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 3500 and is routinely sold out. It was discovered that a total of 300 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

Answer 1

Solution:

Given,

n = 500 ....... Sample size

x = 300 .......no. of successes in the sample

Let denotes the sample proportion.

     = x/n = 300/500 = 0.6

Our aim is to construct 95% confidence interval.

c = 0.95

= 1- c = 1- 0.95 = 0.05

  /2 = 0.025 and 1- /2 = 0.975

Search the probability 0.975 in the Z table and see corresponding z value

= 1.96

Now , the margin of error is given by

E = /2 *  

= 1.96 * [0.6 *(1 - 0.6)/500]

= 0.043

Now the confidence interval is given by

( - E)   ( + E)

(0.6 - 0.043)   (0.6 + 0.043)

0.557 0.643

i.e.

(0.557 , 0.643)

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

0.557 0.643

i.e.

(0.557 , 0.643)