1. ## Arcsine Confidence Interval

From previous parts of a rather long question, I have shown that $\displaystyle \sqrt{n}(2 arcsin \sqrt{\hat{p}} - 2arcsin \sqrt{p})$ converges in distribution to the standard normal distribution. The next part of the question is:

Use this fact to derive an approximate con dence interval for $\displaystyle p$ known as the arcsine-interval, and compute this interval for some arbitrary confidence level, $\displaystyle n = 15, \hat{p}=0$.

It's just algebra, but I'm totally unsure about how to manipulate all these arcsines all over the place.

2. Hello,

So for a large n, we have $\displaystyle \mathbb{P}(2\sqrt{n}|\arcsin(\hat p)-\arcsin(p)|<\epsilon) \approx P(|Z|<\epsilon)$, where Z ~ N(0,1)

But we let $\displaystyle \hat p=0$ (it's weird that you're given n=15, let's just ignore it)
recall that $\displaystyle \arcsin(0)=0$ and it gives $\displaystyle P\left(-\tfrac{\epsilon}{2\sqrt{n}}<\arcsin(p)<\tfrac{\eps ilon}{2\sqrt{n}}\right)\approx P(|Z|<\epsilon)$

But now recall that the arcsine function is strictly increasing. So it's bijective. So $\displaystyle \left\{-\tfrac{\epsilon}{2\sqrt{n}}<\arcsin(p)<\tfrac{\eps ilon}{2\sqrt{n}}\right\}=\left\{-\sin\left(\tfrac{\epsilon}{2\sqrt{n}}\right)<p<\si n\left(\tfrac{\epsilon}{2\sqrt{n}}\right)\right\}$

Satisfied ?

3. Thanks!. But I think you forgot the squareroot over $\displaystyle p$ and $\displaystyle \hat{p}$. If we square the expression you get we get a single point for an interval, is that alright?

4. Yeah, I forgot the square roots, sorry...
It changes the result, look

The arcsine function is positive for any x in [0,1]. So since $\displaystyle \sqrt{p}\geq 0$, we actually have $\displaystyle \left\{-\tfrac{\epsilon}{2\sqrt{n}}<\arcsin(\sqrt{p})<\tfr ac{\epsilon}{2\sqrt{n}}\right\}=\left\{0<\arcsin(\ sqrt{p})<\tfrac{\epsilon}{2\sqrt{n}}\right\}$

and then finish it off