A-LevelMathematicsProbability and statistics9709/62 · Feb/Mar 2025 · Q29 Marks

A researcher records the time, T seconds, taken by adults to complete a questionnaire. The results for a random sample of 60 adults who completed the questionnaire this year are summarised as follows. n = 60 ΣT = 3678 ΣT² = 226313.36

(a)Find an unbiased estimate of E(T), and show that an unbiased estimate of Var(T) is 14.44.[3]
(b)Test at the 2% significance level whether the population mean time for this year is less than 62.4 seconds.[5]
(c)State, with a reason, whether it was necessary to use the Central Limit Theorem in your answer to part (b).[1]

Mark scheme

Marking points from the official mark scheme (Feb/Mar 2025). The mark or mark type is shown in brackets.

[B1] μ = 61.3 = 3678/60 = 613/10
[M1] σ² = (1/59) (226313.36 -'61.3'²) = 1/59 (226313.36 -3678²/60) [=361/25]
[A1] = 14.44 AG from correct expression must see 14.44.
[B1] Ho: Population mean = 62.4 H1: Population mean < 62.4 Allow 'μ' but not just 'mean'.
[M1] '61.3'-62.4 / √(14.44/60) Standardise with their mean. Ignore cc for M1. Must have √60.
[A1] = -2.242 (accept ±) Accept 3sf if nothing better seen.
[M1] 2.242 > 2.054 or -2.242 < -2.054 (accept 2.055) or compare 1 – ф("2.242") with 0.02. i.e. 10.9876 = 0.0124 or 0.0125 < 0.02.
[A1ft] [Reject Ho] There is sufficient evidence to suggest that (at 2% level) that the (mean) time has decreased. OE. Not definite, e.g. not 'The mean time has decreased' No contradictions. In context. Accept cv method 61.392 M1 A1 61.392>61.3 M1 A1 OR 62.308 M1 A1 62.4>62.308 M1 A1 (3sf accuracy) SC For 2-tail test B0 M1 A1 M1(with 2.326 OE) A0
[B1] Yes, because population distribution of times is unknown OE. Allow ‘... is not normal' (Accept ‘parent' dist Accept underlying distribution).

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