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AS & A Level Mathematics (9709)

Hypothesis testing Banker

59 questions · about 4% of the marks we indexed · examined in 9 of 9 sessions.

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2026

  • 2026 May/June · Paper 61 · Q2 (a, b, c)5 marksHypothesis test (binomial) · Type I / Type II errors
  • 2026 May/June · Paper 61 · Q4 (d)5 marksHypothesis test (normal mean)
  • 2026 May/June · Paper 62 · Q3 (a, b, c)6 marksHypothesis test (Poisson) · Type I / Type II errors
  • 2026 May/June · Paper 62 · Q6 (a, c)8 marksHypothesis test (normal mean)
  • 2026 May/June · Paper 63 · Q2 (a, b)6 marksHypothesis test (binomial)
  • 2026 May/June · Paper 63 · Q5 (b, c, d)11 marksHypothesis test (normal mean) · Type I / Type II errors
  • 2026 May/June · Paper 65 · Q5 (a, b, c)7 marksHypothesis test (binomial) · Type I / Type II errors
  • 2026 May/June · Paper 65 · Q6 (b(i))5 marksHypothesis test (normal mean)

2025

  • 2025 Oct/Nov · Paper 61 · Q28 marksHypothesis test (normal mean)
  • 2025 Oct/Nov · Paper 61 · Q5 (a, b, c)6 marksHypothesis test (binomial) · Type I / Type II errors
  • 2025 Oct/Nov · Paper 62 · Q4 (a, b)7 marksHypothesis test (binomial)
  • 2025 Oct/Nov · Paper 62 · Q6 (a, b, c)12 marksHypothesis test (normal mean) · Type I / Type II errors
  • 2025 Oct/Nov · Paper 63 · Q28 marksHypothesis test (normal mean)
  • 2025 Oct/Nov · Paper 63 · Q5 (a, b, c)6 marksHypothesis test (binomial) · Type I / Type II errors
  • 2025 Oct/Nov · Paper 65 · Q2 (a, b, c)4 marksHypothesis test (binomial)
  • 2025 Oct/Nov · Paper 65 · Q3 (a, b)8 marksHypothesis test (Poisson) · Type I / Type II errors
  • 2025 May/June · Paper 61 · Q3 (b)5 marksHypothesis test (normal mean)
  • 2025 May/June · Paper 61 · Q6 (a, b)4 marksHypothesis test (binomial) · Type I / Type II errors
  • 2025 May/June · Paper 62 · Q54 marksHypothesis test (normal mean)
  • 2025 May/June · Paper 62 · Q8 (a, b, c)10 marksHypothesis test (binomial) · Type I / Type II errors
  • 2025 May/June · Paper 63 · Q2 (a(i), a(ii))6 marksHypothesis test (normal mean)
  • 2025 May/June · Paper 63 · Q7 (a, b, c)10 marksHypothesis test (binomial) · Type I / Type II errors
  • 2025 May/June · Paper 65 · Q6 (a, b, c)8 marksHypothesis test (binomial) · Type I / Type II errors
  • 2025 May/June · Paper 65 · Q7 (a, b, c)5 marksHypothesis test (normal mean)

2024

  • 2024 Oct/Nov · Paper 61 · Q5 (a, b)6 marksHypothesis test (normal mean)
  • 2024 Oct/Nov · Paper 61 · Q7 (a, b, c)10 marksHypothesis test (Poisson) · Type I / Type II errors
  • 2024 Oct/Nov · Paper 62 · Q3 (a, b)6 marksHypothesis test (binomial)
  • 2024 Oct/Nov · Paper 62 · Q7 (a, b, c)9 marksHypothesis test (normal mean) · Type I / Type II errors
  • 2024 Oct/Nov · Paper 63 · Q5 (a, b)6 marksHypothesis test (normal mean)
  • 2024 Oct/Nov · Paper 63 · Q7 (a, b, c)10 marksHypothesis test (Poisson) · Type I / Type II errors
  • 2024 May/June · Paper 61 · Q4 (b(i), b(ii))6 marksOfficial paper ↗Hypothesis test (normal mean)
  • 2024 May/June · Paper 61 · Q7 (a, b)7 marksOfficial paper ↗Hypothesis test (Poisson) · Type I / Type II errors
  • 2024 May/June · Paper 62 · Q6 (a, b)10 marksHypothesis test (normal mean) · Type I / Type II errors
  • 2024 May/June · Paper 63 · Q45 marksHypothesis test (binomial)
  • 2024 May/June · Paper 63 · Q6 (b, c, d)6 marksHypothesis test (normal mean) · Type I / Type II errors

