| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2017 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of a Poisson distribution |
| Type | One-tailed test (increase or decrease) |
| Difficulty | Standard +0.3 This is a straightforward application of Poisson hypothesis testing with a standard one-tailed test setup. Part (i) requires summing Poisson distributions (5 days gives Po(3.5)) and comparing P(X≥5) to 10% significance level—routine calculation. Part (ii) involves adding independent Poisson distributions (Po(2.1) + Po(1.8) = Po(3.9)) and finding a single probability—both are standard textbook exercises requiring only direct application of formulas with no novel insight. |
| Spec | 2.05b Hypothesis test for binomial proportion5.02n Sum of Poisson variables: is Poisson |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(H_0\): mean no. sales \(= 3.5\) | B1 | or "… \(= 0.7\) (per day)" |
| \(H_1\): mean no. sales \(> 3.5\) | M1 | allow '\(\lambda\)' or '\(\mu\)' but not just 'mean' |
| \(P(X \geq 5) = 1 - e^{-3.5}\left(1 + 3.5 + \frac{3.5^2}{2!} + \frac{3.5^3}{3!} + \frac{3.5^4}{4!}\right)\) | M1 | |
| \(= 0.275\) | A1 | allow \(0.274\) |
| Comp with \(0.10\) | M1 | valid comparison using Poisson |
| No evidence (at 10%) to believe that sales per day have increased | A1 FT | correct conclusion FT; no contradictions |
| Total: 6 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\lambda = 3.9\) | B1 | |
| \(e^{-3.9} \times \frac{3.9^2}{2!}\) | M1 | any \(\lambda\) (\(\neq 0.7\) or \(0.6\)), single term |
| \(= 0.154\) (3 sf) | A1 | |
| Total: 3 |
## Question 7(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $H_0$: mean no. sales $= 3.5$ | B1 | or "… $= 0.7$ (per day)" |
| $H_1$: mean no. sales $> 3.5$ | M1 | allow '$\lambda$' or '$\mu$' but not just 'mean' |
| $P(X \geq 5) = 1 - e^{-3.5}\left(1 + 3.5 + \frac{3.5^2}{2!} + \frac{3.5^3}{3!} + \frac{3.5^4}{4!}\right)$ | M1 | |
| $= 0.275$ | A1 | allow $0.274$ |
| Comp with $0.10$ | M1 | valid comparison using Poisson |
| No evidence (at 10%) to believe that sales per day have increased | A1 FT | correct conclusion FT; no contradictions |
| **Total: 6** | | |
# Question 7(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\lambda = 3.9$ | B1 | |
| $e^{-3.9} \times \frac{3.9^2}{2!}$ | M1 | any $\lambda$ ($\neq 0.7$ or $0.6$), single term |
| $= 0.154$ (3 sf) | A1 | |
| **Total: 3** | | |
---7 In the past the number of cars sold per day at a showroom has been modelled by a random variable with distribution $\operatorname { Po } ( 0.7 )$. Following an advertising campaign, it is hoped that the mean number of sales per day will increase. In order to test at the $10 \%$ significance level whether this is the case, the total number of sales during the first 5 days after the campaign is noted. You should assume that a Poisson model is still appropriate.\\
(i) Given that the total number of cars sold during the 5 days is 5 , carry out the test.\\
The number of cars sold per day at another showroom has the independent distribution $\operatorname { Po } ( 0.6 )$. Assume that the distribution for the first showroom is still $\operatorname { Po } ( 0.7 )$.\\
(ii) Find the probability that the total number of cars sold in the two showrooms during 3 days is exactly 2 .\\
\hfill \mbox{\textit{CAIE S2 2017 Q7 [9]}}