| Exam Board | AQA |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2006 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Probability Definitions |
| Type | Basic probability calculation |
| Difficulty | Moderate -0.3 This is a straightforward application of the standard confidence interval formula for proportions using normal approximation. It requires only direct substitution into a formula (p̂ ± 1.96√(p̂(1-p̂)/n)) and a simple comparison to the claimed value. The interpretation in part (b) is routine. Slightly easier than average due to being a standard textbook procedure with no conceptual challenges. |
| Spec | 5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\hat{p} = \frac{209}{250} = 0.836\) | B1 | CAO |
| 95% CI \(\Rightarrow z = 1.96\) | B1 | CAO |
| \(\hat{p} \pm z\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\) | M1 | Variance term |
| Use of: \(\hat{p} \pm z \times \sqrt{\text{Var}(\hat{p})}\) | M1 | |
| \(0.836 \pm 1.96 \times \sqrt{\frac{0.836 \times 0.164}{250}}\) | A1\(\checkmark\) | \(\checkmark\) on \(\hat{p}\) and \(z\); not on \(n\) |
| \(0.836 \pm 0.046\) or \((0.790, 0.882)\) | A1 | AWRT; accept 0.79 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Value of 0.8 (80%) is within CI | B1\(\checkmark\) \(\uparrow\) dep | \(\checkmark\) on CI |
| Council's claim is supported (at 5% level) | B1\(\checkmark\) | \(\checkmark\) on CI |
# Question 1:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\hat{p} = \frac{209}{250} = 0.836$ | B1 | CAO |
| 95% CI $\Rightarrow z = 1.96$ | B1 | CAO |
| $\hat{p} \pm z\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$ | M1 | Variance term |
| Use of: $\hat{p} \pm z \times \sqrt{\text{Var}(\hat{p})}$ | M1 | |
| $0.836 \pm 1.96 \times \sqrt{\frac{0.836 \times 0.164}{250}}$ | A1$\checkmark$ | $\checkmark$ on $\hat{p}$ and $z$; not on $n$ |
| $0.836 \pm 0.046$ or $(0.790, 0.882)$ | A1 | AWRT; accept 0.79 |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Value of 0.8 (80%) **is within** CI | B1$\checkmark$ $\uparrow$ dep | $\checkmark$ on CI |
| Council's claim **is supported** (at 5% level) | B1$\checkmark$ | $\checkmark$ on CI |
---1 A council claims that 80 per cent of households are generally satisfied with the services it provides.
A random sample of 250 households shows that 209 are generally satisfied with the council's provision of services.
\begin{enumerate}[label=(\alph*)]
\item Construct an approximate $95 \%$ confidence interval for the proportion of households that are generally satisfied with the council's provision of services.
\item Hence comment on the council's claim.
\end{enumerate}
\hfill \mbox{\textit{AQA S3 2006 Q1 [8]}}