-2LL comparison of two nested models PreviousNext
Mplus Discussion > Multilevel Data/Complex Sample >
 Eva van de Weijer posted on Monday, September 23, 2013 - 7:13 am
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?

kind regards, Eva
 Bengt O. Muthen posted on Monday, September 23, 2013 - 3:54 pm
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

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 - 6:58 am
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!
 Bengt O. Muthen posted on Friday, September 18, 2015 - 5:56 pm
Yes, as long as the models have the same random effect variances.
 fangfang posted on Friday, September 18, 2015 - 6:14 pm
Many thanks to you! Dr.Muthen
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