| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2018 |
| Session | Specimen |
| Marks | 7 |
| 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 clear structure: part (a) is routine probability calculation from tables, and part (b) follows a standard one-tailed test procedure with explicit guidance to state hypotheses. The question requires only direct application of learned techniques with no novel problem-solving or conceptual challenges, making it slightly easier than average. |
| Spec | 5.02i 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 |
| VIIIV SIHI NI JIIIM ION OC | VIIV SIHI NI JAHM ION OO | VI4V SIHI NI JIIIM I ON OO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | 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 \(\frac{e^{-\lambda}\lambda^5}{5!} + \frac{e^{-\lambda}\lambda^6}{6!}\) |
| \(= 0.6063 - 0.2851\) | ||
| \(= 0.3212\) awrt 0.321 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0: \lambda = 9\), \(H_1: \lambda < 9\) | B1 | Both hypotheses correct (\(\lambda\) or \(\mu\)), allow 0.5 instead of 9 |
| \(X \sim \text{Po}(9)\) therefore \(P(X \leq 4) = 0.05496...\) or CR \(X \leq 3\) | B1 | Either 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 | Cso requires correct contextual conclusion with underlined words and all previous marks scored |
## Question 4:
### Part (a):
| Answer/Working | Mark | 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 $\frac{e^{-\lambda}\lambda^5}{5!} + \frac{e^{-\lambda}\lambda^6}{6!}$ |
| $= 0.6063 - 0.2851$ | | |
| $= 0.3212$ awrt 0.321 | A1 | |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0: \lambda = 9$, $H_1: \lambda < 9$ | B1 | Both hypotheses correct ($\lambda$ or $\mu$), allow 0.5 instead of 9 |
| $X \sim \text{Po}(9)$ therefore $P(X \leq 4) = 0.05496...$ or CR $X \leq 3$ | B1 | Either 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 | Cso requires correct contextual conclusion with underlined words and all previous marks scored |
---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.
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VIIIV SIHI NI JIIIM ION OC & VIIV SIHI NI JAHM ION OO & VI4V SIHI NI JIIIM I ON OO \\
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\hfill \mbox{\textit{Edexcel S2 2018 Q4 [7]}}