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The goal is to find if one factor is stronger than the other in the models I have considered. I am using the information-theoretic approach. Since $n/K>40$, I am using AIC. Firstly the model is standardized using the arm package so that input variables are on the same scale.

s1 <- glm(R.1~SU+W, data = s, 
 family = tweedie(var.power = 1.05, link.power = 0))
display(s1)

standardizing input variables.

sm <- standardize(s1)
display(sm)

Then, candidate models are obtained after dredging (using the dredge() function) the global model and are selected based on delta AIC > 2 criterion.

#dredging the global model
model.set<-dredge(sm, rank = AICtweedie)
model.set

#Candidate model set
top.model<-get.models(model.set, subset=delta<2)
top.model

However, this also leads to spurious variables in the candidate model set as AIC only penalizes by 2 units. Thus I went for model averaging. Here I used the zero-based (shrinkage) method rather than the natural average method. The difficulty is ahead.

When I average the model, I get a summary table. This has the standard error (SE) and adjusted standard error (Adjusted SE).

Model averaging using the zero method.

u<-summary(model.avg(top.model), revised.var = T)
u

Call:
model.avg(object = top.model)

Component model call: 
glm(formula = R.1 ~ <2 unique rhs>, family = tweedie(var.power = 1.05, 
     link.power = 0), data = s)

Component models: 
   df logLik AICtweedie delta weight
1   2           -825.17  0.00   0.66
12  3           -823.87  1.29   0.34

Term codes: 
c.SU  z.W 
   1    2 

Model-averaged coefficients:  
(full average) 
            Estimate Std. Error Adjusted SE z value Pr(>|z|)    
(Intercept) -2.11780    0.02390     0.02396  88.398   <2e-16 ***
c.SU         0.51066    0.06334     0.06348   8.044   <2e-16 ***
z.W          0.01609    0.03527     0.03532   0.456    0.649    



(conditional average) 
            Estimate Std. Error Adjusted SE z value Pr(>|z|)    
(Intercept) -2.11780    0.02390     0.02396  88.398   <2e-16 ***
c.SU         0.51066    0.06334     0.06348   8.044   <2e-16 ***
z.W          0.04680    0.04670     0.04681   1.000    0.317    
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Relative variable importance: 
                     c.SU z.W 
Importance:          1.00 0.34
N containing models:    2    1

According to Grueber et al 2011, we have to report the standardized effect sizes and also the unconditional standard error. Can I report the SE values (I will use the full model values only) for the parameters from this summary table? Are these the same effect sizes? How do I obtain the unconditional standard error?

Because I could not find the upper and lower confidence interval and the unconditional standard error using par.avg(). I tried obtaining the confidence intervals using the conf.int() function, but this still does not give the unconditional SE.

#calculating the confidence intervals
conf.int(u)

> confint(u)
                  2.5 %     97.5 %
(Intercept) -2.16475590 -2.0708437
c.SU         0.38623184  0.6350796
z.W         -0.04494773  0.1385478

Further is it right to infer that since the confidence interval for factor W includes zero, hence it does not do not have an effect on the response variable while the factor SU has?

In short, I want to ask, how to find the unconditional standard error? How to create the vector of parameters if we use the par.avg function? Are the calculated confidence intervals correct? Is the conclusion based on the CI correct?

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1 Answer 1

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  1. How to find the unconditional standard error? Unconditional = Full
  2. How to create the vector of parameters if we use the par.avg function? I'm not sure what you're asking, the parameters are right there in the output. Do you want to know how to extract them in R?
  3. Are the calculated confidence intervals correct? Is the conclusion based on the CI correct? We discuss the theory of creating confidence intervals for a model average in Dormann, Carsten F., et al. "Model averaging in ecology: a review of Bayesian, information‐theoretic, and tactical approaches for predictive inference." Ecological Monographs 88.4 (2018): 485-504. This is for predictive averages, for parameter averaging there is the additional problem of collinearity. In short - these intervals are at best an approximation, but it could be a pretty bad one.
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  • $\begingroup$ The review is worth thinking over. My doubt stems from the concern that the package "MuMIn" does not mention the unconditional standard error in its output of the summary. Hence will the adjusted standard error be the same as unconditional one? Or we go for the AICmodavg package. $\endgroup$
    – Harshad
    Mar 31, 2019 at 18:44
  • $\begingroup$ As I said, unconditional = full $\endgroup$ Apr 5, 2019 at 21:19
  • $\begingroup$ @FlorianHartig By this you mean, confint(u, full=TRUE) I presume? $\endgroup$ Nov 28, 2019 at 11:41

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