OCR A Level Mathematics A: Hypothesis testing: language, binomial tests and the sample mean (H240) – 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 OCR A Level Mathematics A (H240) specification, Version 3.1 (August 2026), for first assessment in 2018, sections 2.05a to 2.05e. Questions 1 to 6 and 10 are Stage 1 (common with AS). Questions 7, 8, 9 and 11 are A Level only. Give conclusions in context.

Links: OCR Mathematics A hub, printable checklist, study guide, revision notes and free diagnostics.

Questions

1. A courier says that 8% of its parcels arrive late. A customer thinks the true percentage is higher and checks a random sample of 50 parcels. Write down suitable null and alternative hypotheses, defining any symbol you use. [3]

2. A sports club says that 35% of its members cycle to training. A coach thinks the proportion is lower. In a random sample of 16 members, 2 cycle to training. Test the coach’s belief at the 5% significance level. [5]

3. A manufacturer says that 10% of its phone screens have a flaw. An inspector tests a random sample of 40 screens, using a 5% significance level, to see whether the proportion is greater than 10%. Find the critical region and state the actual significance level of the test. [5]

4. A quiz setter says that 40% of players answer a particular question correctly. A sample of 18 players is taken to test, at the 5% level, whether the proportion differs from 40%. Twelve answer correctly. Carry out the test. [6]

5. This question uses the test in question 3. (a) State the probability of making a Type I error. [1] (b) The true proportion of flawed screens is in fact 0.2. Find the probability of a Type II error. [2] (c) Describe a Type II error in this context. [1] (d) State, with a reason, whether the probability of a Type II error would be greater or smaller if the true proportion were 0.3. [1]

6. A one-tail test of whether the mean journey time of a bus route has increased gives a p-value of 0.034. (a) State the conclusion at the 5% significance level, in context. [2] (b) State the conclusion at the 1% significance level. [1] (c) Explain what the value 0.034 represents. [1]

7. The mass X of a type of parcel is normally distributed with mean 120 g and standard deviation 15 g. A random sample of 25 parcels is taken, with sample mean X̄. (a) State the distribution of X̄. [2] (b) Find P(X̄ < 115). [2]

8. The battery life of a tablet is normally distributed with standard deviation 0.8 hours. The maker claims the mean is 12.5 hours. A reviewer thinks it is lower and finds that 36 tablets have a mean battery life of 12.2 hours. Test the reviewer’s belief at the 5% significance level. [6]

9. Reaction times are normally distributed with standard deviation 6 ms. It is claimed that the mean is 75 ms. A sample of 16 people is used to test, at the 5% level, whether the mean differs from 75 ms. (a) Find the critical region for the sample mean. [4] (b) The sample mean is 77.9 ms. State the conclusion of the test, in context. [3]

10. A school says that 30% of its Year 11 pupils study triple science. A governor thinks the true proportion is different. A random sample of 25 Year 11 pupils is taken and the number studying triple science is X. The test is at the 5% significance level. (a) Find the critical region for the test. [4] (b) Find the actual significance level. [1] (c) In the sample, 14 pupils study triple science. State the conclusion. [2] (d) Find the probability of a Type II error when the true proportion is 0.45. [2]

11. The lifetime of a type of bulb is normally distributed with standard deviation 40 hours. The maker says the mean lifetime is 800 hours. A tester thinks it is lower, and uses a random sample of 25 bulbs and a 5% significance level. (a) Write down the hypotheses and the distribution of the sample mean under the null hypothesis. [2] (b) Find the critical region for the sample mean. [3] (c) The sample mean is 785.2 hours. State the conclusion in context. [2] (d) The true mean lifetime is 778 hours. Find the probability of a Type II error. [2]

Answers

1. Define p as the probability that a parcel arrives late (the population proportion) [1]. H₀: p = 0.08 [1]. H₁: p > 0.08 [1]. [3] Examiner insight: Define the parameter in words, and keep sample results out of the hypotheses.

2. H₀: p = 0.35, H₁: p < 0.35, where p is the probability that a member cycles to training [1]. Under H₀, X ~ B(16, 0.35) [1]. P(X ≤ 2) = 0.0451 [1]. 0.0451 < 0.05 [1]. Reject H₀. At the 5% level the evidence suggests that the proportion who cycle is less than 0.35 [1]. [5] Examiner insight: Finish with a sentence in context that sounds tentative; a bare “reject H₀” or “the coach is right” leaves the conclusion incomplete.

3. H₀: p = 0.1, H₁: p > 0.1. Under H₀, X ~ B(40, 0.1) [1]. P(X ≥ 8) = 1 − P(X ≤ 7) = 0.0419 ≤ 0.05 [1]. P(X ≥ 7) = 0.0995 > 0.05 [1]. Critical region: X ≥ 8 [1]. Actual significance level: 0.0419 (4.19%) [1]. [5] Examiner insight: Show both boundary probabilities, one inside the level and one outside it. A region quoted alone leaves the reasoning unseen.

4. H₀: p = 0.4, H₁: p ≠ 0.4, where p is the probability that a player answers correctly [1]. Under H₀, X ~ B(18, 0.4) [1]. P(X ≥ 12) = 1 − P(X ≤ 11) = 0.0203 [1]. This is the upper tail of a two-tail test, so compare with 0.025 (or double to 0.0406 and compare with 0.05) [1]. 0.0203 < 0.025, so reject H₀ [1]. At the 5% level the evidence suggests that the proportion answering correctly differs from 0.4; it is likely to be higher [1]. [6] Examiner insight: In a two-tail test, say which comparison you are making. A single-tail probability set against the full 5% is a frequent slip.

