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Math Help - Which confidence interval is better?

  1. #1
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    Which confidence interval is better?

    For a proportion, is  \hat{p} \pm z_{\alpha/2} \sqrt{\hat{p} \hat{q}/n} better than

     \frac{\hat{p} + \frac{z_{\alpha/2}^{2}}{2n} \pm z_{\alpha/2} \sqrt{\frac{\hat{p} \hat{q}}{n} + \frac{z_{\alpha/2}^{2}}{4n^2}}}{1 + (z_{\alpha/2}^{2})/n} in terms of precision and reliability?

    Wasn't the first used a lot in the past?
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  2. #2
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    Quote Originally Posted by heathrowjohnny View Post
    For a proportion, is  \hat{p} \pm z_{\alpha/2} \sqrt{\hat{p} \hat{q}/n} better than

     \frac{\hat{p} + \frac{z_{\alpha/2}^{2}}{2n} \pm z_{\alpha/2} \sqrt{\frac{\hat{p} \hat{q}}{n} + \frac{z_{\alpha/2}^{2}}{4n^2}}}{1 + (z_{\alpha/2}^{2})/n} in terms of precision and reliability?

    Wasn't the first used a lot in the past?
    The first is using the normal approximation to the binaomial where means and
    variances are matched between the binomial and the normal approximation.

    This is assuming that np and nq are large.

    The lower formula looks to use a better approximation, and so have a wider
    range of validity.

    Now I would expect that the lower formula might be used in software packages,
    but for hand calculation I would use the upper formula unless
    np and or nq are small, then I suspect I would use the bootstrap or the
    exact binomial calculations.

    RonL
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