OxfordAQA International AS and A-level Mathematics: Hypothesis testing for a proportion and a Poisson mean (9660) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions follow the OxfordAQA International AS and A-level Mathematics (9660) specification, Version 5.2, section S2.6 (first part): tests for a binomial proportion and a Poisson mean, and Type I and Type II errors. This content is part of International A-level (A2) Unit S2 (Statistics), not International AS. Probabilities are given to 4 decimal places.

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Questions

1. A manufacturer says that 12% of its bolts are faulty. An inspector suspects that the true proportion is higher. Write down H₀ and H₁, saying what any symbol means. [3]

2. Under the current process, the mean number of bacterial colonies in a water sample is 6. A new filter is claimed to reduce this mean, and a test of H₀: λ = 6 against H₁: λ < 6 is carried out. Describe, in context, (a) a Type I error [2], (b) a Type II error [2].

3. Last year 45% of the members of a gym renewed their membership. The manager thinks that, after a fee rise, a smaller proportion will renew. Of 18 randomly chosen members, 4 renew. Carry out a test of the manager’s belief at the 5% level. [5]

Given: for X ~ B(18, 0.45), P(X ≤ 4) = 0.0411 and P(X ≤ 5) = 0.1077.

4. A random variable X has distribution B(20, 0.25). A test of H₀: p = 0.25 against H₁: p > 0.25 is to be carried out at the 5% level. Determine the critical region and the actual significance level. [5]

Given: P(X ≥ 8) = 0.1018, P(X ≥ 9) = 0.0409, P(X ≥ 10) = 0.0139.

5. A coin is suspected of being biased. It is tossed 16 times, and the number of heads X is recorded. (a) State suitable hypotheses for a two-tailed test. [2] (b) At the 10% level, determine the critical region. [3] (c) State the actual significance level of the test. [1] (d) The coin lands heads 12 times. State the conclusion. [1]

Given: for X ~ B(16, 0.5), P(X ≤ 4) = 0.0384, P(X ≤ 5) = 0.1051. The distribution is symmetrical about 8.

6. A help desk receives emails at a mean rate of 1.75 per hour. After a newsletter, the manager believes the rate has increased. In 4 hours, 12 emails arrive. (a) State the distribution of the number of emails in 4 hours, under the null hypothesis. [1] (b) State suitable hypotheses. [1] (c) Calculate the p-value. [2] (d) Conclude at the 5% level. [1] (e) State whether your conclusion changes at the 10% level. [1]

Given: for X ~ Po(7), P(X ≤ 11) = 0.9467.

7. Glass tiles have a mean of 6 bubbles per tile. After a furnace change, a manager uses one tile to test, at the 5% level, whether the mean has changed. (a) State suitable hypotheses. [1] (b) Find the critical region. [4] (c) What is the probability of a Type I error? [1]

Given: for X ~ Po(6), P(X ≤ 1) = 0.0174, P(X ≤ 2) = 0.0620, P(X ≥ 11) = 0.0426, P(X ≥ 12) = 0.0201.

8. A seed company says 10% of its seeds fail to germinate. A gardener believes the rate is higher. She plants 40 seeds; X seeds fail. (a) Find the critical region for a 5% test of her belief. [4] (b) State the actual significance level and explain what it means. [2] (c) In her sample 7 seeds fail. State the conclusion of the test in (a). [1] (d) She repeats the test at the 10% level. State the conclusion. [2]

Given: for X ~ B(40, 0.1), P(X ≥ 7) = 0.0995, P(X ≥ 8) = 0.0419, P(X ≥ 9) = 0.0155.

9. Flaws occur singly and at random along a ribbon, with a mean of 1.8 per metre. A new process is said to reduce the number of flaws. (a) State two conditions under which a Poisson distribution is a suitable model. [2] (b) State suitable hypotheses. [2] (c) A 5-metre length made by the new process has 3 flaws. Test the claim, using a 5% level. [3] (d) What would be the conclusion had there been 4 flaws? [1]

Given: for X ~ Po(9), P(X ≤ 3) = 0.0212, P(X ≤ 4) = 0.0550.

10. A student tests H₀: p = 0.5 against H₁: p ≠ 0.5 at the 10% level. For her observed value in the upper tail she finds a probability of 0.07 and writes “0.07 < 0.10, so reject H₀”. (a) Explain the error. [2] (b) State the correct conclusion. [2]

11. A nursery says that half of its seedlings flower in their first year. A rival believes fewer do. A sample of 10 seedlings is checked at the 5% level. Show that the critical region is X ≤ 1, and comment on the actual significance level. [5]

Given: for X ~ B(10, 0.5), P(X ≤ 1) = 0.0107 and P(X ≤ 2) = 0.0547.

Answers

1. Let p be the proportion of bolts that are faulty. [1] H₀: p = 0.12 [1] H₁: p > 0.12 [1] [3] Examiner insight: Define the parameter in words. H₁ follows the inspector’s suspicion.

2. (a) A Type I error is concluding that the mean number of colonies has fallen below 6 [1] when in fact it has not (H₀ is true). [1] (b) A Type II error is concluding that there is no evidence the mean has fallen [1] when in fact it has (H₀ is false). [1] [4] Examiner insight: Mention the decision made, the true situation and the context.

3. Let p be the probability that a member renews. H₀: p = 0.45, H₁: p < 0.45. [1] Under H₀, X ~ B(18, 0.45). [1] P(X ≤ 4) = 0.0411 [1] 0.0411 < 0.05, so reject H₀. [1] There is sufficient evidence at the 5% level that the renewal proportion has fallen. [1] [5] Examiner insight: Use the cumulative probability up to and including the observed value, and finish with a sentence about renewals.

