| Exam Board | OCR |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2007 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Z-tests (known variance) |
| Type | Confidence interval for proportion |
| Difficulty | Standard +0.3 This is a standard two-proportion hypothesis test with a confidence interval calculation. It requires routine application of normal approximation formulas for proportions and a straightforward two-sample z-test. The question is slightly above average difficulty due to the multi-part structure and the conceptual question about approximation, but all techniques are standard S3 material with no novel problem-solving required. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\hat{p} = 62/200 = 0.31\) | B1 | aef |
| Use \(\hat{p}_a \pm z\sqrt{\frac{\hat{p}_a(1-\hat{p}_a)}{200}}\) | M1 | With 200 or 199 |
| \(z = 1.96\). Correct variance estimate \((0.2459, 0.3741)\) | B1 | Seen |
| A1√ | 5 marks; ft \(\hat{p}\); art (0.246, 0.374) | |
| (ii) EITHER: Sample proportion has an approximate normal distribution OR: Variance is an estimate | B1 | 1 mark; Not \(\hat{p}\) is an estimate, unless variance mentioned |
| (iii) \(H_0: p_a = p_a\), \(H_1: p_a \neq p_a\); \(\hat{p} = (62+35)/(200+150)\) | B1 | aef |
| EITHER: \(z = (\pm)\frac{62/200-35/150}{\sqrt{\hat{p}q(200^{-1}+150^{-1})}}\) | M1 | \(s^2\) with, \(\hat{p}\), 200, 150 (or 199, 149) |
| B1√ | Evidence of correct variance estimate. Ft \(\hat{p}\) | |
| \(= 1.586\) | A1 | |
| \((-1.96 <) 1.586 (< 1.96)\) | M1 | Correct comparison with ± 1.96 |
| Do not reject \(H_0\): there is insufficient evidence of a difference in proportions. | A1 | |
| OR: \(p_a - p_a = zs\); \(s = \sqrt{(0.277 \times 0.723)(200^{-1}+150^{-1}))\) | M1 | |
| B1√ | Ft \(\hat{p}\) | |
| CV of \(p_a - p_a = 0.0948\) or \(0.095\) | A1 | |
| Compare \(p_a - p_a = 0.0767\) with their 0.0948 | M1 | |
| Do not reject \(H_0\) and accept that there is insufficient evidence of a difference in proportions | A1 | |
| 6 marks total; Conditional on z=1.96 |
(i) $\hat{p} = 62/200 = 0.31$ | B1 | aef
Use $\hat{p}_a \pm z\sqrt{\frac{\hat{p}_a(1-\hat{p}_a)}{200}}$ | M1 | With 200 or 199
$z = 1.96$. Correct variance estimate $(0.2459, 0.3741)$ | B1 | Seen
| A1√ | 5 marks; ft $\hat{p}$; art (0.246, 0.374)
(ii) EITHER: Sample proportion has an approximate normal distribution OR: Variance is an estimate | B1 | 1 mark; Not $\hat{p}$ is an estimate, unless variance mentioned
(iii) $H_0: p_a = p_a$, $H_1: p_a \neq p_a$; $\hat{p} = (62+35)/(200+150)$ | B1 | aef
EITHER: $z = (\pm)\frac{62/200-35/150}{\sqrt{\hat{p}q(200^{-1}+150^{-1})}}$ | M1 | $s^2$ with, $\hat{p}$, 200, 150 (or 199, 149)
| B1√ | Evidence of correct variance estimate. Ft $\hat{p}$
$= 1.586$ | A1 |
$(-1.96 <) 1.586 (< 1.96)$ | M1 | Correct comparison with ± 1.96
Do not reject $H_0$: there is insufficient evidence of a difference in proportions. | A1 |
OR: $p_a - p_a = zs$; $s = \sqrt{(0.277 \times 0.723)(200^{-1}+150^{-1}))$ | M1 |
| B1√ | Ft $\hat{p}$
CV of $p_a - p_a = 0.0948$ or $0.095$ | A1 |
Compare $p_a - p_a = 0.0767$ with their 0.0948 | M1 |
Do not reject $H_0$ and accept that there is insufficient evidence of a difference in proportions | A1 |
| | 6 marks total; Conditional on z=1.966 Random samples of 200 'Alpha' and 150 'Beta' vacuum cleaners were monitored for reliability. It was found that 62 Alpha and 35 Beta cleaners required repair during the guarantee period of one year. The proportions of all Alpha and Beta cleaners that require repair during the guarantee period are $p _ { \alpha }$ and $p _ { \beta }$ respectively.\\
(i) Find a $95 \%$ confidence interval for $p _ { \alpha }$.\\
(ii) Give a reason why, apart from rounding, the interval is approximate.\\
(iii) Test, at the $5 \%$ significance level, whether $p _ { \alpha }$ differs from $p _ { \beta }$.
\hfill \mbox{\textit{OCR S3 2007 Q6 [12]}}