Hypothesis test for the mean of a Poisson distributionEdexcel A-Level Further Maths: Revision notes

The Poisson distribution $X\sim\mathrm{Po}(\lambda)$ models the number of events in a fixed interval of time or space when events occur randomly, independently and at a constant average rate. The parameter $\lambda$ is the mean number of events in the interval. If the rate is stated for one interval but the sample covers several, scale the mean: a rate of $1.2$ errors per page gives $\mathrm{Po}(6)$ for $5$ pages. A hypothesis test for the mean of a Poisson distribution uses a single observed count to decide whether the population mean $\lambda$ has changed from its stated value.

The null hypothesis $\mathrm{H}_0$ says the parameter has its stated value: $\mathrm{H}_0:\lambda=\lambda_0$. The alternative hypothesis $\mathrm{H}_1$ says how it differs. Write hypotheses in terms of the population parameter $\lambda$ (or $\mu$), not the random variable $X$ or the observed value. Use a one-tailed test when the question says the rate has increased ($\mathrm{H}_1:\lambda>\lambda_0$) or decreased ($\mathrm{H}_1:\lambda<\lambda_0$). Use a two-tailed test when it says only that the rate has changed ($\mathrm{H}_1:\lambda\neq\lambda_0$). Define $\lambda$ in context, for example 'the mean number of calls per hour'. Then state the distribution of the test statistic under $\mathrm{H}_0$, for instance $X\sim\mathrm{Po}(3)$.

Assume $\mathrm{H}_0$ is true, so $X\sim\mathrm{Po}(\lambda_0)$. For $\mathrm{H}_1:\lambda>\lambda_0$, find the probability of a result at least as extreme as the one observed, $\mathrm{P}(X\geq x)=1-\mathrm{P}(X\leq x-1)$, and compare it with the significance level. If it is smaller, reject $\mathrm{H}_0$. For $\mathrm{H}_1:\lambda<\lambda_0$ use $\mathrm{P}(X\leq x)$. Worked example: calls have mean $3$ per hour and $7$ are observed, $\mathrm{H}_1:\lambda>3$ at $5\%$. $\mathrm{P}(X\geq7)=1-0.9665=0.0335<0.05$, so reject $\mathrm{H}_0$. The critical region is the set of values that leads to rejecting $\mathrm{H}_0$. For $X\sim\mathrm{Po}(6)$ and $\mathrm{H}_1:\lambda<6$ at $5\%$: $\mathrm{P}(X\leq1)=0.0174<0.05$ but $\mathrm{P}(X\leq2)=0.0620>0.05$, so the critical region is $X\leq1$. The actual significance level is the probability of being in the critical region, here $0.0174$.

For $\mathrm{H}_1:\lambda\neq\lambda_0$, split the significance level equally between the two tails. At $10\%$ use $5\%$ in each tail: the lower tail $\mathrm{P}(X\leq c_1)\leq0.05$ and the upper tail $\mathrm{P}(X\geq c_2)\leq0.05$. For an observed value in the upper tail, compare $\mathrm{P}(X\geq x)$ with $0.05$ (or double it and compare with $0.10$). Worked example: $X\sim\mathrm{Po}(10)$ at $10\%$. $\mathrm{P}(X\leq4)=0.0293$ and $\mathrm{P}(X\leq5)=0.0671$, so the lower tail is $X\leq4$. $\mathrm{P}(X\geq16)=0.0487$ and $\mathrm{P}(X\geq15)=0.0835$, so the upper tail is $X\geq16$. The critical region is $X\leq4$ or $X\geq16$ and the actual significance level is $0.0293+0.0487=0.0780$.

Finish with a conclusion in two steps: a statement about $\mathrm{H}_0$ ('reject $\mathrm{H}_0$' or 'do not reject $\mathrm{H}_0$') and a sentence in context. For example, 'There is sufficient evidence, at the $5\%$ level, that the mean number of calls per hour has increased.' If the result is not significant, say there is insufficient evidence that the rate has changed, rather than stating that the rate is unchanged. A test can fail to show a change that has really happened, so a non-significant result does not prove $\mathrm{H}_0$. You may be asked to comment on the model. The Poisson distribution needs events that are independent and occur at a constant average rate. If events come in bursts or the rate varies during the day, the counts vary more than the model expects and the test may be unreliable.