Statistical Methods for Psychology

(Michael S) #1
In our case,

Like all contrasts, this contrast has a single degree of freedom, and therefore

. As you probably suspect from what you already know, we can convert
this mean square for the contrast to an Fby dividing by :


This is an Fon 1 and 25 degrees of freedom, and from Appendix Fwe find that

. Because the Ffor the linear component (11.87) exceeds 4.245, we will
reject and conclude that there is a significant linear trend in our means. In other words,
we will conclude that attractiveness varies linearly with increasing levels of Composite.
Notice here that a significant Fmeans that the trend component we are testing is signifi-
cantly different from 0.^13
It is conceivable that we could have a significant linear trend in our data and still have
residual variance that can be explained by a higher-order term. For example, it is possible
that we might have both linear and quadratic, or linear and cubic, components. In fact, it
would be reasonable to expect a quadratic component in addition to a linear one, because it
seems unlikely that judged attractiveness will keep increasing indefinitely as we increase
the number of individual photographs we average to get the composite. There will presum-
ably be some diminishing returns, and the curve should level off.
The next step is to ask whether the residual variance remaining after we fit the linear
component is significantly greater than the error variance that we already know is present.
If accounted for virtually all of , there would be little or nothing left over
for higher-order terms to explain. On the other hand, if were a relatively small part
of , then it would make sense to look for higher-order components. From our
previous calculations we obtain


= 3


= 421


dfresidual=dfComposite 2 dflinear

=0.1156


=2.1704 2 2.0548


SSresidual=SSComposite 2 SSlinear

SSComposite

SSlinear

SSlinear SSComposite

H 0


F.05(1,25)=4.245


=11.8706


=


2.0548


0.1731


F=


MSlinear
MSerror

MSerror

SSlinear=MSlinear

=2.0548


SSlinear=

nc^2
ga^2 j

=


6(1.8506^2 )


10


=1.8506


clinear=(-2)2.6047 1 (-1)2.6450 1 (0)2.8900 1 (1)3.1850 1 (2)3.2600

406 Chapter 12 Multiple Comparisons Among Treatment Means


(^13) I recently received a message from someone with similar data. He was studying the experimental hypothesis
that drug effects increased with dosage. He had obtained a nonsignificant overall F, but when he computed a test
on linear trend, the result was “highly significant.” He wanted to know what to do. Because the linear trend
tested his hypothesis directly, whereas the overall Fdid not, my recommendation was to rely solely on the test
for trend.

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