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

Solution

For observed proportions

y_{i}

and fitted proportions

p_{i}

D = 2 \sum_{i} m_{i} [y_{i} lo g \frac{y _{i}}{p _{i}} + (1 - y_{i}) lo g \frac{1 - y _{i}}{1 - p _{i}}] .

y_{i} = p_{i}

gives

D \approx \sum_{i} \frac{m _{i} ( y _{i} - p _{i} ) ^{2}}{p _{i} ( 1 - p _{i} )},

the generalized Pearson chi-squared statistic. Under the fitted binomial model this is approximately

χ_{4 - 3}^{2} = χ_{1}^{2}

. The observed deviance is

1.6102

, whose upper-tail probability is about

0.20

. Since this exceeds

0.05

, there is no significant evidence of overdispersion.

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