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I konw that a statistical test of size (level) $\alpha$ , $0<\alpha<1$, for testing a compound hypothesis $H_0$:$\theta \in \Theta_0 \subset \Theta$ againt a compound alternative $H_1$:$\theta \in \Theta_1=\Theta \backslash \Theta_0$ whose power function $\beta(\theta)$ satisfies: $$\beta(\theta)\leq\alpha \hspace{0.3cm}\text{if}\hspace{0.3cm}\theta \in \Theta_0$$ $$\beta(\theta)\geq\alpha \hspace{0.3cm}\text{if}\hspace{0.3cm}\theta \in \Theta_1$$ I understand the definition but I do not understand what does this mean for the test, does it mean the test have more power thatn it should? and how do you compare the power of 2 tests, one biased and the other is unbaised?

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It means that the probability that the test rejects (its power) is always higher when the alternative is true than when the null is true.

Suppose, for example, that you use a standard t-test for the null $\theta\leq0$ against the alternative $\theta>0$. The standard rejection rule at $\alpha=0.05$ would be to reject if $t>1.645$ (for either a sample from a normal distribution or asymptotically, when a central limit theorem applies).

Now, suppose you were to use that rule (reject if $t>1.645$) to test $\theta=0$ against $\theta\neq0$. The probability that the test will reject will decrease the more negative the true $\theta$, as we shall rarely observe large positive t-ratios in that case. In particular, this test is be biased, as $\beta(\theta)<\alpha$ when $\theta\in\Theta_1\cap(-\infty,0)$.

For concreteness, we may compute this probability explicitly in the normal case, $X_i\sim N(\theta,1)$, with $\sigma^2=1$ assumed known for simplicity. Then, the t-statistic for $\theta=0$ simply is $t=\sqrt{n}\bar{X}$ and $$\sqrt{n}(\bar{X}-\theta)\sim N(0,1)$$ Thus, \begin{align*} \beta(\theta)&=P(t>1.645)\\ &=1-P(t<1.645)\\ &=1-P(\sqrt{n}(\bar{X}-\theta)<1.645-\sqrt{n}\theta)\\ &=1-\Phi(1.645-\sqrt{n}\theta), \end{align*} which tends to 0 as $\theta\to-\infty$.

Graphically: enter image description here

theta.grid <- seq(-.8,.8,by=.01)
n <- seq(10,90,by=20)
power <- 1-pnorm(qnorm(.95)-outer(theta.grid,sqrt(n),"*"))
colors <- c("#DB2828", "#40AD64", "#E0B43A", "#2A49A1", "#7A7969")
matplot(theta.grid,power, type="l", lwd=2, lty=1, col=colors)
legend("topleft", legend=paste0("n=",n), col=colors, lty=1, lwd=2)
abline(h=0.05)
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  • $\begingroup$ I see, but if a test is biased, is there any corrections to do to make it unbaised? is a unbiased always better than a biased test? $\endgroup$ – Enthusiastic May 12 '17 at 14:43
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    $\begingroup$ See my comment in your other question. $\endgroup$ – Christoph Hanck May 12 '17 at 15:50
  • $\begingroup$ It could be nice to provide a link to this question. $\endgroup$ – Martijn Weterings Aug 31 '17 at 13:29
  • $\begingroup$ @MartijnWeterings, right: stats.stackexchange.com/questions/277740/… $\endgroup$ – Christoph Hanck Aug 31 '17 at 13:50

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