Inference using normal and t-distributions

1(a) 56.56 B1 Accept unsimplified. x = =8.08 7 s2 = 1 457.275− 56.562   = 0.045033  = 1351  M1 Accept unsimplified.   6  7   30000 8.08−8.22 M1 A1 1.74 – 1.75 t = = − 1.745 s2 7 Tabular value 1.440 M1 Comparison with 1.440 and correct FT conclusion. 1.745 > 1.440 , so reject H 0 There is insufficient evidence to support the hypothesis that the A1 Correct conclusion, in context, following correct work. Level (population) mean is 8.22 OR sufficient evidence to suggest mean of uncertainty in language is used. is less than 8.22 OR sufficient evidence to suggest μ<8.22 OE 6 1(b) Underlying distribution is normal/ population is normal B1 X is normal distributed. 1 © UCLES 2021 Page 6 of 12

4 s 2 = 1  1596− 2822   = 0.11265  = 138  M1 One correct, unsimplified. x 49  50    1225   s y 2 = 5 1 9    1808− 3 6 2 0 82    = 0.25311    = 8 2 8 2 5 4   A1 Both correct to 3sf. 0.11265 0.25312 M1 A1 May be implied. s2 = + =0.006471… 50 60 282 328 M1 − z= 50 60 their s [z] = 2.155 A1 Tolerance 2.15 – 2.16 Comparison with tabular value 1.96 OE M1 Comparison with 1.96 OE and conclusion. 2.155 > 1.96 Reject H 0 Sufficient evidence to reject population means are equal OR A1 Correct conclusion following correct work. Level of sufficient evidence to accept population means are not equal OR uncertainty in language is used. sufficient evidence to suggest μ ≠μ OE A B 8 © UCLES 2021 Page 9 of 12

1(a) 56.56 B1 Accept unsimplified. x = =8.08 7 s2 = 1 457.275− 56.562   = 0.045033  = 1351  M1 Accept unsimplified.   6  7   30000 8.08−8.22 M1 A1 1.74 – 1.75 t = = − 1.745 s2 7 Tabular value 1.440 M1 Comparison with 1.440 and correct FT conclusion. 1.745 > 1.440 , so reject H 0 There is insufficient evidence to support the hypothesis that the A1 Correct conclusion, in context, following correct work. Level (population) mean is 8.22 OR sufficient evidence to suggest mean of uncertainty in language is used. is less than 8.22 OR sufficient evidence to suggest μ<8.22 OE 6 1(b) Underlying distribution is normal/ population is normal B1 X is normal distributed. 1 © UCLES 2021 Page 6 of 12

4 s 2 = 1  1596− 2822   = 0.11265  = 138  M1 One correct, unsimplified. x 49  50    1225   s y 2 = 5 1 9    1808− 3 6 2 0 82    = 0.25311    = 8 2 8 2 5 4   A1 Both correct to 3sf. 0.11265 0.25312 M1 A1 May be implied. s2 = + =0.006471… 50 60 282 328 M1 − z= 50 60 their s [z] = 2.155 A1 Tolerance 2.15 – 2.16 Comparison with tabular value 1.96 OE M1 Comparison with 1.96 OE and conclusion. 2.155 > 1.96 Reject H 0 Sufficient evidence to reject population means are equal OR A1 Correct conclusion following correct work. Level of sufficient evidence to accept population means are not equal OR uncertainty in language is used. sufficient evidence to suggest μ ≠μ OE A B 8 © UCLES 2021 Page 9 of 12

1 1 1522  M1 13 s2 = 2313−  =0.28889 , accept unsimplified. 45 9 10   H : μ = 14.8 H : μ > 14.8 B1 If μ not used, ‘population mean’ required. 0 1 Must see 14.8, must be = and >. 152 M1 Using unbiased estimate. −14.8 [ t= ] 10 p-value is 0.0215. s/ 10 [ ] t= 2.35 A1 Rounds to 2.35. Tabular value = 1.833 M1 Comparison with 1.833 and correct FT conclusion. ‘2.35’ > 1.833 Reject H 0 Sufficient evidence to accept Farmer B’s claim oe A1 Correct conclusion, in context, following correct work. Level of uncertainty in language is used. Allow ±1 difference in third significant figure of their t for this mark. 6 © UCLES 2021 Page 6 of 13

3 89 67.2 B1 Implied by 0.1 in the CI formula. x = =1.78 y = =1.68 50 40 σ x 2 = 4 1 9    159.4− 8 5 9 0 2    = 0.02    = 5 1 0    M1 One variance correct unsimplified. σ y 2 = 3 1 9    113.1− 67 4 . 0 22    = 0.0052308    = 3 1 2 7 50    A1 Both correctly calculated. 0.02 0.0052308  69  M1 Unsimplified expression must be seen, 40 and 50 in correct σ2 = + =   places. 50 40 130000 [σ2 =] 0.0005308 A1 May be implied by correct final answer. CI : 1.78 – 1.68 ± z × 0.0005308 M1 Correct form for CI with a z-value. with z = 1.96 A1 0.10 ± 0.0452 or [0.0548, 0.145] A1 Allow either form. 8 SC Assuming equal variances award B1M0A0, M1A1 (for use of pooled variance formula, gives 0.0135), M1A1A0. © UCLES 2021 Page 8 of 13

1(a) 78.5 B1 157 s2 = =8.7222 9 18 s2 M1 Correct expression with a t value. CI: 25.6±t 10 With t = 1.833 A1 With correct t value. 25.6±1.71 ( 2 ) or [23.9, 27.3] A1 Accept in either form, ISW. Accept inequality form. 4 1(b) 25.6−k M1 t s2 10 With t = −1.383 A1 Allow 1.383. 26.9 A1 CAO 3 © UCLES 2021 Page 6 of 14

4 H :μ −μ =2 and H :μ −μ >2 B1 Or use of μ. 0 B A 1 B A d Differences: 5 8 2 0 4 7 −2 1 10 −3 M1 Allow one error. d =32, d2 =272 M1 Sample mean and variance. 1 322   848 d =3.2, s2 = 272−  =18.84  =  9  10   45  3.2−2 M1 A1 t = =0.874 s2 10 Compare with tabular value 1.833: 0.874 < 1.833. M1 Compare their value with 1.833 and conclusion. Accept H . 0 Insufficient evidence to support claim. A1 Correct conclusion, in context, following correct work. Level of uncertainty in language is used. Assumption: population differences are normally distributed B1 8 © UCLES 2021 Page 10 of 14

1 1  7522  304 M1 One correct unsimplified. s 2 = 14320−  =4.677 = x 39 40  65 1  15482  1308 A1 Both correct to at least 3sf. s 2 = 40200−  =4.434 = y 59 60  295 4.6769 4.4339 M1 s2 = + 40 60  3659  A1 = 0.1908    19175 752 1548 M1A1 Correct form with a z-value with 1.96 s can be unsimplified. CI =  −  ±1.96s  40 60  − 7.0 ±0.856 = [−7.86, −6.14] A1 Accept in either form, ISW. Accept [6.14, 7.86]. Accept inequality form. 7 © UCLES 2021 Page 6 of 13

6 H : μ = μ and H : μ > μ B1 0 A B 1 A B x = 21.2 x2 = 47.92 M1 A1 1 21.22  124 s 2 = 47.92−  =0.33067 = x 9 10  375 1  24.482  9527 s 2 = 53.75−  =0.34644 = y 11 12  27500 9×0.33067+11×0.34644 M1 A1 s2 = = 0.3393 10+12−2 2.12−2.04 M1 A1 Accept 0.321 – 0.322. t = =0.321 1 1 s + 10 12 Critical value: 1.325 M1 Compare their value with 1.325 and correct FT conclusion. 0.321 < 1.325 accept H . 0 Insufficient evidence that mean of A is greater than mean of B A1 Correct conclusion, in context, following correct work. Level of uncertainty in language is used. EITHER: Distributions are normal and equal variances B1 Assumptions consistent with method used. OR: Distributions are normal Accept ‘population is normal’. ‘Distribution of difference of means is normal’. Alternative method for question 6 H : μ = μ and H : μ > μ B1 0 A B 1 A B © UCLES 2021 Page 12 of 13 9231/42 Cambridge International AS & A Level – Mark Scheme October/November 2021 PUBLISHED Question Answer Marks Guidance x = 21.2 x2 = 47.92 M1 For one correct unsimplified. 1 21.22  124 s 2 = 47.92−  =0.33067 = A1 For both correct. x 9 10  375 1  24.482  9527 s 2 = 53.75−  =0.34644 = y 11 12  27500 0.33067 0.34644 M1 A1 s2 = + = 0.061936 10 12 t = 2.12−2.04 =0.321 ( 5 ) M1 A1 Accept 0.321 – 0.322. s Critical value: 1.325 M1 Compare their value with 1.325 and correct FT conclusion. 0.321 < 1.325 accept H . 0 Insufficient evidence that mean of A is greater than mean of B. A1 Correct conclusion, in context, following correct work. Level of uncertainty in language is used. EITHER: Distributions are normal and equal variances B1 Assumptions consistent with method used. OR: Distributions are normal. Accept ‘population is normal’. ‘Distribution of difference of means is normal’. 10 © UCLES 2021 Page 13 of 13

