| Exam Board | Edexcel |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2013 |
| Session | June |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of a Poisson distribution |
| Type | Sequential or two-stage test design |
| Difficulty | Challenging +1.8 This is a sophisticated S4 hypothesis testing question involving sequential test design, power functions, and Type II error calculations with Poisson distributions. While parts (a)-(b) are standard S4 fare, parts (c)-(g) require deriving power functions for a two-stage test, comparing test procedures graphically and algebraically, and making justified recommendations—demanding extended reasoning beyond routine application. This is challenging for A-level but represents expected S4 material rather than requiring truly novel insight. |
| Spec | 2.05b Hypothesis test for binomial proportion2.05c Significance levels: one-tail and two-tail |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(X \sim \text{Po}(2)\) (since 20ml at rate \(\lambda\) per 10ml, so rate \(2\lambda\) per 20ml, with \(\lambda=1\) gives Po(2)) | M1 | Must use Po(2) |
| \(P(X \geqslant 4) = 1 - P(X \leqslant 3) = 1 - e^{-2}(1 + 2 + 2 + \frac{4}{3}) = 0.1429...\) awrt 0.143 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| When \(\lambda = 2.5\), \(X \sim \text{Po}(5)\) | M1 | Must use Po(5) |
| \(P(\text{Type II error}) = P(X \leqslant 3) = e^{-5}(1 + 5 + \frac{25}{2} + \frac{125}{6})\) = awrt 0.265 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Let \(Y \sim \text{Po}(\lambda)\) for 10ml sample. \(P(\text{reject on 1st sample}) = P(Y \geqslant 2) = 1 - e^{-\lambda} - \lambda e^{-\lambda}\) | M1 | |
| \(P(\text{exactly 1 on 1st sample}) = \lambda e^{-\lambda}\) | B1 | |
| \(P(\text{reject on 2nd sample} \mid \text{1 on first}) = P(Y \geqslant 2) = 1 - e^{-\lambda} - \lambda e^{-\lambda}\) | M1 | |
| Power \(= (1 - e^{-\lambda} - \lambda e^{-\lambda}) + \lambda e^{-\lambda}(1 - e^{-\lambda} - \lambda e^{-\lambda})\) \(= 1 - e^{-\lambda} - \lambda e^{-\lambda} + \lambda e^{-\lambda} - \lambda e^{-2\lambda} - \lambda^2 e^{-2\lambda}\) \(= 1 - e^{-\lambda} - \lambda(1+\lambda)e^{-2\lambda}\) | A1 | Fully correct derivation shown |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Substitute \(\lambda = 3\): \(1 - e^{-3} - 3(4)e^{-6}\) = awrt 0.92, so \(r = \) 0.92 | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Correct shape — curve starting below scientist's curve and crossing/above it | B1 | Passes through or near correct table values |
| Points plotted correctly from Table 1 and joined smoothly | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Expected time for statistician's test \(= 15 + 15 \times P(\text{1 organism in first sample})\) \(= 15 + 15\lambda e^{-\lambda}\) | M1 A1 | |
| Statistician's test slower when \(15 + 15\lambda e^{-\lambda} > 20\) | M1 | |
| \(15\lambda e^{-\lambda} > 5 \Rightarrow \lambda e^{-\lambda} > \frac{1}{3}\) | A1 | Fully shown |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| At \(\lambda = 1\): times are equal (boundary condition), statistician's test has higher power from graph | B1 | |
| At \(\lambda = 2\): scientist's test is faster (since \(\lambda e^{-\lambda} < \frac{1}{3}\)) but statistician's test has higher power (0.75 vs less). Conclude statistician's test preferred as higher power, or scientist's test with valid reasoning about speed | B1 | Accept either conclusion with valid reason |
# Question 5:
## Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $X \sim \text{Po}(2)$ (since 20ml at rate $\lambda$ per 10ml, so rate $2\lambda$ per 20ml, with $\lambda=1$ gives Po(2)) | M1 | Must use Po(2) |
| $P(X \geqslant 4) = 1 - P(X \leqslant 3) = 1 - e^{-2}(1 + 2 + 2 + \frac{4}{3}) = 0.1429...$ awrt **0.143** | A1 | |
## Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| When $\lambda = 2.5$, $X \sim \text{Po}(5)$ | M1 | Must use Po(5) |
| $P(\text{Type II error}) = P(X \leqslant 3) = e^{-5}(1 + 5 + \frac{25}{2} + \frac{125}{6})$ = awrt **0.265** | A1 | |
## Part (c):
| Answer | Mark | Guidance |
|--------|------|----------|
| Let $Y \sim \text{Po}(\lambda)$ for 10ml sample. $P(\text{reject on 1st sample}) = P(Y \geqslant 2) = 1 - e^{-\lambda} - \lambda e^{-\lambda}$ | M1 | |
