Test a Poisson increase at a fixed exposure
Calculate an inclusive upper tail with the null expected count for the stated observation period.
Exam scope and assessment notes
Probability & Statistics 2 (Paper 6, A Level) · Mathematics 9709, 2026–2027, section 6.5. Paper 6 builds on Paper 5 and assumes Paper 3 calculus. Specify population hypotheses, significance level, test statistic and tail direction before inspecting the result. Calculate tail probabilities under H₀, including the observed discrete value. A p-value is not the probability that H₀ is true. Reject H₀ when the specified tail probability is at most the stated significance threshold; otherwise do not reject. The syllabus term acceptance region means the non-rejection region, not proof of H₀. For two-sided discrete examples here, use an explicitly stated equal-tail allocation; asymmetric discrete models admit different conventions, so do not treat twice the smaller tail as a universal definition. Integer critical regions can have actual Type I probability below the nominal level. Normal approximations need a suitable model and continuity correction for counts; check exact probabilities when approximation could change a close decision. Mean tests use a normal population with known variance or an adequate large independent sample with estimated variance. Small samples with unknown variance do not automatically justify the z method. Type I means rejecting a true null; Type II means not rejecting a false null at a specified alternative. Calculate error probabilities using the fixed decision region under the relevant parameter, not by changing the test after seeing data. Power is one minus Type II probability at that alternative. Quoted normal critical values and table probabilities have rounding limits. Interpret evidence in context without claiming certainty, causation or practical importance. Written practice is self-checked; cumulative assessment and human subject-teacher review remain pending.
Before you begin
- Use Poisson masses and cumulative complements.
- Scale a constant rate by exposure.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What null mean applies to half an hour at four events per hour?
What you will learn
- Form a Poisson count test.
- Keep rate and exposure consistent.
Null parameter is an expected count
A homogeneous independent-event model gives mean four arrivals per hour under H₀. For one fixed hour, X ~ Po(4). Test an increased rate using observed X = 9 and P₀(X ≥ 9) = 1 − Σ from k = 0 to 8 of e⁻⁴4ᵏ/k! ≈ 0.0213634345.
At 5%, reject the stated hourly rate model in favour of an increase. P₀(X ≥ 8) ≈ 0.0511336158 is too large; the upper critical region is X ≥ 9.
Fix exposure before counting
For half an hour, the null expected count is two, not four. Counts and model parameters must refer to the same period. A higher count across a longer observation does not by itself indicate a higher rate.
The model assumes a suitable constant-rate process and independent increments. A test result is conditional on those assumptions and does not explain why the rate changed.
Pause and explain
What upper region fits the 5% hourly Po(4) test?
Worked example
Test an hourly rate of four against an increase using nine arrivals in one hour.
Show the worked solution
- Under H₀, X ~ Po(4).
- P₀(X ≥ 9) ≈ 0.0213634345.
- Reject at 5%; evidence of an increased rate.
Answer Reject H₀ at 5%.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Test the four-per-hour model with nine arrivals in one hour at 5%.
Give me a hint
Complement the cumulative probability through eight.
Compare my reasoning
- X ~ Po(4) under H₀.
- P₀(X ≥ 9) ≈ 0.0213634345.
- Reject in favour of an increased arrival rate.
0.0213634345; reject at 5%.
Look for these in your work
- Use one-hour exposure.
- Include observed nine.
Would eight arrivals reject under the same rule?
Give me a hint
Keep the exact tail near the 5% threshold.
Compare my reasoning
- P₀(X ≥ 8) ≈ 0.0511336158.
- This exceeds 0.05.
- Do not reject H₀.
No; tail 0.0511336158 exceeds 0.05.
Look for these in your work
- Avoid rounding away the comparison.
- Keep the pre-specified rule.
State the null distribution for a half-hour observation at the same claimed rate.
Give me a hint
Multiply the rate by the new exposure duration.
Compare my reasoning
- Claimed rate is four per hour.
- Exposure is one half-hour.
- Null mean is two, giving Po(2).
Po(2) for the half-hour count.
Look for these in your work
- Convert the duration.
- Distinguish rate from expected count.
Common mistakes
- Use the null expected count for the same exposure.
- Complement through eight for ≥ 9.
- Rate and count are different quantities.
What can you explain now?
For ≥ 9 should the complemented cumulative sum include nine?
Compare with the explanation
No.
The complement is X ≤ 8.
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Make it stick
A Poisson test compares counts over matching exposures with the null expected count.
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