Question

According to a poll, 83% of California adults (418 out of 506 surveyed) feel that education is one of the top issues facing California. We wish to construct a 90% confidence interval for the true proportion of California adults who feel that education is one of the top issues facing California.
What is a point estimate for the true population proportion?

Answer 1

Answer:

A point estimate for the true population proportion is $\hat{p} = 0.83$ 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 $\hat{p}$ = X/n = 418/506 = 0.83, the standard deviation is given by $\sigma_{\hat{p}} = \sqrt{p(1-p)/n}\approx\sqrt{\hat{p}(1-\hat{p})/n} = \sqrt{(0.83)(0.17)/506} = 0.0167$. Therefore, the 90% confidence interval for the true population proportion is $\hat{p}\pm z_{0.1/2}\sqrt{\hat{p}(1-\hat{p})/n}$, i.e., $0.83\pm z_{0.05}0.0167$, i.e., $0.83\pm (1.6448)(0.0167)$, (0.8025, 0.8575).