統計学実践ワークブックの行間埋め　第20章

帰無仮説を考えているため、\(y_{i,j}\sim N(\mu,\sigma^2)\)を仮定する。 \(S_T\)の期待値を求めると、 \begin{eqnarray} E[S_T] &=& E\left[\displaystyle \sum_{i=1}^a \displaystyle \sum_{j=1}^n (y_{i,j}-\overline{y})^2 \right] \\ \\ &=& E\left[\displaystyle \sum_{i=1}^a \displaystyle \sum_{j=1}^n ((y_{i,j}-\mu)-(\overline{y}-\mu))^2 \right] \\ \\ &=& E\left[\displaystyle \sum_{i=1}^a \displaystyle \sum_{j=1}^n \left((y_{i,j}-\mu)^2-2(y_{i,j}-\mu)(\overline{y}-\mu)+(\overline{y}-\mu)^2\right) \right] \\ \\ &=& E\left[\displaystyle \sum_{i=1}^a \displaystyle \sum_{j=1}^n (y_{i,j}-\mu)^2\right]-2E \left[(\overline{y}-\mu)\displaystyle \sum_{i=1}^a \displaystyle \sum_{j=1}^n (y_{i,j}-\mu)\right]+E \left[\displaystyle \sum_{i=1}^a \displaystyle \sum_{j=1}^n (\overline{y}-\mu)^2\ \right] \\ \\ &=& E\left[\displaystyle \sum_{i=1}^a \displaystyle \sum_{j=1}^n (y_{i,j}-\mu)^2\right]-2E \left[(\overline{y}-\mu)\displaystyle \sum_{i=1}^a \displaystyle \sum_{j=1}^n (\overline{y}-\mu)\right]+E \left[\displaystyle \sum_{i=1}^a \displaystyle \sum_{j=1}^n (\overline{y}-\mu)^2\ \right] \\ \\ &=& V\left[\displaystyle \sum_{i=1}^a \displaystyle \sum_{j=1}^ny_{i,j}\right]-E \left[\displaystyle \sum_{i=1}^a \displaystyle \sum_{j=1}^n (\overline{y}-\mu)^2\ \right] \\ \\ &=& an\sigma^2-V\left[\displaystyle \sum_{i=1}^a \displaystyle \sum_{j=1}^n\overline{y}\right] \\ \\ &=& an\sigma^2-\displaystyle \sum_{i=1}^a \displaystyle \sum_{j=1}^nV\left[\overline{y}\right] \\ \\ &=& an\sigma^2-\displaystyle \sum_{i=1}^a \displaystyle \sum_{j=1}^nV\left[\frac{1}{an}\displaystyle \sum_{i=1}^a \displaystyle \sum_{j=1}^ny_{i,j}\right] \\ \\ &=& an\sigma^2-\frac{1}{(an)^2}\displaystyle \sum_{i=1}^a \displaystyle \sum_{j=1}^nV\left[\displaystyle \sum_{i=1}^a \displaystyle \sum_{j=1}^ny_{i,j}\right] \\ \\ &=& an\sigma^2-\frac{1}{(an)^2}\displaystyle \sum_{i=1}^a \displaystyle \sum_{j=1}^nan\sigma^2 \\ \\ &=& an\sigma^2-\frac{1}{(an)^2}(an)^2\sigma^2 \\ \\ &=& \sigma^2(an-1) \end{eqnarray} よって、\(S_T \) の自由度は \( \phi_T=an-1 \) 。

