S2 4.6 - Testing a Poisson mean

S2 4.6 - Testing a Poisson mean

A rate claim becomes testable only after it is matched to the time or area actually observed. You will scale the null Poisson mean, find exact one-tailed and two-tailed critical regions, calculate actual significance levels and draw conclusions about the rate in context.

Translating a rate into a sample mean

If a workshop normally receives four repair requests per hour and receives ten in two hours, should you compare ten with four? The observation covers twice the time, so its null mean is eight. Comparing counts over mismatched intervals would create misleading evidence.

For a Poisson process, assume events occur independently and singly at a constant mean rate. A rate rr per unit time gives mean rtrt in tt units of time. Define the parameter's interval explicitly; the same hypothesis can be written about the original rate or the mean over the observed interval, provided both hypotheses use the same convention.

You already know

The sum of independent Poisson counts is Poisson with the sum of their means. Therefore doubling the observation time doubles its null mean, under a constant-rate model. The Poisson mean and variance are in the formula booklet.

The observed sample count and its model mean are different. The count varies between samples; its mean describes the population process over the chosen interval. Use the claimed population rate to build the null model, rather than replacing it with the rate observed in this sample.

Worked example 1: a half-hour count

A desk normally receives a mean of 6 calls per hour. It is suspected that the rate has increased. State hypotheses about the hourly rate and the null model for the count in 30 minutes.

Plan. Keep hypotheses in calls per hour, but scale the count's null mean to half an hour.

H0:λ=6,H1:λ>6,H_0:\lambda=6,\qquad H_1:\lambda>6,

where λ\lambda is the mean calls per hour. Under H0H_0,

X∼Po⁡(6×3060)=Po⁡(3).X\sim\operatorname{Po}\left(6\times\frac{30}{60}\right)=\operatorname{Po}(3).

Answer. Hypotheses about the hourly mean 6; a half-hour test count with null mean 3.

Check. If instead μ\mu denoted the half-hour mean, the equivalent hypotheses would be H0:μ=3H_0:\mu=3, H1:μ>3H_1:\mu>3.

Change only the observation time from half an hour to two hours. The hourly hypotheses remain unchanged.

Worked example 2: a two-hour count

For the same desk and suspected increase, state the hourly hypotheses and null count model for a two-hour observation.

Plan. Multiply the hourly null mean by two before selecting the count distribution.

H0:λ=6,H1:λ>6,X∼Po⁡(2×6)=Po⁡(12).H_0:\lambda=6,\qquad H_1:\lambda>6, \qquad X\sim\operatorname{Po}(2\times6)=\operatorname{Po}(12).

Answer. The same hourly hypotheses, with null count mean 12.

Check. Two hours contains four half-hours. Four independent Po⁡(3)\operatorname{Po}(3) counts sum to Po⁡(12)\operatorname{Po}(12).

A count of 12 does not mean that the parameter is 12 per hour: always connect the numerical value to its interval.

Exact Poisson critical regions

Once the observation interval is fixed, construct a critical region using the Poisson distribution under the null hypothesis. For an increased mean, reject for sufficiently large counts; for a decreased mean, reject for sufficiently small counts.

Poisson cumulative tables give lower tails. To include a count cc and everything above it, use

P(X≥c)=1−P(X≤c−1).\mathrm P(X\ge c)=1-\mathrm P(X\le c-1).

Keep enough probability digits to show which side of the significance level a boundary lies on. If the required mean is not tabulated, calculate the Poisson probabilities instead.

Worked example 3: an upper-tail region

Under H0H_0, a count has distribution X∼Po⁡(6)X\sim\operatorname{Po}(6). Find the largest upper-tail critical region with probability at most 5% for an increased mean.

Plan. Compare the inclusive tail at a proposed cutoff with the tail at the next lower integer.

P(X≥10)=1−P(X≤9)=0.0839240P(X≥11)=1−P(X≤10)=0.0426209.\begin{aligned} \mathrm P(X\ge10)&=1-\mathrm P(X\le9)=0.0839240\\ \mathrm P(X\ge11)&=1-\mathrm P(X\le10)=0.0426209. \end{aligned}

Answer. Critical region X≥11X\ge11, with actual significance approximately 4.26%.

Check. Ten cannot be included because the enlarged tail would have probability greater than 5%.

Change only the alternative from an increase to a decrease. The null mean stays 6, but the relevant tail changes.

Worked example 4: a lower-tail region

For X∼Po⁡(6)X\sim\operatorname{Po}(6) under H0H_0, find the largest lower-tail critical region with probability at most 5% for a decreased mean.

Plan. Use the lower cumulative probabilities directly.

P(X≤1)=0.0173513,P(X≤2)=0.0619688.\mathrm P(X\le1)=0.0173513, \qquad\mathrm P(X\le2)=0.0619688.

Answer. Critical region X≤1X\le1, with actual significance approximately 1.74%.

Check. The region includes both 0 and 1, not just the single outcome 1. Adding 2 would exceed 5%.

The asymmetry of these two regions is natural: the Poisson distribution is not symmetric about its mean.

Tail probabilities and actual significance

For a two-tailed test of H0:λ=λ0H_0:\lambda=\lambda_0 against H1:λ≠λ0H_1:\lambda\ne\lambda_0, unusual counts in either direction count as evidence. With at most half the nominal level in each tail, determine each cutoff separately: Poisson upper and lower tails generally have different probabilities.

The actual significance is the sum of those disjoint tail probabilities under H0H_0. Do not simply report the nominal level, or double one tail as if symmetry were guaranteed.

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