| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2024 |
| Session | November |
| Marks | 14 |
| 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 testing question requiring understanding of Type I and Type II errors, critical region determination for a two-year period (Po(6.6)), and normal approximation. While the individual techniques are standard for Further Maths Statistics, the combination of error probability calculations and the need to work with a two-year aggregated Poisson parameter makes this moderately challenging, above typical A-level but not requiring exceptional insight. |
| Spec | 2.04d Normal approximation to binomial5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\lambda = 6.6\) | B1 | |
| \(P(X \leq 2) = e^{-6.6}(1 + 6.6 + \frac{6.6^2}{2})\) [= 0.0400] [< 0.05] or \(e^{-6.6}(1 + 6.6 + 21.78)\) or \(0.001360 + 0.008978 + 0.02963\) | M1 | Expression must be seen. No end errors. Allow use of 3.3 here |
| \(P(X \leq 3) = e^{-6.6}(1 + 6.6 + \frac{6.6^2}{2} + \frac{6.6^3}{3!})\) or \(0.0400 + e^{-6.6} \times \frac{6.6^3}{3!} = 0.105\ [> 0.05]\) | B1 | Condone unsupported 0.105 |
| P(Type I error) = 0.0400 (3 sf) | A1 | Allow 0.040 or 0.04 AWRT. SC unsupported ans of 0.0400 can score max B1B1B1 |
| Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(H_0: \lambda = 6.6,\ H_1: \lambda < 6.6\) | B1 | May be seen in part (a) and award B1 mark here. Accept \(\mu\) or \(\lambda\). Accept 3.3 or 6.6 |
| \([P(X \leq 2) = 0.0400]\ \text{'0.04'} < 0.05\) | M1 | For comparing their \(P(X \leq 2)\) any \(\lambda\) with 0.05 |
| [Reject \(H_0\)] There is evidence to suggest that mean number of accidents has decreased | A1 | In context, not definite. No contradictions. CWO |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(X > 2)\) attempted, with any \(\lambda\) | M1 | |
| \(P(X > 2) = 1 - e^{-1.2}(1 + 1.2 + \frac{1.2^2}{2})\) or \(= 1 - e^{-1.2}(1 + 1.2 + 0.72)\) or \(= 1 - (0.3012 + 0.3614 + 0.2169)\) | M1 | Expression must be seen. Correct \(\lambda\). No end errors. |
| \(0.121\) (3 sf) or \(0.120\) | A1 | SC unsupported answer scores B2. |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(N(18, 18)\) seen or implied | B1 | |
| \(\frac{10.5 - 18}{\sqrt{18}} \left[= -1.768\right]\) | M1 | Allow with no or incorrect continuity correction. Their 18. |
| \(P(X > \text{`}-1.768\text{'}) = \Phi(\text{`}1.768\text{'})\) | M1 | ft *their* standardised value. Area consistent with their values. |
| \(= 0.961\) or \(0.962\) (3 sf) | A1 | |
| 4 |
## Question 7(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\lambda = 6.6$ | B1 | |
| $P(X \leq 2) = e^{-6.6}(1 + 6.6 + \frac{6.6^2}{2})$ [= 0.0400] [< 0.05] or $e^{-6.6}(1 + 6.6 + 21.78)$ or $0.001360 + 0.008978 + 0.02963$ | M1 | Expression must be seen. No end errors. Allow use of 3.3 here |
| $P(X \leq 3) = e^{-6.6}(1 + 6.6 + \frac{6.6^2}{2} + \frac{6.6^3}{3!})$ or $0.0400 + e^{-6.6} \times \frac{6.6^3}{3!} = 0.105\ [> 0.05]$ | B1 | Condone unsupported 0.105 |
| P(Type I error) = 0.0400 (3 sf) | A1 | Allow 0.040 or 0.04 AWRT. SC unsupported ans of 0.0400 can score max **B1B1B1** |
| **Total: 4** | | |
## Question 7(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $H_0: \lambda = 6.6,\ H_1: \lambda < 6.6$ | B1 | May be seen in part (a) and award **B1** mark here. Accept $\mu$ or $\lambda$. Accept 3.3 or 6.6 |
| $[P(X \leq 2) = 0.0400]\ \text{'0.04'} < 0.05$ | M1 | For comparing their $P(X \leq 2)$ any $\lambda$ with 0.05 |
| [Reject $H_0$] There is evidence to suggest that mean number of accidents has decreased | A1 | In context, not definite. No contradictions. CWO |
| **Total: 3** | | |
## Question 7(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(X > 2)$ attempted, with any $\lambda$ | **M1** | |
| $P(X > 2) = 1 - e^{-1.2}(1 + 1.2 + \frac{1.2^2}{2})$ or $= 1 - e^{-1.2}(1 + 1.2 + 0.72)$ or $= 1 - (0.3012 + 0.3614 + 0.2169)$ | **M1** | Expression must be seen. Correct $\lambda$. No end errors. |
| $0.121$ (3 sf) or $0.120$ | **A1** | SC unsupported answer scores **B2**. |
| | **3** | |
## Question 7(d):
| Answer | Mark | Guidance |
|--------|------|----------|
| $N(18, 18)$ seen or implied | **B1** | |
| $\frac{10.5 - 18}{\sqrt{18}} \left[= -1.768\right]$ | **M1** | Allow with no or incorrect continuity correction. Their 18. |
| $P(X > \text{`}-1.768\text{'}) = \Phi(\text{`}1.768\text{'})$ | **M1** | ft *their* standardised value. Area consistent with their values. |
| $= 0.961$ or $0.962$ (3 sf) | **A1** | |
| | **4** | |7 The number of accidents per year on a certain road has the distribution $\operatorname { Po } ( \lambda )$. In the past the value of $\lambda$ was 3.3 . Recently, a new speed limit was imposed and the council wishes to test whether the value of $\lambda$ has decreased. The council notes the total number, $X$, of accidents during two randomly chosen years after the speed limit was introduced and it carries out a test at the $5 \%$ significance level.
\begin{enumerate}[label=(\alph*)]
\item Calculate the probability of a Type I error.
\item Given that $X = 2$, carry out the test.\\
\includegraphics[max width=\textwidth, alt={}, center]{acd6f1c9-bbaf-40ca-b5cb-8466ddb9f596-10_2716_40_109_2010}\\
\includegraphics[max width=\textwidth, alt={}, center]{acd6f1c9-bbaf-40ca-b5cb-8466ddb9f596-11_2716_29_107_22}
\item The council decides to carry out another similar test at the $5 \%$ significance level using the same hypotheses and two different randomly chosen years.
Given that the true value of $\lambda$ is 0.6 , calculate the probability of a Type II error.
\item Using $\lambda = 0.6$ and a suitable approximating distribution, find the probability that there will be more than 10 accidents in 30 years.\\
If you use the following page to complete the answer to any question, the question number must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2024 Q7 [14]}}