Dear all, 

I used "sem" function from the package SEM to fit a model. However, I cannot say if the model is correspondent to the data or not (chisquare test).
I used the commands:

model1 <- specifyModel()
estadio -> compflora, a1, NA
estadio -> compfauna, a2, NA
estadio -> interacoesobs, a3, NA
compflora -> compfauna, b1, NA
compflora -> interacoesobs, b2, NA
compfauna -> interacoesobs, c1, NA
estadio <-> estadio, e1, NA
compflora <-> compflora, e2, NA
compfauna <-> compfauna, e3, NA
interacoesobs <-> interacoesobs, e4, NA

sem1 <- sem(model1, cov.matrix, length(samples))
summary(sem1)

and I got the result:

Model Chisquare =  -2.873188e-13   Df =  0 Pr(>Chisq) = NA AIC =  20 BIC =  -2.873188e-13 Normalized Residuals Min.   1st Qu.    Median      Mean   3rd Qu.      Max. 
0.000e+00 0.000e+00 2.957e-16 3.193e-16 5.044e-16 8.141e-16  R-square for Endogenous Variables compflora     compfauna interacoesobs  0.0657        0.1056        0.2319  Parameter Estimates Estimate     Std Error    z value    Pr(>|z|)                                     
a1 3.027344e-01 1.665395e-01 1.81779316 6.909575e-02 compflora <--- estadio          
a2 2.189427e-01 1.767404e-01 1.23878105 2.154266e-01 compfauna <--- estadio          
a3 7.314192e-03 1.063613e-01 0.06876742 9.451748e-01 interacoesobs <--- estadio      
b1 2.422906e-01 1.496290e-01 1.61927587 1.053879e-01 compfauna <--- compflora        
b2 3.029933e-01 9.104901e-02 3.32780446 8.753328e-04 interacoesobs <--- compflora    
c1 4.863368e-02 8.638177e-02 0.56300857 5.734290e-01 interacoesobs <--- compfauna    
e1 6.918133e+04 1.427102e+04 4.84767986 1.249138e-06 estadio <--> estadio            
e2 9.018230e+04 1.860319e+04 4.84767986 1.249138e-06 compflora <--> compflora        
e3 9.489661e+04 1.957568e+04 4.84767986 1.249138e-06 compfauna <--> compfauna        
e4 3.328072e+04 6.865289e+03 4.84767986 1.249138e-06 interacoesobs <--> interacoesobs Iterations =  0 

I understand the results, but I do not know how to interpret the first line that tells me about the model:
Model Chisquare =  -2.873188e-13   Df =  0 Pr(>Chisq) = NA

How can DF be zero, if the number of observations I used in sem funcition was 48 and I have only 4 variables? What is the p value?

Thanks in advance.
Bernardo Niebuhr
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