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Math Help - To obtain an estimate for the variance of the population mean

  1. #1
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    To obtain an estimate for the variance of the population mean

    To obtain an estimate for the variance of the population mean,
    sigma^2, a natural idea would be to calculate

    1/n(sum i=1 to n)(xi - (x bar))^2
    .
    It turns
    out that this estimate has a mean of (n-1/n) sigma^2. This can be corrected
    by multiplying 1/n(sum i=1 to n)(xi - (x bar))^2

    by a factor so that the estimate is
    unbiased.
    What is the factor?
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  2. #2
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    I'm not sure what "factor" you are looking for. The estimator that you have for the sample variance is indeed unbiased, demonstrated by the following:

    Let V be a chi-square random variable. It can easily be shown that V=\frac{(n-1)}{\sigma^2}S^2\implies S^2=\frac{\sigma^2}{(n-1)}V.

    Then it follows that:
    E(S^2)=E\left(\frac{\sigma^2}{(n-1)}V\right)=\frac{\sigma^2}{(n-1)}E(V)=\frac{\sigma^2}{(n-1)}(n-1)=\sigma^2.

    Since E(S^2)=\sigma^2, it is an unbiased estimator of \sigma^2.
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by yus2k View Post
    To obtain an estimate for the variance of the population mean,
    sigma^2, a natural idea would be to calculate

    1/n(sum i=1 to n)(xi - (x bar))^2
    .
    It turns
    out that this estimate has a mean of (n-1/n) sigma^2. This can be corrected
    by multiplying 1/n(sum i=1 to n)(xi - (x bar))^2

    by a factor so that the estimate is
    unbiased.
    What is the factor?
    You seem to be talking about the population variance not the sample mean's variance.

    In which case the question is asking you:

    What do you need to multiply:

    \left(n+\frac{1}{n}\right) \sigma^2

    by to obtain \sigma^2? An unbiased estimator has expectation equal to what you are trying to estimate.

    CB
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  4. #4
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    Quote Originally Posted by CaptainBlack View Post
    You seem to be talking about the population variance not the sample mean's variance.

    In which case the question is asking you:

    What do you need to multiply:

    \left(n+\frac{1}{n}\right) \sigma^2

    by to obtain \sigma^2? An unbiased estimator has expectation equal to what you are trying to estimate.

    CB
    yer i got the factor as sumthing similar
    but i had n/n-1
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