I wanted to compare $2$ models and I did anova then get this outcome. I need a p-value but I got this. I cannot figure it out how I need to interpret Values of Chisq ($1.1215$) and Pr(>Chisq) ($0.5708$)

         npar   AIC     BIC logLik deviance  Chisq Df Pr(>Chisq)
Modell_2   12 -1918.6 -1857.1 971.31  -1942.6                     
Modell_1   14 -1915.8 -1843.9 971.88  -1943.8 1.1215  2     0.5708

Could you help me to interpret these values so that I can get how can I know from this table whether the difference between Modell 2 and Modell 1 significant is.


1 Answer 1


When you do this kind of analysis you should pick a significance level $\alpha$ that will help you decide wether to reject the null hypothesis (models have same variance, etc.). Generally this value is set to 0.05 and you will reject the null hypothesis if p < $\alpha$. In this case $0.57 >> 0.05$ so I would conclude that there is no statistically significant difference between the two.

  • $\begingroup$ Thanks a lot! :) 1. So I can always interpret Pr(>Chisq) as a p-value? /////// 2. Could you please why I got Pr(>Chisq) not p-value, although I did write in console "anova (Model_1, Model_2)" /////// 3. what is the difference Chisq vs. Pr(>Chisq)?? $\endgroup$
    – Taede17
    Jul 10, 2021 at 9:36
  • $\begingroup$ All these tests work in a similar way: they compute some sort of value (statistic) that was demonstrated to behave according to a given distribution. Here this statistic is Chisq. Then you check what's the probability of having values of Chisq at least as extreme as the one you found. That's exactly Pr > Chisq and that's by definition the p-value. Now the intuition is: if it is quite common to find extreme results, then H0 is true; if it is uncommon (p-value very small) I should reject H0. $\endgroup$
    – rusiano
    Jul 10, 2021 at 16:35
  • $\begingroup$ In this test, the statistic that is computed has a chi-square distribution with 2 DoF. So you simpy compute the value, which is 1.1215. Now, with the density function of this distribution (which is known), you can compute the probability of obtaining results at least as extreme as the value you just computed, assuming null hypothesis holds true. That's Pr > Chisq, equal to 0.5708. This means that the probability of finding such extreme results as the ones you observed is pretty common (57%) and you "should not worry". If it was lower than, say, 5% (0.05), then you'd be supposed to reject H0. $\endgroup$
    – rusiano
    Jul 10, 2021 at 16:38
  • $\begingroup$ Thank you so much for your explanation!! Then as you said, Pr(>Chisq) value is bigger than 0.05 so it is clear through AIC and BIC-Test that Model 2 is fitter than Model 1 but this is not significantly fitter? Am I right? $\endgroup$
    – Taede17
    Jul 10, 2021 at 21:30
  • 1
    $\begingroup$ Yeah. It is basically saying that model1 has more parameters and the tradeoff between model complexity and goodness of fit does not pay off. The decrease of deviance of the residuals is not significant (p-value >> 0.05), plus AIC is higher. $\endgroup$
    – rusiano
    Jul 12, 2021 at 6:43

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