1(a) 78.5 B1 157 s2 = =8.7222 9 18 s2 M1 Correct expression with a t value. CI: 25.6±t 10 With t = 1.833 A1 With correct t value. 25.6±1.71 ( 2 ) or [23.9, 27.3] A1 Accept in either form, ISW. Accept inequality form. 4 1(b) 25.6−k M1 t s2 10 With t = −1.383 A1 Allow 1.383. 26.9 A1 CAO 3 © UCLES 2021 Page 6 of 14

4 H :μ −μ =2 and H :μ −μ >2 B1 Or use of μ. 0 B A 1 B A d Differences: 5 8 2 0 4 7 −2 1 10 −3 M1 Allow one error. d =32, d2 =272 M1 Sample mean and variance. 1 322   848 d =3.2, s2 = 272−  =18.84  =  9  10   45  3.2−2 M1 A1 t = =0.874 s2 10 Compare with tabular value 1.833: 0.874 < 1.833. M1 Compare their value with 1.833 and conclusion. Accept H . 0 Insufficient evidence to support claim. A1 Correct conclusion, in context, following correct work. Level of uncertainty in language is used. Assumption: population differences are normally distributed B1 8 © UCLES 2021 Page 10 of 14

1(a) Differences: M1 Attempt at differences, allow one error. 0.6 1.3 0.0 0.7 0.6 0.6 1.8 –0.2 –0.2 0.4 d 5.6 d2 6.74 M1 1 5.62  s2  6.74 0.4004 9 10   H :   0.4 and H :   0.4 B1 Allow use of  0 B A 1 B A d 0.560.4 M1 t  (0.800) 0.4004 10 t =0.7996 (0.800) A1 Critical value = t (9) = 1.383 M1 Compare calculated value with 1.383 and correct FT conclusion. 0.90 Compare: ‘0.7996’ < 1.383 Accept H 0 Insufficient evidence to support manager’s claim. A1 Correct conclusion, in context. Level of uncertainty in language is used. 7 1(b) Distribution of population differences is normal B1 1 © UCLES 2022 Page 6 of 13

5(a) x44.4, x2 247.48 B1 1 44.42   53  M1 s2  247.48 0.1514    7  8   350 M1 Correct formula with a t value. 44.4 0.1514 CI: 1.895 8 8 1.895 B1 1.895 used. 5.550.2615.29, 5.81 A1 Either form. 5 5(b) H : 5.9 and H : 5.9 B1 0 1 5.555.9 M1 t  2.544 s2 8 Critical value t (7) = 2.365 M1 Compare calculated value with 2.365 and correct FT 0.975 Compare ‘– 2.544’ < – 2.365, reject H . conclusion. 0 Sufficient evidence to support Raman’s claim. A1 Correct conclusion, in context. Level of uncertainty in language is used. 4 © UCLES 2022 Page 11 of 13

1(a) Differences: M1 Attempt at differences, allow one error. 0.6 1.3 0.0 0.7 0.6 0.6 1.8 –0.2 –0.2 0.4 d 5.6 d2 6.74 M1 1 5.62  s2  6.74 0.4004 9 10   H :   0.4 and H :   0.4 B1 Allow use of  0 B A 1 B A d 0.560.4 M1 t  (0.800) 0.4004 10 t =0.7996 (0.800) A1 Critical value = t (9) = 1.383 M1 Compare calculated value with 1.383 and correct FT conclusion. 0.90 Compare: ‘0.7996’ < 1.383 Accept H 0 Insufficient evidence to support manager’s claim. A1 Correct conclusion, in context. Level of uncertainty in language is used. 7 1(b) Distribution of population differences is normal B1 1 © UCLES 2022 Page 6 of 13

5(a) x44.4, x2 247.48 B1 1 44.42   53  M1 s2  247.48 0.1514    7  8   350 M1 Correct formula with a t value. 44.4 0.1514 CI: 1.895 8 8 1.895 B1 1.895 used. 5.550.2615.29, 5.81 A1 Either form. 5 5(b) H : 5.9 and H : 5.9 B1 0 1 5.555.9 M1 t  2.544 s2 8 Critical value t (7) = 2.365 M1 Compare calculated value with 2.365 and correct FT 0.975 Compare ‘– 2.544’ < – 2.365, reject H . conclusion. 0 Sufficient evidence to support Raman’s claim. A1 Correct conclusion, in context. Level of uncertainty in language is used. 4 © UCLES 2022 Page 11 of 13

1 94.5 B1 s2  13.5 7 s2 M1 Correct formula with a t value, must be 33.8. CI: 33.82.365 8 B1 2.365 used. 33.83.072 30.7, 36.9 A1 Either form. 4 © UCLES 2022 Page 6 of 13

6(a) 1  11202  M1 40 s 2  31400 1.0256 x 39 40 39   1  13702  s y 2  49 37600 50 1.2653 62 Both required   49 1.0256 1.2653 M1 2434 s2   40 50 47775 1.03 1.27 Condone s2   or 0.05115. 40 50 s2 0.050947 A1 May be implied by 2.66 as z value. H :   0 and H :   0 B1 0 a b 1 a b 1120 1370 M1  40 50 0.6 z  s 0.050947 z = 2.66 A1 Critical value of z = 2.326 M1 Compare their z value with 2.326 and correct FT conclusion. 2.66 > 2.326 Reject H 0 Sufficient evidence to support difference in population means oe A1 Correct conclusion, in context. Level of uncertainty in language is used. 8 © UCLES 2022 Page 12 of 13 6(b) 0.60.25 M1 Condone 1.54. z s z 1 .55 A1 0.9392 – 0.9395 or 0.0605 – 0.0608 A1 6.05 A1 Condone 6.05 – 6.08, accept  4 © UCLES 2022 Page 13 of 13

1  1625  1  16252  M1 775   x = 50 =32.5   s x 2 = 49   53200− 50   =7.908 98  1854  1  18542  A1 3057   y = 60 =30.9   s y 2 = 59   57900− 60   =10.3627 295 Both correct 7.908 10.3627 M1 A1 s2 = + [=0.33087] 50 60 CI: (32.5−30.9)1.96s M1 A1 Correct form with a z-value. Correct with 1.96. 1.61.1274=0.473, 2.73 A1 At least 3sf. 7 Pooled variance used: M1A1 M0A0 M1A1A0 max 4/7 © UCLES 2022 Page 6 of 12

6 1  6.242  M1 19 s 2 = 3.26− =0.001382 x 11 12  13750 19 1 5.12  s 2 = 2.6124− =0.001267 15000 y 9 10  A1 Both correct. 110.001382+90.001267 M1 A1 Find pooled variance. s2 = =0.001330 12+10−2 6.24 5.1 M1 A1 Use correct formula for t. − 12 10 t = =0.640 1 1 s + 12 10 H :  = H :   B1 0 a b 1 a b Critical value is 2.086 (t (20)) M1 Compare their calculated value with 2.086 and conclusion. 0.975 0.640 < 2.086 Accept H 0 There is insufficient evidence to support inspector’s claim A1 All correct except possibly B1, in context, level of uncertainty in language. ‘Prove’ scores A0 9 Pooled variance not used: M1A1 M0A0 M0A0 B1 M1A0 max 4/9 © UCLES 2022 Page 12 of 12

1 M1 s2 2.02=x+1.833 10 s2 1.78=x−1.833 10 2.02+1.78 Add, x = =1.90 Add. 2 x=19 A1 M1 Allow 1.372, 1.383, 1.812 instead of 1.833. s2 Subtract: 2.02−1.78=21.833 10 s2 =0.042859 A1 May be implied. s = 0.20702 1 (x)2 M1 But s2 = (x2 − ) 9 10 192 A1 36.486 x2 =9s2 + =36.5 10 Using 1.645: maximum M1A1 M0A0 M1A0 3/6 6 © UCLES 2022 Page 6 of 13

3 1  75.52  M1 239 s 2 = 115.2− =0.02439 x 49 50 9800   259 1  116.82  s 2 = 172.6− =0.02623 9875 y 79 80   A1 Both correct 0.02439 0.02623 M1 A1 May be implied by 1.75 for z s2 = + =0.00081565 50 80 75.5 116.8 M1 − 50 80 z= s 1.75 A1 Compare with 1.645: 1.75 >1.645 M1 Reject null hypothesis Using areas, 0.04 < 0.05. Sufficient evidence to suggest that population mean in country X A1 Correct conclusion in context, following correct work, level of is greater than population mean in country Y uncertainty in language. ‘Prove’ gives A0. 8 Pooled variance: z=1.74 M1A1 M0A0M0A0 M1A0 max 3/8 © UCLES 2022 Page 8 of 13

