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 600 and is routinely sold out. It was discovered that a total of 210 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.

Step-by-step explanation:

Confidence interval is written as

Sample proportion ± margin of error

Margin of error = z × √pq/n

Where

z represents the z score corresponding to the confidence level

p = sample proportion. It also means probability of success

q = probability of failure

q = 1 - p

p = x/n

Where

n represents the number of samples

x represents the number of success

From the information given,

n = 500

x = 210

p = 210/500 = 0.42

q = 1 - 0.42 = 0.58

To determine the z score, we subtract the confidence level from 100% to get α

α = 1 - 0.95 = 0.05

α/2 = 0.05/2 = 0.025

This is the area in each tail. Since we want the area in the middle, it becomes

1 - 0.025 = 0.975

The z score corresponding to the area on the z table is 1.96. Thus, the z score for a confidence level of 95% is 1.96

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

0.42 ± 1.96√(0.42)(0.58)/500

= 0.42 ± 0.043

it's both, now can you answer my most recent question?