| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2015 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of a Poisson distribution |
| Type | Two-tailed test setup or execution |
| Difficulty | Challenging +1.2 This is a reverse critical region problem requiring students to work backwards from given critical values to find the parameter k, then calculate actual significance level. While it involves multiple Poisson probability calculations and understanding of two-tailed tests, it's a standard S2 exercise with clear structure and routine application of formulas—moderately above average difficulty due to the reverse-engineering aspect but not requiring novel insight. |
| Spec | 2.05a Hypothesis testing language: null, alternative, p-value, significance2.05b Hypothesis test for binomial proportion2.05c Significance levels: one-tail and two-tail5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(X=0\ | k=3)=0.0498\); \(P(X=0\ | k=4)=0.0183\); \(P(X=0\ |
| \(P(X\leqslant 8\ | k=3)=0.9962\), \(P(X\geqslant 9\ | k=3)=0.0038\); \(P(X\leqslant 8\ |
| Both tails less than 2.5% when \(k=4\) | B1 | Final answer given as \(k=4\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Actual sig. level \(= 0.0214 + 0.0183\) | M1 | See notes |
| \(= 0.0397\) | A1 cao |
# Question 4:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(X=0\|k=3)=0.0498$; $P(X=0\|k=4)=0.0183$; $P(X=0\|k=5)=0.0067$; $\{e^{-k}<0.025 \Rightarrow k>\}\ 3.688...$ | B1 | At least one of these 9 probabilities or awrt 3.7 seen in working |
| $P(X\leqslant 8\|k=3)=0.9962$, $P(X\geqslant 9\|k=3)=0.0038$; $P(X\leqslant 8\|k=4)=0.9786$, $P(X\geqslant 9\|k=4)=0.0214$; $P(X\leqslant 8\|k=5)=0.9319$, $P(X\geqslant 9\|k=5)=0.0681$ | B1 | Both $P(X=0)=0.0183$ or awrt 3.7 **and** either $P(X\geqslant 9)=0.0214$ or $P(X\leqslant 8)=0.9786$ |
| Both tails less than 2.5% when $k=4$ | B1 | Final answer given as $k=4$ |
**[3]**
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Actual sig. level $= 0.0214 + 0.0183$ | M1 | See notes |
| $= 0.0397$ | A1 cao | |
**[2]**
### Question 4 Notes:
- **1st B1**: For any of 0.0498, 0.0183, 0.0067, 0.9962, 0.9786, 0.9319, 0.0038, 0.0214, 0.0681 or awrt 3.7 seen in working
- **2nd B1**: For both $P(X=0)=0.0183$ **or** awrt 3.7 **and** either $P(X\geqslant 9)=0.0214$ **or** $P(X\leqslant 8)=0.9786$. These must be written as probability statements.
- **3rd B1**: Final answer given as $k=4$. Also allow $\lambda=4$
- **Note**: Do not recover working for part (a) in part (b)
- **(b) M1**: For the addition of two probabilities for two tails, where each tail $< 0.05$
- **(b) A1**: 0.0397 cao
---\begin{enumerate}
\item A single observation $x$ is to be taken from a Poisson distribution with parameter $\lambda$ This observation is to be used to test, at a $5 \%$ level of significance,
\end{enumerate}
$$\mathrm { H } _ { 0 } : \lambda = k \quad \mathrm { H } _ { 1 } : \lambda \neq k$$
where $k$ is a positive integer.\\
Given that the critical region for this test is $( X = 0 ) \cup ( X \geqslant 9 )$\\
(a) find the value of $k$, justifying your answer.\\
(b) Find the actual significance level of this test.\\
\hfill \mbox{\textit{Edexcel S2 2015 Q4 [5]}}