| Exam Board | OCR |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2006 |
| Session | January |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Confidence intervals |
| Type | Calculate CI for proportion |
| Difficulty | Moderate -0.3 This is a straightforward application of the standard confidence interval formula for a proportion with a simple interpretation. Part (i) requires routine calculation using p̂ ± 1.96√(p̂(1-p̂)/n), and part (ii) asks only whether 0.5 lies in the interval—both are textbook exercises requiring no problem-solving insight or novel reasoning. |
| Spec | 5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(p_s \pm z\sigma_{est}\) | M1 | Use formula, \(\sigma\) involving \(p_s\) and |
| \(p_s=186/400(0.465)\) | A1 | |
| \(\sigma_{est}=\sqrt{\frac{0.465 \times 0.535}{400}}\) | B1 | |
| \(z=1.96\) | A1 | |
| \((0.416,0.514)\) | A1 | 5 |
| (ii) Councillor statement implies \(\rho=0.5\). CI does contain 0.5 but only just so councillor probably correct assertive | B1 | 1 |
(i) $p_s \pm z\sigma_{est}$ | M1 | Use formula, $\sigma$ involving $p_s$ and
$p_s=186/400(0.465)$ | A1 |
$\sigma_{est}=\sqrt{\frac{0.465 \times 0.535}{400}}$ | B1 |
$z=1.96$ | A1 |
$(0.416,0.514)$ | A1 | 5
(ii) Councillor statement implies $\rho=0.5$. CI does contain 0.5 but only just so councillor probably correct assertive | B1 | 1 | Any justifiable comment *Not too assertive*
---1 In order to judge the support for a new method of collecting household waste, a city council arranged a survey of 400 householders selected at random. The results showed that 186 householders were in favour of the new method.\\
(i) Calculate a 95\% confidence interval for the proportion of all householders who are in favour of the new method.
A city councillor said he believed that as many householders were in favour of the new method as were against it.\\
(ii) Comment on the councillor's belief.
\hfill \mbox{\textit{OCR S3 2006 Q1 [6]}}