Cambridge A Level Mathematics 9709 — 2023 Oct/Nov Paper 6 · Variant 2

2 The length, in minutes, of mathematics lectures at a certain college has mean - and standard deviation 8.3. (a) The total length of a random sample of 85 lectures was 4590 minutes. Calculate a 95% confidence interval for -. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ The length, in minutes, of history lectures at the college has mean m and standard deviation s. (b) Using a random sample of 100 history lectures, a 95% confidence interval for m was found to have width 2.8 minutes. Find the value of s. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 2(a) 4590 8.3 M1 For expression of correct form. Any z (but not ϕ(z)). z 85 85 z = 1.96 B1 52.2 to 55.8 (3 sf) A1 Must be an interval. 3 Question Answer Marks Guidance 2(b) s s M1 Equation of correct form (any z). 1.96  = 1.4 or 2  1.96  = 2.8 Allow factor of 2 error (i.e. first equation = 2.8). 100 100 50 A1 s = 7.14 (3 sf) or 7 2 Question Answer Marks Guidance

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4 The height H, in metres, of mature trees of a certain variety is normally distributed with standard deviation 0.67. In order to test whether the population mean of H is greater than 4.23, the heights of a random sample of 200 trees are measured. (a) Write down suitable null and alternative hypotheses for the test. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ The sample mean height, h metres, of the 200 trees is found and the test is carried out. The result of the test is to reject the null hypothesis at the 5% significance level. (b) Find the set of possible values of h. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) Ajit said, ‘In (b) we had to assume that H is normally distributed, so it was necessary to use the Central Limit Theorem.’ Explain whether you agree with Ajit. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 4(a) H0: population mean [of H ] = 4.23 B1 Allow μ = 4.23 or population mean of h = 4.23 H1: population mean [of H ] > 4.23 but NOT h = 4.23 or H = 4.23 or h = 4.23 or H = 4.23. 1 4(b) h − 4.23 M1 For standardising and forming an equation. Must have = 1.645 0.67 200 . Allow ± 1.645 or ± 1.96. 200 Accept ‘>’ and ‘<’. Allow H or any letter instead of h . h = 4.31 (3 sf) A1 May be implied by h > 4.31. Allow h < 4.31 for this A1 only, condone 4.15 also seen. h > 4.31 or h ⩾ 4.31 (3 sf) A1 Condone any letter instead of h . 3 4(c) Incorrect, because the population of H is given as normally distributed B1 Allow h instead of H or just ‘The population is normal.’ [with known variance]. Must use ‘population’ or ‘underlying distribution’. 1 Question Answer Marks Guidance

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