Sums of Poisson variables and Poisson hypothesis testsAQA A-Level Further Maths: Revision notes

If $X\sim\mathrm{Po}(\lambda)$ and $Y\sim\mathrm{Po}(\mu)$ are independent, then $X+Y\sim\mathrm{Po}(\lambda+\mu).$ The result extends to any number of independent Poisson variables: the means add. It also covers counts over several time periods, so four independent weeks of $\mathrm{Po}(4)$ give $\mathrm{Po}(16)$.

Example: server A gets $\mathrm{Po}(3)$ emails per minute and server B gets $\mathrm{Po}(2)$. The total is $\mathrm{Po}(5)$, so $\mathrm{P}(\text{total}=4)=\frac{e^{-5}5^4}{4!}=0.1755$.

Once the combined distribution is found, use the ordinary Poisson formula or calculator functions.

Example: $X\sim\mathrm{Po}(2.1)$ and $Y\sim\mathrm{Po}(1.4)$ independent. The total is $\mathrm{Po}(3.5)$ and $\mathrm{P}(X+Y=2)=\frac{e^{-3.5}\times3.5^2}{2!}=0.185$.

State the combined distribution explicitly, with its parameter, before calculating: this is usually a method mark.

A hypothesis test about a Poisson mean $\lambda$ uses a single observation $x$ of the count.

• Null hypothesis $\mathrm{H}_0:\lambda=\lambda_0$, the value being tested.
• Alternative hypothesis $\mathrm{H}_1$: $\lambda<\lambda_0$ or $\lambda>\lambda_0$ (one-tailed), or $\lambda\ne\lambda_0$ (two-tailed), decided by the wording of the claim before looking at the data.

Define $\lambda$ in context in your hypotheses, for example 'let $\lambda$ be the mean number of complaints per week'. The significance level is the probability, assuming $\mathrm{H}_0$, below which the result is considered too unlikely to have occurred by chance.

Assume $\mathrm{H}_0$ is true and find the probability of a result at least as extreme as the one observed.

• For $\mathrm{H}_1:\lambda<\lambda_0$ with observation $x$: find $\mathrm{P}(X\le x)$.
• For $\mathrm{H}_1:\lambda>\lambda_0$ with observation $x$: find $\mathrm{P}(X\ge x)=1-\mathrm{P}(X\le x-1)$.

If this probability is less than or equal to the significance level, reject $\mathrm{H}_0$.

Example: $\mathrm{H}_0:\lambda=4$, $\mathrm{H}_1:\lambda<4$, observed 1. $\mathrm{P}(X\le1)=e^{-4}(1+4)=0.0916>0.05$, so do not reject $\mathrm{H}_0$.

Conclusion in context: 'There is insufficient evidence at the 5% level that the mean number of complaints has fallen.'

When $\mathrm{H}_1:\lambda\ne\lambda_0$ the change could be in either direction, so the significance level is split: half in each tail.

Find the tail probability on the side of the observation, then compare with half the significance level. At the 5% level compare with $0.025$.

Example: $\mathrm{H}_0:\lambda=2.5$, $\mathrm{H}_1:\lambda\ne2.5$, observed 7 (above the mean). $\mathrm{P}(X\ge7)=0.0142<0.025$, so reject $\mathrm{H}_0$. There is evidence that the mean has changed.