2023

  • 2023 Oct/Nov · Paper 61 · Q55 marksHypothesis test (Poisson)
  • 2023 Oct/Nov · Paper 61 · Q7 (a, b)12 marksHypothesis test (Poisson) · Type I / Type II errors
  • 2023 Oct/Nov · Paper 62 · Q3 (a(i), a(ii), b, c)9 marksHypothesis test (binomial) · Type I / Type II errors
  • 2023 Oct/Nov · Paper 62 · Q4 (a, b)4 marksHypothesis test (normal mean)
  • 2023 Oct/Nov · Paper 63 · Q55 marksHypothesis test (Poisson)
  • 2023 Oct/Nov · Paper 63 · Q7 (a, b)12 marksHypothesis test (Poisson) · Type I / Type II errors
  • 2023 May/June · Paper 61 · Q36 marksHypothesis test (normal mean)
  • 2023 May/June · Paper 61 · Q7 (a, b, c, d)10 marksHypothesis test (Poisson) · Type I / Type II errors
  • 2023 May/June · Paper 62 · Q3 (b)5 marksHypothesis test (normal mean)
  • 2023 May/June · Paper 62 · Q6 (a, b)5 marksHypothesis test (binomial) · Type I / Type II errors
  • 2023 May/June · Paper 63 · Q5 (b)5 marksHypothesis test (normal mean)
  • 2023 May/June · Paper 63 · Q8 (a, b, c, d)11 marksHypothesis test (Poisson) · Type I / Type II errors

2022

  • 2022 Oct/Nov · Paper 61 · Q2 (a, b)8 marksHypothesis test (binomial)
  • 2022 Oct/Nov · Paper 61 · Q7 (a, b)10 marksHypothesis test (normal mean) · Type I / Type II errors
  • 2022 Oct/Nov · Paper 62 · Q25 marksHypothesis test (normal mean)
  • 2022 Oct/Nov · Paper 62 · Q4 (a, b, c, d)8 marksHypothesis test (Poisson) · Type I / Type II errors
  • 2022 Oct/Nov · Paper 63 · Q2 (a, b)8 marksHypothesis test (binomial)
  • 2022 Oct/Nov · Paper 63 · Q7 (a, b)10 marksHypothesis test (normal mean) · Type I / Type II errors
  • 2022 May/June · Paper 61 · Q25 marksHypothesis test (binomial)
  • 2022 May/June · Paper 61 · Q7 (a, b, c)6 marksHypothesis test (normal mean) · Type I / Type II errors
  • 2022 May/June · Paper 62 · Q25 marksHypothesis test (normal mean)
  • 2022 May/June · Paper 62 · Q4 (a, b, c, d)8 marksHypothesis test (Poisson) · Type I / Type II errors
  • 2022 May/June · Paper 63 · Q2 (a, b, c)6 marksType I / Type II errors
  • 2022 May/June · Paper 63 · Q3 (b)5 marksHypothesis test (normal mean)

References only, not question text. Open Official paper ↗ where Cambridge publishes the paper, or see the official Cambridge past-papers page; for other papers, ask your school. “Paper 42” means Paper 4, variant 2.

Practice questions

Original questions written by HomiLabs in the style of the exam, with answers and mark schemes. They are not Cambridge past paper questions and have not been reviewed by Cambridge. Try each one before opening the answer.

