Inference using normal and t-distributions — Paper 4 · A Level Mathematics - Further 9231

4 A company has two different machines, X and Y, each of which fills empty cups with coffee. The manager is investigating the volumes of coffee, x and y, measured in appropriate units, in the cups filled by machines X and Y respectively. She chooses a random sample of 50 cups filled by machine X and a random sample of 40 cups filled by machine Y. The volumes are summarised as follows. / x = 15.2 / x 2 = 5.1 / y = 13.4 / y 2 = 4.8 The manager claims that there is no difference between the mean volume of coffee in cups filled by machine X and the mean volume of coffee in cups filled by machine Y. Test the manager’s claim at the 10% significance level. [9] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 4 H0: μ μ = x y H1: μ μ ≠ x y B1 sx2 = 2 1 15.2 5.1 49 50   −     = 0.0097796: sy2 = 2 1 13.4 4.8 0.007974 39 40   − =     M1A1 2 0.00977959 0.007974 0.0003949 50 40 = + = s M1A1 0.304 0.335 0.0003949 − = z = ( )1.56 − M1A1 Compare with 1.645 M1 Accept H0: insufficient evidence to reject manager’s claim A1 9

This question in 9231/41 May/June 2020

5 A large number of children are competing in a throwing competition. The distances, in metres, thrown by a random sample of 8 children are as follows. 19.8 22.1 24.4 21.5 20.8 26.3 23.7 25.0 (a) Assuming that distances are normally distributed, test, at the 5% significance level, whether the population mean distance thrown is more than 22.0 metres. [7] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … (b) Find a 95% confidence interval for the population mean distance thrown. [3] … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 5(a) 2 183.6, 4249.08,  =  = x x 22.95 = x B1 2 2 1 183.6 4249.08 5.066 7 8   = − =     s M1 H0: 22.0, μ = H1: 22.0 μ > B1 2 22.95 22.0 1.194 8 − = = t s M1A1 Compare t with correct tabular value 1.895 M1 Accept H0: mean distance thrown is not more than 22.0 m A1 7 5(b) 2 22.95 8 ± s t M1 With t = 2.365 B1 [21.1, 24.8] A1 3

This question in 9231/41 May/June 2020

4 A company has two different machines, X and Y, each of which fills empty cups with coffee. The manager is investigating the volumes of coffee, x and y, measured in appropriate units, in the cups filled by machines X and Y respectively. She chooses a random sample of 50 cups filled by machine X and a random sample of 40 cups filled by machine Y. The volumes are summarised as follows. / x = 15.2 / x 2 = 5.1 / y = 13.4 / y 2 = 4.8 The manager claims that there is no difference between the mean volume of coffee in cups filled by machine X and the mean volume of coffee in cups filled by machine Y. Test the manager’s claim at the 10% significance level. [9] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 4 H0: μ μ = x y H1: μ μ ≠ x y B1 sx2 = 2 1 15.2 5.1 49 50   −     = 0.0097796: sy2 = 2 1 13.4 4.8 0.007974 39 40   − =     M1A1 2 0.00977959 0.007974 0.0003949 50 40 = + = s M1A1 0.304 0.335 0.0003949 − = z = ( )1.56 − M1A1 Compare with 1.645 M1 Accept H0: insufficient evidence to reject manager’s claim A1 9

This question in 9231/42 May/June 2020

5 A large number of children are competing in a throwing competition. The distances, in metres, thrown by a random sample of 8 children are as follows. 19.8 22.1 24.4 21.5 20.8 26.3 23.7 25.0 (a) Assuming that distances are normally distributed, test, at the 5% significance level, whether the population mean distance thrown is more than 22.0 metres. [7] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … (b) Find a 95% confidence interval for the population mean distance thrown. [3] … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 5(a) 2 183.6, 4249.08,  =  = x x 22.95 = x B1 2 2 1 183.6 4249.08 5.066 7 8   = − =     s M1 H0: 22.0, μ = H1: 22.0 μ > B1 2 22.95 22.0 1.194 8 − = = t s M1A1 Compare t with correct tabular value 1.895 M1 Accept H0: mean distance thrown is not more than 22.0 m A1 7 5(b) 2 22.95 8 ± s t M1 With t = 2.365 B1 [21.1, 24.8] A1 3

This question in 9231/42 May/June 2020

2 A random sample of 40 observations of a random variable X and a random sample of 50 observations of a random variable Y are taken. The resulting values for the sample means, x and y , and the unbiased estimates, sx2 and s 2,y for the population variances are as follows. x = 24 .4 y = 17 .2 s 2x 2 = 10 .2 s y = 11 .1 Find a 90% confidence interval for the difference between the population means of X and Y. [5] … … … … … … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 2 2 10.2 11.1 0.477 40 50 = + = s M1A1 CI = ( ) 24.4 17.2 − ± zs M1 = ( ) 24.4 17.2 1.645 0.477 − ± A1 = [6.06, 8.34] A1 5

This question in 9231/43 May/June 2020

5 Students at two colleges, A and B, are competing in a computer games challenge. (a) The time taken for a randomly chosen student from college A to complete the challenge has a normal distribution with mean n minutes. The times taken, x minutes, are recorded for a random sample of 10 students chosen from college A. The results are summarised as follows. / x = 828 / x 2 = 68622 A test is carried out on the data at the 5% significance level and the result supports the claim that n 2 k . Find the greatest possible value of k. [4] … … … … … … … … … … … (b) A random sample of 8 students is chosen from college B. Their times to complete the same challenge give a sample mean of 79.8 minutes and an unbiased variance estimate of 9.966 minutes2. Use a 2-sample test at the 5% significance level to test whether the mean time for students at college B to complete the challenge is the same as the mean time for students at college A to complete the challenge. You should assume that the two distributions are normal and have the same population variance. [7] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

11 marks

Mark scheme: 5(a) 2 2 828 1 828 82.8 68622 7.0667 10 9 10   = = = − =     x s 2 82.8 10 −k t s  where t = 1.833 (M1 if equality, A1 for inequality and correct t value) M1 A1 81.259 81.3 k k   A1 4 5(b) H0: μ μ = A B H1: μ μ ≠ A B B1 Pooled variance = 9 7.0667 7 9.966 10 8 2 × + × + − = 2 ps M1 = 8.335 A1 82.8 79.8 1 1 10 8 − = + p t s = 2.19 M1A1 Compare with 2.12 (t(16, 0.975)) and reject H0 M1 Population means are not the same A1 7

This question in 9231/43 May/June 2020

1 Kayla is investigating the lengths of the leaves of a certain type of tree found in two forests X and Y. She chooses a random sample of 40 leaves of this type from forest X and records their lengths, x cm. She also records the lengths, y cm, for a random sample of 60 leaves of this type from forest Y. Her results are summarised as follows. / x = 242.0 / x 2 = 1587.0 / y = 373.2 / y 2 = 2532.6 Find a 90% confidence interval for the difference between the population mean lengths of leaves in forests X and Y. [7] … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 1 2 2 1 242.0 1587.0 3.15128 39 40 xs   = − =     1229 390   =     2 2 1 373.2 2532.6 3.58129 59 60 ys   = − =     26412 7375   =     Both correct 2 3.15128 3.58129 0.13847 40 60 s = + = M1 A1 Pooled variance is M0A0 or 0.3721 s = 6.05 6.22 zs − ± M1 FT their s, must be a z value = 0.17 1.645 0.13847 ± A1 With 1.645 = [− 0.442, 0.782] or [− 0.782, 0.442] A1 7

This question in 9231/41 Oct/Nov 2020

4 Members of the Sprints athletics club have been taking part in an intense training scheme, aimed at reducing their times taken to run 400 m. For a random sample of 9 athletes from the club, the times taken, in seconds, before and after the training scheme are given in the following table. Athlete A B C D E F G H I Time before 48.8 48.2 50.3 49.6 49.4 48.9 47.6 50.3 48.4 Time after 47.9 47.8 49.6 49.1 49.6 48.9 47.7 49.1 48.1 The organiser of the training scheme claims that on average an athlete’s time will be reduced by at least 0.3 seconds. Test at the 10% significance level whether the organiser’s claim is justified, stating any assumption that you make. [8] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 4 Assume (population) differences are normally distributed B1 H0: 0.3 X Y μ μ − = H1: 0.3 X Y μ μ − > B1 Diff: 0.9 0.4 0.7 0.5 0.2 0 0.1 1 .2 0.3 − − M1 Signed differences 2 3.7, 3.29 d d  =  = 2 2 1 3.7 0.411, 3.29 0.2211 8 9 d s   = = − =     M1 0.4702 s = 0.411 0.3 0.2211 9 t − = 0.708 = or 0.709 M1 A1 ‘0.708’ < 1.397 M1 Compare their 0.708 with 1.397 Accept H0 Insufficient evidence to support claim A1 FT In context, except possibly hypotheses Level of uncertainty in language used. No contradictions 8

This question in 9231/41 Oct/Nov 2020

1 The heights of the members of a large sports club are normally distributed. A random sample of 11 members of the club is chosen and their heights, x cm, are measured. The results are summarised as follows, where x denotes the sample mean of x. x = 176.2 / ( x - x ) 2 = 313. 1 Test, at the 5% significance level, the null hypothesis that the population mean height for members of this club is equal to 172.5 cm against the alternative hypothesis that the mean differs from 172.5 cm. [5] … … … … … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 1 2 313.1 31.3 10 s = = B1 Can be implied 176.2 172.5 / 11 t s − = M1 = 2.19 A1 ‘2.19’ < 2.228 M1 Comparison of their t value with 2.228 accept H0: insufficient evidence to reject mean height is 172.5 OR sufficient evidence to accept mean height is 172.5 A1 FT Correct conclusion in context, level of uncertainty in language used. No contradictions. CWO Do not accept: ‘mean height is 172.5’ Do not accept use of symbols that are undefined e.g. insufficient evidence to reject μ = 172.5 5

