D.19 Answers: CIs for one proportion

Answers to exercises in Sect. 20.11.

Answer to Exercise 20.1: $\hat{p} = 2182/6882 = 0.317059$ and $n=6882$ . So:

$\text{s.e.}(\hat{p}) = \sqrt{ \frac{0.317059 \times(1 - 0.317059)}{6882} } = 0.005609244.$ The CI is $0.317059 \pm (2\times 0.005609244)$ , or $0.317059\pm 0.01121849$ .

Rounding sensibly:

$0.317\pm 0.011$ (notice we keep lots of decimal places in the working, but round the final answer).

Answer to Exercise 20.2:

$\hat{p} = 8/154 = 0.05194805$ ;

$\text{s.e.}(\hat{p}) = 0.0017833$ ; approximate 95% CI is

$0.05194 \pm (2 \times 0.0017833)$ , or

$0.0519\pm 0.0358$ , equivalent to 0.016 to 0.088. The CI is statistically valid.

Answer to Exercise 20.3: Use

$\hat{p} = 708/864 = 0.8194444$ and

$n=864$ . Standard error:

$\text{s.e.}(\hat{p}) = 0.01308604$ ; approximate 95% CI is

$0.8194444 \pm (2\times 0.01308604)$ . The CI is statistically valid.

FIGURE D.6: The sampling distribution of the proportion of males in samples of 864 people with hiccups

Answer to Exercise 20.4: 1. Approximately

$n = 1/(0.05^2) = 400$ . 2. Approximately

$n = 1/(0.025^2) = 1600$ . 3. To halve the width of the interval, four times as many people are needed.

Answer to Exercise 20.5: After 3000 hours: $\hat{p} = 0.2143$ ; $\text{s.e.}(\hat{p}) = 0.06331$ . The CI is from 0.088 to 0.341. The statistical validity conditions are satisfied.

After 400 hours:

$\hat{p} = 0$ ;

$\text{s.e.}(\hat{p}) = 0$ . The CI is from 0 to 0: clearly silly (implies no sampling variation). This is because the statistical validity conditions are not satisfied.