| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2006 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Poisson hypothesis test |
| Difficulty | Standard +0.8 This is a multi-part Poisson hypothesis test requiring understanding of significance levels, critical regions, and Type II errors. While the calculations are straightforward (finding P(X≥5) under H₀ and P(X<4) under H₁), it requires conceptual understanding of hypothesis testing framework and careful interpretation across different significance levels and error types, placing it moderately above average difficulty. |
| Spec | 2.05a Hypothesis testing language: null, alternative, p-value, significance2.05b Hypothesis test for binomial proportion |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(H_0: \lambda = 2.4\) | B1 | For both \(H_0\) and \(H_1\) |
| \(H_1: \lambda > 2.4\) (or \(0.8\) per hectare) | M1 | For recognisable Poisson expression, any mean, or finding critical region |
| Under \(H_0\): \(P(X \geq 5) = 1 - P(0,1,2,3,4) = 1 - P(0,1,2,3,4) = 1 - 0.904 = 0.0959\) | M1\(^*\)dep | For evaluating \(P(X \geq 5)\) or finding critical region |
| \(0.0959\) is less than \(0.10\) so in critical region ploughing has increased number of metal pieces found | A1 | For \(0.0959\) or \(0.096\) or critical region is \(X \geq 5\) |
| M1\(^*\) | For comparing their \(P(X \geq 5)\) with \(10\%\) or saying \(5\) is in critical region o.e.(o.e. comparison consistent with their \(H_1\)) | |
| A1ft | Correct conclusion, must relate to question, ft on their critical value or their \(P(X \geq 5)\) | |
| (ii) No significant increase at the \(5\%\) level | B1\(^*\)dep | 1 mark |
| (iii) \(P(X < 4) = e^{-4.2} \times (1 + 4.2 + 4.2^2/2 + 4.2^3/6)\) | M1\(^*\) | Using \(\lambda = 4.2\) (or 1.4) in a Poisson expression |
| M1\(^*\)dep | Finding \(P(X < 4)\) | |
| \(= 0.395\) | A1 | Correct answer(as final answer) |
**(i)** $H_0: \lambda = 2.4$ | B1 | For both $H_0$ and $H_1$
$H_1: \lambda > 2.4$ (or $0.8$ per hectare) | M1 | For recognisable Poisson expression, any mean, or finding critical region
Under $H_0$: $P(X \geq 5) = 1 - P(0,1,2,3,4) = 1 - P(0,1,2,3,4) = 1 - 0.904 = 0.0959$ | M1$^*$dep | For evaluating $P(X \geq 5)$ or finding critical region
$0.0959$ is less than $0.10$ so in critical region ploughing has increased number of metal pieces found | A1 | For $0.0959$ or $0.096$ or critical region is $X \geq 5$ | 6 marks total
| M1$^*$ | For comparing their $P(X \geq 5)$ with $10\%$ or saying $5$ is in critical region o.e.(o.e. comparison consistent with their $H_1$)
| A1ft | Correct conclusion, must relate to question, ft on their critical value or their $P(X \geq 5)$
**(ii)** No significant increase at the $5\%$ level | B1$^*$dep | 1 mark
**(iii)** $P(X < 4) = e^{-4.2} \times (1 + 4.2 + 4.2^2/2 + 4.2^3/6)$ | M1$^*$ | Using $\lambda = 4.2$ (or 1.4) in a Poisson expression
| M1$^*$dep | Finding $P(X < 4)$
$= 0.395$ | A1 | Correct answer(as final answer) | 3 marks total
---6 Pieces of metal discovered by people using metal detectors are found randomly in fields in a certain area at an average rate of 0.8 pieces per hectare. People using metal detectors in this area have a theory that ploughing the fields increases the average number of pieces of metal found per hectare. After ploughing, they tested this theory and found that a randomly chosen field of area 3 hectares yielded 5 pieces of metal.\\
(i) Carry out the test at the $10 \%$ level of significance.\\
(ii) What would your conclusion have been if you had tested at the $5 \%$ level of significance?
Jack decides that he will reject the null hypothesis that the average number is 0.8 pieces per hectare if he finds 4 or more pieces of metal in another ploughed field of area 3 hectares.\\
(iii) If the true mean after ploughing is 1.4 pieces per hectare, calculate the probability that Jack makes a Type II error.
\hfill \mbox{\textit{CAIE S2 2006 Q6 [10]}}