| Exam Board | AQA |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2012 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Sample size determination |
| Difficulty | Standard +0.3 This is a standard two-part confidence interval question requiring routine application of normal approximation formulas. Part (a) involves calculating a confidence interval for a proportion using the standard formula with z=2.326. Part (b) requires rearranging the width formula to find sample size, which is a textbook exercise. Both parts are computational with no conceptual challenges beyond S3 syllabus expectations, making it slightly easier than average. |
| Spec | 5.05b Unbiased estimates: of population mean and variance5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\hat{p} = \frac{68}{125} = 0.544\) | B1 | Correct value of \(\hat{p}\) |
| \(z = 2.3263\) (for 98% CI) | B1 | Correct z-value |
| \(\hat{p} \pm z\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\) | M1 | Correct structure of CI formula |
| \(= 0.544 \pm 2.3263\sqrt{\frac{0.544 \times 0.456}{125}}\) | A1 | Correct substitution |
| \(= 0.544 \pm 2.3263 \times 0.04457...\) | ||
| \(= 0.544 \pm 0.10366...\) | A1 | Correct working |
| \((0.440, 0.648)\) or \((44.0\%, 64.8\%)\) | A1 | Correct interval (allow awrt) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Width \(\leq 0.10\), so half-width \(\leq 0.05\) | ||
| \(z\sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \leq 0.05\) | M1 | Correct inequality structure |
| Using \(p = 0.544\) (or \(p = 0.5\) for maximum): \(2.3263\sqrt{\frac{0.544 \times 0.456}{n}} \leq 0.05\) | M1 | Substituting values |
| \(n \geq \frac{2.3263^2 \times 0.544 \times 0.456}{0.05^2}\) | A1 | Correct rearrangement |
| \(n \geq \frac{5.4117 \times 0.248064}{0.0025} = 537.0...\) | ||
| \(n = 540\) (to nearest 5) | A1 | Must round up to nearest 5 |
# Question 5:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\hat{p} = \frac{68}{125} = 0.544$ | B1 | Correct value of $\hat{p}$ |
| $z = 2.3263$ (for 98% CI) | B1 | Correct z-value |
| $\hat{p} \pm z\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$ | M1 | Correct structure of CI formula |
| $= 0.544 \pm 2.3263\sqrt{\frac{0.544 \times 0.456}{125}}$ | A1 | Correct substitution |
| $= 0.544 \pm 2.3263 \times 0.04457...$ | | |
| $= 0.544 \pm 0.10366...$ | A1 | Correct working |
| $(0.440, 0.648)$ or $(44.0\%, 64.8\%)$ | A1 | Correct interval (allow awrt) |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Width $\leq 0.10$, so half-width $\leq 0.05$ | | |
| $z\sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \leq 0.05$ | M1 | Correct inequality structure |
| Using $p = 0.544$ (or $p = 0.5$ for maximum): $2.3263\sqrt{\frac{0.544 \times 0.456}{n}} \leq 0.05$ | M1 | Substituting values |
| $n \geq \frac{2.3263^2 \times 0.544 \times 0.456}{0.05^2}$ | A1 | Correct rearrangement |
| $n \geq \frac{5.4117 \times 0.248064}{0.0025} = 537.0...$ | | |
| $n = 540$ (to nearest 5) | A1 | Must round **up** to nearest 5 |
---5 A random sample of 125 people was selected from a council's electoral roll. Of these, 68 were in favour of a proposed local building plan.
\begin{enumerate}[label=(\alph*)]
\item Construct an approximate 98\% confidence interval for the percentage of people on the council's electoral roll who were in favour of the proposal.
\item Calculate, to the nearest 5, an estimate of the minimum sample size necessary in order that an approximate $98 \%$ confidence interval for the percentage of people on the council's electoral roll who were in favour of the proposal has a width of at most 10 per cent.
\end{enumerate}
\hfill \mbox{\textit{AQA S3 2012 Q5 [10]}}