Non-parametric tests

5(a) Underlying distribution or population of differences is unknown B1 Not known to be normal. Underlying distributions or population of scores for both X and Y Condone ‘scores for X and Y cannot be assumed to be normally are unknown distributed’. 1 5(b) H :difference of population medians=0 B1 Correct hypotheses, allow m but not μ or mean. 0 H :difference of population median≠0 B1 ‘population’ included. 1 Diff: 4 −1 −21 3 −7 −9 −5 8 −10 −12 −11 M1 Differences, allow one error. Rank: 3 −1 −11 2 −5 −7 −4 6 −8 −10 −9 M1 Signed ranks, allow one sign error. [ ] Q= −55, P=11 A1 P = 11 Critical tabular value = 13 M1 Comparison with 13 and correct ft conclusion. ‘11’ < 13 so reject H 0 Sufficient evidence of a preference for one of the uniforms OE A1 Correct conclusion, in context, following correct work, except possibly second B1. Level of uncertainty in language is used. 7 © UCLES 2021 Page 10 of 12

5(a) Underlying distribution or population of differences is unknown B1 Not known to be normal. Underlying distributions or population of scores for both X and Y Condone ‘scores for X and Y cannot be assumed to be normally are unknown distributed’. 1 5(b) H :difference of population medians=0 B1 Correct hypotheses, allow m but not μ or mean. 0 H :difference of population median≠0 B1 ‘population’ included. 1 Diff: 4 −1 −21 3 −7 −9 −5 8 −10 −12 −11 M1 Differences, allow one error. Rank: 3 −1 −11 2 −5 −7 −4 6 −8 −10 −9 M1 Signed ranks, allow one sign error. [ ] Q= −55, P=11 A1 P = 11 Critical tabular value = 13 M1 Comparison with 13 and correct ft conclusion. ‘11’ < 13 so reject H 0 Sufficient evidence of a preference for one of the uniforms OE A1 Correct conclusion, in context, following correct work, except possibly second B1. Level of uncertainty in language is used. 7 © UCLES 2021 Page 10 of 12

2 H : difference in population medians = 0 (m = m ) B1 Correct hypotheses, allow m . Do not allow μ or mean. 0 a b d H : difference in population medians ≠ 0 (m ≠ m ) B1 ‘population’ included. 1 a b Diff: 12 −14 −1 3 −13 8 −5 15 6 M1 Allow one error. Rank: 6 −8 −1 2 −7 5 −3 9 4 M1 Allow one error. Smaller sum 19 A1 (Other sum is 26.) Critical tabular value = 8 M1 Comparison with 8 and correct FT conclusion. ‘19’ > 8 so accept H 0 There is insufficient evidence that marks differ A1 Correct conclusion, in context, following correct work, except possibly the hypotheses. Level of uncertainty in language is used. 7 © UCLES 2021 Page 7 of 13

6 M1 Attempt at ranking. 51 1 72 3 55 2 86 5 75 4 170 10 98 6 175 11 137 7 220 12 152 8 260 17 167 9 262 18 235 13 302 19 238 14 311 20 242 15 320 21 256 16 333 22 351 23 Total ranks: 95 A1 H :m =m and H :m ≠m B1 Allow words, must include ‘population’ median. 0 x y 1 x y Use normal approximation with attempts at mean and variance. M1 Mean =132, variance = 264 A1 95.5−132 M1 Allow no or wrong continuity correction for M1 only. 264 © UCLES 2021 Page 13 of 14 6 – 2.246 A1 CAO Critical value is – 1.96. M1 Compare their value with – 1.96. – 2.246 < – 1.96 reject H . Or area comparison 0.0123 or 0.0124 with 0.025 and FT 0 conclusion. There is sufficient evidence of a difference in levels. A1 Correct conclusion, in context, following correct work. Level of uncertainty in language is used. 9 © UCLES 2021 Page 14 of 14

4(a) M1 Attempt at ranking. 12 1 28 4 18 2 31 6 25 3 35 9 29 5 36 10 32 7 37 11 33 8 41 13 40 12 48 15 46 14 49 16 Test statistic: 52 A1 H :m =m and H :m ≠m B1 Allow in words but ‘population’ must be included. 0 x y 1 x y Critical value for (8, 8) is 49. *B1 Allow 51 if clearly one-tail test in hypotheses. 52 > 49 DM1 Compare their calculated value with 49 and correct FT conclusion. Accept H 0 Insufficient evidence of difference in medians. A1 Correct conclusion, in context, following correct work. Level of uncertainty in language is used. 6 4(b) Not a paired sample. B1 Underlying distribution/population not (known to be) normal B1 B0 for ‘data is not normally distributed’ Underlying distribution/population unknown B0 for ‘marks are not normally distributed’ 2 © UCLES 2021 Page 10 of 13

6 M1 Attempt at ranking. 51 1 72 3 55 2 86 5 75 4 170 10 98 6 175 11 137 7 220 12 152 8 260 17 167 9 262 18 235 13 302 19 238 14 311 20 242 15 320 21 256 16 333 22 351 23 Total ranks: 95 A1 H :m =m and H :m ≠m B1 Allow words, must include ‘population’ median. 0 x y 1 x y Use normal approximation with attempts at mean and variance. M1 Mean =132, variance = 264 A1 95.5−132 M1 Allow no or wrong continuity correction for M1 only. 264 © UCLES 2021 Page 13 of 14 6 – 2.246 A1 CAO Critical value is – 1.96. M1 Compare their value with – 1.96. – 2.246 < – 1.96 reject H . Or area comparison 0.0123 or 0.0124 with 0.025 and FT 0 conclusion. There is sufficient evidence of a difference in levels. A1 Correct conclusion, in context, following correct work. Level of uncertainty in language is used. 9 © UCLES 2021 Page 14 of 14

