Chi-squared tests

2 M1 At least 3 correct to 4sf. 72 (61.425) 42 (44.46) 52 (54.405) 68 (73.71) 33 (43.575) 34 (31.54) 41 (38.595) 58 (52.29) Calculates E values (in brackets above) A1 All correct to 4sf (if 6.037 not found). ( O−E )2 M1 Correct formula used. Test value:  = 1.8206 + 0.1361 + 0.1063 +0.4423 + E 2.5664 + 0.1919 + 0.1499 + 0.6235 = 6.037 A1 Correct to 3 sf. H : (driving) test success is independent of instructor B1 0 Tabular value: 3 df 10% = 6.251 M1 Compare with correct tabular value and conclusion. 6.04 < 6.251 Accept H 0 Sufficient evidence to suggest that (driving) test success is A1 Correct conclusion, in context, following correct work. Level independent of instructor of uncertainty in language is used. Allow ±1 in the third significant figure of test value. 7 © UCLES 2021 Page 7 of 12

2 M1 At least 3 correct to 4sf. 72 (61.425) 42 (44.46) 52 (54.405) 68 (73.71) 33 (43.575) 34 (31.54) 41 (38.595) 58 (52.29) Calculates E values (in brackets above) A1 All correct to 4sf (if 6.037 not found). ( O−E )2 M1 Correct formula used. Test value:  = 1.8206 + 0.1361 + 0.1063 +0.4423 + E 2.5664 + 0.1919 + 0.1499 + 0.6235 = 6.037 A1 Correct to 3 sf. H : (driving) test success is independent of instructor B1 0 Tabular value: 3 df 10% = 6.251 M1 Compare with correct tabular value and conclusion. 6.04 < 6.251 Accept H 0 Sufficient evidence to suggest that (driving) test success is A1 Correct conclusion, in context, following correct work. Level independent of instructor of uncertainty in language is used. Allow ±1 in the third significant figure of test value. 7 © UCLES 2021 Page 7 of 12

5(a) (Mean of distribution = 2.4) M1 Allow 1.52. Variance = 2.25 Mean approx. equal to variance (so Poisson might be suitable) A1 Must have ‘approximately’ oe, not equality. Must have 2.25. 2 5(b) Po(2.5) leads to frequencies: M1 At least 3 correct to 3sf. 14.77[5], 36.938, 46.173, 38.477, 24.048, 12.024, 5.010, 2.554 A1 All correct to 4sf. Combine last two values: 7.564 M1 FT their table values if final figure is less than 5. ( O−E )2 M1 Apply correct formula to their frequencies.  =2.6227+0.4198+0.00065+1.4529+0.04570 E +1.3148+0.7845 6.64 A1 Correct to 3sf. 5(b) H : distribution fits the data B1 Must mention data and distribution . 0 Tabular value, 5 degrees of freedom, is 9.236 M1 Compare with correct tabular value and conclusion. ‘6.64’ < 9.236, so accept H 0 Insufficient evidence to show that data does not follow a Poisson A1 Correct conclusion, in context, following correct work. distribution Level of uncertainty in language is used. Allow ±1 difference in third significant figure. 8 © UCLES 2021 Page 10 of 13

3(a) B1 3 dp or better. x 0 1 2 3 4 5 6 or more O 35 48 43 15 6 3 0 E 40.874 57.704 35.641 12.579 2.775 0.392 0.0352 p = 57.704 q = 0.035 or 0.036 B1 Accept numbers which round to 0.035 or 0.036. 2 © UCLES 2021 Page 8 of 14 3(b) Combine frequencies less than 5: last 4 columns give 24 / 15.7812 M1 Allow last 2 or 3 columns combined for this M1. ( O−E )2 M1  =0.8444+1.6319+1.5199+4.2803 E 8.28 A1 Accept 8.27 – 8.28. SC: 25.60, if values not combined, scores M1A0. SC: 14.96 if last 3 combined, scores M1A0. SC: if last 2 combined, scores M1A0. H : B(8, 0.15) fits the data B1 Must mention distribution and data. 0 3 degrees of freedom, tabular value = 7.815. M1 Compare their value with 7.815 and correct FT conclusion 8.28 > 7.815 3 columns combined compared with 9.488 and correct Reject H . conclusion 0 Insufficient evidence to support manager’s claim. A1 Correct conclusion, in context, following correct work. Level of uncertainty in language is used. 6 © UCLES 2021 Page 9 of 14

2(a) P(130x<140) = P  130−125 ≤Z< 140−125   M1 Must be seen.  12 12  Must see 0.4167 and 1.25, may be implied by 0.8944 and 0.6616 P( 0.4167<Z <1.25) Accept 0.417. 0.8944−0.6616=0.2328 A1 AG Multiply by 60 leads to 13.97 Accept 0.6615. Need to see 0.2328 or 0.2329 or 13.968 or ( 0.8944−0.6616 )×60. 2 2(b) M1 Combine first two values. 0 3 15 23 11 8 1.12 5.22 13.97 19.38 13.97 6.34 Test stat = 1.7588 + 0.0761 + 0.6743 + 0.6309 + 0.4352 M1 3.58 A1 Accept 3.57 – 3.58 . SC: 3.88, if values not combined, scores M1A0. H :distribution fits data B1 Must mention distribution and data. 0 N(125, 122) is a good model for the data oe 4 degrees of freedom, so tabular value = 9.488 M1 Compare their value with 9.488 (9.49) and correct FT conclusion. 3.58 < 9.488 Or, if values not combined, compare their value with 11.07 and Accept H . correct FT conclusion for M1A0. 0 There is sufficient evidence that the normal distribution fits the A1 Correct conclusion, in context, following correct work. Level of data uncertainty in language is used. There is sufficient evidence that the claim is supported by the data 6 © UCLES 2021 Page 7 of 13

3(a) B1 3 dp or better. x 0 1 2 3 4 5 6 or more O 35 48 43 15 6 3 0 E 40.874 57.704 35.641 12.579 2.775 0.392 0.0352 p = 57.704 q = 0.035 or 0.036 B1 Accept numbers which round to 0.035 or 0.036. 2 © UCLES 2021 Page 8 of 14 3(b) Combine frequencies less than 5: last 4 columns give 24 / 15.7812 M1 Allow last 2 or 3 columns combined for this M1. ( O−E )2 M1  =0.8444+1.6319+1.5199+4.2803 E 8.28 A1 Accept 8.27 – 8.28. SC: 25.60, if values not combined, scores M1A0. SC: 14.96 if last 3 combined, scores M1A0. SC: if last 2 combined, scores M1A0. H : B(8, 0.15) fits the data B1 Must mention distribution and data. 0 3 degrees of freedom, tabular value = 7.815. M1 Compare their value with 7.815 and correct FT conclusion 8.28 > 7.815 3 columns combined compared with 9.488 and correct Reject H . conclusion 0 Insufficient evidence to support manager’s claim. A1 Correct conclusion, in context, following correct work. Level of uncertainty in language is used. 6 © UCLES 2021 Page 9 of 14

