1 - p_2 \leq 0 \\
H_1 &: p_1 - p_2 \gt 0
\end{align}
\]
两样本率分别用 \(\hat{p}_1\) 和 \(\hat{p}_2\) 表示。
\[
E(\hat{p}_1 - \hat{p}_2) = p_1 - p_2, \ \ Var(\hat{p}_1 - \hat{p}_2) = \frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}
\]
以下推导过程在边界条件 \(p_1 - p_2 = \delta\) 下进行。
Z-Test Pooled¶
假设 \(\bar{p}\) 表示合并总体率，则：
\[
\bar{p} = \frac{n_1 \hat{p}_1 + n_2 \hat{p}_2}{n_1 + n_2}
\]
在 \(H_0\) 成立时：
两样本的方差可以用 \(\bar{p}\) 来表示：
\[
Var(\hat{p}_1 - \hat{p}_2) = \frac{\bar{p}(1-\bar{p})}{n_1} + \frac{\bar{p}(1-\bar{p})}{n_2} = \bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)
\]
构建 \(z\) 统计量：
\[
z = \frac{\hat{p}_1 - \hat{p}_2}{\sqrt{\bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}} \sim N(0,1)
\]
在 \(H_1\) 成立时，可构建 \(z'\) 统计量：
\[
z' = \frac{\hat{p}_1 - \hat{p}_2}{\sqrt{\bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}}
\]
根据中心极限定理，当 \(n_1\) 和 \(n_2\) 较大时，满足：
\[
\frac{(\hat{p}_1 - \hat{p}_2) - (p_1 - p_2)}{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}} \sim N(0,1)
\]
进而有：
\[
\begin{align}
z' & = \frac{\left(\hat{p}_1 - \hat{p}_2\right) - \left(p_1 - p_2\right) + \left(p_1 - p_2\right)}
{\sqrt{\bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}} \\
& = \frac{\left(\hat{p}_1 - \hat{p}_2\right) - \left(p_1 - p_2\right) + \left(p_1 - p_2\right)}
{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}}
\cdot
\frac{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}}
{\sqrt{\bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}} \\
& = \frac{\left(\hat{p}_1 - \hat{p}_2\right) - \left(p_1 - p_2\right)}
{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}}
\cdot
\frac{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}}
{\sqrt{\bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}}
+
\frac{p_1 - p_2}
{\sqrt{\bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}} \\
& \xrightarrow{d}
N\left(\frac{p_1 - p_2}
{\sqrt{\bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}}, \
\frac{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}
{\bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}
\right)
\end{align}
\]
\[
\begin{align}
Power & = P\left(z' < -z_{1-\alpha}\right) \\
& = \Phi\left(\frac{-z_{1-\alpha} - \frac{p_1 - p_2}{\sqrt{\bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}}}
{\sqrt{\frac{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}{\bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}}}
\right) \\
& = 1 - \Phi\left(\frac{z_{1-\alpha} + \frac{p_1 - p_2}{\sqrt{\bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}}}
{\sqrt{\frac{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}{\bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}}}
\right) \\
& = 1 - \Phi\left(\frac{z_{1-\alpha} \sqrt{\bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)} + \left(p_1-p_2\right)}
{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}}
\right) \\
& = 1 - \beta
\end{align}
\]
样本量公式推导
根据标准正态分布分位数的定义：
\[
\frac{z_{1-\alpha} \sqrt{\bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)} + (p_1-p_2)}{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}} = z_\beta
\]
设 \(n_1 = kn_2\)，由上式可解出：
\[
n_2 = \frac{\left(z_{1-\alpha} \sqrt{\left(\frac{kp_1+p_2}{k+1}\right) \left(1-\frac{kp_1+p_2}{k+1}\right) \left(\frac{1}{k}+1\right)}
+ z_{1-\beta} \sqrt{\frac{1}{k}p_1(1-p_1) + p_2(1-p_2)}\right)^2}{(p_1-p_2)^2}
\]
\[
n_1 = k n_2
\]
\[
\begin{align}
Power & = P\left(z' > z_{1-\alpha}\right) \\
