Cambridge A Level Mathematics - Further 9231 — 2022 Oct/Nov Paper 4 · Variant 3

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. 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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

More questions on Inference using normal and t-distributions

2 An organisation runs courses to train students to become engineers. These students are taught in groups of 8. The director of the organisation claims that on average 60% of the students in a group achieve a pass. A random sample of 150 groups of 8 students is chosen. The following table shows the observed frequencies together with some of the expected frequencies using the appropriate binomial distribution. Number of passes per 0 1 2 3 4 5 6 7 8 group Observed 0 0 8 24 45 36 26 10 1 frequency Expected p 1.180 6.193 18.579 34.836 q r 13.437 2.519 frequency (a) Find the values of p, q and r giving your answers correct to 3 decimal places. 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(b) Carry out a goodness of fit test, at the 10% significance level, to test whether there is evidence to reject the director’s claim. 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Mark scheme: 2(a) Using B(8, 0.6) B1 One correct. p = 0.098, q = 41.804, r = 31.353 B1 Other two correct. If B0B0 scored, SCB1 if all 3 correct but not rounded to 3dp. 2 2(b) H0: the data fits the binomial distribution B(8, 0.6) B1 H1: the data does not fit the binomial distribution B(8, 0.6) Add first 3 columns: O value 8, E value 7.471 M1 AND last two columns: O value 11, E value 15.956 Both. Chi-squared sum: M1 0.0374 + 1.5817 + 2.9655 + 0.8058 + 0.9139 + 1.5394 Method must be seen. 7.84 A1 If M0 awarded SCB1 for 7.84. Critical value, 5df = 9.236 M1 Compare calculated value with 9.236 OR 7.84 < 9.236 13.36 if no columns added. Accept H0 12.02 adding only last 2 columns. 10.64 adding only first 3 columns. There is insufficient evidence to reject the director’s claim. A1 Correct work only except possibly B1 for hypotheses, in context, level of uncertainty in language. ‘prove’ scores A0 6

More questions on χ²-tests

3 A large college is holding a piano competition. Each student has played a particular piece of music and two judges have each awarded a mark out of 80. The marks awarded to a random sample of 14 students are shown in the following table. Student A B C D E F G H I J K L M N Judge 1 79 54 63 74 69 52 50 57 55 42 63 55 56 48 Judge 2 75 62 60 73 76 41 31 51 45 55 49 50 65 36 (a) One of the students claims that on average Judge 1 is awarding higher marks than Judge 2. Carry out a Wilcoxon matched-pairs signed-rank test at the 5% significance level to test whether the data supports the student’s claim. 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(b) Give a reason why it is preferable to use a Wilcoxon matched-pairs signed-rank test in this situation rather than a paired sample t-test. 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Mark scheme: 3(a) Differences: M1 4 -8 3 1 –7 11 19 6 10 –13 14 5 –9 12 Attempt at signed differences, allow 2 errors. Signed Ranks: M1 3 –7 2 1 –6 10 14 5 9 –12 13 4 –8 11 Ranks, allow 4 errors. [P = 72, Q = 33] T = 33 A1 33 clearly identified. H0: difference in population medians = 0 B1 H1: difference in population medians > 0 Correct, allow m for this mark. Critical tabular value = 25 B1 33 > 25 M1 Compares their value of sum of ranks with their 25 and accept H0 conclusion. Insufficient evidence to support claim A1 All correct except possibly first B1, in context, level of uncertainty in language. ‘Prove’ scores A0. 7 3(b) Underlying/population distribution of differences is unknown (not B1 known to be normal) 1

More questions on Non-parametric tests

4 Jason has three biased coins. For each coin the probability of obtaining a head when it is thrown is 2.3 Jason throws all three coins. The number of heads obtained is denoted by X. (a) Find the probability generating function G X ( t) of X. 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Jason also has two unbiased coins. He throws all five coins. The number of heads obtained from the = + 4 2 t + 4 t . The random variable Z is the two unbiased coins is denoted by Y. It is given that G Y ( t) 1 1 1 2 total number of heads obtained when Jason throws all five coins. (b) Find the probability generating function of Z, expressing your answer as a polynomial. 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(c) Find E(Z). 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Mark scheme: 4(a) 1 1 1 1 2 1 1 6 M1 P(0 H) =   = P(1 H) =   =3 At least 2 probabilities correct. 3 3 3 27 3 3 3 27 1 2 2 12 2 2 2 8 P(2 H) =   =3 P(3 H) =   = 3 3 3 27 3 3 3 27 1 6 12 2 8 3 M1A1 Correct form, ft their probabilities. GX (t) = + t + t + t 27 27 27 27 1 2 M1 OR: PGF for one coin is + t 3 3 3 M1A1  1 2  For 3 coins, PGF is + t    3 3  3 4(b) 1 2 2 3 M1 Correct method. GZ (t) = 1 + 2t + t 1 + 6t + 12t + 8t ( )( ) 108 1 2 3 4 5 M1 Obtain quintic polynomial. 1 + 8t + 25t + 38t + 28t + 8t ( ) 108 1 2 25 2 19 3 7 4 2 5 A1 + t + t + t + t + t 108 27 108 54 27 27 3 4(c) 1 2 3 4 M1 GZ’ (t) = 8 + 50t + 114t + 112t + 40t ( ) Differentiate and evaluate at t = 1 108 1 E(Z) = ( 8 + 50 + 114 + 112 + 40 ) 108 324 A1 = 3 108 Alternative method for questions 4(c) 2 2 2 1 1 M1 Summing expected values for each coin + + + + 3 3 3 2 2 3 A1 2

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5 The continuous random variable X has cumulative distribution function F given by 0 x 1 0, - x) 0 G x G 12, F ( x) = 1 - 1441 ( 12 2 * 1 x 2 12. (a) Find the upper quartile of X. 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(b) Find V ar (X 2 ) . 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The random variable Y is given by Y = X . (c) Find the probability density function of Y. 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Mark scheme: 5(a) 1 2 3 M1 F(U) = 0.75, so 1 − (12 − U ) = 144 4 (12 − U ) 2 = 36, 12 − U = 6, A1 U = 6 only 2 5(b) 1 B1 Differentiate given PDF. [f(x) =] (12 −x ) 72 12 1 2 3 1  3 1 4  M1 Integrate x 2 f ( x ) with their f ( x ) . 2 E( X ) =  (12 x − x ) dx =  4 x − x  [= 24] 0 72 72  4  4 12 1 4 5 1  12 5 1 6  M1 Integrate x 4 f ( x ) with their f ( x ) . E X = 12 x − x dx = x − x [= 1382.4] ( )  ( )   0 72 72  5 6  Var X 2 = E X 4 − E 2 X 2 [ = 1382.4 − 24 2 ] M1 Using correct formula with numerical values possibly ( ) ( ) ( ) unsimplified. 1382.4 − 576 = 806 (.4 ) A1 5 25(c)  0 y  0, M1 1 2 Change the variable: 1 − 12 −y ( ) 2 144  1 2 12 − y 0 y 12, Allow domain unchanged at this stage. G ( y ) =  1 − ( )  144  1 y  12. 1 3 M1 Differentiate their G ( y ) . 12 y − y ( ) 36  1 3 A1 Fully correct. 12 y − y 0 y 12  ( ) g ( y ) =  36  0 otherwise

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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. 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Additional page If you use the following page to complete the answer to any question, the question number must be clearly shown. ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................

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

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