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 230 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 (____,___)

Answer 1

Solution :

Given that,

Point estimate = sample proportion = = x / n = 230 / 500 = 0.46

Z_/2 = 1.96

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

= 1.96 * (((0.46 * 0.54) / 500)

= 0.044

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.46 - 0.044 < p < 0.46 + 0.044

0.416 < p < 0.504

(0.416 , 0.504)