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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.
Q1Two-sample tests for means · Core · Paper 4 style · 6 marks
(a)The lifetimes of AA batteries of brand A are normally distributed with standard deviation 4.2 hours, and those of brand B are normally distributed with standard deviation 3.8 hours. A random sample of 30 brand A batteries has mean lifetime 52.3 hours, and a random sample of 25 brand B batteries has mean lifetime 50.1 hours.Test, at the 5% significance level, whether the mean lifetime of brand A batteries is greater than that of brand B batteries.[6]
Show answer
(a) z = 2.038 > 1.645: reject H₀; there is sufficient evidence that brand A batteries last longer on average.
B1: H₀: μ_A = μ_B, H₁: μ_A > μ_BM1: standard error √(4.2²/30 + 3.8²/25) = 1.080A1: z = 2.2/1.080 = 2.038B1: critical value 1.645M1: compare: 2.038 > 1.645, reject H₀A1: sufficient evidence at 5% that brand A has a greater mean lifetime
Q2Two-sample tests for means · Core · Paper 4 style · 9 marks
(a)An agricultural researcher in Punjab measures the heights, x cm, of wheat plants grown with two fertilisers, P and Q. Summary results are:Fertiliser P: n = 50, Σx = 1285, Σx² = 33410.5Fertiliser Q: n = 60, Σx = 1497, Σx² = 37720.8Find unbiased estimates of the population mean and variance for each fertiliser.[3]
(b)Test, at the 5% significance level, whether there is a difference between the mean heights for the two fertilisers.[5]
(c)Explain why it was not necessary to assume that the heights are normally distributed.[1]
B1: means 25.7 and 24.95M1: s² = (Σx² − (Σx)²/n)/(n − 1)A1: 7.878 and 6.282
(b) z = 1.465; |z| < 1.96, so do not reject H₀: there is insufficient evidence of a difference in mean heights.
B1: H₀: μ_P = μ_Q, H₁: μ_P ≠ μ_QM1: z = (25.7 − 24.95)/√(7.878/50 + 6.282/60)A1: z = 1.465B1: critical values ±1.96A1: do not reject H₀; insufficient evidence of a difference
(c) The samples are large, so by the Central Limit Theorem the sample means are approximately normally distributed.
B1
Q3Two-sample tests for means · Core · Paper 4 style · 7 marks
(a)Random samples of sizes 12 and 14 are taken from two normal populations that are believed to have the same variance. The unbiased estimates of the population variances from the samples are 2.4 and 3.1 respectively.Calculate the pooled estimate of the common population variance.[2]
(b)The sample means are 15.2 and 13.9. Find the value of the test statistic for a test of whether the population means are equal, and state the number of degrees of freedom.[3]
(c)Carry out the test at the 5% significance level (two-tailed). The critical value is 2.064.[2]
(c) 1.982 < 2.064: do not reject H₀; insufficient evidence that the means differ.
M1: compare with critical valueA1: conclusion with doubt
Q4Two-sample tests for means · Standard · Paper 4 style · 7 marks
(a)A sports scientist records the reaction times, in seconds, of two groups of cricketers. Group A trained without a ball machine and Group B trained with one.Group A: 0.42, 0.38, 0.45, 0.51, 0.40, 0.47, 0.44Group B: 0.36, 0.41, 0.35, 0.39, 0.43, 0.37, 0.34, 0.38Assuming that reaction times in both populations are normally distributed with a common variance, test at the 5% significance level whether the mean reaction time of players who train with the machine is less than that of players who do not.[7]
Show answer
(a) t = 3.109 > 1.771: reject H₀; there is sufficient evidence that training with the machine reduces the mean reaction time.
