Hello, I currently get anova results out of the aov function (see below) I use the model.tables and I believe it gives me back the model parameters of the fit (betas), however I don't see the intercept (beta_0) and don't understand what the "rep" output means and there is no description in the documentation.
Another question: is there a function that outputs the results in a more meaningful way e.g. show the percentage of variation of each factor towards the response I believe the formula would be something like: for factor X: (Sum_Sq_X / Sum_Sq_Total)*100 Thanks in advance, Best regards, Giovanni > #throughput.aov <- > aov(Throughput~No_databases*Partitioning*No_middlewares*Queue_size,data=throughput) > throughput.aov Call: aov(formula = Throughput ~ No_databases + Partitioning + No_middlewares + Queue_size, data = throughput) Terms: No_databases Partitioning No_middlewares Queue_size Residuals Sum of Squares 43146975 7394 9061130 20710 195504055 Deg. of Freedom 1 1 2 1 433 Residual standard error: 671.9453 Estimated effects may be unbalanced > summary(throughput.aov) Df Sum Sq Mean Sq F value Pr(>F) No_databases 1 43146975 43146975 95.5614 < 2.2e-16 *** Partitioning 1 7394 7394 0.0164 0.8982 No_middlewares 2 9061130 4530565 10.0342 5.497e-05 *** Queue_size 1 20710 20710 0.0459 0.8305 Residuals 433 195504055 451511 --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 > model.tables(throughput.aov,type="effects",se=TRUE) Design is unbalanced - use se.contrast() for se's Tables of effects No_databases 1 4 -317.1 310 rep 217.0 222 Partitioning sharding replication 4.303 -3.91 rep 209.000 230.00 No_middlewares 1 2 4 -97.93 -108.2 199 rep 139.00 150.0 150 Queue_size 40 100 -6.852 6.883 rep 220.000 219.000 > ______________________________________________ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.