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

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