Q1Hypothesis test (normal mean) · Core · Paper 6 style · 5 marks
  1. (a) A rice mill in Sindh fills bags with a nominal mass of 5 kg. The masses are normally distributed with standard deviation 0.08 kg. An inspector suspects that the mean mass is less than 5 kg. A random sample of 20 bags has mean mass 4.962 kg. Test at the 5% significance level whether the inspector's suspicion is justified. [5]
Show answer
  1. (a) z = −2.12 < −1.645, so reject H₀: there is evidence that the mean mass is less than 5 kg.
    B1 H₀: μ = 5, H₁: μ < 5; M1 standardise with 0.08/√20; A1 z = −2.12 (or p = 0.0168); M1 compare with −1.645 (or 0.0168 < 0.05); A1 reject H₀, conclusion in context, not over-assertive.
Q2Hypothesis test (normal mean) · Core · Paper 6 style · 5 marks
  1. (a) The heights of seedlings of a certain plant after 4 weeks are normally distributed with mean 12.4 cm and standard deviation 1.8 cm. A gardener uses a new fertiliser on 40 randomly chosen seedlings and their mean height after 4 weeks is 12.95 cm. Assuming the standard deviation is unchanged, test at the 5% significance level whether the fertiliser increases the mean height. [5]
Show answer
  1. (a) z = 1.93 > 1.645, so reject H₀: evidence that the mean height has increased.
    B1 H₀: μ = 12.4, H₁: μ > 12.4; M1 (12.95 − 12.4)/(1.8/√40); A1 1.93; M1 compare with 1.645; A1 conclusion in context.
Q3Hypothesis test (normal mean) · Core · Paper 6 style · 6 marks
  1. (a) A health magazine claims that university students sleep 8 hours per night on average. A researcher believes the true mean is lower. The number of hours of sleep is normally distributed with standard deviation 1.2 hours. A random sample of 50 students has mean 7.6 hours. Explain why a one-tailed test is appropriate. [1]
  2. (b) Carry out the test at the 5% significance level. [5]
Show answer
  1. (a) The researcher is only looking for evidence that the mean is lower than 8.
    B1.
  2. (b) z = −2.36 < −1.645, so reject H₀: evidence that the mean sleep time is less than 8 hours.
    B1 H₀: μ = 8, H₁: μ < 8; M1 (7.6 − 8)/(1.2/√50); A1 −2.36; M1 compare with −1.645; A1 conclusion in context.
Q4Hypothesis test (normal mean) · Standard · Paper 6 style · 9 marks
  1. (a) The lifetimes, x hours, of a random sample of 50 batteries of a certain brand were recorded, giving Σx = 1218 and Σx² = 29 900. Calculate unbiased estimates of the population mean and variance. [3]
  2. (b) The manufacturer claims that the mean lifetime is 25 hours. Test at the 2% significance level whether the mean lifetime differs from 25 hours. [5]
  3. (c) Explain why it was not necessary to assume that the lifetimes are normally distributed. [1]
Show answer
  1. (a) x̄ = 24.36; s² = 4.68 (4.684)
    B1 24.36; M1 (29900 − 1218²/50)/49; A1 4.684.
  2. (b) z = −2.09; |z| < 2.326, so do not reject H₀: insufficient evidence that the mean lifetime differs from 25 hours.
    B1 H₀: μ = 25, H₁: μ ≠ 25; M1 standardise using √(4.684/50); A1 z = −2.09; M1 compare with ±2.326 (two-tailed 2%); A1 do not reject H₀, conclusion in context.
  3. (c) The sample is large, so by the Central Limit Theorem the sample mean is approximately normally distributed.
    B1.
Q5Hypothesis test (normal mean) · Standard · Paper 6 style · 7 marks
  1. (a) A food delivery company says its delivery times have mean 30 minutes and standard deviation 6 minutes. After a change of routes, a random sample of 36 deliveries has mean 32.1 minutes. Assuming the standard deviation is unchanged, test at the 1% significance level whether the mean delivery time has increased. [5]
  2. (b) Find the p-value of the test and state the smallest significance level, to the nearest 0.1%, at which H₀ would be rejected. [2]
