# What is the null hypothesis of a MANOVA?

Background

In order to analyze differences in some continuous variable between different groups (given by a categorical variable), one can perform a one-way ANOVA. If there are several explanatory (categorical) variables, one can perform a factorial ANOVA. If one wants to analyze differences between groups in several continuous variables (i.e., several response variables), one has to perform a multivariate ANOVA (MANOVA).

Question

I hardly understand how one can perform an ANOVA-like test on several response variables and more importantly, I don't understand what the null hypothesis could be. Is the null hypothesis:

• "For each response variable, the means of all groups are equal",

or is it

• "For at least one response variable, the means of all groups are equal",

or is $H_0$ something else?

• I can't tell, are you also asking how an ANOVA works? In the context of discussing what a standard error is, I essentially explain the basic idea behind an ANOVA here: How does the standard error work? – gung Jan 13 '15 at 21:33
• Neither of your two statements. H0 of MANOVA is that there is no difference in multivariate space. The multivariate case is considerably more complex than univariate because we have to deal with covariances, not just variances. There exist several ways to formulate the H0-H1 hypotheses in MANOVA. Read Wikipedia. – ttnphns Jan 13 '15 at 21:38
• @ttnphns: Why neither? The $H_0$ of ANOVA is that the means of all groups are equal. The $H_0$ of MANOVA is that the multivariate means of all groups are equal. This is exactly alternative 1 in the OP. Covariances etc. enter the assumptions and the computations of MANOVA, not the null hypothesis. – amoeba Jan 13 '15 at 21:41
• @amoeba, I didn't like For each response variable. To me it sounds like (or I read it as) "testing is done univarietly on each" (and then somehow combined). – ttnphns Jan 13 '15 at 21:47

The null hypothesis $H_0$ of a one-way ANOVA is that the means of all groups are equal: $$H_0: \mu_1 = \mu_2 = ... = \mu_k.$$ The null hypothesis $H_0$ of a one-way MANOVA is that the [multivariate] means of all groups are equal: $$H_0: \boldsymbol \mu_1 = \boldsymbol \mu_2 = ... = \boldsymbol \mu_k.$$ This is equivalent to saying that the means are equal for each response variable, i.e. your first option is correct.

In both cases the alternative hypothesis $H_1$ is the negation of the null. In both cases the assumptions are (a) Gaussian within-group distributions, and (b) equal variances (for ANOVA) / covariance matrices (for MANOVA) across groups.

### Difference between MANOVA and ANOVAs

This might appear a bit confusing: the null hypothesis of MANOVA is exactly the same as the combination of null hypotheses for a collection of univariate ANOVAs, but at the same time we know that doing MANOVA is not equivalent to doing univariate ANOVAs and then somehow "combining" the results (one could come up with various ways of combining). Why not?

The answer is that running all univariate ANOVAs, even though would test the same null hypothesis, will have less power. See my answer here for an illustration: How can MANOVA report a significant difference when none of the univariate ANOVAs reaches significance? Naive method of "combining" (reject the global null if at least one ANOVA rejects the null) would also lead to a huge inflation of type I error rate; but even if one chooses some smart way of "combining" to maintain the correct error rate, one would lose in power.

### How the testing works

ANOVA decomposes the total sum-of-squares $T$ into between-group sum-of-squares $B$ and within-group sum-of-squares $W$, so that $T=B+W$. It then computes the ratio $B/W$. Under the null hypothesis, this ratio should be small (around $1$); one can work out the exact distribution of this ratio expected under the null hypothesis (it will depend on $n$ and on the number of groups). Comparing the observed value $B/W$ with this distribution yields a p-value.

MANOVA decomposes the total scatter matrix $\mathbf T$ into between-group scatter matrix $\mathbf B$ and within-group scatter matrix $\mathbf W$, so that $\mathbf T = \mathbf B + \mathbf W$. It then computes the matrix $\mathbf W^{-1} \mathbf B$. Under the null hypothesis, this matrix should be "small" (around $\mathbf{I}$); but how to quantify how "small" it is? MANOVA looks at the eigenvalues $\lambda_i$ of this matrix (they are all positive). Again, under the null hypothesis, these eigenvalues should be "small" (all around $1$). But to compute a p-value, we need one number (called "statistic") in order to be able to compare it with its expected distribution under the null. There are several ways to do it: take the sum of all eigenvalues $\sum \lambda_i$; take maximal eigenvalue $\max\{\lambda_i\}$, etc. In each case, this number is compared with the distribution of this quantity expected under the null, resulting in a p-value.