This question in 9231/42 Oct/Nov 2020

6 Nassa is researching the lengths of a particular type of snake in two countries, A and B. (a) He takes a random sample of 10 snakes of this type from country A and measures the length, x m, of each snake. He then calculates a 90% confidence interval for the population mean length, n m, for snakes of this type, assuming that snake lengths have a normal distribution. This confidence interval is 3.36 G n G 4.22. Find the sample mean and an unbiased estimate for the population variance. [4] … … … … … … … … … … … … … (b) Nassa also measures the lengths, y m, of a random sample of 8 snakes of this type taken from country B. His results are summarised as follows. / y = 27.86 / y 2 = 98.02 Nassa claims that the mean length of snakes of this type in country B is less than the mean length of snakes of this type in country A. Nassa assumes that his sample from country B also comes from a normal distribution, with the same variance as the distribution from country A. Test at the 10% significance level whether there is evidence to support Nassa’s claim. [8] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … Additional Page If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown. … … … … … … … … … … … … … … … … … … … … … …

12 marks

Mark scheme: 6(a) ( ) 1 4.22 3.36 2 x = + = 3.79 B1 2 4.22 3.36 10 ts − = M1 Using z value implies M0 With t = 1.833 A1 ( 0.7418) s = variance = 0.55(0) A1 4 6(b) H0: A B μ μ = H1: A B μ μ > B1 Not x ( ) 2 2 1 27.86 98.02 0.142 507 7 8 s   = − =     M1 = ( ) 1 98.02 97.02 7 − Pooled estimate = 9 0.550316 7 0.142507 16 × + × M1 0.372 = A1 ( ) 27.86 3.79 8 1.063 1 1 0.372 10 8 t − = = − + M1A1 ‘1.063’ < 1.337 oe M1 Compare their value with 1.337 Accept H0 Nassa’s claim is not supported A1ft Correct conclusion in context, level of uncertainty in language used. No contradictions. 8

This question in 9231/42 Oct/Nov 2020

1 Kayla is investigating the lengths of the leaves of a certain type of tree found in two forests X and Y. She chooses a random sample of 40 leaves of this type from forest X and records their lengths, x cm. She also records the lengths, y cm, for a random sample of 60 leaves of this type from forest Y. Her results are summarised as follows. / x = 242.0 / x 2 = 1587.0 / y = 373.2 / y 2 = 2532.6 Find a 90% confidence interval for the difference between the population mean lengths of leaves in forests X and Y. [7] … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 1 2 2 1 242.0 1587.0 3.15128 39 40 xs   = − =     1229 390   =     2 2 1 373.2 2532.6 3.58129 59 60 ys   = − =     26412 7375   =     Both correct 2 3.15128 3.58129 0.13847 40 60 s = + = M1 A1 Pooled variance is M0A0 or 0.3721 s = 6.05 6.22 zs − ± M1 FT their s, must be a z value = 0.17 1.645 0.13847 ± A1 With 1.645 = [− 0.442, 0.782] or [− 0.782, 0.442] A1 7

This question in 9231/43 Oct/Nov 2020

4 Members of the Sprints athletics club have been taking part in an intense training scheme, aimed at reducing their times taken to run 400 m. For a random sample of 9 athletes from the club, the times taken, in seconds, before and after the training scheme are given in the following table. Athlete A B C D E F G H I Time before 48.8 48.2 50.3 49.6 49.4 48.9 47.6 50.3 48.4 Time after 47.9 47.8 49.6 49.1 49.6 48.9 47.7 49.1 48.1 The organiser of the training scheme claims that on average an athlete’s time will be reduced by at least 0.3 seconds. Test at the 10% significance level whether the organiser’s claim is justified, stating any assumption that you make. [8] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 4 Assume (population) differences are normally distributed B1 H0: 0.3 X Y μ μ − = H1: 0.3 X Y μ μ − > B1 Diff: 0.9 0.4 0.7 0.5 0.2 0 0.1 1 .2 0.3 − − M1 Signed differences 2 3.7, 3.29 d d  =  = 2 2 1 3.7 0.411, 3.29 0.2211 8 9 d s   = = − =     M1 0.4702 s = 0.411 0.3 0.2211 9 t − = 0.708 = or 0.709 M1 A1 ‘0.708’ < 1.397 M1 Compare their 0.708 with 1.397 Accept H0 Insufficient evidence to support claim A1 FT In context, except possibly hypotheses Level of uncertainty in language used. No contradictions 8

This question in 9231/43 Oct/Nov 2020

1 A random sample of 7 observations of a variable X are as follows. 8.26 7.78 7.92 8.04 8.27 7.95 8.34 The population mean of X is n. (a) Test, at the 10% significance level, the null hypothesis n = .822 against the alternative hypothesis n 1 8.22. [6] … … … … … … … … … … … … … … … … … … (b) State an assumption necessary for the test in part (a) to be valid. [1] … … … … …

7 marks

Mark scheme: 1(a) 56.56 8.08 7 x = = B1 Accept unsimplified. 2 2 1 56.56 457.275 6 7 s   = −     = 0.045033 1351 30000       = M1 Accept unsimplified. 2 8.08 8.22 7 t s − = = − 1.745 M1 A1 1.74 – 1.75 Tabular value 1.440 1.745 > 1.440 , so reject 0 H M1 Comparison with 1.440 and correct FT conclusion. There is insufficient evidence to support the hypothesis that the (population) mean is 8.22 OR sufficient evidence to suggest mean is less than 8.22 OR sufficient evidence to suggest 8.22 μ < OE A1 Correct conclusion, in context, following correct work. Level of uncertainty in language is used. 6 1(b) Underlying distribution is normal/ population is normal B1 X is normal distributed. 1

This question in 9231/41 May/June 2021

4 A scientist is investigating the lengths of the leaves of birch trees in different regions. He takes a random sample of 50 leaves from birch trees in region A and a random sample of 60 leaves from birch trees in region B. He records their lengths in cm, x and y, respectively. His results are summarised as follows. / x = 282 / x 2 = 1596 / y = 328 / y 2 = 1808 n The population mean lengths of leaves from birch trees in regions A and B are n A cm and B cm respectively. Carry out a test at the 5% significance level to test the null hypothesis n A = n B against the alternative hypothesis n A ! n B . [8] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 4 2 2 1 282 1596 49 50 xs   = −     = 0.11265 138 1225       = M1 One correct, unsimplified. 2 2 1 328 1808 59 60 ys   = −     = 0.25311 224 885   =     A1 Both correct to 3sf. 2 0.11265 0.25312 0.006471 50 60 s = + = … M1 A1 May be implied. 282 328 50 60 their z s − = M1 [z] = 2.155 A1 Tolerance 2.15 – 2.16 Comparison with tabular value 1.96 OE 2.155 > 1.96 Reject 0 H M1 Comparison with 1.96 OE and conclusion. Sufficient evidence to reject population means are equal OR sufficient evidence to accept population means are not equal OR sufficient evidence to suggest A B μ μ ≠ OE A1 Correct conclusion following correct work. Level of uncertainty in language is used. 8

This question in 9231/41 May/June 2021

1 A random sample of 7 observations of a variable X are as follows. 8.26 7.78 7.92 8.04 8.27 7.95 8.34 The population mean of X is n. (a) Test, at the 10% significance level, the null hypothesis n = .822 against the alternative hypothesis n 1 8.22. [6] … … … … … … … … … … … … … … … … … … (b) State an assumption necessary for the test in part (a) to be valid. [1] … … … … …

7 marks

Mark scheme: 1(a) 56.56 8.08 7 x = = B1 Accept unsimplified. 2 2 1 56.56 457.275 6 7 s   = −     = 0.045033 1351 30000       = M1 Accept unsimplified. 2 8.08 8.22 7 t s − = = − 1.745 M1 A1 1.74 – 1.75 Tabular value 1.440 1.745 > 1.440 , so reject 0 H M1 Comparison with 1.440 and correct FT conclusion. There is insufficient evidence to support the hypothesis that the (population) mean is 8.22 OR sufficient evidence to suggest mean is less than 8.22 OR sufficient evidence to suggest 8.22 μ < OE A1 Correct conclusion, in context, following correct work. Level of uncertainty in language is used. 6 1(b) Underlying distribution is normal/ population is normal B1 X is normal distributed. 1

This question in 9231/42 May/June 2021

4 A scientist is investigating the lengths of the leaves of birch trees in different regions. He takes a random sample of 50 leaves from birch trees in region A and a random sample of 60 leaves from birch trees in region B. He records their lengths in cm, x and y, respectively. His results are summarised as follows. / x = 282 / x 2 = 1596 / y = 328 / y 2 = 1808 n The population mean lengths of leaves from birch trees in regions A and B are n A cm and B cm respectively. Carry out a test at the 5% significance level to test the null hypothesis n A = n B against the alternative hypothesis n A ! n B . [8] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 4 2 2 1 282 1596 49 50 xs   = −     = 0.11265 138 1225       = M1 One correct, unsimplified. 2 2 1 328 1808 59 60 ys   = −     = 0.25311 224 885   =     A1 Both correct to 3sf. 2 0.11265 0.25312 0.006471 50 60 s = + = … M1 A1 May be implied. 282 328 50 60 their z s − = M1 [z] = 2.155 A1 Tolerance 2.15 – 2.16 Comparison with tabular value 1.96 OE 2.155 > 1.96 Reject 0 H M1 Comparison with 1.96 OE and conclusion. Sufficient evidence to reject population means are equal OR sufficient evidence to accept population means are not equal OR sufficient evidence to suggest A B μ μ ≠ OE A1 Correct conclusion following correct work. Level of uncertainty in language is used. 8

This question in 9231/42 May/June 2021

1 Farmer A grows apples of a certain variety. Each tree produces 14.8 kg of apples, on average, per year. Farmer B grows apples of the same variety and claims that his apple trees produce a higher mass of apples per year than Farmer A’s trees. The masses of apples from Farmer B’s trees may be assumed to be normally distributed. A random sample of 10 trees from Farmer B is chosen. The masses, x kg, of apples produced in a year are summarised as follows. / x = 152.0 / x 2 = 2313.0 Test, at the 5% significance level, whether Farmer B’s claim is justified. [6] … … … … … … … … … … … … … … … … … … … … … … …

