Testing hypotheses with the Poisson distribution is very similar to testing them with the binomial distribution. If the probability is greater than $\alpha$, the level of significance, then the null hypothesis is accepted. If it is less than $\alpha$, we accepted the alternative hypothesis.
An existing make of car is known to break down on average one and a half times per year. A new model is introduced and the manufacturer claims that this model is less likely to break down. Ten randomly selected cars break down a total of eight times within the first year. Test the manufacturer's claim at a $5$% significance level.
Let $X$ be the number of break downs of the new model of car in a year. Since we have an average rate and the data is discrete, we need to use a Poisson distribution. So $X \sim \mathrm{Poisson}(\lambda)$ with $\lambda=1.5$. The null and alternative hypotheses will be
\begin{align} H_0: & \lambda = 1.5\text{,} \\ H_1: & \lambda < 1.5. \end{align}
We need to decide whether $\mathrm{P}[X\leq 8] < \alpha$, where $\alpha = 0.05$ is the significance level. Firstly, the expected number of breakdowns $\lambda_t = 1.5\times 10 = 15$. We use the cumulative tables with $\lambda_t=15$ and $x=8$ to see $\mathrm{P}[X\leq 8] = 0.0374$. \[\mathrm{P}[X\leq 8] = 0.0374 < 0.05 = \alpha\] so we accept the alternative hypothesis. The average rate of breakdowns has decreased.
In this video, Daniel Organisciak performs a hypothesis test using the Poisson Distribution