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Is the error term $\epsilon_{ij}$ in the linear model $ Y_{ij} = \mu + \alpha_i + \beta_j + \epsilon_{ij} $ the same as the "within samples/groups/treatment error" in the ANOVA model?

I would appreciate it if people could please take the time to clarify this.

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Yes. I will demonstrate with a 2x2 ANOVA example (with dichotomous groupings and no interaction).

The ANOVA model proposes that there is some group difference based on the row (which we can relate to $\alpha$) and based on the column (which we can relate to $\beta$). Thus, if I am in row 1 ($i=1$) and column 1 ($j=1$), then the group mean for that cell would be $$Y_{11}=\mu + \alpha_1 + \beta_1$$ and the fact that values in that cell have some variability is captured with the error term. However, to capture this notation fully, we should use $i$ to indicate the row, $j$ to indicate the column, and $k$ to indicate that particular score in the cell: $$Y_{ijk} = \mu + \alpha_i + \beta_j + \epsilon_{ijk}$$

In the multiple regression model, we would create two dummy variables. Let $x_1$ represent the row ($x_1=0$ if in the first row, associated with $\alpha_1$, and $x_1=1$ if in the second row...change of indices adds a bit of confusion, but this is not uncommon convention with these two different approaches). Likewise, we will create a variable $x_2$ to represent the column. Now the equation for this model can be written as $$Y_k = \gamma_0 + \gamma_1x_1 + \gamma_2x_2 + \epsilon_k$$ Notice, we can drop the $i$ and $j$ indices, as the dummy variables capture the cell in which the observation is located.

Comparing the two frameworks: $$\begin{align}Y_{11} & =\mu + \alpha_1 + \beta_1 = \gamma_0 \\ Y_{21} & = \mu + \alpha_2 + \beta_1 = \gamma_0 + \gamma_1 \\ Y_{12} & = \mu + \alpha_1 + \beta_2 = \gamma_0 + \gamma_2 \\ Y_{22} & = \mu + \alpha_2 + \beta_2 = \gamma_0 + \gamma_1 + \gamma_2 \end{align}$$ With the following equivalences: $$\begin{align}\gamma_0 & = \mu + \alpha_1 + \beta_1 \\ \gamma_1 & = \alpha_2 - \alpha_1 \\ \gamma_2 & = \beta_2 - \beta_1 \end{align}$$ we end up with identical models. (Well, identical if the errors are the same...which they are.)

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  • $\begingroup$ Thank you for taking the time to post such an enlightening answer. I understand now. $\endgroup$ Commented Apr 21, 2018 at 4:22

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