Mathematics 9709 · Probability & Statistics 2

Hypothesis testing with the Poisson

Same machine as the Chapter 1 binomial test — only the distribution under $H_0$ changes. Assume the claimed rate $\lambda_0$ is true, work out how freakish the observed count is (a tail probability), and reject $\lambda_0$ only if that tail falls below your significance level. The hypotheses are about the rate $\lambda$ , never the count you happened to see.

Test a claimed rate λ

State $H_0{:}\ \lambda=\lambda_0$ and the direction; compute the tail under $\text{Po}(\lambda_0)$ in the direction of $H_1$ ; reject if it's below the level. For $\lambda_0>15$ evaluate the tail with the normal approximation.

$H_0{:}\ \lambda=\lambda_0$ ; direction from the wording.
Tail under $\text{Po}(\lambda_0)$ toward $H_1$ ( $P(X\ge c)=1-P(X\le c-1)$ ).
$\lambda_0>15$ ? evaluate the tail with $N(\lambda_0,\lambda_0)$ + c.c.
Compare with the level; conclude in context, non-definitely.

Worked example

Start simple: a direct Poisson test

1% of a population reacts positively; in a village of 120, four react. Test at 5% for an increase.

$X\sim B(120,0.01)\approx\text{Po}(1.2)$ ( $n$ large, $p$ small).
$H_0{:}\ \lambda=1.2$ , $H_1{:}\ \lambda>1.2$ , one-tailed at 5%.
Upper tail: $P(X\ge4)=1-P(X\le3)=0.0338=3.38\%$ .
$3.38\%<5\%$ → reject $H_0$ .

Answer

3.38\%<5\%\ \Rightarrow\ \text{reject }H_0

There is evidence of an increase in the positive-reaction rate.

Worked example

Accidents: one calculation, two verdicts

Accidents run at 7 per month. New measures aim to cut them.

(a) One month, 2 accidents. $\text{Po}(7)$ , $H_1{:}\ \lambda<7$ : $P(X\le2)=0.0296=2.96\%<5\%$ → reject; evidence of a fall.
(b) Over 6 months, 32 accidents (baseline $\lambda=42>15$ ), claim “no longer reducing”. $\text{Po}(42)\approx N(42,42)$ : $P(X\le32)\to P(Y\le32.5)$ , $z=\tfrac{32.5-42}{\sqrt{42}}=-1.466$ , $\Phi(-1.466)=0.0713$ .
$7.13\%>5\%$ → accept at 5%; but $7.13\%<10\%$ → reject at 10%.

Answer

7.13\%{:}\ \text{do not reject at }5\%,\ \text{reject at }10\%

The same tail flips the verdict with the level.

The significance level is a choice made before testing — quoting both verdicts shows exactly how borderline the evidence is.

Hypotheses are about the RATE λ, never the count. Test statistic is a tail under Po(λ₀); for λ₀ > 15 use the normal approximation.

The mode of a Poisson

$\dfrac{P(X=r+1)}{P(X=r)}=\dfrac{\lambda}{r+1}$ , above 1 while $r+1<\lambda$ — so the bars rise then fall and the mode is $\lfloor\lambda\rfloor$ (two modes $\lambda-1,\lambda$ when $\lambda$ is an integer).

Examiner note

P(Type I) for a Poisson test = the achieved tail under

\text{Po}(\lambda_0)

; P(Type II) uses the TRUE rate. (9709/62/M/J/22 Q4) (9709/62/O/N/22 Q4)

Common mistake

Hypotheses about the count not the rate

\lambda

; comparing

P(X=c)

instead of the tail; the off-by-one

P(X\ge c)=1-P(X\le c-1)

; two-tailed comparing with

\alpha/2

; and (normal route) the missing continuity correction.

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Hypothesis testing with the Poisson

Test a claimed rate λ

$H_0{:}\ \lambda=\lambda_0$ ; direction from the wording.
Tail under $\text{Po}(\lambda_0)$ toward $H_1$ ( $P(X\ge c)=1-P(X\le c-1)$ ).
$\lambda_0>15$ ? evaluate the tail with $N(\lambda_0,\lambda_0)$ + c.c.
Compare with the level; conclude in context, non-definitely.

Worked example

Start simple: a direct Poisson test

1% of a population reacts positively; in a village of 120, four react. Test at 5% for an increase.

$X\sim B(120,0.01)\approx\text{Po}(1.2)$ ( $n$ large, $p$ small).
$H_0{:}\ \lambda=1.2$ , $H_1{:}\ \lambda>1.2$ , one-tailed at 5%.
Upper tail: $P(X\ge4)=1-P(X\le3)=0.0338=3.38\%$ .
$3.38\%<5\%$ → reject $H_0$ .

Answer

3.38\%<5\%\ \Rightarrow\ \text{reject }H_0

There is evidence of an increase in the positive-reaction rate.

Worked example

Accidents: one calculation, two verdicts

Accidents run at 7 per month. New measures aim to cut them.

(a) One month, 2 accidents. $\text{Po}(7)$ , $H_1{:}\ \lambda<7$ : $P(X\le2)=0.0296=2.96\%<5\%$ → reject; evidence of a fall.
(b) Over 6 months, 32 accidents (baseline $\lambda=42>15$ ), claim “no longer reducing”. $\text{Po}(42)\approx N(42,42)$ : $P(X\le32)\to P(Y\le32.5)$ , $z=\tfrac{32.5-42}{\sqrt{42}}=-1.466$ , $\Phi(-1.466)=0.0713$ .
$7.13\%>5\%$ → accept at 5%; but $7.13\%<10\%$ → reject at 10%.

Answer

7.13\%{:}\ \text{do not reject at }5\%,\ \text{reject at }10\%

The same tail flips the verdict with the level.

The significance level is a choice made before testing — quoting both verdicts shows exactly how borderline the evidence is.

Hypotheses are about the RATE λ, never the count. Test statistic is a tail under Po(λ₀); for λ₀ > 15 use the normal approximation.

The mode of a Poisson

Examiner note

P(Type I) for a Poisson test = the achieved tail under

\text{Po}(\lambda_0)

; P(Type II) uses the TRUE rate. (9709/62/M/J/22 Q4) (9709/62/O/N/22 Q4)

Common mistake

Hypotheses about the count not the rate

\lambda

; comparing

P(X=c)

instead of the tail; the off-by-one

P(X\ge c)=1-P(X\le c-1)

; two-tailed comparing with

\alpha/2

; and (normal route) the missing continuity correction.