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1 March, 22:13

What is the confidence level (written as a percentage) for the interval x overbar minus 2.14 sigma divided by startroot n endroot less-than-or-equal-to mu less-than-or-equal-to x overbar plus 2.14 sigma divided by startroot n endroot?

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  1. 1 March, 22:34
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    The interval notation is: m + / - z · (σ / √n)

    where:

    m = mean

    z = z-score

    σ = standard deviation

    n = sample

    In your case, you have: x + / - 2.14 · (σ / √n)

    Comparing the two formulae, you get z = + / - 2.14

    Therefore, you are looking for P (-2.14 < z < + 2.14) = P (z < 2.14) - P (z < - 2.14)

    Now, look at a z-table to find that

    P (z < 2.14) = 0.9838 and that

    P (z < - 2.14) = 0.0162

    Therefore:

    P (-2.14 < z < + 2.14) = 0.9838 - 0.0162 = 0.9676

    Hence, the confidence level written as a percentage is 96.76%
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