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Math Help - Estimate of the population mean

  1. #1
    Newbie
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    Estimate of the population mean

    Hello,everybody! I'll be glad if you help me!

    Using the following data
    30, 40, 50, 20, 20, 40.
    construct 95% confidence interval estimate of the popolation mean.

    What is the best formula for solving this problem?
    E=t * S/sqrN (1)
    or 1/2

    E=Z * O/sqrN (2)
    1/2
    WHY?
    Thanks for your time.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by busyboy View Post
    Hello,everybody! I'll be glad if you help me!

    Using the following data
    30, 40, 50, 20, 20, 40.
    construct 95% confidence interval estimate of the popolation mean.
    The mean is m = (30+40+50+20+20+40)/6 = 33.33

    Now if the distribution of the data is normal the sample mean has a
    t-distribution with 5 degrees of freedom. But to use the t-distribution
    we need the population standard deviation, which we will estimate with:

    s = sqrt[1/(n-1) sum (x_1 - m)^2] = 12.11

    Then the interval is:

    (m - cv*s/sqrt(n), m + cv*s/sqrt(n))

    where cv is the critical value for a 95% interval for a t-distribution with n-1
    degrees of freedom.

    The critical t-value for a 95% interval with 5 degrees of freedom is 2.57,
    so the 95% interval would then be:

    (33.33-2.57*12.11/sqrt(6), 33.33+2.57*12.11/sqrt(6))~= (20.62, 46.04)

    We use the t-distribution because we have a sample size n=6, which is
    a small sample so the use of the normal distribution is inappropriate here
    as that requires large samples.

    Note the use of the t-distribution is also inappropriate unless we have some
    reason to believe the underlying distribution that the data is sampled from
    is normal.

    RonL
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