Show answer
  1. (a) z = 2.1 < 2.326, so do not reject H₀: insufficient evidence at 1% that the mean has increased.
    B1 H₀: μ = 30, H₁: μ > 30; M1 (32.1 − 30)/(6/√36); A1 z = 2.1; M1 compare with 2.326; A1 conclusion in context.
  2. (b) p = 0.0179; about 1.8%
    M1 1 − Φ(2.1); A1 0.0179, so reject at levels above 1.79%.
Q6Hypothesis test (normal mean) · Standard · Paper 6 style · 6 marks
  1. (a) A machine fills bottles with mineral water. The volume is normally distributed with mean 500 ml and standard deviation 4 ml when the machine is working correctly. A random sample of 15 bottles has mean volume 498.1 ml. Test at the 10% significance level whether the mean volume has changed. [5]
  2. (b) State one assumption about the sample needed for the test to be valid. [1]
Show answer
  1. (a) z = −1.84; |z| > 1.645, so reject H₀: evidence that the mean volume has changed.
    B1 H₀: μ = 500, H₁: μ ≠ 500; M1 (498.1 − 500)/(4/√15); A1 −1.84; M1 compare with ±1.645 (5% in each tail); A1 reject H₀, in context.
  2. (b) The sample is random (the bottles are chosen independently).
    B1.
Q7Hypothesis test (normal mean) · Standard · Paper 6 style · 8 marks
  1. (a) The journey time, x minutes, of a bus route in Karachi was recorded for a random sample of 60 journeys, giving Σ(x − 40) = 90 and Σ(x − 40)² = 1505. Find unbiased estimates of the population mean and variance of the journey time. [3]
  2. (b) The bus company states that the mean journey time is 40 minutes. Test at the 1% significance level whether the mean journey time is greater than 40 minutes. [5]
Show answer
  1. (a) x̄ = 41.5; s² = 23.2 (23.22)
    B1 40 + 90/60 = 41.5; M1 (1505 − 90²/60)/59; A1 23.22.
  2. (b) z = 2.41 > 2.326, so reject H₀: evidence that the mean journey time is greater than 40 minutes.
    B1 H₀: μ = 40, H₁: μ > 40; M1 1.5/√(23.22/60); A1 2.41; M1 compare with 2.326; A1 conclusion in context.
Q8Hypothesis test (normal mean) · Challenge · Paper 6 style · 5 marks
  1. (a) The scores in an aptitude test are normally distributed with mean 60 and standard deviation 4.5. A coaching centre suspects that students from a certain region score lower on average. A random sample of 25 such students is taken. Find the critical region for the sample mean for a test at the 5% significance level. [4]
  2. (b) The sample mean is found to be 58.7. State, with a reason, the conclusion of the test. [1]
Show answer
  1. (a) x̄ < 58.52 (58.5195)
    B1 H₀: μ = 60, H₁: μ < 60; M1 60 − 1.645 × 4.5/√25; A1 58.52; B1 critical region stated as x̄ < 58.52.
  2. (b) 58.7 is not in the critical region, so do not reject H₀: insufficient evidence that students from the region score lower.
    B1.
Q9Hypothesis test (normal mean) · Challenge · Paper 6 style · 4 marks
  1. (a) The masses of mangoes from an orchard are normally distributed with mean 80 g and standard deviation 10 g. After a new irrigation system is installed, a random sample of n mangoes has mean mass 82.5 g. A test at the 5% significance level of H₀: μ = 80 against H₁: μ > 80 leads to rejection of H₀. Find the least possible value of n. [4]
Show answer
  1. (a) n = 44
    M1 2.5/(10/√n) > 1.645; M1 √n > 6.58; A1 n > 43.3; A1 least n = 44.
Q10Hypothesis test (normal mean) · Challenge · Paper 6 style · 5 marks
  1. (a) A test of H₀: μ = 15 against H₁: μ ≠ 15 is carried out on a normal population with standard deviation 2.4. A random sample of size 30 gives a sample mean of 15.82. Find the p-value of the test. [4]
  2. (b) Hence state the set of significance levels, α%, for which H₀ would be rejected. [1]
Show answer
  1. (a) p = 0.0613 (z = 1.87)
