삼원배치법 (모수모형) (반복없음)

요인 $A$는 모수인자

요인 $B$는 모수인자

요인 $C$는 모수인자

$$ y_{ijk} = \mu + a_{i} + b_{j} + c_{k} + (ab)_{ij} + (ac)_{ik} + (bc)_{jk} + e_{ijk} $$

• $y_{ijk}$ : $A_{i}$ 와 $B_{j}$ , 그리고 $C_{k}$ 에서 얻은 측정값
• $\mu$ : 실험전체의 모평균
• $a_{i}$ : $A_{i}$ 가 주는 효과
• $b_{j}$ : $B_{j}$ 가 주는 효과
• $c_{k}$ : $C_{k}$ 가 주는 효과
• $(ab)_{ij}$ : $A_{i}$ 와 $B_{j}$ 의 교호작용 효과
• $(ac)_{ik}$ : $A_{i}$ 와 $C_{k}$ 의 교호작용 효과
• $(bc)_{jk}$ : $B_{j}$ 와 $C_{k}$ 의 교호작용 효과
• $e_{ijk}$ : $A_{i}$ 와 $B_{j}$ , 그리고 $C_{k}$ 에서 얻은 측정값의 오차 ( $e_{ijk} \sim N(0, \sigma_{E}^{ \ 2})$ 이고 서로 독립)

• $i$ : 인자 $A$ 의 수준 수 $( i = 1,2, \cdots ,l )$
• $j$ : 인자 $B$ 의 수준 수 $( j = 1,2, \cdots ,m )$
• $k$ : 인자 $C$ 의 수준 수 $( k = 1,2, \cdots ,n )$

||<|2> [인자] $$B$$ ||<|2> [인자] $$C$$ |||||||| [인자] $$A$$ || || $$A_{1}$$ || $$A_{2}$$ || $$\cdots$$ || $$A_{l}$$ || |||||||||||| || ||<|4> $$B_{1}$$ || $$C_{1}$$ || $$y_{111}$$ || $$y_{211}$$ || $$\cdots$$ || $$y_{l11}$$ || || $$C_{2}$$ || $$y_{112}$$ || $$y_{212}$$ || $$\cdots$$ || $$y_{l12}$$ || || $$\vdots$$ || $$\vdots$$ || $$\vdots$$ || || $$\vdots$$ || || $$C_{n}$$ || $$y_{11n}$$ || $$y_{21n}$$ || $$\cdots$$ || $$y_{l1n}$$ || ||<|4> $$B_{2}$$ || $$C_{1}$$ || $$y_{121}$$ || $$y_{221}$$ || $$\cdots$$ || $$y_{l21}$$ || || $$C_{2}$$ || $$y_{122}$$ || $$y_{222}$$ || $$\cdots$$ || $$y_{l22}$$ || || $$\vdots$$ || $$\vdots$$ || $$\vdots$$ || || $$\vdots$$ || || $$C_{n}$$ || $$y_{12n}$$ || $$y_{22n}$$ || $$\cdots$$ || $$y_{l2n}$$ || |||| $$\vdots$$ |||||||| $$\vdots$$ || ||<|4> $$B_{m}$$ || $$C_{1}$$ || $$y_{1m1}$$ || $$y_{2m1}$$ || $$\cdots$$ || $$y_{lm1}$$ || || $$C_{2}$$ || $$y_{1m2}$$ || $$y_{2m2}$$ || $$\cdots$$ || $$y_{lm2}$$ || || $$\vdots$$ || $$\vdots$$ || $$\vdots$$ || || $$\vdots$$ || || $$C_{n}$$ || $$y_{1mn}$$ || $$y_{2mn}$$ || $$\cdots$$ || $$y_{lmn}$$ ||

