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2026 May/June · Paper 42 · Q1 (a, b)7 marksChoice of test & assumptions · Sign test
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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.
Q1Wilcoxon signed-rank test · Core · Paper 4 style · 7 marks
(a)A clinic in Rawalpindi records the systolic blood pressure (mm Hg) of 8 adults before and after a 6-week walking programme.Patient: 1, 2, 3, 4, 5, 6, 7, 8Before: 142, 156, 138, 149, 161, 147, 152, 158After: 137, 148, 139, 141, 150, 146, 145, 160Use a Wilcoxon signed-rank test, at the 5% significance level, to test whether the programme reduces median blood pressure.[7]
Show answer
(a) T = 4.5; critical value 5; reject H₀: evidence that the median blood pressure is reduced
B1 H₀: population median difference (before − after) = 0, H₁: > 0; M1 differences 5, 8, −1, 8, 11, 1, 7, −2; M1 rank |differences|: −1→1.5, 1→1.5, −2→3, 5→4, 7→5, 8→6.5, 8→6.5, 11→8; A1 P = 31.5, Q = 4.5; A1 T = 4.5; B1 one-tailed 5% critical value for n = 8 is 5; A1 4.5 ≤ 5, reject H₀; there is evidence that the programme reduces median blood pressure.
Q2Wilcoxon signed-rank test · Core · Paper 4 style · 8 marks
(a)A bakery in Karachi claims that the median mass of its naan is 50 g. A random sample of 10 naan has masses (g):47.2, 51.8, 44.9, 53.6, 49.1, 42.3, 55.4, 46.7, 48.5, 43.1Test the claim using a Wilcoxon signed-rank test at the 5% significance level.[7]
(b)State an assumption needed for this test to be valid.[1]
Show answer
(a) T = 17; critical value 8; do not reject H₀: no evidence against the claim
B1 H₀: median = 50, H₁: median ≠ 50; M1 differences from 50; M1 ranks: −0.9→1, −1.5→2, 1.8→3, −2.8→4, −3.3→5, 3.6→6, −5.1→7, 5.4→8, −6.9→9, −7.7→10; A1 P = 17, Q = 38; A1 T = 17; B1 two-tailed 5% critical value 8; A1 do not reject H₀; insufficient evidence that the median differs from 50 g.
(b) The masses come from a population whose distribution is symmetric (about the median)
B1 symmetric distribution (random sample also acceptable as additional).
Q3Wilcoxon signed-rank test · Core · Paper 4 style · 5 marks
(a)A Wilcoxon signed-rank test is carried out on 13 paired observations. One pair has a zero difference. The sum of the ranks of the positive differences is 61. Find the value of the test statistic T.[2]
(b)Carry out the test at the 5% significance level, using a two-tailed alternative, stating your hypotheses.[3]
Show answer
(a) T = 17
M1 n = 12, total of ranks 12 × 13/2 = 78; A1 Q = 17, T = min(61, 17) = 17.
(b) Critical value 13; do not reject H₀
B1 H₀: median difference 0, H₁: median difference ≠ 0; B1 critical value 13; A1 17 > 13: do not reject H₀; insufficient evidence of a difference in medians.
Q4Wilcoxon signed-rank test · Core · Paper 4 style · 6 marks
(a)Seven students in Kuala Lumpur sit a mental-arithmetic quiz, then practise for a week and sit a similar quiz. Their scores are:First quiz: 64, 71, 58, 80, 67, 75, 62Second quiz: 70, 74, 57, 88, 73, 84, 68State suitable hypotheses for a test of whether practice increases the median score.[1]
(b)Carry out a Wilcoxon signed-rank test at the 5% significance level.[5]
Show answer
(a) H₀: median of (second − first) = 0; H₁: median of (second − first) > 0
B1 both, in terms of the population median difference.
