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.