Statistics Review - Mathematics and Statistics for the Environment - page 150

3

∑ ( )
lots perlevel of nitrogen

2

Nitrogentotals

Nitrogen S df = − CT

No.ofp

2

( ) −= −=

2 2 2

329 + 185 + 16 9

171 027
9

,

= CT CT 1725 6296 .

9

Species-nitrogeninteration S df = Treatment SS − Species SS − Nitrogen SS = 134 81 .49

4 8 df 2 df 2 df

The analysis now becomes:

Source of Variation Degrees of Freedom Sums of Squares Mean Squares F

Blocks 2 11.6296 5.8148 —

Treatments 8 1970.2963 246.2870 13.417 b

Species 2 a 109.8518 a 54.9259 2.992 c

Nitrogen 2 a 1725.6296 a 862.8148 47.003 b

Species-nitrogen 4 a 134.8149 a 33.7037 1.836 c

Error 16 293.7037 18.3565 —

Total 26 2275.6296 — —

a These figures are a partitioning of the degrees of freedom and sum of squares for treatments and
are therefore not included in the total at the bottom of the table.
b Significant at the 0.01 level.
c Not significant.

The degrees of freedom for simple interactions can be obtained in two ways. The first way is by
subtracting the degrees of freedom associated with the component factors (in this case, two for spe-
cies and two for nitrogen levels) from the degrees of freedom associated with all possible treatment
combinations (eight in this case). The second way is to calculate the interaction degrees of freedom
as the product of the component factor degrees of freedom (in this case, 2 × 2 = 4). Do it both ways
as a check. The F values for species, nitrogen, and the species-nitrogen interaction are calculated by
dividing their mean squares by the mean square for error. The analysis indicates a significant differ-
ence among levels of nitrogen, but no difference between species and no species-nitrogen interaction.
As before, a prespecified comparison among treatment means can be tested by breaking out the
sum of squares associated with that comparison. To illustrate the computations, we will test nitro-
gen vs. no nitrogen and also 100 lb vs. 200 lb of nitrogen.

2

92 329
9

185
9

169
9

− −

1 1

Nitrogenvs. no nitrogen S df =

2 2 2

211 ++

1

[ ]

2

2 (329 ) −− 185 169

= = 1711 4074 .

96 ()

In the numerator, the mean for the zero level of nitrogen is multiplied by 2 to give it equal weight
with the mean of levels 1 and 2 with which it is compared. The 9 is the number of plots on which
each mean is based. The (2 2 + 1 2 + 1 2 ) in the denominator is the sum of squares of the coefficients
used in the numerator.