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      F(calc) =23.53

      F(0.01)(2,33 df) is between 5.18 and 5.39    Therefore reject H_o

This could be summarised as follows:  The three training methods resulted in very different test results.  The lowest test score was from method C (4.75), with the highest from method B (9.00).  Method A was roughly halfway between with a mean of 6.42.   This difference was highly statistically significant at the 1% level. Given the number and variance in each group, if the null hypothesis had been correct, these differences could not have been expected to occur more than 1% of the time.   The null hypothesis is therefore rejected. 

   F= 23.53,      2, 33 df,  P<0.01

**  The actual probability was P<0.0001, although you needed a greater selection of F-ratio tables to find this. The result P<0.01 is perfectly acceptable given the information you had in your notes.  If you HAD access to the better tables, the results could be summarized as follows:  " ... This difference was highly statistically significant at the 0.0001 level. Given the number and variance in each group, if the null hypothesis had been correct, these differences could not have been expected to occur more than once in ten thousand times.  The null hypothesis is therefore rejected.

   F=23.53    2, 33 df, P<0.0001  ... "