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-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 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 |
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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. |
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fangfang posted on Friday, September 18, 2015 - 12:58 pm
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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! |
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Yes, as long as the models have the same random effect variances. |
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fangfang posted on Saturday, September 19, 2015 - 12:14 am
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Many thanks to you! Dr.Muthen |
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