6 Differences/rank: M1 A1 Attempt at differences, allow 4 errors. 0.2 – 0.5 –1.6 –2.9 2.5 8.5 1.2 5.2 2 –5 –11 –18 17 24 9 21 –0.1 0.8 7.8 1.9 –0.3 –1.8 –0.7 –1.4 –1 7 23 13 –3 –12 –6 –10 –0.9 2.1 2.3 –5.4 –2.0 3.4 4.7 0.4 –8 15 16 –22 –14 19 20 4 Q = 110 P = 190 A1 Either sum correct. H : population median =18.0 and H : population median 18.0 B1 Must be ‘population’, allow m. 0 1 1 1 B1 Normal approximation: mean. Mean = nn1 2425150 4 4 1 1 B1 Variance = nn12n1 2425491225 24 24 110.5150 M1 Attempt at test statistic, allow incorrect or missing cc, FT their 110. 1225 1.129 A1 © UCLES 2022 Page 12 of 13 6 10% 2-tail, critical value is 1.645 M1 Compare with 1.645, correct FT conclusion. 1.1291.645 or 0.1296 > 0.05, accept H . 0 Insufficient evidence to suggest that median is not 18.0/ A1 Correct conclusion, in context. Insufficient evidence to reject teacher’s claim. Level of uncertainty in language is used. 10 © UCLES 2022 Page 13 of 13

6 Differences/rank: M1 A1 Attempt at differences, allow 4 errors. 0.2 – 0.5 –1.6 –2.9 2.5 8.5 1.2 5.2 2 –5 –11 –18 17 24 9 21 –0.1 0.8 7.8 1.9 –0.3 –1.8 –0.7 –1.4 –1 7 23 13 –3 –12 –6 –10 –0.9 2.1 2.3 –5.4 –2.0 3.4 4.7 0.4 –8 15 16 –22 –14 19 20 4 Q = 110 P = 190 A1 Either sum correct. H : population median =18.0 and H : population median 18.0 B1 Must be ‘population’, allow m. 0 1 1 1 B1 Normal approximation: mean. Mean = nn1 2425150 4 4 1 1 B1 Variance = nn12n1 2425491225 24 24 110.5150 M1 Attempt at test statistic, allow incorrect or missing cc, FT their 110. 1225 1.129 A1 © UCLES 2022 Page 12 of 13 6 10% 2-tail, critical value is 1.645 M1 Compare with 1.645, correct FT conclusion. 1.1291.645 or 0.1296 > 0.05, accept H . 0 Insufficient evidence to suggest that median is not 18.0/ A1 Correct conclusion, in context. Insufficient evidence to reject teacher’s claim. Level of uncertainty in language is used. 10 © UCLES 2022 Page 13 of 13

5 Signs: + – – + + + + – + + + and use Binomial M1 11 signs (values not needed) allow one error. Consider B(11, 0.5) with P(X 8) or P(X 3) M1 Use of correct binomial. Compare 0.227 with 0.1 or 0.113 with 0.05 A1 Differences: M1 Allow one error. 0.06 –0.05 –0.03 0.08 0.04 0.02 0.10 –0.15 0.09 0.01 0.12 Signed Ranks: M1 Allow one swap. 6 -5 -3 7 4 2 9 -11 8 1 10 (P = 47) Q = 19 A1 Critical value from table = 13 B1 19 > 13 M1 FT their (19) compared with 13 and correct FT conclusion Accept H 0 Insufficient evidence to support median not being 5.50 A1 BOTH conclusions correct from correct working, in context. Level of uncertainty in language. 9 © UCLES 2022 Page 11 of 13

3(a) Differences: M1 4 -8 3 1 –7 11 19 6 10 –13 14 5 –9 12 Attempt at signed differences, allow 2 errors. Signed Ranks: M1 3 –7 2 1 –6 10 14 5 9 –12 13 4 –8 11 Ranks, allow 4 errors. [P = 72, Q = 33] T = 33 A1 33 clearly identified. H : difference in population medians = 0 B1 0 H : difference in population medians > 0 Correct, allow m for this mark. 1 Critical tabular value = 25 B1 33 > 25 M1 Compares their value of sum of ranks with their 25 and accept H conclusion. 0 Insufficient evidence to support claim A1 All correct except possibly first B1, in context, level of uncertainty in language. ‘Prove’ scores A0. 7 3(b) Underlying/population distribution of differences is unknown (not B1 known to be normal) 1 © UCLES 2022 Page 8 of 12

6(a) M1 Wrong test: max B1 for hypotheses, B1B1 for correct mean and 359 3 342 1 variance 374 5 355 2 398 7 361 4 Attempt at rankings (allow up to 4 errors) 415 10 384 6 427 11 401 8 452 14 414 9 461 16 437 12 482 18 443 13 502 20 454 15 512 22 472 17 545 23 491 19 612 24 506 21 Test statistic = 127 A1 Clearly identified H : population medians are equal B1 0 H : population median for X is greater than population median for 1 Y 1 B1 Mean = 1225=150 2 1 B1 Variance = 121225=300 12 © UCLES 2022 Page 12 of 13 6(a) 127.5−150 M1 Allow incorrect or no continuity correction. 300 −1.299 A1 Compare with −1.645: −1.299−1.645, or 0.097 > 0.05 M1 Valid comparison with 1.645 or 0.05 and reach correct ft Accept H conclusion. 0 Insufficient evidence to support manager’s claim A1 Correct conclusion in context, following correct work, level of uncertainty in language. ‘Prove’ is A0. 9 6(b) Not appropriate/no, not the same people B1 OE No and reason needed, e.g. individuals in the samples cannot be paired up. 1 © UCLES 2022 Page 13 of 13

3(a) Differences: M1 4 -8 3 1 –7 11 19 6 10 –13 14 5 –9 12 Attempt at signed differences, allow 2 errors. Signed Ranks: M1 3 –7 2 1 –6 10 14 5 9 –12 13 4 –8 11 Ranks, allow 4 errors. [P = 72, Q = 33] T = 33 A1 33 clearly identified. H : difference in population medians = 0 B1 0 H : difference in population medians > 0 Correct, allow m for this mark. 1 Critical tabular value = 25 B1 33 > 25 M1 Compares their value of sum of ranks with their 25 and accept H conclusion. 0 Insufficient evidence to support claim A1 All correct except possibly first B1, in context, level of uncertainty in language. ‘Prove’ scores A0. 7 3(b) Underlying/population distribution of differences is unknown (not B1 known to be normal) 1 © UCLES 2022 Page 8 of 12

