OxfordAQA International AS and A-level Mathematics: Hypothesis tests for means: z-tests and t-tests (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 cover S2.6 (tests for a mean using z and t) from the OxfordAQA International AS and A-level Mathematics (9660) specification, Version 5.2. The content is part of International A-level Unit S2 (Statistics) and is not examined at International AS. Critical values are given in each question. See the study guide for methods, the revision notes for recall, the hub and checklist. The free 10-minute diagnostics are also available.

Questions

1. A population is normal with standard deviation 4. A random sample of 25 values is taken. Under the null hypothesis the population mean is 50. State the distribution of the sample mean X̄ under this hypothesis. [3]

2. A tram timetable states that a journey takes 14.0 minutes on average. Journey times are normally distributed with standard deviation 2.1 minutes. A passenger group believes the journey takes longer. They time 20 journeys and find a mean of 15.1 minutes. Test the belief at the 1% significance level. The critical value for a one-tailed 1% z-test is 2.326. [5]

3. Rice is sold in bags labelled 500 g. The mass of a bag is normally distributed with standard deviation 12 g. An inspector tests, at the 5% significance level, whether the mean mass has changed from 500 g, using a random sample of 9 bags.

(a) Find the critical region for the sample mean x̄. The two-tailed 5% critical values of z are ±1.960. [4] (b) The sample mean is 506.5 g. State the conclusion of the test. [2]

4. Lamp lifetimes are positively skewed and the population standard deviation is unknown. A sample of 100 lamps is tested.

(a) Explain why a z-test for the mean can still be used. [2] (b) State what is used in place of the population standard deviation, and why this is acceptable here. [2]

5. A kiosk owner believes that the mean spend per customer is 33.00. A random sample of 120 transactions gives Σx = 4032 and Σx² = 136 550, where x is the spend in the local currency. Test, at the 5% significance level, whether the mean spend differs from 33.00. The two-tailed 5% critical values of z are ±1.960. [7]

6. The curing time of a glue is claimed to have mean 5.0 hours. A technician suspects a new batch takes longer. Six randomly chosen batches take 4.8, 5.6, 5.1, 6.0, 5.4 and 5.9 hours. Assume the population is normal. Test the suspicion at the 5% significance level. The critical value of t with 5 degrees of freedom for a one-tailed 5% test is 2.015. [6]

7. State the test statistic for a test of the population mean in each case, giving any degrees of freedom.

(i) 10 observations from a normal population, variance unknown. [1] (ii) 15 observations from a normal population with known standard deviation 4. [1] (iii) 200 observations from a skewed population, variance unknown. [1]

8. A bakery says its loaves have a mean mass of 61 g above a baseline. For a random sample of 18 loaves, the differences x (in grams) from the baseline give Σx = 1047.6 and Σx² = 61 400. Assuming the population is normal, test at the 5% level whether the mean differs from 61. The two-tailed 5% critical values of t with 17 degrees of freedom are ±2.110. [6]

9. A test of H₀: μ = 80 against H₁: μ > 80 uses a random sample of 16 from a normal population with standard deviation 5. H₀ is rejected if x̄ > 81.8.

(a) Find the probability of a Type I error. [3] (b) Describe in words what a Type II error would mean here. [1]

10. A random sample of 10 values from a normal population has mean 60.3 and standard deviation s = 4.2. Test at the 5% level H₀: μ = 58 against H₁: μ > 58. Critical values for a one-tailed 5% test: t with 9 degrees of freedom is 1.833, and z is 1.645.

(a) Calculate the standard error estimate and the value of the test statistic. [3] (b) State the conclusion of the correct test. [2] (c) A student compares the same statistic with 1.645. Say what the student would conclude and explain why this is not appropriate. [3]

11. Component masses are normal with standard deviation 6 g. Test H₀: μ = 50 against H₁: μ > 50 at the 5% level, with one-tailed critical value 1.645.

(a) A random sample of 16 components has mean 52.0 g. Carry out the test. [5] (b) Find the smallest sample size for which a sample mean of 52.0 g would lead to H₀ being rejected. [4]

Answers

1. X̄ is normally distributed [1] with mean 50 [1] and variance 4²/25 = 0.64 (standard deviation 0.8) [1]. So X̄ ~ N(50, 0.64). [3] Examiner insight: State whether the second parameter is a variance or a standard deviation.

2. H₀: μ = 14.0, H₁: μ > 14.0, where μ is the mean journey time [1]. Standard error = 2.1/√20 = 0.4696 [1]. z = (15.1 − 14.0)/0.4696 = 2.343 [1]. 2.343 > 2.326 (p-value 0.0096 < 0.01) [1], so reject H₀. There is evidence at the 1% level that the mean journey time exceeds 14.0 minutes [1]. [5] Examiner insight: The result is close to the critical value, so avoid rounding the statistic early.

3. (a) Standard error = 12/√9 = 4 [1]. Cut-off distance = 1.960 × 4 = 7.84 [1]. Cut-offs are 500 − 7.84 = 492.16 and 500 + 7.84 = 507.84 [1]. Critical region: x̄ ≤ 492.16 or x̄ ≥ 507.84 [1]. (b) 506.5 is not in the critical region (z = 6.5/4 = 1.625, inside ±1.960) [1]. Do not reject H₀: insufficient evidence at the 5% level that the mean mass has changed [1]. [6] Examiner insight: Write both tails of the critical region as inequalities in x̄, not in z.

