| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2023 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of a Poisson distribution |
| Type | One-tailed test (increase or decrease) |
| Difficulty | Moderate -0.3 This is a straightforward application of a one-tailed Poisson hypothesis test with clearly stated context and parameter. Part (d) requires standard hypothesis test structure (H₀: λ=4, H₁: λ>4, find P(X≥7), compare to 5%), while parts (a)-(c) involve basic Poisson probability calculations. Slightly easier than average due to explicit guidance and routine procedures, though the multi-part structure and hypothesis test formulation keep it near typical A-level standard. |
| Spec | 2.05b Hypothesis test for binomial proportion2.05c Significance levels: one-tail and two-tail5.02i 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 | Mark | Guidance |
| \(\text{Po}(\lambda = 4)\) | B1 | For Po or Poisson and 4 must be seen in part (a). Do not allow P(4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Pairs of shoes (are sold) singly/randomly/independently/at a constant (average) rate | B1 | For one of the given assumptions in context (must have context of shoes or sales). Ignore extraneous non-contradictory comments. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(X > 4) = 1 - P(X \leq 4)\) | M1 | For writing or using \(P(X > 4) = 1 - P(X \leq 4)\) |
| \(= 0.3712\) | A1 | awrt 0.371 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \((\text{'0.371...'})^3\) | M1 | 'part (i)'^3 |
| \(= 0.051147...\) | A1 | 0.05115 or awrt 0.0511 (Calculator gives 0.051132...) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(H_0: \lambda = 4\), \(H_1: \lambda > 4\) | B1ft | Both hypotheses correct. Must be in terms of \(\lambda\) or \(\mu\). Must be attached to \(H_0\) and \(H_1\) |
| \(P(X \geq 7) = 1 - P(X \leq 6)\) or \(P(X \geq 9) = 1 - P(X \leq 8) = 0.0214\) | M1 | For writing or using \(P(X \geq 7) = 1 - P(X \leq 6)\); if CR approach award M1 for \(P(X \geq 9) = 1 - P(X \leq 8)\) written or used |
| \(= 0.1107\) or CR \(X \geq 9\) | A1 | awrt 0.111 or CR \(X \geq 9\) |
| Not significant / Do not reject \(H_0\) / Not in the critical region | M1 | Any correct ft statement consistent with their \(p\)-value and 0.05 or their CR and 7 |
| There is insufficient evidence of an increase in sales following the appearance of the advert / manager's belief is not supported | dA1 | Dependent on 1st M1. Correct conclusion in context which must be not rejecting \(H_0\). Must include the underlined words (oe). |
# Question 1:
## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\text{Po}(\lambda = 4)$ | B1 | For Po or Poisson and 4 must be seen in part (a). Do not allow P(4) |
## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| Pairs of **shoes** (are **sold**) singly/randomly/independently/at a **constant** (average) **rate** | B1 | For one of the given assumptions in context (must have context of **shoes** or **sales**). Ignore extraneous non-contradictory comments. |
## Part (c)(i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(X > 4) = 1 - P(X \leq 4)$ | M1 | For writing or using $P(X > 4) = 1 - P(X \leq 4)$ |
| $= 0.3712$ | A1 | awrt 0.371 |
## Part (c)(ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $(\text{'0.371...'})^3$ | M1 | 'part (i)'^3 |
| $= 0.051147...$ | A1 | 0.05115 or awrt 0.0511 (Calculator gives 0.051132...) |
## Part (d)
| Answer | Mark | Guidance |
|--------|------|----------|
| $H_0: \lambda = 4$, $H_1: \lambda > 4$ | B1ft | Both hypotheses correct. Must be in terms of $\lambda$ or $\mu$. Must be attached to $H_0$ and $H_1$ |
| $P(X \geq 7) = 1 - P(X \leq 6)$ or $P(X \geq 9) = 1 - P(X \leq 8) = 0.0214$ | M1 | For writing or using $P(X \geq 7) = 1 - P(X \leq 6)$; if CR approach award M1 for $P(X \geq 9) = 1 - P(X \leq 8)$ written or used |
| $= 0.1107$ or CR $X \geq 9$ | A1 | awrt 0.111 or CR $X \geq 9$ |
| Not significant / Do not reject $H_0$ / Not in the critical region | M1 | Any correct ft statement consistent with their $p$-value and 0.05 or their CR and 7 |
| There is insufficient evidence of an increase in sales following the appearance of the advert / manager's belief is not supported | dA1 | Dependent on 1st M1. Correct conclusion in context which must be **not rejecting** $H_0$. Must include the underlined words (oe). |
---\begin{enumerate}
\item A shop sells shoes at a mean rate of 4 pairs of shoes per hour on a weekday.\\
(a) Suggest a suitable distribution for modelling the number of sales of pairs of shoes made per hour on a weekday.\\
(b) State one assumption necessary for this distribution to be a suitable model of this situation.\\
(c) Find the probability that on a weekday the shop sells\\
(i) more than 4 pairs of shoes in a one-hour period,\\
(ii) more than 4 pairs of shoes in each of 3 consecutive one-hour periods.
\end{enumerate}
The area manager visits the shop on a weekday, the day after an advert for the shop appears in a local paper.
In a one-hour period during the manager's visit, the shop sells 7 pairs of shoes. This leads the manager to believe that the advert has increased the shop's sales of pairs of shoes.\\
(d) Stating your hypotheses clearly, test at the $5 \%$ level of significance whether or not there is evidence of an increase in sales of pairs of shoes following the appearance of the advert.
\hfill \mbox{\textit{Edexcel S2 2023 Q1 [11]}}