Question

A newsgroup is interested in constructing a 90% confidence interval for the proportion of all Americans who are in favor of a new Green initiative. Of the 536 randomly selected Americans surveyed, 407 were in favor of the initiative. Round answers to 4 decimal places where possible.

a. With 90% confidence the proportion of all Americans who favor the new Green initiative is between answer _____ and answer _____.

b. If many groups of 536 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, = 0.7593
sample size, n = 536
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.7593 * (1 - 0.7593)/536) = 0.018

Given CI level is 90%, hence α = 1 - 0.9 = 0.1
α/2 = 0.1/2 = 0.05, Zc = Z(α/2) = 1.64

Margin of Error, ME = zc * SE
ME = 1.64 * 0.018
ME = 0.03

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.7593 - 1.64 * 0.018 , 0.7593 + 1.64 * 0.018)
CI = (0.7298 , 0.7888)

With 90% confidence the proportion of all Americans who favor the new Green initiative is between 0.7298 and 0.7888

b)

If many groups of 536 randomly selected Americans were surveyed, then a different confidence interval would be produced from each group. About 90 percent of these confidence intervals will contain the true population proportion of Americans who favor the Green initiative and about 10 percent will not contain the true population proportion.