| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2023 |
| Session | June |
| 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 +0.3 This is a standard hypothesis testing question covering routine procedures: stating hypotheses, finding critical regions from tables, defining Type I error, and making a conclusion. Part (e) adds a normal approximation which is straightforward. All steps are textbook applications with no novel problem-solving required, 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 probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(H_0: \lambda = 7.6\) [or 1.9], \(H_1: \lambda < 7.6\) [or 1.9] | B1 | Or Population mean \(= 7.6\) or \(\mu\) (not just 'mean'). Or Population mean \(< 7.6\) or \(\mu\). |
| Total: 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Mean \(= 7.6\) | B1 | Seen. |
| \(P(X \leqslant 2) = e^{-7.6}\left(1 + 7.6 + \frac{7.6^2}{2}\right)\) \([= 0.0188\) or \(0.0187]\) | M1 | OE. |
| \(P(X \leqslant 3) = e^{-7.6}\left(1 + 7.6 + \frac{7.6^2}{2} + \frac{7.6^3}{3!}\right)\) \([= 0.0554\) or \(0.0553]\) | M1 | OE. Expression must be seen in at least one probability calculation. |
| \(0.0188\) or \(0.0187\) and \(0.0554\) or \(0.0553\) | A1 | A1 for both values. |
| Critical region is \(X \leqslant 2\) | A1 | Dep on both M marks. SC No Poisson expression seen in either prob scores B1 for 0.0188 or 0.0187 and B1 for 0.0554 or 0.0553 and B1 for CR. |
| \(P(\text{Type I error}) = P(X \leqslant 2) = 0.0188\) or \(0.0187\) (3 sf) | B1FT | FT *their* \(P(X \leqslant 2)\) or *their* CR. |
| Total: 6 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Concluding that the (mean) no. of accidents has reduced when it has not. | B1 | OE. Must be in context. Accept: 'It is believed that the booklet has helped to improve safety when actually it has not'. |
| Total: 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| 3 not in critical region. | M1 | FT their CR or \(P(X \leq 3) = 0.0554 > 0.05\). |
| No evidence mean number of accidents has decreased. | A1FT | In context. Cannot be a definite statement, e.g., 'mean number accidents has not decreased'. |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(N(98.8, 98.8)\) | B1 | May be implied. |
| \(\dfrac{100.5 - 98.8}{\sqrt{98.8}} \left[= 0.171\right]\) | M1 | For standardising (could be implied by correct answer). Allow with wrong or no continuity correction. |
| \(1 - \Phi(0.171)\) | M1 | For probability area consistent with their working. |
| \(= 0.432\) (3 sf) | A1 | |
| Total: 4 |
## Question 7(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0: \lambda = 7.6$ [or 1.9], $H_1: \lambda < 7.6$ [or 1.9] | B1 | Or Population mean $= 7.6$ or $\mu$ (not just 'mean'). Or Population mean $< 7.6$ or $\mu$. |
| **Total: 1** | | |
## Question 7(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Mean $= 7.6$ | B1 | Seen. |
| $P(X \leqslant 2) = e^{-7.6}\left(1 + 7.6 + \frac{7.6^2}{2}\right)$ $[= 0.0188$ or $0.0187]$ | M1 | OE. |
| $P(X \leqslant 3) = e^{-7.6}\left(1 + 7.6 + \frac{7.6^2}{2} + \frac{7.6^3}{3!}\right)$ $[= 0.0554$ or $0.0553]$ | M1 | OE. Expression must be seen in at least one probability calculation. |
| $0.0188$ or $0.0187$ and $0.0554$ or $0.0553$ | A1 | A1 for both values. |
| Critical region is $X \leqslant 2$ | A1 | Dep on both M marks. SC No Poisson expression seen in either prob scores B1 for 0.0188 or 0.0187 and B1 for 0.0554 or 0.0553 and B1 for CR. |
| $P(\text{Type I error}) = P(X \leqslant 2) = 0.0188$ or $0.0187$ (3 sf) | B1FT | FT *their* $P(X \leqslant 2)$ or *their* CR. |
| **Total: 6** | | |
## Question 7(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Concluding that the (mean) no. of accidents has reduced when it has not. | B1 | OE. Must be in context. Accept: 'It is believed that the booklet has helped to improve safety when actually it has not'. |
| **Total: 1** | | |
## Question 7(d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| 3 not in critical region. | M1 | FT their CR or $P(X \leq 3) = 0.0554 > 0.05$. |
| No evidence mean number of accidents has decreased. | A1FT | In context. Cannot be a definite statement, e.g., 'mean number accidents has not decreased'. |
| **Total: 2** | | |
## Question 7(e):
| Answer | Mark | Guidance |
|--------|------|----------|
| $N(98.8, 98.8)$ | **B1** | May be implied. |
| $\dfrac{100.5 - 98.8}{\sqrt{98.8}} \left[= 0.171\right]$ | **M1** | For standardising (could be implied by correct answer). Allow with wrong or no continuity correction. |
| $1 - \Phi(0.171)$ | **M1** | For probability area consistent with their working. |
| $= 0.432$ (3 sf) | **A1** | |
| **Total: 4** | | |7 The number of accidents per week at a certain factory has a Poisson distribution. In the past the mean has been 1.9 accidents per week. Last year, the manager gave all his employees a new booklet on safety. He decides to test, at the 5\% significance level, whether the mean number of accidents has been reduced. He notes the number of accidents during 4 randomly chosen weeks this year.
\begin{enumerate}[label=(\alph*)]
\item State suitable null and alternative hypotheses for the test.
\item Find the critical region for the test and state the probability of a Type I error.
\item State what is meant by a Type I error in this context.
\item During the 4 randomly chosen weeks there are a total of 3 accidents.
State the conclusion that the manager should reach. Give a reason for your answer.
\item Assuming that the mean remains 1.9 accidents per week, use a suitable approximation to calculate the probability that there will be more than 100 accidents during a 52-week period.\\
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 Q7 [14]}}