統計検定1級の過去問解答解説を行います。目次は以下をご覧ください。
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目次
問題
統計検定1級の過去問からの出題になります。統計検定の問題の著作権は日本統計学会に帰属していますので，本稿にて記載することはできません。「演習問題を俯瞰する」で詳しく紹介している公式の過去問題集をご購入いただきますようお願い致します。
解答
$F$検定と測定法の比較検討に関する出題でした。
(1)
測定法(1)の計画行列を$X_{(1)}$，測定法(2)の計画行列を$X_{(2)}$とおくと，
\begin{align}
X_{(1)} =
\begin{pmatrix}
1 & 0 & 0 & 0 & 0\\
1 & 0 & 0 & 0 & 0\\
0 & 1 & 0 & 0 & 0\\
0 & 1 & 0 & 0 & 0\\
0 & 0 & 1 & 0 & 0\\
0 & 0 & 1 & 0 & 0\\
0 & 0 & 0 & 1 & 0\\
0 & 0 & 0 & 1 & 0\\
0 & 0 & 0 & 0 & 1\\
0 & 0 & 0 & 0 & 1
\end{pmatrix},\quad
X_{(2)} =
\begin{pmatrix}
1 & 1 & 0 & 0 & 0\\
1 & 0 & 1 & 0 & 0\\
1 & 0 & 0 & 1 & 0\\
1 & 0 & 0 & 0 & 1\\
0 & 1 & 1 & 0 & 0\\
0 & 1 & 0 & 1 & 0\\
0 & 1 & 0 & 0 & 1\\
0 & 0 & 1 & 1 & 0\\
0 & 0 & 1 & 0 & 1\\
0 & 0 & 0 & 1 & 1
\end{pmatrix}
\end{align}
X_{(1)} =
\begin{pmatrix}
1 & 0 & 0 & 0 & 0\\
1 & 0 & 0 & 0 & 0\\
0 & 1 & 0 & 0 & 0\\
0 & 1 & 0 & 0 & 0\\
0 & 0 & 1 & 0 & 0\\
0 & 0 & 1 & 0 & 0\\
0 & 0 & 0 & 1 & 0\\
0 & 0 & 0 & 1 & 0\\
0 & 0 & 0 & 0 & 1\\
0 & 0 & 0 & 0 & 1
\end{pmatrix},\quad
X_{(2)} =
\begin{pmatrix}
1 & 1 & 0 & 0 & 0\\
1 & 0 & 1 & 0 & 0\\
1 & 0 & 0 & 1 & 0\\
1 & 0 & 0 & 0 & 1\\
0 & 1 & 1 & 0 & 0\\
0 & 1 & 0 & 1 & 0\\
0 & 1 & 0 & 0 & 1\\
0 & 0 & 1 & 1 & 0\\
0 & 0 & 1 & 0 & 1\\
0 & 0 & 0 & 1 & 1
\end{pmatrix}
\end{align}
と表される。正規方程式は$X^{\prime}X\theta=X^{\prime}x$であるため，$\theta$の最小二乗推定量は$X^{\prime}X$が正則のとき$(X^{\prime}X)^{-1}X^{\prime}x$となる。二つの測定法それぞれについて，この最小二乗推定量を計算する。測定法(1)については，
\begin{align}
X_{(1)}^{\prime}X_{(1)} =
\begin{pmatrix}
1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 1 & 1 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1\\
\end{pmatrix}
\begin{pmatrix}
1 & 0 & 0 & 0 & 0\\
1 & 0 & 0 & 0 & 0\\
0 & 1 & 0 & 0 & 0\\
0 & 1 & 0 & 0 & 0\\
0 & 0 & 1 & 0 & 0\\
0 & 0 & 1 & 0 & 0\\
0 & 0 & 0 & 1 & 0\\
0 & 0 & 0 & 1 & 0\\
0 & 0 & 0 & 0 & 1\\
0 & 0 & 0 & 0 & 1
\end{pmatrix}
= 2I_{5}
\end{align}
X_{(1)}^{\prime}X_{(1)} =
\begin{pmatrix}
1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 1 & 1 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1\\
\end{pmatrix}
\begin{pmatrix}
1 & 0 & 0 & 0 & 0\\
1 & 0 & 0 & 0 & 0\\
0 & 1 & 0 & 0 & 0\\
0 & 1 & 0 & 0 & 0\\
0 & 0 & 1 & 0 & 0\\
0 & 0 & 1 & 0 & 0\\
0 & 0 & 0 & 1 & 0\\
0 & 0 & 0 & 1 & 0\\
0 & 0 & 0 & 0 & 1\\
0 & 0 & 0 & 0 & 1
\end{pmatrix}
= 2I_{5}
\end{align}
となることに注意すると，
\begin{align}
\hat{\theta}_{1}
&= (2I_{5})^{-1}X_{(1)}^{\prime}x
= \frac{1}{2}
\begin{pmatrix}
1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 1 & 1 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1\\
\end{pmatrix}
\begin{pmatrix}
x_{1}\\
x_{2}\\
x_{3}\\
x_{4}\\
x_{5}\\
x_{6}\\
x_{7}\\
x_{8}\\
x_{9}\\
x_{10}
\end{pmatrix}
= \frac{1}{2}
\begin{pmatrix}
x_{1}+x_{2}\\
x_{3}+x_{4}\\
x_{5}+x_{6}\\
x_{7}+x_{8}\\
x_{9}+x_{10}
\end{pmatrix}
\end{align}
\hat{\theta}_{1}
&= (2I_{5})^{-1}X_{(1)}^{\prime}x
= \frac{1}{2}
\begin{pmatrix}
1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 1 & 1 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1\\
\end{pmatrix}
\begin{pmatrix}
x_{1}\\
x_{2}\\
x_{3}\\
x_{4}\\
x_{5}\\
x_{6}\\
x_{7}\\
x_{8}\\
x_{9}\\
x_{10}
\end{pmatrix}
= \frac{1}{2}
\begin{pmatrix}
x_{1}+x_{2}\\
x_{3}+x_{4}\\
x_{5}+x_{6}\\
x_{7}+x_{8}\\
x_{9}+x_{10}
\end{pmatrix}
\end{align}
が得られる。測定法(2)については，
\begin{align}
X_{(2)}^{\prime}X_{(2)} {=}
\begin{pmatrix}
1 & 1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0\\
1 & 0 & 0 & 0 & 1 & 1 & 1 & 0 & 0 & 0\\
0 & 1 & 0 & 0 & 1 & 0 & 0 & 1 & 1 & 0\\
0 & 0 & 1 & 0 & 0 & 1 & 0 & 1 & 0 & 1\\
0 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 1 & 1
\end{pmatrix}
\!
