| Exam Board | Edexcel |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2013 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of a Poisson distribution |
| Type | Find power function or power value |
| Difficulty | Challenging +1.2 This is a structured multi-part question on hypothesis testing with Poisson distributions from Further Maths Statistics. Part (a) requires basic probability calculation, part (b) is a guided algebraic derivation using the complement rule, parts (c-e) involve straightforward substitution and graph reading. While it covers advanced content (power functions), the question provides significant scaffolding and requires mainly procedural application rather than novel insight. |
| Spec | 2.05b Hypothesis test for binomial proportion2.05c Significance levels: one-tail and two-tail |
| \(\lambda\) | 2.5 | 3.0 | 3.5 | 4.0 | 5.0 | 7.0 |
| Power | 0.46 | \(r\) | 0.68 | \(s\) | 0.88 | 0.97 |
| \(\lambda\) | 2.5 | 3.0 | 3.5 | 4.0 | 5.0 | 7.0 |
| Power | 0.46 | \(r\) | 0.68 | \(s\) | 0.88 | 0.97 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Size \(= P(X \geq 3 \mid \lambda = 2)\) | M1 | Correct probability under \(H_0\) |
| \(= 1 - P(X \leq 2) = 1 - e^{-2}(1 + 2 + 2) = 1 - 5e^{-2}\) | ||
| \(= 0.3233...\) awrt \(\mathbf{0.323}\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Power \(= P(X \geq 3 \mid \lambda)\) | M1 | |
| \(= 1 - P(X \leq 2) = 1 - e^{-\lambda}\left(1 + \lambda + \frac{\lambda^2}{2}\right)\) | M1 | Correct Poisson sum for \(x=0,1,2\) |
| \(= 1 - \frac{1}{2}e^{-\lambda}(2 + 2\lambda + \lambda^2)\) | A1 | Correct factorisation shown |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(r\): substitute \(\lambda = 3.0\): \(1 - \frac{1}{2}e^{-3}(2+6+9) = 1 - \frac{17}{2}e^{-3} = \mathbf{0.5768...}\) awrt \(\mathbf{0.58}\) | A1 | |
| \(s\): substitute \(\lambda = 4.0\): \(1 - \frac{1}{2}e^{-4}(2+8+16) = 1 - 13e^{-4} = \mathbf{0.7619...}\) awrt \(\mathbf{0.76}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Correct points plotted from table | B1 | At least 5 correct points |
| Smooth increasing curve through points | B1 | Appropriate curve |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\lambda > 3.6\) (reading from graph where power \(> 0.6\)) | B1 | ft from graph, awrt |
# Question 3:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Size $= P(X \geq 3 \mid \lambda = 2)$ | M1 | Correct probability under $H_0$ |
| $= 1 - P(X \leq 2) = 1 - e^{-2}(1 + 2 + 2) = 1 - 5e^{-2}$ | | |
| $= 0.3233...$ awrt $\mathbf{0.323}$ | A1 | cao |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Power $= P(X \geq 3 \mid \lambda)$ | M1 | |
| $= 1 - P(X \leq 2) = 1 - e^{-\lambda}\left(1 + \lambda + \frac{\lambda^2}{2}\right)$ | M1 | Correct Poisson sum for $x=0,1,2$ |
| $= 1 - \frac{1}{2}e^{-\lambda}(2 + 2\lambda + \lambda^2)$ | A1 | Correct factorisation shown |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $r$: substitute $\lambda = 3.0$: $1 - \frac{1}{2}e^{-3}(2+6+9) = 1 - \frac{17}{2}e^{-3} = \mathbf{0.5768...}$ awrt $\mathbf{0.58}$ | A1 | |
| $s$: substitute $\lambda = 4.0$: $1 - \frac{1}{2}e^{-4}(2+8+16) = 1 - 13e^{-4} = \mathbf{0.7619...}$ awrt $\mathbf{0.76}$ | A1 | |
## Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Correct points plotted from table | B1 | At least 5 correct points |
| Smooth increasing curve through points | B1 | Appropriate curve |
## Part (e):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\lambda > 3.6$ (reading from graph where power $> 0.6$) | B1 | ft from graph, awrt |3. The number of houses sold per week by a firm of estate agents follows a Poisson distribution with mean 2 . The firm believes that the appointment of a new salesman will increase the number of houses sold. The firm tests its belief by recording the number of houses sold, $x$, in the week following the appointment. The firm sets up the hypotheses $\mathrm { H } _ { 0 } : \lambda = 2$ and $\mathrm { H } _ { 1 } : \lambda > 2$, where $\lambda$ is the mean number of houses sold per week, and rejects the null hypothesis if $x \geqslant 3$
\begin{enumerate}[label=(\alph*)]
\item Find the size of the test.
\item Show that the power function for this test is
$$1 - \frac { 1 } { 2 } e ^ { - \lambda } \left( 2 + 2 \lambda + \lambda ^ { 2 } \right)$$
The table below gives the values of the power function to 2 decimal places.
\begin{table}[h]
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | }
\hline
$\lambda$ & 2.5 & 3.0 & 3.5 & 4.0 & 5.0 & 7.0 \\
\hline
Power & 0.46 & $r$ & 0.68 & $s$ & 0.88 & 0.97 \\
\hline
\end{tabular}
\captionsetup{labelformat=empty}
\caption{Table 1}
\end{center}
\end{table}
\item Calculate the values of $r$ and $s$.
\item Draw a graph of the power function on the graph paper provided on page 6
\item Find the range of values of $\lambda$ for which the power of this test is greater than 0.6
For your convenience Table 1 is repeated here.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | }
\hline
$\lambda$ & 2.5 & 3.0 & 3.5 & 4.0 & 5.0 & 7.0 \\
\hline
Power & 0.46 & $r$ & 0.68 & $s$ & 0.88 & 0.97 \\
\hline
\end{tabular}
\end{center}
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Table 1}
\includegraphics[alt={},max width=\textwidth]{4f096806-33da-453f-a4c1-12be20d1a96d-06_2125_1603_614_166}
\end{center}
\end{figure}
\includegraphics[max width=\textwidth, alt={}, center]{4f096806-33da-453f-a4c1-12be20d1a96d-07_72_47_2615_1886}
\end{enumerate}
\hfill \mbox{\textit{Edexcel S4 2013 Q3 [10]}}