[R-sig-ME] chisq = 0 and p = 1 in glmer model comparison result
Ben Bolker
bbo|ker @end|ng |rom gm@||@com
Tue Jun 4 01:32:54 CEST 2019
The reason might exactly be to test a NH of zero effect at the population
level. I agree about the rest though. If you provide data/reproducible
example, someone might take a look ...
On Tue, Jun 4, 2019, 8:46 AM Paul Johnson <paul.johnson using glasgow.ac.uk>
wrote:
> Hi Becky,
>
> The model with the two additional parameters has a lower log likelihood
> (-480.3, df=15) than the smaller model (-475.8, df=13). This shouldn’t be
> possible, because adding an unconstrained parameter makes the model more
> flexible, allowing a closer fit to the data. This anomalous difference in
> likelihoods has caused the chi-squared statistic in the LRT to be negative,
> giving a p-value of 1, because all of the chi-squared distribution is
> positive, i.e. P(chi-squared > test statistic) = 1 when the test statistic
> is <= 0. Either one or both of the models haven’t converged, or (less
> likely) there is missing data in the added variable so that the data
> differs between the two models.
>
> In addition, it’s very unusual to fit random slopes without also including
> a fixed effect, as you’ve done for targetWordFactor. The null model allows
> the WW-W and NW-W differences in the log odds of being correct to differ
> randomly between subjects and between items, but forces the mean
> differences (across subjects and items) to be zero. You’d need a good
> reason to fit such a model.
>
> Best wishes,
> Paul
>
>
> > On 3 Jun 2019, at 14:25, Guillaume Adeux <guillaumesimon.a2 using gmail.com>
> wrote:
> >
> > Hello Becky,
> >
> > Even though I cannot directly answer your question... a Chisq of 0 with
> > such a difference in AIC seems indeed suspicious.
> >
> > To test the effect of your predictor in a GLMM context through LRT tests,
> > you could (should?) consider using the test_terms function from the
> {monet}
> > package or its little brother function mixed() from the {afex} package.
> > These packages are specifically designed for such tests.
> >
> > I hope this helps.
> >
> > Sincerely,
> >
> > Guillaume ADEUX
> >
> >
> >
> > Le lun. 3 juin 2019 à 15:14, Becky Gilbert <beckyannegilbert using gmail.com>
> a
> > écrit :
> >
> >> Dear list
> >>
> >> I have two glmer models, one with a fixed factor (targetWordFactor) and
> one
> >> without, and I am comparing them using the anova function to get the LRT
> >> results for the fixed effect of targetWordFactor. The anova results are
> >> showing a chi-square value of 0 and p value of 1. Is this result
> possible,
> >> or is it perhaps a sign that I've done something wrong?
> >>
> >> Here are the anova results:
> >>
> >> anova(accModelNullWord,accModelWord)
> >> # Df AIC BIC logLik deviance
> Chisq
> >> Chi Df Pr(>Chisq)
> >> # accModelNullWord 13 977.59 1067.0 -475.8 951.59
> >> # accModelWord 15 990.61 1093.8 -480.3 960.61 0 2
> >> 1
> >>
> >> The targetWordFactor fixed factor has 3 levels (2 contrasts), so the
> >> degrees of freedom in the anova result look correct to me. Here are the
> >> model specifications:
> >>
> >> contrasts(pauseDetValidNoFillersExcluded$targetWordFactor)
> >> # WW NW
> >> # W 0 0
> >> # WW 1 0
> >> # NW 0 1
> >>
> >> accModelNullWord <- glmer(correct ~ 1 +
> >> (1 + targetWordFactor|subject) +
> >> (1 + targetWordFactor|item),
> >> data = pauseDetValidNoFillersExcluded,
> >> family = binomial(link = "logit"),
> >> control = glmerControl(optimizer="bobyqa",
> >> optCtrl =
> >> list(maxfun=2e5)))
> >>
> >> accModelWord <- glmer(correct ~ 1 + targetWordFactor +
> >> (1 + targetWordFactor|subject) +
> >> (1 + targetWordFactor|item),
> >> data = pauseDetValidNoFillersExcluded,
> >> family = binomial(link = "logit"),
> >> control = glmerControl(optimizer="bobyqa",
> >> optCtrl = list(maxfun=2e5)))
> >>
> >> Apologies if this question has been asked before - I did search the list
> >> but couldn't find anything.
> >>
> >> Many thanks,
> >> Becky
> >>
> >> [[alternative HTML version deleted]]
> >>
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> >>
> >
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