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I don't know if this belongs here or in StackExchange, it is a mixed but probably pretty simple question. How do I normally report a Likelihood Ratio Test? I would love a good reference in your answer, I have searched but could not find any good answers.

> glmm0 <- glmer(yngel ~ (1|nest), data, family=poisson(link="log"))
> glmm <- glmer(yngel ~ age.level + (1|nest) + 0, data, family=poisson(link="log"))
> anova(glmm0, glmm)
Data: data
Models:
glmm0: yngel ~ (1 | nest)
glmm: yngel ~ age.level + (1 | nest) + 0
      Df    AIC    BIC  logLik deviance  Chisq Chi Df Pr(>Chisq)    
glmm0  2 682.33 689.38 -339.16   678.33                             
glmm   3 672.37 682.95 -333.18   666.37 11.959      1   0.000544 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

My best guess so far is: I used likelihood ratio test to compare the model with the fixed effect to a model without it. The model including the fixed effect (age-level) was a better fit ($\chi^2(df=?)=11.96$, $p=0.00054$).

And I cannot actually figure out how many df to report from that. The there is 2 for one model and 3 for the other, and 1 between them.

Thank you for your help.

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    $\begingroup$ Any guesses on what the "Chi Df" column is about? $\endgroup$
    – John
    Feb 13, 2014 at 21:13
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    $\begingroup$ "How do I report a likelihood ratio test?" belongs here (though I am not sure where else you meant - here is part of stackexchange; were you referring to stackoverflow perhaps?). The df for the chi-square test is the "1" in the table that's right beside the p-value. $\endgroup$
    – Glen_b
    Feb 13, 2014 at 21:57

1 Answer 1

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For a likelihood ratio test, the degrees of freedom are equal to the difference in number of parameters for the two models. In this case, df = 1, and so $\chi^2(1)=11.96$, $p=0.0005$.

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    $\begingroup$ Thanks (+1). If I perform (for consitency) a LRT on linear models (lm) where do I find the χ2 value? Is it the value given under Sum of Sq? $\endgroup$
    – bee guy
    Dec 22, 2017 at 8:29
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    $\begingroup$ You can compare two linear models using the anova() function, so anova(mod1, mod2). The X^2 value is under the "Chisq" column, much like the output in the original question $\endgroup$
    – dmartin
    Dec 23, 2017 at 17:39
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    $\begingroup$ Hi dmartin, I don't obtain a X^2 value when I run anova(mod1, mod2) on linear models (without random effects). This returns an F-statistic. Even if I run anova(mod1, mod2, test = "Chisq"), I don't see a column Chisq, but a column Sum of sq. $\endgroup$
    – bee guy
    Mar 12, 2018 at 8:40
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    $\begingroup$ Is the absolute value of this the difference in sum of squares the X^2 value? $\endgroup$
    – bee guy
    Mar 12, 2018 at 8:46
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    $\begingroup$ I've already tried to use the function glm to fit a lm (with gaussian error distribution). It returns deviance, which corresponds however to the sum of squares not to chisq values. Thanks for referring me to lrtest. :) $\endgroup$
    – bee guy
    Mar 13, 2018 at 8:25

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