Power of Hypothesis Test for Variance Ratio from Two Normal Samples 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 $H_1:\sigma_2^2/\sigma_1^2&gt;1$, 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:
$$P(S_1^2/S_2^2&gt;F_\alpha(8,8)|_{\sigma_2^2/\sigma_1^2=1.33})$$
Where $F$ is just the fisher percentile and the bar represents &quot;given (conditional probability).&quot;  How do I eliminate this conditional probability so that I can actually calculate a numerical value here?
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.
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.