| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2014 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Poisson hypothesis test |
| Difficulty | Standard +0.3 This is a standard S2 Poisson hypothesis test with routine application of tables and normal approximation. Part (a) requires finding critical regions from tables (straightforward lookup), part (b) reads off actual significance, and part (c) applies the standard normal approximation to Poisson with continuity correction. All techniques are textbook exercises with no novel insight required, though slightly easier than average due to clear structure and standard methods. |
| Spec | 5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.05b Unbiased estimates: of population mean and variance5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(H_0: \lambda = \frac{1}{8}\) (or \(\lambda=5\)); \(H_1: \lambda \neq \frac{1}{8}\) (or \(\lambda \neq 5\)); allow \(\lambda\) or \(\mu\) | B1 | For suitable hypotheses |
| \(X \sim Po(5)\); \(P(X \leq 1) = 0.0404\) or \(P(X \geq 10) = 0.0318\) or \(P(X \geq 9) = 0.0681\) | M1 | For correct use of \(Po(5)\). Award if one relevant probability seen or correct CR as probability statement |
| Critical Regions: \(X \leq 1\) or \(X \geq 10\) | A1, A1 (4) | 1st A1 for \(X \leq 1\) or \(X<2\) etc. 2nd A1 for \(X \geq 10\) or \(X>9\) etc. Ignore \(\cup\) or \(\cap\) signs |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0.0404 + 0.0318 = 0.0722\) (or 7.22% significance level) | M1A1 (2) | M1 for adding probabilities of critical regions if sum gives probability less than 1, or if correct answer given. A1 for awrt 0.0722 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(H_0: \lambda = \frac{1}{8}\) (or \(\lambda=25\)); \(H_1: \lambda < \frac{1}{8}\) (or \(\lambda < 25\)); allow \(\lambda\) or \(\mu\) | B1 | For suitable hypotheses |
| \([Y = \text{no. of defects in 200m of wallpaper}]\ Y \sim Po(25)\ Y \approx N\!\left(25, \sqrt{25}^2\right)\) | M1A1 | 1st M1 for normal approximation. A1 for mean \(=25\) and variance \(=25\) or sd \(=5\) |
| \(P(Y \leq 19) \approx P\!\left(Z < \frac{19.5-25}{\sqrt{25}}\right)\) or \(\pm\frac{x - 0.5 - 25}{5} = 1.96\) | M1M1 | 2nd M1 for continuity correction (\(19 \pm 0.5\)). 3rd M1 for standardising using mean and sd |
| \(= [P(Z < -1.1)] = 0.1357\) (or 0.13566… from calc); \(x = 35.3\) | A1 | For awrt 0.136 or 35.3 or \(-1.1 > -1.96\) |
| \([{>}0.05]\) not significant, insufficient evidence to support Thomas' claim. Or: the number/rate/amount of defects is not decreased/less/reduced | A1cso (7) | For correct contextualised conclusion. cao for one-tailed test, must come from correct working |
# Question 5:
## Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0: \lambda = \frac{1}{8}$ (or $\lambda=5$); $H_1: \lambda \neq \frac{1}{8}$ (or $\lambda \neq 5$); allow $\lambda$ or $\mu$ | B1 | For suitable hypotheses |
| $X \sim Po(5)$; $P(X \leq 1) = 0.0404$ or $P(X \geq 10) = 0.0318$ or $P(X \geq 9) = 0.0681$ | M1 | For correct use of $Po(5)$. Award if one relevant probability seen or correct CR as probability statement |
| Critical Regions: $X \leq 1$ or $X \geq 10$ | A1, A1 (4) | 1st A1 for $X \leq 1$ or $X<2$ etc. 2nd A1 for $X \geq 10$ or $X>9$ etc. Ignore $\cup$ or $\cap$ signs |
## Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.0404 + 0.0318 = 0.0722$ (or 7.22% significance level) | M1A1 (2) | M1 for adding probabilities of critical regions if sum gives probability less than 1, or if correct answer given. A1 for awrt 0.0722 |
## Part (c)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0: \lambda = \frac{1}{8}$ (or $\lambda=25$); $H_1: \lambda < \frac{1}{8}$ (or $\lambda < 25$); allow $\lambda$ or $\mu$ | B1 | For suitable hypotheses |
| $[Y = \text{no. of defects in 200m of wallpaper}]\ Y \sim Po(25)\ Y \approx N\!\left(25, \sqrt{25}^2\right)$ | M1A1 | 1st M1 for normal approximation. A1 for mean $=25$ and variance $=25$ or sd $=5$ |
| $P(Y \leq 19) \approx P\!\left(Z < \frac{19.5-25}{\sqrt{25}}\right)$ or $\pm\frac{x - 0.5 - 25}{5} = 1.96$ | M1M1 | 2nd M1 for continuity correction ($19 \pm 0.5$). 3rd M1 for standardising using mean and sd |
| $= [P(Z < -1.1)] = 0.1357$ (or 0.13566… from calc); $x = 35.3$ | A1 | For awrt 0.136 or 35.3 or $-1.1 > -1.96$ |
| $[{>}0.05]$ not significant, insufficient evidence to support **Thomas' claim**. Or: the **number/rate/amount** of **defects** is not **decreased/less/reduced** | A1cso (7) | For correct contextualised conclusion. cao for one-tailed test, must come from correct working |
---\begin{enumerate}
\item Sammy manufactures wallpaper. She knows that defects occur randomly in the manufacturing process at a rate of 1 every 8 metres. Once a week the machinery is cleaned and reset. Sammy then takes a random sample of 40 metres of wallpaper from the next batch produced to test if there has been any change in the rate of defects.\\
(a) Stating your hypotheses clearly and using a $10 \%$ level of significance, find the critical region for this test. You should choose your critical region so that the probability of rejection is less than 0.05 in each tail.\\
(b) State the actual significance level of this test.
\end{enumerate}
Thomas claims that his new machine would reduce the rate of defects and invites Sammy to test it. Sammy takes a random sample of 200 metres of wallpaper produced on Thomas' machine and finds 19 defects.\\
(c) Using a suitable approximation, test Thomas' claim. You should use a $5 \%$ level of significance and state your hypotheses clearly.\\
\hfill \mbox{\textit{Edexcel S2 2014 Q5 [13]}}