[R-sig-ME] Advice on GLMM with verIdent
Thierry Onkelinx
thierry.onkelinx at inbo.be
Tue Dec 20 22:03:43 CET 2016
Dear Diego,
The linear trend is required otherwise the optimum of the parabola is fixed
a x = 0.
You could try to fit the phylogeny effect as a random effect instead of a
fixed effect. poly(Period, 2) | Phyloge ny and add species as a nested
random intercept.
If you have the individual latitudes you can try to use those instead of
using only their average.
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
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
2016-12-20 17:49 GMT+01:00 Diego Pavon <diego.pavonjordan op gmail.com>:
> Dear all,
>
> I write you because I need some advice about the model I want to fit to my
> data, which I suspect it is not too 'correct'...
>
> My response variable is continuous (mean weighted latitude) of 24 species.
> I have these mean latitudes calculated for 4 periods (95-99, 00-04, 05-09,
> 10-13), for each species. Therefore I have 96 observations. I want to see
> if there is a trend of the mean latitude over time (indicating a shift in
> the population... all the species together). In this case, I would use
> Period as a continuous varaible and not a factor. I want to fit a GLMM with
> random = SpeciesID. However, data exploration revealed a quadratic
> relationship between my response (mean latitude) and Period (my continuous
> covariate for time). But also, I am interested also to see whether there
> are differences between functional groups (some species may move faster
> than others). Thus I include the variable Phylogeny, which is a factor with
> 5 levels. Thus, the model in nlme notation is:
>
> lme(NEnessKM ~ Period_std + I(Period_std^2) + factor(Phylogeny)
> + Period_std : factor(Phylogeny) + I(Period_std^2) :
> factor(Phylogeny),
> random = ~1| factor(Species),
> weights = varIdent(form =~ 1 | factor(Species)),
> control = ctrl2,
> method = "REML",
> data = NEness5y)
>
>
> My concern is that this model might be too complicated. I have only 96
> observation. If I follow the rule of 15 observation per parameter
> estimated, this model i way to complex. I understand that in order to
> include quadratic terms, one must include also the linear effect... and
> that would also apply for the interactions. If the relationship between my
> response (NEness = Mean Latitude) and Period is a parabola, I guess I
> should include also the Period^2 in the interaction (Period^2 * Phylogeny)?
> But in that case, I have to include the interaction of Period * Phylogany?
>
> Is there a way to reduce the complexity? Is it totally wrong if I do not
> include the linear effects and keep only Period^2?
>
> Thanks for sharing your knowledge and for the advice.
>
> Best
>
> Diego
>
>
>
>
> --
> *Diego Pavón Jordán*
>
> *Finnish Museum of Natural History*
> *PO BOX 17 *
>
> *Helsinki. Finland*
>
> *0445061210https://www.researchgate.net/profile/Diego_Pavon-jordan
> <https://www.researchgate.net/profile/Diego_Pavon-jordan>*
>
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>
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