(b) T = 1; critical value 3; reject H₀: evidence that practice increases the median score
M1 differences (second − first): 6, 3, −1, 8, 6, 9, 6; M1 ranks: −1→1, 3→2, 6→4, 6→4, 6→4, 8→6, 9→7; A1 P = 27, Q = 1, T = 1; B1 critical value 3; A1 conclusion in context.
Q5Wilcoxon signed-rank test · Standard · Paper 4 style · 7 marks
(a)Ten table-tennis players at a club in Islamabad each have their reaction time (seconds) measured before and after a caffeine-free energy drink.Player: A, B, C, D, E, F, G, H, I, JBefore: 0.42, 0.38, 0.51, 0.47, 0.36, 0.44, 0.49, 0.40, 0.45, 0.53After: 0.39, 0.38, 0.45, 0.48, 0.30, 0.37, 0.41, 0.42, 0.44, 0.44Test at the 5% significance level whether the drink reduces median reaction time.[7]
Show answer
(a) n = 9 (one zero discarded); T = 4.5; critical value 8; reject H₀: evidence of a reduction
B1 H₀: median difference (before − after) = 0, H₁: > 0; M1 differences, discarding the zero (player B), n = 9; M1 ranks: −0.01→1.5, 0.01→1.5, −0.02→3, 0.03→4, 0.06→5.5, 0.06→5.5, 0.07→7, 0.08→8, 0.09→9; A1 P = 40.5, Q = 4.5; A1 T = 4.5; B1 critical value 8; A1 reject H₀; evidence that the drink reduces median reaction time.
Q6Wilcoxon signed-rank test · Standard · Paper 4 style · 8 marks
(a)A soil scientist in Faisalabad believes that the median pH of soil in a district is greater than 7.5. She measures 12 randomly chosen samples:6.8, 7.9, 8.4, 7.1, 8.4, 9.2, 6.5, 8.7, 7.6, 9.0, 8.1, 7.4Use a Wilcoxon signed-rank test at the 2.5% significance level to test her belief. (Give tied absolute differences the mean of their ranks.)[8]
Show answer
(a) T = 20; critical value 13; do not reject H₀: insufficient evidence that the median pH exceeds 7.5
B1 H₀: median = 7.5, H₁: median > 7.5; M1 differences from 7.5; M1 ranks with ties averaged: 0.1→1.5, −0.1→1.5, 0.4→3.5, −0.4→3.5, 0.6→5, −0.7→6, 0.9→7.5, 0.9→7.5, −1→9, 1.2→10, 1.5→11, 1.7→12; A1 P = 58; A1 Q = 20, T = 20; B1 critical value for n = 12 at 2.5% one-tailed: 13; M1 comparison; A1 conclusion in context.
Q7Wilcoxon signed-rank test · Standard · Paper 4 style · 7 marks
(a)A Wilcoxon signed-rank test compares the times taken by 30 delivery riders in Lahore to complete a route using an old map app and a new one (old time − new time). There are no zero differences. The sum of the ranks of the positive differences is 314. Using a suitable normal approximation, test at the 5% significance level whether the new app reduces the median time.[7]
Show answer
(a) T = 151; z = −1.666; reject H₀: evidence that the new app reduces median time
B1 H₀: median difference 0, H₁: median difference > 0; M1 Q = 465 − 314 = 151, T = 151; B1 E(T) = 30 × 31/4 = 232.5, Var(T) = 30 × 31 × 61/24 = 2363.75; M1 z = (151 + 0.5 − 232.5)/√2363.75 with continuity correction; A1 z = −1.666; M1 compare with −1.645; A1 reject H₀; evidence that the new app reduces median time.
