The number of accidents per week at a certain factory has a Poisson distribution. In the past the mean has been 1.9 accidents per week. Last year, the manager gave all his employees a new booklet on safety. He decides to test, at the 5% significance level, whether the mean number of accidents has been reduced. He notes the number of accidents during 4 randomly chosen weeks this year.

Mark scheme

Marking points from the official mark scheme (May/June 2023). The mark or mark type is shown in brackets.

1. [B1] H₀: λ = 7.6 [or 1.9] H₁: λ < 7.6 [or 1.9] (Or Population mean = 7.6 or μ (not just 'mean'). Or Population mean < 7.6 or μ.)
2. [B1] Mean = 7.6 (Seen.)
3. [M1] P(X < 2) = e^-7.6 (1 + 7.6 + 7.6^2/2!) (OE. [= 0.0188 or 0.0187])
4. [M1] P(X < 3) = e^-7.6 (1 + 7.6 + 7.6^2/2! + 7.6^3/3!) (OE. Expression must be seen in at least one probability calculation. [= 0.0554 or 0.0553])
5. [A1] 0.0188 or 0.0187 and 0.0554 or 0.0553 (A1 for both values.)
6. [A1] Critical region is X ≤ 2 (Dep on both M marks. SC No Poisson expression seen in either prob scores B1 for 0.0188 or 0.0187 and B1 for 0.0554 or 0.0553 and B1 for CR.)
7. [B1FT] P(Type I error) = P(X < 2) = 0.0188 or 0.0187 (3 sf) (FT their P(X < 2) or their CR.)
8. [B1] Concluding that the (mean) no. of accidents has reduced when it has not. (OE. Must be in context. Accept: 'It is believed that the booklet has helped to improve safety when actually it has not'.)
9. [M1] 3 not in critical region. (FT their CR or P(X ≤ 3) = 0.0554 > 0.05 .)
10. [A1FT] No evidence mean number of accidents has decreased. (In context. Cannot be a definite statement, e.g., 'mean number accidents has not decreased'.)
11. [B1] N(98.8, 98.8) (May be implied.)
12. [M1] (100.5-98.8) / √98.8 (For standardising (could be implied by correct answer). Allow with wrong or no continuity correction. [= 0.171])
13. [M1] 1 – Φ('0.171') (For probability area consistent with their working.)
14. [A1] = 0.432 (3 sf)