Finding the P-value of a 2-sided F-test

Question

So I believe that since I am given variances, I use F-distribution in order to test statistic. But I am a little bit confused how to find p-values for this question.. correct me if I am wrong in the first place Thanks!

Answer 1

First I repeat the computation of the test statistic. With sample sizes of 9 in each group you have $S_R^2 = 6.335$ and $S_W^2 = 10.8375,$ so $F = 6.335/10.8375 = 0.5845.$ Under $H_0: \sigma_R^2/\sigma_W^2 = 1.$ the variance ratio $F \sim \mathsf{F}(8,8).$

The graph below shows the situation. (Notice that the observed F-ratio is near the mode of the F-distribution, so we cannot expect to reject $H_0.)$

First, let's get the P-value using R. Then figure out how R did the computation.

r = c(3.5, 8.1, 7.4, 4.0, 0.7, 4.9, 8.4, 7.0, 5.5)
w = c(3.1, 0.5, -3.8, 4.1,-0.6, 2.7, 1.9, -5.9, 0.1)
var.test(r, w)

        F test to compare two variances

data:  r and w
F = 0.58454, num df = 8, denom df = 8, p-value = 0.4643
alternative hypothesis: true ratio of variances is not equal to 1
95 percent confidence interval:
 0.1318543 2.5914373
sample estimates:
ratio of variances 
         0.5845444 

So the P-value is 0.4643. The P-value is defined as a the probability of getting a result that is 'as or more extreme' than the observed F-ratio. Because this is a two-sided test, 'extreme' values can be found in both tails of the F-distribution.

Using R as a statistical calculator, we have $P(F < .5845) = 0.2321.$ This is the area under the density curve above to the left of the vertical black bar.

pf(.5845, 8, 8)
## 0.2321101

You can find this value on many statistical calculators, but not in an ordinary printed F-table, because (in order to save space) such table show only upper tail probabilities.

But this is a two-sided test, so there is another part to the P-value. We might have formed the variance ratio as $F' = 10.8375/6.335 = 1.711.$ Depending on the printed table you use, you might be able to estimate the other part of the P-value $P(F > 1.711) = 0.2321.$ Certainly, you could get this probability on a statistical calculator programmed to give F probabilities. This is the probability to the right of the vertical red broken line in the figure below.

1 - pf(1.711, 8, 8)
## 0.2320774

Thus the P-value is the sum of the two tail probabilities: $0.2321 + 0.2321 = 0.4642,$ which is essentially the same as in the R output from the variance test.

Note: If sample sizes are unequal, components of the P-value in the two tails may not be exactly equal, as they were here. Perhaps see this Q & A for more on this topic.