| Exam Board | AQA |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2009 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Confidence interval for difference of proportions |
| Difficulty | Standard +0.3 This is a standard confidence interval question for difference of proportions using normal approximation. It requires straightforward application of the formula with given sample data and interpretation, but involves more calculation steps than basic inference questions. Slightly above average difficulty due to the two-sample nature and need for careful arithmetic, but remains a textbook exercise with no novel problem-solving required. |
| Spec | 5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(\hat{p}_1 = \frac{102}{150} = 0.68\) and \(\hat{p}_2 = \frac{36}{80} = 0.45\) | B1 | Both CAO |
| \(99\% \Rightarrow z = 2.57\) to \(2.58\) | B1 | AWFW (2.5758) |
| CI for \((p_1 - p_2)\): \((\hat{p}_1 - \hat{p}_2) \pm z \times \sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \frac{\hat{p}_2(1-\hat{p}_2)}{n_2}}\) | M1, m1 | Use of \((\hat{p}_1 - \hat{p}_2) \pm z \times \sqrt{\text{attempted variance}}\); Use of correct expression for variance |
| \((0.68 - 0.45) \pm 2.5758 \times \sqrt{\frac{0.68 \times 0.32}{150} + \frac{0.45 \times 0.55}{80}}\) | A1F | F on \(\hat{p}_1\), \(\hat{p}_2\) and \(z\) |
| \(0.23 \pm (0.173 \text{ to } 0.174)\) or \((0.056 \text{ to } 0.057,\ 0.403 \text{ to } 0.404)\) | A1 | CAO & AWFW (accept 0.17); AWFW (accept 0.06 & 0.4). Note: Pooling of variances maximum B1 B1 M1 |
| Total | 6 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| Whole of confidence interval is above 0 or zero | B1F | F on (a); Or equivalent |
| Disagree with claim / claim appears doubtful | B1F | F on (a); Or equivalent; Dependent on previous B1F |
| Total | 2 |
# Question 1:
## Part (a):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $\hat{p}_1 = \frac{102}{150} = 0.68$ and $\hat{p}_2 = \frac{36}{80} = 0.45$ | B1 | Both CAO |
| $99\% \Rightarrow z = 2.57$ to $2.58$ | B1 | AWFW (2.5758) |
| CI for $(p_1 - p_2)$: $(\hat{p}_1 - \hat{p}_2) \pm z \times \sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \frac{\hat{p}_2(1-\hat{p}_2)}{n_2}}$ | M1, m1 | Use of $(\hat{p}_1 - \hat{p}_2) \pm z \times \sqrt{\text{attempted variance}}$; Use of correct expression for variance |
| $(0.68 - 0.45) \pm 2.5758 \times \sqrt{\frac{0.68 \times 0.32}{150} + \frac{0.45 \times 0.55}{80}}$ | A1F | F on $\hat{p}_1$, $\hat{p}_2$ and $z$ |
| $0.23 \pm (0.173 \text{ to } 0.174)$ or $(0.056 \text{ to } 0.057,\ 0.403 \text{ to } 0.404)$ | A1 | CAO & AWFW (accept 0.17); AWFW (accept 0.06 & 0.4). Note: Pooling of variances maximum B1 B1 M1 |
| **Total** | **6** | |
## Part (b):
| Working/Answer | Marks | Guidance |
|---|---|---|
| Whole of confidence interval is above 0 or zero | B1F | F on (a); Or equivalent |
| **Disagree** with claim / claim appears doubtful | B1F | F on (a); Or equivalent; Dependent on previous B1F |
| **Total** | **2** | |
---1 An analysis of a random sample of 150 urban dwellings for sale showed that 102 are semi-detached.
An analysis of an independent random sample of 80 rural dwellings for sale showed that 36 are semi-detached.
\begin{enumerate}[label=(\alph*)]
\item Construct an approximate $99 \%$ confidence interval for the difference between the proportion of urban dwellings for sale that are semi-detached and the proportion of rural dwellings for sale that are semi-detached.
\item Hence comment on the claim that there is no difference between these two proportions.
\end{enumerate}
\hfill \mbox{\textit{AQA S3 2009 Q1 [8]}}