| Exam Board | Edexcel |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2006 |
| Session | June |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of a Poisson distribution |
| Type | Find power function or power value |
| Difficulty | Challenging +1.8 This is a sophisticated Further Maths Statistics question requiring understanding of hypothesis testing framework, power functions, Type I/II errors, and Poisson distribution. Parts (a-b) and (e) involve standard power calculations, but parts (c-d) require optimizing a test criterion under constraints, and part (g) demands interpretation of power function intersections—concepts beyond typical A-level. The multi-part structure with interconnected reasoning and the need to work with power functions (not just basic hypothesis tests) places this well above average difficulty. |
| Spec | 2.05a Hypothesis testing language: null, alternative, p-value, significance5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x! |
| \(\lambda\) | 0.4 | 0.5 | 0.6 | 0.7 | 0.8 | 0.9 | 1.0 |
| Power | 0.15 | 0.34 | \(r\) | 0.72 | 0.85 | 0.92 | 0.96 |
| \(\lambda\) | 0.4 | 0.5 | 0.6 | 0.7 | 0.8 | 0.9 | 1.0 |
| Power | 0.21 | 0.38 | 0.55 | 0.70 | \(s\) | 0.88 | 0.93 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(X_i =\) no. of defects in 15m, \(X_i \sim Po(4.5)\) | M1 | Use of \(Po(4.5)\) |
| \(\text{Size} = P(X_i \geq 9) = 1 - P(X \leq 8) = 1 - 0.9597 = 0.0403\) | A1 | Awrt (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(r = P(X_2 \geq 9 \mid X \sim Po(9)) = 1 - P(X_2 \leq 8) = 1 - 0.4557 = 0.54\) | M1A1 | Awrt \(0.54\) (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(Y_i =\) no. of defects in 10m, \(Y_i \sim Po(3)\) | M1 | Use of \(Po(3)\) to find \(P(Y \geq c)\) |
| Require smallest \(c\) so that \(P(Y_i \geq c) < 0.10\). Tables \(Y_i \geq 6\) | A1 | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\text{Size} = P(Y_i \geq 6) = 1 - P(Y_i \leq 5) = 1 - 0.9161 = 0.0839\) | B1 | (1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(s = 1 - P(Y_2 \leq 5 \mid Y_2 \sim Po(8)) = 1 - 0.1912 = 0.8088\) | M1A1 | Awrt \(0.81\) (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| See graph | B1, B1 | (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| (i) \(0.62 \sim 0.67\) | B1 | |
| (ii) Test 1 is more powerful | B1 | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Test 2 has higher \(P(\text{Type I error})\) but cost of this is low | B1 | Test 2 |
| Test 2 is more powerful for \(\lambda < 0.7\) and \(\lambda > 0.7\) is rare | B1 | Reason (2) |
# Question 5:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $X_i =$ no. of defects in 15m, $X_i \sim Po(4.5)$ | M1 | Use of $Po(4.5)$ |
| $\text{Size} = P(X_i \geq 9) = 1 - P(X \leq 8) = 1 - 0.9597 = 0.0403$ | A1 | Awrt (2) |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $r = P(X_2 \geq 9 \mid X \sim Po(9)) = 1 - P(X_2 \leq 8) = 1 - 0.4557 = 0.54$ | M1A1 | Awrt $0.54$ (2) |
## Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $Y_i =$ no. of defects in 10m, $Y_i \sim Po(3)$ | M1 | Use of $Po(3)$ to find $P(Y \geq c)$ |
| Require smallest $c$ so that $P(Y_i \geq c) < 0.10$. Tables $Y_i \geq 6$ | A1 | (2) |
## Part (d)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\text{Size} = P(Y_i \geq 6) = 1 - P(Y_i \leq 5) = 1 - 0.9161 = 0.0839$ | B1 | (1) |
## Part (e)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $s = 1 - P(Y_2 \leq 5 \mid Y_2 \sim Po(8)) = 1 - 0.1912 = 0.8088$ | M1A1 | Awrt $0.81$ (2) |
## Part (f)
| Answer/Working | Mark | Guidance |
|---|---|---|
| See graph | B1, B1 | (4) |
## Part (g)
| Answer/Working | Mark | Guidance |
|---|---|---|
| (i) $0.62 \sim 0.67$ | B1 | |
| (ii) Test 1 is more powerful | B1 | (2) |
## Part (h)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Test 2 has higher $P(\text{Type I error})$ but cost of this is low | B1 | Test 2 |
| Test 2 is more powerful for $\lambda < 0.7$ and $\lambda > 0.7$ is rare | B1 | Reason (2) |
---
5. Rolls of cloth delivered to a factory contain defects at an average rate of $\lambda$ per metre. A quality assurance manager selects a random sample of 15 metres of cloth from each delivery to test whether or not there is evidence that $\lambda > 0.3$. The criterion that the manager uses for rejecting the hypothesis that $\lambda = 0.3$ is that there are 9 or more defects in the sample.
\begin{enumerate}[label=(\alph*)]
\item Find the size of the test.
Table 1 gives some values, to 2 decimal places, of the power function of this test.
\begin{table}[h]
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | c | }
\hline
$\lambda$ & 0.4 & 0.5 & 0.6 & 0.7 & 0.8 & 0.9 & 1.0 \\
\hline
Power & 0.15 & 0.34 & $r$ & 0.72 & 0.85 & 0.92 & 0.96 \\
\hline
\end{tabular}
\captionsetup{labelformat=empty}
\caption{Table 1}
\end{center}
\end{table}
\item Find the value of $r$.
The manager would like to design a test, of whether or not $\lambda > 0.3$, that uses a smaller length of cloth. He chooses a length of 10 m and requires the probability of a type I error to be less than $10 \%$.
\item Find the criterion to reject the hypothesis that $\lambda = 0.3$ which makes the test as powerful as possible.
\item Hence state the size of this second test.
Table 2 gives some values, to 2 decimal places, of the power function for the test in part (c).
\begin{table}[h]
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | c | }
\hline
$\lambda$ & 0.4 & 0.5 & 0.6 & 0.7 & 0.8 & 0.9 & 1.0 \\
\hline
Power & 0.21 & 0.38 & 0.55 & 0.70 & $s$ & 0.88 & 0.93 \\
\hline
\end{tabular}
\captionsetup{labelformat=empty}
\caption{Table 2}
\end{center}
\end{table}
\item Find the value of $s$.
\item Using the same axes, on graph paper draw the graphs of the power functions of these two tests.
\item \begin{enumerate}[label=(\roman*)]
\item State the value of $\lambda$ where the graphs cross.
\item Explain the significance of $\lambda$ being greater than this value.
The cost of wrongly rejecting a delivery of cloth with $\lambda = 0.3$ is low. Deliveries of cloth with $\lambda > 0.7$ are unusual.
\end{enumerate}\item Suggest, giving your reasons, which the test manager should adopt.\\
(2)
\end{enumerate}
\hfill \mbox{\textit{Edexcel S4 2006 Q5 [17]}}