# [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

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
>
```