\begin{pmatrix}
1 & 1 & 0 & 0 & 0\\
1 & 0 & 1 & 0 & 0\\
1 & 0 & 0 & 1 & 0\\
1 & 0 & 0 & 0 & 1\\
0 & 1 & 1 & 0 & 0\\
0 & 1 & 0 & 1 & 0\\
0 & 1 & 0 & 0 & 1\\
0 & 0 & 1 & 1 & 0\\
0 & 0 & 1 & 0 & 1\\
0 & 0 & 0 & 1 & 1
\end{pmatrix}
{=}\!
\begin{pmatrix}
4 & 1 & 1 & 1 & 1 \\
1 & 4 & 1 & 1 & 1 \\
1 & 1 & 4 & 1 & 1 \\
1 & 1 & 1 & 4 & 1 \\
1 & 1 & 1 & 1 & 4
\end{pmatrix}
\end{align}
X_{(2)}^{\prime}X_{(2)} {=}
\begin{pmatrix}
1 & 1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0\\
1 & 0 & 0 & 0 & 1 & 1 & 1 & 0 & 0 & 0\\
0 & 1 & 0 & 0 & 1 & 0 & 0 & 1 & 1 & 0\\
0 & 0 & 1 & 0 & 0 & 1 & 0 & 1 & 0 & 1\\
0 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 1 & 1
\end{pmatrix}
\!
\begin{pmatrix}
1 & 1 & 0 & 0 & 0\\
1 & 0 & 1 & 0 & 0\\
1 & 0 & 0 & 1 & 0\\
1 & 0 & 0 & 0 & 1\\
0 & 1 & 1 & 0 & 0\\
0 & 1 & 0 & 1 & 0\\
0 & 1 & 0 & 0 & 1\\
0 & 0 & 1 & 1 & 0\\
0 & 0 & 1 & 0 & 1\\
0 & 0 & 0 & 1 & 1
\end{pmatrix}
{=}\!
\begin{pmatrix}
4 & 1 & 1 & 1 & 1 \\
1 & 4 & 1 & 1 & 1 \\
1 & 1 & 4 & 1 & 1 \\
1 & 1 & 1 & 4 & 1 \\
1 & 1 & 1 & 1 & 4
\end{pmatrix}
\end{align}
となること，および与えられた公式に$a=4,b=1,k=5$を代入して得られる
\begin{align}
\begin{pmatrix}
4 & 1 & 1 & 1 & 1 \\
1 & 4 & 1 & 1 & 1 \\
1 & 1 & 4 & 1 & 1 \\
1 & 1 & 1 & 4 & 1 \\
1 & 1 & 1 & 1 & 4
\end{pmatrix}^{-1}
&=
\frac{1}{24}
\begin{pmatrix}
7 & -1 & -1 & -1 & -1 \\
-1 & 7 & -1 & -1 & -1 \\
-1 & -1 & 7 & -1 & -1 \\
-1 & -1 & -1 & 7 & -1 \\
-1 & -1 & -1 & -1 & 7
\end{pmatrix}
\end{align}
\begin{pmatrix}
4 & 1 & 1 & 1 & 1 \\
1 & 4 & 1 & 1 & 1 \\
1 & 1 & 4 & 1 & 1 \\
1 & 1 & 1 & 4 & 1 \\
1 & 1 & 1 & 1 & 4
\end{pmatrix}^{-1}
&=
\frac{1}{24}
\begin{pmatrix}
7 & -1 & -1 & -1 & -1 \\
-1 & 7 & -1 & -1 & -1 \\
-1 & -1 & 7 & -1 & -1 \\
-1 & -1 & -1 & 7 & -1 \\
-1 & -1 & -1 & -1 & 7
\end{pmatrix}
\end{align}
を利用すると，
\begin{align}
\hat{\theta}_{2}
&=
\frac{1}{24}
\begin{pmatrix}
7 & -1 & -1 & -1 & -1 \\
-1 & 7 & -1 & -1 & -1 \\
-1 & -1 & 7 & -1 & -1 \\
-1 & -1 & -1 & 7 & -1 \\
-1 & -1 & -1 & -1 & 7
\end{pmatrix}
\begin{pmatrix}
1 & 1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0\\
1 & 0 & 0 & 0 & 1 & 1 & 1 & 0 & 0 & 0\\
0 & 1 & 0 & 0 & 1 & 0 & 0 & 1 & 1 & 0\\
0 & 0 & 1 & 0 & 0 & 1 & 0 & 1 & 0 & 1\\
0 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 1 & 1
\end{pmatrix}
\begin{pmatrix}
x_{1}\\
x_{2}\\
x_{3}\\
x_{4}\\
x_{5}\\
x_{6}\\
x_{7}\\
x_{8}\\
x_{9}\\
x_{10}
\end{pmatrix}\\[0.7em]
&=
\frac{1}{24}
\begin{pmatrix}
6 & 6 & 6 & 6 & -2 & -2 & -2 & -2 & -2 & -2\\
6 & -2 & -2 & -2 & 6 & 6 & 6 & -2 & -2 & -2\\
-2 & 6 & -2 & -2 & 6 & -2 & -2 & 6 & 6 & -2\\
-2 & -2 & 6 & -2 & -2 & 6 & -2 & 6 & -2 & 6\\
-2 & -2 & -2 & 6 & -2 & -2 & 6 & -2 & 6 & 6
\end{pmatrix}
\begin{pmatrix}
x_{1}\\
x_{2}\\
x_{3}\\
x_{4}\\
x_{5}\\
x_{6}\\
x_{7}\\
x_{8}\\
x_{9}\\
x_{10}
\end{pmatrix}\\[0.7em]
&= \frac{1}{12}
\begin{pmatrix}
3(x_{1}+x_{2}+x_{3}+x_{4})-(x_{5}+x_{6}+x_{7}+x_{8}+x_{9}+x_{10})\\
3(x_{1}+x_{5}+x_{6}+x_{7})-(x_{2}+x_{3}+x_{4}+x_{8}+x_{9}+x_{10})\\
3(x_{2}+x_{5}+x_{8}+x_{9})-(x_{1}+x_{3}+x_{4}+x_{6}+x_{7}+x_{10})\\
3(x_{3}+x_{6}+x_{8}+x_{10})-(x_{1}+x_{2}+x_{4}+x_{5}+x_{7}+x_{9})\\
3(x_{4}+x_{7}+x_{9}+x_{10})-(x_{1}+x_{2}+x_{3}+x_{5}+x_{6}+x_{8})
\end{pmatrix}
\end{align}
\hat{\theta}_{2}
&=
\frac{1}{24}