帰無仮説を考えているため、\(y_{i,j}\sim N(\mu,\sigma^2)\)を仮定する。 \(S_A\)の期待値を求めると、 \begin{eqnarray} E[S_A] &=& E\left[n\displaystyle \sum_{i=1}^a (\overline{y}_{A_i}-\overline{y})^2 \right] \\ \\ &=& E\left[n\displaystyle \sum_{i=1}^a ((\overline{y}_{A_i}-\mu)-(\overline{y}-\mu))^2 \right] \\ \\ &=& E\left[n\displaystyle \sum_{i=1}^a \left((\overline{y}_{A_i}-\mu)^2-2(\overline{y}_{A_i}-\mu)(\overline{y}-\mu)+(\overline{y}-\mu)^2\right) \right] \\ \\ &=& E\left[n\displaystyle \sum_{i=1}^a (\overline{y}_{A_i}-\mu)^2\right]-2E \left[n(\overline{y}-\mu)\displaystyle \sum_{i=1}^a (\overline{y}_{A_i}-\mu)\right]+E \left[n\displaystyle \sum_{i=1}^a (\overline{y}-\mu)^2\ \right] \\ \\ &=& E\left[n\displaystyle \sum_{i=1}^a (\overline{y}_{A_i}-\mu)^2\right]-2E \left[n(\overline{y}-\mu)\displaystyle \sum_{i=1}^a (\overline{y}-\mu)\right]+E \left[n\displaystyle \sum_{i=1}^a (\overline{y}-\mu)^2\ \right] \\ \\ &=& nV\left[\displaystyle \sum_{i=1}^a\overline{y}_{A_i}\right]-nE \left[\displaystyle \sum_{i=1}^a (\overline{y}-\mu)^2\ \right] \\ \\ &=& n\displaystyle \sum_{i=1}^aV\left[\overline{y}_{A_i}\right]-n\displaystyle \sum_{i=1}^aV \left[\overline{y} \right] \\ \\ &=& n\displaystyle \sum_{i=1}^aV\left[\frac 1n\sum_{j=1}^ny_{i,j}\right]-n\displaystyle \sum_{i=1}^aV \left[\frac{1}{an}\displaystyle \sum_{i=1}^a \displaystyle \sum_{j=1}^n y_{i,j} \right] \\ \\ &=& n\displaystyle \sum_{i=1}^a\frac 1{n^2}V\left[\sum_{j=1}^ny_{i,j}\right]-n\displaystyle \sum_{i=1}^a\frac{1}{(an)^2}V \left[\displaystyle \sum_{i=1}^a \displaystyle \sum_{j=1}^n y_{i,j} \right] \\ \\ &=& n\displaystyle \sum_{i=1}^a\frac 1{n^2}n\sigma^2-n\displaystyle \sum_{i=1}^a\frac{1}{(an)^2}an\sigma^2 \\ \\ &=& na\frac 1{n^2}n\sigma^2-na\frac{1}{(an)^2}an\sigma^2 \\ \\ &=& a\sigma^2-\sigma^2 \\ \\ &=& \sigma^2(a-1) \\ \\ \end{eqnarray} よって、\(S_A\)の自由度は\(\phi_A=a-1\)。

帰無仮説を考えているため、\(y_{i,j}\sim N(\mu,\sigma^2)\)を仮定する。 \(S_E\)の期待値を求めると、 \begin{eqnarray} E[S_E] &=& E\left[\displaystyle \sum_{i=1}^a \displaystyle \sum_{j=1}^n(y_{i,j}-\overline{y}_{A_i})^2 \right] \\ \\ &=& E\left[\displaystyle \sum_{i=1}^a \displaystyle \sum_{j=1}^n ((y_{i,j}-\mu)-(\overline{y}_{A_i}-\mu))^2 \right] \\ \\ &=& E\left[\displaystyle \sum_{i=1}^a \displaystyle \sum_{j=1}^n \left((y_{i,j}-\mu)^2-2(y_{i,j}-\mu)(\overline{y}_{A_i}-\mu)+(\overline{y}_{A_i}-\mu)^2\right) \right] \\ \\ &=& E\left[\displaystyle \sum_{i=1}^a \displaystyle \sum_{j=1}^n (y_{i,j}-\mu)^2\right]-2E \left[\displaystyle \sum_{i=1}^a(\overline{y}_{A_i}-\mu)\displaystyle \sum_{j=1}^n (y_{i,j}-\mu)\right]+E \left[\displaystyle \sum_{i=1}^a\displaystyle \sum_{j=1}^n (\overline{y}_{A_i}-\mu)^2\ \right] \\ \\ &=& an\sigma^2-2E \left[\displaystyle \sum_{i=1}^a(\overline{y}_{A_i}-\mu)\displaystyle \sum_{j=1}^n (\overline{y}_{A_i}-\mu)\right]+E \left[\displaystyle \sum_{i=1}^a\displaystyle \sum_{j=1}^n (\overline{y}_{A_i}-\mu)^2\ \right] \\ \\ &=& an\sigma^2-2E \left[\displaystyle \sum_{i=1}^a\displaystyle \sum_{j=1}^n (\overline{y}_{A_i}-\mu)^2\right]+E \left[\displaystyle \sum_{i=1}^a\displaystyle \sum_{j=1}^n (\overline{y}_{A_i}-\mu)^2\ \right] \\ \\ &=& an\sigma^2-E \left[\displaystyle \sum_{i=1}^a\displaystyle \sum_{j=1}^n (\overline{y}_{A_i}-\mu)^2\ \right] \\ \\ &=& an\sigma^2-nE \left[\displaystyle \sum_{i=1}^a (\overline{y}_{A_i}-\mu)^2 \right] \\ \\ &=& an\sigma^2-n\displaystyle \sum_{i=1}^aV \left[ \overline{y}_{A_i} \right] \\ \\ &=& an\sigma^2-a\sigma^2 \\ \\ &=& \sigma^2a(n-1) \\ \\ \end{eqnarray} よって、\(S_E\)の自由度は\(\phi_E=a(n-1)\)。\(\phi_A\)を求めるところと同じ式は省略した。

帰無仮説を考えているため、\(y_{i,j,k}\sim N(\mu,\sigma^2)\)を仮定する。 \(S_T\)の期待値を求めると、 \begin{eqnarray} E[S_T] &=& E\left[\displaystyle \sum_{i=1}^a \sum_{j=1}^b \sum_{k=1}^n (y_{i,j,k}-\overline{y})^2 \right] \\ \\ &=& E\left[\displaystyle \sum_{i=1}^a \sum_{j=1}^b \sum_{k=1}^n ((y_{i,j,k}-\mu)-(\overline{y}-\mu))^2 \right] \\ \\ &=& E\left[\displaystyle \sum_{i=1}^a \sum_{j=1}^b \sum_{k=1}^n \left((y_{i,j,k}-\mu)^2-2(y_{i,j}-\mu)(\overline{y}-\mu)+(\overline{y}-\mu)^2\right) \right] \\ \\ &=& E\left[\displaystyle \sum_{i=1}^a \sum_{j=1}^b \sum_{k=1}^n (y_{i,j,k}-\mu)^2\right]-2E \left[(\overline{y}-\mu)\displaystyle \sum_{i=1}^a \sum_{j=1}^b \sum_{k=1}^n (y_{i,j,k}-\mu)\right]+E \left[\displaystyle \sum_{i=1}^a \sum_{j=1}^b\sum_{k=1}^n (\overline{y}-\mu)^2\ \right] \\ \\ &=& E\left[\displaystyle \sum_{i=1}^a \sum_{j=1}^b \sum_{k=1}^n (y_{i,j}-\mu)^2\right]-2E \left[(\overline{y}-\mu)\displaystyle \sum_{i=1}^a\sum_{j=1}^n (\overline{y}-\mu)\right]+E \left[\displaystyle \sum_{i=1}^a \sum_{j=1}^b\sum_{k=1}^n (\overline{y}-\mu)^2\ \right] \\ \\ &=& V\left[\displaystyle \sum_{i=1}^a \sum_{j=1}^b \sum_{k=1}^ny_{i,j}\right]-E \left[\displaystyle \sum_{i=1}^a \sum_{j=1}^b \sum_{k=1}^n (\overline{y}-\mu)^2\ \right] \\ \\ &=& abn\sigma^2-V\left[\displaystyle \sum_{i=1}^a \sum_{j=1}^b \sum_{k=1}^n\overline{y}\right] \\ \\ &=& abn\sigma^2-\displaystyle \sum_{i=1}^a \sum_{j=1}^b\sum_{k=1}^nV\left[\overline{y}\right] \\ \\ &=& abn\sigma^2-\displaystyle \sum_{i=1}^a \sum_{j=1}^b\sum_{k=1}^nV\left[\frac{1}{an}\displaystyle \sum_{i=1}^a \sum_{j=1}^b\sum_{k=1}^ny_{i,j,k}\right] \\ \\ &=& abn\sigma^2-\frac{1}{(abn)^2}\displaystyle \sum_{i=1}^a \sum_{j=1}^b\sum_{k=1}^nV\left[\displaystyle \sum_{i=1}^a \sum_{j=1}^b\sum_{k=1}^ny_{i,j,k}\right] \\ \\ &=& abn\sigma^2-\frac{1}{(abn)^2}\displaystyle \sum_{i=1}^a \sum_{j=1}^b\sum_{k=1}^nabn\sigma^2 \\ \\ &=& abn\sigma^2-\frac{1}{(abn)^2}(abn)^2\sigma^2 \\ \\ &=& \sigma^2(abn-1) \end{eqnarray} よって、\(S_T \) の自由度は \( \phi_T=abn-1 \) 。

帰無仮説を考えているため、\(y_{i,j}\sim N(\mu,\sigma^2)\)を仮定する。 \(S_A\)の期待値を求めると、 \begin{eqnarray} E[S_A] &=& E\left[bn\displaystyle \sum_{i=1}^a (\overline{y}_{A_i}-\overline{y})^2 \right] \\ \\ &=& E\left[bn\displaystyle \sum_{i=1}^a ((\overline{y}_{A_i}-\mu)-(\overline{y}-\mu))^2 \right] \\ \\ &=& E\left[bn\displaystyle \sum_{i=1}^a \left((\overline{y}_{A_i}-\mu)^2-2(\overline{y}_{A_i}-\mu)(\overline{y}-\mu)+(\overline{y}-\mu)^2\right) \right] \\ \\ &=& E\left[bn\displaystyle \sum_{i=1}^a (\overline{y}_{A_i}-\mu)^2\right]-2E \left[bn(\overline{y}-\mu)\displaystyle \sum_{i=1}^a (\overline{y}_{A_i}-\mu)\right]+E \left[bn\displaystyle \sum_{i=1}^a (\overline{y}-\mu)^2\ \right] \\ \\ &=& E\left[bn\displaystyle \sum_{i=1}^a (\overline{y}_{A_i}-\mu)^2\right]-2E \left[bn(\overline{y}-\mu)\displaystyle \sum_{i=1}^a (\overline{y}-\mu)\right]+E \left[bn\displaystyle \sum_{i=1}^a (\overline{y}-\mu)^2\ \right] \\ \\ &=& bnV\left[\displaystyle \sum_{i=1}^a\overline{y}_{A_i}\right]-bnE \left[\displaystyle \sum_{i=1}^a (\overline{y}-\mu)^2\ \right] \\ \\ &=& bn\displaystyle \sum_{i=1}^aV\left[\overline{y}_{A_i}\right]-bn\displaystyle \sum_{i=1}^aV \left[\overline{y} \right] \tag{*}\\ \\ &=& bn\displaystyle \sum_{i=1}^aV\left[\frac {1}{bn}\sum_{j=1}^b\sum_{k=1}^ny_{i,j,k}\right]-bn\displaystyle \sum_{i=1}^aV \left[\frac{1}{abn}\displaystyle \sum_{i=1}^a \sum_{j=1}^b \sum_{k=1}^n y_{i,j,k} \right] \\ \\ &=& bn\displaystyle \sum_{i=1}^a\frac {1}{(bn)^2}V\left[\sum_{j=1}^b\sum_{k=1}^ny_{i,j,k}\right]-bn\displaystyle \sum_{i=1}^a\frac{1}{(abn)^2}V \left[\displaystyle \sum_{i=1}^a\sum_{j=1}^b \sum_{k=1}^n y_{i,j,k} \right] \\ \\ &=& bn\displaystyle \sum_{i=1}^a\frac {1}{(bn)^2}bn\sigma^2-bn\displaystyle \sum_{i=1}^a\frac{1}{(abn)^2}abn\sigma^2 \\ \\ &=& bna\frac 1{(bn)^2}bn\sigma^2-bna\frac{1}{(abn)^2}abn\sigma^2 \\ \\ &=& a\sigma^2-\sigma^2 \\ \\ &=& \sigma^2(a-1) \\ \\ \end{eqnarray} よって、\(S_A\)の自由度は\(\phi_A=a-1\)。
\(S_B\)の自由度に関しても同様に計算をする。上式中の(*)の\(A,B\)に関する要素を入れ替えて、 \begin{eqnarray} E[S_B] &=& an\displaystyle \sum_{j=1}^bV\left[\overline{y}_{B_j}\right]-an\displaystyle \sum_{j=1}^bV \left[\overline{y} \right]\\ \\ &=& an\displaystyle \sum_{j=1}^bV\left[\frac {1}{an}\sum_{i=1}^a\sum_{k=1}^ny_{i,j,k}\right]-an\displaystyle \sum_{j=1}^bV \left[\frac{1}{abn}\displaystyle \sum_{i=1}^a \sum_{j=1}^b \sum_{k=1}^n y_{i,j,k} \right] \\ \\ &=& an\displaystyle \sum_{j=1}^b\frac {1}{(an)^2}V\left[\sum_{i=1}^a\sum_{k=1}^ny_{i,j,k}\right]-an\displaystyle \sum_{j=1}^b\frac{1}{(abn)^2}V \left[\displaystyle \sum_{i=1}^a\sum_{j=1}^b \sum_{k=1}^n y_{i,j,k} \right] \\ \\ &=& an\displaystyle \sum_{j=1}^b\frac {1}{(an)^2}an\sigma^2-an\displaystyle \sum_{j=1}^b\frac{1}{(abn)^2}abn\sigma^2 \\ \\ &=& anb\frac 1{(an)^2}an\sigma^2-bna\frac{1}{(abn)^2}abn\sigma^2 \\ \\ &=& b\sigma^2-\sigma^2 \\ \\ &=& \sigma^2(b-1) \\ \\ \end{eqnarray} よって、\(S_B\)の自由度は\(\phi_B=b-1\)。

帰無仮説を考えているため、\(y_{i,j}\sim N(\mu,\sigma^2)\)を仮定する。 \(S_E\)の期待値を求めると、 \begin{eqnarray} E[S_E] &=& E\left[\displaystyle \sum_{i=1}^a \sum_{j=1}^b \sum_{k=1}^n(y_{i,j,k}-\overline{y}_{A_iB_j})^2 \right] \\ \\ &=& E\left[\displaystyle \sum_{i=1}^a \sum_{j=1}^b \sum_{k=1}^n ((y_{i,j,k}-\mu)-(\overline{y}_{A_iB_j}-\mu))^2 \right] \\ \\ &=& E\left[\displaystyle \sum_{i=1}^a \sum_{j=1}^b \sum_{k=1}^n \left((y_{i,j,k}-\mu)^2-2(y_{i,j,k}-\mu)(\overline{y}_{A_iB_j}-\mu)+(\overline{y}_{A_iB_j}-\mu)^2\right) \right] \\ \\ &=& E\left[\displaystyle \sum_{i=1}^a \sum_{j=1}^b \sum_{k=1}^n (y_{i,j,k}-\mu)^2\right]-2E \left[\displaystyle \sum_{i=1}^a\sum_{j=1}^b(\overline{y}_{A_iB_j}-\mu)\displaystyle \sum_{k=1}^n (y_{i,j,k}-\mu)\right]+E \left[\displaystyle \sum_{i=1}^a \sum_{j=1}^b\sum_{k=1}^n (\overline{y}_{A_iB_j}-\mu)^2\ \right] \\ \\ &=& abn\sigma^2-2E \left[\displaystyle \sum_{i=1}^a\sum_{j=1}^b (\overline{y}_{A_iB_j}-\mu)\displaystyle \sum_{k=1}^n (\overline{y}_{A_iB_j}-\mu)\right]+E \left[\displaystyle \sum_{i=1}^a \sum_{j=1}^b\sum_{k=1}^n (\overline{y}_{A_iB_j}-\mu)^2\ \right] \\ \\ &=& abn\sigma^2-2E \left[\displaystyle \sum_{i=1}^a \sum_{j=1}^b\sum_{k=1}^n (\overline{y}_{A_iB_j}-\mu)^2\ \right]+E \left[\displaystyle \sum_{i=1}^a \sum_{j=1}^b\sum_{k=1}^n (\overline{y}_{A_iB_j}-\mu)^2\ \right] \\ \\ &=& abn\sigma^2-E \left[\displaystyle \sum_{i=1}^a \sum_{j=1}^b\sum_{k=1}^n (\overline{y}_{A_iB_j}-\mu)^2\ \right] \\ \\ &=& abn\sigma^2-nE \left[\displaystyle \sum_{i=1}^a\sum_{j=1}^b (\overline{y}_{A_iB_j}-\mu)^2 \right] \\ \\ &=& abn\sigma^2-n\displaystyle \sum_{i=1}^a\sum_{j=1}^bV \left[ \overline{y}_{A_iB_j} \right] \\ \\ &=& abn\sigma^2-n\displaystyle \sum_{i=1}^a\sum_{j=1}^bV \left[\frac 1n\sum_{k=1}^n y_{i,j,k} \right] \\ \\ &=& abn\sigma^2-n\displaystyle \sum_{i=1}^a\sum_{j=1}^b\frac{1}{n^2}V \left[\sum_{k=1}^n y_{i,j,k} \right] \\ \\ &=& abn\sigma^2-n\displaystyle \sum_{i=1}^a\sum_{j=1}^b \frac{1}{n^2}n\sigma^2 \\ \\ &=& abn\sigma^2-n\displaystyle ab \frac{1}{n^2}n\sigma^2 \\ \\ &=& abn\sigma^2-ab\sigma^2 \\ \\ &=& \sigma^2ab(n-1) \\ \\ \end{eqnarray} よって、\(S_E\)の自由度は\(\phi_E=ab(n-1)\)。

帰無仮説を考えているため、\(y_{i,j}\sim N(\mu,\sigma^2)\)を仮定する。 \(S_{A\times B}\)の期待値を求めると、 \begin{eqnarray} E[S_{A\times B}] &=& nE\left[\displaystyle \sum_{i=1}^a \sum_{j=1}^b (\overline{y}_{A_iB_j}-\overline{y}_{A_i}-\overline{y}_{B_j}+\overline{y})^2 \right] \\ \\ &=& nE\left[\displaystyle \sum_{i=1}^a \sum_{j=1}^b ((\overline{y}_{A_iB_j}-\overline{y})-(\overline{y}_{A_i}-\overline{y})-(\overline{y}_{B_j}-\overline{y}))^2 \right] \\ \\ &=& nE\left[\displaystyle \sum_{i=1}^a \sum_{j=1}^b ((\overline{y}_{A_iB_j}-\overline{y})^2+(\overline{y}_{A_i}-\overline{y})^2+(\overline{y}_{B_j}-\overline{y})^2-2 \left((\overline{y}_{A_iB_j}-\overline{y})(\overline{y}_{A_i}-\overline{y})+(\overline{y}_{A_i}-\overline{y})(\overline{y}_{B_j}-\overline{y})+(\overline{y}_{B_j}-\overline{y})(\overline{y}_{A_iB_j}-\overline{y}) \right)) \right] \\ \\ &=& nE\left[\displaystyle \sum_{i=1}^a \sum_{j=1}^b (\overline{y}_{A_iB_j}-\overline{y})^2\right]+nE\left[\displaystyle \sum_{i=1}^a \sum_{j=1}^b (\overline{y}_{A_i}-\overline{y})^2\right]+nE\left[\displaystyle \sum_{i=1}^a \sum_{j=1}^b (\overline{y}_{B_j}-\overline{y})^2\right]\\ \\ &&-2 nE\left[\displaystyle \sum_{i=1}^a \sum_{j=1}^b(\overline{y}_{A_iB_j}-\overline{y})(\overline{y}_{A_i}-\overline{y}) \right]-2 nE\left[\displaystyle \sum_{i=1}^a \sum_{j=1}^b(\overline{y}_{A_i}-\overline{y})(\overline{y}_{B_j}-\overline{y})\right]-2 nE\left[\displaystyle \sum_{i=1}^a \sum_{j=1}^b(\overline{y}_{B_j}-\overline{y})(\overline{y}_{A_iB_j}-\overline{y}) \right] \\ \\ \\ &=& nE\left[\displaystyle \sum_{i=1}^a \sum_{j=1}^b ((\overline{y}_{A_iB_j}-\mu)-(\overline{y}-\mu))^2\right]+nbE\left[\displaystyle \sum_{i=1}^a (\overline{y}_{A_i}-\overline{y})^2\right]+naE\left[\displaystyle \sum_{j=1}^b (\overline{y}_{B_j}-\overline{y})^2\right]\\ \\ &&-2 nE\left[\displaystyle \sum_{i=1}^a (\overline{y}_{A_i}-\overline{y})\sum_{j=1}^b(\overline{y}_{A_iB_j}-\overline{y}) \right]-2 nE\left[\displaystyle \sum_{i=1}^a(\overline{y}_{A_i}-\overline{y}) \sum_{j=1}^b(\overline{y}_{B_j}-\overline{y})\right]-2 nE\left[\displaystyle \sum_{j=1}^b(\overline{y}_{B_j}-\overline{y})\sum_{i=1}^a (\overline{y}_{A_iB_j}-\overline{y}) \right] \\ \\ \\ &=& nE\left[\displaystyle \sum_{i=1}^a \sum_{j=1}^b ((\overline{y}_{A_iB_j}-\mu)^2-2(\overline{y}_{A_iB_j}-\mu)(\overline{y}-\mu)+(\overline{y}-\mu)^2)\right]+\sigma^2(a-1)+\sigma^2(b-1)\\ \\ &&-2 nE\left[\displaystyle \sum_{i=1}^a (\overline{y}_{A_i}-\overline{y})\sum_{j=1}^b(\overline{y}_{A_i}-\overline{y}) \right]-2 nE\left[\displaystyle \sum_{i=1}^a(\overline{y}-\overline{y}) \sum_{j=1}^b(\overline{y}-\overline{y})\right]-2 nE\left[\displaystyle \sum_{j=1}^b(\overline{y}_{B_j}-\overline{y})\sum_{i=1}^a (\overline{y}_{B_j}-\overline{y}) \right] \\ \\ \\ &=& nE\left[\displaystyle \sum_{i=1}^a \sum_{j=1}^b ((\overline{y}_{A_iB_j}-\mu)^2-2(\overline{y}-\mu)(\overline{y}-\mu)+(\overline{y}-\mu)^2)\right]+\sigma^2(a-1)+\sigma^2(b-1)\\ \\ &&-2 bnE\left[\displaystyle \sum_{i=1}^a (\overline{y}_{A_i}-\overline{y})^2 \right]-0-2 naE\left[\displaystyle \sum_{j=1}^b(\overline{y}_{B_j}-\overline{y})^2 \right] \\ \\ \\ &=& nE\left[\displaystyle \sum_{i=1}^a \sum_{j=1}^b ((\overline{y}_{A_iB_j}-\mu)^2-(\overline{y}-\mu)^2)\right]+\sigma^2(a-1)+\sigma^2(b-1)\\ \\ &&-2 \sigma(a-1)-2\sigma^2(b-1) \\ \\ \\ &=& nE\left[\displaystyle \sum_{i=1}^a \sum_{j=1}^b (\overline{y}_{A_iB_j}-\mu)^2\right]-nE\left[\displaystyle \sum_{i=1}^a \sum_{j=1}^b (\overline{y}-\mu)^2\right]\\ \\ &&- \sigma(a-1)-\sigma^2(b-1) \\ \\ \\ &=& n\displaystyle \sum_{i=1}^a \sum_{j=1}^b E\left[(\overline{y}_{A_iB_j}-\mu)^2\right]-n\displaystyle \sum_{i=1}^a \sum_{j=1}^b E\left[(\overline{y}-\mu)^2\right]\\ \\ &&- \sigma(a-1)-\sigma^2(b-1) \\ \\ \\ &=& n\displaystyle \sum_{i=1}^a \sum_{j=1}^b V\left[\overline{y}_{A_iB_j}\right]-n\displaystyle \sum_{i=1}^a \sum_{j=1}^b V\left[\overline{y}\right]\\ \\ &&- \sigma(a-1)-\sigma^2(b-1) \\ \\ \\ &=& n\displaystyle \sum_{i=1}^a \sum_{j=1}^b V\left[\frac{1}{n}\sum_{k=1}^ny_{i,j,k}\right]-n\displaystyle \sum_{i=1}^a \sum_{j=1}^b V\left[\frac{1}{abn}\sum_{i=1}^a\sum_{j=1}^b\sum_{k=1}^n y_{i,j,k}\right]\\ \\ &&- \sigma(a-1)-\sigma^2(b-1) \\ \\ \\ &=& n\displaystyle \sum_{i=1}^a \sum_{j=1}^b \frac{1}{n^2}n\sigma^2-n\displaystyle \sum_{i=1}^a \sum_{j=1}^b \frac{1}{(abn)^2}abn\sigma^2\\ \\ &&- \sigma(a-1)-\sigma^2(b-1) \\ \\ \\ &=& nab\frac{1}{n^2}n\sigma^2-nab \frac{1}{(abn)^2}abn\sigma^2\\ \\ &&- \sigma(a-1)-\sigma^2(b-1) \\ \\ \\ &=& ab\sigma^2-\sigma^2- \sigma a+\sigma^2-\sigma^2 b+\sigma^2 \\ \\ \\ &=& ab\sigma^2- \sigma a-\sigma^2 b+\sigma^2 \\ \\ \\ &=& \sigma^2(a-1)(b-1)\\ \\ \\ \end{eqnarray} よって、\(S_{A\times B}\)の自由度は\(\phi_{A\times B}=(a-1)(b-1)\)。

第五版までは元の式にミスがある。 \begin{eqnarray} S_{[k]} &=& \displaystyle \sum_{i=1}^{N/2}\sum_{j=1}^2 (\overline{y}_{[k]j}-\overline{y})^2 \\ \\ &=& \displaystyle \sum_{i=1}^{N/2}\left( (\overline{y}_{[k]1}-\overline{y})^2+(\overline{y}_{[k]2}-\overline{y})^2\right) \\ \\ &=& \displaystyle \sum_{i=1}^{N/2} \left((\overline{y}_{[k]1}-\frac{\overline{y}_{[k]1}+\overline{y}_{[k]2} }{2})^2+(\overline{y}_{[k]2}-\frac{\overline{y}_{[k]1}+\overline{y}_{[k]2} }{2})^2\right) \;\;\;...\overline{y}=\frac{\overline{y}_{[k]1}+\overline{y}_{[k]2} }{2}\\ \\ &=& \displaystyle \sum_{i=1}^{N/2} \left((\frac{\overline{y}_{[k]1}}{2}-\frac{\overline{y}_{[k]2} }{2})^2+(\frac{\overline{y}_{[k]2}}{2}-\frac{\overline{y}_{[k]1} }{2})^2\right) \\ \\ &=& \displaystyle \sum_{i=1}^{N/2}\frac{ (\overline{y}_{[k]1}-\overline{y}_{[k]2} )^2 }{2} \\ \\ &=& \frac{ N }{4}(\overline{y}_{[k]1}-\overline{y}_{[k]2} )^2 \\ \\ \end{eqnarray}