| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2023 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of a Poisson distribution |
| Type | One-tailed test (increase or decrease) |
| Difficulty | Standard +0.8 This is a multi-part Poisson hypothesis test requiring understanding of Type I/II errors, critical region determination, and contextual interpretation. While the calculations are straightforward (testing λ=3.03 vs λ>3.03), it requires proper statistical reasoning about error types and test design, placing it moderately above average difficulty for A-level statistics. |
| Spec | 5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(H_0\): Pop mean no. people \(= 3.03\) or \(1.01\) (per 20 min); \(H_1\): Pop mean no. people \(> 3.03\) or \(1.01\) (per 20 min) | B1 | Must not just be 'mean', but allow just '\(\lambda\)' or '\(\mu\)'. |
| Use of \(\text{Po}(3.03)\) | M1 | |
| \(= 1 - e^{-3.03}(1 + 3.03 + \frac{3.03^2}{2} + \frac{3.03^3}{3!} + \frac{3.03^4}{4!} + \frac{3.03^5}{5!})\) \(= 1 - e^{-3.03}(1 + 3.03 + 4.5905 + 4.6364 + 3.5120 + 2.128)\) \(= 1 - (0.04832 + 0.1464 + 0.2218 + 0.2240 + 0.1697 + 0.1028)\) | M1 | Allow incorrect \(\lambda\). Allow one end error. Must see Poisson expression used. |
| \(= 0.0870\) (3sf) \([0.0869727]\) | A1 | Allow \(0.087\). |
| \(0.0870 > 0.05\) | M1 | For a valid comparison. |
| (Do not reject \(H_0\)) Insufficient evidence to believe (mean) number of people has increased | A1FT | Conclusion stated must be in context, not definite and include no contradictions. |
| Total | 6 | If only \(P(x=6)\) award max 2/6. SC No working: B1 B2 M1 A1. Award maximum 5/6. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\text{"0.0869727"} - e^{-3.03} \times \frac{3.03^6}{6!}\) or \(0.869727 - e^{-3.03}(1.0748)\) or \(0.869727 - 0.05193\) or \(1 - e^{-3.03}(1 + 3.03 + \frac{3.03^2}{2} + \frac{3.03^3}{3!} + \frac{3.03^4}{4!} + \frac{3.03^5}{5!} + \frac{3.03^6}{6!})\) | M1 | OE. Must see Poisson expression (may be in part (a)). |
| \(0.0350\) or \(0.0351\) | A1 | Accept \(0.035\). SC no working seen, award B1 for \(0.0350\), \(0.0351\) or \(0.035\). |
| Total | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Concluding that the (mean) number of people (using the path per 20 mins in the evening) has increased when it has not | B1 | OE. Conclusion must be in context. |
| Total | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| A value for the true mean | B1 | Allow without context for this mark. |
| Number of people using the path per 20 mins in the evening. | B1 | Condone equivalent comment on three randomly chosen 20-minute periods. |
| Total | 2 |
## Question 8(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0$: Pop mean no. people $= 3.03$ or $1.01$ (per 20 min); $H_1$: Pop mean no. people $> 3.03$ or $1.01$ (per 20 min) | B1 | Must not just be 'mean', but allow just '$\lambda$' or '$\mu$'. |
| Use of $\text{Po}(3.03)$ | M1 | |
| $= 1 - e^{-3.03}(1 + 3.03 + \frac{3.03^2}{2} + \frac{3.03^3}{3!} + \frac{3.03^4}{4!} + \frac{3.03^5}{5!})$ $= 1 - e^{-3.03}(1 + 3.03 + 4.5905 + 4.6364 + 3.5120 + 2.128)$ $= 1 - (0.04832 + 0.1464 + 0.2218 + 0.2240 + 0.1697 + 0.1028)$ | M1 | Allow incorrect $\lambda$. Allow one end error. Must see Poisson expression used. |
| $= 0.0870$ (3sf) $[0.0869727]$ | A1 | Allow $0.087$. |
| $0.0870 > 0.05$ | M1 | For a valid comparison. |
| (Do not reject $H_0$) Insufficient evidence to believe (mean) number of people has increased | A1FT | Conclusion stated must be in context, not definite and include no contradictions. |
| **Total** | **6** | If only $P(x=6)$ award max 2/6. **SC** No working: B1 B2 M1 A1. Award maximum 5/6. |
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## Question 8(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\text{"0.0869727"} - e^{-3.03} \times \frac{3.03^6}{6!}$ or $0.869727 - e^{-3.03}(1.0748)$ or $0.869727 - 0.05193$ or $1 - e^{-3.03}(1 + 3.03 + \frac{3.03^2}{2} + \frac{3.03^3}{3!} + \frac{3.03^4}{4!} + \frac{3.03^5}{5!} + \frac{3.03^6}{6!})$ | M1 | OE. Must see Poisson expression (may be in part (a)). |
| $0.0350$ or $0.0351$ | A1 | Accept $0.035$. **SC** no working seen, award B1 for $0.0350$, $0.0351$ or $0.035$. |
| **Total** | **2** | |
---
## Question 8(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Concluding that the (mean) number of people (using the path per 20 mins in the evening) has increased when it has not | B1 | OE. Conclusion must be in context. |
| **Total** | **1** | |
---
## Question 8(d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| A value for the true mean | B1 | Allow without context for this mark. |
| Number of people using the path per 20 mins in the evening. | B1 | Condone equivalent comment on three randomly chosen 20-minute periods. |
| **Total** | **2** | |8 A new light was installed on a certain footpath. A town councillor decided to use a hypothesis test to investigate whether the number of people using the path in the evening had increased.
Before the light was installed, the mean number of people using the path during any 20 -minute period during the evening was 1.01.
After the light was installed, the total number, $n$, of people using the path during 3 randomly chosen 20 -minute periods during the evening was noted.
\begin{enumerate}[label=(\alph*)]
\item Given that the value of $n$ was 6 , use a Poisson distribution to carry out the test at the $5 \%$ significance level.
\item Later a similar test, at the $5 \%$ significance level, was carried out using another 3 randomly chosen 20 -minute periods during the evening.
Find the probability of a Type I error.
\item State what is meant by a Type I error in this context.
\item State, in context, what further information would be needed in order to find the probability of a Type II error. Do not carry out any further calculation.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2023 Q8 [11]}}