| Exam Board | AQA |
|---|---|
| Module | Further Paper 3 Statistics (Further Paper 3 Statistics) |
| Year | 2021 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Mean-variance comparison for Poisson validation |
| Difficulty | Challenging +1.2 This is a structured Further Maths Statistics question testing standard Poisson distribution concepts: mean-variance comparison for model validation, hypothesis testing with a given significance level, and power calculation. While it requires multiple steps and understanding of hypothesis testing framework, each part follows routine procedures with clear guidance (significance level given, test structure standard). The power calculation in part (c) is slightly more sophisticated but still a textbook application once the alternative hypothesis value is provided. |
| Spec | 2.05a Hypothesis testing language: null, alternative, p-value, significance2.05e Hypothesis test for normal mean: known variance5.02i Poisson distribution: random events model5.02l Poisson conditions: for modelling5.02n Sum of Poisson variables: is Poisson |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\bar{x} = \frac{102}{60} = 1.7\) | B1 (1.1b) | Obtains correct sample mean |
| \(s^2 = \frac{103.25}{59} = 1.75\) | B1 (1.1b) | Obtains correct sample variance |
| It is appropriate to model the number of complaints in a day with a Poisson distribution with mean 1.7 as the sample mean is approximately equal to the sample variance | E1 (2.4) | Explains it is appropriate to model with Poisson with mean 1.7; sample mean is approximately equal to sample variance; condone use of \(\sigma^2\) but values of both must be seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0: \lambda = 6.9\), \(H_1: \lambda \neq 6.9\) | B1 (2.5) | States both hypotheses using correct language with \(\lambda = 6.9\), \(1.7\) or \(5.2\) |
| \(X \sim \text{Po}(6.9)\) | M1 (3.3) | Uses Poisson model with \(\lambda = 6.9\) |
| \(P(X \leq 3) = 0.087\) | A1 (3.4) | Uses model to obtain correct value of \(P(X \leq 3)\); AWRT 0.087; PI by comparing subset of \(P(X \leq 3)\) that is greater than 0.05 with 0.05 |
| \(0.087 > 0.05\), Accept \(H_0\) | R1 (3.5a) | Evaluates Poisson model by comparing probability with 0.05 |
| \(H_0\) not rejected | E1 (2.2b) | Infers \(H_0\) not rejected |
| No evidence to suggest that the mean total number of complaints and enquiries received per day has changed | E1F (3.2a) | Concludes in context; conclusion must not be definite; FT their incorrect rejection of \(H_0\) if stated or 'their' comparison if not |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(X \leq 2) = 0.031\), \(P(X \geq 12) = 0.049\) | M1 (1.1a) | Finds correct value of \(P(X \leq 2)\) or \(P(X \geq 12)\); PI |
| Reject hypothesis test if \(X \leq 2\) or \(X \geq 12\); \(Y \sim \text{Po}(6.1)\) | A1 (1.1b) | Identifies correct critical region (or acceptance region) |
| Power of test \(= P(Y \leq 2) + P(Y \geq 12) = 0.05765\ldots + 0.02244\ldots\) | M1 (1.1a) | Uses Poisson model with \(\lambda = 6.1\) to find a probability |
| \(= 0.0801\) | A1 (1.1b) | Obtains correct power of test; AWRT 0.0801 |
## Question 8(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\bar{x} = \frac{102}{60} = 1.7$ | B1 (1.1b) | Obtains correct sample mean |
| $s^2 = \frac{103.25}{59} = 1.75$ | B1 (1.1b) | Obtains correct sample variance |
| It is appropriate to model the number of complaints in a day with a Poisson distribution with mean 1.7 as the sample mean is approximately equal to the sample variance | E1 (2.4) | Explains it is appropriate to model with Poisson with mean 1.7; sample mean is approximately equal to sample variance; condone use of $\sigma^2$ but values of both must be seen |
---
## Question 8(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0: \lambda = 6.9$, $H_1: \lambda \neq 6.9$ | B1 (2.5) | States both hypotheses using correct language with $\lambda = 6.9$, $1.7$ or $5.2$ |
| $X \sim \text{Po}(6.9)$ | M1 (3.3) | Uses Poisson model with $\lambda = 6.9$ |
| $P(X \leq 3) = 0.087$ | A1 (3.4) | Uses model to obtain correct value of $P(X \leq 3)$; AWRT 0.087; PI by comparing subset of $P(X \leq 3)$ that is greater than 0.05 with 0.05 |
| $0.087 > 0.05$, Accept $H_0$ | R1 (3.5a) | Evaluates Poisson model by comparing probability with 0.05 |
| $H_0$ not rejected | E1 (2.2b) | Infers $H_0$ not rejected |
| No evidence to suggest that the mean total number of complaints and enquiries received per day has changed | E1F (3.2a) | Concludes in context; conclusion must not be definite; FT their incorrect rejection of $H_0$ if stated or 'their' comparison if not |
---
## Question 8(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(X \leq 2) = 0.031$, $P(X \geq 12) = 0.049$ | M1 (1.1a) | Finds correct value of $P(X \leq 2)$ or $P(X \geq 12)$; PI |
| Reject hypothesis test if $X \leq 2$ or $X \geq 12$; $Y \sim \text{Po}(6.1)$ | A1 (1.1b) | Identifies correct critical region (or acceptance region) |
| Power of test $= P(Y \leq 2) + P(Y \geq 12) = 0.05765\ldots + 0.02244\ldots$ | M1 (1.1a) | Uses Poisson model with $\lambda = 6.1$ to find a probability |
| $= 0.0801$ | A1 (1.1b) | Obtains correct power of test; AWRT 0.0801 |8 A company records the number of complaints, $X$, that it receives over 60 months. The summarised results are
$$\sum x = 102 \quad \text { and } \quad \sum ( x - \bar { x } ) ^ { 2 } = 103.25$$
8
\begin{enumerate}[label=(\alph*)]
\item Using this data, explain why it may be appropriate to model the number of complaints received by the company per month by a Poisson distribution with mean 1.7\\
8
\item The company also receives enquiries as well as complaints. The number of enquiries received is independent of the number of complaints received.
The company models the number of complaints per month with a Poisson distribution with mean 1.7 and the number of enquiries per month with a Poisson distribution with mean 5.2
The company starts selling a new product.\\
The company records a total of 3 complaints and enquiries in one randomly chosen month.
Investigate if the mean total number of complaints and enquiries received per month has changed following the introduction of the new product, using the $10 \%$ level of significance.\\
8
\item It is later found that the mean total number of complaints and enquiries received per month is 6.1
Find the power of the test carried out in part (b), giving your answer to four decimal places.\\
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\includegraphics[max width=\textwidth, alt={}]{3ef4c3fd-cbf0-4ac0-a072-a07d763fd50a-20_2496_1723_214_148}
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\hfill \mbox{\textit{AQA Further Paper 3 Statistics 2021 Q8 [13]}}