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Designing a Uniformly Most Powerful Test

  1. Jan 7, 2014 #1
    1. The problem statement, all variables and given/known data

    Let ##X_1, ... , X_n## be iid normally distributed random variables ##N(\mu, \sigma^2)##, ##\mu \in \Bbb{R}##, ## \sigma^2 > 0##.

    a) Design a uniformly most powerful test with significance level ##\alpha## for testing ##H_0: \sigma^2 = \sigma^2_0## vs ##H_1: \sigma^2 > \sigma^2_0##.

    b) Give a formula for the power of the test.

    2. Relevant equations



    3. The attempt at a solution

    Let ##\sigma^2_1 > \sigma^2_0##.

    ##\Lambda = \frac{L(\sigma^2_0)}{L(\sigma^2_1)}## = ##\frac{(\frac{1}{\sqrt{2\pi}})^n (1/\sigma^2_0)^{n/2} e^{-\sum (X_i - \mu)^2/\sigma^2_0}}{(\frac{1}{\sqrt{2\pi}})^n (1/\sigma^2_1)^{n/2} e^{-\sum (X_i - \mu)^2/\sigma^2_1}} \leq k## for some ##k < 1##.

    ##\implies e^{\frac{-\sum (X_i - \mu)^2}{2\sigma^2_0} + \frac{\sum (X_i - \mu)^2}{2\sigma^2_1}} \leq (\frac{\sigma^2_0}{\sigma^2_1})^{n/2}k##

    Let ##k_1 = (\frac{\sigma^2_0}{\sigma^2_1})^{n/2}k##.

    We have ## \frac{-\sum (X_i - \mu)^2}{2\sigma^2_0} + \frac{\sum (X_i - \mu)^2}{2\sigma^2_1} \leq \ln(k_1)##

    ##\implies \sum (X_i - \mu)^2[\frac{1}{2\sigma^2_1} - \frac{1}{2\sigma^2_0}] \leq \ln(k_1)##

    ##\implies \sum (X_i - \mu)^2 \geq \ln(k_1)[\frac{1}{2\sigma^2_1} - \frac{1}{2\sigma^2_0}]^{-1}##

    ##\implies \frac{\sum (X_i - \mu)^2}{\sigma^2_0} \geq \ln(k_1)[\frac{1}{2\sigma^2_1} - \frac{1}{2\sigma^2_0}]^{-1}(\frac{1}{\sigma^2_0})##.

    Let ##k_2 = \ln(k_1)[\frac{1}{2\sigma^2_1} - \frac{1}{2\sigma^2_0}]^{-1}(\frac{1}{\sigma^2_0})##.

    We know that ##\frac{\sum (X_i - \mu)^2}{\sigma^2_0}## has chi-square with n degrees of freedom.

    So we choose ##k_2## such that ##P_{H_0}( \frac{\sum (X_i - \mu)^2}{\sigma^2_0} \geq k_2) = \alpha##

    b) ##P_{H_1}( \frac{\sum (X_i - \mu)^2}{\sigma^2_0} \geq k_2)## ##= 1 - \int^{k_2}_0## ##(\frac{1}{\Gamma(n/2)2^{n/2}})x^{n/2-1}e^{-x/2} dx##.

    Are my answers correct?

    Thanks in advance
     
    Last edited: Jan 7, 2014
  2. jcsd
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