Confidence Interval for a function of a parameter

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To find a confidence interval (CI) for p^2, where p is the proportion in a binomial model, one can use the normal approximation for large sample sizes. The sample proportion, denoted as \hat{p}, is approximately normally distributed, allowing for the distribution of \hat{p}^2 to also be approximated as normal. By applying the method of continuous functions with a non-zero derivative, the CI for p^2 can be derived from the CI for p. This involves using the formula that relates the variance of the function of the estimate to the variance of the original estimate. Thus, an approximate confidence interval for p^2 can be constructed based on these principles.
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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