Cambridge A Level Mathematics 9709 — 2022 May/June Paper 6 · Variant 3

2 Anton believes that 10% of students at his college are left-handed. Aliya believes that this is an under- estimate. She plans to carry out a hypothesis test of the null hypothesis p = 0.1 against the alternative hypothesis p > 0.1, where p is the actual proportion of students at the college that are left-handed. She chooses a random sample of 20 students from the college. She will reject the null hypothesis if at least 5 of these students are left-handed. (a) Explain what is meant by a Type I error in this context. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find the probability of a Type I error in the test. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) Given that the true value of p is 0.3, find the probability of a Type II error in the test. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 2(a) Conclude more than 10% of the students are left handed when this is not true B1 OE. Must be in context (accept use of p). Need the context of one tail test. 1 2(b) 1 – (0.920 + 20 × 0.919 × 0.1 + 20C2 × 0.918 × 0.12 + 20C3 × 0.917 × 0.13 + 20C4 × 0.916 × 0.14) M2 M2: fully correct M1: attempt 1 – P(X = 0, 1, 2, 3, 4); allow 1 – P(X = 0,1,2,3,4,5) or 1 – P(X = 0,1,2,3) need 1 – … the method mark cannot be implied 0.0432 (3 s.f.) A1 If M0 awarded allow SC B2 for 0.0432 3 Question Answer Marks Guidance 2(c) 0.720 + 20 × 0.719 × 0.3 + 20C2 × 0.718 × 0.32 + 20C3 × 0.717 × 0.33 + 20C4 × 0.716 × 0.34 M1 Attempt to find ( ) P 4 ≤ using B(20,0.3) Allow one end error The method mark cannot be implied 0.238 or 0.237 (3 s.f.) A1 If M0 awarded allow SC B1for 0.238 or 0.237 2

More questions on Hypothesis tests

3 Batteries of type A are known to have a mean life of 150 hours. It is required to test whether a new type of battery, type B, has a shorter mean life than type A batteries. (a) Give a reason for using a sample rather than the whole population in carrying out this test. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ A random sample of 120 type B batteries are tested and it is found that their mean life is 147 hours, and an unbiased estimate of the population variance is 225 hours2. (b) Test, at the 2% significance level, whether type B batteries have a shorter mean life than type A batteries. [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) Calculate a 94% confidence interval for the population mean life of type B batteries. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 3(a) Batteries unusable after testing or Population too big or too costly or too time consuming to use the whole population oe B1 1 3(b) H0: μ = 150 H1: μ < 150 B1 Or population mean = 150; not just ‘mean’ = 150 147 150 225 120   M1 Allow with continuity correction Need 120 –2.191 A1 Condone – 2.19 –2.191 < –2.054 [or –2.055] M1 OE. For valid comparison with 2.054 or 2.055 Or 0.0143 (or 0.0142) < 0.02 For two tail test allow comp –2.326 OE if H1: μ ≠ 150 (can score B0M1A1M1A0 max 3/5 ) [Reject Ho] There is evidence that the (mean) life of type B is less than type A (or less than 150) A1 FT In context, not definite with no contradictions Accept critical value method 147.19 M1A1 147 < 147.19 M1 conclusion A1 Or 150 > 149.81 5 Question Answer Marks Guidance 3(c) 147  z × 15 120 M1 Expression of correct form must be a z value z = 1.881 [or 1.882] B1 144 to 150 (3 s.f.) A1 Must be an interval Incorrect z value can only score M1B0A0 3

More questions on Hypothesis tests

4 Each box of Seeds & Raisins contains S grams of seeds and R grams of raisins. The weight of a box, when empty, is B grams. S, R and B are independent random variables, where S ∼N 300, 45 , R ∼N 200, 25 and B ∼N 15, 4 . A full box of Seeds & Raisins is chosen at random. (a) Find the probability that the total weight of the box and its contents is more than 500 grams. [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find the probability that the weight of seeds in the box is less than 1.4 times the weight of raisins in the box. [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 4(a) T~N(515, 74) B1 B1 for N(515, ..) give at early stage B1 B1 for Var = 45 + 25 + 4 = 74 give at early stage 500 '515' '74'  [= –1.744] M1 Standardise with their values. No standard deviation/variance mix Need combination for variance. Allow continuity correction. Φ(’1.744’) M1 Area consistent with their working 0.959 or 0.96[0] (3 s.f.) A1 5 Question Answer Marks Guidance 4(b) E(S – 1.4R) = 300-1.4x200 =20 B1 Give at early stage Var(S – 1.4R) = 45 + 1.42×25 = 94 B1 Give at early stage SC: if B0B0 awarded allow SC B1 for 14 and 105.84 0 (20) '94'  [= –2.063] M1 Standardise with their values. No standard deviation/variance mix. Need combination for variance. 1 – Φ(‘2.063’) M1 Area consistent with their working 0.0196 (3 s.f.) A1 5

More questions on Discrete random variables

5 The number of clients who arrive at an information desk has a Poisson distribution with mean 2.2 per 5-minute period. (a) Find the probability that, in a randomly chosen 15-minute period, exactly 6 clients arrive at the desk. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) If more than 4 clients arrive during a 5-minute period, they cannot all be served. Find the probability that, during a randomly chosen 5-minute period, not all the clients who arrive at the desk can be served. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) Use a suitable approximating distribution to find the probability that, during a randomly chosen 1-hour period, fewer than 20 clients arrive at the desk. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 5(a) λ = 6.6 B1 e-6.6 × 6 6.6 6! M1 Any λ 0.156 (3 s.f.) A1 If M0 awarded SC B1 for 0.156 3 5(b) 1 – e-2.2(1 + 2.2 + 2 3 4 2.2 2.2 2.2 2 3! 4!   ) M1 Allow one end error. Need 1 – … Any λ 0.0725 (3 s.f.) A1 If M0 awarded SC B1 for 0.0725 2 Question Answer Marks Guidance 5(c) N(26.4, 26.4) B1 Give at early stage 2.2  12 19.5 '26.4' '26.4'  [= –1.343] M1 Standardising with their values. Allow wrong or no continuity correction ɸ('–1.343') = 1 – ɸ('1.343') M1 Area consistent with their working 0.0897 or 0.0896 (3 s.f.) A1 4

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6 A random sample of 5 values of a variable X is given below. 2 3 3 5 a (a) Find an expression, in terms of a, for the mean of these values. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ It is given that an unbiased estimate of the population variance of X, using these values, is 4. It is also given that a is positive. (b) Find and simplify a quadratic equation in terms of a and hence find the value of a. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 6(a) 13 5 a 3 3 5 5 a + + + + . Do not ignore subsequent working 1 6(b)   2 2 5 47 13 4 4 5 5 a a           or   2 2 13 1 47 4 4 5 a a             M1 Use of correct formula using their value from (a), in terms of a, and equate to 4 2a2 – 13a – 7 = 0 A1 Any correct three-term quadratic equation rearranged to a form ready to solve a = 7 A1 Condone the other value of a ( 1 2  ) 3

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