Question

In a sample of 42 water specimens taken from a construction site, 26 contained detectable levels of lead. Construct 90% confidence interval for the proportion of water specimens that contain detectable levels of lead.

Answer 1

Answer:

(0.4958, 0.7422)

Step-by-step explanation:

Let p be the true proportion of water specimens that contain detectable levels of lead. The point estimate for p is $\hat{p}=26/42=0.6190$. The estimated standard deviation is given by $\sqrt{\hat{p}(1-\hat{p})/n}=\sqrt{(0.6190)(1-0.6190)/42}=0.0749$. Because we have a large sample, the 90% confidence interval for p is given by $0.6190\pm z_{0.05}0.0749$ where $z_{0.05}=1.6448$ is the value that satisfies that above this and under the standard normal density there is an area of 0.05. So, the confidence interval is $0.6190\pm (1.6448)(0.0749)$, i.e., (0.4958, 0.7422).