R help - Jul 2012 - Predicted values when using offset in ZIP GLM

Hi all!

I have built a model to predict interactions with turtles and the model includes
an offset for effort:

ZIP<-zeroinfl(Sturgeon~fMesh+fSeason+offset(LogEffort),dist="poisson",link="logit",data=data)

I wasn't clear about one aspect of the response to a similar question I
recently posted...I apply the predicted model to a new dataset of standard
conditions and take the sum to get the total predicted numbers of bycatch:

Prediction<-predict(ZIP,newdata=effort,type="response")
sum(Prediction)

Do I first need to exponentiate before taking the sum as in:

Prediction<-exp(predict(ZIP,newdata=effort,type="response"))
sum(Prediction)

Thanks again in advance!

Cheers,

Laura




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On Thu, 12 Jul 2012, Lee, Laura wrote:
> Hi all!
>
> I have built a model to predict interactions with turtles and the model 
> includes an offset for effort:
>
>
ZIP<-zeroinfl(Sturgeon~fMesh+fSeason+offset(LogEffort),dist="poisson",link="logit",data=data)
Note that this includes the offset both in the count component and in the 
zero inflation component. The latter is probably not desired. (As Alain 
already previously pointed out in one of hise e-mails.)

For specifying the offset only in the count component you can do either 
one of

Sturgeon ~ fMesh + fSeason + offset(LogEffort) | fMesh + fSeason

or

Sturgeon ~ fMesh + fSeason, offset = LogEffort

However, it may be useful to include LogEffort as a regular variable (not 
an offset) in the inflation component as well...
> I wasn't clear about one aspect of the response to a similar question I
> recently posted...I apply the predicted model to a new dataset of 
> standard conditions and take the sum to get the total predicted numbers 
> of bycatch:
>
> Prediction<-predict(ZIP,newdata=effort,type="response")
> sum(Prediction)
>
> Do I first need to exponentiate before taking the sum as in:
>
> Prediction<-exp(predict(ZIP,newdata=effort,type="response"))
> sum(Prediction)
No. predict(..., type = "response") computes the expected mean of the 
response (as previously pointed out here and as clearly stated in the 
accompanying paper).
Z
> Thanks again in advance!
>
> Cheers,
>
> Laura
>
>
>
>
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>
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