In a two-ways anova we can decompose the total sum of square in its components.


It is a common notation in the field so it should be clear (let me know if it is not).

We then can obtain the related variance dividing each component for its degree of freedom.

We can test the hypothesis $H_0: \alpha_1=\alpha_2=...=\alpha_a=0$ using the F statistics. $F:=\dfrac{\hat\sigma_\alpha^2}{\hat\sigma_e^2}\sim F(a-1,ab(n-1))$.

Suppose now I would like to test the hypothesis:

$H_0^{interaction}$: there are no interaction between $\alpha$ and $\beta$

Can I test this hypothesis using $\hat\sigma_{\alpha\beta}^2=\dfrac{SSAB}{(a-1)(b-1)}$ and the F statistics $F= \dfrac{\hat\sigma_{\alpha\beta}^2}{\sigma_e^2}\sim F((a-1)(b-1),ab(n-1))$ ?

All the books and website report the first example and I did not find anything regarding the second.

  • $\begingroup$ This should be a standard and easy question. Is it possible that no one is answering? Is maybe the question bad written? $\endgroup$
    – Donbeo
    Mar 24, 2014 at 0:34

1 Answer 1


This question is not very clear, maybe that's why it didn't get an answer ... First, twoway anova is really a shorthand, it is not by itself enough to define a model! The decomposistion you cites


is a version of the Pythagorean theorem, so needs an orthogonal design, and morever, without replication, SSE and SSAB are not separately defined, so you really need to give more context and details. For the formulation of the null hypothesis "no interaction", see What is the NULL hypothesis for interaction in a two-way ANOVA?.


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