| Exam Board | AQA |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2009 |
| Session | June |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of a Poisson distribution |
| Type | One-tailed test (increase or decrease) |
| Difficulty | Standard +0.8 This is a multi-part Further Maths Statistics question requiring: (a) normal approximation to Poisson with continuity correction, (b)(i) one-tailed hypothesis test with Poisson distribution, (b)(ii) critical region determination, and (b)(iii) power calculation. While each component uses standard techniques, the combination of multiple statistical concepts (approximation, hypothesis testing, critical values, and power) in one question, plus the need to work with cumulative Poisson probabilities, makes this moderately challenging and above average difficulty. |
| Spec | 2.04d Normal approximation to binomial5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.05b Unbiased estimates: of population mean and variance |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(T = X_{\Sigma D} \sim \text{Po}(144)\) | B1 | CAO |
| \(T \sim \text{approx } N(144, 144)\) | M1 | Normal with \(\mu = \sigma^2\) |
| \(P(T_{\text{Po}} \leq 150) \approx P(T_N < 150.5)\) | B1 | CAO |
| \(= P\!\left(Z < \dfrac{150.5-144}{12}\right)\) | M1 | Standardising (149.5, 150 or 150.5) with \(\mu>24\) and \(\sqrt{\mu}\) |
| \(= P(Z<0.54) = 0.705\) to \(0.71\) | A1 | 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0: \lambda\text{(or mean)} = 2\text{ (or 10)}\); \(H_1: \lambda\text{(or mean)} > 2\text{ (or 10)}\) | B1 | Both; or equivalent |
| \(P(Y \geq 17) = 1 - P(Y \leq 16) = 1 - 0.9730 = 0.027\) | M1, A1 | Accept \(1-P(Y\leq17)\) |
| \(< 0.10\ (10\%)\) | M1 | Comparison of probability with 0.1 |
| \([z = 2.05\) to \(2.38 > 1.2816]\) | M1 | Comparison of \(z\) with 1.2816 or 1.6449 |
| Thus evidence, at 10% level, of increase in mean daily number of requests | A1F | 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| CV of \(Y\) such that \(P(Y \geq \text{CV}) \leq 0.10\) | M1 | Can be implied by 13, 14 or 15; accept \(P(Y=\text{CV})=0.10\) |
| \(P(Y \leq \text{CV}-1) \geq 0.90\) | M1 | Can be implied by 13, 14 or 15; accept \(P(Y=\text{CV})=0.90\) |
| \(\text{CV} = 15\) | A1 | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Power \(= 1 - P(\text{Type II error}) = 1 - P(\text{accept } H_0 \mid H_0 \text{ false}) = P(\text{accept } H_1 \mid H_1 \text{ true})\) | B1 | Or equivalent; stated or implied use |
| \(\lambda = 5\times3 = 15\) | B1 | Stated or implied use of Po(15) |
| Power \(= P(Y \geq \text{CV}) = P(Y \geq 15) = 1 - P(Y \leq 14) = 1 - 0.4657\) | M1 | Attempt at probability based on C's CV from (ii) and Po(15) |
| \(= 0.53\) to \(0.54\) | A1 | 4 |
# Question 7:
**Part (a):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $T = X_{\Sigma D} \sim \text{Po}(144)$ | B1 | CAO |
| $T \sim \text{approx } N(144, 144)$ | M1 | Normal with $\mu = \sigma^2$ |
| $P(T_{\text{Po}} \leq 150) \approx P(T_N < 150.5)$ | B1 | CAO |
| $= P\!\left(Z < \dfrac{150.5-144}{12}\right)$ | M1 | Standardising (149.5, 150 or 150.5) with $\mu>24$ and $\sqrt{\mu}$ |
| $= P(Z<0.54) = 0.705$ to $0.71$ | A1 | 5 | AWFW (0.70598) |
**Part (b)(i):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0: \lambda\text{(or mean)} = 2\text{ (or 10)}$; $H_1: \lambda\text{(or mean)} > 2\text{ (or 10)}$ | B1 | Both; or equivalent |
| $P(Y \geq 17) = 1 - P(Y \leq 16) = 1 - 0.9730 = 0.027$ | M1, A1 | Accept $1-P(Y\leq17)$ |
| $< 0.10\ (10\%)$ | M1 | Comparison of probability with 0.1 |
| $[z = 2.05$ to $2.38 > 1.2816]$ | M1 | Comparison of $z$ with 1.2816 or 1.6449 |
| Thus evidence, at 10% level, of increase in mean daily number of requests | A1F | 5 | F on probability or on $z$ |
**Part (b)(ii):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| CV of $Y$ such that $P(Y \geq \text{CV}) \leq 0.10$ | M1 | Can be implied by 13, 14 or 15; accept $P(Y=\text{CV})=0.10$ |
| $P(Y \leq \text{CV}-1) \geq 0.90$ | M1 | Can be implied by 13, 14 or 15; accept $P(Y=\text{CV})=0.90$ |
| $\text{CV} = 15$ | A1 | 3 | CAO |
**Part (b)(iii):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Power $= 1 - P(\text{Type II error}) = 1 - P(\text{accept } H_0 \mid H_0 \text{ false}) = P(\text{accept } H_1 \mid H_1 \text{ true})$ | B1 | Or equivalent; stated or implied use |
| $\lambda = 5\times3 = 15$ | B1 | Stated or implied use of Po(15) |
| Power $= P(Y \geq \text{CV}) = P(Y \geq 15) = 1 - P(Y \leq 14) = 1 - 0.4657$ | M1 | Attempt at probability based on C's CV from (ii) and Po(15) |
| $= 0.53$ to $0.54$ | A1 | 4 | AWFW (0.5343) |7 The daily number of customers visiting a small arts and crafts shop may be modelled by a Poisson distribution with a mean of 24 .
\begin{enumerate}[label=(\alph*)]
\item Using a distributional approximation, estimate the probability that there was a total of at most 150 customers visiting the shop during a given 6-day period.
\item The shop offers a picture framing service. The daily number of requests, $Y$, for this service may be assumed to have a Poisson distribution.
Prior to the shop advertising this service in the local free newspaper, the mean value of $Y$ was 2. Following the advertisement, the shop received a total of 17 requests for the service during a period of 5 days.
\begin{enumerate}[label=(\roman*)]
\item Using a Poisson distribution, carry out a test, at the $10 \%$ level of significance, to investigate the claim that the advertisement increased the mean daily number of requests for the shop's picture framing service.
\item Determine the critical value of $Y$ for your test in part (b)(i).
\item Hence, assuming that the advertisement increased the mean value of $Y$ to 3, determine the power of your test in part (b)(i).
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA S3 2009 Q7 [17]}}