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Math Help - Arcsine Confidence Interval

  1. #1
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    Arcsine Confidence Interval

    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 con dence 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.
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  2. #2
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    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 ?
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  3. #3
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    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?
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  4. #4
    Moo
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    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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