so there is no significance at the .05 level, and
cultivars 1 and 2 are not significantly different.

Now, doing the comparison for cultivars 1 and 4:

 Sd(1,4)2 = 0.43 [(2 / 4) + (60.25 50.5)2 / 1066]
 = 0.2533


 Sd(1,4) = (0.2533)1/2
 = 0.5033

 tc(l,4) = (4.4896 3.167) / 0.5073
 = 2.628

and since tl4df,.05 = 2.145, then tc = 2.628,
and cultivar 1 is significantly different from cultivar 4.
The same procedure is followed for all possible
comparisons.