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Confidence Interval for a function of a parameter

  1. May 23, 2009 #1
  2. jcsd
  3. May 25, 2009 #2

    statdad

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    Homework Helper

    I'm not sure of your math-stat background in this problem, so bear with me.

    For a big sample size [tex] \hat p [/tex] has approximately a normal distribution, right? You can approximate the distribution of [tex] \hat{p}^2 [/tex] (it will also turn out to be a normal distribution - look in (say) Hogg/Craig or any introductory math stat book for the idea, or write back and I can put the method here), and then you can get an approximate confidence interval for [tex] p^2 [/tex]

    Note - just so I don't have to post it:

    If an estimate [tex] X_n [/tex] for some parameter [tex] \theta [/tex] satisfies

    [tex]
    \sqrt n \left(X_n - \theta \right) \sim n(0, \sigma^2)
    [/tex]

    (the [tex] \sim [/tex] means "tends to a normal distribution as [tex] n \to \infty [/tex] - i.e., it represents convergence in distribution)

    then for a function [tex] f [/tex] that is continuous and has a non-zero derivative at [tex] \theta [/tex] it is true that

    [tex]
    \sqrt{n} \left(f(X_n) - f(\theta)\right) \sim n(0, \sigma^2 f'(\theta) \right)
    [/tex]

    Your statistic is the sample proportion, the parameter is [tex] p [/tex], and the function is [tex] f(x) = x^2 [/tex]
     
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