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20 October, 04:49

8.3 Confidence Intervals for Proportion (Z-table)

Out of 300 people sampled, 27 had kids. Based on this, construct a 95% confidence interval for the true population

proportion of people with kids.

Give your answers as decimals, to three places

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  1. 20 October, 05:00
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    (0.058, 0.122)

    Step-by-step explanation:

    The confidence interval of a proportion is:

    CI = p ± SE * CV,

    where p is the proportion, SE is the standard error, and CV is the critical value (either a t-score or a z-score).

    We already know the proportion:

    p = 27/300

    p = 0.09

    But we need to find the standard error and the critical value.

    The standard error is:

    SE = √ (p (1 - p) / n)

    SE = √ (0.09 * (1 - 0.09) / 300)

    SE = 0.0165

    To find the critical value, we must first find the alpha level and the degrees of freedom.

    The alpha level for a 95% confidence interval is:

    α = (1 - 0.95) / 2 = 0.025

    The degrees of freedom is one less than the sample size:

    df = 300 - 1 = 299

    Since df > 30, we can approximate this with a normal distribution.

    If we look up the alpha level in a z score table, we find the z-score is 1.96. That's our critical value. CV = 1.96.

    Now we can find the confidence interval:

    CI = 0.09 ± 0.0165 * 1.96

    CI = 0.09 ± 0.0324

    CI = (0.058, 0.122)

    So we are 95% confident that the proportion of people with kids is between 0.058 and 0.122.
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