| $P(\text{exactly 1 on 1st sample}) = \lambda e^{-\lambda}$ | B1 | |
| $P(\text{reject on 2nd sample} \mid \text{1 on first}) = P(Y \geqslant 2) = 1 - e^{-\lambda} - \lambda e^{-\lambda}$ | M1 | |
| Power $= (1 - e^{-\lambda} - \lambda e^{-\lambda}) + \lambda e^{-\lambda}(1 - e^{-\lambda} - \lambda e^{-\lambda})$ $= 1 - e^{-\lambda} - \lambda e^{-\lambda} + \lambda e^{-\lambda} - \lambda e^{-2\lambda} - \lambda^2 e^{-2\lambda}$ $= 1 - e^{-\lambda} - \lambda(1+\lambda)e^{-2\lambda}$ | A1 | Fully correct derivation shown |
## Part (d):
| Answer | Mark | Guidance |
|--------|------|----------|
| Substitute $\lambda = 3$: $1 - e^{-3} - 3(4)e^{-6}$ = awrt **0.92**, so $r = $ **0.92** | B1 | |
## Part (e):
| Answer | Mark | Guidance |
|--------|------|----------|
| Correct shape — curve starting below scientist's curve and crossing/above it | B1 | Passes through or near correct table values |
| Points plotted correctly from Table 1 and joined smoothly | B1 | |
## Part (f):
| Answer | Mark | Guidance |
|--------|------|----------|
| Expected time for statistician's test $= 15 + 15 \times P(\text{1 organism in first sample})$ $= 15 + 15\lambda e^{-\lambda}$ | M1 A1 | |
| Statistician's test slower when $15 + 15\lambda e^{-\lambda} > 20$ | M1 | |
| $15\lambda e^{-\lambda} > 5 \Rightarrow \lambda e^{-\lambda} > \frac{1}{3}$ | A1 | Fully shown |
## Part (g):
| Answer | Mark | Guidance |
|--------|------|----------|
| At $\lambda = 1$: times are equal (boundary condition), statistician's test has higher power from graph | B1 | |
| At $\lambda = 2$: scientist's test is faster (since $\lambda e^{-\lambda} < \frac{1}{3}$) but statistician's test has higher power (0.75 vs less). Conclude **statistician's test** preferred as higher power, or **scientist's test** with valid reasoning about speed | B1 | Accept either conclusion with valid reason |5. Water is tested at various stages during a purification process by an environmental scientist. A certain organism occurs randomly in the water at a rate of $\lambda$ every 10 ml . The scientist selects a random sample of 20 ml of water to check whether there is evidence that $\lambda$ is greater than 1 . The criterion the scientist uses for rejecting the hypothesis that $\lambda = 1$ is that there are 4 or more organisms in the sample of 20 ml .
\begin{enumerate}[label=(\alph*)]
\item Find the size of the test.
\item When $\lambda = 2.5$ find P (Type II error).
A statistician suggests using an alternative test. The statistician's test involves taking a random sample of 10 ml and rejecting the hypothesis that $\lambda = 1$ if 2 or more organisms are present but accepting the hypothesis if no organisms are in the sample. If only 1 organism is found then a second random sample of 10 ml is taken and the hypothesis is rejected if 2 or more organisms are present, otherwise the hypothesis is accepted.
\item Show that the power of the statistician's test is given by
$$1 - \mathrm { e } ^ { - \lambda } - \lambda ( 1 + \lambda ) \mathrm { e } ^ { - 2 \lambda }$$
Table 1 below gives some values, to 2 decimal places, of the power function of the statistician's test.
\begin{table}[h]
\end{table}
Table 1
Figure 1 shows a graph of the power function for the scientist's test.\\
(e) On the same axes draw the graph of the power function for the statistician's test.
Given that it takes 20 minutes to collect and test a 20 ml sample and 15 minutes to collect and test a 10 ml sample\\
(f) show that the expected time of the statistician's test is slower than the scientist's test for $\lambda \mathrm { e } ^ { - \lambda } > \frac { 1 } { 3 }$\\
(g) By considering the times when $\lambda = 1$ and $\lambda = 2$ together with the power curves in part (e) suggest, giving a reason, which test you would use.\\
(2)
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{399f7507-4878-45ad-b77e-02ebd807ed75-10_1185_1157_1452_392}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
\includegraphics[max width=\textwidth, alt={}, center]{399f7507-4878-45ad-b77e-02ebd807ed75-11_81_47_2622_1886}
\end{enumerate}
\hfill \mbox{\textit{Edexcel S4 2013 Q5 [17]}}