[R] question about contrast in R for multi-factor linear regression models?
Michael
comtech.usa at gmail.com
Mon Nov 10 03:28:22 CET 2008
Hi all,
I am using "lm" to fit some anova factor models with interactions.
The default setting for my unordered factors is "treatment". I
understand the resultant "lm" coefficients for one factors, but when
it comes to the interaction term, I got confused.
> options()$contrasts
unordered ordered
"contr.treatment" "contr.poly"
Here is my question:
Factor A has 6 levels, B has 2 levels,
> levels(dd$A)=c("A1", "A2", "A3", "A4", "A5", "A6")
> levels(dd$B)=c("b1", "b2")
My question is how to interpret the resultant coefficients. What are
the bases of "dd$AA2:dd$Bb2" and "dd$AA3:dd$Bb2", etc. ?
I am having a hard time to understand the result and making sense out
of the numbers...
Please help me ! Thank you!
> zz=lm(formula = (dd$Y) ~ dd$A * dd$B)
> summary(zz)
Call:
lm(formula = dd$Y~ dd$A * dd$B)
Residuals:
Min 1Q Median 3Q Max
-1.68582 -0.42469 -0.02536 0.20012 3.50798
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 4.40842 0.40295 10.940 5.34e-13 ***
dd$AA2 0.11575 0.56986 0.203 0.8402
dd$AA3 0.01312 0.56986 0.023 0.9818
dd$AA4 -0.06675 0.56986 -0.117 0.9074
dd$AA5 0.10635 0.56986 0.187 0.8530
dd$AA6 0.11507 0.56986 0.202 0.8411
dd$Bb2 -0.58881 0.56986 -1.033 0.3084
dd$AA2:dd$Bb2 0.26465 0.80590 0.328 0.7445
dd$AA3:dd$Bb2 0.40984 0.80590 0.509 0.6142
dd$AA4:dd$Bb2 -0.02918 0.80590 -0.036 0.9713
dd$AA5:dd$Bb2 0.35574 0.80590 0.441 0.6616
dd$AA6:dd$Bb2 1.55424 0.80590 1.929 0.0617 .
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Residual standard error: 0.8059 on 36 degrees of freedom
Multiple R-squared: 0.2642, Adjusted R-squared: 0.03934
F-statistic: 1.175 on 11 and 36 DF, p-value: 0.3378
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