Reading the deviance

Topic: Logistic regression · more questions on this topic
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In the heart-failure output, the Wald statistic for hbp is z=1.016. For a single coefficient the likelihood-ratio statistic for dropping it is approximately z^2. What residual deviance would you therefore expect for the age-only model, and with how many degrees of freedom? Compare with the \chi^2_1 critical value 3.84 at the 5\% level and state the conclusion. Separately, test the full model against the null model using the two deviances printed.

Worked answer (click to show)

The supplied Wald value gives z^2=1.016^2\approx 1.03. Under the usual large-sample regularity conditions, this can approximate the likelihood-ratio statistic for removing high blood pressure, suggesting a reduced-model deviance near

354.97+1.03=356.0

on 297 degrees of freedom, one more than the full model’s 296. It is not a calculation of that model’s actual likelihood ratio: refitting the reduced model is needed for the exact observed difference. The approximate one-degree-of-freedom comparison is not significant at 5\%, because 1.03<3.84.

For the full model against the intercept-only model, both reported deviances are available. Their difference is 375.35-354.97=20.38 on two degrees of freedom, giving the regular asymptotic \chi^2_2 comparison and p\approx 3.8\times 10^{-5}. This supports improved fit from the included predictors jointly. It is not a causal conclusion about age or blood pressure.