FS1 4.1 - Testing a Poisson mean

FS1 4.1 - Testing a Poisson mean

A surprisingly large or small count can challenge a claimed Poisson mean. You will choose hypotheses, calculate the correct tail, construct discrete critical regions and check their actual significance levels (FS1 4.1).

What would count as evidence?

A machine used to stop four times per hour. After an adjustment, it runs for a whole hour without stopping. Is that enough evidence, at the 5%5\% level, that its mean stopping rate has fallen?

Zero certainly looks encouraging. A test asks a more precise question: how likely is a result this favourable to a reduction if the old rate still applies? Keep the observation separate from the unknown population rate.

You already know

From A-level Mathematics, a significance test compares sample evidence with a null hypothesis; its alternative determines the tail. From earlier Poisson lessons, independent interval counts add their parameters, and a constant rate λ\lambda per hour gives mean tλt\lambda in tt hours.

Define λ\lambda as the population mean number of stops per hour after adjustment. The opening claim gives H0:λ=4H_0:\lambda=4, H1:λ<4H_1:\lambda<4. The observed count is not λ\lambda: it is one outcome of a random variable.

Claim being investigatedAlternativeOutcomes supporting it
The mean has decreasedH1:λ<λ0H_1:\lambda<\lambda_0Small counts
The mean has increasedH1:λ>λ0H_1:\lambda>\lambda_0Large counts
The mean has changedH1:λ≠λ0H_1:\lambda\ne\lambda_0Either extreme

Choose the alternative from the question before inspecting the result. Choosing the direction afterwards would invalidate the stated significance level.

Null model and critical region

The null model is the distribution of the test statistic assuming H0H_0 is true. A critical region is a set of its possible outcomes for which H0H_0 is rejected. For example, X≥9X\ge9 describes outcomes; P(X≥9)=0.0214\mathrm P(X\ge9)=0.0214 describes a probability, not the critical region itself.

If XX is the total over tt hours, use X∼Po⁡(tλ0)X\sim\operatorname{Po}(t\lambda_0) under H0H_0. Likewise, a total SS from nn independent one-hour observations has null model Po⁡(nλ0)\operatorname{Po}(n\lambda_0). The hypotheses still concern the population hourly mean λ\lambda, not the observed total or sample mean. These are applications of parameter scaling and independent addition, neither printed explicitly in the booklet.

The conclusion is conditional on a suitable Poisson model: events occur independently and singly, with a constant average rate over the observation period. A machine that tends to stop repeatedly in clusters may violate these assumptions. A significant count can reveal a problem with the claimed rate or with the model supporting the test.

Key idea

A small tail probability measures how surprising the data are under H0H_0. It is not the probability that H0H_0 is true. If the result is not significant, say there is insufficient evidence for the alternative; do not claim to have proved the null hypothesis.

The variable's definition fixes the unit before any probability is calculated.

Testing an observed count in the correct tail

Include the observed count and all outcomes further in the direction of the alternative when calculating a one-sided tail probability.

One-sided Poisson tail tests

For an observed count xx, calculate under the null model:

H1:λ<λ0P(X≤x)H1:λ>λ0P(X≥x)=1−P(X≤x−1)\begin{array}{c|c} H_1:\lambda<\lambda_0&\mathrm P(X\le x)\\ H_1:\lambda>\lambda_0&\mathrm P(X\ge x)=1-\mathrm P(X\le x-1) \end{array}

At significance level α\alpha, reject H0H_0 if the relevant tail probability is at most α\alpha. Here 5%5\% means α=0.05\alpha=0.05.

Not in the formula booklet: learn it.

The Poisson probability formula and cumulative tables are supplied. Tables give P(X≤k)\mathrm P(X\le k); quote table entries in full, or calculator values to equivalent accuracy. State the distribution and event before using a calculator, retain its unrounded value for the decision, and do not introduce a continuity correction.

Worked example 1: just outside the critical region

Question. A machine's population mean number of stops per hour is denoted by λ\lambda. It previously had mean 44. An engineer suspects an increase and observes 88 stops in one hour. Assuming a Poisson model, test the claim at the 5%5\% level.

Plan. State the hypotheses and calculate the upper tail under the old hourly mean.

H0:λ=4,H1:λ>4test the claimed increaseX∼Po⁡(4) under H0one-hour countP(X≥8)=1−P(X≤7)include count 8=1−0.9488663…=0.0511336…evaluate the upper tail0.0511336…>0.05compare before rounding\begin{aligned} H_0&:\lambda=4,\qquad H_1:\lambda>4 &&\text{test the claimed increase}\\ X&\sim\operatorname{Po}(4)\text{ under }H_0 &&\text{one-hour count}\\ \mathrm P(X\ge8)&=1-\mathrm P(X\le7) &&\text{include count }8\\ &=1-0.9488663\ldots=0.0511336\ldots &&\text{evaluate the upper tail}\\ 0.0511336\ldots&>0.05 &&\text{compare before rounding} \end{aligned}

Why this step? Counts 8,9,10,…8,9,10,\ldots all support an increase. The point probability of exactly 88 ignores the more extreme results.

Answer. Do not reject H0H_0. At the 5%5\% level there is insufficient evidence that the population mean number of stops per hour has increased.

Check. Using the table entry 0.94890.9489 gives tail 0.05110.0511, still above 0.050.05. The evidence is close to the boundary but does not cross it.

Worked example 2 changes only the observed count, from 88 to 99. The hypotheses, observation time and significance level remain the same.

Worked example 2: one additional stop

Question. Carry out the same test if 99 stops are observed instead.

Plan. Keep the null model and move the inclusive upper-tail threshold to 99.

H0:λ=4,H1:λ>4same population hypothesesX∼Po⁡(4) under H0same count modelP(X≥9)=1−P(X≤8)exclude counts below 9=1−0.9786365…=0.0213634…evaluate0.0213634…<0.05significant result\begin{aligned} H_0&:\lambda=4,\qquad H_1:\lambda>4 &&\text{same population hypotheses}\\ X&\sim\operatorname{Po}(4)\text{ under }H_0 &&\text{same count model}\\ \mathrm P(X\ge9)&=1-\mathrm P(X\le8) &&\text{exclude counts below }9\\ &=1-0.9786365\ldots=0.0213634\ldots &&\text{evaluate}\\ 0.0213634\ldots&<0.05 &&\text{significant result} \end{aligned}

Why this step? Raising the observed count removes the probability at 88 from the tail, making the result more surprising under the old rate.

Answer. Reject H0H_0. There is sufficient evidence, at the 5%5\% level, that the population mean number of stops per hour has increased.

Check. The tail is smaller than the preceding 0.0511336…0.0511336\ldots, as an increase in the threshold requires.

For a reduction, reverse the tail: an observed zero in the opening problem uses P(X≤0)\mathrm P(X\le0), not P(X≥0)=1\mathrm P(X\ge0)=1.

Watch out

Wrong: P(X≥9)=1−P(X≤9)\mathrm P(X\ge9)=1-\mathrm P(X\le9). That omits 99. Right: subtract the cumulative probability through 88. Also, a conclusion about the sample count alone does not answer a question about the population mean.

Move the observed count once more and give the full contextual decision.

Continue this lesson.

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