Ask Question
15 September, 23:17

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.

+5
Answers (1)
  1. 16 September, 01:26
    0
    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
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “A university would like to estimate the proportion of fans who purchase concessions at the first basketball game of the season. The ...” in 📘 Mathematics if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers