| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2021 |
| Session | June |
| Marks | 6 |
| 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 a one-tailed Poisson hypothesis test with standard setup. Part (a) requires stating hypotheses and calculating P(X≥5) for Po(2.4), which is routine. Part (b) tests understanding of the Poisson assumption of constant rate. Slightly above average difficulty due to the three-year aggregation requiring λ=2.4, but otherwise a textbook Further Maths Statistics question with no novel problem-solving required. |
| Spec | 2.05a Hypothesis testing language: null, alternative, p-value, significance5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| \(H_0: \lambda = 2.4\); \(H_1: \lambda > 2.4\) | B1 | Accept \(\lambda\) or \(\mu\); Accept 2.4 or 0.8 (per year) |
| Answer | Marks | Guidance |
|---|---|---|
| \(1 - e^{-2.4}\!\left(1 + 2.4 + \dfrac{2.4^2}{2} + \dfrac{2.4^3}{3!} + \dfrac{2.4^4}{4!}\right)\) | M1 | Any \(\lambda\); allow one end error |
| \(0.0959\) (3 sf) | A1 | SC unsupported answer 0.0959 scores B1 only not M1A1 |
| \(0.0959 > 0.05\) | M1 | Valid comparison; Use of \(0.9041 < 0.95\) can recover either M1A1 or B1 |
| There is evidence that Jane's claim not justified or There is insufficient evidence to support Jane's claim | A1 FT | OE. In context, not definite, e.g. not 'Jane is wrong', no contradictions. Condone omission of Jane. |
| Answer | Marks |
|---|---|
| Mean not constant so Poisson model not valid | B1 |
## Question 3:
### Part 3(a)(i):
$H_0: \lambda = 2.4$; $H_1: \lambda > 2.4$ | **B1** | Accept $\lambda$ or $\mu$; Accept 2.4 or 0.8 (per year)
### Part 3(a)(ii):
$1 - e^{-2.4}\!\left(1 + 2.4 + \dfrac{2.4^2}{2} + \dfrac{2.4^3}{3!} + \dfrac{2.4^4}{4!}\right)$ | **M1** | Any $\lambda$; allow one end error
$0.0959$ (3 sf) | **A1** | SC unsupported answer 0.0959 scores B1 only not M1A1
$0.0959 > 0.05$ | **M1** | Valid comparison; Use of $0.9041 < 0.95$ can recover either M1A1 or B1
There is evidence that Jane's claim not justified or There is insufficient evidence to support Jane's claim | **A1 FT** | OE. In context, not definite, e.g. not 'Jane is wrong', no contradictions. Condone omission of Jane.
### Part 3(b):
Mean not constant so Poisson model not valid | **B1** |
---3 The local council claims that the average number of accidents per year on a particular road is 0.8 . Jane claims that the true average is greater than 0.8 . She looks at the records for a random sample of 3 recent years and finds that the total number of accidents during those 3 years was 5 .
\begin{enumerate}[label=(\alph*)]
\item Assume that the number of accidents per year follows a Poisson distribution.
\begin{enumerate}[label=(\roman*)]
\item State null and alternative hypotheses for a test of Jane's claim.
\item Test at the $5 \%$ significance level whether Jane's claim is justified.
\end{enumerate}\item Jane finds that the number of accidents per year has been gradually increasing over recent years. State how this might affect the validity of the test carried out in part (a)(ii).
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2021 Q3 [6]}}