showing the estimator of $\sigma^2$ Let's take a one way ANOVA Model as $Y_{ij}=\mu + \tau_i +\epsilon_{ij}$    $i=1,2,3$  and $j=1,...,n_i$ and $\epsilon_{ij} \sim N(0,\sigma^2),   \ \ \ \forall (i,j).$
Sample variance for factor level is 
$$S_i^2=\frac{1}{n_i-1} \left( \sum_{j=1}^{n_i} y_{ij}^2- \frac{(\sum_{j=1}^{n_i}y_{ij})^2}{n_i}\right)$$ $i=1,2,3$
How can one show that the estimator of $\sigma^2$ is equal to the following?
$$S^2 =\frac{(n_1 -1)S_1^2 +(n_2-1)S_2^2 +(n_3-1)S_3^2}{n_1+n_2+n_3-3}$$
 A: This is not going to be a full answer but a series of hints. 
Indeed the sample variance is given by the formula you have written. One crucial assumption of ANOVA though is that the error term is homoscedastic, meaning that it follows the same variance for all populations. What this means is that you can combine all the information about the variability across samples into one estimator. This is reminiscent of the situation of the pooled variance in a t-test, again under the assumption of equal population variances. 
The question of course is, why do you need to combine sample variances like that instead of taking simple averages? Try to see what happens when you take expectations to find out. 
I hope this clears it up a bit.
A: Taking the two population case because it is easier to write, recall that
$$ w = \text{average} = \frac{\sum^n_{i=1} x_i + \sum^m_{i=1} y_i}{n+m} = \frac{n\bar{x} + m \bar{y}}{n+m} $$
for two populations $X_i ... X_{N_1}$ with observations $x_i ... x_n$ and $Y_i ... Y_{N_2}$ with observations $y_i ... y_m$. 
For total variance, we can write 
$$ \begin{aligned} (n + m -1) V &= \sum^n_{i=1} (x_i - w)^2 +  \sum^m_{i=1} (y_i - w)^2 \\ &= \sum^n_{i=1} (x_i - \bar{x} + \bar{x} - w)^2 +  \sum^m_{i=1} (y_i - \bar{y} + \bar{y} - w)^2 \end{aligned} $$
Knowing that
$$ \begin{aligned} \bar{x}-w &= \frac{m(\bar{x} - \bar{y})}{n + m} \\ \bar{y}-w &= \frac{m(\bar{y} - \bar{x})}{n + m} \\ \sum^n_{i=1} (x_i - \bar{x}) &= 0 \\ \sum^m_{i=1} (y_i - \bar{y}) &= 0 \end{aligned} $$
How can you expand the above to redefine the variance? How can you extend this to three populations? 
