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

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.