| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2018 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of a Poisson distribution |
| Type | One-tailed test (increase or decrease) |
| Difficulty | Standard +0.3 This is a straightforward S2 hypothesis testing question with standard Poisson applications. Part (a) is routine one-tailed testing with clear hypotheses and critical value lookup. Part (b) requires simple algebraic manipulation of the Poisson probability formula. Part (c) is basic expectation calculation. All parts follow standard textbook procedures with no novel insight required, making it slightly easier than average. |
| Spec | 5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.02l Poisson conditions: for modelling5.02m Poisson: mean = variance = lambda5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0: \lambda = 4\), \(H_1: \lambda > 4\) | B1 | Allow \(\mu\) or \(\lambda\) and 4 or 8 |
| \(X \sim Po(8)\) and \(P(X \geqslant 14) = 1 - P(X \leqslant 13)\) | M1 | For using or writing \(1 - P(X \leqslant 13)\) with \(X \sim Po(8)\) |
| \(= 1 - 0.9658 = 0.0341807..\) awrt \(0.0342\) | A1 | For awrt 0.0342; stating CR is \(X \geqslant 14\) scores the M1A1 |
| There is evidence to reject \(H_0\) | dM1 | Dep on 1st M1; correct conclusion for 1-tail test with no incorrect conclusions |
| There is evidence to support Emma's belief | A1 | Correct contextual conclusion mentioning Emma's belief |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([F =\) no. of faults in piece of cloth of length \(l]\); \(e^{-\frac{4l}{50}} = 0.90\) | M1A1 | For using \(\frac{4l}{50}\) with Poisson; can be scored anywhere in part (b) |
| \(e^{-\frac{4 \times 1.25}{50}} = 0.9048...\), \(e^{-\frac{4 \times 1.35}{50}} = 0.8976...\) or \((\pm)\frac{4l}{50} = \ln 0.9\) | M1 | For substituting suitable values either side of 1.3, or using ln to solve |
| These values are either side of 0.90 \(\therefore\) \(l = 1.3\) to 2 sf or \(l = 1.317...\) \(\therefore\) 1.3 | A1cso | If using "solve" approach must see \(l =\) awrt 1.32 and "\(\therefore = 1.3\)" |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Expected number with no faults \(= 5000 \times 0.9 = 4500\); Expected number with some faults \(= 5000 \times 0.1 = 500\) | M1 | For using either 4500 or 500 or expected profit per sale \(2.5 \times 0.9 - 0.5 \times 0.1 = 2.2\) |
| Expected profit \(= 4500 \times 2.5 - 500 \times 0.5\) | M1 | For \(4500 \times 2.5 - 500 \times 0.5\) or \(5000 \times 2.2\) or \(5000 \times 2.5 - 500 \times 3\) |
| \(= £11000\) | A1 | Correct answer only 3/3 |
## Question 2:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0: \lambda = 4$, $H_1: \lambda > 4$ | B1 | Allow $\mu$ or $\lambda$ and 4 or 8 |
| $X \sim Po(8)$ and $P(X \geqslant 14) = 1 - P(X \leqslant 13)$ | M1 | For using or writing $1 - P(X \leqslant 13)$ with $X \sim Po(8)$ |
| $= 1 - 0.9658 = 0.0341807..$ awrt $0.0342$ | A1 | For awrt 0.0342; stating CR is $X \geqslant 14$ scores the M1A1 |
| There is evidence to reject $H_0$ | dM1 | Dep on 1st M1; correct conclusion for 1-tail test with no incorrect conclusions |
| There is evidence to support Emma's belief | A1 | Correct contextual conclusion mentioning Emma's belief |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[F =$ no. of faults in piece of cloth of length $l]$; $e^{-\frac{4l}{50}} = 0.90$ | M1A1 | For using $\frac{4l}{50}$ with Poisson; can be scored anywhere in part (b) |
| $e^{-\frac{4 \times 1.25}{50}} = 0.9048...$, $e^{-\frac{4 \times 1.35}{50}} = 0.8976...$ or $(\pm)\frac{4l}{50} = \ln 0.9$ | M1 | For substituting suitable values either side of 1.3, or using ln to solve |
| These values are either side of 0.90 $\therefore$ $l = 1.3$ to 2 sf or $l = 1.317...$ $\therefore$ 1.3 | A1cso | If using "solve" approach must see $l =$ awrt 1.32 and "$\therefore = 1.3$" |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Expected number with no faults $= 5000 \times 0.9 = 4500$; Expected number with some faults $= 5000 \times 0.1 = 500$ | M1 | For using either 4500 or 500 or expected profit per sale $2.5 \times 0.9 - 0.5 \times 0.1 = 2.2$ |
| Expected profit $= 4500 \times 2.5 - 500 \times 0.5$ | M1 | For $4500 \times 2.5 - 500 \times 0.5$ or $5000 \times 2.2$ or $5000 \times 2.5 - 500 \times 3$ |
| $= £11000$ | A1 | Correct answer only 3/3 |
---2. John weaves cloth by hand. Emma believes that faults are randomly distributed in John's cloth at a rate of more than 4 per 50 metres of cloth. To check her belief, Emma takes a random sample of 100 metres of the cloth and finds that it contains 14 faults.
\begin{enumerate}[label=(\alph*)]
\item Stating your hypotheses clearly, test, at the $5 \%$ level of significance, Emma's belief.
Armani also weaves cloth by hand. He knows that faults are randomly distributed in his cloth at a rate of 4 per 50 metres of cloth. Emma decides to buy a large amount of Armani's cloth to sell in pieces of length $l$ metres. She chooses $l$ so that the probability of no faults in a piece is exactly 0.9
\item Show that $l = 1.3$ to 2 significant figures.
Emma sells 5000 of these pieces of cloth of length 1.3 metres. She makes a profit of $\pounds 2.50$ on each piece of cloth that does not contain any faults but a loss of $\pounds 0.50$ on any piece that contains at least one fault.
\item Find Emma's expected profit.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2018 Q2 [12]}}