4. Under H₀, X ~ B(20, 0.25). [1] P(X ≥ 8) = 0.1018 > 0.05, so 8 is not in the critical region. [1] P(X ≥ 9) = 0.0409 ≤ 0.05. [1] The critical region is X ≥ 9. [1] The actual significance level is 0.0409 (4.09%). [1] [5] Examiner insight: Show the probabilities either side of the boundary.

5. (a) Let p be the probability of heads. H₀: p = 0.5, H₁: p ≠ 0.5. [1] [1] (b) Each tail has at most 5%. Lower tail: P(X ≤ 4) = 0.0384 ≤ 0.05, but P(X ≤ 5) = 0.1051 > 0.05, so X ≤ 4. [1] By symmetry about 8, the upper tail is X ≥ 12 (since 16 − 4 = 12). [1] Critical region: X ≤ 4 or X ≥ 12. [1] (c) 0.0384 + 0.0384 = 0.0768. [1] (d) 12 is in the critical region, so reject H₀. There is sufficient evidence at the 10% level that the coin is biased. [1] [7] Examiner insight: Split the significance level first. Using 10% in each tail gives too large a test.

6. (a) X ~ Po(7), since 1.75 × 4 = 7. [1] (b) Let λ be the mean number of emails per hour. H₀: λ = 1.75, H₁: λ > 1.75. [1] (c) P(X ≥ 12) = 1 − P(X ≤ 11) [1] = 1 − 0.9467 = 0.0533. [1] (d) 0.0533 > 0.05, so do not reject H₀. There is insufficient evidence at the 5% level that the rate of emails has increased. [1] (e) 0.0533 < 0.10, so at the 10% level H₀ is rejected and the conclusion changes. [1] [6] Examiner insight: When the p-value is this close to the significance level, the “−1” in P(X ≥ 12) = 1 − P(X ≤ 11) decides the outcome.

7. (a) Let λ be the mean number of bubbles per tile. H₀: λ = 6, H₁: λ ≠ 6. [1] (b) Each tail has at most 2.5%. Lower tail: P(X ≤ 1) = 0.0174 ≤ 0.025, but P(X ≤ 2) = 0.0620 > 0.025. [1] So X ≤ 1. [1] Upper tail: P(X ≥ 12) = 0.0201 ≤ 0.025, but P(X ≥ 11) = 0.0426 > 0.025. [1] So X ≥ 12. [1] Critical region: X ≤ 1 or X ≥ 12. (c) P(Type I error) = 0.0174 + 0.0201 = 0.0374 (unrounded values). [1] [6] Examiner insight: Give the region as two parts joined by “or”.

8. (a) Under H₀, X ~ B(40, 0.1), H₁: p > 0.1. [1] P(X ≥ 7) = 0.0995 > 0.05. [1] P(X ≥ 8) = 0.0419 ≤ 0.05. [1] Critical region: X ≥ 8. [1] (b) The actual significance level is 0.0419. [1] It is the probability of a Type I error, that is, of rejecting H₀ when the failure rate is in fact 10%. [1] (c) 7 is not in the critical region, so do not reject H₀. There is insufficient evidence at the 5% level that the failure rate exceeds 10%. [1] (d) At 10%, P(X ≥ 7) = 0.0995 < 0.10. [1] Reject H₀ at the 10% level: sufficient evidence that the failure rate is higher. [1] [9] Examiner insight: Say “probability of rejecting H₀ when it is true”, not “the chance of being wrong”.

9. (a) Flaws occur singly [1] and independently at a constant average rate. [1] (b) Let λ be the mean number of flaws per metre. H₀: λ = 1.8, H₁: λ < 1.8. [1] [1] (c) Over 5 metres, under H₀, X ~ Po(9). [1] P(X ≤ 3) = 0.0212 [1] 0.0212 < 0.05, so reject H₀. There is sufficient evidence at the 5% level that the mean has fallen. [1] (d) P(X ≤ 4) = 0.0550 > 0.05, so do not reject H₀. [1] [8] Examiner insight: State the scaled distribution before calculating.

10. (a) The test is two-tailed, so each tail gets 5%: compare 0.07 with 0.05, not 0.10. [1] Equivalently, the p-value is 2 × 0.07 = 0.14. [1] (b) 0.07 > 0.05 (or 0.14 > 0.10), so do not reject H₀. [1] There is insufficient evidence at the 10% level that p differs from 0.5. [1] [4] Examiner insight: Say whether you double the tail or halve the level, and do not mix the two.

11. Under H₀, X ~ B(10, 0.5), and H₁: p < 0.5. [1] P(X ≤ 2) = 0.0547 > 0.05, so 2 is not in the critical region. [1] P(X ≤ 1) = 0.0107 ≤ 0.05. [1] So the critical region is X ≤ 1. [1] The actual significance level is 0.0107, about 1%, which is well below 5%. The distribution is discrete, and X ≤ 2 would exceed 5%. [1] [5] Examiner insight: In a “show that” question, give both boundary probabilities.

Where marks are usually lost

• Hypotheses with an undefined p or λ, an inequality in H₀, or statements about the sample.
• Using 1 − P(X ≤ c) where 1 − P(X ≤ c − 1) is needed.
• Not splitting the significance level for a two-tailed test.
• Leaving the Poisson mean unscaled.
• A conclusion with no context, or one that treats “do not reject” as proof of H₀.

Next steps

• Tighten your method with the revision notes.
• Re-read the study guide for any unfamiliar step.
• Go to the OxfordAQA A-level Mathematics hub and the printable checklist.
• Take one of the free 10-minute diagnostics.
• Book a free trial class on the trial page.

Official syllabus

OxfordAQA International AS and A-level Mathematics (9660), Version 5.2, published by OxfordAQA: Unit S2 (Statistics), section S2.6, Hypothesis testing.