| Exam Board | AQA |
|---|---|
| Module | Further Paper 3 Statistics (Further Paper 3 Statistics) |
| Year | 2023 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of a Poisson distribution |
| Type | Two-tailed test (change) |
| Difficulty | Standard +0.3 This is a straightforward Further Maths Statistics question testing standard Poisson hypothesis testing procedures. Part (a) requires a routine two-tailed test with clearly given parameters (SD=5 means λ=25), part (b) asks for a standard definition, part (c) tests the basic property that sum of independent Poissons is Poisson, and part (d) requires recall of model assumptions. All parts are textbook-standard with no novel problem-solving required, making it slightly easier than average even for Further Maths. |
| Spec | 2.05a Hypothesis testing language: null, alternative, p-value, significance5.02i Poisson distribution: random events model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\lambda = 5^2 = 25\) | B1 | Correct mean of Poisson distribution |
| \(H_0: \lambda = 25\); \(H_1: \lambda \neq 25\); \(X \sim \text{Po}(25)\) | B1F | States both hypotheses using correct language; FT their value of \(\lambda\) |
| Use Poisson to find one of \(P(X \leq 16)\), \(P(X < 16)\), \(P(X \geq 16)\), \(P(X > 16)\), or \(P(X = 16)\) | M1 | Uses Poisson model with their \(\lambda\) |
| \(P(X \leq 16) = 0.0377\ldots\) AWRT 0.04; condone \(P(X < 16) = 0.0377\ldots\) | A1 | |
| \(0.0377\ldots > 0.025\); or concludes 16 is not in critical region as \(16 > 15\) | R1 | Evaluates Poisson model by correctly comparing probability with 0.025; if correct probability seen, must be used in comparison |
| Accept \(H_0\) | E1F | Infers \(H_0\) not rejected; FT comparison of their p-value using Poisson model with 0.025 |
| Insufficient evidence to suggest that the average number of toys per week that do not pass quality checks has changed | E1F | Conclusion in context; must not be definite; must mention average number of toys per week; FT consistent with conclusion |
| Answer | Marks | Guidance |
|---|---|---|
| To conclude that the average number of toys per week that do not pass quality checks has not changed when it has | E1 | Must mention number of toys per week; interprets Type II error in context |
| Answer | Marks | Guidance |
|---|---|---|
| \(X + Y \sim \text{Po}(43)\) | B1F | Obtains correct distribution for their value of \(\lambda\), \(\lambda + 18\) |
| Answer | Marks | Guidance |
|---|---|---|
| Individual toys passing quality checks may not be independent | E1 | Recognises one limitation of the Poisson model relating to independence in context |
| The number of toys produced over time might vary to meet demand | E1 | Recognises limitation of the Poisson model relating to constant rate over time in context |
## Question 7(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\lambda = 5^2 = 25$ | B1 | Correct mean of Poisson distribution |
| $H_0: \lambda = 25$; $H_1: \lambda \neq 25$; $X \sim \text{Po}(25)$ | B1F | States both hypotheses using correct language; FT their value of $\lambda$ |
| Use Poisson to find one of $P(X \leq 16)$, $P(X < 16)$, $P(X \geq 16)$, $P(X > 16)$, or $P(X = 16)$ | M1 | Uses Poisson model with their $\lambda$ |
| $P(X \leq 16) = 0.0377\ldots$ AWRT 0.04; condone $P(X < 16) = 0.0377\ldots$ | A1 | |
| $0.0377\ldots > 0.025$; or concludes 16 is not in critical region as $16 > 15$ | R1 | Evaluates Poisson model by correctly comparing probability with 0.025; if correct probability seen, must be used in comparison |
| Accept $H_0$ | E1F | Infers $H_0$ not rejected; FT comparison of their p-value using Poisson model with 0.025 |
| Insufficient evidence to suggest that the average number of toys per week that do not pass quality checks has changed | E1F | Conclusion in context; must not be definite; must mention average number of toys per week; FT consistent with conclusion |
**Subtotal: 7 marks**
## Question 7(b):
To conclude that the average number of toys per week that do not pass quality checks has not changed when it has | E1 | Must mention number of toys per week; interprets Type II error in context
---
## Question 7(c):
$X + Y \sim \text{Po}(43)$ | B1F | Obtains correct distribution for their value of $\lambda$, $\lambda + 18$
---
## Question 7(d):
Individual toys passing quality checks may not be independent | E1 | Recognises one limitation of the Poisson model relating to independence in context
The number of toys produced over time might vary to meet demand | E1 | Recognises limitation of the Poisson model relating to constant rate over time in context
---7 Company $A$ uses a machine to produce toys.
The number of toys in a week that do not pass Company $A$ 's quality checks is modelled by a Poisson distribution $X$ with standard deviation 5
The machine producing the toys breaks down.\\
After it is repaired, 16 toys in the next week do not pass the quality checks.\\
7
\begin{enumerate}[label=(\alph*)]
\item Investigate whether the average number of toys that do not pass the quality checks in a week has changed, using the $5 \%$ level of significance.\\
7
\item For the test carried out in part (a), state in context the meaning of a Type II error.
7
\item Company $B$ uses a different machine to produce toys.\\
The number of toys in a week that do not pass Company B's quality checks is modelled by a Poisson distribution $Y$ with mean 18
The variables $X$ and $Y$ are independent.\\
Find the distribution of the total number of toys in a week produced by companies $A$ and $B$ that do not pass their quality checks.
7
\item State two reasons why a Poisson distribution may not be a valid model for the number of toys that do not pass the quality checks in a week.
Reason 1 $\_\_\_\_$\\
Reason 2 $\_\_\_\_$
\end{enumerate}
\hfill \mbox{\textit{AQA Further Paper 3 Statistics 2023 Q7 [11]}}