F(calc)=3.190
F(0.05)(2,12 df)= 3.89
Therefore cannot reject Ho
(Note that
the actual probability calculated by a computer was P=0.077.
This agrees completely with the simplified finding shown above)
This could be summarized as follows: There was a
difference in pH readings between the three methods but the difference was
not statistically significant. Method 1 was the lowest with a mean pH of
2.48, method 2 had a mean pH of 2.7 and the highest pH was from method 3, at
3.22. Given the number and variance in each group, these
differences could have been expected more than 5% of the time by chance
alone. The null hypothesis cannot be rejected. (It
is worth noting that the calculated F ratio was fairly close the critical F
ratio needed for significance; perhaps with a larger sample size, the
results would have been more definitive.)
F= 3.190, 2,12 df, P>0.05