Critical values of $F$-Statistic for two tailed test

Question

I am a bit confused while calculating the critical values of $F$ Statistic for a two tailed alternative.( Two tailed test). I require assistance from the Statistical Community regarding this matter.

In one of my textbooks, it has been clearly stated that the critical values of $F$ for two tailed test, $H_0:{\sigma_1}^2={\sigma_2}^2$ against $H_1:{\sigma_1}^2 \neq {\sigma_2}^2$ are given by $F>F_{n_1-1,n_2-1}(\frac{\alpha}{2})$ and $F<F_{n_1-1,n_2-1}(1-\frac{\alpha}{2})$

However, while looking for some practical examples of the same I found in an another book that the two tailed critical values of $F$ are $F>F_{n_1-1,n_2-1}(\frac{\alpha}{2})$ and $F<F_{n_2-1,n_1-1}(1-\frac{\alpha}{2})$

Which one should be correct ? In the second book, the only difference is that the in left tailed value, the original degrees of freedom get altered. And, I think this should be accurate. (Here $\alpha$ is the desired level of significance)

Answer 1

Suppose you have a random sample of $n_1 = 30$ observations from $\mathsf{Norm}(\mu=100,\sigma=15)$ and $n_2 = 40$ observations from $\mathsf{Norm}(\mu=200,\sigma=15).$ Also suppose $S_1^2 = 244.8387,$ $S_2^2 = 370.9807,$ and $F = S_1^2/S_2^2 = 0.659977.$ We know (because we simulated the data) that $\sigma_1^2 = \sigma_2^2 = 225$ so that we hope not to reject $H_0: \sigma_1^2 = \sigma_2^2$ in favor of $H_1: \sigma_1^2 \ne \sigma_2^2.$ [Computations use R statistical software.]

y1 = rnorm(30, 100, 15);  v1 = var(y1)
y2 = rnorm(40, 200, 15);  v2 = var(y2)
f = v1/v2
v1; v2; f
[1] 244.8387  # 1st sample variance
[1] 370.9807  # 2nd
[1] 0.659977  # F-ratio

The critical values of a test at level 5% are 0.492 and 1.962, cutting 2.5% of the probability from the left and right tails, respectively, of $\mathsf{F}(29,39).$ Because $0.492 < F = 0.660 < 1.962$ we do not reject $H_0.$

qf(c(.025,.975), 29, 39)   # 'qf' is an F quantile function
[1] 0.4919648 1.9618689

The density of $\mathsf{F}(29,39),$ the observed value of $F$ (solid black) and the critical values (broken red) are shown below.

You can find the upper critical value in many printed F-distribution tables. However, to save space these printed tables typically do not show lower critical values (corresponding to quantiles below 0.5).

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I believe that your second statement of critical values is incorrect, and guess it is based on an attempt to show how to find the lower critical values from printed tables.

By reversing the degrees of freedom, cutting $\alpha/2$ from the upper tail, and taking the reciprocal, you can find the lower critical value. In R the procedure for finding quantile 0.025 of $\mathsf{F}(29,39)$ can be illustrated as follows:

qf(.975, 39, 29);  1/qf(.975, 39, 29);  qf(.025, 29, 39) 
[1] 2.032666    # available in many printed tables
[1] 0.4919648   # reciprocal of above
[1] 0.4919648   # available directly in R, but not printed tables

Your second expression reverses the degrees of freedom but doesn't get the rest correct. Part of the difficulty may be that printed tables and books that rely on them use notation such as $F_{29,39;.025}$ to represent quantile 0.975 of $\mathsf{F}(29, 39),$ mentioning the percentage above in the right tail instead of the percentage below (as used for CDFs and quantiles). [One can only hope this '(upper) percentage point' notation falls into richly deserved disuse as we rely increasingly on software instead of printed tables.]

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Addendum F-test in R:

 F test to compare two variances

data:  y1 and y2
F = 0.65998, num df = 29, denom df = 39, p-value = 0.2475
alternative hypothesis: true ratio of variances is not equal to 1
95 percent confidence interval:
  0.3364022 1.3415128
sample estimates:
ratio of variances 
          0.659977