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

$$AC$$ 2원표
||<|2> [인자] $$C$$ |||||||| [인자] $$A$$ ||<|2> 합계 ||
|| $$A_{1}$$ || $$A_{2}$$ || $$\cdots$$ || $$A_{l}$$ ||
|||||||||||| ||
|| $$C_{1}$$ || $$T_{1.1}$$ || $$T_{2.1}$$ || $$\cdots$$ || $$T_{l.1}$$ || $$T_{..1}$$ ||
|| $$C_{2}$$ || $$T_{1.2}$$ || $$T_{2.2}$$ || $$\cdots$$ || $$T_{l.2}$$ || $$T_{..2}$$ ||
|| $$\vdots$$ || $$\vdots$$ || $$\vdots$$ || || $$\vdots$$ || $$\vdots$$ ||
|| $$C_{n}$$ || $$T_{1.n}$$ || $$T_{2.n}$$ || $$\cdots$$ || $$T_{l.n}$$ || $$T_{..n}$$ ||
|||||||||||| ||
|| 합계 || $$T_{1..}$$ || $$T_{2..}$$ || $$\cdots$$ || $$T_{l..}$$ || $$T$$ ||

$$BC$$ 2원표
||<|2> [인자] $$C$$ |||||||| [인자] $$B$$ ||<|2> 합계 ||
|| $$B_{1}$$ || $$B_{2}$$ || $$\cdots$$ || $$B_{m}$$ ||
|||||||||||| ||
|| $$C_{1}$$ || $$T_{.11}$$ || $$T_{.21}$$ || $$\cdots$$ || $$T_{.m1}$$ || $$T_{..1}$$ ||
|| $$C_{2}$$ || $$T_{.12}$$ || $$T_{.22}$$ || $$\cdots$$ || $$T_{.m2}$$ || $$T_{..2}$$ ||
|| $$\vdots$$ || $$\vdots$$ || $$\vdots$$ || || $$\vdots$$ || $$\vdots$$ ||
|| $$C_{n}$$ || $$T_{.1n}$$ || $$T_{.2n}$$ || $$\cdots$$ || $$T_{.mn}$$ || $$T_{..n}$$ ||
|||||||||||| ||
|| 합계 || $$T_{.1.}$$ || $$T_{.2.}$$ || $$\cdots$$ || $$T_{.m.}$$ || $$T$$ ||

 || $$T_{i..} = \sum_{j=1}^{m} \sum_{k=1}^{n} y_{ijk}$$ || $$\overline{y}_{i..} = \frac{T_{i..}}{mn}$$ ||
 || $$T_{.j.} = \sum_{i=1}^{l} \sum_{k=1}^{n} y_{ijk}$$ || $$\overline{y}_{.j.} = \frac{T_{.j.}}{ln}$$ ||
 || $$T_{..k} = \sum_{i=1}^{l} \sum_{j=1}^{m} y_{ijk}$$ || $$\overline{y}_{..k} = \frac{T_{..k}}{lm}$$ ||
 || $$T_{ij.} = \sum_{k=1}^{n} y_{ijk}$$ || $$\overline{y}_{ij.} = \frac{T_{ij.}}{n}$$ ||
 || $$T_{i.k} = \sum_{j=1}^{m} y_{ijk}$$ || $$\overline{y}_{i.k} = \frac{T_{i.k}}{m}$$ ||
 || $$T_{.jk} = \sum_{i=1}^{l} y_{ijk}$$ || $$\overline{y}_{.jk} = \frac{T_{.jk}}{l}$$ ||
 || $$T = \sum_{i=1}^{l} \sum_{j=1}^{m} \sum_{k=1}^{n} y_{ijk}$$ || $$\overline{\overline{y}} = \frac{T}{lmn} = \frac{T}{N}$$ ||
 || $$N = lmn$$ || $$CT = \frac{T^{2}}{lmn} = \frac{T^{2}}{N}$$ ||

개개의 데이터&nbsp&nbsp $$y_{ijk}$$ 와 총편균&nbsp&nbsp $$\overline{\overline{y}}$$ 의 차이는 다음과 같이 7부분으로 나뉘어진다.

