Dear Karla, I suggested last night that you send me further information, but decided this morning to try out a reproducible example of my own:
> set.seed(12345) > A <- factor(sample(c("a1", "a2", "a3"), 100, replace=TRUE)) > B <- factor(sample(c("b1", "b2"), 100, replace=TRUE)) > C <- factor(sample(c("c1", "c2", "c3"), 100, replace=TRUE)) > mu <- array(1:18, c(3,2,3)) > a <- as.numeric(A) > b <- as.numeric(B) > c <- as.numeric(C) > y <- mu[cbind(a,b,c)] + rnorm(100) > mod <- lm(y ~ A*B*C) > library(car) > options(contrasts=c("contr.sum", "contr.poly")) > Anova(mod, type="II") Anova Table (Type II tests) Response: y Sum Sq Df F value Pr(>F) A 65.88 2 38.4098 1.696e-12 *** B 196.47 1 229.0775 < 2.2e-16 *** C 2441.00 2 1423.0809 < 2.2e-16 *** A:B 0.22 2 0.1259 0.8819 A:C 6.92 4 2.0174 0.0996 . B:C 0.87 2 0.5095 0.6027 A:B:C 2.89 4 0.8432 0.5018 Residuals 70.33 82 --- Signif. codes: 0 `***' 0.001 `**' 0.01 `*' 0.05 `.' 0.1 ` ' 1 > Anova(mod, type="III") Anova Table (Type III tests) Response: y Sum Sq Df F value Pr(>F) (Intercept) 7830.2 1 9129.8959 < 2.2e-16 *** A 55.7 2 32.4913 4.059e-11 *** B 189.5 1 221.0076 < 2.2e-16 *** C 2124.0 2 1238.2549 < 2.2e-16 *** A:B 0.2 2 0.0942 0.9102 A:C 5.9 4 1.7323 0.1507 B:C 0.6 2 0.3417 0.7115 A:B:C 2.9 4 0.8432 0.5018 Residuals 70.3 82 --- Signif. codes: 0 `***' 0.001 `**' 0.01 `*' 0.05 `.' 0.1 ` ' 1 I don't have a working copy of SPSS anymore, but here's what SAS does with this example: Source DF Type II SS Mean Square F Value Pr > F A 2 65.884048 32.942024 38.41 <.0001 B 1 196.467384 196.467384 229.08 <.0001 A*B 2 0.215883 0.107942 0.13 0.8819 C 2 2440.998718 1220.499359 1423.08 <.0001 A*C 4 6.920872 1.730218 2.02 0.0996 B*C 2 0.873945 0.436973 0.51 0.6027 A*B*C 4 2.892820 0.723205 0.84 0.5018 Source DF Type III SS Mean Square F Value Pr > F A 2 55.732128 27.866064 32.49 <.0001 B 1 189.546201 189.546201 221.01 <.0001 A*B 2 0.161608 0.080804 0.09 0.9102 C 2 2123.968177 1061.984089 1238.25 <.0001 A*C 4 5.942845 1.485711 1.73 0.1507 B*C 2 0.586168 0.293084 0.34 0.7115 A*B*C 4 2.892820 0.723205 0.84 0.5018 So, as you can see, the results check. It's hard to know what to make of this without more information about what you did. Much as I'm not an admirer of SPSS, I doubt whether it computes type-III sums of squares incorrectly, so I suspect something wrong with either your SPSS commands or your R commands. I hope this helps, John -------------------------------- John Fox Department of Sociology McMaster University Hamilton, Ontario Canada L8S 4M4 905-525-9140x23604 http://socserv.mcmaster.ca/jfox -------------------------------- > -----Original Message----- > From: [EMAIL PROTECTED] > [mailto:[EMAIL PROTECTED] On Behalf Of Karla Sartor > Sent: Saturday, December 18, 2004 6:43 PM > To: [EMAIL PROTECTED] > Subject: [R] Sums of sq in car package Anova function > > Hello R users, > > I am trying to run a three factor ANOVA on a data set with > unequal sample sizes. > > I fit the data to a 'lm' object and used the Anova function > from the 'car' package with the 'type=III' option to get type > III sums of squares. I also set the contrast coding option > to 'options(contrasts = c("contr.sum", "contr.poly"))' as > cautioned in Jon Fox's book "An R and S-plus Companion to > Applied Regression'. > > Is there anything else that I need to consider when using the > type III option with the Anova function? > > When I run the same data set in SPSS with General Linear > Model and type III sums of squares, the sums of squares are > different enough that one of the main effect terms is > significant in the R table and not in the SPSS table. I > found a similar discrepancy with a different data set, only > SPSS showed a significant interaction effect while, while the > 'Anova' function did not. > > I also compared the results from SPSS those from the 'anova' > function in the base package, and the results are nearly > identical. I would expect the two methods with type III sums > of squares to be more similar, does anyone have any ideas as > to why that was not the case? I am hoping to not go back to > SPSS at this point, so am trying to decide which of the two R > functions is most appropriate for me (and defensible, > considering the unequal sample sizes). > > Thank you in advance for any ideas you may have! > > Karla > > Karla Sartor > Montana State University - LRES > [EMAIL PROTECTED] > > ______________________________________________ > [EMAIL PROTECTED] mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide! > http://www.R-project.org/posting-guide.html ______________________________________________ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html