For observed proportions $y_i$ and fitted proportions $\widehat p_i$, the <binomial deviance> is
$$
D=2\sum_i m_i\left[
y_i\log\frac{y_i}{\widehat p_i}
+(1-y_i)\log\frac{1-y_i}{1-\widehat p_i}
\right].
$$
A second-order Taylor expansion around $y_i=\widehat p_i$ gives
$$
D\approx\sum_i\frac{m_i(y_i-\widehat p_i)^2}
{\widehat p_i(1-\widehat p_i)},
$$
the generalized <Pearson chi-squared statistic>. Under the fitted binomial model this is approximately $\chi^2_{4-3}=\chi^2_1$. The observed deviance is $1.6102$, whose upper-tail probability is about $0.20$. Since this exceeds $0.05$, \b[there is no significant evidence of overdispersion].

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 218 / 1 / d / Solution

For independent binomial counts $Y_i\sim\operatorname{Bin}(m_i,p_i)$ with observed proportions $y_i=Y_i/m_i$ and fitted probabilities $\widehat p_i$, the deviance from the saturated model is
$$
D=2\sum_i m_i\left[y_i\log\frac{y_i}{\widehat p_i}+(1-y_i)\log\frac{1-y_i}{1-\widehat p_i}\right],
$$
with the usual convention $0\log0=0$.


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 Incoming links: Binomial deviance