$$\begin{displaymath}\begin{split} (y_{ijk}-\overline{\overline{y}}) &= (\overline{y}_{i..} - \overline{\overline{y}}) + (\overline{y}_{.j.} - \overline{\overline{y}}) + (\overline{y}_{..k} - \overline{\overline{y}}) \\ &+ (\overline{y}_{ij.} - \overline{y}_{i..} - \overline{y}_{.j.} + \overline{\overline{y}}) + (\overline{y}_{i.k} - \overline{y}_{i..} - \overline{y}_{..k} + \overline{\overline{y}}) + (\overline{y}_{.jk} - \overline{y}_{.j.} - \overline{y}_{..k} + \overline{\overline{y}}) \\ &+ (y_{ijk} - \overline{y}_{ij.} - \overline{y}_{i.k} - \overline{y}_{.jk} + \overline{y}_{i..} + \overline{y}_{.j.} + \overline{y}_{..k} - \overline{\overline{y}}) \end{split}\end{displaymath}$$

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

$$\begin{displaymath}\begin{split} \sum_{i=1}^{l}\sum_{j=1}^{m}\sum_{k=1}^{n}(y_{ijk}-\overline{\overline{y}})^{2} &= \sum_{i=1}^{l}\sum_{j=1}^{m}\sum_{k=1}^{n}(\overline{y}_{i..} - \overline{\overline{y}})^{2} + \sum_{i=1}^{l}\sum_{j=1}^{m}\sum_{k=1}^{n}(\overline{y}_{.j.} - \overline{\overline{y}})^{2} + \sum_{i=1}^{l}\sum_{j=1}^{m}\sum_{k=1}^{n}(\overline{y}_{..k} - \overline{\overline{y}})^{2} \\ &+ \sum_{i=1}^{l}\sum_{j=1}^{m}\sum_{k=1}^{n}(\overline{y}_{ij.} - \overline{y}_{i..} - \overline{y}_{.j.} + \overline{\overline{y}})^{2} \\ &+ \sum_{i=1}^{l}\sum_{j=1}^{m}\sum_{k=1}^{n}(\overline{y}_{i.k} - \overline{y}_{i..} - \overline{y}_{..k} + \overline{\overline{y}})^{2} \\ &+ \sum_{i=1}^{l}\sum_{j=1}^{m}\sum_{k=1}^{n}(\overline{y}_{.jk} - \overline{y}_{.j.} - \overline{y}_{..k} + \overline{\overline{y}})^{2} \\ &+ \sum_{i=1}^{l}\sum_{j=1}^{m}\sum_{k=1}^{n}(y_{ijk} - \overline{y}_{ij.} - \overline{y}_{i.k} - \overline{y}_{.jk} + \overline{y}_{i..} + \overline{y}_{.j.} + \overline{y}_{..k} - \overline{\overline{y}})^{2} \end{split}\end{displaymath}$$

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

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

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

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

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

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

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

$$\begin{displaymath}\begin{split} S_{A \times C} &= \sum_{i=1}^{l}\sum_{j=1}^{m}\sum_{k=1}^{n}(\overline{y}_{i.k}-\overline{y}_{i..}-\overline{y}_{..k}+\overline{\overline{y}})^{2} \\ &= S_{AC} - S_{A} - S_{C} \end{split}\end{displaymath}$$

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

$$\begin{displaymath}\begin{split} S_{B \times C} &= \sum_{i=1}^{l}\sum_{j=1}^{m}\sum_{k=1}^{n}(\overline{y}_{.jk}-\overline{y}_{.j.}-\overline{y}_{..k}+\overline{\overline{y}})^{2} \\ &= S_{BC} - S_{B} - S_{C} \end{split}\end{displaymath}$$

