Past exam of the options I course of the University of Oxford / 2004 / O / O1

For equal one-second observation windows, the sample mean estimates the Poisson distribution mean:This matches the expected mean count; it is also the maximum-likelihood estimator for independent Poisson distribution observations. With $N=12169$ windows, fitted expected frequencies are $E_j=N{e}^{-\hat{\lambda }}{\hat{\lambda }}^j/j!$ for $j=0,\dots ,4$, and $E_{\ge 5}=N[1-{\sum }_{j=0}^4{e}^{-\hat{\lambda }}{\hat{\lambda }}^j/j!]$.

Each last-column entry is $(O_j-E_j{)}^2/E_j$, and the Pearson statistic is $\boxed{X^2=\sum _{j\in \{0,1,2,3,4,\ge 5\}}(O_j-E_j{)}^2/E_j}$. The displayed expected frequencies and contributions are rounded, so their printed values need not reproduce every final digit of a calculation using unrounded probabilities.

Separate high-count classes have very small expected frequencies, making the chi-square approximation unreliable. Pooling the upper tail gives a class with expected count about $20.9$, comfortably above the customary small-cell threshold. The pooling protects the asymptotic test approximation, while retaining the tail count as part of the goodness-of-fit comparison.

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.