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Say we have $k$ samples of data, where sample $i$ is of size $n_i$ and we write it as $x_{i1}, ... , x_{in_i}$. Let the total sample size be $N$.

The ANOVA model is $X_{ij} \sim N(\mu_i, \sigma^2)$ independently. The null hypothesis is that the $\mu_i$ are all equal. The alternative hypothesis is that the null hypothesis is not true.

The ANOVA F-statistic is

$$F = \frac{S_2/(k-1)}{S_1/(N - k)},$$

where

$$S_1 = \sum_{i, j}(x_{ij} - \bar{x}_{i\bullet})^2$$

is the within samples sum of squares and

$$S_2 = \sum_in_i(\bar{x}_{i\bullet} - \bar{x}_{\bullet\bullet})^2$$

is the between samples sum of squares.

We know that $S_1$ and $S_2$ are independent and $S_0 = S_1 + S_2$ $(*)$, where

$$S_0 = \sum_{i, j}(x_{ij} - \bar{x}_{\bullet\bullet})^2$$

is the total sum of squares.

It is straightforward to show that under both the null and the alternative hypotheses, $S_1 \sim \sigma^2\chi^2_{N - k}$.

Also, under the null hypothesis, the $X_{ij}$ are identically distributed, and so $S_0 \sim \sigma^2\chi^2_{N-1}$. It follows from $(*)$ that under the null hypothesis $S_2 \sim \sigma^2\chi^2_{k-1}$ and thus $F \sim F_{k-1, N-k}$.

It is claimed that under the alternative hypothesis, $F$ follows a non-central $F$-distribution $F_{k-1, N-k}(\lambda)$, where $\lambda = \sum_in_i(\mu_i - \bar\mu)^2$ and $\bar\mu = \sum_in_i\mu_i/N$ — or equivalently, that $S_2$ follows a (scaled) non-central $\chi^2$ distribution, $\sigma^2\chi^2_{k-1}(\lambda)$.

My tentative approach to proving this is similar to the derivation under the null hypothesis — that is, it is sufficient to prove that $S_0$ follows a (scaled) non-central $\chi^2$ distribution, $\sigma^2\chi^2_{N-1}(\lambda)$.

I've shown that this would follow from a slightly more general statement, namely that if $Y_i \sim N(\mu_i, \sigma^2)$ independently (sample size $N$), then $S_0 = \sum_i(Y_i - \bar{Y})^2 \sim \sigma^2\chi^2_{N-1}(\lambda)$, where $\lambda = \sum_i(\mu_i - \bar\mu)^2$.

Is this the best approach? And what is the simplest proof of the final statement above?

Thanks.

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  • $\begingroup$ How do you know that $S_1$ and $S_2$ are independent, even under the null hypothesis? A complete proof of the distribution of $F$ would need to prove that rather than simply assume it. $\endgroup$ – Gordon Smyth Aug 26 '18 at 10:39

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