 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 $\boxed{2.07<9.49}$, 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.