# Intuition for why the “interaction variance” in an ANOVA is an estimate for the population variance

As I understand it, the general idea of the F-test is to exploit two different ways of estimating the variance in the population if all values actually come from a single population (see my reply here: Reasoning for comparing variance between and within for Anova?). If they do come from a single population the ratio of the variances has to be 1.

The first estimate comes from the average estimated population variances in the groups, which is $$\hat\sigma_\mathrm{within}^2$$ and the second comes from looking at the means and exploiting that the variance of this distribution is a function of the population variance: $$\hat{\sigma}^2_{\bar{x}} = \frac{\hat{\sigma}^2}{n}$$. We can estimate the population variance from this relationship and this is called $$\hat\sigma_\mathrm{between}^2$$. The ratio between both estimates (within and between) has to be 1 because both are estimates of the same population variance (if all values come from a single distribution).

What I have problems with is when it comes to the interaction term. Why is the "variance of the interaction" also an estimate of the population variance? The interaction term is calculated by looking at the difference between the actual mean values and the expected mean values when there is no interaction. These differences are squared, summed up, and multiplied with $$n$$ (sample size in each subgroup; we can assume a balanced design) and divided by the degrees of freedom:

$$\frac{n\sum_j^k\sum_l^m(\bar{x}_{jl_\mathrm{actual}}-\bar{x}_{jl_\mathrm{expected}})^2}{(k-1)(m-1)}$$

Where $$n$$ is the sample size for each group (balanced design), $$j=1, ..., j=k$$ is the first factor, and $$l=1, ..., l=m$$ is the second factor.

The formula must somehow be an estimate for the population variance if there is actually no interaction. I assumed that the part without $$n$$ is the variance of a sampling distribution and multiplying with $$n$$ makes it an estimate for the population variance. But there is no apparent intuition to me why this would be the case. Furthermore, I wonder why the degrees of freedom are calculated this way. It makes sense that without an interaction the differences of the means cannot vary as much, but why exactly $$(k-1)(m-1)$$?