| Exam Board | Edexcel |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2011 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of binomial distributions |
| Type | Calculate Type II error probability |
| Difficulty | Challenging +1.2 This is a multi-part S4 hypothesis testing question requiring calculation of test size, Type II error, and power function values. Parts (a)-(c) involve standard binomial probability calculations with clear decision rules. Part (d) requires careful handling of a two-stage sampling procedure but follows systematic probability tree logic. While requiring multiple calculations and understanding of power/Type I/Type II errors, these are core S4 techniques without novel conceptual challenges. |
| Spec | 2.05b Hypothesis test for binomial proportion2.05c Significance levels: one-tail and two-tail |
| \(p\) | 0.075 | 0.100 | 0.125 | 0.150 | 0.175 | 0.200 | 0.225 |
| Power | 0.35 | \(s\) | 0.75 | 0.87 | 0.94 | 0.97 | 0.99 |
| \(p\) | 0.075 | 0.100 | 0.125 | 0.150 | 0.175 | 0.200 | 0.225 |
| Power | 0.35 | \(s\) | 0.75 | 0.87 | 0.94 | 0.97 | 0.99 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \([X = \text{no. of incorrectly addressed letters}, \; X \sim B(40, 0.05)]\) \(P(X>3) = 1 - P(X \leq 3) = 1 - 0.8619 = 0.1381\) awrt 0.138 | M1, A1 | M1 for \(1 - P(X \leq 3)\) and \(X \sim B(40, 0.05)\) |
| (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(\text{Type II Error}) = P(X \leq 3 \mid p = 0.10) = 0.4231\) awrt 0.423 | M1, A1 | M1 for correct interpretation of P(Type II error) |
| (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Power \(= 1 - P(\text{Type II error})\) so \(s = \mathbf{0.58}\) (0.5769) | B1 | Must be 2dp |
| (1 mark) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(Y \sim B(15, 0.05)\); Size \(= P(Y \geq 2) + P(Y=1) \times P(Y \geq 2)\) | M1 | For a correct strategy |
| \(= [1 - 0.8290] \times [1 + 0.8290 - 0.4633]\) | A1 | For correct numerical expression |
| \(= 0.23353\ldots\) awrt 0.23 | A1 | |
| (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| (use overlay) | B1B1 | 1st B1 for correct points (accept \(\pm\) one 2mm square); 2nd B1 for curve |
| (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(2^{\text{nd}}\) consultant's test is quicker (since it uses fewer letters); \(2^{\text{nd}}\) consultant's test is more powerful for \(p < 0.125\) (and values greater than this should be unlikely) | B1 B1 | 1st B1 for selecting \(2^{\text{nd}}\) test; 2nd B1 for a suitable supporting reason, e.g. more powerful for small values of \(p/p\) around 0.05 |
| (2 marks) | ||
| Total: 12 marks |
# Question 4:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $[X = \text{no. of incorrectly addressed letters}, \; X \sim B(40, 0.05)]$ $P(X>3) = 1 - P(X \leq 3) = 1 - 0.8619 = 0.1381$ **awrt 0.138** | M1, A1 | M1 for $1 - P(X \leq 3)$ and $X \sim B(40, 0.05)$ |
| **(2 marks)** | | |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(\text{Type II Error}) = P(X \leq 3 \mid p = 0.10) = 0.4231$ **awrt 0.423** | M1, A1 | M1 for correct interpretation of P(Type II error) |
| **(2 marks)** | | |
## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Power $= 1 - P(\text{Type II error})$ so $s = \mathbf{0.58}$ (0.5769) | B1 | Must be 2dp |
| **(1 mark)** | | |
## Part (d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $Y \sim B(15, 0.05)$; Size $= P(Y \geq 2) + P(Y=1) \times P(Y \geq 2)$ | M1 | For a correct strategy |
| $= [1 - 0.8290] \times [1 + 0.8290 - 0.4633]$ | A1 | For correct numerical expression |
| $= 0.23353\ldots$ **awrt 0.23** | A1 | |
| **(3 marks)** | | |
## Part (e):
| Answer/Working | Marks | Guidance |
|---|---|---|
| (use overlay) | B1B1 | 1st B1 for correct points (accept $\pm$ one 2mm square); 2nd B1 for curve |
| **(2 marks)** | | |
## Part (f):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $2^{\text{nd}}$ consultant's test is quicker (since it uses fewer letters); $2^{\text{nd}}$ consultant's test is more powerful for $p < 0.125$ (and values greater than this should be unlikely) | B1 B1 | 1st B1 for selecting $2^{\text{nd}}$ test; 2nd B1 for a suitable supporting reason, e.g. more powerful for small values of $p/p$ around 0.05 |
| **(2 marks)** | | |
| **Total: 12 marks** | | |
4. A proportion $p$ of letters sent by a company are incorrectly addressed and if $p$ is thought to be greater than 0.05 then action is taken.\\
Using $\mathrm { H } _ { 0 } : p = 0.05$ and $\mathrm { H } _ { 1 } : p > 0.05$, a manager from the company takes a random sample of 40 letters and rejects $\mathrm { H } _ { 0 }$ if the number of incorrectly addressed letters is more than 3 .
\begin{enumerate}[label=(\alph*)]
\item Find the size of this test.
\item Find the probability of a Type II error in the case where $p$ is in fact 0.10
Table 1 below 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
$p$ & 0.075 & 0.100 & 0.125 & 0.150 & 0.175 & 0.200 & 0.225 \\
\hline
Power & 0.35 & $s$ & 0.75 & 0.87 & 0.94 & 0.97 & 0.99 \\
\hline
\end{tabular}
\captionsetup{labelformat=empty}
\caption{Table 1}
\end{center}
\end{table}
\item Write down the value of $s$.
A visiting consultant uses an alternative system to test the same hypotheses. A sample of 15 letters is taken. If these are all correctly addressed then $\mathrm { H } _ { 0 }$ is accepted. If 2 or more are found to have been incorrectly addressed then $\mathrm { H } _ { 0 }$ is rejected. If only one is found to be incorrectly addressed then a further random sample of 15 is taken and $\mathrm { H } _ { 0 }$ is rejected if 2 or more are found to have been incorrectly addressed in this second sample, otherwise $\mathrm { H } _ { 0 }$ is accepted.
\item Find the size of the test used by the consultant.
\section*{Question 4 continues on page 8}
For your convenience Table 1 is repeated here
\begin{table}[h]
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | c | }
\hline
$p$ & 0.075 & 0.100 & 0.125 & 0.150 & 0.175 & 0.200 & 0.225 \\
\hline
Power & 0.35 & $s$ & 0.75 & 0.87 & 0.94 & 0.97 & 0.99 \\
\hline
\end{tabular}
\captionsetup{labelformat=empty}
\caption{Table 1}
\end{center}
\end{table}
Figure 1 shows the graph of the power function of the test used by the consultant.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{8dfc721d-4782-4482-9976-11189370f3b7-07_1712_1673_660_130}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
\item On Figure 1 draw the graph of the power function of the manager's test.\\
(2)
\item State, giving your reasons, which test you would recommend.\\
(2)
\end{enumerate}
\hfill \mbox{\textit{Edexcel S4 2011 Q4 [12]}}