How to find P(X > F) in ANOVA F-test?

Question

(Apologies for the ASCII tables, Stackexchange doesn't allow HTML tables and since I'm not supposed to link to an image, this is the only way I know of showing the data)

I'm learning about ANOVA F-testing and stumbled upon this problem:

There exists the following set of data regarding four teaching methods and the scores of students who were subject to each teaching method:

Method 1  Method 2  Method 3  Method 4
----------------------------------------
  65        75        59        94
  87        69        78        89
  73        83        78        80
  79        81        62        88
  81        72        83
  69        79        76
            90
------------------------------------

$\bar{x} = 75.67$       $78.43$          $70.83$         $87.75$
$s = 8.17$          $7.11$            $9.58$           $5.80$

Where $\bar{x}$ is the mean of all scores for each teaching method and $s$ is the standard deviation for each method.

After conducting an ANOVA, I derive the following ANOVA table:

Source     | Deg. Freedom |    SS     |    MS     |   F  |
-------------------------------------------------------------
Treatment  |     3        | 712.59    |  237.53   | 3.77 |
Error      |     19       | 1196.63   |  62.98    |  -   |
Total      |     22       | 1909.22   |    -      |  -   |

Now the question is: Test a level $\alpha = 0.05$ The null hypothesis is that there is no difference in mean achievement for the four teaching techniques.

So to restate:

$H_0$: Teaching technique does not have an influence on mean achievement of students
$H_1$: Teaching technique does have an influence

I work out the critical f value to be $f_{3,19;0.95}$ from an F-distribution table to be $3.1274$

The next step is what I don't understand.

We can claim that the teaching technique does have an influence on the mean achievement of the students (with less than 5% chance of being wrong)

The associated p-value is:

$p = P(X>3.77) = 0.0281$

and indeed, p < $0.05$ (hence reject of $H_0$)

But now where is this $0.0281$ from? It looks to be the probability that X > 3.77. If I'm not wrong, in this case $X ~ F_{3,19} = 3.1274$ (as calculated before) and so the p value should be the probability that 3.1274 is greater than 3.77. Now how can 3.1274 ever be greater than 3.77?

Answer 1

The answer can not be retrieved without the aid of computation.

If you look at the F distribution table for F(3,19), you'll see that (for F(3,20):

...

.050 | 3.10

.025 | 3.86

...

Which means 0.025 < p < 0.05.

I'm guessing that they "cheated" with MATLAB or something.

Answer 2

Since the test statistics follows F-ratio distribution, the $\mathbb{P}(X>z)$ is known in closed form.

Let $X \sim F(n,m)$. Then $$ \mathbb{P}(X > z) = \tilde{B}_{\frac{m}{m+ n z}}\left( \frac{m}{2}, \frac{n}{2} \right) $$

For $n=3$ and $m=19$ and $z=3.77$, Wolfram Alpha produces the answer of $0.0280768$.