Q8Wilcoxon signed-rank test · Standard · Paper 4 style · 7 marks
(a)A phone maker claims that the median battery life of its new model under a standard test is 20 hours. A consumer group in Dubai tests 14 phones and records (hours):18.6, 21.4, 17.9, 19.2, 16.8, 20.3, 18.1, 15.7, 19.6, 17.4, 16.2, 18.9, 20.8, 17.1Test, at the 1% significance level, whether the median battery life is less than claimed.[7]
Show answer
(a) T = 11; critical value 15; reject H₀: evidence that the median is less than 20 hours
B1 H₀: median = 20, H₁: median < 20; M1 differences from 20; M1 ranks: 0.3→1, −0.4→2, −0.8→3.5, 0.8→3.5, −1.1→5, −1.4→6.5, 1.4→6.5, −1.9→8, −2.1→9, −2.6→10, −2.9→11, −3.2→12, −3.8→13, −4.3→14; A1 P = 11, Q = 94; A1 T = 11; B1 1% one-tailed critical value for n = 14: 15; A1 reject H₀; strong evidence that the median battery life is below 20 hours.
Q9Wilcoxon signed-rank test · Standard · Paper 4 style · 8 marks
(a)An agricultural station near Multan divides each of 9 fields into two halves. One half is sown with wheat variety X and the other with variety Y. The yields (tonnes per hectare) are:Field: 1, 2, 3, 4, 5, 6, 7, 8, 9X: 3.42, 2.98, 3.75, 3.10, 3.56, 2.87, 3.33, 3.64, 3.05Y: 3.51, 3.12, 3.69, 3.30, 3.71, 3.04, 3.29, 3.88, 3.27Test at the 10% significance level whether there is a difference between the median yields of the two varieties.[7]
(b)Explain why a paired test is appropriate here.[1]
Show answer
(a) T = 3; critical value 8; reject H₀: evidence of a difference
B1 H₀: median difference = 0, H₁: ≠ 0; M1 differences X − Y; M1 ranks: 0.04→1, 0.06→2, −0.09→3, −0.14→4, −0.15→5, −0.17→6, −0.2→7, −0.22→8, −0.24→9; A1 P = 3, Q = 42; A1 T = 3; B1 two-tailed 10% critical value for n = 9: 8; A1 reject H₀; evidence of a difference in median yields.
(b) Each field gives a natural pair, removing differences between fields (soil, water)
B1 same field for both varieties, so field-to-field variation is controlled.
Q10Wilcoxon signed-rank test · Standard · Paper 4 style · 8 marks
(a)Two examiners each mark the same 14 essays out of 100.Examiner 1: 56, 72, 64, 81, 49, 68, 77, 59, 85, 62, 70, 53, 66, 74Examiner 2: 61, 70, 71, 84, 57, 66, 83, 68, 86, 70, 79, 50, 75, 78Use a Wilcoxon signed-rank test at the 2% significance level to test whether there is a difference between the examiners' median marks. Use the mean rank for tied values.[8]
Show answer
(a) T = 9.5; critical value 15; reject H₀: evidence of a difference between the examiners
B1 H₀: median difference = 0, H₁: ≠ 0; M1 differences (E1 − E2); M1 ranks with ties: −1→1, 2→2.5, 2→2.5, −3→4.5, 3→4.5, −4→6, −5→7, −6→8, −7→9, −8→10.5, −8→10.5, −9→13, −9→13, −9→13; A1 P = 9.5; A1 Q = 95.5; B1 2% two-tailed critical value for n = 14 is 15; M1 T = 9.5 compared; A1 reject H₀; evidence that the examiners differ (examiner 2 tends to mark higher).
Q11Wilcoxon signed-rank test · Challenge · Paper 4 style · 7 marks
(a)In a Wilcoxon signed-rank test with 15 non-zero differences, P denotes the sum of the ranks of the positive differences. Find the set of values of P for which H₀: median difference = 0 is rejected in favour of H₁: median difference ≠ 0 at the 5% significance level.[4]
(b)The experimenter actually obtains P = 96. One difference, with rank 7, was later found to have been recorded with the wrong sign: it should have been negative, not positive. Determine whether the conclusion of the test changes.[3]
Show answer
(a) P ≤ 25 or P ≥ 95
B1 P + Q = 120; B1 critical value 25; M1 reject when min(P, 120 − P) ≤ 25; A1 P ≤ 25 or P ≥ 95.
