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Consider independent random samples from two normal distributions, $X_i\sim N(\mu_1,\sigma_1^2)$ and $Y_j\sim N(\mu_2,\sigma_2^2)$ for $i=1,...,n_1$ and $j=1,...,n_2$. Let $n_1=n_2=9,\bar{x}=16,\bar{y}=10,s_1^2=36,$ and $s_2^2=45$.

For testing $H_0:\sigma_2^2/\sigma_1^2\le1$ vs $H_1:\sigma_2^2/\sigma_1^2>1$, what is the power of this test if in fact $\sigma_2^2/\sigma_1^2=1.33?$ Use a significance level of $\alpha=0.05$.

I know that the power of a test is the probability of rejecting $H_0$ given that $H_a$ is actually true, but computing this probability is where I'm getting stuck; essentially, I need to compute:

$$P(S_1^2/S_2^2>F_\alpha(8,8)|_{\sigma_2^2/\sigma_1^2=1.33})$$

Where $F$ is just the fisher percentile and the bar represents "given (conditional probability)." How do I eliminate this conditional probability so that I can actually calculate a numerical value here?

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    $\begingroup$ So it is a good question to help you recall how did the F-distribution arises: The test statistics here is $\displaystyle \frac {\displaystyle \frac {S_1^2} {(n_1-1)\sigma_1^2}} {\displaystyle \frac {S_2^2} {(n_2-1)\sigma_2^2}} \sim F(n_1-1, n_2-1)$. When $\sigma_1^2 = \sigma_2^2$ under $H_0$, they are eliminated and thus does not appear in the test statistic. However in other situation, like this case, the ratio may not be $1$ but you still can apply the above result. $\endgroup$
    – BGM
    Apr 13, 2018 at 8:12
  • $\begingroup$ @BGM so does that imply that the probability of type I error is the Fisher(8,8) survival function evaluated at (the ratio of the variances times the Fisher(8'8) 0.05 percentile)? $\endgroup$
    – ereHsaWyhsipS
    Apr 13, 2018 at 13:44
  • $\begingroup$ The probability you computing is correct. You just need to note that the test statistic $S_1^2/S_2^2$, in the LHS of the inequality, no longer follows the $F$ distribution under this condition. Can you scale it back so that the LHS follow $F$-distribution and thus you can apply the survival function of $F$ as you said? (It seems you got it) $\endgroup$
    – BGM
    Apr 13, 2018 at 15:40
  • $\begingroup$ @BGM, right, but I actually said the wrong thing in my comment above; I meant the probability of rejecting $H_0$ given that $H_a$ is true, not the probability of type 1 error. So in the end I calculate that that the power of this test is approximately 0.9. Does this sound right? $\endgroup$
    – ereHsaWyhsipS
    Apr 13, 2018 at 16:56

1 Answer 1

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The answer is easy. The variance ratio, $V2/V1 = 1.333$, so $L = V1/V2 = 3/4 = 0.75, UT = F^{-1}(1-0.05, 8, 8) = 3.4381, \beta = F(UT/L, 8, 8, True) = 0.899$ and Power $= 1 - \beta = [0.101].$

Minitab confirmed this since $n_1=n_2$ and a VisualBasic $6$ Monte-Carlo simulation showed $0.101$ after $30$ million iterations using the Box-Muller generator.

I don't know why this simple formula for $F$ distribution Power is not well published like the $\chi^2$ power formula is.

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