4(a) p 16.03 B1 q 1.70 B1 Condone 1.7 2 © UCLES 2022 Page 9 of 13 4(b) M1 Add last 2 columns (or 3 columns: 13.06). 0 1 2 3 4 5 6 7 12 20 36 32 13 6 1 0 9.85 24.63 30.78 25.65 16.03 8.02 3.34 1.70 Combine last two columns: O value = 1, E value = 5.04 Calculate values of OE2 : M1 Calculate values. E 0.4693 0.8704 0.8853 1.5720 0.5727 0.5088 3.2384 Test statistic = 8.12 A1 H : data fits a Poisson distribution with mean 2.5 B1 Need data and distribution. 0 H : data does not fit a Poisson distribution with mean 2.5 e.g Data fits the given distribution. 1 Number of butterflies per square metre fits Po(2.5). Critical value of chi-squared = 10.64 M1 Compare their test statistic with 10.64. Compare: ‘8.12’ < 10.64, accept H Note: allow 9.236 if 3 columns combined 0 or 12.02 if none combined. There is sufficient evidence to support the scientist’s claim / there A1 Correct conclusion, in context. is sufficient evidence to suggest that the data fits a Poisson Level of uncertainty in language is used. distribution with mean 2.5 6 © UCLES 2022 Page 10 of 13

4(a) p 16.03 B1 q 1.70 B1 Condone 1.7 2 © UCLES 2022 Page 9 of 13 4(b) M1 Add last 2 columns (or 3 columns: 13.06). 0 1 2 3 4 5 6 7 12 20 36 32 13 6 1 0 9.85 24.63 30.78 25.65 16.03 8.02 3.34 1.70 Combine last two columns: O value = 1, E value = 5.04 Calculate values of OE2 : M1 Calculate values. E 0.4693 0.8704 0.8853 1.5720 0.5727 0.5088 3.2384 Test statistic = 8.12 A1 H : data fits a Poisson distribution with mean 2.5 B1 Need data and distribution. 0 H : data does not fit a Poisson distribution with mean 2.5 e.g Data fits the given distribution. 1 Number of butterflies per square metre fits Po(2.5). Critical value of chi-squared = 10.64 M1 Compare their test statistic with 10.64. Compare: ‘8.12’ < 10.64, accept H Note: allow 9.236 if 3 columns combined 0 or 12.02 if none combined. There is sufficient evidence to support the scientist’s claim / there A1 Correct conclusion, in context. is sufficient evidence to suggest that the data fits a Poisson Level of uncertainty in language is used. distribution with mean 2.5 6 © UCLES 2022 Page 10 of 13

2 H : the size of shell is independent of the location B1 0 H : the size of shell is not independent of the location 1 M1 A1 Attempt at calculation of E values. O 68 69 96 81 E 60.29 77.87 100.48 75.36 O 28 55 64 39 E 35.71 46.13 59.52 44.64 OE2 M1 Attempt at Chi-squared contributions. Calculate values of : Accuracy of 0.1. E 0.9859 1.0104 0.1997 0.4221 1.6646 1.7055 0.3372 0.7126 Test statistic = 7.04 A1 Correct total to 3sf. Tabular value for 3df = 6.251 M1 Compare with 6.251 and correct FT conclusion. 7.04 > 6.251 May be implied by FT correct conclusion in words. Reject H o Insufficient evidence to suggest size of shell is independent of A1 Correct conclusion, in context. location/Evidence suggests that size of shell is dependent on Level of uncertainty in language is used. (beach) location 7 © UCLES 2022 Page 7 of 13

2(a) Using B(8, 0.6) B1 One correct. p=0.098, q=41.804, r =31.353 B1 Other two correct. If B0B0 scored, SCB1 if all 3 correct but not rounded to 3dp. 2 2(b) H : the data fits the binomial distribution B(8, 0.6) B1 0 H : the data does not fit the binomial distribution B(8, 0.6) 1 Add first 3 columns: O value 8, E value 7.471 M1 AND last two columns: O value 11, E value 15.956 Both. Chi-squared sum: M1 0.0374 + 1.5817 + 2.9655 + 0.8058 + 0.9139 + 1.5394 Method must be seen. 7.84 A1 If M0 awarded SCB1 for 7.84. Critical value, 5df = 9.236 M1 Compare calculated value with 9.236 OR 7.84 < 9.236 13.36 if no columns added. Accept H 12.02 adding only last 2 columns. 0 10.64 adding only first 3 columns. There is insufficient evidence to reject the director’s claim. A1 Correct work only except possibly B1 for hypotheses, in context, level of uncertainty in language. ‘prove’ scores A0 6 © UCLES 2022 Page 7 of 12

2 H : performance is independent of region B1 0 H : performance is not independent of region 1 M1 A1 Calculate expected frequencies, allow 1 error. 62 (53.9) 41 (44.1) 44 (49) 102 (106.7) 94 (87.3) 95 (97) 56 (59.4) 45 (48.6) 61 (54) Test stat = 1.217 + 0.218 + 0.510 + 0.207 + 0.514 + 0.041 + 0.195 M1 Calculate test statistic. + 0.267 + 0.907 4.08 A1 SCB2 for 4.08 if totally unsupported. SCM1A1 B1 for 4.08 supported only by expected frequencies. Tabular value, 4df = 7.779 M1 Compare with 7.779 and conclusion. 1.08 < 7.779 Accept H 0 Insufficient evidence that performance is not independent of A1 Correct conclusion in context, following correct work, level of region uncertainty in language. ‘Prove’ gives A0. 7 © UCLES 2022 Page 7 of 13

2(a) Using B(8, 0.6) B1 One correct. p=0.098, q=41.804, r =31.353 B1 Other two correct. If B0B0 scored, SCB1 if all 3 correct but not rounded to 3dp. 2 2(b) H : the data fits the binomial distribution B(8, 0.6) B1 0 H : the data does not fit the binomial distribution B(8, 0.6) 1 Add first 3 columns: O value 8, E value 7.471 M1 AND last two columns: O value 11, E value 15.956 Both. Chi-squared sum: M1 0.0374 + 1.5817 + 2.9655 + 0.8058 + 0.9139 + 1.5394 Method must be seen. 7.84 A1 If M0 awarded SCB1 for 7.84. Critical value, 5df = 9.236 M1 Compare calculated value with 9.236 OR 7.84 < 9.236 13.36 if no columns added. Accept H 12.02 adding only last 2 columns. 0 10.64 adding only first 3 columns. There is insufficient evidence to reject the director’s claim. A1 Correct work only except possibly B1 for hypotheses, in context, level of uncertainty in language. ‘prove’ scores A0 6 © UCLES 2022 Page 7 of 12

3(a) 2 1  4  1  4 x3 M1 Integration with correct powers and correct limits   x2  dx    seen. 1.5 24x2  24  x 3  a500.092014 = 4.6007 A1 AG 53 1325 or or 0.092014 or 4.60069 seen 576 288 b11.4211 B1 Or 11.421(0) 3 3(b) Combine first two columns: 6, 9.0278 M1 Must be seen, or implied by 1.0155. Calculate value of chi-squared: 1.0155 + 0.5715 + 0.7661 + 0.2183 + 0.5885 M1 At least 2 correct values (at least 3sf) or expressions seen. Allow columns not combined or three columns combined. 3.16 A1 CWO. Dependent on M1M1 scored. SCB1 for 3.16 with no working H : f is a good fit for the data B1 0 H : f is not a good fit for the data 1 Tabular value of chi-squared: 7.78 M1 Correct tabular value: allow correct FT value if ‘3.16’ < 7.78 and accept H columns not combined (9.236) or three columns 0 combined (6.251). Insufficient evidence to suggest that f is not a good fit for the data A1 Correct conclusion in context, following correct Condone: sufficient evidence to suggest that f is a good fit for the data. work, level of uncertainty in language (not ‘prove’, not ‘there is no evidence’), no contradictions. A0 if hypotheses reversed 6 © UCLES 2023 Page 8 of 12