6 marks

Mark scheme: 1 2 2 1 152 2313 0.28889 9 10 s   = − =     M1 13 45 , accept unsimplified. H0 : μ = 14.8 H1 : μ > 14.8 B1 If μ not used, ‘population mean’ required. Must see 14.8, must be = and >. [ ] 152 14.8 10 / 10 t s − = M1 Using unbiased estimate. p-value is 0.0215. [ ] t = 2.35 A1 Rounds to 2.35. Tabular value = 1.833 ‘2.35’ > 1.833 Reject 0 H M1 Comparison with 1.833 and correct FT conclusion. 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

This question in 9231/43 May/June 2021

3 The heights, x m, of a random sample of 50 adult males from country A were recorded. The heights, y m, of a random sample of 40 adult males from country B were also recorded. The results are summarised as follows. / x = 89.0 / x 2 = 159.4 / y = 67.2 / y 2 = 113. 1 Find a 95% confidence interval for the difference between the mean heights of adult males from country A and adult males from country B. [8] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 3 89 67.2 1.78 1.68 50 40 x y = = = = B1 Implied by 0.1 in the CI formula. 2 2 1 89 159.4 49 50 x σ   = −     = 0.02 1 50   =     M1 One variance correct unsimplified. 2 2 1 67.2 113.1 39 40 y σ   = −     = 0.0052308 17 3250   =     A1 Both correctly calculated. 2 0.02 0.0052308 50 40 σ = + = 69 130000       M1 Unsimplified expression must be seen, 40 and 50 in correct places. [ 2 σ =] 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.

This question in 9231/43 May/June 2021

1 The times taken for students at a college to run 200 m have a normal distribution with mean n s. The times, x s, are recorded for a random sample of 10 students from the college. The results are summarised as follows, where x is the sample mean. x = 25. 6 / ( x - x ) 2 = 78. 5 (a) Find a 90% confidence interval for n. [4] … … … … … … … … … … … A test of the null hypothesis n = k is carried out on this sample, using a 10% significance level. The test does not support the alternative hypothesis n 1 k . (b) Find the greatest possible value of k. [3] … … … … … … … … … … …

7 marks

Mark scheme: 1(a) 2 78.5 8.7222 9 s = = 157 18 CI: 2 25.6 10 s t ± M1 Correct expression with a t value. 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) 2 25.6 10 k t s −  M1 With t = −1.383 A1 Allow 1.383. 26.9 A1 CAO 3

This question in 9231/41 Oct/Nov 2021

4 Manet has developed a new training course to help athletes improve their time taken to run 800 m. Manet claims that his course will decrease an athlete’s time by more than 2 s on average. For a random sample of 10 athletes the times taken, in seconds, before and after the course are given in the following table. Athlete A B C D E F G H I J Before 150 146 131 135 126 142 130 129 137 134 After 145 138 129 135 122 135 132 128 127 137 Use a t-test, at the 5% significance level, to test whether Manet’s claim is justified, stating any assumption that you make. [8] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 4 0 1 : 2 and : 2 B A B A H H μ μ μ μ − = − > B1 Or use of d μ . Differences: 5 8 2 0 4 7 −2 1 10 −3 M1 Allow one error. 2 32, 272 d d  =  = 2 2 1 32 3.2, 272 18.84 9 10 d s   = = − =     848 45   =     M1 Sample mean and variance. 2 3.2 2 0.874 10 t s − = = M1 A1 Compare with tabular value 1.833: 0.874 < 1.833. Accept 0 H . M1 Compare their value with 1.833 and conclusion. 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

This question in 9231/41 Oct/Nov 2021

1 The number, x, of pine trees was counted in each of 40 randomly chosen regions of equal size in country A. The number, y, of pine trees was counted in each of 60 randomly chosen regions of the same equal size in country B. The results are summarised as follows. / x = 752 / x 2 = 14320 / y = 1548 / y 2 = 40200 Find a 95% confidence interval for the difference between the mean number of pine trees in regions of this size in countries A and B. [7] … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

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

This question in 9231/42 Oct/Nov 2021

2 It is claimed that the heights of a particular age group of boys follow a normal distribution with mean 125 cm and standard deviation 12 cm. Observations for a randomly chosen group of 60 boys in this age group are summarised in the following table. The table also gives the expected frequencies, correct to 2 decimal places, based on the normal distribution with mean 125 cm and standard deviation 12 cm. Height, x 1 100 100 G x 1 110 110 G x 1 120 120 G x 1 130 130 G x 1 140 x H 140 x cm Observed 0 3 15 23 11 8 frequency Expected 1.12 5.22 13.97 19.38 13.97 6.34 frequency (a) Show how the expected frequency for 130 G x 1 140 is obtained. [2] … … … … … … … … … … (b) Carry out a goodness of fit test, at the 5% significance level, to determine whether the claim is supported by the data. [6] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

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

This question in 9231/42 Oct/Nov 2021

4 Applicants for a particular college take a written test when they attend for interview. There are two different written tests, A and B, and each applicant takes one or the other. The interviewer wants to determine whether the medians of the distribution of marks obtained in the two tests are equal. The marks obtained by a random sample of 8 applicants who took test A and a random sample of 8 applicants who took test B are as follows. Test A 46 32 29 12 33 18 25 40 Test B 36 28 49 37 48 35 41 31 (a) Carry out a Wilcoxon rank-sum test at the 5% significance level to determine whether there is a difference in the population median marks obtained in the two tests. [6] … … … … … … … … … … … … … … … … … … … … … … The interviewer considers using the given information to carry out a paired sample t-test to determine whether there is a difference in the population means for the two tests. (b) Give two reasons why it is not appropriate to use this test. [2] … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

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

This question in 9231/42 Oct/Nov 2021

6 A scientist is investigating the masses of a particular type of fish found in lakes A and B. He chooses a random sample of 10 fish of this type from lake A and records their masses, x kg, as follows. 2.1 1.8 0.9 3.0 2.4 2.6 1.8 2.2 1.9 2.5 The scientist also chooses a random sample of 12 fish of this type from lake B, but he only has a summary of their masses, y kg, as follows. / y = 24.48 / y 2 = 53 .75 Test at the 10% significance level whether the mean mass of fish of this type in lake A is greater than the mean mass of fish of this type in lake B. You should state any assumptions that you need to make for the test to be valid. [10] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … Additional Page If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown. … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 6 0 1 : : and A B A B H H μ μ μ μ = > B1 2 21.2 47.92  =  = x x 2 2 1 21.2 47.92 0.33067 9 10   = − =     xs 124 375 = 2 2 1 24.48 53.75 0.34644 11 12   = − =     ys 9527 27500 = M1 A1 2 9 0.33067 11 0.34644 10 12 2 × + × = + − s = 0.3393 M1 A1 2.12 2.04 0.321 1 1 10 12 − = = + t s M1 A1 Accept 0.321 – 0.322. Critical value: 1.325 0.321 < 1.325 accept 0 H . M1 Compare their value with 1.325 and correct FT conclusion. 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 OR: Distributions are normal B1 Assumptions consistent with method used. Accept ‘population is normal’. ‘Distribution of difference of means is normal’. Alternative method for question 6 0 1 : : and A B A B H H μ μ μ μ = > B1 Question Answer Marks Guidance 2 21.2 47.92  =  = x x 2 2 1 21.2 47.92 0.33067 9 10   = − =     xs 124 375 = 2 2 1 24.48 53.75 0.34644 11 12   = − =     ys 9527 27500 = M1 For one correct unsimplified. A1 For both correct. 2 0.33067 0.34644 10 12 = + s = 0.061936 M1 A1 ( ) 2.12 2.04 0.321 5 − = = t s M1 A1 Accept 0.321 – 0.322. Critical value: 1.325 0.321 < 1.325 accept 0 H . M1 Compare their value with 1.325 and correct FT conclusion. 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 OR: Distributions are normal. B1 Assumptions consistent with method used. Accept ‘population is normal’. ‘Distribution of difference of means is normal’. 10

This question in 9231/42 Oct/Nov 2021

1 The times taken for students at a college to run 200 m have a normal distribution with mean n s. The times, x s, are recorded for a random sample of 10 students from the college. The results are summarised as follows, where x is the sample mean. x = 25. 6 / ( x - x ) 2 = 78. 5 (a) Find a 90% confidence interval for n. [4] … … … … … … … … … … … A test of the null hypothesis n = k is carried out on this sample, using a 10% significance level. The test does not support the alternative hypothesis n 1 k . (b) Find the greatest possible value of k. [3] … … … … … … … … … … …

7 marks

Mark scheme: 1(a) 2 78.5 8.7222 9 s = = 157 18 CI: 2 25.6 10 s t ± M1 Correct expression with a t value. 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) 2 25.6 10 k t s −  M1 With t = −1.383 A1 Allow 1.383. 26.9 A1 CAO 3

This question in 9231/43 Oct/Nov 2021

4 Manet has developed a new training course to help athletes improve their time taken to run 800 m. Manet claims that his course will decrease an athlete’s time by more than 2 s on average. For a random sample of 10 athletes the times taken, in seconds, before and after the course are given in the following table. Athlete A B C D E F G H I J Before 150 146 131 135 126 142 130 129 137 134 After 145 138 129 135 122 135 132 128 127 137 Use a t-test, at the 5% significance level, to test whether Manet’s claim is justified, stating any assumption that you make. [8] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 4 0 1 : 2 and : 2 B A B A H H μ μ μ μ − = − > B1 Or use of d μ . Differences: 5 8 2 0 4 7 −2 1 10 −3 M1 Allow one error. 2 32, 272 d d  =  = 2 2 1 32 3.2, 272 18.84 9 10 d s   = = − =     848 45   =     M1 Sample mean and variance. 2 3.2 2 0.874 10 t s − = = M1 A1 Compare with tabular value 1.833: 0.874 < 1.833. Accept 0 H . M1 Compare their value with 1.833 and conclusion. 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