\begin{pmatrix}
7 & -1 & -1 & -1 & -1 \\
-1 & 7 & -1 & -1 & -1 \\
-1 & -1 & 7 & -1 & -1 \\
-1 & -1 & -1 & 7 & -1 \\
-1 & -1 & -1 & -1 & 7
\end{pmatrix}
\begin{pmatrix}
1 & 1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0\\
1 & 0 & 0 & 0 & 1 & 1 & 1 & 0 & 0 & 0\\
0 & 1 & 0 & 0 & 1 & 0 & 0 & 1 & 1 & 0\\
0 & 0 & 1 & 0 & 0 & 1 & 0 & 1 & 0 & 1\\
0 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 1 & 1
\end{pmatrix}
\begin{pmatrix}
x_{1}\\
x_{2}\\
x_{3}\\
x_{4}\\
x_{5}\\
x_{6}\\
x_{7}\\
x_{8}\\
x_{9}\\
x_{10}
\end{pmatrix}\\[0.7em]
&=
\frac{1}{24}
\begin{pmatrix}
6 & 6 & 6 & 6 & -2 & -2 & -2 & -2 & -2 & -2\\
6 & -2 & -2 & -2 & 6 & 6 & 6 & -2 & -2 & -2\\
-2 & 6 & -2 & -2 & 6 & -2 & -2 & 6 & 6 & -2\\
-2 & -2 & 6 & -2 & -2 & 6 & -2 & 6 & -2 & 6\\
-2 & -2 & -2 & 6 & -2 & -2 & 6 & -2 & 6 & 6
\end{pmatrix}
\begin{pmatrix}
x_{1}\\
x_{2}\\
x_{3}\\
x_{4}\\
x_{5}\\
x_{6}\\
x_{7}\\
x_{8}\\
x_{9}\\
x_{10}
\end{pmatrix}\\[0.7em]
&= \frac{1}{12}
\begin{pmatrix}
3(x_{1}+x_{2}+x_{3}+x_{4})-(x_{5}+x_{6}+x_{7}+x_{8}+x_{9}+x_{10})\\
3(x_{1}+x_{5}+x_{6}+x_{7})-(x_{2}+x_{3}+x_{4}+x_{8}+x_{9}+x_{10})\\
3(x_{2}+x_{5}+x_{8}+x_{9})-(x_{1}+x_{3}+x_{4}+x_{6}+x_{7}+x_{10})\\
3(x_{3}+x_{6}+x_{8}+x_{10})-(x_{1}+x_{2}+x_{4}+x_{5}+x_{7}+x_{9})\\
3(x_{4}+x_{7}+x_{9}+x_{10})-(x_{1}+x_{2}+x_{3}+x_{5}+x_{6}+x_{8})
\end{pmatrix}
\end{align}
が得られる。
正規方程式の解は，
\begin{align}
\frac{d}{d\theta}\left(\frac{1}{2}\|x-X\theta\|^{2}\right)
&= \frac{1}{2}\frac{d}{d\theta}(x-X\theta)^{\prime}(x-X\theta)
= \frac{1}{2}\frac{d}{d\theta}\left(x^{\prime}x-2\theta^{\prime}X^{\prime}+\theta^{\prime}X^{\prime}X\theta\right)\\[0.7em]
&= \frac{1}{2}\left(-2X^{\prime}x+2X^{\prime}X\theta\right) = 0
\end{align}
\frac{d}{d\theta}\left(\frac{1}{2}\|x-X\theta\|^{2}\right)
&= \frac{1}{2}\frac{d}{d\theta}(x-X\theta)^{\prime}(x-X\theta)
= \frac{1}{2}\frac{d}{d\theta}\left(x^{\prime}x-2\theta^{\prime}X^{\prime}+\theta^{\prime}X^{\prime}X\theta\right)\\[0.7em]
&= \frac{1}{2}\left(-2X^{\prime}x+2X^{\prime}X\theta\right) = 0
\end{align}
を解くことにより，$X^{\prime}X$が正則のときに$\hat{\theta}=(X^{\prime}X)^{-1}X^{\prime}x$が得られます。
(2)
最小二乗推定量$\hat{\theta}$の標本分散$V[\hat{\theta}]$は$\sigma^{2}(X^{\prime}X)^{-1}$と表される。小問(1)の結果より，
\begin{align}
V[\hat{\theta}_{(1)}] = \frac{\sigma^{2}}{2}I_{5},\quad
V[\hat{\theta}_{(2)}] =
\frac{\sigma^{2}}{24}
\begin{pmatrix}
7 & -1 & -1 & -1 & -1 \\
-1 & 7 & -1 & -1 & -1 \\
-1 & -1 & 7 & -1 & -1 \\
-1 & -1 & -1 & 7 & -1 \\
-1 & -1 & -1 & -1 & 7
\end{pmatrix}
\end{align}
V[\hat{\theta}_{(1)}] = \frac{\sigma^{2}}{2}I_{5},\quad
V[\hat{\theta}_{(2)}] =
\frac{\sigma^{2}}{24}
\begin{pmatrix}
7 & -1 & -1 & -1 & -1 \\
-1 & 7 & -1 & -1 & -1 \\
-1 & -1 & 7 & -1 & -1 \\
-1 & -1 & -1 & 7 & -1 \\
-1 & -1 & -1 & -1 & 7
\end{pmatrix}
\end{align}
が得られるため，測定法(1)では$V[\hat{\theta}_{j}]=\sigma^{2}/2$，測定法(2)では$V[\hat{\theta}_{j}]=7\sigma^{2}/24$となる。
最小二乗推定量$\hat{\theta}$の標本分散$V[\hat{\theta}]$が$\sigma^{2}(X^{\prime}X)^{-1}$となることは，
\begin{align}
V[\hat{\theta}]
&= V[(X^{\prime}X)^{-1}X^{\prime}x]
= V[(X^{\prime}X)^{-1}X^{\prime}(X\theta+\varepsilon)]
= V[\theta+(X^{\prime}X)^{-1}X^{\prime}\varepsilon]
= V[(X^{\prime}X)^{-1}X^{\prime}\varepsilon]\\[0.7em]