& = 1 - \Phi\left(\frac{z_{1-\alpha} - \frac{p_1 - p_2}{\sqrt{\bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}}}
{\sqrt{\frac{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}{\bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}}}
\right) \\
& = 1 - \Phi\left(\frac{z_{1-\alpha} \sqrt{\bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)} - \left(p_1-p_2\right)}
{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}}
\right) \\
& = 1 - \beta
\end{align}
\]
样本量公式推导
根据标准正态分布分位数的定义：
\[
\frac{z_{1-\alpha} \sqrt{\bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)} - (p_1-p_2)}{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}} = z_\beta
\]
设 \(n_1 = kn_2\)，由上式可解出：
\[
n_2 = \frac{\left(z_{1-\alpha} \sqrt{\left(\frac{kp_1+p_2}{k+1}\right) \left(1-\frac{kp_1+p_2}{k+1}\right) \left(\frac{1}{k}+1\right)}
+ z_{1-\beta} \sqrt{\frac{1}{k}p_1(1-p_1) + p_2(1-p_2)}\right)^2}{(p_1-p_2)^2}
\]
\[
n_1 = k n_2
\]
\[
\begin{align}
Power & = P\left(z' < -z_{1-\alpha/2}\right) + P\left(z' > z_{1-\alpha/2}\right) \\
& = 1 - \Phi\left(\frac{z_{1-\alpha/2} \sqrt{\bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)} + \left(p_1-p_2\right)}
{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}}
\right)
+
1 - \Phi\left(\frac{z_{1-\alpha/2} \sqrt{\bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)} - \left(p_1-p_2\right)}
{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}}
\right) \\
& = 1 - \beta
\end{align}
\]
Z-Test Pooled 连续性校正¶
在 \(H_0\) 成立时，可构建 \(z\) 统计量：
\[
z = \frac{\hat{p}_1 - \hat{p}_2 + c}{\sqrt{\bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}} \sim N(0,1)
\]
其中：
\[
c =
\begin{cases}
\frac{1}{2}\left(\frac{1}{n_1}+\frac{1}{n_2}\right), & \text{左单侧检验} \\
- \frac{1}{2}\left(\frac{1}{n_1}+\frac{1}{n_2}\right), & \text{右单侧检验}
\end{cases}
\]
在 \(H_1\) 成立时，可构建 \(z'\) 统计量：
\[
z' = \frac{\hat{p}_1 - \hat{p}_2 + c}{\sqrt{\bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}}
\]
根据中心极限定理，当 \(n_1\) 和 \(n_2\) 较大时，满足：
\[
\frac{(\hat{p}_1 - \hat{p}_2) - (p_1 - p_2)}{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}} \sim N(0,1)
\]
进而有：
\[
\begin{align}
z' & = \frac{\left(\hat{p}_1 - \hat{p}_2\right) - \left(p_1 - p_2\right) + \left(p_1 - p_2 + c\right)}
{\sqrt{\bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}} \\
& = \frac{\left(\hat{p}_1 - \hat{p}_2\right) - \left(p_1 - p_2\right) + \left(p_1 - p_2 + c\right)}
{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}}
\cdot
\frac{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}}
{\sqrt{\bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}} \\
& = \frac{\left(\hat{p}_1 - \hat{p}_2\right) - \left(p_1 - p_2\right)}
{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}}
\cdot
\frac{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}}
{\sqrt{\bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}}
+
\frac{p_1 - p_2 + c}
{\sqrt{\bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}} \\
& \xrightarrow{d}
N\left(\frac{p_1 - p_2 + c}
{\sqrt{\bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}}, \
\frac{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}
{\bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}
\right)
\end{align}
\]
\[
\begin{align}
Power & = P\left(z' < -z_{1-\alpha}\right) \\
& = \Phi\left(\frac{-z_{1-\alpha} - \frac{p_1-p_2 + \frac{1}{2}\left(\frac{1}{n_1} + \frac{1}{n_2}\right)}{\sqrt{\bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}}}
