統計検定 1級 過去問 解答/解答例と解説
2014年11月30日 (日) 試験

$$\begin{equation} L(p) = \dfrac{n!}{\prod_{j=0}^{4}n_j!}\prod_{j=0}^4 \{{}_4C_{j}p^j(1-p)^{4-j}\}^{n_j} \end{equation}$$

$$\begin{align} l(p) &= \log L(p) \\ & = \log \dfrac{n!}{\prod_{j=0}^{4}n_j!}\prod_{j=0}^4 \{{}_4C_{j}p^j(1-p)^{4-j}\}^{n_j} \\ & = \log \dfrac{n!}{\prod_{j=0}^{4}n_j!} + \sum_{j=0}^4 \log \{{}_4C_{j}p^j(1-p)^{4-j}\}^{n_j} \\ & = \log \dfrac{n!}{\prod_{j=0}^{4}n_j!} + \sum_{j=0}^4 n_j\log {}_4C_{j}p^j(1-p)^{4-j} \\ & = \log \dfrac{n!}{\prod_{j=0}^{4}n_j!} + \sum_{j=0}^4 n_j \{j\log p + (4-j)\log (1-p) + \log{}_4C_{j}\} \end{align}$$

$$\begin{align} \frac{\partial}{\partial p}l(p) &= \frac{\partial}{\partial p}\left[\log \dfrac{n!}{\prod_{j=0}^{4}n_j!} + \sum_{j=0}^4 n_j \{j\log p + (4-j)\log (1-p) + \log{}_4C_{j}\}\right] \\ &= \sum_{j=0}^4 n_j\left\{\frac{j}{p} - \frac{(4-j)}{1-p}\right\} \\ &= \sum_{j=0}^4 n_j\frac{j(1-p) - (4-j)p}{p(1-p)} \\ &= \sum_{j=0}^4 \frac{jn_j-jn_jp - 4n_jp + jn_jp}{p(1-p)} \\ &= \sum_{j=0}^4 \frac{jn_j - 4n_jp}{p(1-p)} \\ &= \frac{1}{p(1-p)}\left\{\sum_{j=0}^4 jn_j - 4p \sum_{j=0}^4 n_j\right\} \\ &= \frac{1}{p(1-p)}\left\{\sum_{j=0}^4 jn_j - 4np\right\} \end{align}$$

$$\begin{equation} \hat{p} = \frac{1}{4n}\sum_{j=0}^4 jn_j \end{equation}$$