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17 April, 15:10

a random sample of 510 high school students has a normal distribution. The sample mean average ACT exam score was 21 with a 3.2 standard deviation. Construct a 99% confidence interval estimate of the population mean average ACT exam. Find the critical value

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  1. 17 April, 16:10
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    CI = 21 ± 0.365

    Step-by-step explanation:

    The confidence interval is:

    CI = x ± SE * CV

    where x is the sample mean, SE is the standard error, and CV is the critical value (either t score or z score).

    Here, x = 21.

    The standard error for a sample mean is:

    SE = σ / √n

    SE = 3.2 / √510

    SE = 0.142

    The critical value is looked up in a table or found with a calculator. But first, we must find the alpha level and the critical probability.

    α = 1 - 0.99 = 0.01

    p * = 1 - (α/2) = 1 - (0.01/2) = 0.995

    Using a calculator or a z-score table:

    P (x
    z = 2.576

    Therefore:

    CI = 21 ± 0.142 * 2.576

    CI = 21 ± 0.365

    Round as needed.
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