Hypothesis tests for binomial and PoissonEdexcel International A Level Further Maths: Revision notes

To test a claim about a binomial proportion $p$, the test statistic $X$ is the number of successes in a sample of size $n$. Under $\mathrm{H}_0:p=p_0$, $X\sim\mathrm{B}(n,p_0)$. To test a claim about a Poisson mean $\lambda$, the test statistic $X$ is the number of events in a fixed interval. Under $\mathrm{H}_0:\lambda=\lambda_0$, $X\sim\mathrm{Po}(\lambda_0)$. Always scale a Poisson mean to the interval being observed. A rate of $2.5$ per hour becomes $\lambda=10$ for a $4$-hour period.

Two equivalent methods are used. Critical region method: find the largest region in the relevant tail whose probability is at most the significance level, then see whether the observed value is in it. Probability method: find the probability of a result at least as extreme as the one observed and compare it with the significance level. Example: $X\sim\mathrm{B}(20,0.7)$, $\mathrm{H}_1:p<0.7$ at $5\%$. $\mathrm{P}(X\le10)=0.0480<0.05$ and $\mathrm{P}(X\le11)=0.1133>0.05$, so the critical region is $X\le10$. If $10$ seeds germinate, reject $\mathrm{H}_0$. Cumulative binomial tables usually stop at $p=0.5$. For $p>0.5$, let $Y=n-X\sim\mathrm{B}(n,1-p)$: then $\mathrm{P}(X\le10)=\mathrm{P}(Y\ge10)=1-\mathrm{P}(Y\le9)$ for $Y\sim\mathrm{B}(20,0.3)$, which also gives $0.0480$.

Example: calls arrive at a mean of $2.5$ per hour. In $4$ hours there are $16$ calls and we test whether the rate has increased, at $5\%$. $\mathrm{H}_0:\lambda=10$, $\mathrm{H}_1:\lambda>10$, where $\lambda$ is the mean number of calls in $4$ hours. Under $\mathrm{H}_0$, $X\sim\mathrm{Po}(10)$. $\mathrm{P}(X\ge16)=1-\mathrm{P}(X\le15)=1-0.9513=0.0487<0.05$, so reject $\mathrm{H}_0$. There is sufficient evidence that the mean rate of calls has increased.

For $\mathrm{H}_1:p\ne p_0$ or $\mathrm{H}_1:\lambda\ne\lambda_0$, the significance level is shared between the tails. At $5\%$, each tail has $2.5\%$. Find the probability in the tail on the same side as the observed value and compare it with $0.025$ (not $0.05$). Alternatively double that probability and compare with $0.05$.

For large $n$, such as $n=150$, use $X\approx\mathrm{N}(np_0,np_0q_0)$ under $\mathrm{H}_0$, with a continuity correction. Example: $X\sim\mathrm{B}(150,0.4)$, observed $X=72$, $\mathrm{H}_1:p\ne0.4$. $X\approx\mathrm{N}(60,36)$, so $\mathrm{P}(X\ge72)\approx\mathrm{P}(Y>71.5)=\mathrm{P}(Z>1.917)=0.0276$. This is greater than $0.025$, so do not reject $\mathrm{H}_0$. For a critical region, solve an inequality such as $\frac{c+0.5-\mu}{\sigma}<-1.645$ and then take the largest integer $c$ that satisfies it. Check that $np$ and $nq$ are both greater than $5$.

State whether $\mathrm{H}_0$ is rejected, then explain in the language of the question. Say 'sufficient evidence' or 'insufficient evidence', and name the parameter, for example 'the proportion of seeds that germinate is less than $70\%$'. Do not say the claim is proved or disproved.