Past exam of the options II course of the University of Oxford / 2013 / O5 / d

The final cell is a grouped tail. Thus use twelve probabilitiesMaximize the grouped Grouped Poisson likelihoodthen take $\boxed{{\hat{\pi }}_i=p_i(\hat{\theta })}$ for all twelve cells. For example, the interior likelihood equation isbecause $p_{11}^{\prime }(\theta )=p_{10}(\theta )$. It can be solved numerically, with no numerical work required here.

It would be incorrect to treat the last six observations as exactly eleven. If the underlying ungrouped particle counts were available, their sample mean would be the usual Poisson maximum likelihood estimate. For the data as supplied, the grouped likelihood handles the unknown tail counts directly.

There are twelve cells and one fitted parameter, so the likelihood-ratio reference is approximately ${\chi }_{10}^2$. The supplied value givesThus do not reject the Poisson goodness-of-fit model at the conventional 5% level; it would also not be rejected at 10%. The data do not provide strong evidence of a departure from this model. This is a compatibility conclusion, not proof that the Poisson model is true. A fully specified Poisson mean would instead use eleven degrees of freedom, which is not the fitted situation here.