B1: H₀: μ_A = μ_B, H₁: μ_A > μ_B (B has smaller mean)M1: x̄_A = 0.4386, x̄_B = 0.3787; s_A² = 0.001914, s_B² = 0.0009268M1: pooled s² = (6s_A² + 7s_B²)/13 = 0.001383M1: t = (0.4386 − 0.3787)/√(0.001383(1/7 + 1/8))A1: t = 3.109B1: critical value t₁₃(0.95) = 1.771A1: reject H₀, conclusion in context
Q5Two-sample tests for means · Standard · Paper 4 style · 8 marks
(a)A company in Sialkot makes footballs with two machines. The times, in minutes, to stitch a ball are recorded. For a random sample of 9 balls from machine X, the mean is 31.6 and the unbiased estimate of the standard deviation is 4.2. For a random sample of 11 balls from machine Y, the mean is 27.2 and the unbiased estimate of the standard deviation is 3.6.State two assumptions needed to carry out a two-sample t-test.[2]
(b)Test, at the 1% significance level, whether machine X takes longer on average than machine Y.[6]
Show answer
(a) Both populations are normally distributed; the two populations have equal variances (and the samples are independent and random).
B1: normal populationsB1: common (equal) population variance
(b) t = 2.524 < 2.552: do not reject H₀; insufficient evidence at the 1% level that machine X takes longer on average.
B1: H₀: μ_X = μ_Y, H₁: μ_X > μ_YM1: pooled s² = (8 × 4.2² + 10 × 3.6²)/18 = 15.04M1: t = 4.4/√(15.04(1/9 + 1/11))A1: t = 2.524B1: t₁₈(0.99) = 2.552A1: do not reject H₀; conclusion in context
Q6Two-sample tests for means · Standard · Paper 4 style · 6 marks
(a)A delivery company in Dubai compares two routes. On route R, a random sample of 80 journeys has mean time 14.8 minutes and unbiased standard deviation estimate 3.1 minutes. On route S, a random sample of 70 journeys has mean time 12.1 minutes and unbiased standard deviation estimate 2.7 minutes.The manager claims that, on average, route R takes more than 2 minutes longer than route S. Test this claim at the 5% significance level.[6]
Show answer
(a) z = 1.478 < 1.645: do not reject H₀; there is insufficient evidence to support the manager's claim.
B1: H₀: μ_R − μ_S = 2, H₁: μ_R − μ_S > 2M1: z = (14.8 − 12.1 − 2)/√(3.1²/80 + 2.7²/70)A1: z = 1.478B1: critical value 1.645M1: compareA1: do not reject H₀; insufficient evidence for the claim
Q7Two-sample tests for means · Standard · Paper 4 style · 5 marks
(a)The daily outputs of two factories are normally distributed with known standard deviations 6 units (factory F) and 7 units (factory G). A random sample of 64 days from F has mean 120.5 units and a random sample of 81 days from G has mean ḡ units.A test at the 1% significance level is carried out of H₀: μ_F = μ_G against H₁: μ_F > μ_G. Find the set of values of ḡ for which H₀ is rejected.[5]
Q8Two-sample tests for means · Standard · Paper 4 style · 7 marks
(a)A motoring magazine measures the fuel consumption, in litres per 100 km, of randomly chosen cars of two models.Model H: 6.8, 7.2, 6.5, 7.0, 6.9, 7.4Model K: 7.1, 7.6, 7.3, 6.9, 7.8, 7.5, 7.2It is assumed that fuel consumptions for each model are normally distributed with the same variance. Test, at the 10% significance level, whether model K uses more fuel on average than model H.[7]
Show answer
(a) t = 2.168 > 1.363: reject H₀; sufficient evidence that model K uses more fuel on average.
Q9Two-sample tests for means · Challenge · Paper 4 style · 9 marks
(a)The masses, x grams, of eggs from two poultry farms are recorded.Farm A: n = 8, Σx = 196, Σx² = 4840.5Farm B: n = 10, Σx = 228, Σx² = 5246.4Assuming normal distributions with a common variance, test at the 5% significance level whether the mean masses differ.[7]
(b)Assuming the same pooled variance and sample sizes, find the smallest difference between the sample means that would lead to the conclusion that the population means differ, at the 5% level.[2]
Show answer
(a) t = 1.541; |t| < 2.120: do not reject H₀; insufficient evidence of a difference in mean mass.