This question in 9231/43 Oct/Nov 2021

1 A manager is investigating the times taken by employees to complete a particular task as a result of the introduction of new technology. He claims that the mean time taken to complete the task is reduced by more than 0.4 minutes. He chooses a random sample of 10 employees. The times taken, in minutes, before and after the introduction of the new technology are recorded in the table. Employee A B C D E F G H I J Time before 10.2 9.8 12.4 11.6 10.8 11.2 14.6 10.6 12.3 11.0 new technology Time after 9.6 8.5 12.4 10.9 10.2 10.6 12.8 10.8 12.5 10.6 new technology (a) Test at the 10% significance level whether the manager’s claim is justified. [7] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … (b) State an assumption that is necessary for this test to be valid. [1] … … … … …

8 marks

Mark scheme: 1(a) Differences: 0.6 1.3 0.0 0.7 0.6 0.6 1.8 –0.2 –0.2 0.4 M1 Attempt at differences, allow one error. 2 5.6 6.74 d d      2 2 1 5.6 6.74 0.400 4 9 10 s          M1 H0: 0.4 B A     and H1: 0.4 B A     B1 Allow use of d  0.56 0.4 0.4004 10 t   (0.800) M1 t =0.7996 (0.800) A1 Critical value = t0.90(9) = 1.383 Compare: ‘0.7996’ < 1.383 Accept H0 M1 Compare calculated value with 1.383 and correct FT conclusion. 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

This question in 9231/41 May/June 2022

5 Raman is researching the heights of male giraffes in a particular region. Raman assumes that the heights of male giraffes in this region are normally distributed. He takes a random sample of 8 male giraffes from the region and measures the height, in metres, of each giraffe. These heights are as follows. 5.2 5.8 4.9 6.1 5.5 5.9 5.4 5.6 (a) Find a 90% confidence interval for the population mean height of male giraffes in this region. [5] … … … … … … … … … … … … … … … … … … … … … … … … Raman claims that the population mean height of male giraffes in the region is less than 5.9 metres. (b) Test at the 2.5% significance level whether this sample provides sufficient evidence to support Raman’s claim. [4] … … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 5(a) 2 44.4, 247.48 x x     2 2 1 44.4 53 247.48 0.1514 7 8 350 s                 M1 CI: 44.4 0.1514 1.895 8 8  M1 Correct formula with a t value. 1.895 B1 1.895 used.   5.55 0.261 5.29, 5.81   A1 Either form. 5 5(b) 0 1 : 5.9 and : 5.9 H H     B1 2 5.55 5.9 2.544 8 t s    M1 Critical value t0.975(7) = 2.365 Compare ‘– 2.544’ < – 2.365, reject H0. M1 Compare calculated value with 2.365 and correct FT conclusion. Sufficient evidence to support Raman’s claim. A1 Correct conclusion, in context. Level of uncertainty in language is used. 4

This question in 9231/41 May/June 2022

1 A manager is investigating the times taken by employees to complete a particular task as a result of the introduction of new technology. He claims that the mean time taken to complete the task is reduced by more than 0.4 minutes. He chooses a random sample of 10 employees. The times taken, in minutes, before and after the introduction of the new technology are recorded in the table. Employee A B C D E F G H I J Time before 10.2 9.8 12.4 11.6 10.8 11.2 14.6 10.6 12.3 11.0 new technology Time after 9.6 8.5 12.4 10.9 10.2 10.6 12.8 10.8 12.5 10.6 new technology (a) Test at the 10% significance level whether the manager’s claim is justified. [7] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … (b) State an assumption that is necessary for this test to be valid. [1] … … … … …

8 marks

Mark scheme: 1(a) Differences: 0.6 1.3 0.0 0.7 0.6 0.6 1.8 –0.2 –0.2 0.4 M1 Attempt at differences, allow one error. 2 5.6 6.74 d d      2 2 1 5.6 6.74 0.400 4 9 10 s          M1 H0: 0.4 B A     and H1: 0.4 B A     B1 Allow use of d  0.56 0.4 0.4004 10 t   (0.800) M1 t =0.7996 (0.800) A1 Critical value = t0.90(9) = 1.383 Compare: ‘0.7996’ < 1.383 Accept H0 M1 Compare calculated value with 1.383 and correct FT conclusion. 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

This question in 9231/42 May/June 2022

5 Raman is researching the heights of male giraffes in a particular region. Raman assumes that the heights of male giraffes in this region are normally distributed. He takes a random sample of 8 male giraffes from the region and measures the height, in metres, of each giraffe. These heights are as follows. 5.2 5.8 4.9 6.1 5.5 5.9 5.4 5.6 (a) Find a 90% confidence interval for the population mean height of male giraffes in this region. [5] … … … … … … … … … … … … … … … … … … … … … … … … Raman claims that the population mean height of male giraffes in the region is less than 5.9 metres. (b) Test at the 2.5% significance level whether this sample provides sufficient evidence to support Raman’s claim. [4] … … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 5(a) 2 44.4, 247.48 x x     2 2 1 44.4 53 247.48 0.1514 7 8 350 s                 M1 CI: 44.4 0.1514 1.895 8 8  M1 Correct formula with a t value. 1.895 B1 1.895 used.   5.55 0.261 5.29, 5.81   A1 Either form. 5 5(b) 0 1 : 5.9 and : 5.9 H H     B1 2 5.55 5.9 2.544 8 t s    M1 Critical value t0.975(7) = 2.365 Compare ‘– 2.544’ < – 2.365, reject H0. M1 Compare calculated value with 2.365 and correct FT conclusion. Sufficient evidence to support Raman’s claim. A1 Correct conclusion, in context. Level of uncertainty in language is used. 4

This question in 9231/42 May/June 2022

1 The times taken by members of a large quiz club to complete a challenge have a normal distribution with mean n minutes. The times, x minutes, are recorded for a random sample of 8 members of the club. The results are summarised as follows, where x is the sample mean. x = 33 .8 / ( x - x ) 2 = 94. 5 Find a 95% confidence interval for n. [4] … … … … … … … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: 1   2 94.5 13.5 7   s CI: 2 33.8 2.365 8  s M1 Correct formula with a t value, must be 33.8. B1 2.365 used.   33.8 3.072 30.7, 36.9   A1 Either form. 4

This question in 9231/43 May/June 2022

6 A company has two machines, A and B, which independently fill small bottles with a liquid. The volumes of liquid per bottle, in suitable units, filled by machines A and B are denoted by x and y respectively. A scientist at the company takes a random sample of 40 bottles filled by machine A and a random sample of 50 bottles filled by machine B. The results are summarised as follows. / x = 1120 / x 2 = 31400 / y = 1370 / y 2 = 37600 The population means of the volumes of liquid in the bottles filled by machines A and B are denoted by n A and n B . n (a) Test at the 2% significance level whether there is any difference between n A and B . [8] … … … … … … … … … … … … … … … … … … … … … … (b) Find the set of values of a for which there would be evidence at the a% significance level that n - n A B is greater than 0.25. [4] … … … … … … … … … … … … … … … … … … … … … … … … … … … Additional page If you use the following page to complete the answer to any question, the question number must be clearly shown. … … … … … … … … … … … … … … … … … … … … … …

12 marks

Mark scheme: 6(a) 2 2 1 1120 31400 1.0256 39 40          xs 2 2 1 1370 37600 1.2653 49 50          ys 40 39 62 49 Both required 2 1.0256 1.2653 40 50 s   M1 2434 47775 Condone 2 1.03 1.27 or 0.05115 40 50 s   . 2 0.050947 s  A1 May be implied by 2.66 as z value. 0 1 : 0 and : 0 a b a b H H         B1 1120 1370 0.6 40 50 0.050947    z s M1 z = 2.66 A1 Critical value of z = 2.326 2.66 > 2.326 Reject H0 M1 Compare their z value with 2.326 and correct FT conclusion. Sufficient evidence to support difference in population means oe A1 Correct conclusion, in context. Level of uncertainty in language is used. 8 Question Answer Marks Guidance 6(b) 0.6 0.25 z s   M1 Condone 1.54. 1 .55 z  A1 0.9392 – 0.9395 or 0.0605 – 0.0608 A1 6.05  A1 Condone 6.05 – 6.08, accept  4

This question in 9231/43 May/June 2022

1 Jasmine is researching the heights of pine trees in forests in two regions A and B. She chooses a random sample of 50 pine trees in region A and records their heights, x m. She also chooses a random sample of 60 pine trees in region B and records their heights, y m. Her results are summarised as follows. / x = 1625 / x 2 = 53200 / y = 1854 / y 2 = 57900 Find a 95% confidence interval for the difference between the population mean heights of pine trees in regions A and B. [7] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: Question Answer Marks Guidance 1  1625  2 1  1625 2  M1 775 x = = 32.5 s x =  53200 −  = 7.908   98  50  49  50   1854  2 1  1854 2  A1 3057 y = = 30.9 s y =  57900 −  = 10.3627   295  60  59  60  Both correct 2 7.908 10.3627 M1 A1 s = + [ = 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

This question in 9231/41 Oct/Nov 2022

6 A company manufactures copper pipes. The pipes are produced by two different machines, A and B. An inspector claims that the mean diameter of the pipes produced by machine A is greater than the mean diameter of the pipes produced by machine B. He takes a random sample of 12 pipes produced by machine A and measures their diameters, x cm. His results are summarised as follows. / x = 6.24 / x 2 = 3.26 He also takes a random sample of 10 pipes produced by machine B and measures their diameters in cm. His results are as follows. 0.48 0.53 0.47 0.54 0.54 0.55 0.46 0.55 0.50 0.48 The diameters of the pipes produced by each machine are assumed to be normally distributed with equal population variances. Test at the 2.5% significance level whether the data supports the inspector’s claim. [9] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … Additional page If you use the following page to complete the answer to any question, the question number must be clearly shown. … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 6 2 1  6.24 2  M1 19 xs =  3.26 −  = 0.001382 13750 11  12  19 2 1  5.12  ys =  2.6124 −  = 0.001267 15000 9  10  A1 Both correct. 2 11  0.001382 + 9  0.001267 M1 A1 Find pooled variance. s = = 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 0 : a = b H 1 : a  b B1 Critical value is 2.086 (t0.975 (20)) M1 Compare their calculated value with 2.086 and conclusion. 0.640 < 2.086 Accept H0 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