    M1 standardise: 0.82/(2.4/√30); A1 z = 1.871; M1 two-tailed: 2(1 − Φ(1.871)); A1 0.0613.
  2. (b) α > 6.13
    B1.
Q11Type I / Type II errors · Core · Paper 6 style · 3 marks
  1. (a) A factory tests whether the mean breaking strength of its cables has fallen below 400 N, using a 5% significance level. Explain what is meant by a Type I error in this context. [1]
  2. (b) Explain what is meant by a Type II error in this context. [1]
  3. (c) State the probability of making a Type I error. [1]
Show answer
  1. (a) Concluding that the mean breaking strength has fallen below 400 N when in fact it has not.
    B1 rejecting H₀ when it is true, in context.
  2. (b) Concluding that the mean breaking strength has not fallen when in fact it has fallen below 400 N.
    B1 not rejecting H₀ when it is false, in context.
  3. (c) 0.05
    B1 equal to the significance level (the test statistic has a continuous distribution).
Q12Type I / Type II errors · Core · Paper 6 style · 4 marks
  1. (a) Sara suspects that a coin is biased towards heads. She tosses it 10 times and will conclude that it is biased if she obtains 8 or more heads. Find the probability of a Type I error. [3]
  2. (b) Sara obtains 6 heads and concludes that the coin is not biased. State which type of error she might have made, giving a reason. [1]
Show answer
  1. (a) 0.0547 (= 56/1024)
    M1 P(X ≥ 8) with B(10, 0.5); M1 (45 + 10 + 1)/1024; A1 0.0547.
  2. (b) Type II, because she did not reject H₀ (H₀ could be false).
    B1.
Q13Type I / Type II errors · Core · Paper 6 style · 5 marks
  1. (a) The reaction times of drivers are normally distributed with mean 100 ms and standard deviation 10 ms. A researcher tests H₀: μ = 100 against H₁: μ > 100 using a sample of 25 drivers who have been awake for 20 hours, and rejects H₀ if the sample mean exceeds 103.29 ms. Find the probability of a Type I error. [3]
  2. (b) The sample mean is 104.1 ms. State the conclusion of the test and the type of error that could have been made. [2]
Show answer
  1. (a) 0.05
    M1 standard error 10/√25 = 2; M1 z = 3.29/2 = 1.645; A1 1 − Φ(1.645) = 0.05.
  2. (b) Reject H₀ (104.1 > 103.29): evidence of a longer mean reaction time; a Type I error could have been made.
    B1 conclusion; B1 Type I.
Q14Type I / Type II errors · Standard · Paper 6 style · 6 marks
  1. (a) In the past, 30% of customers at a bakery bought a cake. After the price rises, the owner tests H₀: p = 0.3 against H₁: p < 0.3 using a random sample of 20 customers. He will reject H₀ if 2 or fewer customers buy a cake. Find the probability of a Type I error. [3]
  2. (b) Given that the true proportion is now 0.15, find the probability of a Type II error. [3]
Show answer
  1. (a) 0.0355
    M1 P(X ≤ 2) with B(20, 0.3); M1 0.7²⁰ + 20(0.3)(0.7)¹⁹ + 190(0.3)²(0.7)¹⁸; A1 0.0355.
  2. (b) 0.595
    M1 P(X ≥ 3) with B(20, 0.15); M1 1 − P(X ≤ 2); A1 0.595.
Q15Type I / Type II errors · Standard · Paper 6 style · 5 marks
  1. (a) The number of complaints received by a call centre per day has a Poisson distribution. In the past the mean was 4. A manager will conclude that the mean has increased if 9 or more complaints are received on a randomly chosen day. Find the probability of a Type I error. [3]
  2. (b) Given that the mean is actually 6, find the probability of a Type II error. [2]
Show answer
  1. (a) 0.0214
    M1 P(X ≥ 9) with Po(4); M1 1 − P(X ≤ 8); A1 0.0214.
  2. (b) 0.847
    M1 P(X ≤ 8) with Po(6); A1 0.847.
Q16Type I / Type II errors · Standard · Paper 6 style · 7 marks
  1. (a) The masses of packets of tea are normally distributed with standard deviation 12 g. A test of H₀: μ = 250 against H₁: μ ≠ 250 at the 5% significance level uses a random sample of 16 packets. Find the acceptance region for the sample mean. [3]
