Arcsine Confidence Interval

**h2osprey** · February 27th 2010, 02:02 PM

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

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

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

**Moo** · February 27th 2010, 06:50 PM

Hello,

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

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

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

Satisfied ?

**h2osprey** · February 27th 2010, 10:33 PM

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

**Moo** · February 28th 2010, 09:08 AM

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 $\sqrt{p}\geq 0$ , we actually have $\left\{-\tfrac{\epsilon}{2\sqrt{n}}<\arcsin(\sqrt{p})<\tfr ac{\epsilon}{2\sqrt{n}}\right\}=\left\{0<\arcsin(\ sqrt{p})<\tfrac{\epsilon}{2\sqrt{n}}\right\}$

and then finish it off

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