[R-sig-ME] Posthoc for the glmmTMB package

[Ben Bolker]

  I think the easiest way to do this is with lsmeans

https://stats.stackexchange.com/questions/145765/post-hoc-testing-in-multcompglht-for-mixed-effects-models-lme4-with-interact

  You *should* be able to use lsmeans with glmmTMB objects after running
 source(system.file("other_methods","lsmeans_methods.R",package="glmmTMB"))

  I believe this is all operating/making comparisons based on the
conditional model, not the zero-inflation model ...

On 17-07-18 10:08 AM, Ikponmwosa Egbon wrote:
> Hello All,
> 
> Please, I am a novice to 'glm with mixed effects (glmm)' and need the
> guidance of mixed-model experts on how to conduct a posthoc test after
> using the glmmTMB (http://www.biorxiv.org/content/biorxiv/early/
> 2017/05/01/132753.full.pdf) for a zero-inflated (Poisson) model for a
> count data with repeated measures (over different times, hence time was
> built in as a random effect).
> 
> Although I have run the model, I could not separate the different levels
> (or treatments) within a factor (Genotypes) to know which is similar or
> different, as often seen in the traditional ANOVAs or linear models,
> wherein posthoc family-wise comparisons are usually conducted. Or perhaps
> there are things I am not seeing with the novice spectacles.
> 
> *Please, see the script/output for statistical context below*:
> 
>> multiple<-read.delim("Multiplechoice.txt")
>> str(multiple)
> 'data.frame': 1440 obs. of  3 variables:
>  $ Genotypes: Factor w/ 8 levels "AR3","BR6","BR7",..: 2 2 2 2 2 2 2 2 2 2
> ...
>  $ Time     : Factor w/ 18 levels "10m","15m","20m",..: 16 16 16 16 16 16
> 16 16 16 16 ...
>  $ Insects  : int  2 0 0 0 1 0 0 0 0 0 ...
>> head(multiple)
>   Genotypes Time Insects
> 1       BR6   5m       2
> 2       BR6   5m       0
> 3       BR6   5m       0
> 4       BR6   5m       0
> 5       BR6   5m       1
> 6       BR6   5m       0
>> library("glmmTMB")
>> zipm0 <- glmmTMB(Insects~Genotypes + (1 | Time),
> +                  zi = ~Genotypes,
> +                  data = multiple, family = poisson)
>> summary(zipm0)
>  Family: poisson  ( log )
> Formula:          Insects ~ Genotypes + (1 | Time)
> Zero inflation:           ~Genotypes
> Data: multiple
> 
>      AIC      BIC   logLik deviance df.resid
>   2363.3   2453.0  -1164.7   2329.3     1423
> 
> Random effects:
> 
> Conditional model:
>  Groups Name        Variance Std.Dev.
>  Time   (Intercept) 0.7921   0.89
> Number of obs: 1440, groups:  Time, 18
> 
> Conditional model:
>                         Estimate Std. Error z value Pr(>|z|)
> (Intercept)             -0.05931    0.23378  -0.254 0.799747
> GenotypesBR6            -0.09546    0.13452  -0.710 0.477939
> GenotypesBR7            -1.02963    0.19379  -5.313 1.08e-07 ***
> GenotypesDR3            -0.34788    0.17393  -2.000 0.045491 *
> GenotypesOut group      -0.21968    0.55045  -0.399 0.689827
> GenotypesP. grandifolia -1.64415    0.40947  -4.015 5.94e-05 ***
> GenotypesSA1            -0.57111    0.16067  -3.555 0.000378 ***
> GenotypesVZ2            -0.43942    0.15825  -2.777 0.005492 **
> ---
> Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
> 
> Zero-inflation model:
>                         Estimate Std. Error z value Pr(>|z|)
> (Intercept)              -0.7956     0.2453  -3.244  0.00118 **
> GenotypesBR6             -0.7542     0.5047  -1.494  0.13509
> GenotypesBR7             -2.5021     3.5578  -0.703  0.48189
> GenotypesDR3              1.5065     0.3695   4.077 4.55e-05 ***
> GenotypesOut group        2.8023     0.4860   5.766 8.14e-09 ***
> GenotypesP. grandifolia   1.4009     0.6815   2.056  0.03983 *
> GenotypesSA1             -0.5077     0.5662  -0.897  0.36987
> GenotypesVZ2             -0.3105     0.4580  -0.678  0.49779
> ---
> Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
> 
> 
> I look forward to having some feedbacks, and any other assistance that is
> deemed necessary would be highly appreciated. Thank you for your time and
> your assistance.
> 
> 
> Kind
> regards,
> 
> 
> Ik.
>