1  1625  1  16252  M1 775   x = 50 =32.5   s x 2 = 49   53200− 50   =7.908 98  1854  1  18542  A1 3057   y = 60 =30.9   s y 2 = 59   57900− 60   =10.3627 295 Both correct 7.908 10.3627 M1 A1 s2 = + [=0.33087] 50 60 CI: (32.5−30.9)1.96s M1 A1 Correct form with a z-value. Correct with 1.96. 1.61.1274=0.473, 2.73 A1 At least 3sf. 7 Pooled variance used: M1A1 M0A0 M1A1A0 max 4/7 © UCLES 2022 Page 6 of 12

6 1  6.242  M1 19 s 2 = 3.26− =0.001382 x 11 12  13750 19 1 5.12  s 2 = 2.6124− =0.001267 15000 y 9 10  A1 Both correct. 110.001382+90.001267 M1 A1 Find pooled variance. s2 = =0.001330 12+10−2 6.24 5.1 M1 A1 Use correct formula for t. − 12 10 t = =0.640 1 1 s + 12 10 H :  = H :   B1 0 a b 1 a b Critical value is 2.086 (t (20)) M1 Compare their calculated value with 2.086 and conclusion. 0.975 0.640 < 2.086 Accept H 0 There is insufficient evidence to support inspector’s claim A1 All correct except possibly B1, in context, level of uncertainty in language. ‘Prove’ scores A0 9 Pooled variance not used: M1A1 M0A0 M0A0 B1 M1A0 max 4/9 © UCLES 2022 Page 12 of 12

1 91.2 1  91.22    67  M1 Both. x  12 7.6 s x 2  11   695.8 12     0.2436  275     0.2436 M1 With a t-value. CI: 7.6t 0.975 11 12 2.201 seen B1 7.60.3136=7.29, 7.91 A1 4 © UCLES 2023 Page 6 of 12

2 99.6 21.5 43  M1 Both SOI. x  8.3 s 2  1.9545 or 12 x 11   22   H :   H :   B1 0 P Q 1 P Q 92.16111.9545 M1 FT FT their 1.9545 only Pooled estimate = 10122 2.047 (2.05) A1 May be implied, allow 2.04 9.128.3 M1 FT FT their pooled estimate t 1 1 2.047  10 12 1.339 (or 1.338) A1 Accept 1.34 '1.339'1.725 M1 Compare with correct tabular value 1.725 and Accept H / not significant conclusion without context. Condone ‘reject H ’. 0 1 Insufficient evidence to suggest that the (mean) time taken at P is greater than A1 Correct conclusion in context, following correct the (mean) time taken at Q. work, level of uncertainty in language (not ‘prove’, not ‘there is no evidence’), no contradictions. A0 if H and H reversed. 0 1 8 © UCLES 2023 Page 7 of 12

1 91.2 1  91.22    67  M1 Both. x  12 7.6 s x 2  11   695.8 12     0.2436  275     0.2436 M1 With a t-value. CI: 7.6t 0.975 11 12 2.201 seen B1 7.60.3136=7.29, 7.91 A1 4 © UCLES 2023 Page 6 of 12

2 99.6 21.5 43  M1 Both SOI. x  8.3 s 2  1.9545 or 12 x 11   22   H :   H :   B1 0 P Q 1 P Q 92.16111.9545 M1 FT FT their 1.9545 only Pooled estimate = 10122 2.047 (2.05) A1 May be implied, allow 2.04 9.128.3 M1 FT FT their pooled estimate t 1 1 2.047  10 12 1.339 (or 1.338) A1 Accept 1.34 '1.339'1.725 M1 Compare with correct tabular value 1.725 and Accept H / not significant conclusion without context. Condone ‘reject H ’. 0 1 Insufficient evidence to suggest that the (mean) time taken at P is greater than A1 Correct conclusion in context, following correct the (mean) time taken at Q. work, level of uncertainty in language (not ‘prove’, not ‘there is no evidence’), no contradictions. A0 if H and H reversed. 0 1 8 © UCLES 2023 Page 7 of 12

2 M1 SOI s2 s2 x 1.771 1.14 or x 1.771 1.11 Allow incorrect t-value, not z-value. 14 14 1 B1 Does not depend on use of a t-value. [Add: x  1.141.11 1.125] x15.75 2 s2 11.141.11 M1 Allow incorrect t-value, but not a z-value. Subtract or substitute:    14 2 1.771  s2 0.001004 or s = 0.0316[9] A1 450 , implied by correct final answer. 448063 x2 M1 OE x2 13s2  14 x2 17.7(3) A1 CWO 6 © UCLES 2023 Page 7 of 13

4(a) H :   H :   B1 0 x y 1 x y 1  1642  B1 s 2  338.1  0.02405 and x 79 80   19 379 Both. Implied by , or 3sf. 1  124.82  790 14750 s 2  261.1  0.02569 y 59 60   0.02405 0.02569 M1 s2   80 60 s2 0.0007289 or s 0.026998 or 0.0270 A1 Implied by z  1.11 164.0 124.8 M1 FT their value for s.  80 60 z  s 1.11 A1 ‘1.11’ < 1.282 M1 Compare with correct z-value 1.282 and consistent signs. Condone Accept H / not significant ‘reject H ’. 0 1 Using probabilities, P(Z >1.11) = 0.1333 > 0.1. Insufficient evidence to support the inspector’s suspicion/ A1 Correct conclusion in context, following correct work, level of Insufficient evidence that the (mean) lengths of rods from uncertainty in language (not ‘prove’, not ‘there is no evidence’), machine X are shorter than the (mean) lengths of rods from no contradictions. machine Y e.g. proves that the inspector is incorrect scores A0 A0 if hypotheses wrong way round. 8 © UCLES 2023 Page 9 of 13 4(b) Large sample sizes OR central limit theorem applies. B1 1 Question Answer Marks Guidance

1 H : =4.22 H : 4.22 B1 0 1 42.05 1 42.052  13  M1 x = =4.205, s2 = 176.83−  = = 0.0010833  10 9  10  12000  4.205−4.22 M1 Their 4.205, their s2. t = s2 10 t=−1.44 A1 Condone sign. Tabular value = 1.833: ‘1.44’ < 1.833, accept H M1 Compare their value with correct tabular value, signs 0 consistent, allow ‘not significant’. There is insufficient evidence to support Maya’s belief. A1 Correct conclusion in context, following correct work, level of uncertainty in language. A0 if hypotheses wrong way round or missing. 6 © UCLES 2023 Page 6 of 13

3(a) H :  − =2 H :  − 2 B1 May use  , but must be consistent with direction of 0 B E 1 B E d differences found. M1 Differences, at most 2 errors. 8 8 15 –2 –3 20 –3 0 5 4 1 522  2728  M1 Their values but must see 9 and 10 used correctly. d =52, d2 =816, s2 = 816−  = =60.62  9  10  45  52 M1 −2 10 t = s2 10 t=1.29967, 1 .30 A1 Tabular value = 2.262. ‘1.30’ < 2.262, accept H M1 Allow ‘not significant’. 0 Insufficient evidence to suggest that cholesterol level has reduced by more A1 CWO. Correct conclusion in context, following correct than 2. work, level of uncertainty in language. A0 if hypotheses wrong way round or missing. 7 3(b) Population differences are normally distributed B1 1 © UCLES 2023 Page 8 of 13

1 243.8 1219  384.9 3849  B1 Implied by correct sor pooled estimate s2 = = =4.132 , s2 = = =5.201 x 59   295   y 74   740   243.8+384.9 =4.727 60+75−2 4.132 5.201 M1 Using their sample variances. s2 = + =0.1382 or s=0.3718 Pooled estimate M0. 60 75 A1 6345 7614 M1 With a z-value CI: − 1.96'0.3718' 60 75 A1 With 1.96 (with their s) 6345 7614 1 1 Pooled − 1.962.174 + 60 75 60 75 3.50, 4.96 or (3.5, 4.96) A1 4.230.729 is A0, condone 4.96, 3.5 etc.   6 © UCLES 2023 Page 6 of 13

