| Exam Board | AQA |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2008 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Confidence interval for single proportion |
| Difficulty | Standard +0.8 This S3 question requires understanding that Poisson distributions can be approximated by normal distributions for large λ, then constructing a confidence interval for the difference of two means. Students must calculate sample means, apply the variance property of Poisson distributions, use the normal approximation, and correctly handle the difference of independent random variables. While methodical, it requires synthesis of multiple statistical concepts beyond routine application. |
| Spec | 5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Part (a) | ||
| \(\hat{\lambda}_A = \frac{3312}{184} = 18\); \(\hat{\lambda}_B = \frac{2760}{184} = 15\) | B1 | CAO both |
| \(95\% \Rightarrow z = 1.96\) | B1 | CAO |
| CI for \((\lambda_A - \lambda_B)\): \((\hat{\lambda}_A - \hat{\lambda}_B) \pm z\sqrt{\frac{\hat{\lambda}_A}{\mu_A} + \frac{\hat{\lambda}_B}{\mu_B}}\) | M1 | Variance term |
| CI expression used | M1 | |
| ie \((18 - 15) \pm 1.96 \times \sqrt{\frac{18}{184} + \frac{15}{184}}\) | A1↑ | ft on \(\hat{\lambda}_A\), \(\hat{\lambda}_B\) and \(z\) |
| ie \(3 \pm 0.83\) or \((2.17, 3.83)\) | A1 | AWRT; 6 marks |
| Part (b) | ||
| Calls from A and B are independent | B1 | Or equivalent; 1 mark |
| Part (a) Alternative Solution | ||
| \((3312 - 2760) \pm 1.96 \times \sqrt{3312 + 2760} = \) | (M2); (B1); (A1) | 1.96 |
| ie \(552 \pm 152.73\) | (M1) | |
| Dividing by 184 | ||
| ie \(3 \pm 0.83\) or \((2.17, 3.83)\) | (A1) | AWRT |
| **Part (a)** |
|---|
| $\hat{\lambda}_A = \frac{3312}{184} = 18$; $\hat{\lambda}_B = \frac{2760}{184} = 15$ | B1 | CAO both |
| $95\% \Rightarrow z = 1.96$ | B1 | CAO |
| CI for $(\lambda_A - \lambda_B)$: $(\hat{\lambda}_A - \hat{\lambda}_B) \pm z\sqrt{\frac{\hat{\lambda}_A}{\mu_A} + \frac{\hat{\lambda}_B}{\mu_B}}$ | M1 | Variance term |
| CI expression used | M1 | |
| ie $(18 - 15) \pm 1.96 \times \sqrt{\frac{18}{184} + \frac{15}{184}}$ | A1↑ | ft on $\hat{\lambda}_A$, $\hat{\lambda}_B$ and $z$ |
| ie $3 \pm 0.83$ or $(2.17, 3.83)$ | A1 | AWRT; 6 marks |
| **Part (b)** |
|---|
| Calls from A and B are independent | B1 | Or equivalent; 1 mark |
| **Part (a) Alternative Solution** |
|---|
| $(3312 - 2760) \pm 1.96 \times \sqrt{3312 + 2760} = $ | (M2); (B1); (A1) | 1.96 |
| ie $552 \pm 152.73$ | (M1) | |
| Dividing by 184 | | |
| ie $3 \pm 0.83$ or $(2.17, 3.83)$ | (A1) | AWRT |
**Total: 7 marks**
---5 The daily number of emergency calls received from district A may be modelled by a Poisson distribution with a mean of $\lambda _ { \mathrm { A } }$.
The daily number of emergency calls received from district B may be modelled by a Poisson distribution with a mean of $\lambda _ { \mathrm { B } }$.
During a period of 184 days, the number of emergency calls received from district A was 3312, whilst the number received from district B was 2760.
\begin{enumerate}[label=(\alph*)]
\item Construct an approximate $95 \%$ confidence interval for $\lambda _ { \mathrm { A } } - \lambda _ { \mathrm { B } }$.
\item State one assumption that is necessary in order to construct the confidence interval in part (a).
\end{enumerate}
\hfill \mbox{\textit{AQA S3 2008 Q5 [7]}}