The number of accidents per year on a certain road has the distribution Po(λ). In the past the value of λ was 3.3. Recently, a new speed limit was imposed and the council wishes to test whether the value of λ has decreased. The council notes the total number, X, of accidents during two randomly chosen years after the speed limit was introduced and it carries out a test at the 5% significance level.
- (a)Calculate the probability of a Type I error.[4]
- (b)Given that X = 2, carry out the test.[3]
- (c)The council decides to carry out another similar test at the 5% significance level using the same hypotheses and two different randomly chosen years. Given that the true value of λ is 0.6, calculate the probability of a Type II error.[3]
- (d)Using λ = 0.6 and a suitable approximating distribution, find the probability that there will be more than 10 accidents in 30 years.[4]
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2024). The mark or mark type is shown in brackets.
- [B1] For two years, λ = 3.3 * 2 = 6.6
- [M1] P(X < 2) = e⁻⁶.⁶(1 + 6.6 + 6.6²/2!) [= 0.0400] [< 0.05 ] or e⁻⁶.⁶(1 + 6.6 + 21.78) or 0.001360 + 0.008978 + 0.02963
- [B1] P(X ≤ 3) = e⁻⁶.⁶(1 + 6.6 + 6.6²/2 + 6.6³/3!) or 0.0400 + e⁻⁶.⁶ 6.6³/3! = 0.105 [ > 0.05 ]
- [A1] P(Type I error) = 0.0400 (3 sf)
- [B1] H₀: λ = 6.6, H₁: λ < 6.6
- [M1] [P(X ≤ 2) = 0.0400], 0.04 < 0.05
- [A1] [Reject H₀] There is evidence to suggest that mean number of accidents has decreased
- [M1] P(X > 2) attempted, with any λ
- [M1] P(X > 2) = 1 - e⁻¹.²(1 + 1.2 + 1.2²/2!) or = 1 - e⁻¹.²(1 + 1.2 +0.72) or = 1- (0.3012 +0.3614 +0.2169 )
- [A1] 0.121 (3 sf) or 0.120
- [B1] For 30 years, λ = 0.6 * 30 = 18. N(18, 18) seen or implied
- [M1] P(Y > 10.5) = (10.5 - 18) / √18 [= -1.768]
- [M1] P(X > '-1.768') = Φ('1.768')
- [A1] = 0.961 or 0.962 (3 sf)
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