*(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.