(b) Yes: corrected P = 89, so H₀ is no longer rejected
M1 corrected P = 96 − 7 = 89, Q = 31; A1 T = 31 > 25; A1 originally T = 24 ≤ 25 (reject), now do not reject: conclusion changes.
Q12Wilcoxon signed-rank test · Challenge · Paper 4 style · 6 marks
(a)A Wilcoxon signed-rank test is to be carried out on 40 non-zero paired differences, using a one-tailed test at the 1% significance level and a normal approximation with continuity correction. State the mean and variance of the test statistic T under H₀.[2]
(b)Find the largest value of T that leads to rejection of H₀.[4]
Show answer
(a) Mean 410, variance 5535
B1 40 × 41/4 = 410; B1 40 × 41 × 81/24 = 5535.
(b) T = 236
M1 (T + 0.5 − 410)/√5535 < −2.326; M1 √5535 = 74.398; A1 T < 236.45; A1 largest integer T = 236.
Q13Wilcoxon signed-rank test · Challenge · Paper 4 style · 8 marks
(a)A Wilcoxon signed-rank test is used to test H₀: median = 12 against H₁: median ≠ 12 for the data12.4, 15.1, 11.2, 13.8, 10.6, 14.3, 9.9Show that, for these 7 values, P = 18 and Q = 10, and carry out the test at the 5% significance level.[5]
(b)An eighth value x is added, where x > 16. Find the new value of T and determine whether the conclusion changes.[3]
Show answer
(a) T = 10; critical value 2; do not reject H₀
M1 differences: 0.4, 3.1, −0.8, 1.8, −1.4, 2.3, −2.1; M1 ranks: 0.4→1, −0.8→2, −1.4→3, 1.8→4, −2.1→5, 2.3→6, 3.1→7; A1 P = 18, Q = 10; B1 two-tailed 5% critical value for n = 7 is 2; A1 T = 10 > 2, do not reject H₀.
(b) T = 10; critical value 3; conclusion unchanged: do not reject H₀
M1 x − 12 > 4 has the largest absolute difference, rank 8, others keep their ranks; A1 P = 26, Q = 10, T = 10; A1 critical value for n = 8 is 3: do not reject H₀; conclusion unchanged.
Q14Wilcoxon signed-rank test · Challenge · Paper 4 style · 7 marks
(a)Researchers in Nairobi compare the daily water use (litres) of 27 households before and after fitting water-saving taps (before − after). One household shows no change. The sum of the ranks of the negative differences is 98. Find the test statistic and, using a normal approximation with continuity correction, find the p-value for a test of H₀: median difference = 0 against H₁: median difference > 0.[6]
(b)State the conclusion at the 1% significance level.[1]
Show answer
(a) T = 98; z = −1.956; p-value = 0.0253
B1 n = 26 after discarding the zero; M1 P = 351 − 98 = 253, T = 98; B1 E(T) = 175.5, Var(T) = 1550.25; M1 z = (98 + 0.5 − 175.5)/√1550.25; A1 z = −1.956; A1 p = Φ(−1.956) = 0.0253.
(b) p > 0.01: do not reject H₀
B1 comparison and conclusion in context.
Q15Sign test · Core · Paper 4 style · 5 marks
(a)A tea company in Lahore asks 12 randomly chosen customers to taste its usual blend and a new blend. 9 prefer the new blend and 3 prefer the usual one. Use a sign test at the 5% significance level to test whether customers in general prefer the new blend.[5]
Show answer
(a) P(X ≥ 9) = 0.0730; do not reject H₀: insufficient evidence of a preference
B1 H₀: p = 0.5, H₁: p > 0.5, where p is the probability a customer prefers the new blend; M1 X ~ B(12, 0.5); M1 P(X ≥ 9) = (220 + 66 + 12 + 1)/4096; A1 = 0.0730; A1 0.0730 > 0.05, do not reject H₀.
