# Non-omnibus ANOVA test

Let's say I have $N$ groups that are mutually independent ($X_1,\ldots,X_N$). What I am looking for is a p-value to test the null hypothesis "$\mu_1=\mu_2\vee\mu_1=\mu_3\vee\ldots\vee\mu_1=\mu_N$", if that makes any sense (my apologies if that's an abuse of notation; I'm a bit of a newbie when it comes to statistics).

What I mean is, I would like to test whether the mean of $X_1$ in particular is significantly different from the means of $X_2,\ldots,X_N$. Actually, it would be ideal if I could test whether it's not just "different", but less than the rest (i.e., $\mu_1<\mu_2\wedge\ldots\wedge\mu_1<\mu_N$).

My problem is that I can't assume $\mu_2=\mu_3=\ldots=\mu_N$, or else I could just perform an ANOVA on the $X_1,\ldots,X_N$. My Wikipedia searching has told me that my problem comes from the fact that ANOVA is an "omnibus test", when I really need some type of "contrast test" or something. I'm sure there's a standard test of for this problem, but I just haven't found the right thing to search for yet. If anyone could point me in the right direction I'd really appreciate it (if that direction happens to involve a Matlab function, that would be amazing). Thanks!

• You can certainly frame a set of hypotheses each tested by contrasts (non-orthogonal) and test each of them, though many people would take account of the multiple testing in some way so as to reduce the overall type I error rate. – Glen_b May 8 '14 at 22:19

## 1 Answer

If H$_{0}\text{: }\mu_{1} = \mu_{2}, \dots, \mu_{N}$, then it must also logically be the case that under H$_{0}$ that $\mu_{2} = \mu_{3}$, etc. Transitivity, right?

Otherwise, it sounds like you need to simply perform $N-1$ pairwise tests and approrpiately control for the family-wise error rate, or false discovery rate given the multiple comparisons.

• Thanks for the comment! As for your first point: that's definitely true. It's just that $\mu_1=\mu_2,\ldots,\mu_N$ is a stronger hypothesis than the one I would like to test. I don't care whether they're all the same, just that $\mu_1$ is different than each individually. For your second point, that would definitely work, but then I would end up with $N-1$ separate p-values, right? Is there a way to, in a sense, "combine" them? – Chuck N May 8 '14 at 15:00
• OK, I think that you are indeed just talking about a garden-variety oneway ANOVA, even though you are substantively only interested in group 1 versus groups 2 through $N$. If you like, you can think of the the extra pair-wise comparisons that you get with the null hypothesis H$_{0}\text{: }\mu_{1} = \mu_{2} = \dots \mu_{N}$ as freebies due to transitivity. For the second part of your question, I would again direct you to family-wise error rate or false discovery rate methods for multiple comparisons. (See links in edited answer). – Alexis May 8 '14 at 17:16
• Ah, okay, I think I get it now, and I'll definitely check out the links you added. Thanks for the help! – Chuck N May 8 '14 at 17:31