I have measured nest building (building / not building) 5 times over the breeding season and want to see if there is an effect of my treatment (treated / control). After selecting random effect with likelihood ratio tests (anova(model1,model2) ), I have a model like this:

glmer(nest ~ treatment*time + (1|ID), family=binomial)

Now I want to select my fixed effects. According to Bolker et al. TREE 2009 I should use Wald Z or Chi squared tests to select my fixed effects. How do I do this? Anova{car} won't test 2 glmer's against each other. Can I use wald.test {aod}? If so, how?

Thanks for your help!



1 Answer 1


First of all: do you really need to do a model selection? MS with subsequent naive regression tests inflates type I error, so if you want to test for effects of your explanatory variables and you have enough data (no issue with power), there is no reason to run a MS.

If you must do a MS, I don't think anova(model1,model2) is appropriate to select on the random effects - likely this function will use naive degrees of freedom (1 or k, not sure) for the random effects. I would recommend fixing the random structure a priori and stick with it. If you must select, simulated LRTs would probably be safe.

Once the random effect structure is fixed, degrees of freedom are less problematic, so it should usually be OK to run a normal AICc basesd model selection on the fixed effects, e.g. with the MuMIn package - all the caveats that I noted first remain though.

  • $\begingroup$ Thank you Florian for your reply! I have 27 control and 28 treated individuals, I don't know what that means in terms of my power because I don't know effect sizes. The random structure is hard to fix a priori because the treatment could have an effect on the intercept and slope of nest ~ time which is why I chose to include both main effects and the intercept. I will use AICs then, thanks for the tip. The MuMln package, would that be the same as using drop1(model, test="Chisq"), which is a LRT, and only looking at the AIC? $\endgroup$ Commented Mar 3, 2016 at 7:47
  • $\begingroup$ If you have around 60 individuals with (n?) repeated observations per individual and two treatments interacting with one continuous variable time, I don't see the reason for a MS on either fixed or random effects. If you still do one, LRT is fundamentally different from AIC because one is a hypothesis test and the other isn't. If you want a p-value use LRT, but if you do stepwise tests via LRT, the p-values are screwed up anyway, so there's not much point to it imo. $\endgroup$ Commented Mar 3, 2016 at 15:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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