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8 October, 10:47

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.

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  1. 8 October, 11:08
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    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
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