I am trying to use R lm() with quantitative and qualitative predictors, but am
getting different results than those that I get in SAS.
In the R ANOVA table documentation I see that "Type-II tests corresponds to the
tests produced by SAS for analysis-of-variance models, where all of the
predictors are factors, but not more generally (i.e., when there are
quantitative predictors)." Is this the problem? Is there a way around this so
that the output matches the results that I am getting in SAS? Is there a
better/more appropriate way to handle quantitative predictors?
For example, output from SAS for one of my response variables:
Source DF Type III SS Mean Square F Value Pr > F
Treatment 5 6.92081345 1.38416269 1.17 0.3379
Plantation 2 20.67107509 10.33553754 8.73 0.0006
Treat*Plant 10 22.23762534 2.22376253 1.88 0.0718
Light 1 2.52841104 2.52841104 2.14 0.1504
Litter 1 0.25885309 0.25885309 0.22 0.6421
Output from R
Df Sum Sq Mean Sq F value Pr(>F)
daniela2$Treatment 5 6.982 1.396 1.1799 0.332914
daniela2$Plantation 2 18.905 9.453 7.9869 0.001013 **
daniela2$Light 1 8.037 8.037 6.7909 0.012167 *
daniela2$Litter 1 1.912 1.912 1.6154 0.209860
Treatment:Plantation 10 22.238 2.224 1.8790 0.071831 .
Residuals 48 56.808 1.184
For example, multivariate output from SAS for one of my predictors:
Statistic Value F Value Num DF Den DF Pr > F
Wilks' Lambda 0.69901456 1.18 15 127.39 0.2984
Pillai's Trace 0.32156466 1.15 15 144 0.3159
Hotelling 0.40147329 1.21 15 81.797 0.2835
Roy's GRoot 0.31281143 3.00 5 48 0.0194
>From R:
Df test stat approx F num Df den Df Pr(>F)
Pillai 5.0000 0.401037 1.481343 15.0000 144.0000 0.1193910
Wilks 5.0000 0.636571 1.509595 15.0000 127.3871 0.1109964
Hotelling-Lawley 5.0000 0.513082 1.527844 15.0000 134.0000 0.1037152
Roy 5.0000 0.368760 3.540095 5.0000 48.0000 0.0083553 **
Thank you,
Ali
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