5(a) H : population medians are equal or m = m B1 Do not accept ‘difference between population 0 x y H : population median for X < population median for Y or m < m medians < 0’ without X or Y oe specified. 1 x y M1 Rankings, allow at most 3 errors. X Y 4.0 1 4.5 2 4.6 3 4.9 6 4.7 4 5.1 8 4.8 5 5.3 10 5.0 7 5.4 11 5.2 9 5.7 13 5.6 12 5.9 15 5.8 14 6.3 16 6.4 17 Sum: 55 98 Test statistic = 55 A1 Tabular value for m=8, n=9 is 54 B1 ‘55’ > ‘54’, accept H /not significant M1 Ft their ‘55’ Must come from ranks. 0 Ft their ‘54’, must come from table. Insufficient evidence to support manager’s claim. A1 Correct conclusion in context, following correct Insufficient evidence to suggest that the median time of machine X is less than the work, level of uncertainty in language. median time of machine Y. A0 if hypotheses the wrong way round or missing. 6 © UCLES 2023 Page 12 of 13 5(b) H : 𝜇 = 𝜇 H : 𝜇 < 𝜇 B1 0 𝑥 𝑦 1 𝑥 𝑦 x=39.7 x2 =199.33 y=49.5 y2 =275.47 B1 1 39.72  s x 2 = 7   199.33− 8   =0.33125 1 49.52  B1 s2 y = 8   275.47− 9   =0.4025 70.33125+80.4025 M1 Pooled variance s2 = 8+9−2 0.36925 A1 39.7 −49.5 M1 t = 8 9 s 1+1 8 9 t=−1.82 A1 Tabular value = 1.753: 1.82 > 1.753 M1 Reject H , sufficient evidence that mean for machine X is less than mean for machine A1 CWO 0 Y. Correct conclusion in context, following correct work, level of uncertainty in language. 9 5(c) t-test is assuming a normal distribution, and with equal variances. This may not be B1 Not specific to data in question. true. So, no reason to expect results to be the same. Mention of normal distribution is not enough. Outliers affect part (b) but not part (a). 1 © UCLES 2023 Page 13 of 13

1 H : =4.22 H : 4.22 B1 0 1 42.05 1 42.052  13  M1 x = =4.205, s2 = 176.83−  = = 0.0010833  10 9  10  12000  4.205−4.22 M1 Their 4.205, their s2. t = s2 10 t=−1.44 A1 Condone sign. Tabular value = 1.833: ‘1.44’ < 1.833, accept H M1 Compare their value with correct tabular value, signs 0 consistent, allow ‘not significant’. There is insufficient evidence to support Maya’s belief. A1 Correct conclusion in context, following correct work, level of uncertainty in language. A0 if hypotheses wrong way round or missing. 6 © UCLES 2023 Page 6 of 13

3(a) H :  − =2 H :  − 2 B1 May use  , but must be consistent with direction of 0 B E 1 B E d differences found. M1 Differences, at most 2 errors. 8 8 15 –2 –3 20 –3 0 5 4 1 522  2728  M1 Their values but must see 9 and 10 used correctly. d =52, d2 =816, s2 = 816−  = =60.62  9  10  45  52 M1 −2 10 t = s2 10 t=1.29967, 1 .30 A1 Tabular value = 2.262. ‘1.30’ < 2.262, accept H M1 Allow ‘not significant’. 0 Insufficient evidence to suggest that cholesterol level has reduced by more A1 CWO. Correct conclusion in context, following correct than 2. work, level of uncertainty in language. A0 if hypotheses wrong way round or missing. 7 3(b) Population differences are normally distributed B1 1 © UCLES 2023 Page 8 of 13

1 s2  941.5 B1 72.423. 13 s2 M1 A1 Correct formula with a t value. CI: 42.82.160 2.160 seen in a CI formula for A1. 14 [37.9, 47.7] A1 (37.9, 47.7) A1 37.9'x'47.7 A1 (ignore symbol used inside the inequality) Condone [47.7, 37.9] A1 Final answer 42.84.91 A0 4 © Cambridge University Press & Assessment 2024 Page 6 of 15

6(a) Signed differences: M1 Allow one error, must be signed differences. 4.1 1 .1 0.4 0.2 3.9 2.6 0.6 3.5 0.1 0 1 14.52  M1 d 14.5, d2 52.81 s2  52.81 3.532 9 10   H :   1 B1 0 B A Accept H :  1, H :  1. H 1 :  B  A 1 0 d 1 d 1.451 M1 The – 1 must be present in the numerator. t  s2 10 0.757 A1 Critical value is 1.383: ‘0.757’ < 1.383 Accept H M1 ‘0.757’ must come from a paired sample t- 0 calculation, 1.383 must be correct Correct ft conclusion for their 0.757 and 1.383 Condone Reject H 1 ‘Accept H ’ can be implied by a conclusion that is 0 consistent with their 0.757 and 1.383 Insufficient evidence to support Jade’s claim 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…..’ 7 © Cambridge University Press & Assessment 2024 Page 13 of 15 6(b) Distribution of population differences is normal B1 Underlying distribution of differences normal B1 ‘data’ implies B0. 1

1 s2  941.5 B1 72.423. 13 s2 M1 A1 Correct formula with a t value. CI: 42.82.160 2.160 seen in a CI formula for A1. 14 [37.9, 47.7] A1 (37.9, 47.7) A1 37.9'x'47.7 A1 (ignore symbol used inside the inequality) Condone [47.7, 37.9] A1 Final answer 42.84.91 A0 4 © Cambridge University Press & Assessment 2024 Page 6 of 15

6(a) Signed differences: M1 Allow one error, must be signed differences. 4.1 1 .1 0.4 0.2 3.9 2.6 0.6 3.5 0.1 0 1 14.52  M1 d 14.5, d2 52.81 s2  52.81 3.532 9 10   H :   1 B1 0 B A Accept H :  1, H :  1. H 1 :  B  A 1 0 d 1 d 1.451 M1 The – 1 must be present in the numerator. t  s2 10 0.757 A1 Critical value is 1.383: ‘0.757’ < 1.383 Accept H M1 ‘0.757’ must come from a paired sample t- 0 calculation, 1.383 must be correct Correct ft conclusion for their 0.757 and 1.383 Condone Reject H 1 ‘Accept H ’ can be implied by a conclusion that is 0 consistent with their 0.757 and 1.383 Insufficient evidence to support Jade’s claim 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…..’ 7 © Cambridge University Press & Assessment 2024 Page 13 of 15 6(b) Distribution of population differences is normal B1 Underlying distribution of differences normal B1 ‘data’ implies B0. 1

2(a) 1 M1 x  68.6664.2266.44 2 x1266.44797.28 A1 May come from use of incorrect t value. Accept AWRT 797. ts 1 M1 Must be with a t value (not z value).  68.6664.22 12 2 With t  2.718 gives s2 8.0055 A1 Accept AWRT 8.01; may be recovered by final answer exactly 53059(.34). s 2.8294. 1  x2  M1 Must see an expression for x2. Use s2  x2   to find an 11 12    expression for x2 x2 53059 A1 AWRT 53100 but not from wrong working. 6 2(b) Population is normally distributed B1 Underlying distribution is normal B1. Pulse rates are normal B1. Population mean is normally distributed B0. (Underlying) data is normal B0. 1 © Cambridge University Press & Assessment 2024 Page 8 of 15

6(a) 1  10802  M1 152 s 2  23480  (= 3.102) 9 x 49  50  10 3 1  9402  Both. s 2  22220  (= 3.333) y 39 40   3.102 3.333 M1 s2   0.1454 50 40 s2  2137 0.1454 A1 14700 1080 940 M1 B1 Use of correct formula with a z value. CI:  1.645 s2 or B1 for 1.645 ( accept 1.64 or 1.65) used in CI formula. 50 40 940 1080  1.645 s2 40 50 1.27, 2.53 or [–2.53, –1.27] A1 (1.27, 2.53) A1. 1.27  2.53 A1 (ignore symbol used inside the inequality). Y X Condone [2.53, 1.27] or [–1.27, –2.53] A1. Final answer 1.90.627 A0. Pooled estimate: max score M1 M0A0 M1B1A0. First M1 for correct form of pooled estimate. 6 © Cambridge University Press & Assessment 2024 Page 14 of 15 6(b) 1.91.1 M1 1.91.1 z  z  or 7.87 implies M1. s s 2.098 A1 AWRT 2.10 ‘2.098’ > 1.28(2) Reject H M1 ‘2.098’ from their attempt at z, 1.28(2) must be correct. 0 Consistent signs in comparison. or Φ2.098 0.9821 > 0.9 Reject H FT conclusion from their 2.098. 0 Condone ‘Accept H ’. 1 or 1  Φ2.098 = 0.0179 < 0.1 Reject H 0 ‘Reject H 0 ’ can be implied by a conclusion in context that is consistent with their 2.098. Sufficient evidence to suggest that the lengths of A1 Correct work only, conclusion in context with level of uncertainty in language. Not tails of wallabies in region Y are, on average, ‘prove’ or equivalent. more than 1.1 cm longer than the tails of wallabies in region X Accept ‘enough evidence’. Condone ‘some evidence’. 4 Note that it is possible to use part (a) but it must be made clear explicitly that rejection at 5% level implies rejection at 10% level. © Cambridge University Press & Assessment 2024 Page 15 of 15

1 1  16542  B1 3392 s 2 = 54850−  =2.769 x 49 50 1225   4.95 2.769 M1 A1 s2 = + =0.1379 60 50 1654 M1 Correct formula with a z value. CI: 35.6− 1.96s 50 B1 1.96 seen. 1.79, 3.25 A1 Accept inequality. 6 Question Answer Marks Guidance

