Statistical Methods for Psychology

Because Ffor the residual is less than 1, we know automatically that it is not significant.
This tells us that there is no significant variability left to be explained over and above that
accounted for by the linear component. We would, therefore, normally stop here. However,
for purposes of an example I will go ahead and calculate the quadratic component. The cal-
culations will be shown without discussion, because the discussion would essentially be
the same as above with the word quadraticsubstituted for linear.

As our test on the residual suggested, there is no significant quadratic component on
our plot of the group means. Thus there is no indication, over the range of values used in
this study, that the means are beginning to level off. Therefore, we would conclude from
these data that attractiveness increases linearly with Composite, at least given the defini-
tion of Composite used here.
A word of caution is in order at this point. You might be tempted to go ahead and apply
the cubic and quartic coefficients that you find in Appendix Polynomial. You might also ob-
serve that having done this, the four sums of squares ( , ..., ) will sum to
, and be very impressed that you have accounted for all of the sums of squares be-
tween groups. Before you get too impressed, think about how proud you would be if you
showed that you could draw a straight line that exactly fit two points. The same idea applies
here. Regardless of the data, you know before you begin that a polynomial of order k 2 1 will
exactly fit kpoints. That is one reason why I was not eager to go much beyond fitting the lin-
ear components to the data at hand. A quadratic was stretching things a bit. Moreover, if you
were to fit a fourth-order polynomial and found that the quartic component was significant,

SSComposite

SSlinear SSquartic

61

=

0.0061

0.1731

F=

MSquadratic

MSerror

=0.0061

=

6 A0.1194^2 B

14

SSquadratic=

nc^2

ab

2

j

=0.1194

cquadratic=(2)2.6047 1 (-1)2.6450 1 (-2)2.8900 1 (-1)3.1850 1 (2)3.2600

61

=

0.0385

0.1731

Fresidual=

MSresidual

MSerror

=0.0385

=

0.1156

3

MSresidual=

SSresidual

dfresidual

12.10 Trend Analysis 407