checking normality in repeated ANOVA (residuals vs differences)

Question

If I check for the assumptions of a repeated ANOVA (one IV with 3 measurements) I need to check for the normality of differences.

From t-test for paired data I know, I calculate the difference and run and hist(), qqplot() or shapiro.test() over the differences. Now, in ANOVA it is the same assumption. However, everybody seems to check the normality of residuals instead. As I asked in class I got the answer: it is faster. My question outside the speed-argument is, why is it the same?

Normally I would calculate 3 differences for 3 groups and then check the normality of each difference. Residuals are the deviation of the value to the mean(s). How is this the same?

Thanks for your time. I hope for helpful comments.

Answer 1

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 $c^2Var(X)$, given that $a$ and $c$ are constants.).

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 $\epsilon_{ij}$.