6 1 64.22  M1 851 s 2 = 519.46−  =0.607857 x 7 8 1400   17 1 71.12  , both. s 2 = 567.13−  =0.680 25 y 8 9   80.680 + 70.607857 M1 A1 1939 s2 = =0.64633 Correct formula for pooled variance used, . 9+8−2 3000 H :  − =0 B1 0 X Y H :  − 0 1 X Y 8.025−7.9 0.125 M1 A1 Dependent on pooled variance being used t = = =0.320 1 1 0.3906 s + 8 9 ‘0.320’ < 1.341 accept H M1 Compare with 1.341 and conclusion 0 Insufficient evidence to support Ansal’s suspicion A1 Correct work only, except possibly B1, in Insufficient evidence to suggest that wingspan from region X is greater than context, level of uncertainty in language wingspan from region Y. 8 © Cambridge University Press & Assessment 2024 Page 11 of 11

1 x=30.6 x2 =120.52 M1 139 1 30.62  With their x, x2, . s2 = 120.52−  =0.49643 280 7 8   30.6 s2 M1 Use correct formula with a t value. CI = 1.895 8 8 B1 1.895 seen. 3.35, 4.30 A1 Accept with inequality signs or open brackets. . 4 Question Answer Marks Guidance

5 1  30.82  B1 237 s 2 = 29.0−  =0.055765 x 34 35 4250   1488 1  62.42  s 2 = 76.8−  =0.201763 7375 y 59 60   Allow B1 for correct expression for pooled variance. 0.055765 0.201763 M1 Pooled variance M0A0. s2 = + =0.004956 A1 35 60 H :  = B1 0 x y H   1: x y 30.8 62.4 M1 − 35 60 z= A1 Pooled variance M0A0. s 0.25 − = −2.27 0.004956 Corresponding probability is 0.9885 M1 FT their z . One tail: 100(1−'0.9885')=1.15 M1 Allow for 1 – ‘0.9885’. 1.15 A1 Allow ⩾. 9 © Cambridge University Press & Assessment 2024 Page 11 of 13

1 1  16542  B1 3392 s 2 = 54850−  =2.769 x 49 50 1225   4.95 2.769 M1 A1 s2 = + =0.1379 60 50 1654 M1 Correct formula with a z value. CI: 35.6− 1.96s 50 B1 1.96 seen. 1.79, 3.25 A1 Accept inequality. 6 Question Answer Marks Guidance

6 1 64.22  M1 851 s 2 = 519.46−  =0.607857 x 7 8 1400   17 1 71.12  , both. s 2 = 567.13−  =0.680 25 y 8 9   80.680 + 70.607857 M1 A1 1939 s2 = =0.64633 Correct formula for pooled variance used, . 9+8−2 3000 H :  − =0 B1 0 X Y H :  − 0 1 X Y 8.025−7.9 0.125 M1 A1 Dependent on pooled variance being used t = = =0.320 1 1 0.3906 s + 8 9 ‘0.320’ < 1.341 accept H M1 Compare with 1.341 and conclusion 0 Insufficient evidence to support Ansal’s suspicion A1 Correct work only, except possibly B1, in Insufficient evidence to suggest that wingspan from region X is greater than context, level of uncertainty in language wingspan from region Y. 8 © Cambridge University Press & Assessment 2024 Page 11 of 11

1 H : = H :   B1 If in words, must contain ‘population means’. 0 x y 1 x y Allow H : =0, H :  0. 0 d 1 d s2 =103.218... M1 SOI, one unbiased estimate correct to 3sf or x better. s2 =171.139... y Do not accept biased estimates (102.4, 169.0). s2 s2 A1 A0 for pooled s . x + y =2.999 Accept 3, 3.0, 3.00. 120 80 May be implied by z=3.03. 10470−6560 M1 Value of z obtained using correct form with their z= 120 80 =3.03 s2 s2 s x 2 and their s y 2. x + y Use of pooled s scores M0 A0. 120 80 A1 Use of pooled s scores M0 A0. '3.03' > 2.576, reject H (significant). M1 Compare their z with 2.576 and appropriate 0 OR conclusion (may be in terms of H ). 1 P(Z '3.03')=0.001220.005, reject H (significant). Reject H 0 can be implied by an attempt at an 0 appropriate conclusion in context. May see 0.002440.01, reject H (significant). 0 Allow M1 for comparison of their 3.03 with 2.326 and appropriate conclusion. There is sufficient evidence to suggest the mean spend of customers is different on A1 Correct conclusion in context from correct weekends compared to weekdays. working ignoring hypotheses. Level of uncertainty in language used (for example, not ‘prove’). Must refer to mean or average spend. 7 © Cambridge University Press & Assessment 2025 Page 9 of 16

4(a) (x−x)2 =40104− 1 6122 =2649.6 M1 Either expression, may be embedded. 10 May be implied, for example, by 294.4 or (y−y)2 =27460−14442 =2818 402.57 seen. 8 2649.6+2818 A1 SOI s2 = =341.725 p 10+8−2 (61.2−55.5)2.120 's2' (1 +1)=5.72.128.7686 B1 2.120, SOI. p 10 8 M1 Correct form, must be a t value. A1 Correct un-simplified confidence interval, SOI. −12.9, 24.3 A1 Accept with inequality signs or open brackets or −24.3, 12.9 from Y −X . Condone 24.3, −12.9. Do not accept 5.718.6. 6 4(b) Because zero lies inside the confidence interval. B1 1 © Cambridge University Press & Assessment 2025 Page 13 of 16

1 H : = H :   B1 If in words, must contain ‘population means’. 0 x y 1 x y Allow H : =0, H :  0. 0 d 1 d s2 =103.218... M1 SOI, one unbiased estimate correct to 3sf or x better. s2 =171.139... y Do not accept biased estimates (102.4, 169.0). s2 s2 A1 A0 for pooled s. x + y =2.999 Accept 3, 3.0, 3.00. 120 80 May be implied by z=3.03. 10470−6560 M1 Value of z obtained using correct form with their z= 120 80 =3.03 s2 s2 s x 2 and their s y 2. x + y Use of pooled s scores M0 A0. 120 80 A1 Use of pooled s scores M0 A0. '3.03' > 2.576, reject H (significant). M1 Compare their z with 2.576 and appropriate 0 OR conclusion (may be in terms of H ). 1 P(Z '3.03')=0.001220.005, reject H (significant). Reject H 0 can be implied by an attempt at an 0 appropriate conclusion in context. May see 0.002440.01, reject H (significant). 0 Allow M1 for comparison of their 3.03 with 2.326 and appropriate conclusion. There is sufficient evidence to suggest the mean spend of customers is different on A1 Correct conclusion in context from correct weekends compared to weekdays. working ignoring hypotheses. Level of uncertainty in language used (for example, not ‘prove’). Must refer to mean or average spend. 7 © Cambridge University Press & Assessment 2025 Page 9 of 15

4(a) (x−x)2 =40104− 1 6122 =2649.6 M1 Either expression, may be embedded. 10 May be implied, for example, by 294.4 or (y−y)2 =27460−14442 =2818 402.57 seen. 8 2649.6+2818 A1 SOI s2 = =341.725 p 10+8−2 (61.2−55.5)2.120 's2' (1 +1)=5.72.128.7686 B1 2.120, SOI. p 10 8 M1 Correct form, must be a t value. A1 Correct un-simplified confidence interval, SOI. −12.9, 24.3 A1 Accept with inequality signs or open brackets or −24.3, 12.9 from Y −X . Condone 24.3, −12.9. Do not accept 5.718.6. 6 4(b) Because zero lies inside the confidence interval. B1 1 © Cambridge University Press & Assessment 2025 Page 13 of 15

2 s2 = 1 ( 720−1682) =0.36923 = 24 M1 Correct method to find one unbiased estimate, x 39 40  65 may be un-simplified. s2 = 1 ( 900−2282) =0.56949 =168 y 59 60  295 A1 Both unbiased estimates correct to 3sf or better. s2 = '0.36923'+ '0.56949' M1 M0 for pooled estimate. 40 60 s2 =0.0187 = 359  A1 SOI   19175 CI= (168−228) 1.645's' M1 Correct form with any z-value, FT their s. 40 60 B1 1.645 (accept 1.64 or 1.65) used in CI formula. 0.175, 0.625 A1 Accept with inequality signs or open brackets. Condone 0.625,0.175. Do not accept 0.40.225. 7 © Cambridge University Press & Assessment 2025 Page 10 of 16

5(a) H :  − = k , H :  − k B1 Allow H :  = k , H :  k if defined or 0 B A 1 B A 0 d 1 d consistent with working. Differences, d : 19, 17, 8, 16, 9, 23, 6, 5. M1 Allow one error. d =103, d2 =1641, M1 Sample mean and variance for differences. d =12.875, s2 = 1 ( 1641−1 ( 1032)) =44.982 = 2519 d 7 8  56  12.875−k M1 Correct form for t with their d and their s2. t= d s2 M0 if k is omitted. d 8 A1 12.875−k M1 Accept ≥, > or =. ≥1.895 s2 d 8 0≤ k≤8.38 A1 Accept k8.38. 7 5(b) Population of differences are normally distributed. B1 Accept underlying distribution or population. B0 for "population mean normally distributed". B0 for "(underlying) data is normal". 1 © Cambridge University Press & Assessment 2025 Page 14 of 16