{\sqrt{\frac{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}{\bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}}}
\right) \\
& = 1 - \Phi\left(\frac{z_{1-\alpha} + \frac{p_1-p_2 + + \frac{1}{2}\left(\frac{1}{n_1} + \frac{1}{n_2}\right)}{\sqrt{\bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}}}
{\sqrt{\frac{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}{\bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}}}
\right) \\
& = 1 - \Phi\left(\frac{z_{1-\alpha} \sqrt{\bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)} + (p_1-p_2) + \frac{1}{2}\left(\frac{1}{n_1} + \frac{1}{n_2}\right)}
{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}}
\right) \\
& = 1 - \beta
\end{align}
\]
\[
\begin{align}
Power & = P\left(z' > z_{1-\alpha}\right) \\
& = 1 - \Phi\left(\frac{z_{1-\alpha} - \frac{p_1-p_2 - \frac{1}{2}\left(\frac{1}{n_1} + \frac{1}{n_2}\right)}{\sqrt{\bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}}}
{\sqrt{\frac{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}{\bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}}}
\right) \\
& = 1 - \Phi\left(\frac{z_{1-\alpha} \sqrt{\bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)} - (p_1-p_2) + \frac{1}{2}\left(\frac{1}{n_1} + \frac{1}{n_2}\right)}
{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}}
\right) \\
& = 1 - \beta
\end{align}
\]
\[
\begin{align}
Power & = P\left(z' < -z_{1-\alpha/2}\right) + P\left(z' > z_{1-\alpha/2}\right) \\
& = 1 - \Phi\left(\frac{z_{1-\alpha/2} \sqrt{\bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)} + (p_1-p_2) + \frac{1}{2}\left(\frac{1}{n_1} + \frac{1}{n_2}\right)}
{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}}
\right)
+
1 - \Phi\left(\frac{z_{1-\alpha/2} \sqrt{\bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)} - (p_1-p_2) + \frac{1}{2}\left(\frac{1}{n_1} + \frac{1}{n_2}\right)}
{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}}
\right) \\
& = 1 - \beta
\end{align}
\]
Z-Test Unpooled¶
在 \(H_0\) 成立时，可构建 \(z\) 统计量：
\[
z = \frac{\hat{p}_1 - \hat{p}_2}{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}} \sim N(0,1)
\]
在 \(H_1\) 成立时，可构建 \(z'\) 统计量：
\[
z' = \frac{\hat{p}_1 - \hat{p}_2}{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}}
\]
根据中心极限定理，当 \(n_1\) 和 \(n_2\) 较大时，满足：
\[
\frac{(\hat{p}_1 - \hat{p}_2) - (p_1 - p_2)}{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}} \sim N(0,1)
\]
进而有：
\[
\begin{align}
z' & = \frac{\left(\hat{p}_1 - \hat{p}_2\right) - \left(p_1 - p_2\right) + \left(p_1 - p_2\right)}
{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}} \\
& = \frac{\left(\hat{p}_1 - \hat{p}_2\right) - \left(p_1 - p_2\right)}
{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}}
+
\frac{\left(p_1 - p_2\right)}
{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}} \\
& \xrightarrow{d}
N\left(\frac{p_1 - p_2}{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}}, \ 1\right)
\end{align}
\]
\[
\begin{align}
Power & = P\left(z' < -z_{1-\alpha}\right) \\
& = \Phi\left(-z_{1-\alpha} - \frac{p_1 - p_2}{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}}\right) \\
& = 1 - \Phi\left(z_{1-\alpha} + \frac{p_1 - p_2}{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}}\right) \\
& = 1 - \beta
\end{align}
\]
样本量公式推导
根据标准正态分布分位数的定义：
\[
z_{1-\alpha} + \frac{p_1 - p_2}{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}} = z_\beta
