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Hypothesis tests using approximations

Edexcel IAL Mathematics · WST02 Statistics 2

When the sample is large, the test statistic's distribution may not be in the tables. You then carry out the test with a Poisson or normal approximation.

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Binomial with small p: use Poisson

2% of items are usually faulty. In a sample of 200 after a machine is serviced, 9 are faulty. Test at 5% whether the proportion has increased.

H0: p = 0.02, H1: p > 0.02. Under H0, X ~ B(200, 0.02) ≈ Po(4) (n large, p small).

P(X ≥ 9) ≈ 1 - P(X ≤ 8) = 1 - 0.9786 = 0.0214 < 0.05: reject H0. Evidence that the proportion of faulty items has increased.

Poisson with large λ: use normal

Calls average 30 per day. On a day after an advert there are 42 calls. Test at 5% whether the rate has increased.

H0: λ = 30, H1: λ > 30. Under H0, X ~ Po(30) ≈ N(30, 30).

P(X ≥ 42) ≈ P(Y > 41.5) = P(Z > 2.10) = 0.0179 < 0.05: reject H0.

Critical regions with a normal approximation

For the same test, find the smallest c with P(X ≥ c) ≤ 0.05: P(Y > c - 0.5) ≤ 0.05 needs (c - 0.5 - 30)/sqrt(30) ≥ 1.6449, so c ≥ 39.51. The critical region is X ≥ 40.

Binomial tests for p may also use a normal approximation when n is large and p is near 0.5, with N(np, np(1 - p)) and a continuity correction.

Worked example

H0: p = 0.4, H1: p > 0.4. In a sample of 150, 72 are successes. Test at 5% using a normal approximation.

  1. Under H0, X ~ B(150, 0.4) ≈ N(60, 36).
  2. P(X ≥ 72) ≈ P(Y > 71.5)
  3. z = 11.5/6 = 1.9167; probability = 0.0276
  4. Less than 0.05, so reject H0.

Answer: There is evidence that the proportion is greater than 0.4.

Key idea: Choose the approximation from the sizes of n, p or λ, apply it under H0 (with a continuity correction for normal), then test as usual.

Check yourself

  1. Under H0, X ~ B(200, 0.02). Using a Poisson approximation, find P(X ≥ 9). Give your answer to 4 decimal places.
    Show answer

    0.0214. Po(4): 1 - P(X ≤ 8) = 1 - 0.9786 = 0.0214.

  2. Under H0, X ~ Po(30). Using a normal approximation, find P(X ≥ 42). Give your answer to 3 significant figures.
    Show answer

    0.0179. N(30, 30): P(Y > 41.5), z = 2.0996, probability 0.0179 (table 0.0179).

  3. Under H0, X ~ Po(30). Using a normal approximation, find the smallest value c such that the critical region X ≥ c gives a significance level of at most 5%.
    Show answer

    40. (c - 0.5 - 30)/sqrt(30) ≥ 1.6449 gives c ≥ 39.51, so c = 40.

  4. Under H0, X ~ B(150, 0.4). Using a normal approximation, find P(X ≥ 72). Give your answer to 3 significant figures.
    Show answer

    0.0276. N(60, 36): P(Y > 71.5), z = 1.9167, probability 0.0276 (table 0.0274).

  5. A test at the 5% level gives P(X ≥ 9) ≈ 0.0214 for the number of faulty items. What is the conclusion?
    • Do not reject H0: insufficient evidence of an increase
    • Reject H0: there is evidence that the proportion of faulty items has increased
    • Reject H0: the proportion has decreased
    • The approximation makes the test invalid
    Show answer

    Reject H0: there is evidence that the proportion of faulty items has increased. 0.0214 < 0.05.

  6. Under H0, X ~ B(500, 0.008). Which approximation is most suitable for the test?
    • N(4, 3.968)
    • N(500, 4)
    • Po(500)
    • Po(4)
    Show answer

    Po(4). n is large and p is small, so Po(np) = Po(4).

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