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Thread: Basic power function

  1. #1
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    Basic power function

    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 $\displaystyle X$ has a normal distribution with unknown mean $\displaystyle \theta$ and standard deviation $\displaystyle \sigma=8$.

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


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

    My work:

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

    So,

    $\displaystyle 1-0.0228= \bf{0.9772}$

    Am I understanding this "power function" thing correctly?
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  2. #2
    MHF Contributor matheagle's Avatar
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    Sorry, you reject the null for $\displaystyle \bar X<98$

    $\displaystyle P( \bar X<98)=P \left( Z< \frac{98-99}{8/ \sqrt{64}} \right) =P(Z<-1)$
    Last edited by matheagle; Feb 5th 2010 at 02:24 PM.
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  3. #3
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    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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