A university would like to estimate the proportion of fans who purchase concessi

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, 500 and is routinely sold out. It was discovered that a total of 220 fans out of a random sample of 400 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.)

Explanation / Answer

Observed proportion ,p = 220/400=0.55

z for 95% confidence level is 1.96

sigma=sqrt(pq)=sqrt(0.55*0.45/n) = sqrt(0.55*0.45/400)= 0.02487

Hence, the 95% confidenc interval is (p-z*sigma,p+z*sigma) = (0.55-1.96*0.02487,0.55+1.96*0.02487) = (0.5012,0.5987)

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