5. (a) P(Type I error) = 0.0419 [1]. (b) Type II error: X does not lie in the critical region, so X ≤ 7 when X ~ B(40, 0.2) [1]. P(X ≤ 7) = 0.437 [1]. (c) Finding no evidence that more than 10% of screens are flawed when the true figure is 20% [1]. (d) Smaller: 0.3 is further from 0.1, so P(X ≤ 7) falls to 0.0553 [1]. [5] Examiner insight: Describe errors in the words of the situation. A general phrase with no context is a weaker answer.

6. (a) 0.034 < 0.05, so reject H₀ [1]. At the 5% level the evidence suggests that the mean journey time has increased [1]. (b) 0.034 > 0.01, so the data do not support rejecting H₀ at the 1% level [1]. (c) How likely a sample this extreme (or more so) would be if the null hypothesis were true [1]. [4] Examiner insight: The comparison is the decision; the sentence in context is a separate part of the answer. State both.

7. (a) X̄ ~ N(120, 15²/25) [1], so X̄ ~ N(120, 3²) [1]. (b) z = (115 − 120)/3 = −1.667 [1]. P(X̄ < 115) = 0.0478 [1]. [4] Examiner insight: A common loss is using 15 rather than 15/5 = 3. A single parcel gives 0.369, so the answer for a mean should be smaller.

8. H₀: μ = 12.5, H₁: μ < 12.5, where μ is the population mean battery life in hours [1]. Under H₀, X̄ ~ N(12.5, (0.8/6)²), standard deviation 0.1333 [1]. z = (12.2 − 12.5)/0.1333 = −2.25 [1]. p-value = P(Z < −2.25) = 0.0122 (or compare −2.25 with the critical value −1.645) [1]. 0.0122 < 0.05, so reject H₀ [1]. At the 5% level the evidence suggests that the mean battery life is less than 12.5 hours [1]. [6] Examiner insight: Define μ in context and state the distribution of X̄ under H₀. A stated distribution shows your method even if a later value slips.

9. (a) Under H₀, X̄ ~ N(75, 6²/16) [1]. Standard deviation = 6/4 = 1.5 [1]. Critical values 75 ± 1.960 × 1.5 [1]. Critical region: X̄ < 72.06 or X̄ > 77.94 (2 d.p.) [1]. (b) 77.9 < 77.94, so it lies in the acceptance region [1]. There is no evidence at the 5% level to reject H₀ [1]. There is no reason to suppose the mean reaction time differs from 75 ms [1]. [7] Examiner insight: The sample mean is very close to the boundary, so keep enough decimal places. Trimming the boundary to 77.8 would wrongly put 77.9 in the critical region.

10. (a) H₀: p = 0.3, H₁: p ≠ 0.3, X ~ B(25, 0.3), 2.5% in each tail [1]. Lower: P(X ≤ 2) = 0.0090 ≤ 0.025, P(X ≤ 3) = 0.0332 > 0.025 [1]. Upper: P(X ≥ 13) = 0.0175 ≤ 0.025, P(X ≥ 12) = 0.0442 > 0.025 [1]. Critical region: X ≤ 2 or X ≥ 13 [1]. (b) 0.0090 + 0.0175 = 0.0264 [1]. (c) 14 ≥ 13, so it lies in the critical region: reject H₀ [1]. At the 5% level the evidence suggests that the proportion of pupils studying triple science is not 0.3; it is likely to be higher [1]. (d) X ~ B(25, 0.45) and a Type II error is 3 ≤ X ≤ 12 [1]. P(X ≤ 12) − P(X ≤ 2) = 0.694 [1]. [9] Examiner insight: In part (d) the acceptance region has two boundaries. Using only one is a common source of lost marks.

11. (a) H₀: μ = 800, H₁: μ < 800, where μ is the population mean lifetime in hours [1]. Under H₀, X̄ ~ N(800, 40²/25) = N(800, 8²) [1]. (b) Critical z-value is −1.645 [1]. 800 − 1.645 × 8 [1]. Critical region: X̄ < 786.84 (2 d.p.) [1]. (c) 785.2 < 786.84, so it lies in the critical region: reject H₀ [1]. At the 5% level the evidence suggests that the mean lifetime is less than 800 hours [1]. (d) Type II error: X̄ ≥ 786.84 when X̄ ~ N(778, 8²) [1]. z = (786.84 − 778)/8 = 1.105, so probability = 0.135 [1]. [9] Examiner insight: Part (d) centres a new distribution on the true mean, 778, not the null value 800.

Where marks are usually lost

• Writing hypotheses with a sample value, or not defining p or μ; also not stating the distribution under H₀.
• Using P(X = x) or P(X > x) instead of P(X ≥ x), so the critical value is out by one.
• Comparing a single-tail probability with the full level in a two-tail test.
• Forgetting to divide σ by √n for a sample mean.
• Giving a conclusion such as “accept H₀” or “the claim is true”, or a conclusion that is not in context.
• Finding a Type II error by using H₀’s parameter instead of the true value.

Next steps

• Review the revision notes and the study guide.
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Official syllabus

OCR A Level Mathematics A (H240) specification, Version 3.1 (August 2026), for first assessment in 2018, published by OCR. This page covers section 2.05, parts a to e.