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Math Help - SD from confidence interval

  1. #1
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    Question SD from confidence interval

    Hi

    Wanting help with dissertation - performing meta analysis as part of my medical degree - hope I am posting on correct line.

    Need to work out Standard deviation from confidence interval for both placebo and treatment group think the equation is:

    SD = SQRT of N x (upper limit - lower limit)/3.92.

    CI = 95%

    Is this correct?


    Treatment group: Number = 52 - upper limit 55 - lower limit 24.3

    Control group: Number 57 - upper limit 28.1 - lower limit 13.7

    So are the answers: Treatment = 56
    Control = 28

    Any help would be great

    Cheers
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  2. #2
    MHF Contributor matheagle's Avatar
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    Kinda, what you have here is a one sample CI for a mean.
    If n is small you have a problem.
    And I do not know if this SD is the sample st dev, known as s or the population sd, known as \sigma.
    If its s the correct CI is...

    (\bar X-t_{n-1,\alpha/2}s/\sqrt{n}, \bar X+t_{n-1,\alpha/2}s/\sqrt{n}).
    where we are assuming normality of our sample.

    And if we do know \sigma, then the correct CI is...
    (\bar X-z_{\alpha/2}\sigma/\sqrt{n}, \bar X+z_{\alpha/2}\sigma/\sqrt{n}).
    where we are assuming normality of our sample, or approximating via the Central Limit Theorem.

    Now 2z_{.025}=(2)(1.96)=3.92, so if we take the second interval here and subract the lower limit from the upper limit
    we have...

    2z_{\alpha/2}\sigma/\sqrt{n}. Multiplying this by \sqrt{n}/3.92 does reveal the \sigma.
    Last edited by matheagle; March 16th 2009 at 05:30 PM.
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