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Thread: Distribution of variance

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    Distribution of variance




    Hello,

    I need help on a particular issue:
    I have performed a number of replicate experiments to determine the variance of my setup. None of the tests had significantly different variances, and my data points are normally distributed. The means were occasionally different from the rest.
    Now I want to make a probability distribution of the variance in this setup. Hence, in what area will 95% of my measured variances lie.
    I understand that variance follows a chi squared distribution, but I can't figure out how to pool my standard deviations (without combining all the data directly) to make a 95% confidence interval of the standard deviation.
    Any ideas?

  2. #2
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    Re: Distribution of variance


    And as a follow-up on my own question: would it be alright to simply sum the contributions from each experiment so:

    0.95 CI =
    ((n1-1)rho_1^2+(n2-1)rho_2^2+.....(nn-1)rho_n^2 )
    ------------------------------------------------
    chisq(0.95, df=(n_1-1)+(n_2-1)+...(n_n-1)

    And then taking the square root of the result to yield the total contribution on the one-sided 95% CI of the SD?

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