= Solution
There are twelve cells and one fitted parameter, so the likelihood-ratio reference is approximately $\chi_{10}^2$. The supplied value gives
$$
\boxed{p\text{-value}\approx1-F_{10}(14)=0.18.}
$$
Thus \b[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.
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