| Exam Board | WJEC |
|---|---|
| Module | Further Unit 2 (Further Unit 2) |
| Session | Specimen |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of a Poisson distribution |
| Type | Comment on test validity or assumptions |
| Difficulty | Standard +0.3 This is a standard chi-squared goodness of fit test with a Poisson distribution. Students must calculate expected frequencies using the given mean, combine categories to meet the rule of 5, compute the test statistic, and compare to critical values. While it requires multiple steps and careful calculation, it follows a well-rehearsed procedure with no novel insight needed, making it slightly easier than average for Further Maths. |
| Spec | 5.06b Fit prescribed distribution: chi-squared test |
| Goals scored | 0 | 1 | 2 | 3 | 4 or more |
| Frequency | 6 | 11 | 15 | 10 | 8 |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(H_0\): The data can be modelled by the Poisson distribution with mean 2. \(H_1\): The data cannot be modelled by the Poisson distribution with mean 2. | B1 | AO3 |
| Answer | Marks | Guidance |
|---|---|---|
| Goals scored | 0 | 1 |
| Obs | 6 | 11 |
| Exp | 6.767 | 13.534 |
| B1, B1 | AO3, AO3 | For at least 1 correct; For all correct |
| Use of \(\chi^2\) stat = \(\sum \frac{O^2}{E} - N = \frac{6^2}{6.767} + \frac{11^2}{13.534} + ... + \frac{8^2}{7.144} - 50 = 0.93\) | M1, A1, A1 | AO3, AO2, AO1 |
| DF = 4; 5% crit val = 9.488; Since 0.93 < 9.488 (Accept \(H_0\)). We conclude that the data can be modelled by the Poisson distribution with mean 2. | B1, B1, B1, B1 | AO1, AO1, AO2, AO3 |
**(a)** $H_0$: The data can be modelled by the Poisson distribution with mean 2. $H_1$: The data cannot be modelled by the Poisson distribution with mean 2. | B1 | AO3
**(b)** The expected frequencies are:
| Goals scored | 0 | 1 | 2 | 3 | 4 or more |
|---|---|---|---|---|---|
| Obs | 6 | 11 | 15 | 10 | 8 |
| Exp | 6.767 | 13.534 | 13.534 | 9.022 | 7.144 |
| B1, B1 | AO3, AO3 | For at least 1 correct; For all correct
Use of $\chi^2$ stat = $\sum \frac{O^2}{E} - N = \frac{6^2}{6.767} + \frac{11^2}{13.534} + ... + \frac{8^2}{7.144} - 50 = 0.93$ | M1, A1, A1 | AO3, AO2, AO1
DF = 4; 5% crit val = 9.488; Since 0.93 < 9.488 (Accept $H_0$). We conclude that the data can be modelled by the Poisson distribution with mean 2. | B1, B1, B1, B1 | AO1, AO1, AO2, AO3 | [10]
---The manager of a hockey team studies last season's results and puts forward the theory that the number of goals scored per match by her team can be modelled by a Poisson distribution with mean 2.0. The number of goals scored during the season are summarised below.
\begin{tabular}{|l|c|c|c|c|c|}
\hline
Goals scored & 0 & 1 & 2 & 3 & 4 or more \\
\hline
Frequency & 6 & 11 & 15 & 10 & 8 \\
\hline
\end{tabular}
\begin{enumerate}[label=(\alph*)]
\item State suitable hypotheses to carry out a goodness of fit test. [1]
\item Carry out a $\chi^2$ goodness of fit test on this data set, using a 5% level of significance and draw a conclusion in context. [9]
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 2 Q5 [10]}}