Return to Revisions

An ANOVA is a general linear model of the form (this is just one way of writing the model called the cell-means model, and I do so without loss of generality):

$Y_{ij}=\mu_{i}+\epsilon_{ij}$

where $\mu_i$ are parameters (treatment means), $Y_{ij}$ is the value of the response variable for the $j$th trial for the $i$th treatment, and $i=1,...,r$ and $j=1..., n_i$, and $\epsilon_{ij}$ are independent $N(0, \sigma^2)$ as specified by the Gauss-Markov theorem.

Note that the error term is assumed to be Normal with mean 0 and variance $\sigma^2$ and that $\mu_i$ is considered a constant. One can validate the "normality assumption" by two methods:

1. One can directly examine the errors $\epsilon_{ij}$ (residulas in the fitted model) which represent the error terms in the above model to verify that they are in fact distributed as $N(0, \sigma^2)$; or
2. One can look at the $Y_{ij}$ terms directly since $Y_{ij}-\mu_i=\epsilon_{ij}.$ Because $\mu_{ij}$ is a constant:

\begin{eqnarray*} E(Y_{ij}) & = & E(\mu_{ij}-\epsilon_{ij})\\ & = & E(\mu_{ij})-E(\epsilon_{ij})\\ & = & \mu_{ij}-0\\ & & \mu_{ij} \end{eqnarray*}

and

\begin{eqnarray*} V(Y_{ij}) & = & V(\mu_{ij}-\epsilon_{ij})\\ & = & V(\mu_{ij})+V(\epsilon_{ij})\\ & = & 0+V(\epsilon_{ij})\\ & & \sigma^{2} \end{eqnarray*}

And because $Y_{ij}$ is a linear function of a normally distributed random variable, $\epsilon_{ij}$, $Y_{ij}$ itself is a normally distributed random variable (Recall that if $X$ is a random variable, then $Z=a+cX$ is normally distributed with mean $a+cE(X)$ and variance $c^2Var(X)$).

So, this implies $Y_{ij}$ are independent $N(\mu_{ij}, \sigma^2)$ in the cell-means ANOVA model, and, as a result, implies that one can verify normality be examining either (1) the residuals or by (2) examining the $Y_{ij}$ since they too are assumed normal.

So to summarize:

Because the $\epsilon_{ij}$ term is Normal and any linear combination of $\epsilon_{ij}$ is Normal, then the residuals must also be normal as well as the $Y_{ij}$ themselves since they are a linear combination of the $\epsilon_{ij}$.