Analysis of Variance will be used here to test the null hypothesis that the means of three or more populations are the same against the alternative fact not all population means are the same. (For two populations t-tests are more efficient.)

One-way ANOVA is used under the following assumptions:

1. 1. The populations from which the samples are drawn are (approximately) normally distributed.
2. 2. The populations have the same variance.
3. 3. The samples are random and independent.

We can test $ H_0: \mu_1=\mu_2,\,\, H_0: \mu_2=\mu_3,\,\, $and $H_0: \mu_1=\mu_3, $ separately, but their Type I error values add up.

The test proceeds by calculating two estimates of the common variance $\sigma^2 $.
Let $\overline{X}$ be the mean of all observations, let $\overline X_i$ be the mean in the sample of population $i$, and let $X_{ij}$ be the $i^{th}$ measurement in the $j^{th}$ sample. Then the between samples sum of squares is $$ MSB=\frac{1}{k-1} \sum_{i=1}^{k}(\overline{X}_i-\overline{X})^2 $$and the within samples sum of squares is$$ MSW=\frac{1}{n-k} \sum_{i=1}^{k} \sum_{j=1}^{n_i} (\overline{X}_{ij}-\overline{X}_i)^2 $$Here $k$ is the number of different populations (or treatments), $n_i$ is the size of the sample from population $i$ and$$ n =n_1+n_2+\cdots +n_k $$The test statistic is$$ f =\frac{MSB}{MSW} $$with$ (k-1, n-1) $degrees of freedom.

$MSB $ overestimates $\sigma^2$ if $H_0$ is false. So this is a one tailed test with a critical region in the right tail of the F-distribution.