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

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

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

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

$$\nu_{C}=n-1$$

$$\nu_{A \times B}=\nu_{A} \times \nu_{B}=(l-1)(m-1)$$

$$\nu_{A \times C}=\nu_{A} \times \nu_{C}=(l-1)(n-1)$$

$$\nu_{B \times C}=\nu_{B} \times \nu_{C}=(m-1)(n-1)$$

$$\nu_{E}=\nu_{T}-(\nu_{A}+\nu_{B}+\nu_{C}+\nu_{A \times B}+\nu_{A \times C}+\nu_{B \times C})=(l-1)(m-1)(n-1)$$

$$\nu_{T}=lmn-1=N-1$$

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

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

$$V_{C}=\frac{S_{C}}{\nu_{C}}$$

$$V_{A \times B}=\frac{S_{A \times B}}{\nu_{A \times B}}$$

$$V_{AB}=\frac{S_{AB}}{\nu_{AB}}$$

$$V_{A \times C}=\frac{S_{A \times C}}{\nu_{A \times C}}$$

$$V_{AC}=\frac{S_{AC}}{\nu_{AC}}$$

$$V_{B \times C}=\frac{S_{B \times C}}{\nu_{B \times C}}$$

$$V_{BC}=\frac{S_{BC}}{\nu_{BC}}$$

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

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

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

$$E(V_{C})=\sigma_{E}^{ \ 2} +lm \sigma_{C}^{ \ 2}$$

$$E(V_{A \times B})=\sigma_{E}^{ \ 2} +n \sigma_{A \times B}^{ \ 2}$$

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

$$E(V_{B \times C})=\sigma_{E}^{ \ 2} +l \sigma_{A \times B}^{ \ 2}$$

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

인자 $A$에 대한 분산분석

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

기각역 : $F_{0} > F_{1-\alpha}(\nu_{_{A}},\nu_{_{E}})$

인자 $B$에 대한 분산분석

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

기각역 : $F_{0} > F_{1-\alpha}(\nu_{_{B}},\nu_{_{E}})$

인자 $C$에 대한 분산분석

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

기각역 : $F_{0} > F_{1-\alpha}(\nu_{_{C}},\nu_{_{E}})$

인자 $A , \ B$의 교호작용 대한 분산분석

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

기각역 : $F_{0} > F_{1-\alpha}(\nu_{_{A \times B}},\nu_{_{E}})$

인자 $A , \ C$의 교호작용 대한 분산분석

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

기각역 : $F_{0} > F_{1-\alpha}(\nu_{_{A \times C}},\nu_{_{E}})$

인자 $B , \ C$의 교호작용 대한 분산분석

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

기각역 : $F_{0} > F_{1-\alpha}(\nu_{_{B \times C}},\nu_{_{E}})$

주효과인 인자&nbsp $$A, B, C$$ 만이 유의한 경우 [교호작용]들이 모두 오차항에 [풀링]되어 버린다.

(단,&nbsp&nbsp $$S_{E}\acute{}=S_{E}+S_{A \times B}+S_{A \times C}+S_{B \times C}, \ \nu_{E}\acute{}=\nu_{E}+\nu_{A \times B}+\nu_{A \times C}+\nu_{B \times C}, \ V_{E}\acute{}=S_{E}\acute{}/\nu_{E}\acute{}$$ 이다.)

* '[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.

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

—- * '[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.

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

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

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

 $$\hat{\mu}(C_{k})=\widehat{\mu + c_{k}} = \overline{y}_{..k}$$

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

 $$\hat{\mu}(C_{k})= \left( \overline{y}_{..k} - t_{\alpha/2}(\nu_{E}\acute{} \ ) \sqrt{\frac{V_{E}\acute{}}{lm}} \ , \ \overline{y}_{..k} + t_{\alpha/2}(\nu_{E}\acute{} \ ) \sqrt{\frac{V_{E}\acute{}}{lm}} \right)$$

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

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

 $$\hat{\mu}(A_{i}B_{j}C_{k})=\widehat{\mu+a_{i}+b_{j}+c_{k}}=\overline{y}_{i..} + \overline{y}_{.j.} + \overline{y}_{..k} - 2 \overline{\overline{y}}$$

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

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

 단,&nbsp&nbsp $$n_{e}$$ 는 [유효반복수]이고&nbsp&nbsp $$n_{e} = \frac{lmn}{l+m+n-2}$$ 이다.

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