| Exam Board | OCR |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2007 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Proportion confidence interval |
| Difficulty | Standard +0.3 This is a straightforward application of standard proportion confidence interval formulas from S3. Parts (i)-(iii) are routine calculations and bookwork definitions, while part (iv) requires rearranging the margin of error formula—a standard textbook exercise requiring no novel insight, making it slightly easier than average overall. |
| Spec | 5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(978/1200 = 0.815\) | B1 | 1 |
| (ii) Use \(\hat{p} \pm z_{\sqrt{\frac{p(1-p)}{1200}}}\) | M1 | Reasonable variance |
| \(z = 1.645\) | B1 | |
| \(\sqrt{(0.815 \times 0.185/1200)}\); \((0.797, 0.833)\) | A1√, A1 | ft \(\hat{p}\); Allow 1199; 4 Interval |
| (iii) If a large number of such samples were taken, \(p\) would be contained in about 90% of the confidence intervals. | B2 | 2 B1 if idea correct but badly expressed. |
| (iv) \(1.645\sqrt{(0.815 \times 0.185/n)} = 0.01\); \(n = 1.645^2(0.815 \times 0.185)/0.01^2 = 4080\) | M1, M1, A1, A1 | Allow one error; \(>\) or \(<\); Correct procedure for sim equ; 4 Integer rounding to 4100 |
**(i)** $978/1200 = 0.815$ | B1 | 1
**(ii)** Use $\hat{p} \pm z_{\sqrt{\frac{p(1-p)}{1200}}}$ | M1 | Reasonable variance
$z = 1.645$ | B1 |
$\sqrt{(0.815 \times 0.185/1200)}$; $(0.797, 0.833)$ | A1√, A1 | ft $\hat{p}$; Allow 1199; 4 Interval
**(iii)** If a large number of such samples were taken, $p$ would be contained in about 90% of the confidence intervals. | B2 | 2 B1 if idea correct but badly expressed.
**(iv)** $1.645\sqrt{(0.815 \times 0.185/n)} = 0.01$; $n = 1.645^2(0.815 \times 0.185)/0.01^2 = 4080$ | M1, M1, A1, A1 | Allow one error; $>$ or $<$; Correct procedure for sim equ; 4 Integer rounding to 41005 Each person in a random sample of 1200 people was asked whether he or she approved of certain proposals to reduce atmospheric pollution. It was found that 978 people approved. The proportion of people in the whole population who would approve is denoted by $p$.\\
(i) Write down an estimate $\hat { p }$ of $p$.\\
(ii) Find a 90\% confidence interval for $p$.\\
(iii) Explain, in the context of the question, the meaning of a $90 \%$ confidence interval.\\
(iv) Estimate the sample size that would give a value for $\hat { p }$ that differs from the value of $p$ by less than 0.01 with probability $90 \%$.
\hfill \mbox{\textit{OCR S3 2007 Q5 [11]}}