| Exam Board | AQA |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2015 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Confidence interval for single proportion |
| Difficulty | Standard +0.3 This is a straightforward application of confidence intervals for the difference of two Poisson means using normal approximation. Students need to recognize the Poisson setup, calculate sample means (507/26 and 416/26), apply the standard formula for difference of means with variance sum, and use z=2.576 for 99% CI. The interpretation in part (b) is trivial. Slightly above average difficulty due to being S3 material and requiring Poisson distribution knowledge, but the execution is mechanical with no novel insight required. |
| Spec | 5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\hat{\lambda}_M = \frac{507}{26}\), \(\hat{\lambda}_F = \frac{416}{26}\) | B1 | Both estimates stated or implied |
| \(\hat{\lambda}_M - \hat{\lambda}_F = \frac{507-416}{26} = \frac{91}{26} = 3.5\) | B1 | Correct point estimate |
| \(\text{Var}(\hat{\lambda}_M - \hat{\lambda}_F) = \frac{\hat{\lambda}_M}{26} + \frac{\hat{\lambda}_F}{26} = \frac{507/26}{26} + \frac{416/26}{26} = \frac{507+416}{676}\) | M1 | Correct variance formula |
| \(= \frac{923}{676} = 1.3653...\) | A1 | Correct value |
| \(99\%\) CI: \(3.5 \pm 2.5758\sqrt{\frac{923}{676}}\) | M1 | Using \(z = 2.5758\) |
| \(= 3.5 \pm 2.5758 \times 1.1686... = 3.5 \pm 3.010...\) | A1 | |
| \((0.490, 6.510)\) awrt \((0.49, 6.51)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Since the interval \((0.49, 6.51)\) does not contain \(0\) (both limits positive) | M1 | Reference to interval not containing 0 |
| There is evidence to support Emilia's belief that she receives more orders on Mondays than Fridays | A1 | Conclusion in context |
# Question 2:
## Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\hat{\lambda}_M = \frac{507}{26}$, $\hat{\lambda}_F = \frac{416}{26}$ | B1 | Both estimates stated or implied |
| $\hat{\lambda}_M - \hat{\lambda}_F = \frac{507-416}{26} = \frac{91}{26} = 3.5$ | B1 | Correct point estimate |
| $\text{Var}(\hat{\lambda}_M - \hat{\lambda}_F) = \frac{\hat{\lambda}_M}{26} + \frac{\hat{\lambda}_F}{26} = \frac{507/26}{26} + \frac{416/26}{26} = \frac{507+416}{676}$ | M1 | Correct variance formula |
| $= \frac{923}{676} = 1.3653...$ | A1 | Correct value |
| $99\%$ CI: $3.5 \pm 2.5758\sqrt{\frac{923}{676}}$ | M1 | Using $z = 2.5758$ |
| $= 3.5 \pm 2.5758 \times 1.1686... = 3.5 \pm 3.010...$ | A1 | |
| $(0.490, 6.510)$ awrt $(0.49, 6.51)$ | | |
## Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Since the interval $(0.49, 6.51)$ does not contain $0$ (both limits positive) | M1 | Reference to interval not containing 0 |
| There is evidence to support Emilia's belief that she receives more orders on Mondays than Fridays | A1 | Conclusion in context |
---2 Emilia runs an online perfume business from home. She believes that she receives more orders on Mondays than on Fridays.
She checked this during a period of 26 weeks and found that she received a total of 507 orders on the Mondays and a total of 416 orders on the Fridays.
The daily numbers of orders that Emilia receives may be modelled by independent Poisson distributions with means $\lambda _ { \mathrm { M } }$ for Mondays and $\lambda _ { \mathrm { F } }$ for Fridays.
\begin{enumerate}[label=(\alph*)]
\item Construct an approximate $99 \%$ confidence interval for $\lambda _ { \mathrm { M } } - \lambda _ { \mathrm { F } }$.
\item Hence comment on Emilia's belief.
\end{enumerate}
\hfill \mbox{\textit{AQA S3 2015 Q2 [8]}}