Different choices of the test statistic lead to slightly different p-values, but it is important to realize that in each case the same null hypothesis is being tested.

• Also, if you don't correct for multiple testing, the all-univariate-ANOVAs approach will yield type I error inflation as well. – gung Jan 13 '15 at 22:20
• @gung: Yes, that is true as well. However, one can be smarter in "combining" than just rejecting the null as soon as at least one of the ANOVAs rejects the null. My point was that however smart one tries to be in "combining", one will still lose in power as compared to MANOVA (even if one manages to maintain the size of the test without inflating the error rate). – amoeba Jan 13 '15 at 22:24
• But isn't now that "power" directly related to the notion of the covariance? The moral is that with a (series of) univariate test we test only for marginal effect which is SSdifference/SSerror scalar. In MANOVA the multivariate effect is SSCPerror^(-1)SSCPdifference matrix (covariances total and within-groups accounted for). But since there are several eigenvalues in it which could be "combined" not in a single manner in a test statistic, several possible alternative hypotheses exist. More power - more theoretical complexity. – ttnphns Jan 14 '15 at 7:03
• @ttnphns, yes, this is all correct, but I think does not change the fact that the null hypothesis is what I wrote it is (and that's what the question was about). Whatever test statistic is used (Wilks/Roy/Pillai-Bartlett/Lawley-Hotelling), they are trying to test the same null hypothesis. I might expand my answer later to discuss this in more detail. – amoeba Jan 14 '15 at 9:18
• @gung asked me to chime in (not sure why... I taught MANOVA some 7 years ago, and never applied it) -- I would say that amoeba is right in saying that $H_1$ is a full negation of the null $H_0: \mu_{\mbox{group }1} = \ldots = \mu_{\mbox{group }k}$, which is a $p$-dimensional hyperspace in $kp$ dimensional space of parameters (if $p$ is the dimension that nobody bothered defining so far). And it is option 1 given by the OP. Option 2 is significantly more difficult to test. – StasK Jan 20 '15 at 4:28

It is the former.

However, the way it does it isn't literally to compare the means of each of the original variables in turn. Instead the response variables are linearly transformed in a way that is very similar to principal components analysis. (There is an excellent thread on PCA here: Making sense of principal component analysis, eigenvectors & eigenvalues.) The difference is that PCA orients your axes so as to align with the directions of maximal variation, whereas MANOVA rotates your axes in the directions that maximize the separation of your groups.

To be clear though, none of the tests associated with a MANOVA is testing all the means one after another in a direct sense, either with the means in the original space or in the transformed space. There are several different test statistics that each work in a slightly different way, nonetheless they tend to operate over the eigenvalues of the decomposition that transforms the space. But as far as the nature of the null hypothesis goes, it is that all means of all groups are the same on each response variable, not that they can differ on some variables but are the same on at least one.

• Ooh...So Manova makes a linear discriminant analysis (to maximize the distance between the mean of the groups) and then, it runs a standard anova using the first axis as response variable? So, $Ho$ is "the means - in term of PC1 - of all groups are the same". Is that right? – Remi.b Jan 13 '15 at 21:19
• There are several different possible tests. Testing only the the 1st axis is essentially using Roy's largest root as your test. This will often be the most powerful test, but it is also more limited. I gather there is ongoing discussion over which test is 'best'. – gung Jan 13 '15 at 21:23
• I guess we use MANOVA rather than several ANOVAs in order to avoid multiple testing issues. But if, by doing an MANOVA we just make an ANOVA on PC1 of a LDR, then we still have a multiple testing issue to consider when looking at the Pvalue. Is this right? (Hope that makes more sense. I deleted my previous unclear comment) – Remi.b Jan 13 '15 at 21:28
• That's an insightful point, but there are two issues: 1) the axes are now orthogonal, & that can change the issues w/ multiple testing; 2) the sampling distributions of the MANOVA test statistics take the multiple axes into account. – gung Jan 13 '15 at 21:32
• @Remi.b: These are good questions, but just to be clear: MANOVA is not equivalent to a ANOVA on the first discriminant axis of LDA! See here for a the relation between MANOVA and LDA: How is MANOVA related to LDA? – amoeba Jan 13 '15 at 21:34