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$k'$ is supposed to be $\mu_0+ \frac{Z_\alpha}{\sqrt n}$, but I don't know how to get there.

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    @GEdgar And maybe they should not.2012-10-03

2 Answers 2

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The change of variable $\bar x=\sqrt{n}(s-\mu_0)$ yields $ Z_\alpha=\int_u^{+\infty}\frac1{\sqrt{2\pi}}\mathrm e^{-s^2/2}\mathrm ds, $ with $u=\sqrt{n}\cdot(k'-\mu_0)$, hence $ 2Z_\alpha=1+\mathrm{erf}(u/\sqrt2), $ where $\mathrm{erf}$ denote the error function. In particular, $ k'=\mu_0+\sqrt{2/n}\cdot\mathrm{erf}^{-1}(2Z_\alpha-1). $ Thus, $k'=\mu_0+Y_\alpha/\sqrt{n}$, as required, but for $Y_\alpha=\sqrt{2}\cdot\mathrm{erf}^{-1}(2Z_\alpha-1)$.

In particular, $Y_\alpha=Z_\alpha$ if and only if $Z_\alpha=z^*$ with $z^*\approx0.783264\ldots$, $Y_\alpha\lt Z_\alpha$ for every $Z_\alpha\lt z^*$ and $Y_\alpha\gt Z_\alpha$ for every $Z_\alpha\gt z^*$.

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Take a look here: Proving $\int_{0}^{\infty} \mathrm{e}^{-x^2} dx = \dfrac{\sqrt \pi}{2}$

You can use the same technique as Ross Millikan.

To be more specific:

Call $I=\frac{n}{2\pi}\int_{k'}^{\infty}\int_{k'}^{\infty}e^{\frac{-n(x-\mu_{0})^{2}}{2}}dx$

So $I^{2}=Z_{\alpha}^{2}$. But $I^{2}=\frac{n}{2\pi}\int_{k'}^{\infty}\int_{k'}^{\infty}e^{-\frac{\bigl[\sqrt\frac{n}{2}(x-\mu_{0})\bigr]^{2}}{2}-\frac{\bigl[\sqrt\frac{n}{2}(y-\mu_{0})\bigr]^{2}}{2}}dxdy$

Define $u=\sqrt\frac{n}{2}(x-\mu_{0})$ and $v=\sqrt\frac{n}{2}(y-\mu_{0})$, then

$I^{2}=\frac{n^{2}}{4\pi}\int_{\sqrt\frac{n}{2}(k'-\mu_{0})}^{\infty}\int_{\sqrt\frac{n}{2}(k'-\mu_{0})}^{\infty}e^{-u^2-v^2}dudv$

Now use polar coordinates and conclude.

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    Sorry my answer for this question is not correct. I had did some mistakes here. I cant continue now from the last point...2012-10-02