Test a Poisson decrease with an exact cumulative tail
Use low-count evidence under a specified Poisson rate model.
Section 6.5 · Hypothesis tests
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
- Calculate Poisson probabilities.
- Choose a lower alternative.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which event is relevant for observed one in a decrease test?
What you will learn
- Evaluate a low-count tail.
- Interpret a decrease without proving a zero rate.
Sum low counts
Under an expected count of five per observation period, X ~ Po(5). For H₁: λ < 5 and observed X = 1, the lower tail is P₀(X ≤ 1) = e⁻⁵(1 + 5) ≈ 0.0404276820. Reject at 5% with evidence of a decreased mean count.
At X = 2 the cumulative tail is e⁻⁵(1 + 5 + 25/2) ≈ 0.1246520195, so do not reject. The critical region is X ≤ 1 for this level.
No events does not imply no process
A zero count has positive probability for every finite positive Poisson mean. Observing no events may be evidence against a particular rate, but it does not establish that the true rate is zero.
The null probabilities retain λ = 5. Do not substitute the observed count into the null mean or treat the model as verified merely because it is not rejected.
Pause and explain
Does observed two reject the stated 5% Po(5) decrease test?
Worked example
Test λ = 5 against λ < 5 with observed count one at 5%.
Show the worked solution
- Use Po(5) under H₀.
- Sum P(X = 0) + P(X = 1) ≈ 0.0404276820.
- Reject at 5% in favour of a decrease.
Answer Reject H₀; evidence of a decrease.
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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 λ = 5 against λ < 5 using X = 1 at 5%.
Give me a hint
Sum the probabilities at zero and one.
Compare my reasoning
- Use e⁻⁵(1 + 5).
- The result is approximately 0.0404276820.
- Reject H₀; evidence of a smaller expected count.
0.0404276820; reject.
Look for these in your work
- Include zero.
- Interpret the expected count.
Use the same test with X = 2.
Give me a hint
Add the mass at two to the previous cumulative sum.
Compare my reasoning
- Use e⁻⁵(1 + 5 + 25/2).
- The result is approximately 0.1246520195.
- Do not reject H₀ at 5%.
0.1246520195; do not reject.
Look for these in your work
- Use the lower cumulative event.
- Keep the null model.
Why does observing no events fail to prove that the event rate is zero?
Give me a hint
Inspect P(X = 0) for a positive Poisson mean.
Compare my reasoning
- P(X = 0) = e⁻λ.
- For a finite positive λ this is positive.
- A zero observation is therefore possible under a nonzero rate.
Because e⁻λ > 0 for every finite positive λ.
Look for these in your work
- Use model support.
- Avoid certainty from one observation.
Common mistakes
- Include zero in the lower cumulative sum.
- Keep the null mean fixed.
- Zero observed events does not prove zero rate.
What can you explain now?
Does a zero count prove λ = 0?
Compare with the explanation
No.
Positive means can still produce zero observations.
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Make it stick
A decrease test sums the observed count and all smaller nonnegative counts.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.