이원배치법 (모수모형) (반복없음)

[인자]&nbsp&nbsp $$A$$ 는 [모수인자]

[인자]&nbsp&nbsp $$B$$ 는 [모수인자]

$$ y_{ij} = \mu + a_{i} + b_{j} + e_{ij} $$

 $$y_{ij}$$ &nbsp&nbsp : &nbsp&nbsp $$A_{i}$$ 와&nbsp&nbsp $$B_{j}$$ 에서 얻은 [측정값]

 $$\mu$$ &nbsp&nbsp : 실험전체의 [모평균]

 $$a_{i}$$ &nbsp&nbsp : &nbsp&nbsp $$A_{i}$$ 가 주는 효과

 $$b_{j}$$ &nbsp&nbsp : &nbsp&nbsp $$B_{j}$$ 가 주는 효과

 $$e_{ij}$$ &nbsp&nbsp : &nbsp&nbsp $$A_{i}$$ 와&nbsp&nbsp $$B_{j}$$ 에서 얻은 [측정값]의 [오차]  ( $$e_{ij} \sim N(0, \sigma_{E}^{ \ 2})$$ 이고 서로 [독립])

  $$i$$ &nbsp&nbsp : 인자&nbsp&nbsp $$A$$ 의 [수준] 수&nbsp&nbsp $$( i = 1,2, \cdots ,l )$$

  $$j$$ &nbsp&nbsp : 인자&nbsp&nbsp $$B$$ 의 [수준] 수&nbsp&nbsp $$( j = 1,2, \cdots ,m )$$

||<|2> [인자] $$B$$ |||||||| [인자] $$A$$ ||<|2> 합계 ||<|2> [평균] || || $$A_{1}$$ || $$A_{2}$$ || $$\cdots$$ || $$A_{l}$$ || |||||||||||||| || || $$B_{1}$$ || $$y_{11}$$ || $$y_{21}$$ || $$\cdots$$ || $$y_{l1}$$ || $$T_{.1}$$ || $$\overline{y}_{.1}$$ || || $$B_{2}$$ || $$y_{12}$$ || $$y_{22}$$ || $$\cdots$$ || $$y_{l2}$$ || $$T_{.2}$$ || $$\overline{y}_{.2}$$ || || $$\vdots$$ || $$\vdots$$ || $$\vdots$$ || || $$\vdots$$ || $$\vdots$$ || $$\vdots$$ || || $$B_{m}$$ || $$y_{1m}$$ || $$y_{2m}$$ || $$\cdots$$ || $$y_{lm}$$ || $$T_{.m}$$ || $$\overline{y}_{.m}$$ || |||||||||||||| || || 합계 || $$T_{1.}$$ || $$T_{2.}$$ || $$\cdots$$ || $$T_{l.}$$ || $$T$$ || || || [평균] || $$\overline{y}_{1.}$$ || $$\overline{y}_{2.}$$ || $$\cdots$$ || $$\overline{y}_{l.}$$ || || $$\overline{\overline{y}}$$ ||

|| $$T_{i.} = \sum_{j=1}^{m} y_{ij}$$ || $$\overline{y}_{i.} = \frac{T_{i.}}{m}$$ ||
|| $$T_{.j} = \sum_{i=1}^{l} y_{ij}$$ || $$\overline{y}_{.j} = \frac{T_{.j}}{l}$$ ||
|| $$T = \sum_{i=1}^{l} \sum_{j=1}^{m} y_{ij}$$ || $$\overline{\overline{y}} = \frac{T}{lm} = \frac{T}{N}$$ ||
|| $$N = lm$$ || $$CT = \frac{T^{2}}{lm} = \frac{T^{2}}{N}$$ ||

개개의 데이터&nbsp&nbsp $$y_{ij}$$ 와 총 [평균]&nbsp&nbsp $$\overline{\overline{y}}$$ 의 차이는 다음과 같이 세 부분으로 나뉘어진다.

$$(y_{ij} - \overline{\overline{y}}) = (\overline{y}_{i.} - \overline{\overline{y}}) + (\overline{y}_{.j} - \overline{\overline{y}}) + (y_{ij} - \overline{y}_{i.} - \overline{y}_{.j} +  \overline{\overline{y}})$$

