<h4>尤度比検定</h4><p>正規分布における母平均の検定を尤度比検定で行う方法について説明します。</p><h4>尤度比検定の理論的基礎</h4><p class='step'><strong>Step 1: 尤度関数の定義</strong></p><p>正規分布$N(\mu, \sigma^2)$の尤度関数：</p><div class='formula'>$L(\mu) = \prod_{i=1}^n \frac{1}{\sqrt{2\pi\sigma^2}} \exp\left(-\frac{(x_i-\mu)^2}{2\sigma^2}\right)
lt;/div><div class='formula'>$= \frac{1}{(2\pi\sigma^2)^{n/2}} \exp\left(-\frac{1}{2\sigma^2}\sum_{i=1}^n (x_i-\mu)^2\right)
lt;/div><p class='step'><strong>Step 2: 対数尤度関数</strong></p><p>対数尤度関数：</p><div class='formula'>$\ell(\mu) = \log L(\mu) = -\frac{n}{2}\log(2\pi\sigma^2) - \frac{1}{2\sigma^2}\sum_{i=1}^n (x_i-\mu)^2
lt;/div><p class='step'><strong>Step 3: 尤度比統計量の定義</strong></p><p>尤度比統計量：</p><div class='formula'>$\Lambda = \frac{\max_{\mu \in \Theta_0} L(\mu)}{\max_{\mu \in \Theta} L(\mu)}
lt;/div><p>ここで$\Theta_0 = \{\mu_0\}$、$\Theta = \mathbb{R}$です。</p><h4>データの整理と基本統計量</h4><p class='step'><strong>Step 4: 観測値の整理</strong></p><p>与えられた観測値：</p><ul><li>$x_1 = 10.5, x_2 = 9.2, x_3 = 11.8, x_4 = 8.9, x_5 = 10.1
lt;/li><li>標本サイズ：$n = 5
lt;/li><li>既知分散：$\sigma^2 = 4
lt;/li><li>帰無仮説：$H_0: \mu = 10
lt;/li></ul><p class='step'><strong>Step 5: 標本平均の計算</strong></p><p>標本平均：</p><div class='formula'>$\bar{x} = \frac{1}{n}\sum_{i=1}^n x_i = \frac{10.5 + 9.2 + 11.8 + 8.9 + 10.1}{5} = \frac{50.5}{5} = 10.1
lt;/div><p class='step'><strong>Step 6: 平方和の計算</strong></p><p>帰無仮説の下での平方和：</p><div class='formula'>$S_0 = \sum_{i=1}^n (x_i - \mu_0)^2 = \sum_{i=1}^n (x_i - 10)^2
lt;/div><div class='formula'>$= (10.5-10)^2 + (9.2-10)^2 + (11.8-10)^2 + (8.9-10)^2 + (10.1-10)^2
lt;/div><div class='formula'>$= 0.25 + 0.64 + 3.24 + 1.21 + 0.01 = 5.35
lt;/div><p>最尤推定量での平方和：</p><div class='formula'>$S_1 = \sum_{i=1}^n (x_i - \bar{x})^2 = \sum_{i=1}^n (x_i - 10.1)^2
lt;/div><div class='formula'>$= (10.5-10.1)^2 + (9.2-10.1)^2 + (11.8-10.1)^2 + (8.9-10.1)^2 + (10.1-10.1)^2
lt;/div><div class='formula'>$= 0.16 + 0.81 + 2.89 + 1.44 + 0 = 5.30
lt;/div><h4>尤度比統計量の計算</h4><p class='step'><strong>Step 7: 尤度比の導出</strong></p><p>尤度比：</p><div class='formula'>$\Lambda = \frac{L(\mu_0)}{L(\hat{\mu})} = \frac{\exp\left(-\frac{S_0}{2\sigma^2}\right)}{\exp\left(-\frac{S_1}{2\sigma^2}\right)} = \exp\left(-\frac{S_0 - S_1}{2\sigma^2}\right)
lt;/div><p class='step'><strong>Step 8: 対数尤度比統計量</strong></p><p>対数尤度比統計量：</p><div class='formula'>$-2\log\Lambda = -2\log\left(\exp\left(-\frac{S_0 - S_1}{2\sigma^2}\right)\right) = \frac{S_0 - S_1}{\sigma^2}
lt;/div><div class='formula'>$= \frac{5.35 - 5.30}{4} = \frac{0.05}{4} = 0.0125
lt;/div>