# Confidence Interval for a function of a parameter

1. May 23, 2009

### alexhleb366

2. May 25, 2009

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

$$\sqrt n \left(X_n - \theta \right) \sim n(0, \sigma^2)$$

(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

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

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