1 10.24−10.6 M1 0.52 t= Must be in correct form, allow in denominator. 0.52 12 12 ()1.729 A1 CWO, AWRT ()1.73. '−1.729 ' −1.363 (or 1.729 1.363), reject H (significant). M1 Compare their t with 1.363 (consistent signs) and appropriate 0 conclusion (may be in terms of H ). 1 Reject H can be implied by an attempt at an appropriate 0 conclusion in context. Allow M1 for comparison of their t with 1.356 , 1.796, or 1.782 (consistent signs) and appropriate conclusion. Sufficient evidence to support/suggest that the [population] mean A1 Correct conclusion in context from correct working ignoring is less than 10.6. hypotheses. Level of uncertainty in language used (for example, not ‘prove’). Allow  or mean for population mean. 4 © Cambridge University Press & Assessment 2025 Page 9 of 15

6(a) x=30.4 B1 2.262 B1 2.26 or 2.262 seen. 10(3.13−2.95) M1 Subtracting, or substituting, to find an equation involving s . x Subtract: s = x (22.262) Allow M1 for any of the following t values seen instead of 2.262: 1.812, 1.833, 2.228 or 3.690. s2 = 0.0158 or s =0.126 A1 FT FT their 2.262 following M1. x x Implied by AWRT 92.6 following M1. 30.42 A1 FT FT their 2.262. x2 =90.0158307+ =92.6 CWO 10 5 © Cambridge University Press & Assessment 2025 Page 14 of 15 6(b) H :  = B1 If in words, must contain ‘population means’. 0 Y X Allow H :  =0, H :  0 if defined or consistent with H :   0 d 1 d 1 Y X working. 24.82 M1 May be embedded, for example s2 = 1 ( 76.98−1 ( 24.82)) . 76.98− y 7 8 8 9s2 +7s2 M1 Correct form using their s2 and their s2, may be embedded. s2 = x y =0.015155 x y p 10+8−2 A1 SOI, can be implied by any t that rounds to 1.02 or 1.03. 3.04−3.1 M1 Correct form using their s . t= =−1.028 p s 1 +1 p 10 8 A1 AWRT 1.02 or 1.03implies M1 A1. [18 degrees of freedom, so critical value is 1.746.] M1 Compare their t with 1.746 (consistent signs) and appropriate '1.028 '1.746, accept H (not significant). conclusion (may be in terms of H ). 0 1 Accept H can be implied by an attempt at an appropriate 0 conclusion in context. Allow M1 for comparison of their t with 1.740 or 2.120and appropriate conclusion. Insufficient evidence to support Mona’s claim. A1 Correct conclusion in context from correct working ignoring hypotheses. Level of uncertainty in language used (for example, not ‘prove’). 8 © Cambridge University Press & Assessment 2025 Page 15 of 15

1(a) s2 =1 ( '71314'−'844'2) =134 =8.933 M1 Correct expression, implied by AWRT 8.93.   9 10 15 2.262 B1 2.262 or 2.26 seen. M1 Correct form, must be a t-value. '844''2.262' '8.933' 10 10 [82.3, 86.5] A1 Accept with inequality signs or open brackets. Condone [86.5, 82.3]. Do not accept 84.42.1. The distribution of estimates of angles is normal. B1 OR The estimates are a random sample from some population. OR The estimates are independent. OR Underlying distribution is normal. 5 1(b) Population unlikely to be normal as  is close to right-angle / data B1 Must refer to the context. is skewed. OR Estimates may not be independent, for example due to collusion. OR No indication that sample is random. 1 © Cambridge University Press & Assessment 2025 Page 9 of 16

3 (x−x)2 =90.36=3.24 B1 '16.6'2 M1 Can be implied by 0.172 ( = 43 ) or 0.688 ( = 86). '55.8'− 250 125 5 3.24+0.688 M1 Correct use of formula for pooled variance using their values. (=0.302) 10+5−2 s=0.550 A1 CAO 4 Question Answer Marks Guidance

6(a) H : = H:   B1 If in words, must contain ‘population means’. 0 B A 1 B A Allow H : =0,H:  0 if defined or consistent with working. 0 d 1 d Differences: 5, 0, −1, 6,1 , −2, 8, 7, 15 M1 Attempt at signed differences. s2 =1 ( '405'−'39'2) =29.5 M1 Correct form for s2, note that s =5.43. d 8 9 d d '39' M1 Correct form. 9 =2.393 '29.5' 9 A1 AWRT 2.39. '2.393'1.860, reject H / significant. M1 Compare their 2.393 with 1.860 and appropriate result (may be in 0 terms of H ). 1 Allow M1 for comparison from 2-sample test using 1.746. Test result may be implied by an attempt at an appropriate conclusion in context ignoring hypotheses. Sufficient evidence to suggest new training programme results in A1 Correct conclusion from correct working ignoring hypotheses. reduced times. In context. OR Level of uncertainty in language is used (for example, not ‘prove’). Sufficient evidence to support the coach’s belief. Do not accept statements such as “there is insufficient evidence to suggest…”. 7 6(b) The population of differences may not be normally distributed. B1 Must mention population/distribution and differences. The population of differences may not be symmetrical (and hence Accept “the sample may not be random” or “times may not be not normal). independent”. 1 6(c) A Wilcoxon matched-pairs signed-rank test. B1 Must refer to pairs / paired. OR A paired-sample sign test. Accept “paired Wilcoxon test” or “paired sign test”. 1 © Cambridge University Press & Assessment 2025 Page 14 of 16

2(a) s2 = 1 ( '476117'−'2390'2) = 326 =9.8788 M1 Correct expression, implied by AWRT 9.88.   11 12 33 3.106 B1 3.106 or 3.11 seen. '2390''3.106' '9.8788' M1 Correct form, must be a t-value. 12 12 196, 202 A1 Accept with inequality signs or open brackets. Condone [202, 196]. Do not accept 1993. 4 2(b) −1.796 B1 1.796 or 1.80 seen. '199.17'−k '199.17'−k M1 Correct form, must be a t-value, consistent signs. '−1.796' or '1.796' Condone non-strict inequality. '9.8788' '9.8788' 12 12 200.8 A1 Accept 200.7, 200 or 201 from correct working. Condone final answer of the form k < 200.8 OE. Condone non-strict inequalities. 3 © Cambridge University Press & Assessment 2025 Page 10 of 15

5 H :  = H:   B1 If in words, must have population means. 0 A B 1 A B Allow H : =0 , H : 0. 0 d 1 d '1331848'−'3264'2 [=136] M1 OE, may be implied by s2 =19.43. 8 A Biased estimate of 17 earns M0 unless recovered. '964964'−'2406'2 [=158] M1 OE, may be implied by s2 =31.6. 6 B Biased estimate of 26 earns M0 unless recovered. '136'+'158' M1 Correct form using their values, may be embedded. s2 = = 24.5 p 8+6−2 A1 OE, for example 49 . 2 '3264'−'2406' M1 Correct form using their values. t= 8 6 =2.62 's ' 1+1 p 8 6 A1 AWRT 2.62. '2.62'2.179, reject H / significant. M1 Compare their 2.62 with 2.179 and appropriate result (may be in terms 0 of H ). 1 Test result may be implied by an attempt at an appropriate conclusion in context ignoring hypotheses. Sufficient evidence to suggest that there is a difference in the A1 Correct conclusion from correct working ignoring hypotheses. mean tensile strengths of steel rods produced by the two In context. machines. Level of uncertainty in language is used (for example, not ‘prove’). Allow “(Some/enough) evidence that / to show / conclude that…”. Do not accept statements such as “there is insufficient evidence to suggest…”. 9 © Cambridge University Press & Assessment 2025 Page 13 of 15

1(a) s2 =1 ( '71314'−'844'2) =134 =8.933 M1 Correct expression, implied by AWRT 8.93.   9 10 15 2.262 B1 2.262 or 2.26 seen. M1 Correct form, must be a t-value. '844''2.262' '8.933' 10 10 [82.3, 86.5] A1 Accept with inequality signs or open brackets. Condone [86.5, 82.3]. Do not accept 84.42.1. The distribution of estimates of angles is normal. B1 OR The estimates are a random sample from some population. OR The estimates are independent. OR Underlying distribution is normal. 5 1(b) Population unlikely to be normal as  is close to right-angle / data B1 Must refer to the context. is skewed. OR Estimates may not be independent, for example due to collusion. OR No indication that sample is random. 1 © Cambridge University Press & Assessment 2025 Page 9 of 16

3 (x−x)2 =90.36=3.24 B1 '16.6'2 M1 Can be implied by 0.172 ( = 43 ) or 0.688 ( = 86). '55.8'− 250 125 5 3.24+0.688 M1 Correct use of formula for pooled variance using their values. (=0.302) 10+5−2 s=0.550 A1 CAO 4 Question Answer Marks Guidance