3(a) 2 1  4  1  4 x3 M1 Integration with correct powers and correct limits   x2  dx    seen. 1.5 24x2  24  x 3  a500.092014 = 4.6007 A1 AG 53 1325 or or 0.092014 or 4.60069 seen 576 288 b11.4211 B1 Or 11.421(0) 3 3(b) Combine first two columns: 6, 9.0278 M1 Must be seen, or implied by 1.0155. Calculate value of chi-squared: 1.0155 + 0.5715 + 0.7661 + 0.2183 + 0.5885 M1 At least 2 correct values (at least 3sf) or expressions seen. Allow columns not combined or three columns combined. 3.16 A1 CWO. Dependent on M1M1 scored. SCB1 for 3.16 with no working H : f is a good fit for the data B1 0 H : f is not a good fit for the data 1 Tabular value of chi-squared: 7.78 M1 Correct tabular value: allow correct FT value if ‘3.16’ < 7.78 and accept H columns not combined (9.236) or three columns 0 combined (6.251). Insufficient evidence to suggest that f is not a good fit for the data A1 Correct conclusion in context, following correct Condone: sufficient evidence to suggest that f is a good fit for the data. work, level of uncertainty in language (not ‘prove’, not ‘there is no evidence’), no contradictions. A0 if hypotheses reversed 6 © UCLES 2023 Page 8 of 12

6(a) H : grade is independent of age B1 Condone ‘Ability to remember’ instead of ‘grade’. 0 H : grade is not independent of age 1 Calculate expected values, shown in table in bold M1 At least 2 correct values or expressions. 25 21.2 16 24.4 19 14.4 60 A1 6 correct values or expressions. 28 31.8 45 36.6 17 21.6 90 53 61 36 150 Calculate chi-squared values: M1 At least 2 correct values (at least 3sf) or expressions seen. 0.6811 + 2.8918 + 1.4694 + 0.4541 + 1.9279 + 0.9796 8.40[4] A1 Tabular value, 2 degrees of freedom = 7.378 M1 Compare their value with 7.378 and conclusion without context. ‘8.404’ > 7.378 and reject H /significant Condone ‘accept H ’. 0 1 Sufficient evidence to suggest that grade is not independent of A1 Correct conclusion in context, following correct work, level of age. uncertainty in language (not ‘prove’, not ‘there is no evidence’), no contradictions. A0 if hypotheses wrong way round. 7 6(b) Value, 4 degrees of freedom = 11.14 M1 Condone ‘reject H ’. 1 10.91 < 11.14 Accept H /not significant 0 Insufficient evidence to suggest that grade is not independent of A1 CAO age. 2 © UCLES 2023 Page 12 of 13 6(c) For example, result in part (b) because the table contains more B1 Any appropriate comment to support part (a) or part (b). information Allow ‘more specific’. Not ‘more data’ or ‘more accurate’ on its e.g. More degrees of freedom, more groups, more detail own. 1 © UCLES 2023 Page 13 of 13

2 H : opinion is independent of street H : opinion is not independent of street B1 0 1 M1 At least 2 correct values or expressions. 33 32.24 54 57.66 42 43.4 26 21.7 155 A1 6 correct values or expressions. 19 19.76 39 35.34 28 26.6 9 13.3 95 52 93 70 35 250 Calculate chi-squared values: M1 Correct to 3 decimal places. 0.0179 + 0.2323 + 0.0452 + 0.8521 + 0.0292 + 0.3790 + 0.0737 + 1.3902 At least 2 correct values or expressions. 3.02 A1 SC B1 3.02 following M1M0 SC B2 3.02 following M0M0 Tabular value, 3 degrees of freedom 7.815: ‘3.02’ < 7.815, accept H M1 Allow ‘not significant’. 0 Insufficient evidence to suggest that opinion depends on street. A1 CWO. Correct conclusion in context, following correct work, level of uncertainty in language. A0 if hypotheses wrong way round or missing. 7 © UCLES 2023 Page 7 of 13

2(a) B1 Each. 0 5 13 17 21 16 9 5 4 B1 2.718 9.512 16.646 19.421 16.993 11.895 6.939 3.469 2.407 2 2(b) H : Po(3.5) is a good fit to the data B1 0 H : Po(3.5) is not a good fit to the data 1 Combine first 2 columns: 5, 12.23 M1 Both. And last 2 columns: 9, 5.876 Chi-squared values: M1 Allow if no or incorrect number of columns 4.274 + 0.799 + 0.302 + 0.945 + 1.417 + 0.612 + 1.661 added. At least two ‘correct’ values (3 sf) or expressions seen from their grouping (or lack of). 2.718 + 2.140 + … + 0.6757 + 1.054 10.0 A1 AWRT 10.0 If M0 awarded then SC B1 for 10.0. Tabular value: 10.64 ‘10.0’ < 10.64, accept H /not significant M1 Allow equivalent to 10.64 if columns not 0 combined or only one pair combined (12.02 one pair combined, 13.36 none combined). Insufficient evidence to suggest that Po(3.5) is not a good fit to the data A1 Correct conclusion in context, following correct work, level of uncertainty in language. A0 if hypotheses the wrong way round or missing. 6 © UCLES 2023 Page 7 of 13

2 H : opinion is independent of street H : opinion is not independent of street B1 0 1 M1 At least 2 correct values or expressions. 33 32.24 54 57.66 42 43.4 26 21.7 155 A1 6 correct values or expressions. 19 19.76 39 35.34 28 26.6 9 13.3 95 52 93 70 35 250 Calculate chi-squared values: M1 Correct to 3 decimal places. 0.0179 + 0.2323 + 0.0452 + 0.8521 + 0.0292 + 0.3790 + 0.0737 + 1.3902 At least 2 correct values or expressions. 3.02 A1 SC B1 3.02 following M1M0 SC B2 3.02 following M0M0 Tabular value, 3 degrees of freedom 7.815: ‘3.02’ < 7.815, accept H M1 Allow ‘not significant’. 0 Insufficient evidence to suggest that opinion depends on street. A1 CWO. Correct conclusion in context, following correct work, level of uncertainty in language. A0 if hypotheses wrong way round or missing. 7 © UCLES 2023 Page 7 of 13