4(a) M1 Differences (allow up to 3 errors) 2 5 1 –6 42 11 –3 7 20 12 –8 –10 50 A1 Correct rank order, ignore signs 2 4 1 –5 12 9 –3 6 11 10 –7 –8 13 P68, Q23 A1 CWO H : difference in population medians = 0 B1 ‘Population’ required. Accept use of m, not . 0 H 1 : population median in 2022 > population median in 2018 Do not accept ‘difference between population medians > 0’ without 2022, 2018 OE specified Critical value 21 B1 ‘23’ > ‘21’ and accept H M1 Compare their 23 with their 21 and FT conclusion. 0 Their 23 must be less than 46. Insufficient evidence to support researcher’s belief / insufficient evidence that A1 Correct conclusion in context, following correct the (median) number of employees in 2022 is greater than the (median) number work, level of uncertainty in language (not ‘prove’, of employees in 2018 not ‘there is no evidence’), no contradictions. A0 if hypotheses reversed 7 4(b) Rank for G will now be +3, giving Q = 20 which is < 21 M1 Must include numbers, ft their 21 from part (a) and ft their 20 (23 – 3) Change in conclusion A1 CWO Condone ‘reject H ’ OE. 0 2 © UCLES 2023 Page 9 of 12

4(a) M1 Differences (allow up to 3 errors) 2 5 1 –6 42 11 –3 7 20 12 –8 –10 50 A1 Correct rank order, ignore signs 2 4 1 –5 12 9 –3 6 11 10 –7 –8 13 P68, Q23 A1 CWO H : difference in population medians = 0 B1 ‘Population’ required. Accept use of m, not . 0 H 1 : population median in 2022 > population median in 2018 Do not accept ‘difference between population medians > 0’ without 2022, 2018 OE specified Critical value 21 B1 ‘23’ > ‘21’ and accept H M1 Compare their 23 with their 21 and FT conclusion. 0 Their 23 must be less than 46. Insufficient evidence to support researcher’s belief / insufficient evidence that A1 Correct conclusion in context, following correct the (median) number of employees in 2022 is greater than the (median) number work, level of uncertainty in language (not ‘prove’, of employees in 2018 not ‘there is no evidence’), no contradictions. A0 if hypotheses reversed 7 4(b) Rank for G will now be +3, giving Q = 20 which is < 21 M1 Must include numbers, ft their 21 from part (a) and ft their 20 (23 – 3) Change in conclusion A1 CWO Condone ‘reject H ’ OE. 0 2 © UCLES 2023 Page 9 of 12

3(a) M1 Differences, allow at most 2 errors. 5 7 -6 1 –4 2 8 26 –10 4 6 -5 1 –3 2 7 9 –8 A1 Correct rank order, ignore signs. [P = 29] Q = 16 A1 Condone P not excluded. H : population medians equal or m m B1 ‘Population’ required. Accept use of m, not . 0 1 2 Do not accept ‘difference between population medians > 0’ H : population median X > population median Y or m m 1 1 2 without X, Y OE specified. Critical value = 8 B1 16 > 8 and accept H / not significant M1 Compare their ‘16’ with their ‘8’ and conclusion. o Their ‘16’ must be less than 23. Ignore their hypotheses. Condone ‘reject H ’. 1 Insufficient evidence to support teacher’s belief A1 Correct conclusion in context, following correct work, level of or uncertainty in language (not ‘prove’, not ‘there is no evidence’), insufficient evidence that the marks/median in Paper 1 are/is no contradictions. higher than the marks/median in Paper 2 e.g. Proves that the teacher is incorrect scores A0. A0 if hypotheses wrong way round. 7 3(b) The population differences are symmetrical (about the median B1 Words in bold, or their equivalent, are required. difference) 1 © UCLES 2023 Page 8 of 13

6 H : population medians are equal B1 Do not accept ‘differences between population medians < 0 H : population median after > population median before 0’ unless difference defined. 1 Allow H :m=22, H :m22.0 0 1 M1 Signed differences, allow at most 4 errors –0.8 -7 1.5 12 0.9 8 –3.4 –19 –2.6 –18 0.1 1 4.5 21 –1.8 –14 3.7 20 –1.4 –11 M1 Attempt at ranks (ignore signs). 0.3 3 –4.6 –22 0.2 2 5 23 1.9 15 6.2 25 0.6 5 5.2 24 1 9 1.7 13 –2.2 -16 0.7 6 1.3 10 –0.5 –4 2.3 17 (W = 214) W = 111 A1 Cao identified, or used, as test statistic. + − 1 1 B1 Normal: mean = n(n+1)= 2526=162.5 4 4 1 1 B1 Variance = n(n+1)(2n+1)= 252651=1381.25 24 24 111.5−162.5 M1 Allow missing or incorrect continuity correction. z- value: Their 111 must come from ranks. 1381.25 −1.37 A1 CAO Tabular value is −1.645: ‘−1.37'−1.645, or 0.915<0.95 oe, accept H , M1 Consistent signs. Allow ‘not significant’. 0 Insufficient evidence to support the teacher’s claim A1 All correct. Correct conclusion in context, following Insufficient evidence to suggest that the distances thrown are further when it correct work, level of uncertainty in language. is hot A0 if hypotheses wrong way round or missing 10 © UCLES 2023 Page 13 of 13

