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 200 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.)

Answer 1

Solution :

Given that,

n = 400

x = 200

Point estimate = sample proportion = = x / n = 200 / 400 = 0.500

1 - = 1 - 0.500 = 0.500

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 1.96 * (((0.500 * 0.500) / 400)

= 0.049

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.500 - 0.049 < p < 0.500 + 0.049

0.451 < p < 0.549

The 95% confidence interval for the population proportion p is : (0.451 , 0.549)