| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2022 |
| 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 +0.3 This is a standard Further Maths Statistics hypothesis testing question covering routine Poisson test procedures (parts a-d) and normal approximation (part e). While it requires understanding of multiple concepts (critical regions, Type I errors, normal approximation), each part follows textbook methods with no novel problem-solving required. The 2-minute period adjustment and normal approximation are standard techniques at this level, making it slightly easier than average. |
| Spec | 2.04d Normal approximation to binomial2.05a 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 |
|---|---|---|
| Answer | Marks | Guidance |
| \(H_0\): Pop mean \(= 4.6\) [or \(9.2\)]; \(H_1\): Pop mean \(< 4.6\) [or \(9.2\)] | B1 | or \(\lambda = 4.6\) or \(\mu\) (not just 'mean'); or \(\lambda < 4.6\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Use of Poisson with \(\lambda = 9.2\) | B1 | SOI |
| \(P(X \leq 3) = e^{-9.2}(1 + 9.2 + \frac{9.2^2}{2} + \frac{9.2^3}{3!}) = 0.0184\) or \(0.018\ [< 0.02]\) | M1 | At least one of these attempted correct \(\lambda\) (with Poisson expression seen not implied) |
| \(P(X \leq 4) = 0.0184 + e^{-9.2} \times \frac{9.2^4}{4!} = 0.0486\) or \(0.049\ [> 0.02]\) | *A1 | Both correct; SC use of \(\lambda = 4.6\) scores B1 for \(P(X=0) = 0.01[0][1]\) and \(P(X \leq 1) = 0.056[3]\) only |
| CR is \(X \leq 3\) | DA1 | From CWO and at least one comparison seen; SC if M0 awarded allow *B1 for both \(0.018\) and \(0.049\) or better and DB1 for correct critical region from CWO and at least one comparison seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(5\) is not in critical region OR \(P(X \leq 5) = 0.104 > 0.02\), so [not reject \(H_0\)] no evidence that number of cars arriving is now fewer | M1, A1 FT | For a comparison (i.e. \(5 > 3\)) OE; in context, not definite, no contradictions; e.g. not 'No. of cars arriving is not fewer'; ft *their* critical region if used (but must be from Poisson and integers) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| No, because \(H_0\) was not rejected | B1 FT | OE, FT *their* (c) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(N(276, 276)\) | B1 | SOI |
| \(\frac{300.5 - 276}{\sqrt{276}}\ [= 1.475]\) | M1 | Standardising with *their* values; allow with wrong or no continuity correction |
| \(1 - \phi('1.475') = 0.0701\) (3 s.f.) | A1 | SC use of Poisson: B1 for answer \(0.0727\) (3 sf) |
## Question 4(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0$: Pop mean $= 4.6$ [or $9.2$]; $H_1$: Pop mean $< 4.6$ [or $9.2$] | B1 | or $\lambda = 4.6$ or $\mu$ (not just 'mean'); or $\lambda < 4.6$ |
---
## Question 4(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use of Poisson with $\lambda = 9.2$ | B1 | SOI |
| $P(X \leq 3) = e^{-9.2}(1 + 9.2 + \frac{9.2^2}{2} + \frac{9.2^3}{3!}) = 0.0184$ or $0.018\ [< 0.02]$ | M1 | At least one of these attempted correct $\lambda$ (with Poisson expression seen not implied) |
| $P(X \leq 4) = 0.0184 + e^{-9.2} \times \frac{9.2^4}{4!} = 0.0486$ or $0.049\ [> 0.02]$ | *A1 | Both correct; SC use of $\lambda = 4.6$ scores B1 for $P(X=0) = 0.01[0][1]$ and $P(X \leq 1) = 0.056[3]$ only |
| CR is $X \leq 3$ | DA1 | From CWO and at least one comparison seen; SC if M0 awarded allow *B1 for both $0.018$ and $0.049$ or better and DB1 for correct critical region from CWO and at least one comparison seen |
---
## Question 4(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $5$ is not in critical region OR $P(X \leq 5) = 0.104 > 0.02$, so [not reject $H_0$] no evidence that number of cars arriving is now fewer | M1, A1 FT | For a comparison (i.e. $5 > 3$) OE; in context, not definite, no contradictions; e.g. not 'No. of cars arriving is not fewer'; ft *their* critical region if used (but must be from Poisson and integers) |
---
## Question 4(d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| No, because $H_0$ was not rejected | B1 FT | OE, FT *their* (c) |
---
## Question 4(e):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $N(276, 276)$ | B1 | SOI |
| $\frac{300.5 - 276}{\sqrt{276}}\ [= 1.475]$ | M1 | Standardising with *their* values; allow with wrong or no continuity correction |
| $1 - \phi('1.475') = 0.0701$ (3 s.f.) | A1 | SC use of Poisson: B1 for answer $0.0727$ (3 sf) |
---4 The number of cars arriving at a certain road junction on a weekday morning has a Poisson distribution with mean 4.6 per minute. Traffic lights are installed at the junction and council officer wishes to test at the $2 \%$ significance level whether there are now fewer cars arriving. He notes the number of cars arriving during a randomly chosen 2 -minute period.
\begin{enumerate}[label=(\alph*)]
\item State suitable null and alternative hypotheses for the test.
\item Find the critical region for the test.\\
The officer notes that, during a randomly chosen 2 -minute period on a weekday morning, exactly 5 cars arrive at the junction.
\item Carry out the test.
\item State, with a reason, whether it is possible that a Type I error has been made in carrying out the test in part (c).\\
The number of cars arriving at another junction on a weekday morning also has a Poisson distribution with mean 4.6 per minute.
\item Use a suitable approximating distribution to find the probability that more than 300 cars will arrive at this junction in an hour.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2022 Q4 [11]}}