From nist for the test statistic in one-way ANOVA:
The mean squares are formed by dividing the sum of squares by the associated degrees of freedom.
Let $N = \sum_i n_i$. Then, the degrees of freedom for treatment, $DFT = k - 1$, and the degrees of freedom for error, $DFE = N - k$.
The corresponding mean squares are:
$MST = SST / DFT$
$MSE = SSE / DFE$
I was wondering why the degrees of freedom for the two sums are given as $k-1$ and $N-k$, where $N$ is the sample size and $k$ is the number of groups? For the definitions of the two sums $SST$ (sum of squares of treatments) and $SSE$ (sum of squares of errors), see another page.
So dividing a sum of squares by its degree of freedom isn't "mean square", isn't it? Why does it say "the mean squares are formed by dividing the sum of squares by the associated degrees of freedom"?
Thanks and regards!