The likelihood ratio test of comparing reduced model with full model differs by fixed factor result to chi-square distribution of zero degree of freedom.
/* reduced model */
proc mixed method = ml;
class block gen;
model rtwt = /ddfm = kr;
random block gen;
run;
/* full model*/
proc mixed method = ml;
class block gen;
model rtwt = prop_hav/ddfm = kr;
random block gen;
run;
There are 3 degree of freedom from reduced model - block variance, genotype variance, and residual variance. The same degree of freedom for full model with includes prop_hav as covariate. The difference in their -2 loglikelihood has zero degree of freedom under chi-distribution. Please could anyone guide me on how to compare these model to ascertain if the full modelis significantly different from reduced model.
LRT only works when one model is a simplification of another. You can’t take out terms and add in with one go, do it in two stages.
The models to be compared must be nested
Please have a look at this paper:
https://www.researchgate.net/publication/299598766_Testing_non-nested_regression_models_Distinguishing_additive_from_multiplicative_statistical_interactions
That is a case where you can use AIC and BIC to 'compare' models. It will give you an idea of which one is best, but it will NOT indicate if they are significantly different.
Salvador is correct - but the difference in AIC between models can be converted to a posterior model probability. First the difference can be converted to an odd scale:
e^(0.5 x D)
where D is the difference in AIC
Then odds are converted to probabilities. For comparing two models:
p = o / (o+1)
Thus D = 5 is equivalent to odds of 12.2 favouring one model over the other or a posterior model probability of .92 (92%).
I find this scaling to be more intuitive and easier to interpret than a p value. You can also easily extend the logic to multiple models.
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