6(a) H : = H:   B1 If in words, must contain ‘population means’. 0 B A 1 B A Allow H : =0,H:  0 if defined or consistent with working. 0 d 1 d Differences: 5, 0, −1, 6,1 , −2, 8, 7, 15 M1 Attempt at signed differences. s2 =1 ( '405'−'39'2) =29.5 M1 Correct form for s2, note that s =5.43. d 8 9 d d '39' M1 Correct form. 9 =2.393 '29.5' 9 A1 AWRT 2.39. '2.393'1.860, reject H / significant. M1 Compare their 2.393 with 1.860 and appropriate result (may be in 0 terms of H ). 1 Allow M1 for comparison from 2-sample test using 1.746. Test result may be implied by an attempt at an appropriate conclusion in context ignoring hypotheses. Sufficient evidence to suggest new training programme results in A1 Correct conclusion from correct working ignoring hypotheses. reduced times. In context. OR Level of uncertainty in language is used (for example, not ‘prove’). Sufficient evidence to support the coach’s belief. Do not accept statements such as “there is insufficient evidence to suggest…”. 7 6(b) The population of differences may not be normally distributed. B1 Must mention population/distribution and differences. The population of differences may not be symmetrical (and hence Accept “the sample may not be random” or “times may not be not normal). independent”. 1 6(c) A Wilcoxon matched-pairs signed-rank test. B1 Must refer to pairs / paired. OR A paired-sample sign test. Accept “paired Wilcoxon test” or “paired sign test”. 1 © Cambridge University Press & Assessment 2025 Page 14 of 16

1(a) x=723, x2 =42261 B1 x=723 implied by AWRT 55.6 for x. x2 =42261 implied by AWRT 171 for s2(or 158 if biased). s2 = 1 ( '42261'−'723'2) = 2222 =170.9231 M1 Correct form, must be unbiased, s=13.07.   12 13 13 Implied by AWRT 171. 1.782 B1 1.782 or 1.78 seen. '723' '170.9231' M1 Correct form, must be a t-value. '1.782' 13 13 49.2, 62.1 A1 Accept with inequality signs or open brackets. Condone 62.1,49.2. Do not accept 55.66.5. 5 1(b) The population (of car speeds) is normally distributed. B1 Population mean normally distributed earns B0. OR (Underlying) data is normal earns B0. Underlying distribution is normal. It is normal distribution earns B0. 1 © Cambridge University Press & Assessment 2025 Page 9 of 17

3(a) H : = H:   B1 If in words, must have population means. 0 Y X 1 Y X Allow H : =0, H:  0 if d defined or consistent with 0 d 1 d working. Differences: 3, –1, 6, 2, 3, 3, –1, –2, 3, 3 M1 Attempt at signed differences, SOI. s2 = 1 ( '91'−'19'2) =6.1 M1 Correct form, unbiased, implied by s2 =6.1 or s =2.47 seen. d 9 10 d d '19' M1 Correct form using their s2. 10 =2.43 d '6.1' '696'−'677' 10 Allow M1 for 10 10 =0.666 from 2-sample t-test. 's ' 1 + 1 p 10 10 A1 AWRT 2.43. '2.43'2.821, accept H / do not reject H / not significant. M1 Compare their 2.43 with 2.821 and appropriate result (may be in 0 0 terms of H ). 1 Allow M1 for comparing their 0.666 with 2.552 and appropriate result from 2-sample t-test. Test result may be implied by an attempt at an appropriate conclusion in context ignoring hypotheses. Their 2.43 must come from an attempt at standardising. Insufficient evidence to support the manufacturer’s claim. A1 Correct conclusion from correct working ignoring hypotheses. OR In context. Insufficient evidence to support that (on average) machine Y Level of uncertainty in language is used (for example, not ‘prove’). extracts more juice than machine X. 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…’. 7 © Cambridge University Press & Assessment 2025 Page 12 of 17 3(b) Population of differences is normally distributed. B1 Population mean normally distributed earns B0. OR (Underlying) data is normal earns B0. Underlying distribution of differences is normal. Must mention population/distribution and differences. Following 2-sample test in part 3(a), allow either of the following: Populations/underlying distributions are normal. Populations have equal variances. Must refer to both populations. 1 3(c) The difference remains the same for pair 4. The set of differences *B1 Comment that recognises that the differences remain unchanged. has not changed. OR OR Redo the test: Compare their 2.43 from part 3(a) with 2.821. 2.432.821 Following 2-sample test in part 3(a), compare their 0.653 with 2.552. The conclusion remains the same. DB1 2 © Cambridge University Press & Assessment 2025 Page 13 of 17

3 (x−x)2 =29144−1 ( 4452) =854.7 M1 Either expression, may be embedded. 7 May be implied by (y−y)2 =11622−1 ( 3102) =944.2 9 © Cambridge University Press & Assessment 2026 Page 12 of 17 5 94 82 3 or 4 23 46 9 or 142.45 or 118.03 seen. s 2p = '8 5 4 .7 ' 7 + + 9 '9 − 4 2 4 .2 ' M1 Correct form for pooled variance. 128.495 A1 1 1 3 3 3 8 8 2 3 , SOI, may be implied by s = 1 1 .3 3 6 . (445−310) 2.145 '128.5' (1 +1) B1 B1 for 2.145 seen. 7 9 7 9 M1 Correct form, must be a t-value. FT their s. If variances combined, must use ( 4 47 5 − 3 19 0 )  2 .1 4 5 5 94 82 3 + 4 23 46 9 . 1 6 .9 , 4 1 .4  A1 Accept with inequality signs or open brackets or  − 4 1 .4 , − 1 6 .9  from Y − X . Condone  4 1 .4 , 1 6 .9  . Do not accept 2 9 .1 5  1 2 .2 5 . 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/

5(a) H : 0 © Cambridge University Press & Assessment 2026 Page 15 of 17 d 5  = , H : 1 d 5   OR H : 0 A B 5   − = , H : 1 A B 5   −  B1 If  not used, must have population means. Allow use of  in place of  d . M1 Calculates differences, at most 2 errors. d 6 24 12 8 5 26 13 2 s 2d = 1 7  '1 6 9 4 '− '9 6 8 2'   = 5 4 7 2 = 7 7 .4 3  M1 Must be unbiased. s=8.799  t =  '9 6 8'7 ' − 7 .4 8 5 3 ' M1 M0 if 5 omitted. 2.25 A1 AWRT 2.25 ‘2.25’ > 1.415, reject H / significant. M1 Compare their 2.25 with 1.415 and appropriate 0 conclusion (may be in terms of H ). Their 2.25 1 must come from an attempt at standardising. Test result can be implied by an attempt at an appropriate conclusion in context ignoring hypotheses. For 2-sample t-test allow comparison with 1.345. Sufficient evidence to support Margaret’s claim. A1 Correct conclusion from correct working, ignoring OR hypotheses. Sufficient evidence to suggest that the course will improve a student’s score by more In context. Level of uncertainty in language is than 5 marks (on average). used (for example, not ‘prove’). Do not accept statements such as “there is insufficient evidence to suggest…”. 7 Biased variance: max 4/7 (B1M1M0M1A0M1A0) 2-sample: max 2/7 (B1M0M0M0A0M1A0) 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/ 5(b) Population of differences are normally distributed. B1 B1 for underlying/parent distribution of differences is normal. B0 for population mean is normally distributed. B0 for (underlying) data is normal. 1 Question Answer Marks Guidance

3 (x−x)2 =29144−1 ( 4452) =854.7 7 © Cambridge University Press & Assessment 2026 Page 12 of 17  ( y − y ) 2 = 1 1 6 2 2 − 19 ( 3 1 0 2 )  = 9 4 4 .2  M1 Either expression, may be embedded. May be implied by 5 94 82 3 or 4 23 46 9 or 142.45 or 118.03 seen. s 2p = '8 5 4 .7 ' 7 + + 9 '9 − 4 2 4 .2 ' M1 Correct form for pooled variance. 128.495 A1 1 1 3 3 3 8 8 2 3 , SOI, may be implied by s = 1 1 .3 3 6 . ( 4 47 5 − 3 19 0 )  2 .1 4 5 '1 2 8 .5 (' 17 + 19 ) B1 B1 for 2.145 seen. M1 Correct form, must be a t-value. FT their s. If variances combined, must use ( 4 47 5 − 3 19 0 )  2 .1 4 5 5 94 82 3 + 4 23 46 9 . 1 6 .9 , 4 1 .4  A1 Accept with inequality signs or open brackets or  − 4 1 .4 , − 1 6 .9  from Y − X . Condone  4 1 .4 , 1 6 .9  . Do not accept 2 9 .1 5  1 2 .2 5 . 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/

