| Exam Board | WJEC |
|---|---|
| Module | Further Unit 5 (Further Unit 5) |
| Session | Specimen |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Confidence intervals |
| Type | Calculate CI for proportion |
| Difficulty | Standard +0.3 This is a straightforward confidence interval question testing standard formulas for proportions. Part (a) requires direct application of the normal approximation formula with given sample data. Part (b) involves working backwards from a confidence interval to find sample proportion and confidence level—routine algebraic manipulation of the same formulas. While it's a Further Maths statistics question with multiple parts worth 12 marks total, it requires no novel insight or complex problem-solving, just careful application of memorized procedures. |
| Spec | 5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| \(\hat{p} = \frac{1242}{1800} = 0.69\) | B1 | AO3 |
| \(\text{ESE} = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\) | M1 | AO1 |
| \(= \sqrt{\frac{0.69 \times 0.31}{1800}}\) | A1 | AO1 |
| \(= 0.0109(0.0107\ldots)\) | M1 | AO3 |
| 95% confidence limits are \(\hat{p} \pm z \times \text{ESE}\) | A1 | AO2 |
| \(0.69 \pm 1.96 \times 0.0109\) | A1 | AO1 |
| giving \([0.669, 0.711]\) | B1 | AO3 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\hat{p} = \frac{0.672 + 0.732}{2} = 0.702\) | B1 | AO3 |
| Number of people = \(0.702 \times 1000 = 702\) | B1 | AO1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(0.732 - 0.672 = 2z\sqrt{\frac{0.702 \times 0.298}{1000}}\) | M1 | AO3 |
| \(z = 2.07417\ldots\) | A1 | AO1 |
| Prob from tables = 0.98077 or 0.98097 from calc | A1 | AO1 |
| Confidence level = 96.2% | A1 | AO2 |
| [12] |
## 4(a)
$\hat{p} = \frac{1242}{1800} = 0.69$ | B1 | AO3 |
$\text{ESE} = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$ | M1 | AO1 |
$= \sqrt{\frac{0.69 \times 0.31}{1800}}$ | A1 | AO1 |
$= 0.0109(0.0107\ldots)$ | M1 | AO3 |
95% confidence limits are $\hat{p} \pm z \times \text{ESE}$ | A1 | AO2 |
$0.69 \pm 1.96 \times 0.0109$ | A1 | AO1 |
giving $[0.669, 0.711]$ | B1 | AO3 |
## 4(b)(i)
$\hat{p} = \frac{0.672 + 0.732}{2} = 0.702$ | B1 | AO3 |
Number of people = $0.702 \times 1000 = 702$ | B1 | AO1 |
## 4(b)(ii)
$0.732 - 0.672 = 2z\sqrt{\frac{0.702 \times 0.298}{1000}}$ | M1 | AO3 |
$z = 2.07417\ldots$ | A1 | AO1 |
Prob from tables = 0.98077 or 0.98097 from calc | A1 | AO1 |
Confidence level = 96.2% | A1 | AO2 |
| [12] | |\begin{enumerate}[label=(\alph*)]
\item In an opinion poll of 1800 people, 1242 said that they preferred red wine to white wine. Calculate a 95% confidence interval for the proportion of people in the population who prefer red wine to white wine. [6]
\item In another opinion poll of 1000 people on the same subject, the following confidence interval was calculated.
$[0.672, 0.732]$.
Determine
\begin{enumerate}[label=(\roman*)]
\item the number of people in the sample who stated that they prefer red wine to white wine,
\item the confidence level of the confidence interval, giving your answer as a percentage correct to three significant figures. [6]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 5 Q4 [12]}}