Question

A newsgroup is interested in constructing a 95% confidence interval for the proportion of all Americans who are in favor of a new Green initiative. Of the 549 randomly selected Americans surveyed, 361 were in favor of the initiative. Round answers to 4 decimal places where possible.

a. With 95% confidence the proportion of all Americans who favor the new Green initiative is between answer ________ and answer _______.

b. If many groups of 549 randomly selected Americans were surveyed, then a different confidence interval would be produced from each group. About answer _____ percent of these confidence intervals will contain the true population proportion of Americans who favor the Green initiative and about answer _______ percent will not contain the true population proportion.

Answer 1

a)
sample proportion, pcap = 361/549 = 0.6576
sample size, n = 549
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.6576 * (1 - 0.6576)/549) = 0.0203

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, Zc = Z(α/2) = 1.96

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.6576 - 1.96 * 0.0203 , 0.6576 + 1.96 * 0.0203)
CI = (0.6178 , 0.6974)

With 95% confidence the proportion of all Americans who favor the new Green initiative is between 0.6178 and 0.6974

b)
About 95 percent of these confidence intervals will contain the true population proportion of Americans who favor the Green initiative and about 5 percent will not contain the true population proportion.