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I have a binomial distribution where the estimate for p is 0.03 out of 1000 sample trials.

Using the normal approximation and Chi-square distribution for the square of normal distribution, how can I find a confidence interval for the square of p?

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$$\hat p \sim N\left(p, \frac {p(1-p)}n\right)$$ $$\frac {\sqrt n(\hat p -p)}{\sqrt{p(1-p)}} \sim N\left(0,1\right)$$ $$\frac {n(\hat p -p)^2}{p(1-p)} \sim \chi^2(df=1)$$

Here $\sim$ means "asymptotically following"

Find the percentile of $\chi^2$ distribution. Let $X$ be its 2.5 percentile. $$X<\frac {n(\hat p -p)^2}{p(1-p)} $$ $$-(X+n)p^2 + (X+2n\hat p)p -n\hat p^2 <0 $$ You can get the solution following the quadratic formula. Similarly, you can get other one using 97.5% percentile.

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There are better methods for your case: First, there are exact methods for a confidence interval for a binomial $p$, see the many relevant posts here.

Use one such method and get a confidence interval $(l, u)$ for $p$. Then $(l^2, u^2)$ is a confidence interval for $p^2$.

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