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D In ' 6:: 0.233? via) a. Identify the experimental design used here. Be very speciﬁc, and make sure
that you identify which factors are treatment factors and which are blocking
factors. Answer: The experimental design here is a replicated Latin square design, with rows
(subjects) nested in squares. Two 5 x 5 Latin squares are used to form the design. One
could also identify the design as a crossover design formed from two replicated Latin
squares. The blocking factors are subjects and graders, with 10 and 5 levels, respectively.
The treatment factor is work basket (ﬁve levels). It is probably more appropriate to regard
subjects and graders as random effects. However, since the model we will use for analysis
is a completely additive one with no interactions between ﬁxed and random factors, the
ﬁxedeﬂ'ects analysis will be no different from what we’d get if we regarded subjects and
graders as random. b. Obtain the appropriate ANOVA table for these data and test the signiﬁcance
of each source of variability in the table. State the appropriate conclusions. Answer: See ﬁles hwk54.sas and hwk5—4.lst. In hwk5—4.sas we ﬁt the model
yijk = M+ 0% + ﬁj + ’71.: +81%, where a, is the ith basket (treatment) effect, 5, is the jth subject effect (j = 1, . . . , 10), and 7;, is the kth grader effect. We could have broken the subject effects into subjects Within
squares (ﬁve levels per square) and squares (two levels). If we had done so, we would have
gotten an equivalent analysis, but there seems little point to doing so in this case in which
the replicated Latin squares don’t correspond to the levels of a nuisance variable (source
of variation). In hwk54.sas I used both PROC GLM to ﬁt the ﬁxedeﬁects version of
this model. I also used PROC MIXED to ﬁt the mixedeffects version in which subject
effects and grader effects are regarded as random. The latter is probably more appropriate
here. However, since we have a completely additive model with no interactions involving
subjects or graders, it does not affect the basic analysis whether or not we regard subjects
and graders as random. The only difference between the results with subjects and graders
ﬁxed and the analysis With subjects and graders random is in the standard errors for the
basket (treatment) means. Since these standard errors don’t enter into any of the requested
result for this problem, either analysis can be used. To test Whether or not there are diﬂ'erences in the mean score across baskets we use
F = MSBasket/MSE = 33.96, which has apvalue < .0001. Therefore, we reject H0 : M1 =
M2 = [i3 = M4 = M5 and conclude that the mean response is not the same for all 5 baskets.
Since subjects and graders are blocking factors, we don’t perform and formal inference on
these factors. It does appear, however, from the magnitude of the F statistics for these
factors that there is considerable variability from subject to subject and from grader to
grader. Therefore, it was a good thing that we blocked on these factors. c. Form and test appropriate contrasts for comparing each of the two new work
baskets to the old work baskets. State your conclusions. Answer: The appropriate contrasts are 1/)1 = 3n1—u3—u4—u5 and $2 = 3n2—u3—u4—u5.
These contrasts yield F test statistics of 35.37 (p < .0001) and 51.78 (p < .0001), so in
each case we conclude that the new work basket has a signiﬁcantly different mean score
than the avergae score on the old work baskets. Rom the Is means for baskets, we conclude
that new work basket A is easier than the old baskets, and new work basket B is harder
than the old work baskets. .. ..~w~me——————————_——— ...
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 Fall '08
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