Theory and Methods of Statistics

Example 11.4.4

In the balanced two-factor ANOVA model (ie, n_ij ≡ n₀), the estimates are much simpler when one uses the following notations

$\begin{array}{ll}\hfill {\stackrel{-}{Y}}_{\cdot \cdot \cdot }& ={\left(n_{0}ab\right)}^{-1}\sum _i\sum _j\sum _k{Y}_{ijk},\hfill \\ \hfill {\stackrel{-}{Y}}_{i\cdot \cdot }& ={\left(n_{0}b\right)}^{-1}\sum _j\sum _k{Y}_{ijk},{\stackrel{-}{Y}}_{\cdot j\cdot }={\left(n_{0}a\right)}^{-1}\sum _i\sum _k{Y}_{ijk},\text{and}\hfill \\ \hfill {\stackrel{-}{Y}}_{ij\cdot }& =n_0^{-1}\sum _k{Y}_{ijk}.\hfill \end{array}$

With these notations

$\begin{array}{l}\widehat{\mu }={\stackrel{-}{Y}}_{\cdot \cdot \cdot },{\widehat{\alpha }}_i={\stackrel{-}{Y}}_{i\cdot \cdot }-{\stackrel{-}{Y}}_{\cdot \cdot \cdot },{\widehat{\beta }}_j={\stackrel{-}{Y}}_{\cdot j\cdot }-{\stackrel{-}{Y}}_{\cdot \cdot \cdot },\hfill \\ {\widehat{\left(\alpha \beta \right)}}_{ij}={\stackrel{-}{Y}}_{ij\cdot }-{\stackrel{-}{Y}}_{i\cdot \cdot }-{\stackrel{-}{Y}}_{\cdot j\cdot }+{\stackrel{-}{Y}}_{\cdot \cdot \cdot },\text{and}\hfill \\ Y_{ijk}-\widehat{\mu }={\widehat{\alpha }}_i+{\widehat{\beta }}_j+{\widehat{\left(\alpha \beta \right)}}_{ij}+{\widehat{\epsilon }}_{ijk},\hfill \end{array}$

where $\left\{{\widehat{\epsilon }}_{ijk}=Y_{ijk}-{\stackrel{-}{Y}}_{ij\cdot }\right\}$ are the residuals. If both sides are squared and summed over i, j, and