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Math Help - confidence interval chi square probability search

  1. #1
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    confidence interval chi square probability search

    1000 samples each of size n=121, where observations Xi ~ iid N ( m, sigma^2), for each sample we calculate Confidence Interval: I = [ 0.75 s^2, s^2 + s^2/4 ] where s^2=(1/(n-1))SIGMA(Xi - X bar)^2

    how many samples include the true value of sigma^2, ie 1000 * prob.

    we know that (n-1) s^2/sigma^2 ~ Chi-sqrd with (n-1) degrees of freedom

    so i started:

    P (0.75s^2 < sigma^2 < s^2 + s^2/4) = P (0.75 < sigma^2/s^2 < 5/4) =
    =P (0.75/(n-1) < sigma^2/(s^2(n-1)) < 1.25/(n-1)) =
    =P (1.25/(n-1) < (n-1)s^2/sigma^2 < 0.75/(n-1)) =
    =P (1.25/120 < Chi-sqrd with (n-1) df < 0.75/120) =
    = chi n-1 df (0.75/120) - chi n-1 df (1.25/120) =
    = chi n-1 df (0.00625) - chi n-1 df (0.0104167) ..... =
    =1 -alpha

    and now im stuck, i am not sure if i am allowed to deduct them, and whether the numbers represent the quantiles of chi-squared, and if yes, i have a complications to find them in the table, was also thinking to get each separaetely the p value for v=120 and then deduct but dont like it....

    how do i finish it now to get the probability...

    thank you, if any

    nicky
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  2. #2
    MHF Contributor matheagle's Avatar
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    It would be a lot easier if you used TeX.

     P(3S^2/4<\sigma^2<5S^2/4)=P(3 (120)S^2/(4\sigma^2)<120< 5(120)S^2/(4\sigma^2))

    = P(3 C/4<120<5C/4) where  C\sim\chi^2_{120}

    = P(96<C<160)=1-(.0524+.0085)
    Last edited by matheagle; April 16th 2010 at 11:15 PM.
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