| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2017 |
| 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 straightforward application of Poisson hypothesis testing with standard procedures: conducting a one-tailed test with given data, defining Type II error contextually, and calculating its probability. While it requires understanding of hypothesis testing concepts and Poisson distribution properties, all steps follow routine procedures taught in S2 with no novel problem-solving required. The multi-part structure and Type II error calculation elevate it slightly above average difficulty. |
| Spec | 2.05a 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 | Mark | Guidance |
| \(H_0\): Pop mean no. accidents \(= 5.64\); \(H_1\): Pop mean no. accidents \(< 5.64\) | B1 | or "\(= 0.47\) (per month)"; not just "mean", but allow just "\(\lambda\)" or "\(\mu\)" |
| Use of \(\lambda = 5.64\) | B1 | used in a Poisson calculation |
| \(= e^{-5.64}\left(1 + 5.64 + \frac{5.64^2}{2}\right)\) | M1 | Allow incorrect \(\lambda\) in otherwise correct |
| \(= 0.08(0)\) | A1 | |
| Comp with 0.05 | M1 | Valid comparison (Poisson only), no contradictions |
| No evidence to believe mean no. of accidents has decreased; accept \(H_0\) (if correctly defined) | A1FT | Normal distribution: M0M0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Mean \(< 0.47\) but conclude that this is not so | B1 | (Mean) no. of accidents reduced, but conclude not reduced. Must be in context. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| (Need greatest \(x\) such that \(P(X \leqslant x) < 0.05\)); \(P(X \leqslant 1) = e^{-5.64}(1 + 5.64) = 0.024\); \(P(X \leqslant 2) = 0.08\) | B1 | Both, could be seen in (i) |
| Hence rejection region is \(X \leqslant 1\) | B1 | Can be implied |
| With \(\lambda = 12 \times 0.05 = 0.6\), \(1 - P(X \leqslant 1) = 1 - e^{-0.6}(1 + 0.6)\) | M1 | \(\lambda = 0.6\) and \(1 - P(X \leqslant 1)\) |
| \(= 0.122\) (3 sf) | A1 | Normal scores 0 |
## Question 7(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $H_0$: Pop mean no. accidents $= 5.64$; $H_1$: Pop mean no. accidents $< 5.64$ | B1 | or "$= 0.47$ (per month)"; not just "mean", but allow just "$\lambda$" or "$\mu$" |
| Use of $\lambda = 5.64$ | B1 | used in a Poisson calculation |
| $= e^{-5.64}\left(1 + 5.64 + \frac{5.64^2}{2}\right)$ | M1 | Allow incorrect $\lambda$ in otherwise correct |
| $= 0.08(0)$ | A1 | |
| Comp with 0.05 | M1 | Valid comparison (Poisson only), no contradictions |
| No evidence to believe mean no. of accidents has decreased; accept $H_0$ (if correctly defined) | A1FT | Normal distribution: **M0M0** |
**Total: 6**
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## Question 7(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Mean $< 0.47$ but conclude that this is not so | B1 | (Mean) no. of accidents **reduced**, but conclude not reduced. Must be in context. |
**Total: 1**
---
## Question 7(iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| (Need greatest $x$ such that $P(X \leqslant x) < 0.05$); $P(X \leqslant 1) = e^{-5.64}(1 + 5.64) = 0.024$; $P(X \leqslant 2) = 0.08$ | B1 | Both, could be seen in (i) |
| Hence rejection region is $X \leqslant 1$ | B1 | Can be implied |
| With $\lambda = 12 \times 0.05 = 0.6$, $1 - P(X \leqslant 1) = 1 - e^{-0.6}(1 + 0.6)$ | M1 | $\lambda = 0.6$ and $1 - P(X \leqslant 1)$ |
| $= 0.122$ (3 sf) | A1 | Normal scores 0 |
**Total: 4**7 In the past the number of accidents per month on a certain road was modelled by a random variable with distribution $\operatorname { Po } ( 0.47 )$. After the introduction of speed restrictions, the government wished to test, at the 5\% significance level, whether the mean number of accidents had decreased. They noted the number of accidents during the next 12 months. It is assumed that accidents occur randomly and that a Poisson model is still appropriate.\\
(i) Given that the total number of accidents during the 12 months was 2 , carry out the test.\\
(ii) Explain what is meant by a Type II error in this context.\\
It is given that the mean number of accidents per month is now in fact 0.05 .\\
(iii) Using another random sample of 12 months the same test is carried out again, with the same significance level. Find the probability of a Type II error.\\
\hfill \mbox{\textit{CAIE S2 2017 Q7 [11]}}