I have a linear mixed-effect model $$ y=\beta_0+\beta_1x_1+\beta_2x_2+\beta_3x_3+Zb+e, $$ where $[x_1\ x_2\ x_3]$ represents the fixed effects, and $Z$ represents the random effects. Now, I test the null hypothesis that $\beta_1=\beta_2=\beta_3=0$. The $p$-value turns out to be $2\times10^{-6}$. So I conclude, $\beta_{1,2,3}$ are statistically different.

Now, I wish to further compare the three $\beta$'s with three pairwise comparisons.

  1. The $p$-value for "$\beta_1=\beta_2$" is $0.03$. OK, they are not very different.
  2. The $p$-value for "$\beta_1=\beta_3$" is $0.09$. OK, they are not different.
  3. The $p$-value for "$\beta_2=\beta_3$" is $0.57$. OK, they are VERY not different.

How can this happen?

I mean, the first overall hypothesis testing tells me with $p=2\times10^{-6}$ that the three $\beta$'s are different, then the pairwise comparisons tell me there are no pairwise differences?!

  • 2
    $\begingroup$ You were testing that they all equal zero versus testing equalities between pairs but not necessarily that each pair is equal to 0. $\endgroup$
    – jld
    Commented May 8, 2015 at 15:09
  • $\begingroup$ @Chaconne Oh yes, you are right! But I wonder how I can test $\beta_1=\beta_2=\beta_3$? Testing $\beta_1=\beta_2=\beta_3=0$ is easy: just use contrast matrix $$\begin{pmatrix}0&1&0&0&\ldots\\0&0&1&0&\ldots\\0&0&0&1&\ldots\end{pmatrix}$$ $\endgroup$ Commented May 8, 2015 at 15:22

1 Answer 1


As I said in my comment, what you're seeing is because your overall test tests that they are all equal to zero, not just that they are equal.

If you want to test that the three are all equal, i.e. test $H_0: \beta_1 = \beta_2 = \beta_3$ vs $H_A: \mathrm{not} \ H_0$, then observe that this is equivalent to testing $H_0: \beta_1 - \beta_2 = 0 \ \mathrm{and} \ \beta_2 - \beta_3 = 0$ so we can use the contrast $$ \begin{pmatrix} 0&1&-1&0&\ldots \\ 0&0&1&-1&\ldots \\ \end{pmatrix} $$

And just a general comment, always be careful to keep your model parameterization in mind when testing coefficients. What you end up doing may be different if $\beta_2$ represents the mean of group 2 vs. the deviation of the mean of group 2 from the mean of a baseline group, for example.


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