Home » Writing » Writing a null hypothesis for anova

# Writing a null hypothesis for anova

Let’s say we have two factors (A and B), each with two levels (A1, A2 and B1, B2) and a response variable (y).

The when performing a two way ANOVA of the type:

We are testing three null hypothesis:

1. There is no difference in the means of factor A
2. There is no difference in means of factor B
3. There is no interaction between factors A and B

When written down, the first two hypothesis are easy to formulate (for 1 it is $H_0:\; \mu_=\mu_$)

But how should hypothesis 3 be formulated?

edit. and how would it be formulated for the case of more then two levels?

I think it’s important to clearly separate the hypothesis and its corresponding test. For the following, I assume a balanced, between-subjects CRF-$pq$ design (equal cell sizes, Kirk’s notation: Completely Randomized Factorial design).

$Y_$ is observation $i$ in treatment $j$ of factor $A$ and treatment $k$ of factor $B$ with $1 \leq i \leq n$, $1 \leq j \leq p$ and $1 \leq k \leq q$. The model is $Y_ = \mu_ + \epsilon_, \quad \epsilon_ \sim N(0, \sigma_^2)$

B 1 \ldots B k \ldots B q

\\\hline A 1 \mu_ \ldots \mu_ \ldots \mu_ \mu_\\ \ldots \ldots \ldots \ldots \ldots \ldots \ldots\\ A j \mu_ \ldots \mu_ \ldots \mu_ \mu_\\ \ldots \ldots \ldots \ldots \ldots \ldots \ldots\\ A p \mu_ \ldots \mu_ \ldots \mu_ \mu_\\\hline

\mu_ \ldots \mu_ \ldots \mu_ \mu \end\mu_$is the expected value in cell$jk$,$\epsilon_$is the error associated with the measurement of person$i$in that cell. The$()$notation indicates that the indices$jk$are fixed for any given person$i$because that person is observed in only one condition. A few definitions for the effects:$\mu_ = \frac \sum_^ \mu_$(average expected value for treatment$j$of factor$A$)$\mu_ = \frac

\sum_^

\mu_$(average expected value for treatment$k$of factor$B$)$\alpha_ = \mu_ – \mu$(effect of treatment$j$of factor$A$,$\sum_^

\alpha_ = 0$)$\beta_ = \mu_ – \mu$(effect of treatment$k$of factor$B$,$\sum_^ \beta_ = 0$)$(\alpha \beta)_ = \mu_ – (\mu + \alpha_ + \beta_) = \mu_ – \mu_ – \mu_ + \mu$(interaction effect for the combination of treatment$j$of factor$A$with treatment$k$of factor$B$,$\sum_^

(\alpha \beta)_ = 0 \, \wedge \, \sum_^ (\alpha \beta)_ = 0)\alpha_^ = \mu_ – \mu_$(conditional main effect for treatment$j$of factor$A$within fixed treatment$k$of factor$B$,$\sum_^

\alpha_^ = 0 \, \wedge \, \frac \sum_^ \alpha_^ = \alpha_ \quad \forall \, j, k)\beta_^ = \mu_ – \mu_$(conditional main effect for treatment$k$of factor$B$within fixed treatment$j$of factor$A$,$\sum_^ \beta_^ = 0 \, \wedge \, \frac

\sum_^

\beta_^ = \beta_ \quad \forall \, j, k)$With these definitions, the model can also be written as:$Y_ = \mu + \alpha_ + \beta_ + (\alpha \beta)_ + \epsilon_$This allows us to express the null hypothesis of no interaction in several equivalent ways:$H_>: \sum_\sum_ (\alpha \beta)^_ = 0$(all individual interaction terms are$0$, such that$\mu_ = \mu + \alpha_ + \beta_ \, \forall j, k$. This means that treatment effects of both factors – as defined above – are additive everywhere.)$H_>: \alpha_^ – \alpha_^ = 0 \quad \forall \, j \, \wedge \, \forall \, k, k’; \quad (k \neq k’;)$(all conditional main effects for any treatment$j$of factor$A$are the same, and therefore equal$\alpha_$. This is essentially Dason’s answer.)$H_>: \beta_^ – \beta_^ = 0 \quad \forall \, j, j’; \, \wedge \, \forall \, k \quad (j \neq j’;)$(all conditional main effects for any treatment$k$of factor$B$are the same, and therefore equal$\beta_$.)$H_>$: In a diagramm which shows the expected values$\mu_$with the levels of factor$A$on the$x$-axis and the levels of factor$B$drawn as separate lines, the$q$different lines are parallel. An interaction tells us that the levels of factor A have different effects based on what level of factor B you’re applying. So we can test this through a linear contrast. Let C = (A1B1 – A1B2) – (A2B1 – A2B2) where A1B1 stands for the mean of the group that received A1 and B1 and so on. So here we’re looking at A1B1 – A1B2 which is the effect that factor B is having when we’re applying A1. If there is no interaction this should be the same as the effect B is having when we apply A2: A2B1 – A2B2. If those are the same then their difference should be 0 so we could use the tests:$H_0: C = 0\quad\text\quad H_A: C \neq 0.\$

answered Dec 18 ’10 at 14:14

chl ♦
37.4k ● 6 ● 124 ● 243

Thanks Dason, that helped. Also, after reading your reply, it suddenly became clear to me that I am not fully sure how this generalizes in case we are having more factors. Could you advise? Thanks again. Tal Tal Galili Dec 18 ’10 at 14:21

You can test multiple contrasts simultaneously. So for example if A had three levels and B had 2 we could use the two contrasts: C1 = (A1B1 – A2B1) – (A2B1 – A2B2) and C2 = (A2B1 – A2B2) – (A3B1 – A3B2) and use a 2 degree of freedom test to simultaneously test if C1 = C2 = 0. It’;s also interesting to note that C2 could equally have been (A1B1 – A1B2) – (A3B1 – A3B2) and we would come up with the same thing. Dason Dec 18 ’10 at 14:23