Basic power function

**MartyJ** · February 5th 2010, 12:03 PM

I'm trying to re-learn statistics but the terminology used in my class is different than what I'm used to. Can somebody tell me if I'm on the right track?

Suppose units per day $X$ has a normal distribution with unknown mean $\theta$ and standard deviation $\sigma=8$ .

I want to test $H_0:\theta \ge 100$ vs. $H_1:\,\theta < 100$ .

Suppose $n=64$ days and $H_0$ will be rejected if the sample mean daily units satisfies $\bar{x}<98$ . What is the value of the power function if $\theta=99$ ?

My work:

$P( \bar{x} >97)=P \left( Z> \frac{99-97}{8/ \sqrt{64}} \right) =P(Z=2)=0.0228$

So,

$1-0.0228= \bf{0.9772}$

Am I understanding this "power function" thing correctly?

**matheagle** · February 5th 2010, 02:00 PM

Sorry, you reject the null for $\bar X<98$

$P( \bar X<98)=P \left( Z< \frac{98-99}{8/ \sqrt{64}} \right) =P(Z<-1)$

**MartyJ** · February 5th 2010, 02:49 PM

I know its basic, but I'm still trying to get my head around this.

So does that mean the value of the power function is 0.1587 and the probabilty of a type 2 error is 0.8413?

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