4. (a) The sample is large [1], so by the Central Limit Theorem the sample mean is approximately normally distributed whatever the shape of the population [1]. (b) The sample standard deviation s is used as an estimate of σ [1]. With n = 100, s is a reliable estimate, so the statistic is close to N(0, 1) [1]. [4] Examiner insight: Refer to the sampling distribution of the mean, not to the population being normal.

5. H₀: μ = 33, H₁: μ ≠ 33, where μ is the mean spend [1]. x̄ = 4032/120 = 33.6 [1]. s² = [136 550 − 4032²/120]/119 = 9.032 [1], so s = 3.005. Standard error estimate = 3.005/√120 = 0.2743 [1]. z = (33.6 − 33)/0.2743 = 2.19 [1]. 2.19 > 1.960 (two-tailed p-value 0.029), so reject H₀ [1]. Using the normal approximation, there is evidence at the 5% level that the mean spend differs from 33.00 [1]. [7] Examiner insight: Show the s² line explicitly, and say that the conclusion is approximate.

6. H₀: μ = 5.0, H₁: μ > 5.0, assuming a normal population [1]. Σx = 32.8, so x̄ = 5.467 [1]. Σx² = 180.38, so s² = (180.38 − 32.8²/6)/5 = 0.2147 [1], and s = 0.4633. Standard error estimate = 0.4633/√6 = 0.1892 [1]. t = (5.467 − 5.0)/0.1892 = 2.47 with 5 degrees of freedom [1]. 2.47 > 2.015, so reject H₀. There is evidence at the 5% level that the new batch cures for longer than 5.0 hours [1]. [6] Examiner insight: Quote the degrees of freedom alongside the t value.

7. (i) t statistic with 9 degrees of freedom (normal, σ unknown, small sample) [1]. (ii) z statistic, (x̄ − μ₀)/(4/√15), compared with N(0, 1) [1]. (iii) z statistic using s in place of σ, approximate by the Central Limit Theorem because n is large [1]. [3] Examiner insight: Give a reason with each choice, such as “σ unknown and n small”.

8. H₀: μ = 61, H₁: μ ≠ 61, assuming a normal population [1]. x̄ = 1047.6/18 = 58.2 [1]. s² = (61 400 − 1047.6²/18)/17 = 25.28 [1], so s = 5.027 and the standard error estimate is 5.027/√18 = 1.185. t = (58.2 − 61)/1.185 = −2.36 [1]. −2.36 < −2.110, so it is in the critical region [1]: reject H₀. There is evidence at the 5% level that the mean differs from 61 g (the p-value is about 0.030) [1]. [6] Examiner insight: For a negative statistic, state which tail it is in.

9. (a) Under H₀, X̄ ~ N(80, 5²/16), standard error 1.25 [1]. P(Type I error) = P(X̄ > 81.8 | μ = 80) = P(Z > 1.8/1.25) = P(Z > 1.44) [1]. = 0.0749 [1]. (b) Failing to reject H₀ (concluding the mean is not above 80) when the mean really is above 80 [1]. [4] Examiner insight: A Type I error probability is found assuming H₀ is true, so use μ = 80.

10. (a) Standard error estimate = 4.2/√10 = 1.328 [1]. t = (60.3 − 58)/1.328 = 1.73 [1], with 9 degrees of freedom [1]. (b) The population is normal and σ is unknown with n small, so use t [1]. 1.73 < 1.833, so do not reject H₀: insufficient evidence at the 5% level that the mean exceeds 58 [1]. (c) 1.73 > 1.645, so the student would reject H₀ [1]. This is inappropriate because s from 10 values is an unreliable estimate of σ [1], so the statistic follows a t-distribution with heavier tails, not N(0, 1) [1]. [8] Examiner insight: When asked to explain, mention both that σ is estimated and that the sample is small.

11. (a) H₀: μ = 50, H₁: μ > 50 [1]. Standard error = 6/√16 = 1.5 [1]. z = (52.0 − 50)/1.5 = 1.333 [1]. 1.333 < 1.645 (p-value 0.091) [1], so do not reject H₀: insufficient evidence at the 5% level that the mean mass exceeds 50 g [1]. (b) Require (52.0 − 50)/(6/√n) > 1.645 [1]. √n > 1.645 × 3 = 4.935 [1], so n > 24.35 [1]. Check: n = 24 gives z = 1.633 and n = 25 gives z = 1.667, so n = 25 [1]. [9] Examiner insight: With an inequality in n, round up to the next whole number and confirm with the neighbouring value.

Where marks are usually lost

• Using σ in place of the standard error σ/√n in the denominator.
• Calculating s² with divisor n instead of n − 1 from Σx and Σx².
• Taking the degrees of freedom as n rather than n − 1, or using z for a small sample with unknown variance.
• Giving a critical region in terms of z when the question asks for x̄.
• Comparing with a positive critical value in a lower-tail test.
• Concluding with “H₀ is true” or “the claim is proved”, or omitting the context.
• Not stating the normality assumption for t, or the approximation for z with s.

Next steps

• Revisit the revision notes and study guide.
• Tick off the topic on the printable checklist and see the course hub.
• Take one of the free 10-minute diagnostics to find your next gap.
• Book a free trial class if you want help with this topic.

Official syllabus

OxfordAQA International AS and A-level Mathematics (9660), Version 5.2 (International AS exams from May/June 2018, A-level from May/June 2019), published by OxfordAQA. Content: S2.6 Hypothesis testing, tests for the mean (z-statistic and t-statistic), within Unit S2 (Statistics).