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14 March, 05:51

In a survey of 7200 T. V. viewers, 40% said they watch network news programs. Find the margin of error for this survey if we want 95% confidence in our estimate of the percent of T. V. viewers who watch network news programs.

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  1. 14 March, 08:40
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    Margin of Error=M. E = ± 0.0113

    Explanation:

    Margin of Error = M. E=?

    Probability that watched network news programs = p = 0.4

    α = 95%

    Margin of Error = M. E = zₐ/₂√p (1-p) / n

    Margin of Error=M. E = ±1.96 √0.4 (1-0.4) / 7200

    Margin of Error=M. E = ±1.96√0.24/7200

    Margin of Error=M. E - ±1.96 * 0.005773

    Margin of Error=M. E = ±0.0113

    The Margin of Error is the estimate of how much error is possible as a result of random sampling.
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