| Exam Board | Edexcel |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2006 |
| Session | January |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of a Poisson distribution |
| Type | Find power function or power value |
| Difficulty | Standard +0.3 This is a straightforward application of hypothesis testing for a Poisson distribution requiring calculation of probabilities from tables and understanding of power function definition. While it involves multiple parts, each step follows standard procedures: finding significance level P(X>4|λ=3), then calculating power values P(X>4|λ=4,6) using Poisson probabilities. The conceptual demand is moderate for Further Maths Statistics, with no novel problem-solving required beyond applying definitions correctly. |
| Spec | 2.05a Hypothesis testing language: null, alternative, p-value, significance2.05b Hypothesis test for binomial proportion |
| \(\lambda\) | 4 | 5 | 6 | 7 |
| Power | \(r\) | 0.56 | \(s\) | 0.83 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| Size of test \(= P(X > 4 \mid \lambda = 3)\) | M1 | |
| \(= 1 - P(X \leq 4 \mid \lambda = 3) = 1 - 0.8153\) | A1 | |
| \(= 0.1847\) | A1 | (3) awrt 0.185 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(r = 1 - 0.6268 = 0.3712 = 0.37\) (2 d.p.) | B1 | |
| \(s = 1 - 0.2851 = 0.7149 = 0.71\) (2 d.p.) | B1 | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| When \(\lambda = 4\), power \(= 0.27 < 0.5\). Probability of coming to correct conclusion is less than probability of coming to wrong conclusion. Not suitable. | B1 | (1) |
| TOTAL | 6 |
## Question 4:
### Part (a):
| Answer/Working | Marks | Notes |
|---|---|---|
| Size of test $= P(X > 4 \mid \lambda = 3)$ | M1 | |
| $= 1 - P(X \leq 4 \mid \lambda = 3) = 1 - 0.8153$ | A1 | |
| $= 0.1847$ | A1 | (3) awrt 0.185 |
### Part (b):
| Answer/Working | Marks | Notes |
|---|---|---|
| $r = 1 - 0.6268 = 0.3712 = 0.37$ (2 d.p.) | B1 | |
| $s = 1 - 0.2851 = 0.7149 = 0.71$ (2 d.p.) | B1 | (2) |
### Part (c):
| Answer/Working | Marks | Notes |
|---|---|---|
| When $\lambda = 4$, power $= 0.27 < 0.5$. Probability of coming to correct conclusion is less than probability of coming to wrong conclusion. Not suitable. | B1 | (1) |
| **TOTAL** | **6** | |
---4. The number of accidents that occur at a crossroads has a mean of 3 per month. In order to improve the flow of traffic the priority given to traffic is changed. Colin believes that since the change in priority the number of accidents has increased. He tests his belief by recording the number of accidents $x$ in the month following the change. Colin sets up the hypotheses $\mathrm { H } _ { 0 } : \lambda = 3$ and $\mathrm { H } _ { 1 } : \lambda > 3$, where $\lambda$ is the mean number of accidents per month, and rejects the null hypothesis if $x > 4$.
\begin{enumerate}[label=(\alph*)]
\item Find the size of the test.
The table gives the values of the power function of the test to two decimal places.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | }
\hline
$\lambda$ & 4 & 5 & 6 & 7 \\
\hline
Power & $r$ & 0.56 & $s$ & 0.83 \\
\hline
\end{tabular}
\end{center}
\item Calculate the value of $r$ and the value of $s$.
\item Comment on the suitability of the test when $\lambda = 4$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S4 2006 Q4 [6]}}