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Help nubcke Learn Statistics! :D


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#1 nubcke

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Posted 10 April 2011 - 02:40 PM

Okay, this is the first thing I have no idea how to do. Finding the confidence interval in the problem in the attached image. Help? D;

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    #2 luvsmyncis

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    Posted 10 April 2011 - 04:03 PM

    You need to find the confidence interval of your LIFE first.

    #3 Scot

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    Posted 10 April 2011 - 05:18 PM

    First you need to find p1 and p2.

    -Let Women who find female bosses more harsh(26%, 8525) be x1 and 32787 be n1 since it's half of the sample size.
    -Let Men who find female bosses more harsh(24%, 7869) be x2 and 32787 is n2.

    -p1 is given as x1/n1 and p2 is x2/n2.

    p1 = .26
    p2 = .24
    alpha = 1 - .95 = .05
    So alpha/2 = .025

    In general, a confidence interval is created using a value plus or minus the Z-score times the Standard Error. In the case of a proportion, the equation is

    (p1-p2) +/- Za/2 * SQRT((p1(1-p1)/n1 + p2(1-p2)/n2)

    Looking at the cumulative normal distribution chart, the z score for .025 is -1.96, if you don't have the chart, in excel do =NORMSINV(.025).

    The standard error is SQRT((.26*.74)/32787 + (.24*.76)/32787) = .00338

    So plug all that good shit in and the answer is

    (p1-p2) + Za/2 * SQRT((p1(1-p1)/n1 + p2(1-p2)/n2)
    .02 + (-1.96 )* .00338 = 0.0134

    (p1-p2) - Za/2 * SQRT((p1(1-p1)/n1 + p2(1-p2)/n2)
    .02 - (-1.96) * .00338 = 0.0266


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