  2. (b) State the probability of a Type I error. [1]
  3. (c) Given that the true mean is 256 g, find the probability of a Type II error. [3]
Show answer
  1. (a) 244.12 ≤ x̄ ≤ 255.88
    M1 σ/√n = 3; M1 250 ± 1.96 × 3; A1.
  2. (b) 0.05
    B1.
  3. (c) 0.484
    M1 P(244.12 ≤ X̄ ≤ 255.88 | μ = 256); M1 Φ(−0.04) − Φ(−3.96); A1 0.484.
Q17Type I / Type II errors · Challenge · Paper 6 style · 6 marks
  1. (a) The masses of bags of sugar filled by a machine are normally distributed with standard deviation 6 g. A test of H₀: μ = 1000 against H₁: μ < 1000 uses a random sample of 36 bags and rejects H₀ if the sample mean is less than 998.355 g. Show that the significance level of the test is 5%. [2]
  2. (b) Given that the true mean is 997 g, find the probability of a Type II error. [3]
  3. (c) Without further calculation, state how the probability of a Type II error would change if the true mean were 995 g, and explain why. [1]
Show answer
  1. (a) P(X̄ < 998.355 | μ = 1000) = Φ((998.355 − 1000)/1) = Φ(−1.645) = 0.05
    M1 standardise with 6/√36 = 1; A1 z = −1.645, probability 0.05 (5%).
  2. (b) 0.0877
    M1 P(X̄ ≥ 998.355 | μ = 997); M1 z = (998.355 − 997)/1 = 1.355; A1 1 − Φ(1.355) = 0.0877 (accept 0.0869–0.0885).
  3. (c) It would decrease, because the true mean is further below the critical value, so the sample mean is more likely to fall in the critical region.
    B1 decrease with reason.
Q18Type I / Type II errors · Challenge · Paper 6 style · 6 marks
  1. (a) The number of breakdowns of a fleet of buses has a Poisson distribution with mean 2.5 per week. After new servicing is introduced, the manager records the number of breakdowns in a 4-week period. He will conclude that the mean has decreased if there are 4 or fewer breakdowns in the 4 weeks. Find the probability of a Type I error. [3]
  2. (b) Given that the mean has actually fallen to 1.5 per week, find the probability of a Type II error. [3]
Show answer
  1. (a) 0.0293
    B1 mean for 4 weeks = 10; M1 P(X ≤ 4) with Po(10); A1 0.0293.
  2. (b) 0.715
    B1 mean 6; M1 P(X ≥ 5) = 1 − P(X ≤ 4) with Po(6); A1 0.715.
Q19Type I / Type II errors · Challenge · Paper 6 style · 6 marks
  1. (a) A hospital claims that 40% of its patients are seen within an hour. A patients' group believes the true proportion is higher. They plan to ask 12 randomly chosen patients and test H₀: p = 0.4 against H₁: p > 0.4 at the 5% significance level. Find the critical region and the probability of a Type I error. [4]
  2. (b) Find the probability of a Type II error if the true proportion is 0.7. [2]
Show answer
  1. (a) X ≥ 9; P(Type I) = 0.0153
    M1 B(12, 0.4) upper-tail probabilities; A1 P(X ≥ 8) = 0.0573 > 0.05; A1 P(X ≥ 9) = 0.0153 < 0.05, so critical region X ≥ 9; B1 P(Type I) = 0.0153.
  2. (b) 0.507
    M1 P(X ≤ 8) with B(12, 0.7); A1 0.507.
Q20Hypothesis test (binomial) · Core · Paper 6 style · 5 marks
  1. (a) Imran thinks a coin is biased towards heads. He tosses it 15 times and obtains 11 heads. Test Imran's belief at the 5% significance level. [5]
Show answer
  1. (a) P(X ≥ 11) = 0.0592 > 0.05, so do not reject H₀: insufficient evidence that the coin is biased towards heads.
    B1 H₀: p = 0.5, H₁: p > 0.5; M1 B(15, 0.5); M1 P(X ≥ 11) = (1365 + 455 + 105 + 15 + 1)/32768; A1 0.0592; A1 compare with 0.05 and conclude in context.
Q21Hypothesis test (binomial) · Standard · Paper 6 style · 5 marks
  1. (a) A seed company claims that 80% of its seeds germinate. A farmer in Punjab thinks the claim is too high. He plants 25 randomly chosen seeds and 16 germinate. Test the company's claim at the 5% significance level. [5]
