= Solution
There are six pooled classes, one normalization constraint, and one fitted <Poisson distribution> parameter. Thus the reference <degrees of freedom> are $\boxed{6-1-1=4}$. Under the null model the Pearson statistic has approximately the <Chi-squared distribution> with four <degrees of freedom>. Its supplied 5% upper critical value is $9.49$.
Because $\boxed{2.07<9.49}$, \b[do not reject the Poisson count distribution at the 5% level]. This is consistency with that distribution, not proof that it is exactly true. A test of the marginal count frequencies also does not by itself verify independence of successive time windows.
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