6 H : population medians are equal B1 Do not accept ‘differences between population medians < 0 H : population median after > population median before 0’ unless difference defined. 1 Allow H :m=22, H :m22.0 0 1 M1 Signed differences, allow at most 4 errors –0.8 -7 1.5 12 0.9 8 –3.4 –19 –2.6 –18 0.1 1 4.5 21 –1.8 –14 3.7 20 –1.4 –11 M1 Attempt at ranks (ignore signs). 0.3 3 –4.6 –22 0.2 2 5 23 1.9 15 6.2 25 0.6 5 5.2 24 1 9 1.7 13 –2.2 -16 0.7 6 1.3 10 –0.5 –4 2.3 17 (W = 214) W = 111 A1 Cao identified, or used, as test statistic. + − 1 1 B1 Normal: mean = n(n+1)= 2526=162.5 4 4 1 1 B1 Variance = n(n+1)(2n+1)= 252651=1381.25 24 24 111.5−162.5 M1 Allow missing or incorrect continuity correction. z- value: Their 111 must come from ranks. 1381.25 −1.37 A1 CAO Tabular value is −1.645: ‘−1.37'−1.645, or 0.915<0.95 oe, accept H , M1 Consistent signs. Allow ‘not significant’. 0 Insufficient evidence to support the teacher’s claim A1 All correct. Correct conclusion in context, following Insufficient evidence to suggest that the distances thrown are further when it correct work, level of uncertainty in language. is hot A0 if hypotheses wrong way round or missing 10 © UCLES 2023 Page 13 of 13

2 H 0 : practical results and written results are equal B1 Or H 0 : m P m W 0, H 1 : m P m W 0, oe H : practical results are greater than written results 1 (m is population median). There are 10 + signs; B(15, 0.5) soi M1 (P(X 10)1PX 9 =) 1 – 0.849 = 0.151 A1 P(X ⩽ 5) = 0.151 (implies M1) Compare with 0.1: ‘0.151’ > 0.1 accept H M1 ‘0.151’ must come from a valid binomial 0 calculation Correct ft conclusion for their 0.151 and 0.1 Condone Reject H 1 ‘Accept H ’ can be implied by a conclusion that is 0 consistent with their 0.151 and 0.1 Insufficient evidence to suggest that practical results are greater than written results A1 Correct work only, ignoring their hypotheses, conclusion in context with level of uncertainty in language. Not ‘prove’. Condone ‘no sufficient’ ‘not enough’. Do not accept statements such as ‘there is sufficient evidence to suggest…..’. 5 © Cambridge University Press & Assessment 2024 Page 7 of 15

3(a) Signed differences M1 Allow one error. 0.4 1.1 0.8 0.5 1 .2 0.9 1 .9 0.3 2.2 1 .6 0.6 1 .0 Ranks: 2 8 5 3 9 6 1 1 1 12 1 0 4 7 M1 Attempt at ranking. (P = 54) P = 24 A1 All ranks must be correct. + - H : population median = 22.0 B1 Accept m used as ‘population median’ 0 H : population median  22.0 1 Critical value from table = 17 B1 ‘24’ > ‘17’ Accept H M1 ‘24’ must come from a ranking of signed 0 differences, ‘17’ must be critical value from tables (7, 9, 10, 13, 17 or 21). Correct ft conclusion for their 24 and their 17. Condone Reject H . 1 ‘Accept H ’ can be implied by a conclusion that is 0 consistent with their 24 and their 17. Insufficient evidence to support (population) median not being 22.0 A1 Correct work only, ignoring their hypotheses, conclusion in context with level of uncertainty in Insufficient evidence against the manager’s claim language. Not ‘prove’. Condone ‘no sufficient’ ‘not enough’. Do not accept statements such as ‘there is sufficient evidence to suggest…..’. 7 © Cambridge University Press & Assessment 2024 Page 8 of 15 3(b) Distribution is symmetric about the population median B1 Need to see population or underlying distribution Underlying/population distribution is symmetric about the median mentioned. Underlying distribution is symmetric B0. Population distribution is symmetric B0. ‘data’ implies B0 but condone ‘population data’ ‘mean’ implies B0. 1 Question Answer Marks Guidance

2 H 0 : practical results and written results are equal B1 Or H 0 : m P m W 0, H 1 : m P m W 0, oe H : practical results are greater than written results 1 (m is population median). There are 10 + signs; B(15, 0.5) soi M1 (P(X 10)1PX 9 =) 1 – 0.849 = 0.151 A1 P(X ⩽ 5) = 0.151 (implies M1) Compare with 0.1: ‘0.151’ > 0.1 accept H M1 ‘0.151’ must come from a valid binomial 0 calculation Correct ft conclusion for their 0.151 and 0.1 Condone Reject H 1 ‘Accept H ’ can be implied by a conclusion that is 0 consistent with their 0.151 and 0.1 Insufficient evidence to suggest that practical results are greater than written results A1 Correct work only, ignoring their hypotheses, conclusion in context with level of uncertainty in language. Not ‘prove’. Condone ‘no sufficient’ ‘not enough’. Do not accept statements such as ‘there is sufficient evidence to suggest…..’. 5 © Cambridge University Press & Assessment 2024 Page 7 of 15

3(a) Signed differences M1 Allow one error. 0.4 1.1 0.8 0.5 1 .2 0.9 1 .9 0.3 2.2 1 .6 0.6 1 .0 Ranks: 2 8 5 3 9 6 1 1 1 12 1 0 4 7 M1 Attempt at ranking. (P = 54) P = 24 A1 All ranks must be correct. + - H : population median = 22.0 B1 Accept m used as ‘population median’ 0 H : population median  22.0 1 Critical value from table = 17 B1 ‘24’ > ‘17’ Accept H M1 ‘24’ must come from a ranking of signed 0 differences, ‘17’ must be critical value from tables (7, 9, 10, 13, 17 or 21). Correct ft conclusion for their 24 and their 17. Condone Reject H . 1 ‘Accept H ’ can be implied by a conclusion that is 0 consistent with their 24 and their 17. Insufficient evidence to support (population) median not being 22.0 A1 Correct work only, ignoring their hypotheses, conclusion in context with level of uncertainty in Insufficient evidence against the manager’s claim language. Not ‘prove’. Condone ‘no sufficient’ ‘not enough’. Do not accept statements such as ‘there is sufficient evidence to suggest…..’. 7 © Cambridge University Press & Assessment 2024 Page 8 of 15 3(b) Distribution is symmetric about the population median B1 Need to see population or underlying distribution Underlying/population distribution is symmetric about the median mentioned. Underlying distribution is symmetric B0. Population distribution is symmetric B0. ‘data’ implies B0 but condone ‘population data’ ‘mean’ implies B0. 1 Question Answer Marks Guidance