5(a) M1 A1 Calculate E values to at least 2dp, allow one error. Company Poor Satisfactory Good These must be seen. For A1, working to at least 2dp. P (18) (43) (64) 25 40.625 59.375 Q (22) (22) (31) 15 24.375 35.625 Chi-squared contributions: M1 Company Poor Satisfactory Good At least 2 correct values or expressions seen. Or may be implied by AWRT 6.56 P 1.96 0.1388 0.3603 Q 3.2667 0.2314 0.6004 Test statistic = 6.56 A1 AWRT 6.56. H : Quality (of items) is independent of company B1 Condone ‘no association/relationship between….’ 0 H : Quality (of items) is not independent of company ‘association/relationship between….’. 1 6.56 > 5.991 Reject H M1 ‘6.56’ from their attempt at chi-squared, 5.991 0 must be correct. Consistent signs in comparison. FT conclusion from their 6.56. Condone ‘Accept H ’. 1 ‘Reject H ’ can be implied by a conclusion that is 0 consistent with their 6.56. © Cambridge University Press & Assessment 2024 Page 11 of 15 5(a) There is sufficient evidence to suggest that the quality of brushes is not A1 Correct work only, conclusion in context with independent of company. level of uncertainty in language. Not ‘prove’. Accept ‘enough evidence’. Condone ‘some evidence’. Do not accept ‘there is insufficient…’. 7 5(b) P produces better quality brushes (than Q) B1 Condone P is better (than Q). or Q produces worse quality brushes (than P) P has fewer poor quality brushes than expected AND Q has more poor quality brushes than expected 1 © Cambridge University Press & Assessment 2024 Page 12 of 15

5(a) M1 A1 Calculate E values to at least 2dp, allow one error. Company Poor Satisfactory Good These must be seen. For A1, working to at least 2dp. P (18) (43) (64) 25 40.625 59.375 Q (22) (22) (31) 15 24.375 35.625 Chi-squared contributions: M1 Company Poor Satisfactory Good At least 2 correct values or expressions seen. Or may be implied by AWRT 6.56 P 1.96 0.1388 0.3603 Q 3.2667 0.2314 0.6004 Test statistic = 6.56 A1 AWRT 6.56. H : Quality (of items) is independent of company B1 Condone ‘no association/relationship between….’ 0 H : Quality (of items) is not independent of company ‘association/relationship between….’. 1 6.56 > 5.991 Reject H M1 ‘6.56’ from their attempt at chi-squared, 5.991 0 must be correct. Consistent signs in comparison. FT conclusion from their 6.56. Condone ‘Accept H ’. 1 ‘Reject H ’ can be implied by a conclusion that is 0 consistent with their 6.56. © Cambridge University Press & Assessment 2024 Page 11 of 15 5(a) There is sufficient evidence to suggest that the quality of brushes is not A1 Correct work only, conclusion in context with independent of company. level of uncertainty in language. Not ‘prove’. Accept ‘enough evidence’. Condone ‘some evidence’. Do not accept ‘there is insufficient…’. 7 5(b) P produces better quality brushes (than Q) B1 Condone P is better (than Q). or Q produces worse quality brushes (than P) P has fewer poor quality brushes than expected AND Q has more poor quality brushes than expected 1 © Cambridge University Press & Assessment 2024 Page 12 of 15

3 M1 A1 Calculate expected values, at least 4 correct to 3sf. Bus Bus Bus These must be seen. company company company A1 for all correct to at least 3sf. A B C Arrived 22 (17.28) 22 (21.60) 10 (15.12) early Arrived 30 (39.68) 52 (49.60) 42 (34.72) on time Arrived 28 (23.04) 26 (28.80) 18 (20.16) late M1 At least 4 correct values (to 2sf) or expressions. 1.2893 0.00741 1.7338 May be implied by 8.600 – 8.615. 2.3615 0.1161 1.5265 1.0678 0.2722 0.2314 Test statistic = 8.61 A1 8.600 – 8.615 Unsupported 8.600 – 8.615 scores M0A0 M1A1. H : Reliability of buses is independent of bus B1 Condone ‘No association/relationship between…’ 0 company ‘Association/relationship between…’ H : Reliability of buses is not independent 1 of/dependent on bus company ‘8.61’ < 9.488 Accept H M1 ‘8.61’ from an attempt at chi-squared, 9.488 must be correct. 0 ft conclusion from their 8.61. Condone ‘Reject H ’. 1 ‘Accept H ’ can be implied by a conclusion in context that is consistent with their 0 8.61. © Cambridge University Press & Assessment 2024 Page 9 of 15 3 There is insufficient evidence to support/ suggest A1 Correct work only, ignoring their hypotheses, conclusion in context with level of that reliability of buses is dependent/not uncertainty in language. Not ‘prove’ or equivalent. independent of bus company Condone ‘no sufficient’ ‘not enough’. Do not accept statements such as ‘there is sufficient evidence to suggest…..’ 7 Question Answer Marks Guidance

3(a) 17.098 8.122 3.086 B1 One correct. B1 All correct. 2 3(b) M1 0 1 2 3 4 or more Last two or three columns combined. 10 18 35 21 16 14.957 28.418 26.997 17.098 12.53 Contributions to test statistic are: M1 1.6428 3.8192 2.3724 0.8905 0.9609(7) May be implied by awrt 9.69 Test statistic is 9.69 A1 9.686 H : Po(1.9) is a good fit for the data B1 0 H : Po(1.9) is not a good fit for the data 1 Critical value is 7.779, compare ‘9.69’ > 7.779 reject H M1 4 degrees of freedom 0 Sufficient evidence to suggest that Po(1.9) is a not a good fit for the data/ A1 Correct work only, including hypotheses, in Sufficient evidence to reject/not support the statistician’s claim context, level of uncertainty in language. 6 © Cambridge University Press & Assessment 2024 Page 7 of 11

3(a) 40+86+105+64+20 315 2.1 B1 Must see either 315 or 2.1. x = = =2.1, p= =0.42 AG 150 150 5 1 © Cambridge University Press & Assessment 2024 Page 7 of 13 3(b) B1 Calculate expected frequencies (must be seen) Number of 0 1 2 3 4 5 at least 2 correct to at least 2 decimal places. seeds that germinate B1 At least 4 correct to at least 2 decimal places. Number of 12 40 43 35 16 4 pots Expected 9.846 35.6475 51.627 37.3845 13.536 1.9605 frequency Combine last two columns M1 20, 15.50, may be implied by answer 3.90 – 3.91. Chi-squared contributions: M1 At least 2 correct, may be implied by answer 3.90 0.4712 0.5314 1.4416 0.1521 1.3088 – 3.91. Test statistic = 3.905 A1 accept 3.90 – 3.910 H : Binomial B(5, 0.42) fits the data B1 Allow ‘Binomial’ for ‘B(5, 0.42)’ 0 H : Binomial B(5, 0.42) does not fit the data Allow ‘Number of seeds that germinate can be 1 modelled by B(5, 0.42)’ Critical value is 6.251 B1 Must come from combined columns. Allow 7.779. ‘3.905’ < ‘6.251’ Accept H M1 Reject H , not significant. 0 1 Insufficient evidence to suggest that B(5, 0.42) is a not a good fit (to the data) A1 Correct work only, including hypotheses, level of uncertainty in language. 9 © Cambridge University Press & Assessment 2024 Page 8 of 13

