20
$\begingroup$

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:

y~A+B+A*B

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 $H_0:\; \mu_{A1}=\mu_{A2}$)

But how should hypothesis 3 be formulated?

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

Thanks.

$\endgroup$
  • 3
    $\begingroup$ I don't have the reputation to allow me to edit, but I think you want $H_0 = \mu_{A1}=\mu_{A2}$ (or $\mu_{A_1}$ if you want a double subscript) [oops, it has automatically tex-ified that: H_0 = \mu_{A1}=\mu_{A2} or \mu_{A_1}] $\endgroup$ – Ben Bolker Dec 18 '10 at 15:12
  • 1
    $\begingroup$ Oups, didn't see that you are using capital letters to denote factor name and their levels -- fix it (following @Ben notation). $\endgroup$ – chl Dec 18 '10 at 16:12
18
$\begingroup$

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).

$Y_{ijk}$ 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 $Y_{ijk} = \mu_{jk} + \epsilon_{i(jk)}, \quad \epsilon_{i(jk)} \sim N(0, \sigma_{\epsilon}^2)$

Design: $\begin{array}{r|ccccc|l} ~ & B 1 & \ldots & B k & \ldots & B q & ~\\\hline A 1 & \mu_{11} & \ldots & \mu_{1k} & \ldots & \mu_{1q} & \mu_{1.}\\ \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots\\ A j & \mu_{j1} & \ldots & \mu_{jk} & \ldots & \mu_{jq} & \mu_{j.}\\ \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots\\ A p & \mu_{p1} & \ldots & \mu_{pk} & \ldots & \mu_{pq} & \mu_{p.}\\\hline ~ & \mu_{.1} & \ldots & \mu_{.k} & \ldots & \mu_{.q} & \mu \end{array}$

$\mu_{jk}$ is the expected value in cell $jk$, $\epsilon_{i(jk)}$ 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:

$\mu_{j.} = \frac{1}{q} \sum_{k=1}^{q} \mu_{jk}$ (average expected value for treatment $j$ of factor $A$)

$\mu_{.k} = \frac{1}{p} \sum_{j=1}^{p} \mu_{jk}$ (average expected value for treatment $k$ of factor $B$)

$\alpha_{j} = \mu_{j.} - \mu$ (effect of treatment $j$ of factor $A$, $\sum_{j=1}^{p} \alpha_{j} = 0$)

$\beta_{k} = \mu_{.k} - \mu$ (effect of treatment $k$ of factor $B$, $\sum_{k=1}^{q} \beta_{k} = 0$)

$(\alpha \beta)_{jk} = \mu_{jk} - (\mu + \alpha_{j} + \beta_{k}) = \mu_{jk} - \mu_{j.} - \mu_{.k} + \mu$
(interaction effect for the combination of treatment $j$ of factor $A$ with treatment $k$ of factor $B$, $\sum_{j=1}^{p} (\alpha \beta)_{jk} = 0 \, \wedge \, \sum_{k=1}^{q} (\alpha \beta)_{jk} = 0)$

$\alpha_{j}^{(k)} = \mu_{jk} - \mu_{.k}$
(conditional main effect for treatment $j$ of factor $A$ within fixed treatment $k$ of factor $B$, $\sum_{j=1}^{p} \alpha_{j}^{(k)} = 0 \, \wedge \, \frac{1}{q} \sum_{k=1}^{q} \alpha_{j}^{(k)} = \alpha_{j} \quad \forall \, j, k)$

$\beta_{k}^{(j)} = \mu_{jk} - \mu_{j.}$
(conditional main effect for treatment $k$ of factor $B$ within fixed treatment $j$ of factor $A$, $\sum_{k=1}^{q} \beta_{k}^{(j)} = 0 \, \wedge \, \frac{1}{p} \sum_{j=1}^{p} \beta_{k}^{(j)} = \beta_{k} \quad \forall \, j, k)$

With these definitions, the model can also be written as: $Y_{ijk} = \mu + \alpha_{j} + \beta_{k} + (\alpha \beta)_{jk} + \epsilon_{i(jk)}$

This allows us to express the null hypothesis of no interaction in several equivalent ways:

  1. $H_{0_{I}}: \sum_{j}\sum_{k} (\alpha \beta)^{2}_{jk} = 0$
    (all individual interaction terms are $0$, such that $\mu_{jk} = \mu + \alpha_{j} + \beta_{k} \, \forall j, k$. This means that treatment effects of both factors - as defined above - are additive everywhere.)

  2. $H_{0_{I}}: \alpha_{j}^{(k)} - \alpha_{j}^{(k')} = 0 \quad \forall \, j \, \wedge \, \forall \, k, k' \quad (k \neq k')$
    (all conditional main effects for any treatment $j$ of factor $A$ are the same, and therefore equal $\alpha_{j}$. This is essentially Dason's answer.)

  3. $H_{0_{I}}: \beta_{k}^{(j)} - \beta_{k}^{(j')} = 0 \quad \forall \, j, j' \, \wedge \, \forall \, k \quad (j \neq j')$
    (all conditional main effects for any treatment $k$ of factor $B$ are the same, and therefore equal $\beta_{k}$.)

  4. $H_{0_{I}}$: In a diagramm which shows the expected values $\mu_{jk}$ 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.

$\endgroup$
  • 1
    $\begingroup$ A really impressive answer Caracal - thank you. $\endgroup$ – Tal Galili Dec 19 '10 at 22:05
9
$\begingroup$

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:

$H_0: C = 0\quad\text{vs.}\quad H_A: C \neq 0.$

$\endgroup$
  • 1
    $\begingroup$ 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 $\endgroup$ – Tal Galili Dec 18 '10 at 14:21
  • 2
    $\begingroup$ 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. $\endgroup$ – Dason Dec 18 '10 at 14:23
  • $\begingroup$ Hi @Dason: you seem to have multiple accounts. Could you please complete the form at stats.stackexchange.com/contact and request that they be merged? That will simplify your use of this site (and give you the combined net reputation of both accounts). $\endgroup$ – whuber Sep 25 '13 at 19:23

protected by Glen_b -Reinstate Monica Apr 24 '15 at 9:48

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.