Hello,
I currently run aov in the following way:
> throughput.aov <-
aov(log(Throughput)~No_databases+Partitioning+No_middlewares+Queue_size,data=throughput)
> summary(throughput.aov)
Df Sum Sq Mean Sq F value Pr(>F)
No_databases 1 184.68 184.675 136.6945 < 2.2e-16 ***
Partitioning 1 70.16 70.161 51.9321 2.516e-12 ***
No_middlewares 2 44.22 22.110 16.3654 1.395e-07 ***
Queue_size 1 0.40 0.395 0.2926 0.5888
Residuals 440 594.44 1.351
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
In order to compute the fraction of variation I need to know the total Sum Sq.
and I assume it is like this:
SST = SS-No_databases + SS-Partitioning + SS-No_middlewares + SS-Queue_size
= 184.68 + 70.16 + 44.22 + 0.40
= 299.46
So the fraction of variation explained by the No_databases would be:
SST/SS-No_databases = 184.68/299.46 = 0.6167101 ... and finally I can say that
the No_databases explains 61.6% of the variation in Throughput.
Is this correct? if so, how can I do the same calculations using R? I
haven't found the way to extract the Sum Sq out of the throughput.aov
Object. Is there a function to get this 0.6167101 and 61.6% results without
having to do it manually? even better if I get a table containing all these
fraction of variations?
Since this is a 2^k experiment, I cant see how the Residuals fit in the formula.
When I introduce replications (blocking factor) then I can also include a SSE
term into the SST calculation.
TIA,
Best regards,
Giovanni
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