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 chi-sqaure 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?
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
chi-square = -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£¿