Q16Sign test · Core · Paper 4 style · 6 marks
(a)A brick kiln near Gujranwala claims that the median compressive strength of its bricks is 300 units. A buyer tests 14 bricks:312, 287, 345, 298, 276, 305, 291, 283, 268, 299, 277, 302, 289, 294Use a sign test at the 5% significance level to test whether the median strength is less than 300.[6]
Show answer
(a) 4 above, 10 below; P(X ≤ 4) = 0.0898; do not reject H₀
B1 H₀: median = 300, H₁: median < 300; M1 count signs: 4 '+', 10 '−'; M1 X ~ B(14, 0.5); M1 P(X ≤ 4); A1 0.0898; A1 do not reject H₀; insufficient evidence.
Q17Sign test · Standard · Paper 4 style · 6 marks
(a)A sample of 16 commuters in Mumbai each record their journey times on a weekday by bus and by metro. In 3 cases the bus is faster, in 12 cases the metro is faster and in 1 case the times are equal. Use a sign test at the 5% significance level to test whether there is a difference between median journey times by bus and by metro.[6]
Show answer
(a) P(X ≤ 3) = 0.0176; reject H₀: evidence of a difference
B1 H₀: median difference = 0, H₁: ≠ 0; B1 discard the tie: n = 15; M1 X ~ B(15, 0.5), X = 3; M1 P(X ≤ 3) = (1 + 15 + 105 + 455)/32768; A1 0.0176; A1 compare with 0.025 (two-tailed): reject H₀; evidence of a difference in median times.
Q18Sign test · Standard · Paper 4 style · 6 marks
(a)In a survey of 44 randomly chosen shoppers at a market in Cairo, each rates two brands of dates on a 10-point scale. 27 rate brand A higher, 13 rate brand B higher and 4 give equal ratings. Use a sign test with a suitable normal approximation, at the 5% significance level, to test whether brand A is preferred.[6]
Show answer
(a) z = 2.055; reject H₀: evidence that brand A is preferred
B1 H₀: p = 0.5, H₁: p > 0.5; B1 n = 40 after removing ties; M1 X ~ B(40, 0.5) ≈ N(20, 10); M1 z = (26.5 − 20)/√10 with continuity correction; A1 z = 2.055; A1 2.055 > 1.645: reject H₀, evidence that brand A is preferred.
Q19Sign test · Standard · Paper 4 style · 6 marks
(a)Eleven athletes in Addis Ababa have their resting heart rate (beats per minute) measured before and after an 8-week training camp.Before: 68, 74, 81, 59, 77, 70, 85, 66, 72, 79, 63After: 65, 74, 76, 61, 70, 66, 80, 66, 69, 74, 58Use a sign test at the 1% significance level to test whether training lowers the median resting heart rate.[6]
Show answer
(a) 8 decreases, 1 increase, 2 unchanged; P(X ≥ 8) = 0.0195; do not reject H₀: insufficient evidence at the 1% level
B1 H₀: median (before − after) = 0, H₁: > 0; M1 signs of differences, ignoring 2 zeros: n = 9; M1 X ~ B(9, 0.5), X = 8; M1 P(X ≥ 8); A1 0.0195; A1 do not reject H₀ at 1%; insufficient evidence.
Q20Sign test · Challenge · Paper 4 style · 6 marks
(a)A sign test is to be used with 20 non-zero differences to test H₀: median difference = 0 against H₁: median difference > 0, at the 5% significance level. X is the number of positive differences. Find the critical region and the actual significance level of the test.[4]
(b)In fact 70% of differences in the population are positive. Find the probability that the test leads to rejection of H₀.[2]
Show answer
(a) X ≥ 15; actual significance level 0.0207
M1 X ~ B(20, 0.5); M1 P(X ≥ 15) = 0.0207, P(X ≥ 14) = 0.0577; A1 critical region X ≥ 15; A1 0.0207.