&= E\left[\left\{(X^{\prime}X)^{-1}X^{\prime}\varepsilon\right\}\left\{(X^{\prime}X)^{-1}X^{\prime}\varepsilon\right\}^{\prime}\right]
= E\left[(X^{\prime}X)^{-1}X^{\prime}\varepsilon\varepsilon^{\prime}X\left\{(X^{\prime}X)^{-1}\right\}^{\prime}\right]\\[0.7em]
&= E\left[(X^{\prime}X)^{-1}X^{\prime}\varepsilon\varepsilon^{\prime}X(X^{\prime}X)^{-1}\right]
= (X^{\prime}X)^{-1}X^{\prime}E[\varepsilon\varepsilon^{\prime}]X(X^{\prime}X)^{-1}\\[0.7em]
&= (X^{\prime}X)^{-1}X^{\prime}(\sigma^{2}I_{10})X(X^{\prime}X)^{-1}
= \sigma^{2}(X^{\prime}X)^{-1}
\end{align}
V[\hat{\theta}]
&= V[(X^{\prime}X)^{-1}X^{\prime}x]
= V[(X^{\prime}X)^{-1}X^{\prime}(X\theta+\varepsilon)]
= V[\theta+(X^{\prime}X)^{-1}X^{\prime}\varepsilon]
= V[(X^{\prime}X)^{-1}X^{\prime}\varepsilon]\\[0.7em]
&= E\left[\left\{(X^{\prime}X)^{-1}X^{\prime}\varepsilon\right\}\left\{(X^{\prime}X)^{-1}X^{\prime}\varepsilon\right\}^{\prime}\right]
= E\left[(X^{\prime}X)^{-1}X^{\prime}\varepsilon\varepsilon^{\prime}X\left\{(X^{\prime}X)^{-1}\right\}^{\prime}\right]\\[0.7em]
&= E\left[(X^{\prime}X)^{-1}X^{\prime}\varepsilon\varepsilon^{\prime}X(X^{\prime}X)^{-1}\right]
= (X^{\prime}X)^{-1}X^{\prime}E[\varepsilon\varepsilon^{\prime}]X(X^{\prime}X)^{-1}\\[0.7em]
&= (X^{\prime}X)^{-1}X^{\prime}(\sigma^{2}I_{10})X(X^{\prime}X)^{-1}
= \sigma^{2}(X^{\prime}X)^{-1}
\end{align}
のように示されます。また，小問(1)の結果をそのまま用いて，測定法(1)については
\begin{align}
V[\theta_{j}] &= V\left[\frac{x_{*}+x_{*}}{2}\right] = \frac{2V[x_{*}]}{4} = \frac{\sigma^{2}}{2}
\end{align}
V[\theta_{j}] &= V\left[\frac{x_{*}+x_{*}}{2}\right] = \frac{2V[x_{*}]}{4} = \frac{\sigma^{2}}{2}
\end{align}
と求め，測定法(2)については
\begin{align}
V[\theta_{j}] &= V\left[\frac{3(x_{*}+x_{*}+x_{*}+x_{*})-(x_{*}+x_{*}+x_{*}+x_{*}+x_{*}+x_{*})}{12}\right] = \frac{9\cdot 4\sigma^{2}+6\sigma^{2}}{12^{2}} = \frac{7\sigma^{2}}{24}
\end{align}
V[\theta_{j}] &= V\left[\frac{3(x_{*}+x_{*}+x_{*}+x_{*})-(x_{*}+x_{*}+x_{*}+x_{*}+x_{*}+x_{*})}{12}\right] = \frac{9\cdot 4\sigma^{2}+6\sigma^{2}}{12^{2}} = \frac{7\sigma^{2}}{24}
\end{align}
と求めてもよいです。$x_{*}$は測定値ですが，$x=X\theta+\varepsilon$より$V[x]$は測定誤差の分散$\sigma^{2}$となることに注意してください。
(3)
帰無仮説$H_{0}$に対し，帰無仮説を制約条件として計算した最小二乗推定値のもとでの残差平方和を$RSS_{R}$，何の制約条件も設けずに計算した最小二乗推定値のもとでの残差平方和を$RSS_{UR}$とおくと，以下で定義される
\begin{align}
F &= \frac{(RSS_{R}-RSS_{UR})/q}{RSS_{UR}/(n-K)}
\end{align}
F &= \frac{(RSS_{R}-RSS_{UR})/q}{RSS_{UR}/(n-K)}
\end{align}
は，$F(q,n-K)$に従う。ただし，$q$は帰無仮説における等号の数，$K$はパラメータ数を表している。これを本問題に適用すると，
\begin{align}
RSS_{R} = \|x-X_{0}\hat{\theta}_{0}\|^{2},\quad
RSS_{UR} = \|x-X\hat{\theta}\|^{2},\quad
n=10,\quad
q=4,\quad
K=5
\end{align}
RSS_{R} = \|x-X_{0}\hat{\theta}_{0}\|^{2},\quad
RSS_{UR} = \|x-X\hat{\theta}\|^{2},\quad
n=10,\quad
q=4,\quad
K=5
\end{align}
となる。ただし，$\|\cdot\|$は各成分の二乗和の平方根であるノルムを表し，$\theta_{0}$は帰無仮説$H_{0}$の元での線形モデル$x=X\theta_{0}+\varepsilon$で定義する。なお，公式解答では帰無仮説の下での計画行列$X$を$X_{0}$として定義しているが，帰無仮説は母数に対する仮定であって測定方法に関する仮定ではないため，帰無仮説の有無に応じて$X$の形が変化することはない。ここで，
\begin{align}
RSS_{R}
&= \|x-X\hat{\theta}_{0}\|^{2}
= (x-X\hat{\theta}_{0})^{\prime}(x-X\hat{\theta}_{0})\\[0.7em]
&= \left\{(x-X\hat{\theta})+(X\hat{\theta}-X\hat{\theta}_{0})\right\}^{\prime}\left\{(x-X\hat{\theta})+(X\hat{\theta}-X\hat{\theta}_{0})\right\}\\[0.7em]
