The number of faults in cloth made on a certain machine has a Poisson distribution with mean 2.4 per 10m². An adjustment is made to the machine. It is required to test at the 5% significance level whether the mean number of faults has decreased. A randomly selected 30 m² of cloth is checked and the number of faults is found.

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2022). The mark or mark type is shown in brackets.

1. [B1] 4(a) Ho: Population mean = 7.2 or 2.4 H1: Population mean < 7.2 or 2.4 or λ or μ = 7.2 or 2.4 (Not just 'mean'). or λ or μ < 7.2 or 2.4
2. [B1] 4(b) λ = 7.2 SOI
3. [M1] 4(b) [P(X≤2)] = e^(-7.2) (1+7.2+7.2²/2) or e^(-7.2) (1 + 7.2 + 25.92) or 0.0007465 +0.0053754 + 0.01935 [= 0.0255] [P(X≤3)] = ‘0.0255’+ e^(-7.2) x 7.2³/3! or ‘0.0255’+e^(-7.2) (62.21) or ‘0.0255’+0.04644 [= 0.0719] Both expressions needed, allow any λ If λ ≠ 7.2 allow P(X ≤ n) for 2 consecutive values of n with P(X < n) < 0.05 and P(X ≤ n + 1) > 0.05.
4. [B1] 4(b) P(Type I) = 0.02547 or 0.0255 (3 sf)
5. [M1] 4(c) 3 > 2 or P(X < 3) > 0.05 or '0.0719' > 0.05 For a valid comparison or 3 outside critical region. FT their CR in (b).
6. [A1 FT] 4(c) [Not reject Ho] No evidence that [mean] number of faults has decreased No contradictions. In context, not definite.
7. [M1] 4(d) 1 – e^(-1.5) (1 + 1.5 + 1.5²/2) or 1 – e^(-1.5) (1 + 1.5 + 1.125) or 1 – (0.2231 +0.3347 +0.2510) Must see expression. FT their CR in (b).
8. [A1] 4(d) = 0.191 (3 sf)