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

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