Cambridge A Level Mathematics 9709 — 2010 Oct/Nov Paper 7 · Variant 3

6 A clinic monitors the amount, X milligrams per litre, of a certain chemical in the blood stream of patients. For patients who are taking drug A, it has been found that the mean value of X is 0.336. A random sample of 100 patients taking a new drug, B, was selected and the values of X were found. The results are summarised below. n = 100, Σ x = 43.5, Σx2 = 31.56. (i) Test at the 1% significance level whether the mean amount of the chemical in the blood stream of patients taking drug B is different from that of patients taking drug A. [8] (ii) For the test to be valid, is it necessary to assume a normal distribution for the amount of chemical in the blood stream of patients taking drug B? Justify your answer. [2]

Mark scheme: 6 (i) x = 43.5/100 = 0.435 B1 100 31. 56 2 31. 56 2 s = × − .0 435 (=0.3573) M1 s = − .0 435 M0 99 100 100 or Var (= 0.128) or 1/99(31.56-(43.5)2/100) (= 0.3555), or Var (= 0.126) H0: Pop mean (for B) = 0.336 B1 Undefined mean: B0, but allow just H1: Pop mean (for B) ≠ 0.336 “µ” .0435 − .0336 .0435 − .0336 M1 M1 ".03573" ".03555" 100 100 Or xcrit = 0.336 +/-“2.576”√(0.12765/100) = 2.77 (3 sfs) A1 Or xcrit = (0.244 ) or 0.428 A1 z = 2.785 (3 sfs) A0 Zcrit = 2.576 B1 Or use of area – correct 0.005 (2-tail) (or 2.326 consistent with 1-tail test ) or 0.01 (1-tail) Valid comparison with z-value M1 Valid comp P(z > 2.77) with 0.005 or 0.01 Or comp 0.435 with “0.428” Evidence that B amounts diff from A A1ft [8] No errors seen. Conclusion consistent with their H0/H1.No contradictions. (ii) Must state or imply “No” to score these marks n large B1 X approx normally distr or CLT applies B1 [2] B0 for “No” with invalid (or no) reason SR both reasons correct but wrong conclusion scores SR B1. GCE A LEVEL – October/November 2010 9709 73

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7 In the past, the number of house sales completed per week by a building company has been modelled by a random variable which has the distribution Po(0.8). Following a publicity campaign, the builders hope that the mean number of sales per week will increase. In order to test at the 5% significance level whether this is the case, the total number of sales during the first 3 weeks after the campaign is noted. It is assumed that a Poisson model is still appropriate. (i) Given that the total number of sales during the 3 weeks is 5, carry out the test. [6] (ii) During the following 3 weeks the same test is carried out again, using the same significance level. Find the probability of a Type I error. [3] (iii) Explain what is meant by a Type I error in this context. [1] (iv) State what further information would be required in order to find the probability of a Type II error. [1]

Mark scheme: 7 (i) H0: mean no. sales = 2.4 B1 Or “= 0.8 per week” H1: mean no. sales > 2.4 Accept λ, not µ. P(X > 5) M1* Attempted with or without “1–“. 4.2 2 4.2 3 4.2 4 Allow one end error. = 1 – e-2.4(1 + 2.4 + + + !2 !3 !4 (= 1 – 0.9041) A1 Allow incorrect λ in otherwise correct expression. = 0.0959 A1 Comp with 0.05 M1* Indep M. (Allow recovery of above 3 marks at this point if comparison with 0.95 done.) No evidence to believe mean sales incr A1ft dep [6] Conclusion, no contradictions. 4.2 5 SC: e-2.4 × = 0.0602 > 0.05: !5 max B1M0A0A0M1A0 (ii) Need 1st x such that P(X > x) < 0.05 M1* Attempt sum of at least 3 relevant Poisson terms, with comparison with 0.05 (can be implied). Can be implied, e.g. by P(X < 5) = 0.9643 identified. 4.2 5 P(X > 6) = 1 – e–2.4(1 + 2.4 + . . . + ) M1*dep !5 (= 1 – 0.9643) = 0.0357 A1 [3] (iii) Mean sales still 0.8 per week, but > 6 sales Conclude mean sales have increased in 3 weeks, so reject 0.8. B1 [1] when not true (iv) Value of true (new, changed) mean oe B1 [1]

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