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

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(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?

Best Answer

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.

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