統計検定 1級 過去問 解答/解答例と解説
2015年11月29日 (日) 試験

$$\begin{equation} H_{0}~:~\mu = 0 \end{equation}$$

$$\begin{equation} H_{1}~:~\mu = \mu_{1}~~(> 0) \end{equation}$$

$$\begin{align} f(\bm{x} ; \mu) &= \prod_{i=1}^{n}\frac{1}{\sqrt{2\pi}}\exp\left[-\frac{(x_i - \mu)^2}{2}\right] \\ &= \left(\frac{1}{\sqrt{2\pi}}\right)^n \exp\left[-\frac{1}{2}\sum_{i=1}^{n}(x_i - \mu)^2\right] \\ \end{align}$$

$$\begin{equation} f(\bm{x} ; \mu_1) = \left(\frac{1}{\sqrt{2\pi}}\right)^n \exp\left[-\frac{1}{2}\sum_{i=1}^{n}(x_i - \mu_1)^2\right] \\ \end{equation}$$

$$\begin{equation} f(\bm{x} ; 0) = \left(\frac{1}{\sqrt{2\pi}}\right)^n \exp\left[-\frac{1}{2}\sum_{i=1}^{n}{x_i}^2\right] \\ \end{equation}$$

$$\begin{align} \frac{f(\bm{x} ; \mu_1)}{f(\bm{x} ; 0)} &= \frac{\ds \left(\frac{1}{\sqrt{2\pi}}\right)^n \exp\left[-\frac{1}{2}\sum_{i=1}^{n}(x_i - \mu_1)^2\right]}{\ds \left(\frac{1}{\sqrt{2\pi}}\right)^n \exp\left[-\frac{1}{2}\sum_{i=1}^{n}{x_i}^2\right]} \\ &= \frac{\ds \exp\left[-\frac{1}{2}\sum_{i=1}^{n}(x_i - \mu_1)^2\right]}{\ds \exp\left[-\frac{1}{2}\sum_{i=1}^{n}{x_i}^2\right]} \\ &= \exp\left[-\frac{1}{2}\sum_{i=1}^{n}(x_i - \mu_1)^2 + \frac{1}{2}\sum_{i=1}^{n}{x_i}^2\right] \\ &= \exp\left[\mu_1\sum_{i=1}^{n}x_i -\frac{1}{2}n{\mu_1}^2\right] \\ &= \exp\left[\mu_1 n \bar{x} - \frac{1}{2}n{\mu_1}^2\right] \\ &= \exp\left[\mu_1 n \left(\bar{x} - \frac{1}{2}\mu_1\right)\right] > c \end{align}$$

$$\begin{equation} \Longleftrightarrow~\bar{x} > c' \end{equation}$$

$$\begin{equation} P\left(\bar{X} > c'\right) = \alpha \end{equation}$$

$$\begin{equation} \bar{X} > \frac{z_{\alpha}}{\sqrt{n}} \end{equation}$$