양변을 제곱한 후에 모든&nbsp&nbsp $$i$$ 와&nbsp&nbsp $$j$$ 에 대하여 합하면 아래의 등식을 얻을 수 있다.

$$\sum_{i=1}^{l}\sum_{j=1}^{m}(y_{ij} - \overline{\overline{y}})^{2} = \sum_{i=1}^{l}\sum_{j=1}^{m}(\overline{y}_{i.} - \overline{\overline{y}})^{2} + \sum_{i=1}^{l}\sum_{j=1}^{m}(\overline{y}_{.j} - \overline{\overline{y}})^{2} + \sum_{i=1}^{l}\sum_{j=1}^{m}(y_{ij} - \overline{y}_{i.} - \overline{y}_{.j} +  \overline{\overline{y}})^{2}$$

위 식에서 왼쪽 항은 총변동 $$S_{T}$$ 이고, 오른쪽 항은 차례대로&nbsp&nbsp $$A$$ 의 [변동],&nbsp&nbsp $$B$$ 의 [변동], [오차변동]인&nbsp&nbsp $$S_{A}$$ , $$S_{B}$$ , $$S_{E}$$ 가 된다.

$$\begin{displaymath}\begin{split} S_{T} &= \sum_{i=1}^{l} \sum_{j=1}^{m} (y_{ij}- \overline{\overline{y}})^{2} \\ &= \sum_{i=1}^{l} \sum_{j=1}^{m} y_{ij}^{ \ 2} - CT \end{split}\end{displaymath}$$

$$\begin{displaymath}\begin{split} S_{A} &= \sum_{i=1}^{l} \sum_{j=1}^{m} (\overline{y}_{i.}- \overline{\overline{y}})^{2} \\ &= \sum_{i=1}^{l} \frac{T_{i.}^{ \ 2}}{m} - CT \end{split}\end{displaymath}$$

$$\begin{displaymath}\begin{split} S_{B} &= \sum_{i=1}^{l} \sum_{j=1}^{m} (\overline{y}_{.j}- \overline{\overline{y}})^{2} \\ &= \sum_{j=1}^{m} \frac{T_{.j}^{ \ 2}}{l} - CT \end{split}\end{displaymath}$$

$$\begin{displaymath}\begin{split} S_{E} &= \sum_{i=1}^{l} \sum_{j=1}^{m} (y_{ij} - \overline{y}_{i.} - \overline{y}_{.j} + \overline{\overline{y}})^{2} \\ &= S_{T} - S_{A} - S_{B} \end{split}\end{displaymath}$$

