# Question 4:

## Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\hat{p} = \frac{940}{2000} = 0.47$ | B1 | |
| $ESE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} = \sqrt{\frac{0.47 \times 0.53}{2000}} = 0.01116\ldots$ | M1, A1 | FT their $\hat{p}$ for M1A1 |
| 95% confidence limits are $\hat{p} \pm z \times ESE$ | M1 | FT their $\hat{p}$ and ESE for M1A1 |
| $0.47 \pm 1.96 \times 0.01116\ldots$ | A1 | |
| Giving $[0.448, 0.492]$ | A1 | cao |

## Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Two valid reasons e.g. We have used an approximate value for $p$ (in calculating the standard error); The binomial distribution has been approximated by the normal distribution; No continuity correction has been used | E2 | E1 for one reason |

## Part (c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $2.5758\ldots \times \sqrt{\frac{0.53 \times 0.47}{n}} \leq 0.02$ | M1 | Full FT their $\hat{p}$; Attempt at equation or inequality with 2.5758, $n$ and 0.02 |
| $\frac{2.5758\ldots \times \sqrt{0.53 \times 0.47}}{0.02} \leq \sqrt{n}$ | A1 | Correct equation or inequality; 4132.4295… from $Z_{0.995} = 2.576$ |
| $n \geq 4131.88\ldots$ | A1 | 2133 from $Z_{0.995} = 2.576$ |
| Therefore, an additional $(4132 - 2000 =)\ 2132$ people | A1 | |