Question

A university would like to estimate the proportion of fans who purchase concessions at the first basketball game of the season. The basketball facility has a capacity of 3 comma 500 and is routinely sold out. It was discovered that a total of 290 fans out of a random sample of 500 purchased concessions during the game. Construct a​ 95% confidence interval to estimate the proportion of fans who purchased concessions during the game.

The​ 95% confidence interval to estimate the proportion of fans who purchased concessions during the game is (_, _)

​(Round to three decimal places as​ needed.)

Answer 1

sample proportion, = 0.58
sample size, n = 500
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.58 * (1 - 0.58)/500) = 0.0221

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 * 0.0221
ME = 0.0433

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.58 - 1.96 * 0.0221 , 0.58 + 1.96 * 0.0221)
CI = (0.537 , 0.623)