Question

A newsgroup is interested in constructing a 99% confidence interval for the proportion of all Americans who are in favor of a new Green initiative. Of the 590 randomly selected Americans surveyed, 410 were in favor of the initiative. Round answers to 4 decimal places where possible.

a. With 99% confidence the proportion of all Americans who favor the new Green initiative is between and .

b. If many groups of 590 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, = 0.79
sample size, n = 519
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.79 * (1 - 0.79)/519) = 0.0179

Given CI level is 99%, hence α = 1 - 0.99 = 0.01
α/2 = 0.01/2 = 0.005, Zc = Z(α/2) = 2.58

Margin of Error, ME = zc * SE
ME = 2.58 * 0.0179
ME = 0.0462

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.79 - 2.58 * 0.0179 , 0.79 + 2.58 * 0.0179)
CI = (0.7438 , 0.8362)

b)

If many groups of 590 randomly selected Americans were surveyed, then a different confidence interval would be produced from each group. About 99 percent of these confidence intervals will contain the true population proportion of Americans who favor the Green initiative and about 1 percent will not contain the true population proportion.