[R] aov output question
Giovanni Azua
bravegag at gmail.com
Mon Nov 14 12:01:48 CET 2011
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
>
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