| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Poisson hypothesis test |
| Difficulty | Standard +0.3 Part (a) is direct substitution into the Poisson formula. Part (b) requires conducting a hypothesis test with a Poisson distribution, finding the p-value by calculating P(X ≤ 3) when n=100 observations follow Poisson(5), and comparing to significance levels. While this involves multiple steps and understanding of hypothesis testing, it follows a standard S2 procedure without requiring novel insight or complex reasoning. |
| Spec | 2.05b Hypothesis test for binomial proportion5.02i Poisson distribution: random events model |
| Answer | Marks | Guidance |
|---|---|---|
| (a) If mean = 5, \(X \sim Po(5)\) \(P(X = 0) = 0.0067\) | B1 | |
| (b) \(X \sim Po(\lambda)\) \(H_0: \lambda = 5\) \(H_1: \lambda < 5\) | B1 B1 | |
| Under \(H_0\), no. of 0's in 100 measurements \(\sim Po(0.67)\) | M1 A1 | |
| \(P(X \geq 3) = 1 - e^{-0.67}(1 + 0.67 + 0.67^2/2!) = 0.031 = 3.1\%\) | M1 A1 A1 | |
| Reject \(H_0\) at the 5% significance level, but not at the 1% level. | A1 | 9 marks total |
(a) If mean = 5, $X \sim Po(5)$ $P(X = 0) = 0.0067$ | B1 |
(b) $X \sim Po(\lambda)$ $H_0: \lambda = 5$ $H_1: \lambda < 5$ | B1 B1 |
Under $H_0$, no. of 0's in 100 measurements $\sim Po(0.67)$ | M1 A1 |
$P(X \geq 3) = 1 - e^{-0.67}(1 + 0.67 + 0.67^2/2!) = 0.031 = 3.1\%$ | M1 A1 A1 |
Reject $H_0$ at the 5% significance level, but not at the 1% level. | A1 | 9 marks totalA random variable $X$ has a Poisson distribution with a mean, $\lambda$, which is assumed to equal 5.
\begin{enumerate}[label=(\alph*)]
\item Find P$(X = 0)$. [1 mark]
\item In 100 measurements, the value 0 occurs three times. Find the highest significance level at which you should reject the original hypothesis in favour of $\lambda < 5$. [8 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 Q3 [9]}}