| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 2 Statistics (Further AS Paper 2 Statistics) |
| Year | 2024 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Poisson hypothesis test |
| Difficulty | Standard +0.3 This is a straightforward application of a Poisson hypothesis test with standard parts: (a) perform a two-tailed test with given data, (b) find P(Type I error) which equals the significance level by definition, and (c) comment on model validity. All parts are routine for Further Maths Statistics with no novel problem-solving required, making it slightly easier than average. |
| Spec | 2.05a Hypothesis testing language: null, alternative, p-value, significance5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0: \lambda = 6\), \(H_1: \lambda \neq 6\), \(X \sim \text{Po}(6)\) | B1 (2.5) | States both hypotheses using correct notation; if letter other than \(\lambda\) or \(\mu\) used, must be correctly defined |
| Uses Poisson model \(X \sim \text{Po}(6)\) to calculate one of: \(P(X \geq 11)\) AWRT 0.04, \(P(X \leq 10)\) AWRT 0.96, \(P(X \leq 11)\) AWRT 0.98, or \(P(X > 11)\) AWRT 0.02 | M1 (3.3) | PI by correct upper tail of the critical region; condone mislabelling the probabilities |
| \(P(X \geq 11) = 0.043\) | A1 (3.4) | Uses Poisson model to calculate \(P(X \geq 11)\) AWRT 0.043 or obtains upper tail of critical region \(X \geq 12\); condone mislabelling |
| \(0.043 > 0.025\); correctly comparing probability with 0.025 or 0.05 | M1 (3.5a) | Or by correctly comparing 11 with upper tail of critical region |
| Do not reject \(H_0\) | A1F (2.2b) | FT comparison of \(P(X \geq 11)\) or \(P(X > 11)\) with 0.025 or 0.05; condone Accept \(H_0\) or Reject \(H_1\) |
| There is not sufficient evidence to suggest that the mean number of customers per hour has changed. | R1 (3.2a) | Concludes in context referring to change in mean (number of) customers per hour oe; conclusion must not be definite |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(X \leq 1) = 0.01735\), \(P(X \leq 2) = 0.06197\), \(P(X \geq 11) = 1 - P(X \leq 10) = 0.04262\), \(P(X \geq 12) = 1 - P(X \leq 11) = 0.02009\) | M1 (3.4) | Uses Poisson model to calculate one of \(P(X \leq 1)\), \(P(X \leq 2)\), \(P(X \leq 10)\), \(P(X \leq 11)\), \(P(X \geq 11)\) or \(P(X \geq 12)\); condone mislabelling |
| \(P(X \leq 1) =\) AWRT 0.017 and \(P(X \leq 11) =\) AWRT 0.98 or \(P(X \geq 12) =\) AWRT 0.02 | A1 (1.1b) | Condone mislabelling |
| Probability of Type I error \(= 0.01735 + 0.02009\) | M1 (1.1a) | Adds probability of lower tail to probability of upper tail |
| \(= 0.0374\) | R1 (2.1) | AWRT 0.0374; completes reasoned argument; must see \(P(X \leq 1)\), \(P(X \leq 2)\), \(P(X \geq 11)\) and \(P(X \geq 12)\) or \(P(X \leq 1)\), \(P(X \leq 2)\), \(P(X \leq 10)\) and \(P(X \leq 11)\) with correct labelling |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Some customers are not entering the store independently. | E1 (3.5b) | Explains that the Poisson condition of independence does not hold |
# Question 7(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0: \lambda = 6$, $H_1: \lambda \neq 6$, $X \sim \text{Po}(6)$ | B1 (2.5) | States both hypotheses using correct notation; if letter other than $\lambda$ or $\mu$ used, must be correctly defined |
| Uses Poisson model $X \sim \text{Po}(6)$ to calculate one of: $P(X \geq 11)$ **AWRT** 0.04, $P(X \leq 10)$ **AWRT** 0.96, $P(X \leq 11)$ **AWRT** 0.98, or $P(X > 11)$ **AWRT** 0.02 | M1 (3.3) | PI by correct upper tail of the critical region; condone mislabelling the probabilities |
| $P(X \geq 11) = 0.043$ | A1 (3.4) | Uses Poisson model to calculate $P(X \geq 11)$ **AWRT** 0.043 or obtains upper tail of critical region $X \geq 12$; condone mislabelling |
| $0.043 > 0.025$; correctly comparing probability with 0.025 or 0.05 | M1 (3.5a) | Or by correctly comparing 11 with upper tail of critical region |
| Do not reject $H_0$ | A1F (2.2b) | **FT** comparison of $P(X \geq 11)$ or $P(X > 11)$ with 0.025 or 0.05; condone Accept $H_0$ or Reject $H_1$ |
| There is not sufficient evidence to suggest that the mean number of customers per hour has changed. | R1 (3.2a) | Concludes in context referring to change in mean (number of) customers per hour **oe**; conclusion must not be definite |
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# Question 7(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(X \leq 1) = 0.01735$, $P(X \leq 2) = 0.06197$, $P(X \geq 11) = 1 - P(X \leq 10) = 0.04262$, $P(X \geq 12) = 1 - P(X \leq 11) = 0.02009$ | M1 (3.4) | Uses Poisson model to calculate one of $P(X \leq 1)$, $P(X \leq 2)$, $P(X \leq 10)$, $P(X \leq 11)$, $P(X \geq 11)$ or $P(X \geq 12)$; condone mislabelling |
| $P(X \leq 1) =$ **AWRT** 0.017 and $P(X \leq 11) =$ **AWRT** 0.98 or $P(X \geq 12) =$ **AWRT** 0.02 | A1 (1.1b) | Condone mislabelling |
| Probability of Type I error $= 0.01735 + 0.02009$ | M1 (1.1a) | Adds probability of lower tail to probability of upper tail |
| $= 0.0374$ | R1 (2.1) | **AWRT** 0.0374; completes reasoned argument; must see $P(X \leq 1)$, $P(X \leq 2)$, $P(X \geq 11)$ and $P(X \geq 12)$ or $P(X \leq 1)$, $P(X \leq 2)$, $P(X \leq 10)$ and $P(X \leq 11)$ with correct labelling |
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# Question 7(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Some customers are not entering the store independently. | E1 (3.5b) | Explains that the Poisson condition of independence does not hold |7 Over a period of time, it has been shown that the mean number of customers entering a small store is 6 per hour.
The store runs a promotion, selling many products at lower prices.
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\begin{enumerate}[label=(\alph*)]
\item Luke randomly selects an hour during the promotion and counts 11 customers entering the store.
He claims that the promotion has changed the mean number of customers per hour entering the store.
Investigate Luke's claim, using the $5 \%$ level of significance.\\
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\item Luke randomly selects another hour and carries out the same investigation as in part (a).
Find the probability of a Type I error, giving your answer to four decimal places.\\
Fully justify your answer.\\
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\item When observing the store, Luke notices that some customers enter the store together as a group.
Explain why the model used in parts (a) and (b) might not be valid.\\
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\end{enumerate}
\hfill \mbox{\textit{AQA Further AS Paper 2 Statistics 2024 Q7 [11]}}