SB5 - Testing a Poisson mean | AQA A-Level Further Mathematics

An unusually large count may indicate a higher event rate, but random variation can also produce surprises. You will formulate a claim about a population mean, evaluate an exact Poisson tail, and make a cautious decision in context. Critical regions show how that decision would change for other possible observations.

Null and alternative hypotheses

A fault counter usually records a mean of 4 faults per day. If it records 9 tomorrow, is that evidence that the mean has increased? The question concerns an underlying mean, not whether the observed count is literally equal to 4.

Recall the logic of a hypothesis test: assume a reference claim is true, then assess whether the observed result would be sufficiently unusual under that claim. First define the parameter. Let $\lambda$ be the population mean number of faults per day. Assume daily counts follow a Poisson model.

Statistical hypotheses

The null hypothesis, $H_0$ , states the reference parameter value. The alternative hypothesis, $H_1$ , states the change being investigated. For an increase from 4 per day,

H_0:\lambda=4,\qquad H_1:\lambda>4

A significance level is the chosen upper limit on the probability of rejection when the null model is true.

“Decreased” gives $H_1:\lambda<4$ . “Changed” gives $H_1:\lambda\ne4$ . Choose the direction from the question before looking at the count: choosing whichever tail looks more surprising afterwards changes the test. The null hypothesis uses equality, and both hypotheses refer to the same population parameter and interval.

If the established rate were 2 per hour and the observation covered 3 hours, you could define a mean rate $r$ per hour and write $H_0:r=2$ , then use a count model $X\sim\operatorname{Po}(6)$ under $H_0$ . Alternatively define $\lambda$ per 3 hours and test $\lambda=6$ . These are consistent descriptions; mixing their units is not.

The observed random variable and population parameter have different jobs: $X$ is a possible count; $x$ is the observed value; $\lambda$ is the mean being tested. Writing $H_0:X=4$ wrongly turns the null into a claim about one observation.

Key idea

State what the parameter means, then write $H_0$ and $H_1$ about that parameter. The context determines the direction and observation interval.

Choose hypotheses for a fresh claim.

Testing from a single observation

A single observation here means one count over the stated period. It does not mean only one event occurred. Under $H_0:\lambda=\lambda_0$ , use $X\sim\operatorname{Po}(\lambda_0)$ and evaluate Poisson probabilities directly.

For an upper-tail alternative, the p-value is $\operatorname{P}_{H_0}(X\ge x)$ : the probability of the observed count or a more extreme larger count. For a lower-tail alternative it is $\operatorname{P}_{H_0}(X\le x)$ . The subscript emphasises that the probability is calculated assuming $H_0$ , not using an estimated mean equal to the observation.

Reject $H_0$ when the appropriate p-value is at most the chosen level $\alpha$ . Otherwise do not reject it. Rejection supports the alternative; non-rejection says the evidence is insufficient. Neither proves a population claim.

Worked example 1: eight faults

Question. Faults follow a Poisson model. Test $H_0:\lambda=4$ against $H_1:\lambda>4$ at the 5% level after observing 8 faults in one day.

Plan. Use the null mean 4 and include 8 in the upper tail.

\begin{aligned} X&\sim\operatorname{Po}(4) && \text{under }H_0\\ \operatorname{P}(X\ge8)&=1-\operatorname{P}(X\le7)\\ &=0.0511336\ldots>0.05 \end{aligned}

Answer. Do not reject $H_0$ . There is insufficient evidence at the 5% level that the population mean number of faults per day has increased.

Check. A direct sum of the masses at $8,9,10,\ldots$ gives $0.0511336\ldots$ . Although 8 is twice the null mean, its tail probability is just above the threshold.

Change only the observation, from 8 to 9. This makes the upper tail smaller and answers the opening question.

Worked example 2: one additional fault

Question. Repeat the same 5% test after observing 9 faults in one day.

Plan. Keep the hypotheses and null model, but move the tail boundary to 9.

\operatorname{P}_{H_0}(X\ge9)=1-\operatorname{P}_{H_0}(X\le8) =0.0213634\ldots<0.05

Answer. Reject $H_0$ . There is sufficient evidence at the 5% level to suggest that the population mean number of faults per day has increased.

Check. Removing the mass at 8 from the previous tail gives

0.0511336\ldots-\frac{\mathrm{e}^{-4}4^8}{8!}=0.0213634\ldots

The difference is one whole outcome, which explains the abrupt change in the decision.

A p-value is not the probability that $H_0$ is true. It measures how unusual a result is assuming that model is true. Show the model, correct tail and comparison so that your conclusion is traceable.

Watch out

Wrong: test an increase using $\operatorname{P}(X=9)$ or $\operatorname{P}(X>9)$ . Include the observed value and every more extreme count: $\operatorname{P}(X\ge9)$ . Use the null mean, not 9.

Carry out the same chain with a fresh count model.

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SB5 - Testing a Poisson mean

Null and alternative hypotheses

Testing from a single observation

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