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17 December, 14:39

Editors preparing a report on the economy are trying to estimate the percentage of businesses that plan to hire additional employees in the next 60 days. They are willing to accept a margin of error of 6 % but want 99 % confidence. How many randomly selected employers will they need to contact?

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  1. 17 December, 17:25
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    They will need to contact 462 people

    Explanation:

    In this question, we are asked to calculate the number of employers, randomly selected will the editors need to contact.

    We proceed as follows;

    we note that the confidence level is 99%

    For (1-α) = 0.99

    α=0.01

    α / 2 = 0.005



    From the Standard Normal Table, the required z=0.005 for 99% confidence level is 2.58

    We now proceed to calculate the sample size

    Mathematically, we can obtain the sample size as follows;

    p = 0.5 and q = 1-p = 1-0.5 = 0.5

    n = (z^2 * p * q) / (mE) ^2 where mE is the margin of error

    n = [ (2.58) ^2 * 0,5 * 0.5] / (0.06) ^2

    n = 462
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