1 M1 A1 Attempt at ranking from 1 to 19. 21.4 4 20.9 2 A1 for all correct. 24.6 15 23.5 11 25.3 18 24.8 16 22.7 8 21.9 6 20.8 1 23.4 10 21.5 5 24.0 13 22.9 9 23.8 12 21.3 3 24.1 14 22.3 7 25.1 17 25.8 19 Test statistic: 70 A1 If ranks reversed, then sum of ranks = 110, and then 9(9 + 10 + 1) – 110 = 70. H 0 : population median of X = population median of B1 Accept H 0 : m X m Y H 1 : m X m Y . Y Do not accept any other notation. H population median of X < population median of 1: Y Critical value for m = 9, n = 10 is 61 B1 Using normal approximation implies last 3 marks are B0M0A0, max 4 out of 7. © Cambridge University Press & Assessment 2024 Page 6 of 15 1 ‘70’ > ‘61’ accept H M1 ‘70’ must come from ranks 1 – 19, ‘61’ must be a critical value from the correct 0 table. ft conclusion for their 70 and their 61. Condone Reject H . 1 ‘Accept H ’ can be implied by a conclusion in context that is consistent with their 0 70 and their 61. Insufficient evidence to suggest that population A1 Correct work only, ignoring their hypotheses, conclusion in context with level of medians of X and Y are not equal uncertainty in language. Not ‘prove’ or equivalent. or Condone ‘no sufficient’ ‘not enough’. Insufficient evidence to suggest that population Do not accept statements such as ‘there is sufficient evidence to suggest…..’. median of X is less than median of Y or Insufficient evidence to support the interviewer’s belief 7 © Cambridge University Press & Assessment 2024 Page 7 of 15

2(a) 2 4 −8 5 1 0 7 −3 6 −12 9 1 4 1 1 M1 Signed differences, allow one error. 1 3 −7 4 9 6 −2 5 −11 8 1 2 1 0 M1 Attempt at ranking. (Sum of + ranks = 58) Test statistic = 20 A1 H : population median for teacher 2 = population median for teacher 1 B1 0 H : population median for teacher 2 > population median for teacher 1 Allow use of m for population median. 1 Critical value, from tables, is 17 B1 ‘20’ > 17, accept H M1 0 Insufficient evidence to support the claim A1 Correct work only except possibly hypotheses, in Insufficient evidence that the scores of Teacher 2 are higher than those of Teacher 1 context, level of uncertainty in language. 7 © Cambridge University Press & Assessment 2024 Page 6 of 11 2(b) New test statistic is ‘13’ B1 FT Their 20 minus 7. 13 < 17, so conclusion is now ‘reject H ’, (sufficient evidence to support claim) B1 Must be 13 and 17. 0 2 Question Answer Marks Guidance

6 Signed differences from 51.0 M1 Allow up to 2 errors. 0.3 1.0 2.4 - 1.8 - 1.7 0.1 1.2 - 3.8 2.0 - 2.5 - 1.6 - 0.7 - 0.2 0.6 - 1.9 1.3 0.8 1.4 - 3.1 - 2.1 - 0.4 0.9 M1 Attempt at ranking of signed differences from 3 9 19 -15 -14 1 10 -22 51.0. 17 -20 -13 -6 -2 5 -16 11 7 12 -21 -18 -4 8 Smaller sum is 102 A1 151 H : population median is 51.0 (seconds) B1 Allow m. 0 H population median is less than 51.0 (seconds) 1: Normal approximation: mean =126.5 B1 Both. variance = 948.75 '102'+0.5−126.5 M1 Allow no continuity correction. 948.75 −0.779 A1 Accept 0.779. Accept p-value = 0.218. (A0 for z = – 0.795 or p = 0.213 from no continuity correction). '−0.779' −1.645 accept H M1 Or ‘0.218’ > 0.05 or ‘0.782’ < 0.95 0 © Cambridge University Press & Assessment 2024 Page 12 of 13 6 Insufficient evidence to suggest that (population) median is less than 51.0 seconds/ A1 Correct work only, except possibly hypotheses, in Insufficient evidence to support the claim context, level of uncertainty in language. Alternative approach using critical values. 9 © Cambridge University Press & Assessment 2024 Page 13 of 13

2(a) 2 4 −8 5 1 0 7 −3 6 −12 9 1 4 1 1 M1 Signed differences, allow one error. 1 3 −7 4 9 6 −2 5 −11 8 1 2 1 0 M1 Attempt at ranking. (Sum of + ranks = 58) Test statistic = 20 A1 H : population median for teacher 2 = population median for teacher 1 B1 0 H : population median for teacher 2 > population median for teacher 1 Allow use of m for population median. 1 Critical value, from tables, is 17 B1 ‘20’ > 17, accept H M1 0 Insufficient evidence to support the claim A1 Correct work only except possibly hypotheses, in Insufficient evidence that the scores of Teacher 2 are higher than those of Teacher 1 context, level of uncertainty in language. 7 © Cambridge University Press & Assessment 2024 Page 6 of 11 2(b) New test statistic is ‘13’ B1 FT Their 20 minus 7. 13 < 17, so conclusion is now ‘reject H ’, (sufficient evidence to support claim) B1 Must be 13 and 17. 0 2 Question Answer Marks Guidance