(b) 0.416
M1 X ~ B(20, 0.7), P(X ≥ 15); A1 0.416.
Q21Sign test · Challenge · Paper 4 style · 10 marks
(a)A taxi company in Doha believes the median waiting time for a taxi at a mall is more than 5.8 minutes. A random sample of 10 waiting times (minutes) is:4.6, 5.9, 6.3, 5.2, 7.4, 6.8, 5.5, 8.1, 6.1, 7.0Carry out a sign test at the 5% significance level.[5]
(b)Carry out a Wilcoxon signed-rank test on the same data at the 5% significance level, and comment on the two conclusions.[5]
Show answer
(a) 7 above, 3 below; P(X ≥ 7) = 0.172; do not reject H₀
B1 H₀: median = 5.8, H₁: median > 5.8; M1 X ~ B(10, 0.5), X = 7; M1 P(X ≥ 7); A1 0.172; A1 do not reject H₀; insufficient evidence.
(b) T = 15; critical value 10; do not reject H₀; both tests agree
M1 ranks: 0.1→1, −0.3→2.5, 0.3→2.5, 0.5→4, −0.6→5, 1→6, −1.2→7.5, 1.2→7.5, 1.6→9, 2.3→10; A1 P = 40, Q = 15, T = 15; B1 critical value 10; A1 do not reject H₀; B1 comment: Wilcoxon uses magnitudes as well as signs (more powerful) but needs a symmetric distribution.
Q22Wilcoxon rank-sum test · Core · Paper 4 style · 7 marks
(a)The times (minutes) taken to assemble a flat-pack desk are recorded for 5 people using printed instructions (A) and 6 people using a video guide (B).A: 23.4, 27.1, 21.8, 25.6, 22.9B: 28.3, 26.4, 30.2, 24.7, 29.5, 31.0Use a Wilcoxon rank-sum test at the 5% significance level to test whether people using printed instructions tend to be quicker.[7]
Show answer
(a) R_A = 18; W = 18; critical value 20; reject H₀: evidence that printed instructions are quicker
B1 H₀: the two populations have equal medians, H₁: median time for A is less; M1 rank all 11 times; M1 sum of ranks of the smaller sample A: R_A = 18; A1 m(m + n + 1) − R_A = 60 − 18 = 42, W = 18; B1 critical value (m = 5, n = 6, 5% one-tailed) 20; M1 compare; A1 reject H₀; evidence that printed instructions lead to quicker assembly.
Q23Wilcoxon rank-sum test · Standard · Paper 4 style · 7 marks
(a)A biologist counts the number of earthworms in randomly chosen 1 m² plots in two orchards near Swat.Orchard A (6 plots): 14, 19, 11, 17, 22, 16Orchard B (8 plots): 21, 25, 18, 24, 19, 27, 23, 26Test at the 5% significance level whether there is a difference between the median numbers of earthworms per plot in the two orchards. Give tied values the mean of their ranks.[7]
Show answer
(a) R_A = 25.5; W = 25.5; critical value 29; reject H₀: evidence of a difference
B1 H₀: no difference in population medians, H₁: medians differ; M1 rank all 14 with ties averaged: 11(A)→1, 14(A)→2, 16(A)→3, 17(A)→4, 18(B)→5, 19(A)→6.5, 21(B)→8, 22(A)→9, 23(B)→10, 24(B)→11, 25(B)→12, 26(B)→13, 27(B)→14; A1 R_A = 25.5; M1 W = min(25.5, 90 − 25.5) = 25.5; B1 two-tailed 5% critical value for m = 6, n = 8: 29; A1 reject H₀; evidence that the median numbers differ.