This question in 9231/41 Oct/Nov 2022

1 A basketball club has a large number of players. The heights, x m, of a random sample of 10 of these players are measured. A 90% confidence interval for the population mean height, n m, of players in this club is calculated. It is assumed that heights are normally distributed. The confidence interval is 1.78 G n G 2.02 . Find the values of / x and / x2 for this sample. [6] … … … … … … … … … … … … … … … … … … … … … … … … …

6 marks

Mark scheme: Question Answer Marks Guidance 1 2 M1 s 2.02 = x + 1.833 10 s 2 1.78 = x − 1.833 10 2.02 + 1.78 Add, x = = 1.90 Add. 2  x = 19 A1 2 M1 Allow 1.372, 1.383, 1.812 instead of 1.833. s Subtract: 2.02 − 1.78 = 2  1.833 10 s 2 = 0.042859 A1 May be implied. s = 0.20702 2 1 2 (  x ) 2 M1 But s = (  x − ) 9 10 2 2 19 2 A1 36.486  x = 9 s + = 36.5 10 Using 1.645: maximum M1A1 M0A0 M1A0 3/6 6

This question in 9231/42 Oct/Nov 2022

3 A scientist is investigating the masses of birds of a certain species in country X and country Y. She takes a random sample of 50 birds of this species from country X and a random sample of 80 birds of this species from country Y. She records their masses in kg, x and y, respectively. Her results are summarised as follows. / x = 75.5 / x 2 = 115.2 / y = 116.8 / y 2 = 172.6 n The population mean masses of these birds in countries X and Y are n x kg and y kg respectively. Test, at the 5% significance level, the null hypothesis n x = n y against the alternative hypothesis n 2 n x y . State your conclusion in the context of the question. [8] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 3 2 1  75.52  M1 239 xs =  115.2 −  = 0.02439 49 50 9800   259 2 1  116.8 2  ys =  172.6 −  = 0.02623 9875 79  80  A1 Both correct 2 0.02439 0.02623 M1 A1 May be implied by 1.75 for z s = +  = 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

This question in 9231/42 Oct/Nov 2022

6 The manager of a technology company A claims that his employees earn more per year than the employees at technology company B. The amounts earned per year, in hundreds of dollars, by a random sample of 12 employees from company A and an independent random sample of 12 employees from company B are shown below. Company A 461 482 374 512 415 452 502 427 398 545 612 359 Company B 454 506 491 384 361 443 401 472 414 342 355 437 (a) Carry out a Wilcoxon rank‑sum test at the 5% significance level to test whether the manager’s claim is supported by the data. [9] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … (b) Explain whether a paired sample t‑test would be appropriate to test the manager’s claim if earnings are normally distributed. [1] … … … … … … … … … … … … … … … … Additional page If you use the following page to complete the answer to any question, the question number must be clearly shown. … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 6(a) M1 Wrong test: max B1 for hypotheses, B1B1 for correct mean and 359 3 342 1 variance 374 5 355 2 398 7 361 4 Attempt at rankings (allow up to 4 errors) 415 10 384 6 427 11 401 8 452 14 414 9 461 16 437 12 482 18 443 13 502 20 454 15 512 22 472 17 545 23 491 19 612 24 506 21 Test statistic = 127 A1 Clearly identified H0: population medians are equal B1 H1: population median for X is greater than population median for Y 1 B1 Mean =  12  25 = 150 2 1 B1 Variance =  12  12  25 = 300 12 6(a) 127.5 − 150 M1 Allow incorrect or no continuity correction. 300 −1.299 A1 Compare with −1.645: −1.299 −1.645 , or 0.097 > 0.05 M1 Valid comparison with 1.645 or 0.05 and reach correct ft Accept H0 conclusion. Insufficient evidence to support manager’s claim A1 Correct conclusion in context, following correct work, level of uncertainty in language. ‘Prove’ is A0. 9 6(b) Not appropriate/no, not the same people B1 OE No and reason needed, e.g. individuals in the samples cannot be paired up. 1

This question in 9231/42 Oct/Nov 2022

1 Jasmine is researching the heights of pine trees in forests in two regions A and B. She chooses a random sample of 50 pine trees in region A and records their heights, x m. She also chooses a random sample of 60 pine trees in region B and records their heights, y m. Her results are summarised as follows. / x = 1625 / x 2 = 53200 / y = 1854 / y 2 = 57900 Find a 95% confidence interval for the difference between the population mean heights of pine trees in regions A and B. [7] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: Question Answer Marks Guidance 1  1625  2 1  1625 2  M1 775 x = = 32.5 s x =  53200 −  = 7.908   98  50  49  50   1854  2 1  1854 2  A1 3057 y = = 30.9 s y =  57900 −  = 10.3627   295  60  59  60  Both correct 2 7.908 10.3627 M1 A1 s = + [ = 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

This question in 9231/43 Oct/Nov 2022

6 A company manufactures copper pipes. The pipes are produced by two different machines, A and B. An inspector claims that the mean diameter of the pipes produced by machine A is greater than the mean diameter of the pipes produced by machine B. He takes a random sample of 12 pipes produced by machine A and measures their diameters, x cm. His results are summarised as follows. / x = 6.24 / x 2 = 3.26 He also takes a random sample of 10 pipes produced by machine B and measures their diameters in cm. His results are as follows. 0.48 0.53 0.47 0.54 0.54 0.55 0.46 0.55 0.50 0.48 The diameters of the pipes produced by each machine are assumed to be normally distributed with equal population variances. Test at the 2.5% significance level whether the data supports the inspector’s claim. [9] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … Additional page If you use the following page to complete the answer to any question, the question number must be clearly shown. … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 6 2 1  6.24 2  M1 19 xs =  3.26 −  = 0.001382 13750 11  12  19 2 1  5.12  ys =  2.6124 −  = 0.001267 15000 9  10  A1 Both correct. 2 11  0.001382 + 9  0.001267 M1 A1 Find pooled variance. s = = 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 0 : a = b H 1 : a  b B1 Critical value is 2.086 (t0.975 (20)) M1 Compare their calculated value with 2.086 and conclusion. 0.640 < 2.086 Accept H0 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

This question in 9231/43 Oct/Nov 2022

1 The lengths of the leaves of a particular type of tree are normally distributed with mean μcm. The lengths, x cm, of a random sample of 12 leaves of this type are recorded. The results are summarised as follows. / x = 91.2 / x 2 = 695.8 Find a 95% confidence interval for μ. [4] … … … … … … … … … … … … … … … … … … … … … … … …

4 marks

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

This question in 9231/41 May/June 2023

2 The children at two large schools, P and Q, are all given the same puzzle to solve. A random sample of size 10 is taken from the children at school P. Their individual times to complete the puzzle give a sample mean of 9.12 minutes and an unbiased variance estimate of 2.16 minutes2. A random sample of size 12 is taken from the children at school Q. Their individual times, x minutes, to complete the puzzle are summarised by / x = 99. 6 / ( x - x ) 2 = 21.5 , where x is the sample mean. Times to complete the puzzle are assumed to be normally distributed with the same population variance. Test at the 5% significance level whether the population mean time taken to complete the puzzle by children at school P is greater than the population mean time taken to complete the puzzle by children at school Q. [8] … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 2   2 99.6 21.5 43 8.3 1.9545 or 12 11 22 x x s           H0: P Q    H1: P Q    B1 Pooled estimate = 9 2.16 11 1.9545 10 12 2      M1 FT FT their 1.9545 only 2.047 (2.05) A1 May be implied, allow 2.04 9.12 8.3 1 1 2.047 10 12 t     M1 FT FT their pooled estimate 1.339 (or 1.338) A1 Accept 1.34 '1.339' 1.725  Accept H0 / not significant M1 Compare with correct tabular value 1.725 and conclusion without context. Condone ‘reject H1’. Insufficient evidence to suggest that the (mean) time taken at P is greater than the (mean) time taken at Q. A1 Correct conclusion in context, following correct work, level of uncertainty in language (not ‘prove’, not ‘there is no evidence’), no contradictions. A0 if H0 and H1 reversed. 8

This question in 9231/41 May/June 2023

1 The lengths of the leaves of a particular type of tree are normally distributed with mean μcm. The lengths, x cm, of a random sample of 12 leaves of this type are recorded. The results are summarised as follows. / x = 91.2 / x 2 = 695.8 Find a 95% confidence interval for μ. [4] … … … … … … … … … … … … … … … … … … … … … … … …

4 marks

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

This question in 9231/42 May/June 2023

2 The children at two large schools, P and Q, are all given the same puzzle to solve. A random sample of size 10 is taken from the children at school P. Their individual times to complete the puzzle give a sample mean of 9.12 minutes and an unbiased variance estimate of 2.16 minutes2. A random sample of size 12 is taken from the children at school Q. Their individual times, x minutes, to complete the puzzle are summarised by / x = 99. 6 / ( x - x ) 2 = 21.5 , where x is the sample mean. Times to complete the puzzle are assumed to be normally distributed with the same population variance. Test at the 5% significance level whether the population mean time taken to complete the puzzle by children at school P is greater than the population mean time taken to complete the puzzle by children at school Q. [8] … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 2   2 99.6 21.5 43 8.3 1.9545 or 12 11 22 x x s           H0: P Q    H1: P Q    B1 Pooled estimate = 9 2.16 11 1.9545 10 12 2      M1 FT FT their 1.9545 only 2.047 (2.05) A1 May be implied, allow 2.04 9.12 8.3 1 1 2.047 10 12 t     M1 FT FT their pooled estimate 1.339 (or 1.338) A1 Accept 1.34 '1.339' 1.725  Accept H0 / not significant M1 Compare with correct tabular value 1.725 and conclusion without context. Condone ‘reject H1’. Insufficient evidence to suggest that the (mean) time taken at P is greater than the (mean) time taken at Q. A1 Correct conclusion in context, following correct work, level of uncertainty in language (not ‘prove’, not ‘there is no evidence’), no contradictions. A0 if H0 and H1 reversed. 8

