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assignment #100002

assignment #100002 - of the F ratio and giving a bigger F...

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40. the SS error in the one way ANOVA table was 18061.8 and the sum of the SS row and SS error in the repeated measures table was 18061.8! 41. There is adifference of 1652 between the SS error in the two source tables 42. The SS error is the smallest in the repeated measures source table ,\ 43. My answer to 42 does p.et'support that arepeated measures technique is more powerful than aone way ANOVA with an SS error value being smaller than that of the one way ANOVA 44. The repeated measures design was not more powerful in this specific example because the degrees of freedom taken out of the rows is to large making the degrees of freedom for the error very small. When Ifound the MS error value, it was larger because that number was divided by asmaller df value. Iwould have expected to see more significance because the error term is found in the
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Unformatted text preview: of the F ratio, and giving a bigger F ration would lead to this prediction, but it obviously didn't! When I went to calculate the F ratio, it appeared to be smaller in the repeated measures design than the one way ANOVA. This increases the probability of committing a type 2 error, accepting a false null hypothesis, ultimately making it less powerful. While looking at the repeated measures design, the Tukey value was higher than in the one way. This makes the results more conservative and less powerful. The subject in the study were very similar making very little error among results, however we saw a large degree of freedom value. u S.Se, r=~ N';w-d{ ~ SS\f\J o.{ \N *' So S. [; S:.YV\.l.lU-e V-\ V\ ('Ie l\ t7JJ0\ e 11r\ [LVl \IV \ \VI \ Vl -F Y/}. ..-n O.S VVlO,-d-~ Vlp D +-WI ell0 s q V CLV es . .....
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