3(a) 17.098 8.122 3.086 B1 One correct. B1 All correct. 2 3(b) M1 0 1 2 3 4 or more Last two or three columns combined. 10 18 35 21 16 14.957 28.418 26.997 17.098 12.53 Contributions to test statistic are: M1 1.6428 3.8192 2.3724 0.8905 0.9609(7) May be implied by awrt 9.69 Test statistic is 9.69 A1 9.686 H : Po(1.9) is a good fit for the data B1 0 H : Po(1.9) is not a good fit for the data 1 Critical value is 7.779, compare ‘9.69’ > 7.779 reject H M1 4 degrees of freedom 0 Sufficient evidence to suggest that Po(1.9) is a not a good fit for the data/ A1 Correct work only, including hypotheses, in Sufficient evidence to reject/not support the statistician’s claim context, level of uncertainty in language. 6 © Cambridge University Press & Assessment 2024 Page 7 of 11

3(a) 01844+1143+211+30+41+50+61  7 B1 175 , AWRT 0.01458. = 12000    20006  480 1 3(b) a=2000−(1831.3+6.016+0.119+0.001)=162.6 B1 AG, shown convincingly. Alternative method for question 3(b) 6 B1 AG, shown convincingly. a=  7  ( 1− 7 )5 2000 = 162.6 1 480 480 1 © Cambridge University Press & Assessment 2025 Page 11 of 16 3(c) H : A binomial distribution is a satisfactory model for the data. B1 Must mention binomial and data. 0 H : A binomial distribution is not a satisfactory model for the data. 1 (1844−1831.3)2 (143−162.6)2 (13−6.136)2 M1 Expression of correct form with cells correctly + + =0.08807+2.3626+7.6784 combined. 1831.3 162.6 6.136 May use 6.1 (found by subtraction). OR Implied by 2 correct contributions. 18442 1432 132 + + −2000 1831.3 162.6 6.136 10.1 A1 AWRT 10.1. Allow AWRT 10.3 from using 6.1. SC unsupported 10.1 (or 10.3) scores M0 B1. [1 degree of freedom so critical value is 7.879.] M1 Compare their 10.1 (or 10.3) with 7.879 and '10.1'7.879, reject H (significant). appropriate conclusion (may be in terms of H ). 0 1 Reject H can be implied by an attempt at an 0 appropriate conclusion in context. Allow M1 for comparison of their 10.1 (or 10.3) with 10.6 and appropriate conclusion. Sufficient evidence to suggest that the manager’s model is not satisfactory. A1 Correct conclusion in context from correct working ignoring hypotheses. Level of uncertainty in language used (for example, not ‘prove’). 5 3(d) Not independent or probability of “success” may not be constant or plausible reason B1 Credit any valid comment in context. in context (for example, higher number of boxes with more than 1 broken egg due to boxes being dropped). 1 © Cambridge University Press & Assessment 2025 Page 12 of 16

3(a) 01844+1143+211+30+41+50+61  7 B1 175 , AWRT 0.01458. = 12000    20006  480 1 3(b) a=2000−(1831.3+6.016+0.119+0.001)=162.6 B1 AG, shown convincingly. Alternative method for question 3(b) 6 B1 AG, shown convincingly. a=  7  ( 1− 7 )5 2000 = 162.6 1 480 480 1 © Cambridge University Press & Assessment 2025 Page 11 of 15 3(c) H : A binomial distribution is a satisfactory model for the data. B1 Must mention binomial and data. 0 H : A binomial distribution is not a satisfactory model for the data. 1 (1844−1831.3)2 (143−162.6)2 (13−6.136)2 M1 Expression of correct form with cells correctly + + =0.08807+2.3626+7.6784 combined. 1831.3 162.6 6.136 May use 6.1 (found by subtraction). OR Implied by 2 correct contributions. 18442 1432 132 + + −2000 1831.3 162.6 6.136 10.1 A1 AWRT 10.1. Allow AWRT 10.3 from using 6.1. SC unsupported 10.1 (or 10.3) scores M0 B1. [1 degree of freedom so critical value is 7.879.] M1 Compare their 10.1 (or 10.3) with 7.879 and '10.1'7.879, reject H (significant). appropriate conclusion (may be in terms of H ). 0 1 Reject H can be implied by an attempt at an 0 appropriate conclusion in context. Allow M1 for comparison of their 10.1 (or 10.3) with 10.6 and appropriate conclusion. Sufficient evidence to suggest that the manager’s model is not satisfactory. A1 Correct conclusion in context from correct working ignoring hypotheses. Level of uncertainty in language used (for example, not ‘prove’). 5 3(d) Not independent or probability of “success” may not be constant or plausible reason B1 Credit any valid comment in context. in context (for example, higher number of boxes with more than 1 broken egg due to boxes being dropped). 1 © Cambridge University Press & Assessment 2025 Page 12 of 15

1 H : uniform distribution fits the data. B1 Must mention data and uniform distribution (or 0 H : uniform distribution does not fit the data. uniform distribution clearly described). 1 Expected frequencies are 40, 40, 40. B1 (38−40)2 (52−40)2 (30−40)2 M1 Expression of correct form. + + =0.1+3.6+2.5 May see individual terms not summed. 40 40 40 OR 382 522 302 + + −120 40 40 40 6.2 A1 SC unsupported 6.2 scores M0 B1. Note: 6.208 scores M0 A0 XP. [2 degrees of freedom, so critical value is 5.991.] M1 Compare their 6.2 with 5.991 and appropriate '6.2'5.991, reject H (significant). conclusion (may be in terms of H ). 0 1 Reject H can be implied by an attempt at an 0 appropriate conclusion in context. Sufficient evidence to suggest that the scientist’s claim is incorrect. A1 Correct conclusion in context from correct OR Sufficient evidence eye colour is not uniformly distributed. working ignoring hypotheses. OR Sufficient evidence to suggest that uniform distribution is not a good fit. Level of uncertainty in language is used (for example, not ‘prove’). 6 © Cambridge University Press & Assessment 2025 Page 9 of 16

3 H : type of item sold is independent of salesperson. B1 Allow dependent, association or relationship for not independent. 0 For example: H : No association/relationship between… H : type of item sold is not independent of salesperson. 0 1 H : Association/relationship between… 1 Laptop Camera Television M1 Calculate expected values (must be seen). M1 for at least 4 expected values correct to 3sf or better. Avril 31 (28.5) 40 (41.80) 24 (24.70) Ben 23 (29.1) 45 (42.68) 29 (25.22) A1 All expected values correct to 3sf or better. Charlie 21 (17.4) 25 (25.52) 12 (15.08) (31−28.5)2 (40−41.8)2 (15.08−12)2 M1 At least 2 terms (not including 250) of correct form seen. + ++ May see individual terms not summed. 28.5 41.8 15.08 May be implied by AWRT 3.67 or 3.68. OR 312 402 122 + ++ −250 28.5 41.8 15.08 Test statistic = 3.67 (or 3.68) A1 AWRT 3.67 or 3.68. '3.67' < 7.779, accept H (not significant). M1 Compare their 3.67 (or 3.68) with 7.779 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 3.67 (or 3.68) with 9.488 and appropriate conclusion. Insufficient evidence to suggest that type of item sold and A1 Correct conclusion in context from correct working ignoring salesperson are not independent. hypotheses. Condone ‘no sufficient evidence’. OR Level of uncertainty in language used (for example, not ‘prove’). Insufficient evidence to suggest that there is dependence between Must refer to both item and salesperson. type of item sold and salesperson. Do not accept statements such as ‘there is sufficient evidence to suggest…’. 7 © Cambridge University Press & Assessment 2025 Page 11 of 15

