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Math Help - Finding sample standard deviation

  1. #1
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    Finding sample standard deviation

    If a 90% confidence interval for \sigma^2 is reported to be (51.47,261.90), what is the value of the sample standard deviation.

    Attempt:
    51.47<\sigma^2<261.90

    So
    \frac{(n-1)s^2}{\chi_{.95,n-1}^2}=51.47 and \frac{(n-1)s^2}{\chi_{.05,n-1}^2}=261.90

    Then
    s=\sqrt{\frac{51.47*\chi_{.95,n-1}^2}{n-1}}=\sqrt{\frac{261.90*\chi_{.05,n-1}^2}{n-1}}

    5.0884=\frac{\chi_{.95,n-1}^2}{\chi_{.05,n-1}^2}

    So this looks like it should have an F-distribution with same numerator and denominator degrees of freedom, which should help me figure out what those degrees of freedom are.

    I can't figure out how to do this. Or am I way off?
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  2. #2
    MHF Contributor matheagle's Avatar
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    you need n, and then just look up either of those chi-square percentiles
    then solve for s.
    If, you really don't have n, just go down the tables and see which of those ratio of percentiles
    gives you your number 5.0884
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  3. #3
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    I really don't have n. The question typed is precisely as it is posed in the text.

    Which F-Distribution table should I use? In other words, how do I infer the alpha level from my work so far? Is it just the same as for \sigma^2, which is .10?
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  4. #4
    MHF Contributor matheagle's Avatar
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    no, use the chi-square table.
    Let n=2,3,4,5... find the .05 and .95 percentiles and take that ratio
    see when the ratio is 5.0884
    (But I didn't check the math to see if 5.0884 was correct.)
    Last edited by matheagle; March 9th 2010 at 09:43 AM.
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