| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2015 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Poisson hypothesis test |
| Difficulty | Standard +0.3 This is a straightforward application of Poisson distribution with standard hypothesis testing. Part (a) requires basic probability calculation with given rate (λ=6 for a year). Part (b) is a routine one-tailed hypothesis test following a standard template: state H₀ and H₁, calculate probability under H₀ (λ=9 for 18 months), compare to 5% significance level. The question requires no conceptual insight beyond applying the standard S2 hypothesis testing procedure, making it slightly easier than average. |
| Spec | 2.05a Hypothesis testing language: null, alternative, p-value, significance2.05c Significance levels: one-tail and two-tail5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| \(X \sim \text{Po}(6)\) | M1 | Writing or using Po(6) |
| \(P(5 \leq X < 7) = P(X \leq 6) - P(X \leq 4)\) or \(\frac{e^{-6}6^5}{5!} + \frac{e^{-6}6^6}{6!}\) | M1 | Either \(P(X\leq 6)-P(X\leq 4)\) or direct sum |
| \(= 0.6063 - 0.2851 = 0.3212\) | A1 | awrt 0.321 |
| Answer | Marks | Guidance |
|---|---|---|
| \(H_0: \lambda = 9\), \(H_1: \lambda < 9\) | B1 | Both hypotheses correct (\(\lambda\) or \(\mu\)); allow 0.5 instead of 9 |
| \(X \sim \text{Po}(9)\); \(P(X \leq 4) = 0.05496...\) or CR \(X \leq 3\) | B1 | awrt 0.055 or critical region \(X \leq 3\) |
| Insufficient evidence to reject \(H_0\) or Not Significant or 4 does not lie in the critical region | dM1 | Correct comment dependent on previous B1; contradictory non-contextual statements score M0 |
| There is no evidence that the mean number of accidents at the crossroads has reduced/decreased | A1cso | Correct contextual conclusion with underlined words; all previous marks in (b) required |
## Question 4:
### Part (a):
| $X \sim \text{Po}(6)$ | M1 | Writing or using Po(6) |
| $P(5 \leq X < 7) = P(X \leq 6) - P(X \leq 4)$ or $\frac{e^{-6}6^5}{5!} + \frac{e^{-6}6^6}{6!}$ | M1 | Either $P(X\leq 6)-P(X\leq 4)$ or direct sum |
| $= 0.6063 - 0.2851 = 0.3212$ | A1 | awrt 0.321 |
### Part (b):
| $H_0: \lambda = 9$, $H_1: \lambda < 9$ | B1 | Both hypotheses correct ($\lambda$ or $\mu$); allow 0.5 instead of 9 |
| $X \sim \text{Po}(9)$; $P(X \leq 4) = 0.05496...$ or CR $X \leq 3$ | B1 | awrt 0.055 **or** critical region $X \leq 3$ |
| Insufficient evidence to reject $H_0$ **or** Not Significant **or** 4 does not lie in the critical region | dM1 | Correct comment dependent on previous B1; contradictory non-contextual statements score M0 |
| There is no evidence that the mean number of accidents at the crossroads has reduced/decreased | A1cso | Correct contextual conclusion with underlined words; all previous marks in (b) required |
---4. Accidents occur randomly at a crossroads at a rate of 0.5 per month. A researcher records the number of accidents, $X$, which occur at the crossroads in a year.
\begin{enumerate}[label=(\alph*)]
\item Find $\mathrm { P } ( 5 \leqslant X < 7 )$
A new system is introduced at the crossroads. In the first 18 months, 4 accidents occur at the crossroads.
\item Test, at the $5 \%$ level of significance, whether or not there is reason to believe that the new system has led to a reduction in the mean number of accidents per month. State your hypotheses clearly.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2015 Q4 [7]}}