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:

$$\text{s.e.}\left(\hat{p}\right)=\sqrt{\frac{0.317059\times (1-0.317059)}{6882}}=\mathrm{0.005609244.}$$ 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: $\hat{p}=8/154=0.05194805$; $\text{s.e.}\left(\hat{p}\right)=0.0017833$; 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: $\text{s.e.}\left(\hat{p}\right)=0.01308604$; 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 $n=1/\left({0.05}^2\right)=400$. 2. Approximately $n=1/\left({0.025}^2\right)=1600$. 3. To halve the width of the interval, four times as many people are needed.

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

After 400 hours: $\hat{p}=0$; $\text{s.e.}\left(\hat{p}\right)=0$. The CI is from 0 to 0: clearly silly (implies no sampling variation). This is because the statistical validity conditions are not satisfied.