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Math Help - Confidence inteval

  1. #1
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    Confidence inteval

    Let X be a random variable following a normal distribution.
    A sample is taken from a population.
    If n= 8
    sample mean = 7.2

    (i=1) E (n=8) [x(i) - x-bar ]^2 = 3.44

    Calculate a 99% confidence interval for the population standard deviation?

    Do you do this by first solving for the population variance confidence interval and then taking the square root of the upper and lower limits?
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  2. #2
    MHF Contributor matheagle's Avatar
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    Is that E really a sum?
    You need to use the chi-square distribution for a CI about sigma.
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  3. #3
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    Quote Originally Posted by matheagle View Post
    Is that E really a sum?
    You need to use the chi-square distribution for a CI about sigma.
    Yes, E = sum.

    You mean the confidence interval is
    (n-1)s^2/ X^2 (n-1);(a/2)

    and

    (n-1)s^2/ X^2 (n-1);(1 - a/2)?

    Like (3.44/20.28 ; 3.44/0.989)?

    But I thought the standard deviation CI was being asked for, not the variance CI, or could you maybe explain how this works? Thanks!
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  4. #4
    MHF Contributor matheagle's Avatar
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    I didn't check your CI formula
    BUT IF (a,b) is your CI for \sigma^2
    Then (\sqrt{a},\sqrt{b}) is the CI for \sigma
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  5. #5
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    Quote Originally Posted by matheagle View Post
    I didn't check your CI formula
    BUT IF (a,b) is your CI for \sigma^2
    Then (\sqrt{a},\sqrt{b}) is the CI for \sigma
    Ok, thanks!
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