# Is the deviance statistic in an Analysis of Deviance table a chi squared statistic

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 .
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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")

$$H_0: \text{The additional parameters in the more complex model are 0.}$$
$$H_a: H_0\text{ is false.}$$
Under this setup, the test statistic has an asymptotic $$\chi^2$$ distribution, yes.
• @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.