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I have my2pl as ltm data and my3pl as tpm data. 

anova(my2pl,my3pl)  
Likelihood Ratio Table  
AIC      BIC  log.Lik   LRT df p.value  
my2pl 16317.52 16630.39 -8078.76                   
my3pl 16299.71 16769.00 -8029.85 97.81 40  <0.001

Shall I use ltm or tpm model based on this anova?

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  • $\begingroup$ my3pl should be selected $\endgroup$
    – Chris Kelly
    Commented Apr 9, 2019 at 7:03

1 Answer 1

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Based purely on the likelihood-ratio statistic, without looking at your data and ignoring any theoretical basis for the model structure, the tmp model (my3pl) offers a significantly better fit to the data than the ltm model (my2pl). Thus, you could use this to justify using tmp over ltm.

Essentially, the LRT takes the log likelihood (log.Lik), a measure of goodness of fit, of the ltm model and compares it with the log likelihood for the tmp model. The 97.81 Chi-square value indicates that the tmp model is a better fit to the data than the ltm model.

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  • $\begingroup$ just as i know, lower AIC,BIC means better fit.Then, i am confused. $\endgroup$
    – kittygirl
    Commented May 30, 2017 at 15:46
  • $\begingroup$ AIC is a criterion for model selection, and so is BIC. BIC penalizes the model complexity more than the AIC, and so BIC tends to favour simpler models, whereas AIC tends to favour the more complex models. In both cases, the closer the value to negative infinity the better your model approximates the underlying data. Additionally, both provide the means for model selection, but depending on the field that you come from, you may choose one over another. In contrast to these, the LRT provides a test of comparison telling you whether some additional features of the model improve the fit. $\endgroup$
    – user139190
    Commented May 30, 2017 at 16:48

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