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12 April, 20:43

We have the survey data on the body mass index (BMI) of 664 young women. The mean BMI in the sample was. We treated these data as an SRS from a Normally distributed population with standard deviation 7.4. Give confidence intervals for the mean BMI and the margins of error for 90%, 95%, and 99% confidence. Interval (0.01) ? margins of error (0.0001) ?

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  1. 12 April, 21:21
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    Step-by-step explanation:

    Margin of error = critical value * standard error

    The standard error is:

    s = σ / √n

    s = 7.4 / √664

    s = 0.287

    The population is normally distributed, so the critical value can be found from a z-score table.

    For 90% confidence, z = 1.645.

    For 95% confidence, z = 1.960.

    For 99% confidence, z = 2.576.

    The margins of error are:

    90%: 1.645 * 0.287 = 0.4724

    95%: 1.960 * 0.287 = 0.5629

    99%: 2.576 * 0.287 = 0.7398

    The mean is 26.6, so the confidence intervals are:

    90%: (26.6 - 0.47, 26.6 + 0.47) = (26.13, 27.07)

    95%: (26.6 - 0.56, 26.6 + 0.56) = (26.04, 27.16)

    99%: (26.6 - 0.74, 26.6 + 0.74) = (25.86, 27.34)
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