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5 February, 12:51

How many women must be randomly selected to estimate the mean weight of women in one age group. we want 90% confidence that the sample mean is within 2.7 lb of the population mean, and the population standard deviation is known to be 22 lb?

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  1. 5 February, 13:20
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    That is a very good question ...

    Look Grank

    you know that:

    confidence interval = mean + or - Margin of Error

    and

    Margin of Error = (z) * (standard deviation) / (sqrt of n)

    where n is the number of sample records

    So we need to calculate z-value firstly

    It says: " we want 90% confidence"

    So that:

    confidence90% corresponds to z-value of 1.645

    Plug that into our formula of Margin of Error:

    Margin of Error = (1.645) * (22) / (sqrt of n)

    "The sample mean is within 2.7 lb of the population mean" means that Margin of Error is 2.7

    Margin of Error:

    2.7 = (1.645) * (22) / (sqrt of n)

    Now solve for n:

    n=179.66~180

    SO that 180 women must be randomly selected to estimate the mean weight of women in one age group.
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