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Suppose you have a sample from a normal population with mean mu, and known variance $\sigma^2$. What is the power function for $H_0: \mu = 0$ versus $H_a: \mu \ne 0$ at $\alpha = 0.05$?

Attempt: If we standardize, we get $Z= \bar{x} - \mu/(\sigma/\sqrt{n})$. So $\Pr(Z> c+ (\theta_0-\theta)/(\sigma/\sqrt{n}))$ or $\Pr(Z < -c- (\theta_0-\theta)/(\sigma/\sqrt{n}))$ is the power function.

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pls use math formatting:… – leonbloy Oct 24 '11 at 15:50

The power is $1-\beta$, where $\beta$ is the probability of Type II error, that is, the probability of not rejecting the null hypothesis. For a normal population with known variance, the power function would be

$\pi(\mu) = 1 - P(-z_{\alpha/2}-\dfrac{\mu-\mu_0}{\sigma/\sqrt{n}}<Z<z_{\alpha/2}-\dfrac{\mu-\mu_0}{\sigma/\sqrt{n}})$,

where $Z\sim N(0,1)$, $n$ is the size of the sample, $\sigma$ the standard deviation (sq. root of the variance) and $z_{\alpha/2}=\Phi^{-1}(1-\alpha/2)$, the inverse of standard normal cdf. Also, in your case your hypothesis is that $\mu_0=0$.

The probability is calculated by means of the standard normal cdf $\Phi(\cdot)$ :

$\pi(\mu) = 1 - \Phi(z_{\alpha/2}-\dfrac{\mu-\mu_0}{\sigma/\sqrt{n}}) + \Phi(-z_{\alpha/2}-\dfrac{\mu-\mu_0}{\sigma/\sqrt{n}})$.

If $\alpha=0.05$, that is, confidence intervals at 95%, you have $z_{\alpha/2}=1.9599639845\approx 1.96$

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