| Exam Board | AQA |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2014 |
| Session | June |
| Marks | 2 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Confidence interval for single proportion |
| Difficulty | Moderate -0.8 This is a straightforward application of the normal approximation to the binomial for a confidence interval. Students need to calculate p-hat (23/200), find the critical value for 96% (z=2.054), apply the standard formula, and compare the interval to 4% (2 in 50). All steps are routine with no problem-solving or novel insight required, making it easier than average. |
| Spec | 5.05a Sample mean distribution: central limit theorem5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\hat{p} = \frac{23}{200} = 0.115\) | B1 | Correct sample proportion stated or used |
| \(z = 2.0537...\) (for 96% CI) | B1 | Correct \(z\) value (accept 2.054) |
| \(0.115 \pm 2.054\sqrt{\frac{0.115 \times 0.885}{200}}\) | M1 | Correct structure of CI formula with their \(\hat{p}\) and \(z\) |
| \(= 0.115 \pm 2.054 \times 0.02256...\) | A1 | Correct calculation of SE |
| \((0.0687, 0.1613)\) | A1 | Correct final interval (accept equivalent forms) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Supplier claims \(p \leq \frac{2}{50} = 0.04\) | B1 | Correct identification of claimed value 0.04 |
| Since \(0.04\) is below/outside the confidence interval \((0.0687, 0.1613)\), there is evidence to doubt/reject the supplier's claim | B1 | Correct conclusion with reference to CI and 0.04 |
# Question 1:
## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\hat{p} = \frac{23}{200} = 0.115$ | B1 | Correct sample proportion stated or used |
| $z = 2.0537...$ (for 96% CI) | B1 | Correct $z$ value (accept 2.054) |
| $0.115 \pm 2.054\sqrt{\frac{0.115 \times 0.885}{200}}$ | M1 | Correct structure of CI formula with their $\hat{p}$ and $z$ |
| $= 0.115 \pm 2.054 \times 0.02256...$ | A1 | Correct calculation of SE |
| $(0.0687, 0.1613)$ | A1 | Correct final interval (accept equivalent forms) |
## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| Supplier claims $p \leq \frac{2}{50} = 0.04$ | B1 | Correct identification of claimed value 0.04 |
| Since $0.04$ is below/outside the confidence interval $(0.0687, 0.1613)$, there is evidence to doubt/reject the supplier's claim | B1 | Correct conclusion with reference to CI and 0.04 |
---1 A hotel's management is concerned about the quality of the free pens that it provides in its meeting rooms.
The hotel's assistant manager tests a random sample of 200 such pens and finds that 23 of them fail to write immediately.
\begin{enumerate}[label=(\alph*)]
\item Calculate an approximate $\mathbf { 9 6 \% }$ confidence interval for the proportion of pens that fail to write immediately.
\item The supplier of the pens to the hotel claims that at most 2 in 50 pens fail to write immediately.
Comment, with numerical justification, on the supplier's claim.\\[0pt]
[2 marks]
QUESTION\\
PART Answer space for question 1
\end{enumerate}
\hfill \mbox{\textit{AQA S3 2014 Q1 [2]}}