\]
设 \(n_1 = kn_2\)，由上式可解出：
\[
n_2 = \frac{\left(z_{1-\alpha} + z_{1-\beta}\right)^2 \left[ \frac{1}{k}p_1(1-p_1) + p_2(1-p_2) \right]}{(p_1-p_2)^2}
\]
\[
n_1 = k n_2
\]
\[
\begin{align}
Power & = P\left(z' > z_{1-\alpha}\right) \\
& = 1 - \Phi\left(z_{1-\alpha} - \frac{p_1 - p_2}{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}}\right) \\
& = 1 - \beta
\end{align}
\]
样本量公式推导
根据标准正态分布分位数的定义：
\[
z_{1-\alpha} - \frac{p_1 - p_2}{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}} = z_\beta
\]
设 \(n_1 = kn_2\)，由上式可解出：
\[
n_2 = \frac{\left(z_{1-\alpha} + z_{1-\beta}\right)^2 \left[ \frac{1}{k}p_1(1-p_1) + p_2(1-p_2) \right]}{(p_1-p_2)^2}
\]
\[
n_1 = k n_2
\]
\[
\begin{align}
Power & = P\left(z' < -z_{1-\alpha/2}\right) + P\left(z' > z_{1-\alpha/2}\right) \\
& = 1 - \Phi\left(z_{1-\alpha/2} + \frac{p_1 - p_2}{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}}\right)
+
1 - \Phi\left(z_{1-\alpha/2} - \frac{p_1 - p_2}{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}}\right) \\
& = 1 - \beta
\end{align}
\]
Z-Test Unpooled 连续性校正¶
在 \(H_0\) 成立时，可构建 \(z\) 统计量：
\[
z = \frac{\hat{p}_1 - \hat{p}_2 + c}
{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}} \sim N(0,1)
\]
其中：
\[
c =
\begin{cases}
\frac{1}{2}\left(\frac{1}{n_1}+\frac{1}{n_2}\right), & \text{左单侧检验} \\
- \frac{1}{2}\left(\frac{1}{n_1}+\frac{1}{n_2}\right), & \text{右单侧检验}
\end{cases}
\]
在 \(H_1\) 成立时，可构建 \(z'\) 统计量：
\[
z' = \frac{\hat{p}_1 - \hat{p}_2 + c}
{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}}
\]
根据中心极限定理，当 \(n_1\) 和 \(n_2\) 较大时，满足：
\[
\frac{(\hat{p}_1 - \hat{p}_2) - (p_1 - p_2)}{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}} \sim N(0,1)
\]
进而有：
\[
\begin{align}
z' & = \frac{\left(\hat{p}_1 - \hat{p}_2\right) - \left(p_1 - p_2\right) + \left(p_1 - p_2 + c\right)}
{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}} \\
& = \frac{\left(\hat{p}_1 - \hat{p}_2\right) - \left(p_1 - p_2\right)}
{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}}
+
\frac{\left(p_1 - p_2 + c\right)}
{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}} \\
& \xrightarrow{d}
N\left(\frac{p_1 - p_2 + c}
{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}}, \ 1
\right)
\end{align}
\]
\[
\begin{align}
Power & = P\left(z' < -z_{1-\alpha}\right) \\
& = \Phi\left(-z_{1-\alpha} - \frac{p_1 - p_2 + \frac{1}{2}\left(\frac{1}{n_1} + \frac{1}{n_2}\right)}{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}}\right) \\
& = 1 - \Phi\left(z_{1-\alpha} + \frac{p_1 - p_2 + \frac{1}{2}\left(\frac{1}{n_1} + \frac{1}{n_2}\right)}{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}}\right) \\
& = 1 - \beta
\end{align}
\]
\[
\begin{align}
Power & = P\left(z' > z_{1-\alpha}\right) \\
& = 1 - \Phi\left(z_{1-\alpha} - \frac{p_1 - p_2 - \frac{1}{2}\left(\frac{1}{n_1} + \frac{1}{n_2}\right)}{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}}\right) \\
& = 1 - \beta
\end{align}
\]
\[
\begin{align}
Power & = P\left(z' < -z_{1-\alpha/2}\right) + P\left(z' > z_{1-\alpha/2}\right) \\
& = 1 - \Phi\left(z_{1-\alpha/2} + \frac{p_1 - p_2 + \frac{1}{2}\left(\frac{1}{n_1} + \frac{1}{n_2}\right)}{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}}\right)
+
1 - \Phi\left(z_{1-\alpha/2} - \frac{p_1 - p_2 - \frac{1}{2}\left(\frac{1}{n_1} + \frac{1}{n_2}\right)}{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}}\right) \\
& = 1 - \beta
\end{align}
\]