&= RSS_{UR}{+}(x{-}X\hat{\theta})^{\prime}(X\hat{\theta}{-}X\hat{\theta}_{0}){+}(X\hat{\theta}{-}X\hat{\theta}_{0})^{\prime}(x{-}X\hat{\theta}){+}\|X\hat{\theta}{-}X\hat{\theta}_{0}\|^{2}
\end{align}
RSS_{R}
&= \|x-X\hat{\theta}_{0}\|^{2}
= (x-X\hat{\theta}_{0})^{\prime}(x-X\hat{\theta}_{0})\\[0.7em]
&= \left\{(x-X\hat{\theta})+(X\hat{\theta}-X\hat{\theta}_{0})\right\}^{\prime}\left\{(x-X\hat{\theta})+(X\hat{\theta}-X\hat{\theta}_{0})\right\}\\[0.7em]
&= RSS_{UR}{+}(x{-}X\hat{\theta})^{\prime}(X\hat{\theta}{-}X\hat{\theta}_{0}){+}(X\hat{\theta}{-}X\hat{\theta}_{0})^{\prime}(x{-}X\hat{\theta}){+}\|X\hat{\theta}{-}X\hat{\theta}_{0}\|^{2}
\end{align}
において，
\begin{align}
(x-X\hat{\theta})^{\prime}(X\hat{\theta}-X\hat{\theta}_{0})
&{=} x^{\prime}X\hat{\theta}-x^{\prime}X\hat{\theta}_{0}-\hat{\theta}^{\prime}X^{\prime}X\hat{\theta}+\hat{\theta}^{\prime}X^{\prime}X\hat{\theta}_{0}\\[0.7em]
&{=} x^{\prime}X\hat{\theta}-x^{\prime}X\hat{\theta}_{0}-(X^{\prime}X\hat{\theta})^{\prime}\hat{\theta}+(X^{\prime}X\hat{\theta})^{\prime}\hat{\theta}_{0}\\[0.7em]
&{=} x^{\prime}X\hat{\theta}{-}x^{\prime}X\hat{\theta}_{0}{-}(X^{\prime}X(X^{\prime}X)^{-1}X^{\prime}x)^{\prime}\hat{\theta}{+}(X^{\prime}X(X^{\prime}X)^{-1}X^{\prime}x)^{\prime}\hat{\theta}_{0}\\[0.7em]
&{=} x^{\prime}X\hat{\theta}{-}x^{\prime}X\hat{\theta}_{0}{-}(X^{\prime}x)^{\prime}\hat{\theta}+(X^{\prime}x)^{\prime}\hat{\theta}_{0}\\[0.7em]
&{=} x^{\prime}X\hat{\theta}{-}x^{\prime}X\hat{\theta}_{0}{-}x^{\prime}X\hat{\theta}+x^{\prime}X\hat{\theta}_{0} = 0
\end{align}
(x-X\hat{\theta})^{\prime}(X\hat{\theta}-X\hat{\theta}_{0})
&{=} x^{\prime}X\hat{\theta}-x^{\prime}X\hat{\theta}_{0}-\hat{\theta}^{\prime}X^{\prime}X\hat{\theta}+\hat{\theta}^{\prime}X^{\prime}X\hat{\theta}_{0}\\[0.7em]
&{=} x^{\prime}X\hat{\theta}-x^{\prime}X\hat{\theta}_{0}-(X^{\prime}X\hat{\theta})^{\prime}\hat{\theta}+(X^{\prime}X\hat{\theta})^{\prime}\hat{\theta}_{0}\\[0.7em]
&{=} x^{\prime}X\hat{\theta}{-}x^{\prime}X\hat{\theta}_{0}{-}(X^{\prime}X(X^{\prime}X)^{-1}X^{\prime}x)^{\prime}\hat{\theta}{+}(X^{\prime}X(X^{\prime}X)^{-1}X^{\prime}x)^{\prime}\hat{\theta}_{0}\\[0.7em]
&{=} x^{\prime}X\hat{\theta}{-}x^{\prime}X\hat{\theta}_{0}{-}(X^{\prime}x)^{\prime}\hat{\theta}+(X^{\prime}x)^{\prime}\hat{\theta}_{0}\\[0.7em]
&{=} x^{\prime}X\hat{\theta}{-}x^{\prime}X\hat{\theta}_{0}{-}x^{\prime}X\hat{\theta}+x^{\prime}X\hat{\theta}_{0} = 0
\end{align}
および同様に
\begin{align}
(X\hat{\theta}-X\hat{\theta}_{0})^{\prime}(x-X\hat{\theta})
&= (X\hat{\theta})^{\prime}x-(X\hat{\theta})^{\prime}X\hat{\theta}-(X\hat{\theta}_{0})^{\prime}x+(X\hat{\theta}_{0})^{\prime}X\hat{\theta}\\[0.7em]
&= (X\hat{\theta})^{\prime}x-\hat{\theta}^{\prime}(X^{\prime}X\hat{\theta})-(X\hat{\theta}_{0})^{\prime}x+\theta_{0}^{\prime}(X^{\prime}_{0}X\hat{\theta})\\[0.7em]
&= (X\hat{\theta})^{\prime}x-\hat{\theta}^{\prime}(X^{\prime}x)-(X\hat{\theta}_{0})^{\prime}x+\theta_{0}^{\prime}(X^{\prime}_{0}x)\\[0.7em]
&= \hat{\theta}^{\prime}X^{\prime}x-\hat{\theta}^{\prime}X^{\prime}x-\hat{\theta}_{0}^{\prime}X^{\prime}x-\hat{\theta}_{0}^{\prime}X^{\prime}x
= 0
\end{align}
(X\hat{\theta}-X\hat{\theta}_{0})^{\prime}(x-X\hat{\theta})