This question in 9231/42 May/June 2023

2 Shane is studying the lengths of the tails of male red kangaroos. He takes a random sample of 14 male red kangaroos and measures the length of the tail, x m, for each kangaroo. He then calculates a 90% confidence interval for the population mean tail length, n m, of male red kangaroos. He assumes that the tail lengths are normally distributed and finds that 1.11 G n G 1. 14 . Find the values of / x and / x2 for this sample. [6] … … … … … … … … … … … … … … … … … … … … … … … … …

6 marks

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

This question in 9231/43 May/June 2023

4 An inspector is checking the lengths of metal rods produced by two machines, X and Y. These rods should be of the same length, but the inspector suspects that those made by machine X are shorter, on average, than those made by machine Y. The inspector chooses a random sample of 80 rods made by machine X and a random sample of 60 rods made by machine Y. The lengths of these rods are x cm and y cm respectively. Her results are summarised as follows. / x = 164.0 / x 2 = 338.1 / y = 124 .8 / y 2 = 261 .1 (a) Test at the 10% significance level whether the data supports the inspector’s suspicion. [8] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … (b) Give a reason why it is not necessary to make any assumption about the distributions of the lengths of the rods. [1] … … … … … … … … … …

9 marks

Mark scheme: 4(a) x x y B1   2 2 1 164 338.1 0.02405 79 80          xs and   2 2 1 124.8 261.1 0.02569 59 60          ys B1 Both. Implied by 19 379 , 790 14750 or 3sf. 2 0.02405 0.02569 80 60   s M1 2 0.0007289  s or 0.026998 or 0.0270  s A1 Implied by 1.11  z 164.0 124.8 80 60   z s M1 FT their value for s. 1.11  A1 ‘1.11’ < 1.282 Accept H0/ not significant M1 Compare with correct z-value 1.282 and consistent signs. Condone ‘reject H1’. Using probabilities, P(Z >1.11) = 0.1333 > 0.1. Insufficient evidence to support the inspector’s suspicion/ Insufficient evidence that the (mean) lengths of rods from machine X are shorter than the (mean) lengths of rods from machine Y A1 Correct conclusion in context, following correct work, level of uncertainty in language (not ‘prove’, not ‘there is no evidence’), no contradictions. e.g. proves that the inspector is incorrect scores A0 A0 if hypotheses wrong way round. 8 Question Answer Marks Guidance 4(b) Large sample sizes OR central limit theorem applies. B1 1

This question in 9231/43 May/June 2023

1 Maya is an athlete who competes in 1500-metre races. Last summer her practice run times had mean 4.22 minutes. Over the winter she has done some intense training to try to improve her times. A random sample of 10 of her practice run times, x minutes, this summer are summarised as follows. / x = 42.05 / x 2 = 176.83 Maya’s new practice run times are normally distributed. She believes that on average her times have improved as a result of her training. Test, at the 5% significance level, whether Maya’s belief is supported by the data. [6] … … … … … … … … … … … … … … … … … … … … … … …

6 marks

Mark scheme: Question Answer Marks Guidance 1 H0: = 4.22 H1:  4.22 B1 42.05 2 1  42.052  13  M1 = = 0.0010833 x =  = 4.205 , s =  176.83 −   10 9  10   12000  4.205 − 4.22 M1 Their 4.205, their 2s . t = s 2 10 t = −1.44 A1 Condone sign. Tabular value = 1.833: ‘1.44’ < 1.833, accept H0 M1 Compare their value with correct tabular value, signs 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

This question in 9231/41 Oct/Nov 2023

3 Scientists are studying the effects of exercise on LDL blood cholesterol levels. Over a three-month period, a large group of people exercised for 20 minutes each day. For a randomly chosen sample of 10 of these people, the LDL blood cholesterol levels were measured at the beginning and the end of the three-month period. The results, measured in suitable units, are as follows. Person A B C D E F G H I J Beginning 72 84 120 90 102 135 64 75 80 88 Cholesterol level End 64 76 105 92 105 115 67 75 75 84 (a) Test, at the 2.5% significance level, whether there is evidence that the population mean LDL blood cholesterol level has reduced by more than 2 units after the three-month period. [7] … … … … … … … … … … … … … … … … … … … (b) State any assumption that you have made in part (a). [1] …

8 marks

Mark scheme: 3(a) H0: B − E = 2 H1: B − E  2 B1 May use d , but must be consistent with direction of differences found. M1 Differences, at most 2 errors. 8 8 15 –2 –3 20 –3 0 5 4 2 2 1  52 2  2728  M1 Their values but must see 9 and 10 used correctly.  d = 52,  d = 816 , s =  816 −  = = 60.62  9  10   45  52 M1 − 2 10 t = s 2 10 t = 1.29967, 1.30 A1 Tabular value = 2.262. ‘1.30’ < 2.262, accept H0 M1 Allow ‘not significant’. 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

This question in 9231/41 Oct/Nov 2023

1 A factory produces small bottles of natural spring water. Two different machines, X and Y, are used to fill empty bottles with the water. A quality control engineer checks the volumes of water in the bottles filled by each of the machines. He chooses a random sample of 60 bottles filled by machine X and a random sample of 75 bottles filled by machine Y. The volumes of water, x and y respectively, in millilitres, are summarised as follows. / x = 6345 / ( x - x ) 2 = 243.8 / y = 7614 /( y - y 2) = 384.9 x and y are the sample means of the volume of water in the bottles filled by machines X and Y respectively. Find a 95% confidence interval for the difference between the mean volume of water in bottles filled by machine X and the mean volume of water in bottles filled by machine Y. [6] … … … … … … … … … … … … … … … … … … … … …

6 marks

Mark scheme: Question Answer Marks Guidance 1 2 243.8  1219  2 384.9  3849  B1 Implied by correct s or pooled estimate s x =  = = 4.132  , s y =  = = 5.201 243.8 + 384.9 59  295  74  740  = 4.727 60 + 75 − 2 2 4.132 5.201 M1 Using their sample variances. s = +  = 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 4.96, 3.5  etc.  0 , 4.96  3.5, 4.96 ) A1 4.23 0.729 is A0, condone  3.5  or ( 6

This question in 9231/42 Oct/Nov 2023

5 A company is deciding which of two machines, X and Y, can make a certain type of electrical component more quickly. The times taken, in minutes, to make one component of this type are recorded for a random sample of 8 components made by machine X and a random sample of 9 components made by machine Y. These times are as follows. Machine X 4.0 4.6 4.7 4.8 5.0 5.2 5.6 5.8 Machine Y 4.5 4.9 5.1 5.3 5.4 5.7 5.9 6.3 6.4 The manager claims that on average the time taken by machine X to make one component is less than that taken by machine Y. (a) Carry out a Wilcoxon rank‑sum test at the 5% significance level to test whether the manager’s claim is supported by the data. [6] … … … … … … … … … … … … … … … … … … … … … (b) Assuming that the times taken to produce the components by the two machines are normally distributed with equal variances, carry out a t‑test at the 5% significance level to test whether the manager’s claim is supported by the data. [9] … … … … … … … … … … … … … … … … … … … … … … … … … … Question 5(c) is printed on the next page. … … … … … … … … … … … … … … … (c) In general, would you expect the conclusions from the tests in parts (a) and (b) to be the same? Give a reason for your answer. [1] … … … … … … … … … … …

16 marks

Mark scheme: 5(a) H0: population medians are equal or mx = my B1 Do not accept ‘difference between population H1: population median for X < population median for Y or mx < my medians < 0’ without X or Y oe specified. 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 H0 /not significant M1 Ft their ‘55’ Must come from ranks. 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 5(b) H0: 𝜇𝑥= 𝜇𝑦 H1: 𝜇𝑥< 𝜇𝑦 B1 x = 39.7 x 2 = 199.33 y = 49.5 y 2 = 275.47 B1     2 1  39.7 2  xs =  199.33 −  = 0.33125 7  8  2 1  49.5 2  B1 ys =  275.47 −  = 0.4025 8  9  2 7  0.33125 + 8  0.4025 M1 Pooled variance s = 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 H0, sufficient evidence that mean for machine X is less than mean for machine A1 CWO 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

This question in 9231/42 Oct/Nov 2023

1 Maya is an athlete who competes in 1500-metre races. Last summer her practice run times had mean 4.22 minutes. Over the winter she has done some intense training to try to improve her times. A random sample of 10 of her practice run times, x minutes, this summer are summarised as follows. / x = 42.05 / x 2 = 176.83 Maya’s new practice run times are normally distributed. She believes that on average her times have improved as a result of her training. Test, at the 5% significance level, whether Maya’s belief is supported by the data. [6] … … … … … … … … … … … … … … … … … … … … … … …

6 marks

Mark scheme: Question Answer Marks Guidance 1 H0: = 4.22 H1:  4.22 B1 42.05 2 1  42.052  13  M1 = = 0.0010833 x =  = 4.205 , s =  176.83 −   10 9  10   12000  4.205 − 4.22 M1 Their 4.205, their 2s . t = s 2 10 t = −1.44 A1 Condone sign. Tabular value = 1.833: ‘1.44’ < 1.833, accept H0 M1 Compare their value with correct tabular value, signs 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

This question in 9231/43 Oct/Nov 2023

3 Scientists are studying the effects of exercise on LDL blood cholesterol levels. Over a three-month period, a large group of people exercised for 20 minutes each day. For a randomly chosen sample of 10 of these people, the LDL blood cholesterol levels were measured at the beginning and the end of the three-month period. The results, measured in suitable units, are as follows. Person A B C D E F G H I J Beginning 72 84 120 90 102 135 64 75 80 88 Cholesterol level End 64 76 105 92 105 115 67 75 75 84 (a) Test, at the 2.5% significance level, whether there is evidence that the population mean LDL blood cholesterol level has reduced by more than 2 units after the three-month period. [7] … … … … … … … … … … … … … … … … … … … (b) State any assumption that you have made in part (a). [1] …