Q24Wilcoxon rank-sum test · Standard · Paper 4 style · 7 marks
(a)The ages of 12 randomly chosen customers at a café in Istanbul and 15 at a café in Ankara are ranked together, from youngest (rank 1). The sum of the ranks of the Istanbul customers is 131. Using a normal approximation, test at the 5% significance level whether there is a difference between the median ages of customers at the two cafés.[7]
Show answer
(a) W = 131; z = −1.781; do not reject H₀
B1 H₀: equal population medians, H₁: different; M1 m(m + n + 1) − R = 336 − 131 = 205, W = 131; B1 mean 168, variance 12 × 15 × 28/12 = 420; M1 z = (131 + 0.5 − 168)/√420; A1 z = −1.781; M1 compare with −1.96; A1 do not reject H₀.
Q25Wilcoxon rank-sum test · Standard · Paper 4 style · 8 marks
(a)Fourteen students in Singapore are split at random into two groups of 7. Group A learns vocabulary using flash cards, group B using a mobile app. Their test scores are:A: 52, 61, 48, 57, 66, 55, 59B: 63, 58, 70, 67, 72, 64, 69Use a Wilcoxon rank-sum test at the 5% significance level to test whether there is a difference in median scores.[7]
(b)State one reason why a rank-sum test might be chosen here instead of a two-sample t-test.[1]
Show answer
(a) R_A = 33; W = 33; critical value 36; reject H₀: evidence of a difference
B1 H₀: equal population medians, H₁: medians differ; M1 rank the combined scores: 48(A)→1, 52(A)→2, 55(A)→3, 57(A)→4, 58(B)→5, 59(A)→6, 61(A)→7, 63(B)→8, 64(B)→9, 66(A)→10, 67(B)→11, 69(B)→12, 70(B)→13, 72(B)→14; A1 R_A = 33; M1 7 × 15 − 33 = 72, W = 33; B1 critical value 36; A1 reject H₀; evidence that median scores differ (the app group scores higher).
(b) The scores may not be normally distributed (small samples; distribution unknown)
B1 no assumption of normality needed.
Q26Wilcoxon rank-sum test · Challenge · Paper 4 style · 8 marks
(a)A chemist in Riyadh compares the yield (g) of a reaction with a new catalyst (B, 7 trials) and the old catalyst (A, 4 trials).A: 8.2, 9.7, 7.5, 8.9B: 10.4, 9.1, 11.8, 10.9, 9.9, 12.3, 10.1Test at the 1% significance level whether the new catalyst gives a higher median yield.[6]
(b)Find the smallest possible value of R_A. Given that the critical value for a 0.5% one-tailed test with these sample sizes is 10, state what the data would have to look like for H₀ to be rejected at that level.[2]
Show answer
(a) R_A = 11; W = 11; critical value 11; reject H₀: evidence that the new catalyst gives higher yields
B1 H₀: equal medians, H₁: median for B is greater; M1 ranks: 7.5(A)→1, 8.2(A)→2, 8.9(A)→3, 9.1(B)→4, 9.7(A)→5, 9.9(B)→6, 10.1(B)→7, 10.4(B)→8, 10.9(B)→9, 11.8(B)→10, 12.3(B)→11; A1 R_A = 11; A1 W = 11; B1 critical value (m = 4, n = 7, 1% one-tailed) 11; A1 reject H₀; strong evidence that the new catalyst increases the median yield.
(b) Smallest R_A = 1 + 2 + 3 + 4 = 10; need W = 10, i.e. all four old-catalyst yields lower than every new-catalyst yield
B1 10; B1 W ≤ 10 only when R_A = 10, so the four A values must take ranks 1 to 4.
Q27Wilcoxon rank-sum test · Challenge · Paper 4 style · 7 marks
(a)A Wilcoxon rank-sum test is to compare a sample of 20 tomato plants grown with compost and 25 grown without, using the heights after 8 weeks. The test is one-tailed at the 5% level, H₁ being that compost-grown plants tend to be taller. The plants are ranked from tallest (rank 1). Give the mean and variance of the rank sum R of the compost sample under H₀.[2]
(b)Using a normal approximation with continuity correction, find the largest value of R that leads to H₀ being rejected.[4]
(c)The compost plants have R = 382. State the conclusion.[1]
Show answer
(a) Mean 460, variance 1916.67
B1 20 × 46/2 = 460; B1 20 × 25 × 46/12 = 1916.67.