2(a) e−2.8(2.8)3 B1 Or by subtraction. p= 100=22.248 Allow 22.2 or 22.3. 3! (29−22.25)2 B1 AWRT 2.05. q= =2.05 22.25 2 2(b) H : Po(2.8) fits the data. B1 Must mention distribution and data/number of cars. 0 H : Po(2.8) does not fit the data. 1 OR H : Po(2.8) is a satisfactory model for the number of cars (entering 0 the car park). H : Po(2.8) is not a satisfactory model for the number of cars 1 (entering the car park). 2.739+0.241+2.152+'2.049'+0.425+3.753+0.037 =11.396 M1 Sum of chi-square contributions using their q. May be implied by 11.4. '11.4'12.59, accept H / do not reject H / not significant. M1 Compare their 11.4 with 12.59 and appropriate result (may be in 0 0 terms of H ). 1 Test result may be implied by an attempt at an appropriate conclusion in context ignoring hypotheses. Insufficient evidence to suggest that Po(2.8) is not a good fit to the A1 Correct conclusion from correct working ignoring hypotheses. data. In context. OR Level of uncertainty in language is used (for example, not ‘prove’). Insufficient evidence to suggest that the manager’s claim is false. Allow ‘Not enough evidence that / to show / conclude that…’ Do not accept statements such as ‘there is sufficient evidence to suggest…’ or ‘no evidence…..’ 4 © Cambridge University Press & Assessment 2025 Page 10 of 16

3(a) m=69.522 B1 n=0.156 or 0.157 B1 Subtraction gives 0.156. 2 3(b) H : Po(0.7) fits the data. B1 Must mention distribution and data/number of breakdowns. 0 May see H : Po(0.7) is a satisfactory model for the data. H : Po(0.7) does not fit the data. 0 1 H : Po(0.7) is not a satisfactory model for the data. 1 Combining last 3 columns give 13 and 6.828 or 6.829. M1 Combine expected frequencies less than 5. (88−99.317)2 (73−69.552)2 (13−6.829)2 M1 At least 2 expressions (not including 200) of correct form seen. + ++ May see individual terms not summed. 99.317 69.552 6.829 May be implied by 2 contributions correct to 3sf seen. [1.2896+0.1740+0.1142+5.5764] 6.828 gives 5.5790. OR 882 732 262 132 + + + −200 99.317 69.552 24.333 6.829 7.15 or 7.16 A1 SC B1 for 7.15 or 7.16 following M0. '7.15'7.815, accept H / do not reject H / not significant. M1 Compare their 7.15 with 7.815 (9.488 if 2 columns combined, 11.07 if 0 0 no columns combined) and appropriate result (may be in terms of H ). 1 Test result may be implied by an attempt at an appropriate conclusion in context ignoring hypotheses. Insufficient evidence to reject expert’s claim. A1 Correct conclusion from correct working ignoring hypotheses. OR In context. Insufficient evidence to suggest that Po(0.7) is not a good Level of uncertainty in language is used (for example, not ‘prove’) Allow ‘Not enough evidence that / to show / conclude that…’. fit/model [for the data]. Do not accept statements such as ‘there is sufficient evidence to suggest…’ or ‘no evidence…’. 6 © Cambridge University Press & Assessment 2025 Page 11 of 15

2(a) e−2.8(2.8)3 B1 Or by subtraction. p= 100=22.248 Allow 22.2 or 22.3. 3! (29−22.25)2 B1 AWRT 2.05. q= =2.05 22.25 2 2(b) H : Po(2.8) fits the data. B1 Must mention distribution and data/number of cars. 0 H : Po(2.8) does not fit the data. 1 OR H : Po(2.8) is a satisfactory model for the number of cars (entering 0 the car park). H : Po(2.8) is not a satisfactory model for the number of cars 1 (entering the car park). 2.739+0.241+2.152+'2.049'+0.425+3.753+0.037 =11.396 M1 Sum of chi-square contributions using their q. May be implied by 11.4. '11.4'12.59, accept H / do not reject H / not significant. M1 Compare their 11.4 with 12.59 and appropriate result (may be in 0 0 terms of H ). 1 Test result may be implied by an attempt at an appropriate conclusion in context. Insufficient evidence to suggest that Po(2.8) is not a good fit to the A1 Correct conclusion from correct working ignoring hypotheses. data. In context. OR Level of uncertainty in language is used (for example, not ‘prove’). Insufficient evidence to suggest that the manager’s claim is false. Allow ‘Not enough evidence that / to show / conclude that…’ Do not accept statements such as ‘there is sufficient evidence to suggest…’ or ‘no evidence…..’ 4 © Cambridge University Press & Assessment 2025 Page 10 of 16

5 H : performance is independent of age. B1 Allow dependent, association or relationship for not independent. 0 H : performance is not independent of age. For example: 1 OR H : No association/relationship between… 0 H : passing test is independent of age. H : Association/relationship between… 0 1 H : passing test is not independent of age. Must refer to “performance” and “age” in H . 1 0 34 (27.93) 41 (44.69) 6 (8.38) M1 At least 3 correct to 3sf (must be seen). 16 (22.07) 39 (35.31) 9 (6.62) A1 All correct to 3sf. (34−27.93)2 (41−44.69)2 (9−6.62)2 M1 At least 2 expressions (not including 145) of correct form seen. + ++ May see individual terms not summed. 27.93 44.69 6.62 May be implied by 2 contributions correct to 3sf seen. 1.319+0.305+0.676+1.669+0.386+0.856 May be implied by AWRT 5.21. OR  342 412 92   + ++ −145 27.93 44.69 6.62   5.21 A1 AWRT 5.21. '5.21'4.605, reject H / significant. M1 Compare their 5.21 with 4.605 and appropriate result (may be in 0 terms of H ). 1 Test result may be implied by an attempt at an appropriate conclusion in context ignoring hypotheses. Sufficient evidence to suggest that performance and age are not A1 Correct conclusion from correct working ignoring hypotheses. independent. Must refer to both performance and age. OR Sufficient evidence to suggest that performance and age are In context. dependent. Level of uncertainty in language is used (for example, not ‘prove’). OR Sufficient evidence to suggest that performance and age are Do not accept statements such as ‘there is insufficient evidence to associated. suggest…’. OR Sufficient evidence to support the (driving) instructor’s belief. 7 © Cambridge University Press & Assessment 2025 Page 16 of 17