8 marks

Mark scheme: 3(a) H0: B − E = 2 H1: B − E  2 B1 May use d , but must be consistent with direction of differences found. M1 Differences, at most 2 errors. 8 8 15 –2 –3 20 –3 0 5 4 2 2 1  52 2  2728  M1 Their values but must see 9 and 10 used correctly.  d = 52,  d = 816 , s =  816 −  = = 60.62  9  10   45  52 M1 − 2 10 t = s 2 10 t = 1.29967, 1.30 A1 Tabular value = 2.262. ‘1.30’ < 2.262, accept H0 M1 Allow ‘not significant’. 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

This question in 9231/43 Oct/Nov 2023

1 The times taken by members of a large cycling club to complete a cross-country circuit have a normal distribution with mean n minutes. The times taken, x minutes, are recorded for a random sample of 14 members of the club. The results are summarised as follows, where x is the sample mean. x = 42.8 / ( x - x ) 2 = 941.5 Find a 95% confidence interval for n. [4] … … … … … … … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: 1 941.5 2 13 s  CI: 2 42.8 2.160 14 s  M1 A1 Correct formula with a t value. 2.160 seen in a CI formula for A1. [37.9, 47.7] A1 (37.9, 47.7) A1 37.9 ' ' 47.7 x   A1 (ignore symbol used inside the inequality) Condone [47.7, 37.9] A1 Final answer 42.8 4.91  A0 4

This question in 9231/41 May/June 2024

6 Jade is a swimming instructor at a sports college. She claims that, as a result of an intensive training course, the mean time taken by students to swim 50 metres has reduced by more than 1 second. She chooses a random sample of 10 students. The times taken, in seconds, before and after the training course are recorded in the table. Student A B C D E F G H I J Time before 54.2 47.4 52.1 59.0 55.3 51.0 48.9 52.2 58.4 51.4 course Time after 50.1 46.3 52.5 58.8 51.4 48.4 49.5 48.7 58.3 51.4 course (a) Test, at the 10% significance level, whether Jade’s claim is justified. [7] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … (b) State an assumption that is necessary for this test to be valid. [1] … … …

8 marks

Mark scheme: 6(a) Signed differences: 4.1 1 .1 0.4 0.2 3.9 2.6 0.6 3.5 0.1 0   2 14.5, 52.81 d d     2 2 1 14.5 52.81 3.532 9 10 s          M1 H0: 1 B A     H1 : 1 B A     B1 Accept H0: 1, d  H1 : 1 d . 2 1.45 1 10 t s   M1 The – 1 must be present in the numerator. 0.757 A1 Critical value is 1.383: ‘0.757’ < 1.383 Accept H0 M1 ‘0.757’ must come from a paired sample t- calculation, 1.383 must be correct Correct ft conclusion for their 0.757 and 1.383 Condone Reject H1 ‘Accept H0’ can be implied by a conclusion that is 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 Question Answer Marks Guidance 6(b) Distribution of population differences is normal B1 Underlying distribution of differences normal B1 ‘data’ implies B0. 1 B1

This question in 9231/41 May/June 2024

1 The times taken by members of a large cycling club to complete a cross-country circuit have a normal distribution with mean n minutes. The times taken, x minutes, are recorded for a random sample of 14 members of the club. The results are summarised as follows, where x is the sample mean. x = 42.8 / ( x - x ) 2 = 941.5 Find a 95% confidence interval for n. [4] … … … … … … … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: 1 941.5 2 13 s  CI: 2 42.8 2.160 14 s  M1 A1 Correct formula with a t value. 2.160 seen in a CI formula for A1. [37.9, 47.7] A1 (37.9, 47.7) A1 37.9 ' ' 47.7 x   A1 (ignore symbol used inside the inequality) Condone [47.7, 37.9] A1 Final answer 42.8 4.91  A0 4

This question in 9231/42 May/June 2024

6 Jade is a swimming instructor at a sports college. She claims that, as a result of an intensive training course, the mean time taken by students to swim 50 metres has reduced by more than 1 second. She chooses a random sample of 10 students. The times taken, in seconds, before and after the training course are recorded in the table. Student A B C D E F G H I J Time before 54.2 47.4 52.1 59.0 55.3 51.0 48.9 52.2 58.4 51.4 course Time after 50.1 46.3 52.5 58.8 51.4 48.4 49.5 48.7 58.3 51.4 course (a) Test, at the 10% significance level, whether Jade’s claim is justified. [7] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … (b) State an assumption that is necessary for this test to be valid. [1] … … …

8 marks

Mark scheme: 6(a) Signed differences: 4.1 1 .1 0.4 0.2 3.9 2.6 0.6 3.5 0.1 0   2 14.5, 52.81 d d     2 2 1 14.5 52.81 3.532 9 10 s          M1 H0: 1 B A     H1 : 1 B A     B1 Accept H0: 1, d  H1 : 1 d . 2 1.45 1 10 t s   M1 The – 1 must be present in the numerator. 0.757 A1 Critical value is 1.383: ‘0.757’ < 1.383 Accept H0 M1 ‘0.757’ must come from a paired sample t- calculation, 1.383 must be correct Correct ft conclusion for their 0.757 and 1.383 Condone Reject H1 ‘Accept H0’ can be implied by a conclusion that is 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 Question Answer Marks Guidance 6(b) Distribution of population differences is normal B1 Underlying distribution of differences normal B1 ‘data’ implies B0. 1 B1

This question in 9231/42 May/June 2024

1 Ellie is investigating the heights of two types of beech tree, A and B, in a certain region. She has chosen a random sample of 60 beech trees of type A in the region, recorded their heights, x m, and calculated unbiased estimates for the population mean and population variance as 35.6 m and 4.95 m 2 respectively. Ellie also chooses a random sample of 50 beech trees of type B in the region and records their heights, y m. Her results are summarised as follows. / y = 1654 / y 2 = 54 850 Find a 95% confidence interval for the difference between the population mean heights of type A and type B beech trees in the region. [6] … … … … … … … … … … … … … … … … … … … … … …

6 marks

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

This question in 9231/41 Oct/Nov 2024

6 Ansal is investigating the wingspans of Monarch butterflies in two different regions, X and Y. He takes a random sample of 8 Monarch butterflies from region X and records their wingspans, x cm. His results are as follows. 8.2 7.0 7.3 8.8 7.8 8.5 9.2 7.4 Ansal also takes a random sample of 9 Monarch butterflies from region Y and records their wingspans, y cm. His results are summarised as follows. / y = 71.10 / y 2 = 567.13 Ansal suspects that the mean wingspan of Monarch butterflies from region X is greater than the mean wingspan of Monarch butterflies from region Y. It is known that the wingspans of Monarch butterflies in regions X and Y are normally distributed with equal population variances. Test, at the 10% significance level, whether Ansal’s suspicion is supported by the data. [8] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … Additional page If you use the following page to complete the answer to any question, the question number must be clearly shown. … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 6 2 1  64.2 2  M1 851 xs =  519.46 −   = 0.607857  7 8 1400   17 2 1  71.12  , both. 25 ys =  567.13 −   = 0.680  8  9  2 8  0.680 + 7  0.607857 M1 A1 1939 s =  = 0.64633 Correct formula for pooled variance used, . 9 + 8 − 2 3000 H0: X − Y = 0 B1 H1: X − Y  0 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 H0 M1 Compare with 1.341 and conclusion 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

This question in 9231/41 Oct/Nov 2024

1 Ellie is investigating the heights of two types of beech tree, A and B, in a certain region. She has chosen a random sample of 60 beech trees of type A in the region, recorded their heights, x m, and calculated unbiased estimates for the population mean and population variance as 35.6 m and 4.95 m 2 respectively. Ellie also chooses a random sample of 50 beech trees of type B in the region and records their heights, y m. Her results are summarised as follows. / y = 1654 / y 2 = 54 850 Find a 95% confidence interval for the difference between the population mean heights of type A and type B beech trees in the region. [6] … … … … … … … … … … … … … … … … … … … … … …

6 marks

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

This question in 9231/43 Oct/Nov 2024

6 Ansal is investigating the wingspans of Monarch butterflies in two different regions, X and Y. He takes a random sample of 8 Monarch butterflies from region X and records their wingspans, x cm. His results are as follows. 8.2 7.0 7.3 8.8 7.8 8.5 9.2 7.4 Ansal also takes a random sample of 9 Monarch butterflies from region Y and records their wingspans, y cm. His results are summarised as follows. / y = 71.10 / y 2 = 567.13 Ansal suspects that the mean wingspan of Monarch butterflies from region X is greater than the mean wingspan of Monarch butterflies from region Y. It is known that the wingspans of Monarch butterflies in regions X and Y are normally distributed with equal population variances. Test, at the 10% significance level, whether Ansal’s suspicion is supported by the data. [8] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … Additional page If you use the following page to complete the answer to any question, the question number must be clearly shown. … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 6 2 1  64.2 2  M1 851 xs =  519.46 −   = 0.607857  7 8 1400   17 2 1  71.12  , both. 25 ys =  567.13 −   = 0.680  8  9  2 8  0.680 + 7  0.607857 M1 A1 1939 s =  = 0.64633 Correct formula for pooled variance used, . 9 + 8 − 2 3000 H0: X − Y = 0 B1 H1: X − Y  0 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 H0 M1 Compare with 1.341 and conclusion 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

This question in 9231/43 Oct/Nov 2024

1 A random sample of 12 observations of a normal random variable is taken. The results give unbiased estimates for the population mean and variance as 10.24 and 0.52 respectively. Test, at the 10% significance level, the null hypothesis that the population mean is 10.6 against the alternative hypothesis that the population mean is less than 10.6. [4] … … … … … … … … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: Question Answer Marks Guidance 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 0 (significant). M1 Compare their t with 1.363 (consistent signs) and appropriate conclusion (may be in terms of H1). Reject H0 can be implied by an attempt at an appropriate 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

