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 584 randomly selected Americans surveyed, 360 were in favor of the initiative. Round answers to 2 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 584 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, = 0.6164
sample size, n = 584
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.6164 * (1 - 0.6164)/584) = 0.0201

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 = (0.6164 - 1.96 * 0.0201 , 0.6164 + 1.96 * 0.0201)
CI = (0.577 , 0.6558)

With 95% confidence the proportion of all Americans who favor the new Green initiative is between 0.58 and 0.66

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.