AQA S3 2012 June — Question 4 6 marks

Exam BoardAQA
ModuleS3 (Statistics 3)
Year2012
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicApproximating the Poisson to the Normal distribution
TypeHypothesis test for sum of Poisson observations
DifficultyStandard +0.8 This requires understanding that the sum of independent Poisson distributions is Poisson, then applying a normal approximation with continuity correction to perform a one-tailed hypothesis test. While the individual steps are S3-standard, combining these concepts (sum of Poissons, normal approximation, hypothesis testing framework) in a single question with minimal scaffolding makes it moderately challenging.
Spec2.04d Normal approximation to binomial5.05b Unbiased estimates: of population mean and variance

4 The manager of a medical centre suspects that patients using repeat prescriptions were requesting, on average, more items during 2011 than during 2010. The mean number of items on a repeat prescription during 2010 was 2.6.
An analysis of a random sample of 250 repeat prescriptions during 2011 showed a total of 688 items requested. The number of items requested on a repeat prescription may be modelled by a Poisson distribution. Use a distributional approximation to investigate, at the \(5 \%\) level of significance, the manager's suspicion.

Question 4 text reads:
- Mean number of items on repeat prescription in 2010 = 2.6
- Sample of 250 repeat prescriptions in 2011 showed 688 items total
- Items modelled by Poisson distribution
- Use distributional approximation to test at 5% significance level *(6 marks)*
To provide the mark scheme working for Question 4, the key steps would be:
AnswerMarks Guidance
- \(H_0: \lambda = 2.6\), \(H_1: \lambda > 2.6\)B1 One-tailed test
- \(\bar{x} = \frac{688}{250} = 2.752\)B1
- Under \(H_0\): \(X \sim \text{Po}(250 \times 2.6) = \text{Po}(650)\), approximate by \(N(650, 650)\)M1
- \(z = \frac{688 - 650}{\sqrt{650}} = \frac{38}{\sqrt{650}} \approx 1.491\)M1 A1
- \(1.491 < 1.645\), do not reject \(H_0\)A1 Insufficient evidence manager's suspicion correct
**Question 4** text reads:
- Mean number of items on repeat prescription in 2010 = 2.6
- Sample of 250 repeat prescriptions in 2011 showed 688 items total
- Items modelled by Poisson distribution
- Use distributional approximation to test at 5% significance level *(6 marks)*

To provide the mark scheme working for **Question 4**, the key steps would be:

- $H_0: \lambda = 2.6$, $H_1: \lambda > 2.6$ | B1 | One-tailed test
- $\bar{x} = \frac{688}{250} = 2.752$ | B1 |
- Under $H_0$: $X \sim \text{Po}(250 \times 2.6) = \text{Po}(650)$, approximate by $N(650, 650)$ | M1 |
- $z = \frac{688 - 650}{\sqrt{650}} = \frac{38}{\sqrt{650}} \approx 1.491$ | M1 A1 |
- $1.491 < 1.645$, do not reject $H_0$ | A1 | Insufficient evidence manager's suspicion correct
4 The manager of a medical centre suspects that patients using repeat prescriptions were requesting, on average, more items during 2011 than during 2010.

The mean number of items on a repeat prescription during 2010 was 2.6.\\
An analysis of a random sample of 250 repeat prescriptions during 2011 showed a total of 688 items requested.

The number of items requested on a repeat prescription may be modelled by a Poisson distribution.

Use a distributional approximation to investigate, at the $5 \%$ level of significance, the manager's suspicion.

\hfill \mbox{\textit{AQA S3 2012 Q4 [6]}}