Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 37 / 3 / b / Solution

There are 23 independent vial responses. The intercept-only fit has 22 residual statistical degrees of freedom and deviance 33.631. Adding two day indicators removes two statistical degrees of freedom and reduces the residual deviance by $33.631-33.358=0.273$. Adding the two species indicators then reduces it by $33.358-14.646=18.712$. Thus the six replacements, in their printed order, are $22,33.631,2,0.273,18.712,14.646$.

The residual deviance statistical degrees of freedom are $23-1=22$, $23-3=20$, and $23-5=18$. These sequential analysis-of-deviance rows compare nested generalized linear models in the displayed order. In an unbalanced design, the first day row does not test day after controlling for species.