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And now I have Type III sums of squares for $A$, $B$, and their interaction (A:B) using drop1(model, .~., test="F"). What I am stuck with is how sums of squares is calculated for $B$. … predB<-predict(lm(y~A+A:B)) sum((predFull-predB)^2) #Sums of squares should be 0.15075 (according to anova table) #but calculated to be 2.5e-31 #SS(AB|A,B) predAB<-predict(lm(y~A+B)) sum((predFull-predAB …

One way to obtain the "correct" Type III sums of squares is to change the model parameterization to sum coding, "contr.sum". … III Sums of Squares produced under sum coding These correctly match the Type I Sums of Squares for this balanced case. …

I am trying to replicate output in SPSS, that was computed using Type 3 Sums of Squares, in R. … Can someone help me get Type 3 Sums of Squares for a mixed model in R? Thank you! …

Can someone explain the difference between type 1:4 sums of squares using examples with relatable variables? …

I am trying to understand Type I/II/III sums of squares. Take for instance the 2 way Anova with 2 factors A and B. I do not follow what Type III Sums of Squares are. … In type III sums of squares we account for the main effect (say A) AFTER removing the effect of the other main effect(B) + interaction(AB). …

I have read many times that one has to set orthogonal contrast to get correct type III sums of square. E.g. … John Fox says To compute Type-III tests using incremental F-tests, one needs contrasts that are orthogonal in the row-basis of the model matrix. …

I ran a two-way ANCOVA in R: ancova = lm(DV ~ IV1*IV2 + CV1 + CV2 + CV3, data = Data) summary.aov(ancova) Anybody know if this uses type III sums of squares? … I know that the anova() and aov() functions use type I, but I am trying to get type III. …

Performing a 2 x 2 ANOVA with type II sums of squares: aov_car(value ~ treatment * gender + Error(id), data=dd, type=2) yields the following results: Effect df MSE F ges p.value … II sums of squares imply that they are assessed independently, without the inclusion of their interaction treatment:gender. …

I have a statistical analysis / data analysis problem: I am analyzing data using a factorial three-way ANOVA with a-priori contrasts and type III sums of squares. … (Please don't speak about type I SS vs. type III SS. That's not the point of my question.) I get the contrasts like I need using summary.aov(), however that uses type I SS. …

The ezANOVA() function mostly serves as a wrapper to car::Anova(), but the current version of ezANOVA() implements only type-II sums of squares, whereas car::Anova() permits specification of either type-II … Reasons for not including a "SS_type" argument in ezANOVA(): The difference between type I, II, and III sum squares only crops up when data are unbalanced, in which case I'd say that more benefit is …

Following my short experience in statistics, it seems that the type of sums of squares (type I,II,III,IV...) used in getting ANOVA results could make a dramatic difference in test results (especially of …

library(car) > anova(res.aov3) # type I Analysis of Variance Table Response: len Df Sum Sq Mean Sq F value Pr(>F) supp 1 205.35 205.35 15.572 0.0002312 *** dose 2 2426.43 … =2) # type II Anova Table (Type II tests) Response: len Sum Sq Df F value Pr(>F) supp 205.35 1 15.572 0.0002312 *** dose 2426.43 2 92.000 < 2.2e-16 *** supp:dose 108.32 …

My design is balanced, and I use the same type of sums of squares (Type 3) in both analysis, but still get different F-ratios. Anyone know why? … (VCondition, Stimulus), type =3) and here is the output: $ANOVA Effect DFn DFd F p p<.05 2 VCondition 1 81 0.006498711 0.945754134 …

The key seems to be the coding of the variable interaction (and vs. xor for treatment vs. sum). … When I compare the full model (~ A*B) with a model including only B and xor(A,B), I get the "correct" difference in the sums of squared residuals for A, regardless of the type of contrasts used. …

Is it true that if $H_0$ and $H_a$ are complementary hypotheses of the Binomial trial, i.e., the negation of $H_0$ is $H_a$ then the type-I error $\alpha$ plus type-II error $\beta$ equals 1? … Or is that sum always less then 1, or can it be sometimes even greater then 1? …