5(a) H :m =m B1 If in words, must contain ‘population median’. 0 s c Allow H : m = 0, H : m > 0 (or < 0) if defined H :m m 0 d 1 d 1 s c or consistent with working. d 21 19 –2 22 –10 4 16 27 35 29 M1 Attempt at differences. Rank 6 5 –1 7 –3 2 4 8 10 9 M1 Attempt at (signed) ranks. (P=51, Q=4) T =4 A1 Allow errors in positive ranks. '4'≤10, reject H (significant). M1 Compare their 4 (from rank sum only) with 10 0 & appropriate conclusion (allow in terms of H ). 1 Reject H can be implied by an attempt at an 0 appropriate conclusion in context. Allow M1 for comparison of their 4 (from rank sum only) with 8 and appropriate conclusion. Sufficient evidence to suggest that children, on average, complete the cartoon puzzle A1 CWO ignoring hypotheses. more quickly than the seaside puzzle. In context, level of uncertainty in language. OR Sufficient evidence to support the researcher’s belief. A0 for T =4 from incorrect ranks/differences. 6 5(b)   X ~B(10, 0.5)  P(X ≤2)=0.0547 B1 OR P(X ≥8)=0.0547. '0.0547'0.05, accept H (not significant). M1 Compare their 0.0547 with 0.05 and appropriate 0 conclusion (may be in terms of H ). 1 In part 5(a) H is rejected, in part 5(b) H is accepted, so opposite conclusion reached. A1 CWO, compares with test result in part 5(a). 0 0 3 5(c) The order of the tests means that children may do the second test quicker (or slower) B1 Credit any valid comment in context. because, for example, they have had practice (or are tired). 1 © Cambridge University Press & Assessment 2025 Page 14 of 16

5(a) H :m =m B1 If in words, must contain ‘population median’. 0 s c Allow H : m = 0, H : m > 0 (or < 0) if defined H :m m 0 d 1 d 1 s c or consistent with working. d 21 19 –2 22 –10 4 16 27 35 29 M1 Attempt at differences. Rank 6 5 –1 7 –3 2 4 8 10 9 M1 Attempt at (signed) ranks. (P=51, Q=4) T =4 A1 Allow errors in positive ranks. '4'≤10, reject H (significant). M1 Compare their 4 (from rank sum only) with 10 0 & appropriate conclusion (allow in terms of H ). 1 Reject H can be implied by an attempt at an 0 appropriate conclusion in context. Allow M1 for comparison of their 4 (from rank sum only) with 8 and appropriate conclusion. Sufficient evidence to suggest that children, on average, complete the cartoon puzzle A1 CWO ignoring hypotheses. more quickly than the seaside puzzle. In context, level of uncertainty in language. OR Sufficient evidence to support the researcher’s belief. A0 for T =4 from incorrect ranks/differences. 6 5(b)   X ~B(10, 0.5)  P(X ≤2)=0.0547 B1 OR P(X ≥8)=0.0547. '0.0547'0.05, accept H (not significant). M1 Compare their 0.0547 with 0.05 and appropriate 0 conclusion (may be in terms of H ). 1 In part 5(a) H is rejected, in part 5(b) H is accepted, so opposite conclusion reached. A1 CWO, compares with test result in part 5(a). 0 0 3 5(c) The order of the tests means that children may do the second test quicker (or slower) B1 Credit any valid comment in context. because, for example, they have had practice (or are tired). 1 © Cambridge University Press & Assessment 2025 Page 14 of 15

4 H : population medians are equal for under 25s and over 50s (or m =m ). B1 B0 for defining m as median, not population 0 25 50 median. H : population median for under 25s < population median time for over 50s 1 (or m m ). 25 50 M1 Attempt at ranks using the integers 1 to 14 198 5 178 1 (ranks may be reversed). 212 8 181 2 217 9 183 3 229 11 192 4 235 13 203 6 242 14 209 7 223 10 231 12 Sum of ranks for sample of size 6: R =60 A1 Note: R =30 from reverse ranks. 6 6 6 ( 6+8+1 ) −'R ' [=30] M1 Note:6(6+8+1)−'R '=60 from reverse ranks. 6 6 [W =min(60,30)=] 30 A1 Critical value for n=6, m=8 is 31. B1 '30'≤31, reject H 0 (significant). M1 Compare their W with 31 and appropriate conclusion (may be in terms of H ). 1 Reject H may be implied by attempt at 0 appropriate conclusion in context. Sufficient evidence to suggest that older people take longer to react. A1 Correct conclusion in context from correct OR Sufficient evidence to support researcher’s claim. working ignoring hypotheses. Level of uncertainty in language used (for example, not ‘prove’). 8 © Cambridge University Press & Assessment 2025 Page 13 of 16

2(a) H : population median=8.00 B1 Allow m, but not . If m defined, must see population. 0 H : population median8.00 1 –0.34, 0.48, 0.21, –0.02, 0.01, –0.23, 0.25, 0.11, 0.03, 0.16, –0.08 M1 Attempt at (signed) differences. –10, 11, 7, –2, 1, –8, 9, 5, 3, 6, –4 M1 Attempt at (signed) ranks, ranks may be reversed. (P=42, Q=24) T =24 A1 '24' > 13, accept H (not significant). M1 Compare their 24 (from rank sum only) with 13 and appropriate 0 conclusion (may be in terms of H ). 1 Accept H can be implied by an attempt at an appropriate 0 conclusion in context. Allow M1 for comparison of their 24 (from rank sum only) with 10 and appropriate conclusion. Insufficient evidence to support/suggest that the average sound A1 Correct conclusion in context from correct working ignoring level is more than 8.00. hypotheses. Level of uncertainty in language used (for example, not ‘prove’). Must refer to average/median sound (level). 6 2(b) It is not known if the population is normal. B1 Must refer to population and/or distribution. It is not known if the (underlying) distribution is normal. B0 for data is normal. B0 for (population) differences are normal. 1 © Cambridge University Press & Assessment 2025 Page 10 of 15

4 128(28+1) =203 B1 Both expressions correct. Implied by 203 and 1928.5 seen. 4 1 28(28+1)(56+1) =1928.5 24 T +0.5−'203' M1 Correct form or expression for CI with any z-value. z= Condone T or T−0.5 instead of T+0.5 for M1. '1928.5' T +0.5−'203' M1 Compares their z with 2.576. −2.576 '1928.5' Condone T or T−0.5 instead of T+0.5 for M1. T 89.4 A1 Allow T =89.4. T =90 A1 All correct with no errors seen, including correct continuity correction. 5 © Cambridge University Press & Assessment 2025 Page 11 of 16