B1: H₀: μ_A = μ_B, H₁: μ_A ≠ μ_BM1: x̄_A = 24.5, x̄_B = 22.8M1: s_A² = 5.500, s_B² = 5.333M1: pooled s² = (7s_A² + 9s_B²)/16 = 5.406A1: t = 1.541B1: t₁₆(0.975) = 2.120A1: do not reject H₀; conclusion in context
(b) 2.34 g
M1: difference > 2.120 × √(5.406(1/8 + 1/10))A1: 2.34 g
Q10Two-sample tests for means · Challenge · Paper 4 style · 9 marks
(a)Two classes of students at a school in Peshawar sat the same mathematics test. The marks of random samples from each class are:Class 1: 68, 74, 59, 81, 72, 65, 77, 70, 63Class 2: 61, 66, 58, 70, 64, 69, 60, 62Test at the 5% significance level whether there is a difference between the population mean marks of the two classes. State any assumptions you make.[8]
(b)A teacher suggests that a paired-sample t-test should have been used. Explain why this would not be appropriate.[1]
Show answer
(a) t = 2.149; |t| > 2.131: reject H₀; sufficient evidence of a difference in mean marks. Assumptions: normal populations with equal variances.
B1: assumptions: both populations normal, common varianceB1: H₀: μ₁ = μ₂, H₁: μ₁ ≠ μ₂M1: x̄₁ = 69.89, x̄₂ = 63.75M1: s₁² = 48.61, s₂² = 18.50M1: pooled s² = (8s₁² + 7s₂²)/15 = 34.56A1: t = 2.149B1: t₁₅(0.975) = 2.131A1: reject H₀; conclusion in context
(b) The two samples are independent (different students) and of different sizes, so the data cannot be paired.
B1
Q11Two-sample tests for means · Challenge · Paper 4 style · 6 marks
(a)The masses of 1 kg bags of flour filled by two machines are normally distributed, each with standard deviation 5 g. Samples of equal size n are taken from each machine. It is required that a difference of 1.8 g between the sample means will lead to the conclusion, at the 5% significance level (two-tailed), that the population means differ.Find the least possible value of n.[5]
(b)Explain why the answer would be larger if a 1% significance level were used.[1]
Show answer
(a) n = 60
M1: standard error 5√(2/n)B1: 1.96M1: 1.8/(5√(2/n)) > 1.96A1: n > 2(1.96 × 5/1.8)² = 59.28A1: least n = 60
(b) The critical value increases (to 2.576), so a smaller standard error, and hence a larger n, is needed.
B1
Q12Confidence intervals · Core · Paper 4 style · 5 marks
(a)The masses, in grams, of a random sample of 10 Chaunsa mangoes from an orchard near Multan are:312, 298, 305, 321, 289, 310, 302, 317, 295, 308Assuming that the masses are normally distributed, find a 95% confidence interval for the population mean mass.[5]
Q13Confidence intervals · Core · Paper 4 style · 5 marks
(a)The times, x minutes, taken by a random sample of 120 commuters to travel to work in Lahore are summarised by Σx = 5184 and Σx² = 226710.Calculate a 95% confidence interval for the mean commuting time.[4]
(b)Explain why a normal distribution can be used even though the distribution of commuting times is unknown.[1]
(b) n is large, so by the Central Limit Theorem the sample mean is approximately normally distributed.
B1
Q14Confidence intervals · Core · Paper 4 style · 5 marks
(a)Eight trainee typists each take a typing test before and after a two-week course. Their speeds, in words per minute, are:Typist: 1, 2, 3, 4, 5, 6, 7, 8Before: 72, 65, 80, 58, 69, 75, 61, 70After: 76, 70, 81, 63, 72, 80, 62, 75Assuming that the differences are normally distributed, find a 95% confidence interval for the population mean increase in speed.[5]
Q15Confidence intervals · Standard · Paper 4 style · 5 marks
(a)A machine fills bottles of cooking oil. A random sample of 15 bottles has mean volume 248.6 ml and unbiased standard deviation estimate 5.3 ml. Volumes are normally distributed.Find a 98% confidence interval for the population mean volume.[4]
(b)The label states that the mean volume is 250 ml. Comment on this claim using your interval.[1]
(b) 250 lies inside the interval, so the claim is not contradicted at this level of confidence.
