Answer:
A point estimate for the true population proportion is [tex]\hat{p} = 0.83[/tex] and the 90% confidence interval is (0.8025, 0.8575)
Step-by-step explanation:
We have a large sample of people of size n = 506. Let X be the random variable that represents the number of adults who feel that education is one of the top issues facing California. We are interested in the unknown true proportion p of adults who feel that education is one of the top issues facing California. A point estimate for the true population proportion is given by [tex]\hat{p}[/tex] = X/n = 418/506 = 0.83, the standard deviation is given by [tex]\sigma_{\hat{p}} = \sqrt{p(1-p)/n}\approx\sqrt{\hat{p}(1-\hat{p})/n} = \sqrt{(0.83)(0.17)/506} = 0.0167[/tex]. Therefore, the 90% confidence interval for the true population proportion is [tex]\hat{p}\pm z_{0.1/2}\sqrt{\hat{p}(1-\hat{p})/n}[/tex], i.e., [tex]0.83\pm z_{0.05}0.0167[/tex], i.e., [tex]0.83\pm (1.6448)(0.0167)[/tex], (0.8025, 0.8575).
