hwk5soln

hwk5soln - “4.... __....

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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 fixed-efl'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 hwk5-4.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 hwk5-4.sas I used both PROC GLM to fit the fixed-efiects version of this model. I also used PROC MIXED to fit the mixed-effects 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 ap-value < .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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hwk5soln - “4.... __....

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