&= (X\hat{\theta})^{\prime}x-(X\hat{\theta})^{\prime}X\hat{\theta}-(X\hat{\theta}_{0})^{\prime}x+(X\hat{\theta}_{0})^{\prime}X\hat{\theta}\\[0.7em]
&= (X\hat{\theta})^{\prime}x-\hat{\theta}^{\prime}(X^{\prime}X\hat{\theta})-(X\hat{\theta}_{0})^{\prime}x+\theta_{0}^{\prime}(X^{\prime}_{0}X\hat{\theta})\\[0.7em]
&= (X\hat{\theta})^{\prime}x-\hat{\theta}^{\prime}(X^{\prime}x)-(X\hat{\theta}_{0})^{\prime}x+\theta_{0}^{\prime}(X^{\prime}_{0}x)\\[0.7em]
&= \hat{\theta}^{\prime}X^{\prime}x-\hat{\theta}^{\prime}X^{\prime}x-\hat{\theta}_{0}^{\prime}X^{\prime}x-\hat{\theta}_{0}^{\prime}X^{\prime}x
= 0
\end{align}
より
\begin{align}
RSS_{R}-RSS_{UR}
&= \|X\hat{\theta}{-}X\hat{\theta}_{0}\|^{2}
\end{align}
RSS_{R}-RSS_{UR}
&= \|X\hat{\theta}{-}X\hat{\theta}_{0}\|^{2}
\end{align}
となることに注意すると，$F$統計量は
\begin{align}
F &= \frac{\|X\hat{\theta}-X\hat{\theta}_{0}\|^{2}/4}{\|x-X\hat{\theta}\|^{2}/5}
\end{align}
F &= \frac{\|X\hat{\theta}-X\hat{\theta}_{0}\|^{2}/4}{\|x-X\hat{\theta}\|^{2}/5}
\end{align}
となる。
$F$統計量の詳しい解説は，どこかのタイミングで別記事で解説します。
(4)
問題文で与えられた非心度の定義より，$F$統計量の分子が従うカイ二乗分布の非心度を求めればよい。帰無仮説$H_{0}$の下では$\bar{\theta}=\theta_{1}=\cdots=\theta_{5}$となるため，代表して$\theta_{1}$などを用いることもできるが，ここでは簡単のため
\begin{align}
\hat{\theta}_{0} &=
\begin{pmatrix}
\bar{\theta}\\
\bar{\theta}\\
\bar{\theta}\\
\bar{\theta}\\
\bar{\theta}
\end{pmatrix}
\end{align}
\hat{\theta}_{0} &=
\begin{pmatrix}
\bar{\theta}\\
\bar{\theta}\\
\bar{\theta}\\
\bar{\theta}\\
\bar{\theta}
\end{pmatrix}
\end{align}
と表すことにする。測定法(1)について，小問(1)で求めた$X_{(1)}$を用いれば，
\begin{align}
X_{(1)}\hat{\theta}-X_{(1)}\hat{\theta}_{0} &=
\begin{pmatrix}
\hat{\theta}_{1}-\bar{\theta}\\
\hat{\theta}_{1}-\bar{\theta}\\
\hat{\theta}_{2}-\bar{\theta}\\
\hat{\theta}_{2}-\bar{\theta}\\
\hat{\theta}_{3}-\bar{\theta}\\
\hat{\theta}_{3}-\bar{\theta}\\
\hat{\theta}_{4}-\bar{\theta}\\
\hat{\theta}_{4}-\bar{\theta}\\
\hat{\theta}_{5}-\bar{\theta}\\
\hat{\theta}_{5}-\bar{\theta}
\end{pmatrix}
\end{align}
X_{(1)}\hat{\theta}-X_{(1)}\hat{\theta}_{0} &=
\begin{pmatrix}
\hat{\theta}_{1}-\bar{\theta}\\
\hat{\theta}_{1}-\bar{\theta}\\
\hat{\theta}_{2}-\bar{\theta}\\
\hat{\theta}_{2}-\bar{\theta}\\
\hat{\theta}_{3}-\bar{\theta}\\
\hat{\theta}_{3}-\bar{\theta}\\
\hat{\theta}_{4}-\bar{\theta}\\
\hat{\theta}_{4}-\bar{\theta}\\
\hat{\theta}_{5}-\bar{\theta}\\
\hat{\theta}_{5}-\bar{\theta}
\end{pmatrix}
\end{align}
となるため，ノルムは
\begin{align}
\|X_{(1)}\hat{\theta}-X_{(1)}\hat{\theta}_{0}\|^{2}
&= (\hat{\theta}_{1}-\bar{\theta})^{2}+(\hat{\theta}_{1}-\bar{\theta})^{2}+\cdots+(\hat{\theta}_{5}-\bar{\theta})^{2}+(\hat{\theta}_{5}-\bar{\theta})^{2}\label{ノルム_1}
\end{align}
\|X_{(1)}\hat{\theta}-X_{(1)}\hat{\theta}_{0}\|^{2}
&= (\hat{\theta}_{1}-\bar{\theta})^{2}+(\hat{\theta}_{1}-\bar{\theta})^{2}+\cdots+(\hat{\theta}_{5}-\bar{\theta})^{2}+(\hat{\theta}_{5}-\bar{\theta})^{2}\label{ノルム_1}
\end{align}
となる。一方，$F$統計量の定義より$\|X\hat{\theta}-X\hat{\theta}_{0}\|^{2}/\sigma^{2}$は自由度$4$の非心カイ二乗分布に従う。非心カイ二乗分布は正規分布に従う確率変数の二乗和が従う確率分布であるという非心カイ二乗分布の定義より，式($\ref{ノルム_1}$)の各項の平方根が従っている正規分布の平均を求めればカイ二乗分布の非心度が求められることが分かる。各項の平方根について，最小二乗推定量は不偏性をもつため，
\begin{align}
E[\hat{\theta}_{i}-\bar{\theta}] &= E[\hat{\theta}_{i}]-\bar{\theta} = \theta_{i}-\bar{\theta}