B1: correct comment with reason
Q16Confidence intervals · Standard · Paper 4 style · 7 marks
(a)The lengths of time, in hours, that two brands of mobile phone run on a full charge under standard use were recorded.Brand U: 23.1, 25.4, 24.0, 22.8, 26.1, 24.7Brand V: 21.0, 22.6, 20.4, 23.3, 21.8, 22.1, 20.9Assuming normal distributions with a common variance, find a 95% confidence interval for μ_U − μ_V.[6]
(b)Does your interval suggest that the brands differ in mean running time? Explain.[1]
Q17Confidence intervals · Standard · Paper 4 style · 6 marks
(a)A random sample of 10 observations is taken from a normal distribution. A 95% confidence interval for the population mean, calculated from the sample using the t-distribution, is (41.27, 46.33).Find the sample mean and the unbiased estimate of the population variance.[4]
(b)Find a 90% confidence interval for the population mean based on the same sample.[2]
Q18Confidence intervals · Standard · Paper 4 style · 5 marks
(a)In a study of students in Karachi and Islamabad, a random sample of 100 students from Karachi had a mean score of 68.4 in a reading test with unbiased standard deviation estimate 9.2. A random sample of 90 students from Islamabad had mean 64.9 and unbiased standard deviation estimate 8.1.Find a 99% confidence interval for the difference between the population mean scores (Karachi − Islamabad).[4]
(b)State, with a reason, what the interval suggests about the population mean scores.[1]
(b) The interval lies entirely above 0, so the Karachi mean appears higher (at the 1% level).
B1
Q19Confidence intervals · Challenge · Paper 4 style · 8 marks
(a)From a random sample of 12 observations of a normally distributed variable, a 90% confidence interval for the population mean is calculated as (17.42, 19.86).Without finding the sample variance, calculate a 99% confidence interval for the population mean from the same data.[4]
(b)Explain whether the 99% interval is wider or narrower than the 90% interval, and why.[1]
(c)Find the least number of observations needed for the 90% interval to have width less than 1, assuming the sample standard deviation stays the same and using a normal approximation (z = 1.645).[3]
(b) Wider: a higher confidence level needs a larger critical value, so the interval must be wider.
B1
(c) 60
M1: s = 1.22 × √12/1.796 = 2.353M1: 2 × 1.645 × 2.353/√n < 1, n > 59.93A1: n = 60
Q20Confidence intervals · Challenge · Paper 4 style · 7 marks
(a)The daily water consumption of households in a town is normally distributed with standard deviation 12 litres. A researcher wants a 99% confidence interval for the mean daily consumption whose total width is at most 5 litres.Find the least sample size required.[4]
(b)A sample of this size gives a mean of 143.2 litres. Find the 99% confidence interval.[2]
(c)The researcher calculates 50 such 99% intervals from independent samples. Find the expected number of intervals that do not contain the population mean.[1]
Show answer
(a) 153
B1: z = 2.576M1: 2 × 2.576 × 12/√n ≤ 5A1: n ≥ 152.89A1: least n = 153
(b) (140.7, 145.7)
M1: 143.2 ± 2.576 × 12/√153A1: (140.7, 145.7)
(c) 0.5
B1: 50 × 0.01 = 0.5
Q21Confidence intervals · Challenge · Paper 4 style · 7 marks
(a)Two varieties of tomato plant are grown in a greenhouse. For 10 plants of variety A, the mean yield is 48.3 kg with unbiased variance estimate 12.6 kg². For 12 plants of variety B, the mean yield is 41.2 kg with unbiased variance estimate 10.8 kg². Yields are normally distributed with a common variance.Find a 90% confidence interval for μ_A − μ_B.[5]
(b)A grower claims that variety A yields more than 5 kg more than variety B on average. Use your interval to comment on this claim.[2]
(b) The interval includes values below 5, so the claim is not supported at this level.