$$\nu_{_{A}} = l-1$$

$$\nu_{_{B}} = m-1$$

$$\nu_{_{E}} = (l-1)(m-1)$$

$$\nu_{_{T}} = lm-1=N-1$$

$$V_{A} = \frac{S_{A}}{\nu_{_{A}}}$$

$$V_{B} = \frac{S_{B}}{\nu_{_{B}}}$$

$$V_{E} = \frac{S_{E}}{\nu_{_{E}}}$$

$$E(V_{A}) = \sigma_{_{E}}^{ \ 2} + m \ \sigma_{_{A}}^{ \ 2}$$

$$E(V_{B}) = \sigma_{_{E}}^{ \ 2} + l \ \sigma_{_{B}}^{ \ 2}$$

$$E(V_{E}) = \sigma_{_{E}}^{ \ 2}$$

|| '[요인]' || '[제곱합]' $$SS$$ || '[자유도]' $$DF$$ || '[평균제곱]' $$MS$$ || $$E(MS)$$ || $$F_{0}$$ || '기각치' || '[순변동]' $$ S\acute{} $$ || '[기여율]' $$\rho$$ || |||||||||||||||||| || || $$A$$ || $$S_{_{A}}$$ || $$\nu_{_{A}} = l - 1$$ || $$V_{_{A}} = S_{_{A}} / \nu_{_{A}}$$ || $$\sigma_{_{E}}^{ \ 2} + m \ \sigma_{_{A}}^{ \ 2}$$ || $$V_{_{A}}/V_{_{E}}$$ || $$F_{1-\alpha}(\nu_{_{A}} \ , \ \nu_{_{E}})$$ || $$S_{_{A}}\acute{} = S_{_{A}} - \nu_{_{A}} \ V_{_{E}}$$ || $$S_{_{A}}\acute{} / S_{_{T}} $$ || || $$B$$ || $$S_{_{B}}$$ || $$\nu_{_{B}} = m - 1$$ || $$V_{_{B}} = S_{_{B}} / \nu_{_{B}}$$ || $$\sigma_{_{E}}^{ \ 2} + l \ \sigma_{_{B}}^{ \ 2}$$ || $$V_{_{B}}/V_{_{E}}$$ || $$F_{1-\alpha}(\nu_{_{B}} \ , \ \nu_{_{E}})$$ || $$S_{_{B}}\acute{} = S_{_{B}} - \nu_{_{B}} \ V_{_{E}}$$ || $$S_{_{B}}\acute{} / S_{_{T}} $$ || || $$E$$ || $$S_{_{E}}$$ || $$\nu_{_{E}} = (l - 1)(m - 1)$$ || $$V_{_{E}} = S_{_{E}} / \nu_{_{E}}$$ || $$\sigma_{_{E}}^{ \ 2}$$ || || || $$S_{_{E}}\acute{} = S_{_{T}} - S_{_{A}}\acute{} - S_{_{B}}\acute{}$$ || $$S_{_{E}}\acute{} / S_{_{T}} $$ || |||||||||||||||||| || || $$T$$ || $$S_{_{T}}$$ || $$\nu_{_{T}} = lm - 1$$ || || || || || $$S_{_{T}}$$ || $$1$$ ||

인자&nbsp&nbsp $$A$$ 에 대한 [분산분석]

$$F_{0}=\frac{V_{_{A}}}{V_{_{E}}}$$

[기각역] :&nbsp&nbsp $$F_{0} > F_{a-\alpha}(\nu_{_{A}},\nu_{_{E}})$$

인자&nbsp&nbsp $$B$$ 에 대한 [분산분석]

$$F_{0}=\frac{V_{_{B}}}{V_{_{E}}}$$

[기각역] :&nbsp&nbsp $$F_{0} > F_{a-\alpha}(\nu_{_{B}},\nu_{_{E}})$$

* '[인자]&nbsp&nbsp $$A$$ 의 [모평균]에 관한 [추정]'

$$i$$ [수준]에서의 [모평균]&nbsp&nbsp $$\mu(A_{i})$$ 의 [점추정]값

 $$\hat{\mu}(A_{i})=\widehat{\mu + a_{i}} = \overline{y}_{i.}$$

$$i$$ [수준]에서의 [모평균]&nbsp&nbsp $$\mu(A_{i})$$ 의&nbsp&nbsp $$100(1-\alpha) \% $$ [신뢰구간]은 아래와 같다.

 $$\hat{\mu}(A_{i})= \left( \overline{y}_{i.} - t_{\alpha/2}(\nu_{_{E}}) \sqrt{\frac{V_{E}}{m}} \ , \ \overline{y}_{i.} + t_{\alpha/2}(\nu_{_{E}}) \sqrt{\frac{V_{E}}{m}} \right)$$

—- * '[인자]&nbsp&nbsp $$B$$ 의 [모평균]에 관한 [추정]'

$$j$$ [수준]에서의 [모평균]&nbsp&nbsp $$\mu(B_{j})$$ 의 [점추정]값

 $$\hat{\mu}(B_{j})=\widehat{\mu + b_{j}} = \overline{y}_{.j}$$

$$j$$ [수준]에서의 [모평균]&nbsp&nbsp $$\mu(B_{j})$$ 의&nbsp&nbsp $$100(1-\alpha) \% $$ [신뢰구간]은 아래와 같다.

 $$\hat{\mu}(B_{j})= \left( \overline{y}_{.j} - t_{\alpha/2}(\nu_{_{E}}) \sqrt{\frac{V_{E}}{l}} \ , \ \overline{y}_{.j} + t_{\alpha/2}(\nu_{_{E}}) \sqrt{\frac{V_{E}}{l}} \right)$$