\end{align}
E[\hat{\theta}_{i}-\bar{\theta}] &= E[\hat{\theta}_{i}]-\bar{\theta} = \theta_{i}-\bar{\theta}
\end{align}
となることに注意すれば，
\begin{align}
\frac{\hat{\theta}_{i}-\bar{\theta}}{\sigma} &\sim \N(\theta_{i}-\bar{\theta}, 1)
\end{align}
\frac{\hat{\theta}_{i}-\bar{\theta}}{\sigma} &\sim \N(\theta_{i}-\bar{\theta}, 1)
\end{align}
となる。したがって，問題文で与えられた非心度の定義より，$F$統計量の分子が従う非心度$\lambda_{(1)}$は
\begin{align}
\lambda_{(1)}
&= (\theta_{1}-\bar{\theta})^{2}+(\theta_{1}-\bar{\theta})^{2}+\cdots+(\theta_{5}-\bar{\theta})^{2}+(\theta_{5}-\bar{\theta})^{2}
= 2\sum_{i=1}^{5}(\theta_{i}-\bar{\theta})^{2}
\end{align}
\lambda_{(1)}
&= (\theta_{1}-\bar{\theta})^{2}+(\theta_{1}-\bar{\theta})^{2}+\cdots+(\theta_{5}-\bar{\theta})^{2}+(\theta_{5}-\bar{\theta})^{2}
= 2\sum_{i=1}^{5}(\theta_{i}-\bar{\theta})^{2}
\end{align}
となる。同様に，測定法(2)について，小問(1)で求めた$X_{(2)}$を用いれば，
\begin{align}
X_{(2)}\hat{\theta}-X_{(2)}\hat{\theta}_{0} &=
\begin{pmatrix}
(\hat{\theta}_{1}+\hat{\theta}_{2})-(\bar{\theta}+\bar{\theta})\\
(\hat{\theta}_{1}+\hat{\theta}_{3})-(\bar{\theta}+\bar{\theta})\\
(\hat{\theta}_{1}+\hat{\theta}_{4})-(\bar{\theta}+\bar{\theta})\\
(\hat{\theta}_{1}+\hat{\theta}_{5})-(\bar{\theta}+\bar{\theta})\\
(\hat{\theta}_{2}+\hat{\theta}_{2})-(\bar{\theta}+\bar{\theta})\\
(\hat{\theta}_{2}+\hat{\theta}_{4})-(\bar{\theta}+\bar{\theta})\\
(\hat{\theta}_{2}+\hat{\theta}_{5})-(\bar{\theta}+\bar{\theta})\\
(\hat{\theta}_{3}+\hat{\theta}_{4})-(\bar{\theta}+\bar{\theta})\\
(\hat{\theta}_{3}+\hat{\theta}_{5})-(\bar{\theta}+\bar{\theta})\\
(\hat{\theta}_{4}+\hat{\theta}_{5})-(\bar{\theta}+\bar{\theta})
\end{pmatrix}
=
\begin{pmatrix}
(\hat{\theta}_{1}-\bar{\theta})+(\hat{\theta}_{2}-\bar{\theta})\\
(\hat{\theta}_{1}-\bar{\theta})+(\hat{\theta}_{3}-\bar{\theta})\\
(\hat{\theta}_{1}-\bar{\theta})+(\hat{\theta}_{4}-\bar{\theta})\\
(\hat{\theta}_{1}-\bar{\theta})+(\hat{\theta}_{5}-\bar{\theta})\\
(\hat{\theta}_{2}-\bar{\theta})+(\hat{\theta}_{3}-\bar{\theta})\\
(\hat{\theta}_{2}-\bar{\theta})+(\hat{\theta}_{4}-\bar{\theta})\\
(\hat{\theta}_{2}-\bar{\theta})+(\hat{\theta}_{5}-\bar{\theta})\\
(\hat{\theta}_{3}-\bar{\theta})+(\hat{\theta}_{4}-\bar{\theta})\\
(\hat{\theta}_{3}-\bar{\theta})+(\hat{\theta}_{5}-\bar{\theta})\\
(\hat{\theta}_{4}-\bar{\theta})+(\hat{\theta}_{5}-\bar{\theta})
\end{pmatrix}
\end{align}
X_{(2)}\hat{\theta}-X_{(2)}\hat{\theta}_{0} &=
\begin{pmatrix}
(\hat{\theta}_{1}+\hat{\theta}_{2})-(\bar{\theta}+\bar{\theta})\\
(\hat{\theta}_{1}+\hat{\theta}_{3})-(\bar{\theta}+\bar{\theta})\\
(\hat{\theta}_{1}+\hat{\theta}_{4})-(\bar{\theta}+\bar{\theta})\\
(\hat{\theta}_{1}+\hat{\theta}_{5})-(\bar{\theta}+\bar{\theta})\\
(\hat{\theta}_{2}+\hat{\theta}_{2})-(\bar{\theta}+\bar{\theta})\\
(\hat{\theta}_{2}+\hat{\theta}_{4})-(\bar{\theta}+\bar{\theta})\\
(\hat{\theta}_{2}+\hat{\theta}_{5})-(\bar{\theta}+\bar{\theta})\\
(\hat{\theta}_{3}+\hat{\theta}_{4})-(\bar{\theta}+\bar{\theta})\\
(\hat{\theta}_{3}+\hat{\theta}_{5})-(\bar{\theta}+\bar{\theta})\\
(\hat{\theta}_{4}+\hat{\theta}_{5})-(\bar{\theta}+\bar{\theta})
\end{pmatrix}
=
\begin{pmatrix}
(\hat{\theta}_{1}-\bar{\theta})+(\hat{\theta}_{2}-\bar{\theta})\\
(\hat{\theta}_{1}-\bar{\theta})+(\hat{\theta}_{3}-\bar{\theta})\\