5(a) H : 0 © Cambridge University Press & Assessment 2026 Page 15 of 17 d 5  = , H : 5 1 d OR H : 0 A B 5   − = , H : 1 A B 5   −  B1 If  not used, must have population means. Allow use of  in place of  d . M1 Calculates differences, at most 2 errors. d 6 24 12 8 5 26 13 2 s 2d = 1 7  '1 6 9 4 '− '9 6 8 2'   = 5 4 7 2 = 7 7 .4 3  M1 Must be unbiased. s=8.799  t =  '9 6 8'7 ' − 7 .4 8 5 3 ' M1 M0 if 5 omitted. 2.25 A1 AWRT 2.25 ‘2.25’ > 1.415, reject H / significant. M1 Compare their 2.25 with 1.415 and appropriate 0 conclusion (may be in terms of H ). Their 2.25 1 must come from an attempt at standardising. Test result can be implied by an attempt at an appropriate conclusion in context ignoring hypotheses. For 2-sample t-test allow comparison with 1.345. Sufficient evidence to support Margaret’s claim. A1 Correct conclusion from correct working, ignoring OR hypotheses. Sufficient evidence to suggest that the course will improve a student’s score by more In context. Level of uncertainty in language is than 5 marks (on average). used (for example, not ‘prove’). Do not accept statements such as “there is insufficient evidence to suggest…”. 7 Biased variance: max 4/7 (B1M1M0M1A0M1A0) 2-sample: max 2/7 (B1M0M0M0A0M1A0) 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/ 5(b) Population of differences are normally distributed. B1 B1 for underlying/parent distribution of differences is normal. B0 for population mean is normally distributed. B0 for (underlying) data is normal. 1 Question Answer Marks Guidance

1 © Cambridge University Press & Assessment 2026 Page 9 of 15 x = 1 9 2 9 .2 B1 OE, for example, 21.356 or 9 6 1 4 5 . Accuracy must be 3 significant figures or better. s 2 = 1 8  4 1 2 7 .6 4 − 1 9 2 9 .2 2  M1 Correct form with their values. May be implied by sight of 2.887… or 2 5 9 9 9 0 0 for s2 or 1.69934… for s. t = 1 9 2 9 .2 s 9 − 2 2 0 M1 Correct form with their values. 2.39 A1 AWRT 2.39 or 2.40. Accept ± here. '2 .3 9 '  2 .3 0 6 , reject H / significant. M1 Compare their 2.39 with 2.306 and appropriate conclusion (may 0 be in terms of H ). 1 Test result can be implied by an attempt at an appropriate conclusion in context. Their 2.39 must come from an attempt at standardising. There is sufficient evidence to support that 2 0   . A1 Correct conclusion from correct working. No contradictions. Level of uncertainty in language used (for example, not ‘prove’). Allow , or (population) mean. 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/

5(a) © Cambridge University Press & Assessment 2026 Page 13 of 15 H 0 : x y H :1 x y     =  B1 If  not used, must have population means. 1  6.632  s x 2 = 43 1.21− 44    =4.906510−3    or s 2y = 1 5 1  1 .6 5 − 8 .5 5 6 2 2   = 4 .7 2 3 4  1 0 − 3  M1 Either expression, may be embedded. Implied by 2.02310−4. s2 s2 4.906510−3 4.723410−3 M1 Correct form with their sample variances. x + y = + For pooled s, award this M1 for correct form of pooled estimate. 44 52 44 52 2 .0 2 3  1 0 − 4 A1 May be implied by s = 0 .0 1 4 2 2 . Accept 3 significant figure accuracy or better. A0 for pooled s. z = 6 .6 3 4 4 2 s x 4 4 − + 8 .5 5 2 2 s y 5 2 6 = − 0 .9 8 ( 0 ) M1 Value of z obtained using correct form with their sample variances. Order of numerator may be reversed. A1 Accept –0.977 to –0.980. May see 0.980 for opposite order of numerator. Use of pooled s scores M0 A0. ' − 0 .9 8 0 '  − 1 .2 8 2 , accept H 0 / do not reject H 0 / not significant. OR P ( Z  − 0 .9 8 0 ) = 0 .1 6 3 5  0 .1 , accept H 0 / do not reject H 0 M1 Compare their z with / not significant.  1 .2 8 2 (consistent signs) and appropriate conclusion (may be in terms of H ). 1 Accept 0.1637. Test result can be implied by a conclusion that is consistent with their − 0 .9 8 0 or 0.1635. Insufficient evidence to support Rhiannon’s claim. A1 Correct conclusion from correct working ignoring hypotheses. In context. Level of uncertainty in language used (for example, not ‘prove’). Condone ‘no sufficient evidence’. Do not accept statements such as ‘there is sufficient evidence to suggest…’. 8 Use of pooled s scores max 3/8 B1 M1 M0 A0 M0 A0 M1 A0. 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/ 5(b) M1 Accept their s (may be pooled). 6.63 8.56 s2 s2  − 1.645 x + y Use correct formula with any z-value.  44 52  44 52 May subtract the means in opposite order. B1 1.645 SOI. Allow 1.64 or 1.65. © Cambridge University Press & Assessment 2026 Page 14 of 15 [ − 0 .0 3 7 3 , 0 .0 0 9 4 7 ] A1 Accept 0.00947 to 0.0095. Accept with inequality signs or open brackets or  − 9 .4 7  1 0 − 3 , 0 .0 3 7 3   from Y − X . Condone  9 .4 7  1 0 − 3 , − 0 .0 3 7 3   . Do not accept − 0 .0 1 3 9  0 .0 2 3 4 . 3 Question Answer Marks Guidance

1 934.3 B1 Either expression, may be embedded. s2 = =23.956 x 39 Accuracy must be 3 significant figures or better. May be implied by 756.2 s2 = =12.817 y 59 17.25, © Cambridge University Press & Assessment 2026 Page 9 of 18 6 94 , 0.71875, 23 32 , 18 4 6 or 0.84779 seen. Allow B1 for correct form of pooled estimate. s 2 = '2 3 .9 4 0 5 6 ' + '1 2 .8 6 0 1 7 '[ = 0 .8 1 2 5 ] M1 Correct form using their values. M0 for pooled estimate. A1 For fully correct expression SOI. Implied by AWRT 0.812 or 0.813 or s = 0 .9 0 1 or fully correct interval. 6 7 4 .1 0 − 8 4 6 .1 0  1 .9 6 '0 .8 1 2 5  ' = 0 .2 7 5 8  1 .7 6 6 7  M1 Correct form with any z-value. Condone t = 1.98. FT their s. If pooled must use s2( 1 + 1 ) . p 40 60 B1 1.96 seen.  − 1 .4 9 , 2 .0 4  A1 CWO Accept with inequality signs or open brackets. Condone  2 .0 4 , − 1 .4 9  . Do not accept 0 .2 7 6  1 .7 6 7 . 6 Use of pooled s scores max 3/6 B1 M0 A0 M1 B1 A0 leading to [–1.39, 1.94]. 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/

6(a) H :  =, H:   B1 If in words, must have population means. 0 A B 1 A B © Cambridge University Press & Assessment 2026 Page 17 of 18  ( x − x ) 2 = '9 7 7 8 '− '2 7 8 0 2'  = 6 6 5 .5  B1 For correct sum and sum of squares for either x or y. May be implied, for example, by either 95.1 or 56.75 seen. '225'2 (y− y)2 ='6079'− =454.0 M1 Correct form for both expressions using their 9 values. May be embedded. '665.5'+'454.0' M1 Correct form of pooled variance using their s2 = p 8+9−2 values. 2 23 30 9  = 7 4 .6 3 3  A1 AWRT 74.6, implied by 2.08. ( t = ) '2 '7 7 8 4 0 .6 ' 3 − 3 '2 (' 2 9 18 5 + ' 19 ) M1 Must be in correct form using their s. 2 .0 8 4 A1 AWRT 2.08 '2.084' > 1.753, reject H / significant. M1 Compare their 0  2 .0 8 4 with  1 .7 5 3 (consistent signs) and appropriate conclusion (may be in terms of H 1 ). Their  2 .0 8 4 must come from an attempt at standardising. Test result can be implied by an attempt at an appropriate conclusion in context. 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/ 6(a) Sufficient evidence to suggest that there is a difference in the (average) percentage A1 Correct conclusion from correct working, of waste (that is recycled in City A and City B). ignoring hypotheses. In context. Level of uncertainty in language used (for example, not ‘prove’). Allow ‘(Some/enough) evidence that/to show/conclude that…’. Do not accept statements such as ‘there is insufficient evidence to suggest…’. 9 Combined s max 5/9 B1B1M1M0A0M1A0M1A0 6(b) Population of percentage of household waste recycled in each city are normal. B1 B1 for underlying/parent distributions are normal. B0 for population mean normally distributed. B0 for (population of/underlying) data/sample is normal. Allow singular, for example, population is normal. Allow use of A and B to denote population / random variable. Must not refer to differences. 1 © Cambridge University Press & Assessment 2026 Page 18 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/