The <Pearson residual> is
$$
r_i^{P}=\frac{Y_i-\widehat\mu_i}
{\sqrt{\widehat\phi V(\widehat\mu_i)}}
=\frac{Y_i-\widehat\mu_i}{\sqrt{\widehat\phi\widehat\mu_i^3}},
$$
where the moment estimate of dispersion is
$$
\widehat\phi
=\frac1{n-p}\sum_{i=1}^n
\frac{(Y_i-\widehat\mu_i)^2}{\widehat\mu_i^3}.
$$
If the fitted model is adequate and the deviance residual sum is close to the Pearson statistic, then
$$
\widehat\phi\approx\frac{245.00}{238}=1.029.
$$

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 218 / 1 / c / Solution

{c}
{title2=$r_i^P$}

The Pearson residual is $r_i^P=(Y_i-\widehat\mu_i)/\sqrt{\widehat\phi V(\widehat\mu_i)}$.


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