| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/2 (Pre-U Further Mathematics Paper 2) |
| Year | 2017 |
| Session | June |
| Marks | 8 |
| Topic | Confidence intervals |
| Type | Find minimum sample size |
| Difficulty | Standard +0.3 This is a standard confidence interval question for proportions with straightforward application of formulas. Part (i) requires direct substitution into the normal approximation formula, and part (ii) involves rearranging the width formula to solve for n—both are routine textbook exercises requiring no novel insight, making it slightly easier than average. |
| Spec | 5.05d Confidence intervals: using normal distribution |
Total: 6 marks
**Part (i):**
$\dfrac{32}{100} \pm 2.576\sqrt{\dfrac{32}{100} \times \dfrac{68}{100} \div 100}$ **M1** Correct form incl. 100. Allow $n/(n-1)$ for M1.
2.576 **A1**
$= \text{awrt } (0.200, 0.440)$ **A1** Answer, limits correct to $\geqslant 3$ sf. Condone wrong order.
**Part (ii):**
$2 \times 2.576 \times \sqrt{\dfrac{0.32 \times 0.68}{n}} = 0.04$ o.e. **M1** Correct equation, allow same wrong $z$ or $\sigma^2$ as in (i), or 2 omitted. Allow $n$-1 for M1.
Solve including squaring **M1**
$n_{\min} = 3610$ **A1** 3610 only
Total: **6 marks**3 In a random sample of 100 voters from a constituency, 32 said that they would support the Cyan Party.\\
(i) Find an approximate $99 \%$ confidence interval for the proportion of voters in the constituency who would support the Cyan Party.\\
(ii) Using the given sample proportion, estimate the smallest size of sample needed for the width of a $99 \%$ confidence interval to be less than 0.04 .
\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2017 Q3 [8]}}