Confidence Intervals on Proportions

Proportions = Probability of “success” in Bernoulli

We want to estimate P

$$\hat{P}=\frac{\text{number}\text{of}\text{"successes"}}{\text{number}\text{of}\text{trials}}=\frac{N_n}{n}$$

Confidence Interval on P

$$[L,U{]}_{1-\alpha }=(\hat{P})\pm (Z_{\frac{\alpha }{2}})({\sigma }_{\hat{p}})$$

This is a two-sided confidence interval

Where ${\sigma }_{\hat{P}}^2=Var(\hat{P})=Var(\frac{N_n}{n})=\frac{1}{n^2}Var(N_n)=\frac{npq}{n^2}=\frac{pq}{n}\approx \frac{\hat{p}\hat{q}}{n}$

(Note: In hypothesis tests, we use the assumed p (under H_o))

$$(L,U{)}_{1-\alpha }=(\hat{P})\pm (Z_{\frac{\alpha }{2}})\sqrt{\frac{\hat{p}\hat{q}}{n}}$$ $$\hat{q}=1-\hat{P}$$

Note, if this interval includes negative numbers, or numbers above 1.0, then our approximation is bad.

Rule of thumb, if $n\hat{p}\ge 5$ or $n\hat{q}\ge 5$, you’ve got a problem. Your data will be centred too close to zero, and there will be negative numbers present.

---