Difference between F-test and confidence intervals on variance estimates

Question

Given n samples from a normally-distributed variable X, we estimate variance as $s^2=\frac{1}{n-1}\sum{(x_i - \bar{x})^2}$. We can also get a confidence interval for such a variance estimate as: $$CI_\alpha = [\frac{(n-1)s^2}{\chi^2_{\frac{\alpha}{2},n-1}},\frac{(n-1)s^2}{\chi^2_{1-\frac{\alpha}{2},n-1}}]$$

Now suppose we have two variables $X \sim N(0,\sigma_x), Y \sim N(0,\sigma_y)$, and we want to test whether $\sigma_x = \sigma_y$ for some level of significance $\alpha$. In this case it is common to use the F-distribution to perform an F-test on $\frac{s_x^2}{s_y^2}$.

However, we can also look at the confidence intervals for the two variance estimates.

If have found that it is often the case that if the confidence intervals overlap then the F-test (for the same $\alpha$) does not reject the null hypothesis (which is that $\sigma_x = \sigma_y$), and if the confidence intervals do not overlap then the F-test does reject the null hypothesis. But this is not always the case. Here is a sample set where, for $\alpha$=0.1, the confidence intervals overlap but the F-test rejects the null hypothesis:

 X   Y
 13 -24
-10 -24
-12   7
 10 -51
  2  28
-10  -4
 -6   2
  0   5
 -1 -32
-21 -10

Variance X = 108, 90% CI = [57, 292]
Variance Y = 522, 90% CI = [277, 1412]
F-test = .028

Squinting at the math, it looks like the F-test is considering a ratio where the confidence intervals are considering differences. It's unclear to me why one is better than the other for purposes of this test.

Question: For the purposes of testing the hypothesis $\sigma_x = \sigma_y$, is it better to use the F-test than to use confidence intervals as described above? Why?

Answer 1

If the data from the first sample gives a very tight estimate for variance, $\sigma^2_x$, whilst data from the second sample gives a very wide estimate for second variance, $\sigma^2_y$, then even though confidence intervals may be overlapping now, there is a good chance that if you were to increase the number of samples in the second sample, your estimate for $\sigma_y^2$ would end up converging to a value outside the confidence interval for $\sigma_x^2$. So I would suggest that F-test, since it literally is about ratio of sample variances, is more appropriate.

Another way to approach this is with Bayesian methods. The appropriate distribution of variance given data is Scaled Inverse $\chi^2$. You can then have distributions for $\sigma^2_{x,y}$ and literally compute the probability $P\left[-\epsilon<\left(\sigma^2_x-\sigma^2_y\right)<\epsilon\right]$ for some epsilon which you deem to be sufficiently small to ignore the differences between variables. It might be a good idea to tie $\epsilon$ to uncertainty in the variances themselves.