Hi Thierry,

I used the lmer package, using the following code

FL.glmm <- lmer(RESPONSE ~ fEXPN + fPBN + fEXPN*fPBN + (1|fNID), data =
FL.1, family = binomial)

Here is the summary


Generalized linear mixed model fit by the Laplace approximation
Formula: RESPONSE ~ fEXPN + fPBN + fEXPN * fPBN + (1 | fNID)
   Data: FL.1
   AIC   BIC logLik deviance
 104.1 163.0 -35.04    70.09
Random effects:
 Groups Name        Variance Std.Dev.
 fNID   (Intercept) 0.047168 0.21718
Number of obs: 236, groups: fNID, 15

Fixed effects:
               Estimate Std. Error z value Pr(>|z|)
(Intercept)     -0.4102     0.5304  -0.774    0.439
fEXPN2           1.4329     0.7880   1.818    0.069 .
fEXPN3          21.9715 12840.4523   0.002    0.999
fEXPN4          21.9715 12409.4403   0.002    0.999
fPBN2           21.9715 12409.4408   0.002    0.999
fPBN3            0.5453     0.7390   0.738    0.461
fPBN4           -1.4788     0.9286  -1.593    0.111
fEXPN2:fPBN2    -1.4171 17619.4645  -1e-04    1.000
fEXPN3:fPBN2   -21.9715 21994.2881  -0.001    0.999
fEXPN4:fPBN2   -21.9715 21493.7828  -0.001    0.999
fEXPN2:fPBN3    20.0155 12547.8259   0.002    0.999
fEXPN3:fPBN3    -0.5390 18188.0409   0.000    1.000
fEXPN4:fPBN3    -0.5390 17577.5297   0.000    1.000
fEXPN2:fPBN4   -21.1113 12447.6988  -0.002    0.999
fEXPN3:fPBN4   -41.6658 18265.7721  -0.002    0.998
fEXPN4:fPBN4   -41.6660 17647.6445  -0.002    0.998
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Correlation of Fixed Effects:
            (Intr) fEXPN2 fEXPN3 fEXPN4 fPBN2  fPBN3  fPBN4  fEXPN2:PBN2
fEXPN3:PBN2 fEXPN4:PBN2
fEXPN2      -0.666

fEXPN3       0.000  0.000

fEXPN4       0.000  0.000  0.000

fPBN2        0.000  0.000  0.000  0.000

fPBN3       -0.710  0.478  0.000  0.000  0.000

fPBN4       -0.565  0.380  0.000  0.000  0.000  0.405

fEXPN2:PBN2  0.000  0.000  0.000  0.000 -0.704  0.000  0.000

fEXPN3:PBN2  0.000  0.000 -0.584  0.000 -0.564  0.000  0.000  0.397

fEXPN4:PBN2  0.000  0.000  0.000 -0.577 -0.577  0.000  0.000  0.407
0.326
fEXPN2:PBN3  0.000  0.000  0.000  0.000  0.000  0.000  0.000  0.000
0.000       0.000
fEXPN3:PBN3  0.000  0.000 -0.706  0.000  0.000  0.000  0.000  0.000
0.412       0.000
fEXPN4:PBN3  0.000  0.000  0.000 -0.706  0.000  0.000  0.000  0.000
0.000       0.408
fEXPN2:PBN4  0.000  0.000  0.000  0.000  0.000  0.000  0.000  0.000
0.000       0.000
fEXPN3:PBN4  0.000  0.000 -0.703  0.000  0.000  0.000  0.000  0.000
0.410       0.000
fEXPN4:PBN4  0.000  0.000  0.000 -0.703  0.000  0.000  0.000  0.000
0.000       0.406
            fEXPN2:PBN3 fEXPN3:PBN3 fEXPN4:PBN3 fEXPN2:PBN4 fEXPN3:PBN4
fEXPN2
fEXPN3
fEXPN4
fPBN2
fPBN3
fPBN4
fEXPN2:PBN2
fEXPN3:PBN2
fEXPN4:PBN2
fEXPN2:PBN3
fEXPN3:PBN3  0.000
fEXPN4:PBN3  0.000       0.000
fEXPN2:PBN4  0.000       0.000       0.000
fEXPN3:PBN4  0.000       0.496       0.000       0.000
fEXPN4:PBN4  0.000       0.000       0.496       0.000       0.000

So here I had no F statistic at all, and the results of the fixed effects
don't make sense, when compared against the actual data (where the pattern
between PBN (playback type) and EXPN (experiment) are quite stark.

Because this didn't make sense to me, and because I am not a whiz at R, I
ran the same model in SPSS 20.0, and that is where I got the huge F
statistic. Here is that model summary:

 F = 1, 229, 656.019, df1 = 15, df2=220, P<0.0001
EXPN F = 83.83, df1=3, df2=220, P<0.0001
PBN F=2,2280.08, df1=3, df2=220, P<0.0001
EXPNxPNB F=82.96, df1=9, df2=220, P<0.0001

To test if this result was due to underdispersion of my response variable I
tried tweaking the data so that the variance was not zero.  The model then
ran fine (in SPSS).

Thanks,

Tonya

On Fri, Aug 10, 2012 at 8:44 AM, ONKELINX, Thierry <Thierry.ONKELINX@inbo.be
> wrote:

> Dear Tonya,
>
> Can you give us the output of the model summary?
>
> Best regards,
>
> ir. Thierry Onkelinx
> Instituut voor natuur- en bosonderzoek / Research Institute for Nature and
> Forest
> team Biometrie & Kwaliteitszorg / team Biometrics & Quality Assurance
> Kliniekstraat 25
> 1070 Anderlecht
> Belgium
> + 32 2 525 02 51
> + 32 54 43 61 85
> Thierry.Onkelinx@inbo.be
> www.inbo.be
>
> To call in the statistician after the experiment is done may be no more
> than asking him to perform a post-mortem examination: he may be able to say
> what the experiment died of.
> ~ Sir Ronald Aylmer Fisher
>
> The plural of anecdote is not data.
> ~ Roger Brinner
>
> The combination of some data and an aching desire for an answer does not
> ensure that a reasonable answer can be extracted from a given body of data.
> ~ John Tukey
>
>
> -----Oorspronkelijk bericht-----
> Van: r-sig-mixed-models-bounces@r-project.org [mailto:
> r-sig-mixed-models-bounces@r-project.org] Namens Tonya Haff
> Verzonden: donderdag 9 augustus 2012 18:58
> Aan: r-sig-mixed-models@r-project.org
> Onderwerp: [R-sig-ME] Response variable with zero variance
>
> Hi all,
>
> Ideally, I would like to run a GLMM on my data, which are are binomial (so
> I'm trying to run a model with a binomial distribution and a logit link).
>  I have two fixed effects, fledgling age and playback type.  I have
> fledging ID as a random effect.  The problem I'm running into is zero
> variance in my response variable - ie fledglings always respond the same
> way to some of the playback types.  I can run a model and it converges, but
> it spits out an F value of over a million. So I'm wondering (1) is it
> possible to use a GLMM with a zero variance response; and (2) if so, which
> R package is the most appropriate?
>
>
> Thanks for your help!
>
> Cheers,
>
> Tonya
> --
> Tonya Haff
> PhD candidate
> Evolution, Ecology and Genetics
> Research School of Biology
> Australian National University
> Mobile:+61-4-3331-2908
> Lab: +61-2-6125-5651
>
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