(b) R = 387
M1 (R + 0.5 − 460)/43.780 < −1.645; M1 R < 387.48; A1 R ≤ 387; A1 so the largest R is 387.
(c) Reject H₀: evidence that compost-grown plants tend to be taller
B1 382 ≤ 387.
Q28Choice of test & assumptions · Core · Paper 4 style · 4 marks
(a)A researcher measures the cholesterol level of 9 patients before and after a change of diet. She considers a sign test, a Wilcoxon signed-rank test and a paired-sample t-test. State the assumption about the population of differences needed for (i) the Wilcoxon signed-rank test and (ii) the paired-sample t-test.[2]
(b)Explain why the Wilcoxon signed-rank test is generally preferred to the sign test when its assumption holds.[2]
Show answer
(a) (i) the differences have a symmetric distribution; (ii) the differences are normally distributed
B1 symmetric; B1 normal.
(b) It uses the magnitudes of the differences as well as their signs, so it uses more information and is more powerful
B1 uses sizes (ranks) of differences, not just signs; B1 more powerful / more likely to detect a real difference.
Q29Choice of test & assumptions · Standard · Paper 4 style · 4 marks
(a)For each situation, name the most suitable non-parametric test and give a reason.(i) The weights of 8 randomly chosen goats from one farm and 10 from another farm are compared; nothing is known about the distribution of weights.(ii) 25 people each rate two films as 'better' or 'worse' than the other, and only the preferences are recorded.(iii) The times of 12 runners on two different tracks are compared; the differences can be assumed to have a symmetric distribution.[3]
(b)In (iii), a student instead uses a sign test. Explain why this is valid but less appropriate.[1]
Show answer
(a) (i) Wilcoxon rank-sum test: two independent samples; (ii) sign test: only the direction of each preference is known; (iii) Wilcoxon signed-rank test: paired data with symmetric differences
B1 each test with a valid reason.
(b) The sign test needs no symmetry assumption so is valid, but it ignores the sizes of the differences and is less powerful
B1 valid but wastes information.
Q30Choice of test & assumptions · Challenge · Paper 4 style · 9 marks
(a)The differences (new method − old method) in the yield of a reaction for 10 paired trials are:3.1, −0.4, 2.6, 1.8, −0.7, 4.2, 2.2, 0.9, 3.5, −0.2Carry out a one-tailed sign test and a one-tailed Wilcoxon signed-rank test, each at the 5% significance level, to test whether the new method increases the median yield.[7]
(b)Explain why the two tests can reach different conclusions, and state which you would rely on, with a reason.[2]
Show answer
(a) Sign test: P(X ≥ 7) = 0.172, do not reject H₀; Wilcoxon: T = 6, critical value 10, reject H₀
B1 H₀: median difference 0, H₁: > 0; M1 X ~ B(10, 0.5), X = 7; A1 P(X ≥ 7) = 0.172, do not reject; M1 ranks: −0.2→1, −0.4→2, −0.7→3, 0.9→4, 1.8→5, 2.2→6, 2.6→7, 3.1→8, 3.5→9, 4.2→10; A1 P = 49, Q = 6, T = 6; B1 critical value 10; A1 reject H₀.
(b) The negative differences are all small, which the Wilcoxon test takes into account but the sign test ignores; rely on the Wilcoxon test if the differences are symmetric, otherwise the sign test
B1 Wilcoxon uses magnitudes (small negative ranks); B1 choice with the symmetry condition.
Similar questions
Questions that test the same subtopic in different years, grouped so you can work on one skill at a time.
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