2(a) H : reliability of service is independent of bus company. B1 Allow dependent, association or relationship for 0 H : reliability of service is not independent of bus company. not independent, for example, 1 “H : No association/relationship between…, 0 H : Association/relationship between…”. 1 Must refer to ‘reliability’ and ‘company’ in H . 0 M1 Calculate expected frequencies (must be seen). 7 (12.25) 22 (19.11) 20 (17.64) At least two correct to at least 2 significant figures. 14 (9.25) 11 (14.43) 12 (13.32) A1 All expected frequencies correct to at least 3 4 (3.50) 6 (5.46) 4 (5.04) significant figures. M1 Combine rows (for example rows 2 and 3). 7 (12.25) 22 (19.11) 20 (17.64) OR Combine columns (for example columns A and B) 18 (12.75) 17 (19.89) 16 (18.36) OR 29 (31.36) 20 (17.64) 25 (23.68) 12 (13.32) 10 (8.96) 4 (5.04) © Cambridge University Press & Assessment 2026 Page 9 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/ 2(a) Combining rows 2 and 3: (7−12.25)2 (22−19.11)2 (16−18.36)2 + ++ 12.25 19.11 18.36 © Cambridge University Press & Assessment 2026 Page 10 of 17  = 2 .2 5 + 0 .4 3 7 0 5 + 0 .3 1 5 7 4 + 2 .1 6 1 7 6 + 0 .4 1 9 9 1 + 0 .3 0 3 3 6  OR 1 7 2 2 .2 5 + 1 2 9 2 2 .1 1 +  + 1 1 8 2 6 .3 6 − 1 0 0 Combining columns A and B: (29−31.36)2 (20−17.64)2 (4−5.04)2 + ++ 31.36 17.64 5.04  = 0 .1 7 7 6 0 + 0 .3 1 5 7 4 + 0 .0 7 3 5 8 + 0 .1 3 0 8 1 + 0 .1 2 0 7 1 + 0 .2 1 4 6 0  OR 3 2 1 2 9 .3 6 + 1 2 7 2 0 .6 4 +  + 5 2 4 .0 4 − 1 0 0 No rows / columns combined: ( 7 − 1 1 2 2 .2 .2 5 5 ) 2 + ( 2 2 − 1 1 9 9 .1 .1 1 1 ) 2 +  + ( 4 − 5 5 .0 .0 4 4 ) 2 M1 At least 2 expressions (not including 100) of correct form seen. May see individual terms not summed. May be implied by 2 contributions correct to 3 significant figures seen. May be implied by AWRT 5.89 or AWRT 1.03 or AWRT 6.73. [= 2.250+0.43705+0.31574+2.4392+0.81531+0.13081+0.07143+0.05341+0.21460] Combining rows 2 and 3: Test statistic = 5.888 A1 AWRT 5.89 OR OR Combining columns A and B: Test statistic = 1.033 AWRT 1.03 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/ 2(a) Combining rows 2 and 3: M1 Compare their test statistic with 5.991 (or 9.488 if ‘5.888’ < 5.991, accept H / do not reject H / not significant. no rows/columns combined) and appropriate 0 0 conclusion (may be in terms of H ). OR 1 Test result can be implied by an attempt at an Combining columns A and B: appropriate conclusion in context ignoring ‘1.033’ < 5.991, accept H / do not reject H / not significant. 0 0 hypotheses. OR No cells combined: ‘6.73’ < 9.488, accept H / do not reject H / not significant. 0 0 Insufficient evidence to suggest reliability (of service) is not independent of (bus) A1 Correct conclusion from correct working ignoring company. hypotheses. OR In context. Must refer to both reliability and Insufficient evidence to suggest reliability (of service) is associated with (bus) company. company. OR Level of uncertainty in language is used (for Insufficient evidence to suggest that reliability (of service) is dependent on (bus) example not ‘prove’). company. Allow ‘Not enough evidence / no sufficient evidence that / to show / to conclude that …’ Do not accept ‘No evidence…’. Do not accept statements such as ‘there is sufficient evidence to suggest…’. 8 No cells combined can earn a maximum of 5/8 (B1 M1 A1 M0 M1 A0 M1 A0). 2(b) Company A (appears less reliable). B1 The observed frequencies are higher than the (combined) expected frequencies for DB1 Comparison of observed and expected company A for late and cancelled, but not for company B and company C. frequencies. 2 © Cambridge University Press & Assessment 2026 Page 11 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/

2(a) H : reliability of service is independent of bus company. B1 Allow dependent, association or relationship for 0 H : reliability of service is not independent of bus company. not independent, for example, 1 “H : No association/relationship between…, 0 H : Association/relationship between…”. 1 Must refer to ‘reliability’ and ‘company’ in H . 0 M1 Calculate expected frequencies (must be seen). 7 (12.25) 22 (19.11) 20 (17.64) At least two correct to at least 2 significant figures. 14 (9.25) 11 (14.43) 12 (13.32) A1 All expected frequencies correct to at least 3 4 (3.50) 6 (5.46) 4 (5.04) significant figures. M1 Combine rows (for example rows 2 and 3). 7 (12.25) 22 (19.11) 20 (17.64) OR Combine columns (for example columns A and B) 18 (12.75) 17 (19.89) 16 (18.36) OR 29 (31.36) 20 (17.64) 25 (23.68) 12 (13.32) 10 (8.96) 4 (5.04) © Cambridge University Press & Assessment 2026 Page 9 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/ 2(a) Combining rows 2 and 3: (7−12.25)2 (22−19.11)2 (16−18.36)2 + ++ 12.25 19.11 18.36 © Cambridge University Press & Assessment 2026 Page 10 of 17  = 2 .2 5 + 0 .4 3 7 0 5 + 0 .3 1 5 7 4 + 2 .1 6 1 7 6 + 0 .4 1 9 9 1 + 0 .3 0 3 3 6  OR 1 7 2 2 .2 5 + 1 2 9 2 2 .1 1 +  + 1 1 8 2 6 .3 6 − 1 0 0 Combining columns A and B: (29−31.36)2 (20−17.64)2 (4−5.04)2 + ++ 31.36 17.64 5.04  = 0 .1 7 7 6 0 + 0 .3 1 5 7 4 + 0 .0 7 3 5 8 + 0 .1 3 0 8 1 + 0 .1 2 0 7 1 + 0 .2 1 4 6 0  OR 3 2 1 2 9 .3 6 + 1 2 7 2 0 .6 4 +  + 5 2 4 .0 4 − 1 0 0 No rows / columns combined: (7−12.25)2 (22−19.11)2 (4−5.04)2 + ++ 12.25 19.11 5.04 [ = 2 .2 5 0 + 0 .4 3 7 0 5 + 0 .3 1 5 7 4 + 2 .4 3 9 2 + 0 .8 1 5 3 1 + 0 .1 3 0 8 1 + 0 .0 7 1 4 3 + 0 .0 5 3 4 1 + 0 .2 1 4 6 0 ] M1 At least 2 expressions (not including 100) of correct form seen. May see individual terms not summed. May be implied by 2 contributions correct to 3 significant figures seen. May be implied by AWRT 5.89 or AWRT 1.03 or AWRT 6.73. Combining rows 2 and 3: Test statistic = 5.888 A1 AWRT 5.89 OR OR Combining columns A and B: Test statistic = 1.033 AWRT 1.03 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) Combining rows 2 and 3: M1 Compare their test statistic with 5.991 (or 9.488 if ‘5.888’ < 5.991, accept H / do not reject H / not significant. no rows/columns combined) and appropriate 0 0 conclusion (may be in terms of H ). OR 1 Test result can be implied by an attempt at an Combining columns A and B: appropriate conclusion in context ignoring ‘1.033’ < 5.991, accept H / do not reject H / not significant. 0 0 hypotheses. OR No cells combined: ‘6.73’ < 9.488, accept H / do not reject H / not significant. 0 0 Insufficient evidence to suggest reliability (of service) is not independent of (bus) A1 Correct conclusion from correct working ignoring company. hypotheses. OR In context. Must refer to both reliability and Insufficient evidence to suggest reliability (of service) is associated with (bus) company. company. OR Level of uncertainty in language is used (for Insufficient evidence to suggest that reliability (of service) is dependent on (bus) example not ‘prove’). company. Allow ‘Not enough evidence / no sufficient evidence that / to show / to conclude that …’ Do not accept ‘No evidence…’. Do not accept statements such as ‘there is sufficient evidence to suggest…’. 8 No cells combined can earn a maximum of 5/8 (B1 M1 A1 M0 M1 A0 M1 A0). 2(b) Company A (appears less reliable). B1 The observed frequencies are higher than the (combined) expected frequencies for DB1 Comparison of observed and expected company A for late and cancelled, but not for company B and company C. frequencies. 2 © Cambridge University Press & Assessment 2026 Page 11 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/