—- * '[인자]&nbsp&nbsp $$A$$ 와&nbsp&nbsp $$B$$ 의 [모평균]에 관한 [추정]'

$$A$$ [인자]의&nbsp&nbsp $$i$$ [수준]과&nbsp&nbsp $$B$$ [인자]의&nbsp&nbsp $$j$$ [수준]에서의 [모평균]&nbsp&nbsp $$\mu(A_{i}B_{j})$$ 의 [점추정]값

 $$\hat{\mu}(A_{i}B_{j})=\widehat{\mu+a_{i}+b_{j}}=\overline{y}_{i.} + \overline{y}_{.j} - \overline{\overline{y}}$$

$$A$$ [인자]의&nbsp&nbsp $$i$$ [수준]과&nbsp&nbsp $$B$$ [인자]의&nbsp&nbsp $$j$$ [수준]에서의 [모평균]&nbsp&nbsp $$\mu(A_{i}B_{j})$$ 의&nbsp&nbsp $$100(1-\alpha) \% $$ [신뢰구간]은 아래와 같다.

 $$\hat{\mu}(A_{i}B_{j})= \left( (\overline{y}_{i.} + \overline{y}_{.j} - \overline{\overline{y}}) - t_{\alpha/2}(\nu_{E})\sqrt{\frac{V_{E}}{n_{e}}} \ , \ (\overline{y}_{i.} + \overline{y}_{.j} - \overline{\overline{y}}) + t_{\alpha/2}(\nu_{E})\sqrt{\frac{V_{E}}{n_{e}}} \right)$$

 단,&nbsp&nbsp $$n_{e}$$ 는 [유효반복수]이고&nbsp&nbsp $$n_{e} = \frac{lm}{l+m-1}$$ 이다.

—-

* '[인자]&nbsp&nbsp $$A$$ 의 [모평균]차에 관한 [추정]'

$$i$$ [수준]과&nbsp&nbsp $$j$$ [수준]의 [모평균]차&nbsp&nbsp $$\mu(A_{i})-\mu(A_{j})$$ 의 [점추정]값

 $$\widehat{\mu(A_{i})-\mu(A_{j})} = \overline{y}_{i.} - \overline{y}_{j.}$$

$$i$$ [수준]과&nbsp&nbsp $$j$$ [수준]의 [모평균]차&nbsp&nbsp $$\mu(A_{i})-\mu(A_{j})$$ 의&nbsp&nbsp $$100(1-\alpha) \% $$ [신뢰구간]은 아래와 같다.

 $$\widehat{\mu(A_{i})-\mu(A_{j})}= \left( (\overline{y}_{i.} - \overline{y}_{j.}) - t_{\alpha/2}(\nu_{_{E}}) \sqrt{\frac{2V_{E}}{m}} \ , \ (\overline{y}_{i.} - \overline{y}_{j.}) + t_{\alpha/2}(\nu_{_{E}}) \sqrt{\frac{2V_{E}}{m}} \right)$$

—- * '[인자]&nbsp&nbsp $$B$$ 의 [모평균]차에 관한 [추정]'

$$i$$ [수준]과&nbsp&nbsp $$j$$ [수준]의 [모평균]차&nbsp&nbsp $$\mu(B_{i})-\mu(B_{j})$$ 의 [점추정]값

 $$\widehat{\mu(B_{i})-\mu(B_{j})} = \overline{y}_{.i} - \overline{y}_{.j}$$

$$i$$ [수준]과&nbsp&nbsp $$j$$ [수준]의 [모평균]차&nbsp&nbsp $$\mu(B_{i})-\mu(B_{j})$$ 의&nbsp&nbsp $$100(1-\alpha) \% $$ [신뢰구간]은 아래와 같다.

 $$\widehat{\mu(B_{i})-\mu(B_{j})}= \left( (\overline{y}_{.i} - \overline{y}_{.j}) - t_{\alpha/2}(\nu_{_{E}}) \sqrt{\frac{2V_{E}}{l}} \ , \ (\overline{y}_{.i} - \overline{y}_{.j}) + t_{\alpha/2}(\nu_{_{E}}) \sqrt{\frac{2V_{E}}{l}} \right)$$

• 결측치 추정 (Yates방법)