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Likelihood ratio test between nested ... |
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Jinseok Kim posted on Sunday, November 28, 2010 - 12:38 am
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Hi, I am conducting an LCA using a set of categorical(ordinal) indicators. I am trying to impose a few constraints in the original model (M0) and to test whether the model with the constraints (M1) should be selected. In the following, I presented the outputs from M0 & M1 and wonder if you could direct me regarding which numbers should I use to determine likelihood ratio chi-square value for the test. Also, do I have to consider scaling correction factor in the test? Thanks. Jinseok M0 (no constraints): TESTS OF MODEL FIT Loglikelihood H0 Value -11384.819 H0 Scaling Correction Factor 1.291 Information Criteria Number of Free Parameters 62 ** omitted *** Pearson Chi-Square Value 15932.368 Degrees of Freedom 268435144 P-Value 1.0000 Likelihood Ratio Chi-Square 3892.824 Degrees of Freedom 268435144 P-Value 1.0000 M1 (with constraints): TESTS OF MODEL FIT Loglikelihood H0 Value -11464.515 H0 Scaling Correction Factor 1.320 Information Criteria Number of Free Parameters 59 ** omitted ** Pearson Chi-Squar Value 16608.856 Degrees of Freedom 268435137 P-Value 1.0000 Likelihood Ratio Chi-Square 3940.894 Degrees of Freedom 268435137 P-Value 1.0000 |
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You use the H0 loglikelihood values and their scaling correction factors as described on the website under Chi-square difference test for MLM and MLR. |
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“If the GMM model gives a considerably better log likelihood value for fewer ... parameters than the LCGA, GMM should clearly be chosen over LCGA” (Muthén 2006). I am using http://www.statmodel.com/chidiff.shtml to look into this. Please can I check my understanding of the following: 1. Using the formulas in the link, the following output can be compared by: GMM (2 classes) Number of Free Parameters 11 Loglikelihood H0 Value -537782.542 H0 Scaling Correction Factor 1.3080 for MLR LCGA (4 classes) Number of Free Parameters 14 Loglikelihood H0 Value -544635.032 H0 Scaling Correction Factor 1.5112 for MLR cd = ((11*1.3080) - (14*1.5112)) / (11-14) = 2.256 df = 14 - 11 = 3 TRd = -2*(-537782.542 + 544635.032) / 2.256 = -6074.184 2. The resulting TRd and df can then be compared to a Chi-squared significance table? 3. But in this case, the Chi-squared is negative, so cannot be interpreted? 4. If so, is it sufficient to say that the GMM has a higher log-likelihood value without employing a Chi-squared difference test? Do you have any alternative suggestions? |
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The models are not nested and the likelihood ratio test is not applicable. We recommend using the BIC for comparing such non-nested models (smaller BIC is better). |
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