Analysis of Deviance Table

Model 1: Exercise ~ 1
Model 2: Exercise ~ WaketimeStand
  Resid. Df Resid. Dev Df Deviance Pr(>Chi)  
1       115     124.55                       
2       114     121.51  1   3.0351  0.08148 .

Simple question just to confirm - is the deviance statistic in an Analysis of Deviance table the Chi squared value for the difference between the models?

To make things clearer I add this the code I wrote to get this output;

anova(NullEx,ModEx7,test = "Chi")


1 Answer 1


$$H_0: \text{The additional parameters in the more complex model are 0.}$$

$$H_a: H_0\text{ is false.} $$

Under this setup, the test statistic has an asymptotic $\chi^2$ distribution, yes.

  • $\begingroup$ Would it therefore be correct to state Waketime standardised is non-significant (X2 = 3.0351, p =0.08148) $\endgroup$
    – chris1
    Commented Nov 23, 2020 at 11:58
  • $\begingroup$ @chris1 It looks like the difference between models is just one parameter, which I suppose is Waketime. Your test is giving you a p-value of $0.08$-ish, which is not significant at the $0.05$-level. $\endgroup$
    – Dave
    Commented Nov 23, 2020 at 12:10

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