## Question 7:

**Part (i):**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0$: mean no. sales $= 2.4$; $H_1$: mean no. sales $> 2.4$ | B1 | Or "$= 0.8$ per week". Accept $\lambda$, not $\mu$ |
| $P(X \geq 5) = 1 - e^{-2.4}\left(1 + 2.4 + \frac{2.4^2}{2!} + \frac{2.4^3}{3!} + \frac{2.4^4}{4!}\right)$ | M1* | Attempted with or without "$1-$". Allow one end error |
| $= 1 - 0.9041$ | A1 | Allow incorrect $\lambda$ in otherwise correct expression |
| $= 0.0959$ | A1 | |
| Compare 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 increased | A1ft dep [6] | Conclusion, no contradictions. SC: $e^{-2.4} \times \frac{2.4^5}{5!} = 0.0602 > 0.05$: max B1M0A0A0M1A0 |

**Part (ii):**

| Answer/Working | Mark | Guidance |
|---|---|---|
| Need $1^{\text{st}}\ x$ such that $P(X \geq x) < 0.05$ | M1* | Attempt sum of at least 3 relevant Poisson terms, with comparison with $0.05$ (can be implied). Can be implied, e.g. by $P(X \leq 5) = 0.9643$ identified |
| $P(X \geq 6) = 1 - e^{-2.4}\left(1 + 2.4 + \ldots + \frac{2.4^5}{5!}\right)$ | M1*dep | |
| $= 1 - 0.9643 = 0.0357$ | A1 [3] | |

**Part (iii):**

| Answer/Working | Mark | Guidance |
|---|---|---|
| Mean sales still $0.8$ per week, but $\geq 6$ sales in 3 weeks, so reject $0.8$ | B1 [1] | Conclude mean sales have increased when not true |

**Part (iv):**

| Answer/Working | Mark | Guidance |
|---|---|---|
| Value of true (new, changed) mean | B1 [1] | |