Given two samples from N with known variances how to find probability that one sample's variance is two times larger than the other? I have two samples with sizes 9 and 11 drawn from two independent normal distributions with variances 21 and 37 respectively. How can I find the probability that the variance of the first sample is two times larger than the other?
 A: The first sample comes from a population with a smaller variance
and the second sample from a population with a larger variance.
So it seems unlikely that $S_1^2$ is much larger than $S_2^2,$
and even less likely still that $S_1^2/S_2^2 \ge 2.$ So let's
start with a simulation in R to get an idea just how likely it is
to have $R = S_1^2/S_2^2 \ge 2.$
set.seed(914)
r = replicate(10^6,
  var(rnorm(9,0,sqrt(21)))/var(rnorm(11,0,sqrt(37))))
mean(r >= 2)
[1] 0.033342    # aprx P(R >= 2) = 0.0332 (see below)

We see that we can have $R \ge2,$ but rarely. With a million iterations (pairs of samples), the simulated value $P(R \ge 2)=0.0333$ should be accurate to two or three decimal places.
Now the question
is whether we can express this probability in terms of the F-distribution.
The ratio $F = \frac{S_1^2/\sigma_1^2}{S_2^2/\sigma_2^2}  =
\frac{S_1^2/21}{S_2^2/37}
\sim \mathsf{F}(n_1-1, n_2 - 1) \equiv \mathsf{F}(8, 10).$ 
We need to express this in terms of $R.$ It is easy to see that
$R \ge 2$ is the same as $F \ge 2(37/21) = 74/21.$
So using the CDF pf of the appropriate F-distribution ought to
give us nearly the same answer as we got from the simulation, which
it does. The exact answer to four places is $0.0332.$
1 - pf(74/21, 8,10)
[1] 0.03316164

Note: Most printed F-tables are not quite up to the task of evaluating this probability. My printed table shows probability .05 in the upper tail beyond 3.07 and probability 0.025 beyond 3.83. Because 
$72/21 \approx 3.429$ (between 3.07 and 3.83) we know from my table only that $0.025 < P(R \ge 2) <0.05$--maybe about halfway between.
