| Exam Board | OCR |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2013 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Confidence interval for single proportion |
| Difficulty | Standard +0.3 This is a standard confidence interval question for proportions with straightforward application of the normal approximation formula. Part (i) is routine calculation, part (ii) tests interpretation, and part (iii) requires algebraic manipulation to find sample size from interval width—all textbook exercises requiring no novel insight, though slightly above average due to the reverse-engineering in part (iii). |
| Spec | 2.04d Normal approximation to binomial5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(z = 1.645\) | B1 | |
| \(s^2 = 0.7 \times 0.3/50\) | B1 | or \(/49\) |
| Use \(0.7 \pm zs\) | M1 | |
| to give \((0.593,\ 0.807)\) | A1 | or \((0.592, 0.808)\) using \(/49\); or \(\%\) |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| If a large number of independent 90% confidence intervals were found for the proportion \(p\) of club members who would vote for the amendment then about 90% of the intervals would contain \(p\) | B1 | Allow with specific number of CIs. NOT *the probability of THE interval containing \(p\) is 0.9*. If candidate uses word 'probability', it must be clear they are considering a large number of intervals, not just one. |
| [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Use \(0.7 + 2.576\sqrt{(0.21/n)} < 0.85\) | M1*, A1 | Any \(z\) value. Allow \(\leq\) or \(=\), correct \(z\). Allow \(\geq\) |
| Solve to give \(n = 62\) | *M1A1 | |
| [4] |
# Question 5:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $z = 1.645$ | B1 | |
| $s^2 = 0.7 \times 0.3/50$ | B1 | or $/49$ |
| Use $0.7 \pm zs$ | M1 | |
| to give $(0.593,\ 0.807)$ | A1 | or $(0.592, 0.808)$ using $/49$; or $\%$ |
| | **[4]** | |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| If a large number of independent 90% confidence intervals were found for the proportion $p$ of club members who would vote for the amendment then about 90% of the intervals would contain $p$ | B1 | Allow with specific number of CIs. NOT *the probability of THE interval containing $p$ is 0.9*. If candidate uses word 'probability', it must be clear they are considering a large number of intervals, not just one. |
| | **[1]** | |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use $0.7 + 2.576\sqrt{(0.21/n)} < 0.85$ | M1*, A1 | Any $z$ value. Allow $\leq$ or $=$, correct $z$. Allow $\geq$ |
| Solve to give $n = 62$ | *M1A1 | |
| | **[4]** | |
---5 A constitutional change was proposed for a Golf Club with a large membership. This was to be voted on at the Annual General Meeting. A month before this meeting the secretary asked a random sample of 50 members for their opinions. Out of the 50 members $70 \%$ said they approved.\\
(i) Calculate an approximate $90 \%$ confidence interval for the proportion $p$ of all members who would approve the proposal.\\
(ii) Explain what is meant by a $90 \%$ confidence interval in this context.\\
(iii) Nearer the date of the meeting the secretary asked a random sample of $n$ members, and, as before, $70 \%$ said they approved. This time the secretary calculated an approximate $99 \%$ confidence interval for $p$. It is given that the confidence interval does not include 0.85 . Find the smallest possible value of $n$.
\hfill \mbox{\textit{OCR S3 2013 Q5 [9]}}