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Hypothesis tests for a Poisson mean

Edexcel IAL Mathematics · WST02 Statistics 2

Tests for a Poisson mean follow the same pattern as binomial tests, with λ in the hypotheses. The extra step is scaling λ to the interval in which the data were collected.

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Layout

1. Write H0: λ = λ0 and H1: λ > λ0, λ < λ0 or λ ≠ λ0, where λ is the mean rate (per stated interval) in context.

2. Scale to the observed interval: if the rate is 4 per hour and data cover 2 hours, X ~ Po(8) under H0.

3. Compare P(X ≥ x) or P(X ≤ x) with α (α/2 if two-tailed), or use a critical region.

4. Conclude in context. Mark schemes insist on a rate: 'insufficient evidence that the rate of customers entering the shop has changed', not just 'the number of customers'.

Example

Faults occur at a rate of 4 per hour. After maintenance, 15 faults are seen in 2 hours. Test at 5% whether the rate has increased.

H0: λ = 4, H1: λ > 4 (per hour). Under H0, X ~ Po(8) for 2 hours.

P(X ≥ 15) = 1 - P(X ≤ 14) = 1 - 0.9827 = 0.0173 < 0.05. Reject H0: there is evidence that the rate of faults has increased.

Two-tailed Poisson regions

H0: λ = 9, H1: λ ≠ 9 at 5%: P(X ≤ 3) = 0.0212, P(X ≤ 4) = 0.0550, so lower region X ≤ 3. P(X ≥ 16) = 1 - 0.9780 = 0.0220, P(X ≥ 15) = 0.0415, so upper region X ≥ 16. Actual significance level 0.0432.

Worked example

Accidents at a junction happen at 1.2 per day. After new signs, there is only 1 accident in 5 days. Test at 5% whether the rate has decreased.

  1. H0: λ = 1.2, H1: λ < 1.2 (per day). In 5 days, X ~ Po(6) under H0.
  2. P(X ≤ 1) = 0.0174
  3. 0.0174 < 0.05, so reject H0.

Answer: There is evidence that the rate of accidents at the junction has decreased.

Key idea: Hypotheses use λ. Scale λ to the observation period. Conclude about the rate, in context.

Check yourself

  1. Under H0, faults in 2 hours follow Po(8). 15 faults are observed. Find P(X ≥ 15). Give your answer to 4 decimal places.
    Show answer

    0.0173. 1 - P(X ≤ 14) = 1 - 0.9827 = 0.0173.

  2. Under H0, X ~ Po(8). Find the critical value c for an upper-tail test at the 5% level (critical region X ≥ c).
    Show answer

    14. P(X ≥ 13) = 0.0638 > 0.05; P(X ≥ 14) = 0.0342 ≤ 0.05, so c = 14.

  3. Under H0, X ~ Po(9). The critical region for a two-tailed test at 5% is X ≤ 3 or X ≥ 16. Find the actual significance level. Give your answer to 4 decimal places.
    Show answer

    0.0433. 0.0212 + 0.0220 = 0.0432 from the tables (0.0433 if you add unrounded calculator values). Both are accepted.

  4. Under H0, X ~ Po(9). For a two-tailed test at 5%, find the smallest d for the upper critical region X ≥ d.
    Show answer

    16. P(X ≥ 16) = 0.0220 ≤ 0.025 but P(X ≥ 15) = 0.0415 > 0.025.

  5. Accidents occur at 1.2 per day. Under H0, find the probability of at most 1 accident in 5 days. Give your answer to 4 decimal places.
    Show answer

    0.0174. Po(6): P(X ≤ 1) = 0.0174.

  6. The rate of calls is claimed to be 3 per hour. Calls are counted for 4 hours. What is the distribution of the number of calls under H0?
    • Po(3)
    • Po(0.75)
    • B(4, 3)
    • Po(12)
    Show answer

    Po(12). Scale λ: 3 × 4 = 12.

  7. A two-tailed test of the rate of customers entering a shop is not significant. Which conclusion matches the mark scheme wording?
    • There is insufficient evidence that the number of customers has changed
    • There is insufficient evidence that the rate of customers entering the shop has changed
    • The rate of customers is unchanged
    • There is no evidence that the rate is unchanged
    Show answer

    There is insufficient evidence that the rate of customers entering the shop has changed. Mark schemes require 'rate' and 'insufficient evidence', in context.

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Last updated 8 November 2026. Spotted a mistake? Tell us.