1(a) Data has outliers / is skewed / lacks symmetry, suggesting B1 Must be in context. population/distribution may not be symmetrical. Condone no explicit mention of population/distribution or a definite statement relating to population. 1 1(b) H : population median is [$]32,500. B1 Allow m, but not . If m defined, must see population. 0 H 1 : population median is not [$]32,500. Median not mean or average. Test statistic = 4 (or 11) B1 Can be implied by P(X ≤4) or P(X 4). (For use of Normal approximation, the value of the test statistic can be implied by 4.5 or 3.5 for 4 and 10.5 or 11.5 for 11). [ X ~B(15,0.5)] B1 AWRT 0.0592. Might see 0.1184/0.1185/0.118/0.119 from2P(X ≤4) or 2P(X≥11). P(X≤4)=0.0592 OR P(X≥11)=0.0592 Use of N(7.5,3.75) scores B1 for ±1.549. '0.0592'0.05 , accept H / do not reject H / not significant. M1 Compare their 0.0592 (must be a tail probability) with 0.05 (or 0.1 if 0 0 1-tail test used) OE and appropriate result (may be in terms of H ). 1 Result may be implied by an attempt at an appropriate conclusion in context ignoring hypotheses. Use of Normal: compare their ±1.549 with ±1.645, signs must be consistent (1.282 for 1-tail test). Insufficient evidence to reject [company’s] claim. A1 Correct conclusion from correct working ignoring hypotheses. OR In context. Insufficient evidence to suggest that median [salary] is not Level of uncertainty in language is used (for example, not ‘prove’). [$]32,500. Allow “Not enough evidence / no sufficient evidence that / to show / conclude that…”. Do not accept “No evidence…”. Do not accept statements such as “there is sufficient evidence to suggest…”. 5 © Cambridge University Press & Assessment 2025 Page 9 of 15

4 128(28+1) =203 B1 Both expressions correct. Implied by 203 and 1928.5 seen. 4 1 28(28+1)(56+1) =1928.5 24 T +0.5−'203' M1 Correct form or expression for CI with any z-value. z= Condone T or T−0.5 instead of T+0.5 for M1. '1928.5' T +0.5−'203' M1 Compares their z with 2.576. −2.576 '1928.5' Condone T or T−0.5 instead of T+0.5 for M1. T 89.4 A1 Allow T =89.4. T =90 A1 All correct with no errors seen, including correct continuity correction. 5 © Cambridge University Press & Assessment 2025 Page 11 of 16

6(a) H : population median mark for school A is equal to population B1 Allow m, but not . If m defined, must see population. 0 median mark for school B. H : population median mark for school A is not equal to 1 population median mark for school B. 1123(123+147+1) =16666.5 B1 Both expressions correct. 2 Implied by 16666.5 and 408329.25 seen. 1 123147(123+147+1) =408329.25 12 15355+0.5−'16666.5' M1 OR attempts to find critical value or p-value using their mean and variance. Condone continuity correction omitted or incorrect. '408329.25' −2.05 A1 AWRT −2.05 or CV = 15414 or p=0.0201. Condone continuity correction omitted or incorrect. −2.05−1.96 OR 0.9800.975 OR 0.0200.025 OR M1 Compare their 2.05 with 1.96 (consistent signs) or their 0.980 15355+0.515414, reject H / significant. with 0.975, or their 0.020/0.040 with 0.025/0.05 or their 15355.5 0 with 15414 (condone continuity correction omitted or incorrect). Consistent signs and appropriate result (may be in terms of H ). 1 Test result can be implied by an attempt at an appropriate conclusion in context. Sufficient evidence to suggest that there is a difference in average A1 Correct conclusion from correct working, ignoring hypotheses. marks between the students in school A and B. Must consider continuity correction e.g. 15355.5 seen. OR In context. Sufficient evidence to suggest that, on average, the test scores for Level of uncertainty in language is used (for example, not ‘prove’). students at the two schools are not equal. Do not accept statements such as ‘there is insufficient evidence to suggest…’. 6 6(b) The (underlying) distributions are symmetrical/similar shape. B1 OR The populations are symmetrical/similar shape. OR No repeated marks. 1 © Cambridge University Press & Assessment 2025 Page 17 of 17

1(a) Data has outliers / is skewed / lacks symmetry, suggesting population / distribution may B1 Condone no explicit mention of not be symmetrical. population/distribution or a definite statement relating to population. Allow comment about tied ranks. 1 © Cambridge University Press & Assessment 2026 Page 7 of 17 Trace ID: PC-6B4604D1D5 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/4a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/41 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 1(b) H : population median is 8. B1 Allow m, but not . 0 H 1 : population median is greater than 8. If m defined, must see population. Median, not mean or average. © Cambridge University Press & Assessment 2026 Page 8 of 17 X ~ B ( 1 5 , 0 .5 ) B1 SOI, may be implied by N ( 7 .5 , 3 .7 5 ) . 15 15 15 P(X ≤3)=(0.5)15 +  (0.5)15 +  (0.5)15 +  (0.5)15  1   2  3 OR   3 (  0 .5 3 ) .7 − 5 7 .5  M1 Attempt at using binomial distribution with 4 terms. May calculate P(X ≥12). Implied by AWRT 0.018. OR Attempt to use correct normal approximation, ignoring continuity correction. Implied by AWRT  2 .0 7 ,  2 .3 2 o r  2.58. 0 .0 1 7 6 A1 AWRT 0.018 OR M1A1 for unsupported 0.018 or 2.07 from 2.07 normal approximation. ‘0.0176’ < 0.05, reject H / significant. M1 Compare their 0.0176 with 0.05 and appropriate 0 OR conclusion (may be in terms of H ). 1 2.07 > 1.645, reject H / significant. Their 0.0176 must come from use of binomial. 0 Test result can be implied by an attempt at an appropriate conclusion in context. For normal approximation, compare their  2 .0 7 (must come from attempt at standardising) with 1.645 or 1.64 or 1.65, signs must be consistent. Sufficient evidence to support the manager’s belief. A1 Correct conclusion from correct working, ignoring hypotheses. In context. Level of uncertainty in language is used (for example, not ‘prove’). Do not accept statements such as ‘there is insufficient evidence to suggest…’. 6 Trace ID: PC-6B4604D1D5 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/4a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/41 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/

