Confidence Interval for a function of a parameter

alexhleb366
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I'm not sure of your math-stat background in this problem, so bear with me.

For a big sample size \hat p has approximately a normal distribution, right? You can approximate the distribution of \hat{p}^2 (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 p^2

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

If an estimate X_n for some parameter \theta satisfies

<br /> \sqrt n \left(X_n - \theta \right) \sim n(0, \sigma^2)<br />

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

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

<br /> \sqrt{n} \left(f(X_n) - f(\theta)\right) \sim n(0, \sigma^2 f&#039;(\theta) \right)<br />

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