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 560 randomly selected Americans surveyed, 414 were in favor of the initiative.

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

b. If many groups of 560 randomly selected Americans were surveyed, then a different confidence interval would be produced from each group. About ___ percent of these confidence intervals will contain the true population proportion of Americans who favor the Green initiative and about ___ percent will not contain the true population proportion.

Answer 1

a)

sample proportion, = 0.7393
sample size, n = 560
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.7393 * (1 - 0.7393)/560) = 0.019

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

Margin of Error, ME = zc * SE
ME = 1.96 * 0.019
ME = 0.0372

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.7393 - 1.96 * 0.019 , 0.7393 + 1.96 * 0.019)
CI = (0.7021 , 0.7765)

With 95% confidence the proportion of all Americans who favor the new Green initiative is between 0.7021 and 0.7765

b)

If many groups of 560 randomly selected Americans were surveyed, then a different confidence interval would be produced from each group. 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.