Show answer
  1. (a) P(X ≤ 16) = 0.0468 < 0.05, so reject H₀: evidence that fewer than 80% germinate.
    B1 H₀: p = 0.8, H₁: p < 0.8; M1 B(25, 0.8), lower tail; M1 P(X ≤ 16); A1 0.0468; A1 conclusion in context.
Q22Hypothesis test (binomial) · Standard · Paper 6 style · 5 marks
  1. (a) A six-sided die is thrown 30 times and a six is obtained 10 times. Test at the 5% significance level whether the probability of obtaining a six is different from 1/6. [5]
Show answer
  1. (a) P(X ≥ 10) = 0.0197 < 0.025, so reject H₀: evidence that the probability of a six is not 1/6.
    B1 H₀: p = 1/6, H₁: p ≠ 1/6; M1 B(30, 1/6), upper tail since 10 > 5 (the expected number); M1 P(X ≥ 10); A1 0.0197; A1 compare with 0.025 (two-tailed) and conclude.
Q23Hypothesis test (binomial) · Standard · Paper 6 style · 6 marks
  1. (a) In the past, 40% of a bank's customers used its mobile app. After an advertising campaign, a random sample of 18 customers is taken and 11 of them use the app. Test at the 5% significance level whether the proportion has increased. [5]
  2. (b) Find the least number of app users in the sample that would have led to the conclusion that the proportion has increased. [1]
Show answer
  1. (a) P(X ≥ 11) = 0.0576 > 0.05, so do not reject H₀: insufficient evidence that the proportion has increased.
    B1 H₀: p = 0.4, H₁: p > 0.4; M1 B(18, 0.4); M1 P(X ≥ 11) = 1 − P(X ≤ 10); A1 0.0576; A1 conclusion in context.
  2. (b) 12
    B1 P(X ≥ 12) = 0.0203 < 0.05.
Q24Hypothesis test (binomial) · Challenge · Paper 6 style · 6 marks
  1. (a) It is known that 25% of visitors to a museum in Lahore are tourists from abroad. A guide wishes to test, at the 10% significance level, whether this proportion has changed, using a random sample of 20 visitors. Using a two-tailed test with at most 5% in each tail, find the critical region. [4]
  2. (b) State the actual significance level of the test. [1]
  3. (c) In the sample, 8 visitors are tourists from abroad. State the conclusion of the test. [1]
Show answer
  1. (a) X ≤ 1 or X ≥ 9
    M1 B(20, 0.25); A1 lower: P(X ≤ 1) = 0.0243, P(X ≤ 2) = 0.0913; A1 upper: P(X ≥ 9) = 0.0409, P(X ≥ 8) = 0.1018; A1 critical region X ≤ 1 or X ≥ 9.
  2. (b) 6.52% (0.0652)
    B1 0.0243 + 0.0409.
  3. (c) 8 is not in the critical region, so do not reject H₀: insufficient evidence that the proportion has changed.
    B1.
Q25Hypothesis test (binomial) · Challenge · Paper 6 style · 6 marks
  1. (a) A drug is effective for 35% of patients. A new drug is tested on 10 patients to see whether it is more effective, using a 2.5% significance level. Find the critical region for the test. [4]
  2. (b) The new drug is effective for 7 of the 10 patients. State the conclusion, and comment on the fact that P(X ≥ 7) is very close to the significance level. [2]
Show answer
  1. (a) X ≥ 8
    B1 H₀: p = 0.35, H₁: p > 0.35; M1 B(10, 0.35) upper tail; A1 P(X ≥ 7) = 0.0260 > 0.025 and P(X ≥ 8) = 0.0048; A1 critical region X ≥ 8.
  2. (b) Do not reject H₀ (7 is not in the critical region); the result is only just not significant, so a larger trial would be sensible.
    B1 conclusion in context; B1 sensible comment.
Q26Hypothesis test (Poisson) · Core · Paper 6 style · 5 marks
  1. (a) Calls to a helpline arrive at random at a mean rate of 3.2 per 10 minutes. After a television advert, 7 calls arrive in a randomly chosen 10-minute period. Test at the 5% significance level whether the mean rate has increased. [5]
Show answer
  1. (a) P(X ≥ 7) = 0.0446 < 0.05, so reject H₀: evidence that the rate has increased.
    B1 H₀: λ = 3.2, H₁: λ > 3.2; M1 Po(3.2); M1 1 − P(X ≤ 6); A1 0.0446; A1 conclusion in context.