M1: compare lower limit with 5A1: conclusion with reason
Q22Paired-sample t-test · Core · Paper 4 style · 7 marks
(a)Eight sprinters at an athletics club in Lahore each run 100 m before and after a six-week training programme. Their times, in seconds, are:Sprinter: 1, 2, 3, 4, 5, 6, 7, 8Before: 12.4, 13.1, 11.8, 12.9, 13.5, 12.2, 12.7, 13.0After: 12.1, 12.8, 11.9, 12.4, 13.0, 12.0, 12.6, 12.5Test, at the 5% significance level, whether the programme reduces the mean time. State the assumption you make.[7]
Show answer
(a) t = 3.752 > 1.895: reject H₀; sufficient evidence that the programme reduces the mean time. Assumption: the population of differences is normal.
Q23Paired-sample t-test · Standard · Paper 4 style · 7 marks
(a)Two judges each award marks to the same nine contestants in a qirat competition.Contestant: 1, 2, 3, 4, 5, 6, 7, 8, 9Judge A: 7.5, 8.2, 6.9, 9.1, 7.8, 8.5, 6.4, 7.2, 8.8Judge B: 7.1, 8.4, 6.5, 8.6, 7.0, 8.3, 6.6, 6.8, 8.1Explain why a paired-sample test is appropriate here.[1]
(b)Test, at the 5% significance level, whether there is a difference between the mean marks awarded by the two judges.[6]
Show answer
(a) Each contestant receives a mark from both judges, so the marks come in natural pairs.
B1
(b) t = 2.857; |t| > 2.306: reject H₀; sufficient evidence of a difference in mean marks.
Q24Paired-sample t-test · Standard · Paper 4 style · 7 marks
(a)Twelve plots of land in Sindh are each divided into two halves. One half is treated with a new fertiliser and the other half is not. The differences in yield, d kg (treated − untreated), are summarised by Σd = 30 and Σd² = 151.5.Test, at the 1% significance level, whether the fertiliser increases the mean yield.[6]
(b)State an assumption needed for this test to be valid.[1]
Show answer
(a) t = 3.284 > 2.718: reject H₀; sufficient evidence at the 1% level that the fertiliser increases mean yield.
(b) The differences in yield are normally distributed (in the population).
B1
Q25Paired-sample t-test · Standard · Paper 4 style · 7 marks
(a)A food laboratory measures the vitamin C content (mg per 100 g) of seven samples of guava juice, using two different methods on each sample.Sample: 1, 2, 3, 4, 5, 6, 7Method 1: 24.3, 18.7, 31.2, 27.5, 22.0, 29.8, 20.4Method 2: 24.9, 19.0, 31.8, 27.6, 22.7, 30.1, 21.0Test, at the 1% significance level, whether the two methods give different mean results.[6]
(b)Explain why a two-sample t-test on the two sets of results would not be suitable.[1]
Show answer
(a) t = 5.435; |t| > 3.707: reject H₀; sufficient evidence at the 1% level that the methods differ on average.
(b) The samples are not independent: both methods were applied to the same juice samples, and the large variation between samples would hide the difference between methods.
B1
Q26Paired-sample t-test · Challenge · Paper 4 style · 9 marks
(a)A language school claims that its intensive course raises students' test scores by more than 2 marks on average. Ten students take a test before and after the course.Student: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10Before: 54, 61, 48, 70, 66, 59, 63, 52, 57, 68After: 58, 63, 53, 71, 70, 62, 68, 55, 59, 73Test the school's claim at the 5% significance level, stating your hypotheses clearly.[7]
(b)Find the largest value of k for which the data would give evidence at the 5% level that the mean increase exceeds k marks.[2]
Show answer
(a) t = 3.096 > 1.833: reject H₀; sufficient evidence to support the claim that the mean increase exceeds 2 marks.
