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I am doing a model selection starting from the follwing full model (cma):

vf1<-varIdent(form=~1|Region)

cf<-formula(rrG~plotsize+alt+baseT+baseP+baseN+EIV_R+EIV_F+Overstory+Overstory_diff+SCA+SCA_diff)

cma<- lme(cf,data=dat,random=~1|Region,weights=vf1,control=ctrl,na.action=na.exclude,method="ML")

Estimates are the following:

Fixed effects: list(cf) 
                      Value   Std.Error   DF   t-value p-value
(Intercept)    -0.007018325 0.003443949 1561 -2.037871  0.0417
plotsize        0.009858282 0.002483949 1561  3.968794  0.0001
alt            -0.000319573 0.002196082 1561 -0.145520  0.8843
baseT          -0.001808871 0.003164477   33 -0.571618  0.5715
baseP           0.000613365 0.003659040   33  0.167630  0.8679
baseN          -0.000388065 0.003721218   33 -0.104284  0.9176
EIV_R           0.006280127 0.000986477 1561  6.366215  0.0000
EIV_F          -0.000965895 0.000704349 1561 -1.371331  0.1705
Overstory      -0.000144179 0.001148155 1561 -0.125574  0.9001
Overstory_diff -0.002515170 0.000999187 1561 -2.517217  0.0119
SCA            -0.001903248 0.001116551 1561 -1.704577  0.0885
SCA_diff       -0.002666183 0.000763426 1561 -3.492391  0.0005

I remove the least significant term (baseN) and then compare the models:

cma1<-update(cma,~. -baseN)
anova(cma,cma1)

Which shows the following:

     Model df       AIC       BIC   logLik   Test    L.Ratio p-value
cma      1 50 -6873.897 -6604.822 3486.948                          
cma1     2 49 -6875.886 -6612.192 3486.943 1 vs 2 0.01090057  0.9168

I should I interpret that? The new model as lower AIC and BIC, so I should go for that one, however p-value is >0.05 so the data used in the models do not makes the two models significantly different, right? Does it mean that my model selection stops here and select the full model as the best one?

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If I am reading this correctly, both methods actually gave you the same result.

As you say, AIC tells you that your second (reduced) model is better (as judged by AIC).

The likelihood ratio test (using the anova() function) also tells you that the variable you dropped is not significant (p-value = 0.9168), and can therefore indeed be removed from the model. In other words, the likelihood ratio test tells you that model with baseN is not significantly better than the one without it.

So both methods gave you the same answer.

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  • $\begingroup$ BIC also would lead to the same conclusion. $\endgroup$ – Michael R. Chernick Mar 5 '17 at 18:08
  • $\begingroup$ Thanks. I got confused about the interpretation of p-value. I thought that the two statistics were giving opposite results $\endgroup$ – Gabriele Midolo Mar 5 '17 at 19:53
  • $\begingroup$ @GabrieleMidolo, yes it's easy to get confused about what the output is telling you, I was there too. :) If you found the answer useful and it solved your problem, you may consider accepting it. $\endgroup$ – Tilen Mar 6 '17 at 10:34

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