Q27Hypothesis test (Poisson) · Standard · Paper 6 style · 5 marks
  1. (a) A typist makes errors at random at a mean rate of 1.8 per page. After using new software, a random sample of 5 pages contains 4 errors in total. Test at the 5% significance level whether the mean rate has decreased. [5]
Show answer
  1. (a) P(X ≤ 4) = 0.0550 > 0.05, so do not reject H₀: insufficient evidence that the error rate has decreased.
    B1 H₀: λ = 9 (for 5 pages), H₁: λ < 9; M1 Po(9); M1 P(X ≤ 4); A1 0.0550; A1 conclusion in context.
Q28Hypothesis test (Poisson) · Standard · Paper 6 style · 6 marks
  1. (a) Accidents on a stretch of motorway occur at random at a mean rate of 0.6 per week. Speed cameras are installed, and in the following 12 weeks there are 2 accidents. Test at the 5% significance level whether the mean rate has decreased. [5]
  2. (b) State one assumption needed for the Poisson model to be valid in this context. [1]
Show answer
  1. (a) P(X ≤ 2) = 0.0255 < 0.05, so reject H₀: evidence that the accident rate has decreased.
    B1 H₀: λ = 7.2 (for 12 weeks), H₁: λ < 7.2; M1 Po(7.2); M1 P(X ≤ 2); A1 0.0255; A1 conclusion in context.
  2. (b) Accidents occur independently of each other (and at a constant average rate).
    B1.
Q29Hypothesis test (Poisson) · Challenge · Paper 6 style · 6 marks
  1. (a) Emails arrive at an office inbox at random at a mean rate of 4 per hour. After a new website is launched, 52 emails arrive in a randomly chosen 10-hour period. Use a suitable approximating distribution to test at the 5% significance level whether the mean rate has increased. [6]
Show answer
  1. (a) N(40, 40) with continuity correction: z = (51.5 − 40)/√40 = 1.82 > 1.645, so reject H₀: evidence that the rate has increased.
    B1 H₀: λ = 40, H₁: λ > 40; B1 approximate by N(40, 40) (λ > 15); M1 continuity correction 51.5; M1 standardise; A1 z = 1.82 (p = 0.0345); A1 compare with 1.645 and conclude in context.
Q30Hypothesis test (Poisson) · Challenge · Paper 6 style · 6 marks
  1. (a) A shop sells on average 5 pressure cookers per day, and daily sales have a Poisson distribution. The manager wants to test, at the 5% significance level, whether the mean has changed, using sales on one randomly chosen day. Using a two-tailed test with at most 2.5% in each tail, find the critical region. [4]
  2. (b) Find the actual significance level of the test. [1]
  3. (c) On the chosen day 10 pressure cookers are sold. State the conclusion. [1]
Show answer
  1. (a) X = 0 or X ≥ 11
    M1 Po(5); A1 lower: P(X = 0) = 0.0067, P(X ≤ 1) = 0.0404; A1 upper: P(X ≥ 11) = 0.0137, P(X ≥ 10) = 0.0318; A1 X = 0 or X ≥ 11.
  2. (b) 0.0204 (2.04%)
    B1 0.0067 + 0.0137.
  3. (c) 10 is not in the critical region, so do not reject H₀: insufficient evidence that the mean has changed.
    B1.

Similar questions

Questions that test the same subtopic in different years, grouped so you can work on one skill at a time.

References only: find each question in its paper (official link where available, otherwise through your school).

Hypothesis test (normal mean) Banker27 questions · 9/9 sessions
Type I / Type II errors Banker30 questions · 9/9 sessions
Hypothesis test (binomial) Banker18 questions · 9/9 sessions
Hypothesis test (Poisson) Banker13 questions · 8/9 sessions

Subtopics

Our analysis of how often each subtopic was examined and how many marks it carried, 2022 to May/June 2026.

SubtopicMarks indexedSessions
Hypothesis test (normal mean) Banker1469/9
Type I / Type II errors Banker1219/9
Hypothesis test (binomial) Banker809/9
Hypothesis test (Poisson) Banker718/9

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