M1: (d̄ − k)/√(s²/10) > 1.833 ⇒ k < 3.4 − 1.833√(2.044/10)A1: k < 2.57
Q27Paired-sample t-test · Challenge · Paper 4 style · 9 marks
(a)The grip strengths, in newtons, of the left and right hands of nine left-handed tennis players are measured.Player: 1, 2, 3, 4, 5, 6, 7, 8, 9Left: 142, 155, 138, 160, 149, 151, 145, 158, 147Right: 139, 150, 140, 153, 146, 147, 146, 151, 141Test, at the 5% significance level, whether there is a difference between the mean grip strengths of the left and right hands of left-handed players.[6]
(b)Find a 95% confidence interval for the population mean difference (left − right), and explain how it is consistent with your answer to (a).[3]
Show answer
(a) t = 3.287; |t| > 2.306: reject H₀; sufficient evidence of a difference in mean grip strength.
(b) (1.06, 6.05); it does not contain 0, consistent with rejecting H₀ in (a).
M1: 3.556 ± 2.306√(10.53/9)A1: (1.06, 6.05)B1: 0 is outside the interval, matching the test result
Q28One-sample t-test · Core · Paper 4 style · 6 marks
(a)A pharmacist checks the mass, in grams, of sachets of oral rehydration salts, which should have mean mass 4.9 g. A random sample of 10 sachets gives:4.82, 5.11, 4.95, 5.20, 4.76, 5.05, 4.89, 5.14, 4.98, 5.23Assuming that the masses are normally distributed, test at the 5% significance level whether the mean mass differs from 4.9 g.[6]
Show answer
(a) t = 2.233; |t| < 2.262: do not reject H₀; insufficient evidence that the mean mass differs from 4.9 g.
B1: H₀: μ = 4.9, H₁: μ ≠ 4.9M1: x̄ = 5.013, s² = 0.02560M1: t = (5.013 − 4.9)/√(0.02560/10)A1: t = 2.233B1: t₉(0.975) = 2.262A1: do not reject H₀; conclusion in context
Q29One-sample t-test · Standard · Paper 4 style · 6 marks
(a)A manufacturer claims that its electric kettles boil 1.5 litres of water in 78 seconds on average. The times, x seconds, for a random sample of 14 kettles are summarised by Σx = 1065.4 and Σx² = 81192.3.Test, at the 1% significance level, whether the mean time is less than 78 seconds. Assume the times are normally distributed.[6]
Show answer
(a) t = −2.387; do not reject H₀ (critical value −2.650); insufficient evidence at the 1% level that the mean time is less than 78 s.
B1: H₀: μ = 78, H₁: μ < 78M1: x̄ = 76.10, s² = 8.874M1: t = (76.10 − 78)/√(8.874/14)A1: t = −2.387B1: critical value −2.650 (t₁₃)A1: do not reject H₀; conclusion in context
Q30One-sample t-test · Challenge · Paper 4 style · 7 marks
(a)The lengths of cotton fibres from a new variety are normally distributed. A random sample of 16 fibres has unbiased standard deviation estimate 2.8 mm. A test at the 5% significance level is carried out of H₀: μ = 30 against H₁: μ > 30, where μ mm is the population mean length.Find the set of values of the sample mean x̄ for which H₀ is rejected.[4]
(b)The sample mean is in fact 31.1 mm. A researcher says that if the sample size had been 64 with the same mean and standard deviation, the conclusion would have been different. Using a normal approximation for the larger sample, investigate this claim.[3]
(b) n = 16: t = 1.571 < 1.753, not significant; n = 64: z = 3.143 > 1.645, significant — the claim is correct.
M1: n = 16: t = 1.1/0.7 = 1.571 < 1.753, do not rejectM1: n = 64: z = 1.1/(2.8/8) = 3.143A1: 3.143 > 1.645, reject H₀; conclusion differs, so the claim is correct
Similar questions
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References only: find each question in its paper (official link where available, otherwise through your school).
Two-sample tests for means6 questions · 6/9 sessions