4(a)  2 2   1  2 7  M1 Attempt at integral with correct limits.  dx = − =   x3    x2  36 1.5  1.5  a= 7 200=38.89 A1 AG, convincingly obtained with no errors seen. 36 5.89 or 5.90 B1 Summing to 200 gives 5.89, integration gives 5.90. 3 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(b) H : f is a good fit for the data. B1 Allow ‘satisfactory model for the data’. 0 H : f is not a good fit for the data. Must refer to f and data. 1 Combine 5th and 6th columns: (16, 9.72) B1 SOI OR 5.90 gives 9.73 instead of 9.72. Combine final two columns: (20, 16.33) 5th and 6th columns combined: © Cambridge University Press & Assessment 2026 Page 12 of 15 ( 9 7 − 1 1 1 1 1 1 .1 .1 1 1 ) 2 +  + ( 1 6 − 9 9 .7 .7 2 2 ) 2 + ( 1 4 − 1 1 2 2 .5 .5 0 0 ) 2 =1.792+0.389+1.389+2.786+4.057+0.18 Final 2 columns combined: (97−111.11)2 (10−5.89)2 (20−16.33)2 ++ + 111.11 5.89 16.33  = 1 .7 9 2 + 0 .3 8 9 + 1 .3 8 9 + 2 .7 8 6 + 2 .8 6 8 ( 2 .8 4 9 ) + 0 .8 2 5  M1 At least 2 terms correct. 972 162 142 OE, for example, ++ + −200. 111.11 9.72 12.50 Use of 9.73 instead of 9.72 from using 5.90 for b gives 1 .7 9 2 + 0 .3 8 9 + 1 .3 8 9 + 2 .7 8 6 + 4 .0 4 0 + 0 .1 8 . OE, for example, 1 9 1 7 1 2 .1 1 +  + 1 5 0 .8 2 9 + 1 2 6 2 0 .3 3 − 2 0 0 . Using 5.90 instead of 5.89 for b gives 1 .7 9 2 + 0 .3 8 9 + 1 .3 8 9 + 2 .7 8 6 + 2 .8 4 9 + 0 .8 2 5 . M1 possible if no columns combined, may be implied by answer in the range 10.61 to 10.63. 5th and 6th columns combined: 10.6 A1 SC Unsupported answer of 10.6 or 10.0 or 10.1 scores B1. Final two columns combined: 10.0 or 10.1 Must combine columns to earn this mark. '1 0 .6 '  9 .2 3 6 , reject H / significant. M1 Compare their 10.6 with 9.236 and appropriate conclusion (may 0 be in terms of H ). If no evidence of columns combined, compare 1 their 10.6 with 10.64. Test result can be implied by appropriate conclusion that is consistent with their 10.6. Sufficient evidence to suggest that f is not a good fit (for the data). A1 Correct conclusion from correct working, ignoring hypotheses. Level of uncertainty in language used (for example, not ‘prove’). Do not accept statements such as ‘there is insufficient evidence to suggest…’. 6 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/

3(a) X ~N ( 236.5,1 1.62 )    c = 1 2 0  P ( 2 5 0 ≤ X ≤ 2 6 0 ) =  1 2 .1 0 B1 12.09 or 12.10 or 12.11 Accept 12.1 for 12.10. d =120(1−Φ(2.026))=2.57 or [(120 – total) =] 2.56 B1 2.55, 2.56 or 2.57.   2 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/ 3(b) H : A normal distribution fits the data. B1 0 Condone H : A normal distribution does not fit the data. 1 OR H : A normal distribution is a satisfactory/suitable model for the masses (of this 0 species of frog). H : A normal distribution is not a satisfactory/suitable model for the masses (of this 1 species of frog). © Cambridge University Press & Assessment 2026 Page 12 of 18 N ( 2 3 6 .5 ,1 1 .6 2 ) included in hypotheses. Must mention distribution/model and data/masses. Combines expected frequencies in either first two cells or last two cells. M1 At least one cell must have an expected frequency less than 5 seen. May be implied by a corresponding contribution seen. ( 1 1 − 9 9 .2 .2 8 8 ) 2 + ( 2 0 − 2 2 5 5 .2 .2 2 2 ) 2 +  + ( 1 0 − '1  1 4 4 .6 .6 7 7 ' )' 2 0.3188+1.0804+0.0134+2.5404+1.4866 OR 1 9 1 .2 2 8 + 2 2 5 2 0 .2 2 +  + '1 1 4 2 0 .6 7 ' − 1 2 0 M1 At least 2 expressions (not including 120) of correct form seen. May see individual terms not summed. May be implied by 2 contributions correct to 3 decimal places seen. May be implied by AWRT 5.43 or 5.44 or 5.45. For contribution from combining last two cells allow AWRT 1.48 or 1.49 to 3 significant figures. No combining gives For last two cells: 12.09 gives 1.3836, 2.57 gives 0.3251+0.1415+1.0804+0.0134+2.5404+1.3893+0.1225 0.1264. 5.440 A1 AWRT 5.43 or 5.44 or 5.45 Critical value is 5.991 B1 With both first 2 cells and last 2 cells combined. '5.440' < 5.991, accept H / do not reject H / not significant. M1 Compare their 5.44 with 5.991 or 7.815 or 9.488 0 0 or 11.07 or 12.59 and appropriate conclusion (may be in terms of H ). 1 Test result can be implied by an attempt at an appropriate conclusion in context ignoring hypotheses. 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/ 3(b) Insufficient evidence to suggest the masses (of this species of frog) are not normally A1 Correct conclusion from correct working (using distributed. 5.99) ignoring hypotheses. In context. OR Level of uncertainty in language used (for example, not ‘prove’). Insufficient evidence to reject the biologist’s belief. Allow ‘Not enough evidence / no sufficient evidence that / to show / to conclude that …’. OR Do not accept ‘No evidence…’. Do not accept statements such as “there is Insufficient evidence to suggest that a normal distribution is not a suitable model / sufficient evidence to suggest…”. good fit (for the data). 7 Question Answer Marks Guidance