

2LL comparison of two nested models 

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Hi, I am comparing two nested models using the H0 value. If I understand correctly, I have to multiply this values by 2 to get the 2LL. When I do this I get the next results: model 1: df 10, 2LL = 41738 model 2: df 9, 2LL = 41782 Should I interpret this as model 1 having a better fit? I always thought that losing a degree of freedom would result in decrease in chisqaure or 2LL? In another model comparison the next results are displayed: model 1: df 4, 2LL = 4010 model 2: df 3, 2LL = 3783 In this case, which is the smaller value (the 4010 or the 3783). In other words, should I ignore the minus sign or not? Also in this latest comparison, model 1 gives a scaling correction factor, but model 2 does not. Do I need to use this in any way? kind regards, Eva 


In your first example you have model 1 LL = 41738/2 model 2 LL = 41782/2 The LL value for model 1 is better (is higher  i.e. it is closer to zero). This is a strange outcome when model 1 has one less parameter (one more df) but only if the two models are nested; otherwise not. In your second example, model 2 has the better LL and things are as expected. You compute chisquare = 2*(LL_a  LL_b), where LL_a is for the model with higher df. You should use H0 scaling factors if they are given. 

fangfang posted on Friday, September 18, 2015  12:58 pm



Dear Dr. Muthen, I am comparing the hypothesized structural model with several alternative models. All of these models are random slope models.The only fit indices available are LL,AIC and BIC. Could I use the method you mentioned above£¿ Thank you in advance! 


Yes, as long as the models have the same random effect variances. 

fangfang posted on Saturday, September 19, 2015  12:14 am



Many thanks to you! Dr.Muthen 

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