1(a) Data has outliers / is skewed / lacks symmetry, suggesting population / distribution may B1 Condone no explicit mention of not be symmetrical. population/distribution or a definite statement relating to population. Allow comment about tied ranks. 1 © Cambridge University Press & Assessment 2026 Page 7 of 17 Trace ID: PC-E48FEEF9A8 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/4a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/42 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 1(b) H : population median is 8. B1 Allow m, but not . 0 H 1 : population median is greater than 8. If m defined, must see population. Median, not mean or average. © Cambridge University Press & Assessment 2026 Page 8 of 17 X ~ B ( 1 5 , 0 .5 ) B1 SOI, may be implied by N ( 7 .5 , 3 .7 5 ) . 15 15 15 P(X ≤3)=(0.5)15 +  (0.5)15 +  (0.5)15 +  (0.5)15  1   2  3 OR   3 (  0 .5 3 ) .7 − 5 7 .5  M1 Attempt at using binomial distribution with 4 terms. May calculate P(X ≥12). Implied by AWRT 0.018. OR Attempt to use correct normal approximation, ignoring continuity correction. Implied by AWRT 2.07, 2.32 or 2.58. 0 .0 1 7 6 A1 AWRT 0.018 OR M1A1 for unsupported 0.018 or 2.07 from 2.07 normal approximation. ‘0.0176’ < 0.05, reject H / significant. M1 Compare their 0.0176 with 0.05 and appropriate 0 OR conclusion (may be in terms of H ). 1 2.07 > 1.645, reject H / significant. Their 0.0176 must come from use of binomial. 0 Test result can be implied by an attempt at an appropriate conclusion in context. For normal approximation, compare their  2 .0 7 (must come from attempt at standardising) with 1.645 or 1.64 or 1.65, signs must be consistent. Sufficient evidence to support the manager’s belief. A1 Correct conclusion from correct working, ignoring hypotheses. In context. Level of uncertainty in language is used (for example, not ‘prove’). Do not accept statements such as ‘there is insufficient evidence to suggest…’. 6 Trace ID: PC-E48FEEF9A8 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/4a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/42 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/

2(a) Population of differences may not be normally distributed. B1 Allow underlying/parent distribution for population. B0 for population mean may not be normal. B0 for (underlying) data may not be normal. 1 2(b) H : difference in population medians =0 0 H : difference in population medians 1 © Cambridge University Press & Assessment 2026 Page 10 of 15  0 B1 Allow use of m d or m 1 − m 2 OE. If m defined, must include reference to population. Median not mean or average. Differences: 4, 8, 15, 3, –1, –2, 5, –7, 11, 6 M1 Attempt at differences (at most 2 errors). Signed ranks: 4, 8, 10, 3, –1, –2, 5, –7, 9, 6 M1 Attempt at signed ranks from an attempt at differences.  P = 4 5 , Q = 1 0  T=10 A1 Critical value = 8 B1 '1 0 '  '8 ' , accept H / do not reject H / not significant. M1 Compare their 10 with their 8 (their 8 may be 13, 10, 8, 5 or 3) 0 0 and appropriate conclusion (may be in terms of H ). 1 Their 10 must come from a valid attempt at ranking. Test result can be implied by an attempt at an appropriate conclusion in context. Insufficient evidence to support that there is (on average) a A1 Correct conclusion from correct working ignoring hypotheses. difference between the marks (awarded by the teachers). In context. Level of uncertainty in language used (for example, not ‘prove’). Do not accept ‘No evidence…’. Do not accept statements such as ‘there is sufficient evidence to suggest…’. 7 Trace ID: PC-1E62B2C2D5 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/4a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/43 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/

4(a) Reduces the uncertainty/variability from choosing cyclists of different abilities. B1 Allow for an attempt at an explanation that pairing/differences/matching leads to improved OR accuracy/precision/clarity/sensitivity/certainty. Controls for differences between different cyclists. 1 © Cambridge University Press & Assessment 2026 Page 13 of 18 Trace ID: PC-1430F78DC7 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/4a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/44 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 4(b) H : population median difference= 0 B1 Allow 0 H : population median difference > 0 1 © Cambridge University Press & Assessment 2026 Page 14 of 18 m , m d , m a − m c OE but not . If m defined, must see population. Median not mean or average. 3.9 3 8 –1.4 1.8 3.7 0.1 0.2 –0.3 2.9 M1 Attempt at (signed) differences. At most 2 errors. May be seen with table before 4(a). 9 7 10 –4 5 8 1 2 –3 6 M1 Attempt at (signed) ranks, ranks may be reversed. May be seen with table before 4(a). [P=48] Q = 7 A1 Ignore any errors in positive ranks. '7' < 10, reject H / significant. M1 Compare their 7 with 10 and appropriate 0 conclusion (may be in terms of H ). Note that if 1 their 7 is 10, they should still reject H . 0 Their 7 must come from attempt at ranks and be the smaller of their P and Q. Test result can be implied by an attempt at an appropriate conclusion in context ignoring hypotheses. Sufficient evidence to support the club captains’ belief. A1 Correct conclusion from correct working ignoring hypotheses. OR A0 for Q=7 from incorrect ranks. In context. Level of uncertainty in language used Sufficient evidence to suggest that the anticlockwise route takes longer (to complete (for example, not ‘prove’). on average than the clockwise route). Allow ‘(Some/enough) evidence that/to show/conclude that…’. Do not accept statements such as ‘there is insufficient evidence to suggest…’. 6 4(c) The population of differences is symmetrical. B1 Allow underlying/parent distribution for population. 1 Trace ID: PC-1430F78DC7 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/4a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/44 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/