How do I evaluate $k'$ from $\int_{k'}^{\infty} \sqrt{n/2\pi} \, \, e^{(-n/2 \, (\bar x-\mu_0)^2)} \, d\bar x = Z_\alpha$?

Question

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

Answer 1

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 $Z_\alpha\gt z^*$.

Answer 2

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

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.