This question in 9231/44 May/June 2025

6 Lina and Mona are two statisticians who also write songs. The ‘time’ of a song is the number of minutes for which it lasts. For a random sample of 10 of her songs, Lina calculates a 95% confidence interval for the population mean time, n minutes. This confidence interval is 2.95 G n G 3.13 . The times, x minutes, of Lina’s songs are normally distributed. (a) Find the values of / x and / x2 for the 10 songs in Lina’s sample. [5] … … … … … … … … … … … … … … … … … … … … … … … … … Mona’s songs have times, y minutes, that are normally distributed. The times for a random sample of 8 of Mona’s songs are summarised as follows. / y = 24.8 / y 2 = 76.98 Mona claims that the population mean time of her songs is greater than the population mean time of Lina’s songs. (b) Assuming that the two distributions have the same population variance, test at the 5% significance level whether there is evidence to support Mona’s claim. [8] … … … … … … … … … … … … … … … … … … … … … … … Additional page If you use the following page to complete the answer to any question, the question number must be clearly shown. … … … … … … … … … … … … … … … … … … … … … …

13 marks

Mark scheme: 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 xs . Subtract: xs = ( 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. xs2 = 0.0158 or xs = 0.126 A1 FT FT their 2.262 following M1. Implied by AWRT 92.6 following M1. 2 30.4 2 A1 FT FT their 2.262.  x = 9  0.0158307 + = 92.6 CWO 10 5 6(b) H0 :Y = X B1 If in words, must contain ‘population means’. d = 0, H1 :d  0 if defined or consistent with H1 :Y  X Allow H0:  working. . 24.82 24.82 M1 May be embedded, for example ys 76.98 − 18 ( 2 = 17 ( ) ) 76.98 − 8 2 9 s x2 + 7 s y2 M1 Correct form using their xs2 and their ys2 , may be embedded. s p =  = 0.015155  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 ps . t = = −1.028 s p 101 + 81 A1 AWRT 1.02 or 1.03 implies 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 0 (not significant). conclusion (may be in terms of H1). Accept H0 can be implied by an attempt at an appropriate conclusion in context. Allow M1 for comparison of their t with 1.740 or 2.120 and 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

This question in 9231/44 May/June 2025

1 A group of 10 school children are asked to estimate the size of an angle i° in a given acute angled triangle. These estimates, in degrees, are as follows. 84 85 77 85 84 87 86 88 83 85 (a) Stating any assumptions you make, calculate a 95% confidence interval for i. [5] … … … … … … … … … … … … … … … … … … … … … … (b) Give a reason why the assumptions made in part (a) may not be appropriate in this case. [1] … …

6 marks

Mark scheme: Question Answer Marks Guidance 1(a)   = 13415 = 8.933 M1 Correct expression, implied by AWRT 8.93. s 2 = 19 ( '71314'− '844'10 2 ) 2.262 B1 2.262 or 2.26 seen. '844' 10  '2.262' '8.933'10 M1 Correct form, must be a t-value. [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

This question in 9231/41 Oct/Nov 2025

3 A random sample of 10 newborn baby boys is taken and their masses in kg are recorded. From this sample, the population standard deviation of all newborn baby boys is estimated as 0.6 kg. A random sample of 5 newborn baby girls is taken and their masses in kg are recorded as follows. 3.9 3.1 2.9 3.1 3.6 It is assumed that the masses of newborn baby boys and girls have the same population standard deviation, v kg. By pooling the two samples, calculate an estimate of v. [4] … … … … … … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: 3  ( x − x )2 = 9  0.36 = 3.24 B1 43 86 or 0.688 . ) ) '16.6'2 M1 Can be implied by 0.172 ( = 250 ( = 125 '55.8'− 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

This question in 9231/41 Oct/Nov 2025

6 Nine athletes in a club have a new coach. The coach adopts a new training programme which he believes will reduce the race times of these athletes. Each athlete completes a 1500 m time trial before and after completing the new training programme. Their times, in seconds (s), are recorded. Athlete A B C D E F G H I Time before training (s) 250 251 252 267 276 291 310 320 335 Time after training (s) 245 251 253 261 275 293 302 313 320 (a) Carry out a paired t-test at the 5% significance level to test the coach’s belief. [7] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … Further research suggests that the effects of the training programme tend to reduce the times of the slower athletes by more than those of the faster athletes. (b) Suggest a reason why the paired t-test used in part (a) may not have been an appropriate test in this case. [1] … … … (c) Suggest a suitable alternative test that could have been used instead of a paired t-test. [1] … … … …

9 marks

Mark scheme: 6(a) H 0: B = A H1: B  A B1 If in words, must contain ‘population means’. Allow H 0: d = 0, H1: d 0 if defined or consistent with working. Differences: 5, 0, −1, 6, 1, −2, 8, 7, 15 M1 Attempt at signed differences. = 29.5 M1 Correct form for ds2 , note that ds = 5.43 . ds2 = 18 ( '405'− '39'9 2 )  '39' M1 Correct form. 9 = 2.393 '29.5' 9 A1 AWRT 2.39. '2.393'  1.860 , reject H0 / significant. M1 Compare their 2.393 with 1.860 and appropriate result (may be in terms of H1). 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

This question in 9231/41 Oct/Nov 2025

2 A factory produces packets of biscuits. The total mass of biscuits in a packet has a normal distribution with mean n. A random sample of 12 packets is taken and the mass of the contents of each packet, x g, is recorded. The results are summarised as follows. / x = 2390 / x 2 = 476117 (a) Find a 99% confidence interval for n. [4] … … … … … … … … … … … A test of the null hypothesis n = k is carried out on this sample using a 5% significance level. The test does not support the alternative hypothesis n 1 k . (b) Find the greatest possible value of k. [3] … … … … … … … … … … …

7 marks

Mark scheme: 2(a)   = 32633 = 9.8788  M1 Correct expression, implied by AWRT 9.88. s 2 = 111 ( '476117'− '2390'12 2 ) 3.106 B1 3.106 or 3.11 seen. '2390' 12  '3.106' '9.8788'12 M1 Correct form, must be a t -value. 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

This question in 9231/42 Oct/Nov 2025

5 An engineer is comparing the tensile strengths of steel rods made from two machines, A and B. The engineer randomly selects 8 rods from machine A and 6 rods from machine B. The tensile strengths, in appropriate units, are given in the following table. Machine A 402 403 415 412 409 407 406 410 Machine B 401 398 395 397 410 405 You should assume that the two distributions are normal and have the same population variance. Use a t-test at the 5% significance level to test whether there is any difference in the mean tensile strengths of steel rods from the two machines. [9] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 5 H 0:A = B H1:A  B B1 If in words, must have population means. Allow H0: =d 0 , H1: d 0 . '1331848'− '3264'8 2 [ = 136] M1 OE, may be implied by s A2 = 19.43 . Biased estimate of 17 earns M0 unless recovered. 31.6 . '964964'− '2406'6 2 [ = 158] M1 OE, may be implied by Bs2 = Biased estimate of 26 earns M0 unless recovered. 2 '136'+ '158' M1 Correct form using their values, may be embedded. ps = = 24.5 8 + 6 − 2 A1 OE, for example 492 . '3264' 8 − '2406'6 M1 Correct form using their values. t = = 2.62 ' s p ' 18 + 16 A1 AWRT 2.62. '2.62'  2.179, reject H0 / significant. M1 Compare their 2.62 with 2.179 and appropriate result (may be in terms of H1). 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

This question in 9231/42 Oct/Nov 2025

1 A group of 10 school children are asked to estimate the size of an angle i° in a given acute angled triangle. These estimates, in degrees, are as follows. 84 85 77 85 84 87 86 88 83 85 (a) Stating any assumptions you make, calculate a 95% confidence interval for i. [5] … … … … … … … … … … … … … … … … … … … … … … (b) Give a reason why the assumptions made in part (a) may not be appropriate in this case. [1] … …

6 marks

Mark scheme: Question Answer Marks Guidance 1(a)   = 13415 = 8.933 M1 Correct expression, implied by AWRT 8.93. s 2 = 19 ( '71314'− '844'10 2 ) 2.262 B1 2.262 or 2.26 seen. '844' 10  '2.262' '8.933'10 M1 Correct form, must be a t-value. [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

This question in 9231/43 Oct/Nov 2025

3 A random sample of 10 newborn baby boys is taken and their masses in kg are recorded. From this sample, the population standard deviation of all newborn baby boys is estimated as 0.6 kg. A random sample of 5 newborn baby girls is taken and their masses in kg are recorded as follows. 3.9 3.1 2.9 3.1 3.6 It is assumed that the masses of newborn baby boys and girls have the same population standard deviation, v kg. By pooling the two samples, calculate an estimate of v. [4] … … … … … … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: 3  ( x − x )2 = 9  0.36 = 3.24 B1 43 86 or 0.688 . ) ) '16.6'2 M1 Can be implied by 0.172 ( = 250 ( = 125 '55.8'− 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

This question in 9231/43 Oct/Nov 2025

6 Nine athletes in a club have a new coach. The coach adopts a new training programme which he believes will reduce the race times of these athletes. Each athlete completes a 1500 m time trial before and after completing the new training programme. Their times, in seconds (s), are recorded. Athlete A B C D E F G H I Time before training (s) 250 251 252 267 276 291 310 320 335 Time after training (s) 245 251 253 261 275 293 302 313 320 (a) Carry out a paired t-test at the 5% significance level to test the coach’s belief. [7] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … Further research suggests that the effects of the training programme tend to reduce the times of the slower athletes by more than those of the faster athletes. (b) Suggest a reason why the paired t-test used in part (a) may not have been an appropriate test in this case. [1] … … … (c) Suggest a suitable alternative test that could have been used instead of a paired t-test. [1] … … … …

9 marks

Mark scheme: 6(a) H 0: B = A H1: B  A B1 If in words, must contain ‘population means’. Allow H 0: d = 0, H1: d 0 if defined or consistent with working. Differences: 5, 0, −1, 6, 1, −2, 8, 7, 15 M1 Attempt at signed differences. = 29.5 M1 Correct form for ds2 , note that ds = 5.43 . ds2 = 18 ( '405'− '39'9 2 )  '39' M1 Correct form. 9 = 2.393 '29.5' 9 A1 AWRT 2.39. '2.393'  1.860 , reject H0 / significant. M1 Compare their 2.393 with 1.860 and appropriate result (may be in terms of H1). 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

This question in 9231/43 Oct/Nov 2025