(\hat{\theta}_{1}-\bar{\theta})+(\hat{\theta}_{4}-\bar{\theta})\\
(\hat{\theta}_{1}-\bar{\theta})+(\hat{\theta}_{5}-\bar{\theta})\\
(\hat{\theta}_{2}-\bar{\theta})+(\hat{\theta}_{3}-\bar{\theta})\\
(\hat{\theta}_{2}-\bar{\theta})+(\hat{\theta}_{4}-\bar{\theta})\\
(\hat{\theta}_{2}-\bar{\theta})+(\hat{\theta}_{5}-\bar{\theta})\\
(\hat{\theta}_{3}-\bar{\theta})+(\hat{\theta}_{4}-\bar{\theta})\\
(\hat{\theta}_{3}-\bar{\theta})+(\hat{\theta}_{5}-\bar{\theta})\\
(\hat{\theta}_{4}-\bar{\theta})+(\hat{\theta}_{5}-\bar{\theta})
\end{pmatrix}
\end{align}
となるため，ノルムは
\begin{align}
\|X_{(2)}\hat{\theta}-X_{(2)}\hat{\theta}_{0}\|^{2}
&= 4\sum_{i=1}^{5}(\hat{\theta}_{i}-\bar{\theta})^{2}
+2\sum_{i=1}^{4}\sum_{j=i+1}^{5}(\hat{\theta}_{i}-\bar{\theta})(\hat{\theta}_{j}-\bar{\theta})\\[0.7em]
&= 4\sum_{i=1}^{5}(\hat{\theta}_{i}-\bar{\theta})^{2}
+\left\{\sum_{i=1}^{5}(\hat{\theta}_{i}-\bar{\theta})\sum_{j=1}^{5}(\hat{\theta}_{i}-\bar{\theta})
-\sum_{i=1}^{5}(\hat{\theta}_{i}-\bar{\theta})^{2}\right\}\\[0.7em]
&= 4\sum_{i=1}^{5}(\hat{\theta}_{i}-\bar{\theta})^{2}
+\left\{\sum_{i=1}^{5}(\hat{\theta}_{i}-\bar{\theta})\left(\sum_{j=1}^{5}\hat{\theta}_{i}-5\bar{\theta}\right)
-\sum_{i=1}^{5}(\hat{\theta}_{i}-\bar{\theta})^{2}\right\}\\[0.7em]
&= 4\sum_{i=1}^{5}(\hat{\theta}_{i}-\bar{\theta})^{2}
+\left\{\sum_{i=1}^{5}(\hat{\theta}_{i}-\bar{\theta})\left(5\bar{\theta}-5\bar{\theta}\right)
-\sum_{i=1}^{5}(\hat{\theta}_{i}-\bar{\theta})^{2}\right\}\\[0.7em]
&= 4\sum_{i=1}^{5}(\hat{\theta}_{i}-\bar{\theta})^{2} - \sum_{i=1}^{5}(\hat{\theta}_{i}-\bar{\theta})^{2}
= 3\sum_{i=1}^{5}(\hat{\theta}_{i}-\bar{\theta})^{2}
\end{align}
\|X_{(2)}\hat{\theta}-X_{(2)}\hat{\theta}_{0}\|^{2}
&= 4\sum_{i=1}^{5}(\hat{\theta}_{i}-\bar{\theta})^{2}
+2\sum_{i=1}^{4}\sum_{j=i+1}^{5}(\hat{\theta}_{i}-\bar{\theta})(\hat{\theta}_{j}-\bar{\theta})\\[0.7em]
&= 4\sum_{i=1}^{5}(\hat{\theta}_{i}-\bar{\theta})^{2}
+\left\{\sum_{i=1}^{5}(\hat{\theta}_{i}-\bar{\theta})\sum_{j=1}^{5}(\hat{\theta}_{i}-\bar{\theta})
-\sum_{i=1}^{5}(\hat{\theta}_{i}-\bar{\theta})^{2}\right\}\\[0.7em]
&= 4\sum_{i=1}^{5}(\hat{\theta}_{i}-\bar{\theta})^{2}
+\left\{\sum_{i=1}^{5}(\hat{\theta}_{i}-\bar{\theta})\left(\sum_{j=1}^{5}\hat{\theta}_{i}-5\bar{\theta}\right)
-\sum_{i=1}^{5}(\hat{\theta}_{i}-\bar{\theta})^{2}\right\}\\[0.7em]
&= 4\sum_{i=1}^{5}(\hat{\theta}_{i}-\bar{\theta})^{2}
+\left\{\sum_{i=1}^{5}(\hat{\theta}_{i}-\bar{\theta})\left(5\bar{\theta}-5\bar{\theta}\right)
-\sum_{i=1}^{5}(\hat{\theta}_{i}-\bar{\theta})^{2}\right\}\\[0.7em]
&= 4\sum_{i=1}^{5}(\hat{\theta}_{i}-\bar{\theta})^{2} - \sum_{i=1}^{5}(\hat{\theta}_{i}-\bar{\theta})^{2}
= 3\sum_{i=1}^{5}(\hat{\theta}_{i}-\bar{\theta})^{2}
\end{align}
となる。測定法(1)との対称性に注意すれば，$F$統計量の分子が従う非心度$\lambda_{(2)}$は
\begin{align}
\lambda_{(2)} &= 3\sum_{i=1}^{5}(\theta_{i}-\bar{\theta})^{2}
\end{align}
\lambda_{(2)} &= 3\sum_{i=1}^{5}(\theta_{i}-\bar{\theta})^{2}
\end{align}
となる。
非心分布の解説も併せてご参照ください。
(5)
小問(2)の結果より，測定法(2)の方が測定法(1)と比べて推定量の分散が小さいため有効であり，小問(4)の結果より，測定法(2)の方が測定法(1)と比べて$F$統計量の非心度が大きいため効率的である。なお，$F$統計量の非心度が大きいと，対立仮説が真の場合に$F$統計量の分子が大きくなるため検出力が大きくなることを利用した。
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