> Df <- data.frame(x=1:9, y=rep(c(-1,1), length=9)) > anova(lm(y~x, Df)) Analysis of Variance Table
Response: y
Df Sum Sq Mean Sq F value Pr(>F)
x 1 2.861e-34 2.861e-34 2.253e-34 1
Residuals 7 8.8889 1.2698
> anova(lm(y~x+I(x^2), Df))
Analysis of Variance TableResponse: y
Df Sum Sq Mean Sq F value Pr(>F)
x 1 2.861e-34 2.861e-34 2.065e-34 1.0000
I(x^2) 1 0.5772 0.5772 0.4167 0.5425
Residuals 6 8.3117 1.3853
>
> Df <- data.frame(x=1:9, y=rep(c(-1,1), length=9))
> anova(lm(y~x, Df))
Analysis of Variance TableResponse: y
Df Sum Sq Mean Sq F value Pr(>F)
x 1 2.861e-34 2.861e-34 2.253e-34 1
Residuals 7 8.8889 1.2698
> anova(lm(y~x+I(x^2), Df))
Analysis of Variance TableResponse: y
Df Sum Sq Mean Sq F value Pr(>F)
x 1 2.861e-34 2.861e-34 2.065e-34 1.0000
I(x^2) 1 0.5772 0.5772 0.4167 0.5425
Residuals 6 8.3117 1.3853
> anova(lm(y~I(x^2)+x, Df))
Analysis of Variance TableResponse: y
Df Sum Sq Mean Sq F value Pr(>F)
I(x^2) 1 0.0282 0.0282 0.0203 0.8912
x 1 0.5490 0.5490 0.3963 0.5522
Residuals 6 8.3117 1.3853
>
In S-Plus 6.1, the ANOVA table is preceeded by a statement, "Terms added sequentially (first to last)". From these examples, it certainly looks like this is what it is doing. Apart from round off error, the sum of squares and mean squares are identical for the models without and with I(x^2). In an example with a nonzero sum of squares for x, the F value would be different, because the mean square for residuals would be different, and the Pr(>F) would also be affected by differing degrees of freedom.
The third example here puts I(x^2) before x in the model statement and gets a clearly different anova. (The coefficients should be not change when the order of the terms is modified, though they could change if other terms are addeed. I didn't check that for this example, but I've done this before and would be surprised if they were different.)
Best Wishes, Spencer
Bjørn-Helge Mevik wrote:
I've used Anova() from the car package to get marginal (aka type II) sum-of-squares and tests for linear models with categorical variables. Is it possible to get marginal SSs also for continuous variables, when the model includes powers of the continuous variables?
For instance, if A and B are categorical ("factor"s) and x is continuous ("numeric"),
Anova (lm (y ~ A*B + x, ...))
will produce marginal SSs for all terms (A, B, A:B and x). However,
with
Anova (lm (y ~ A*B + x + I(x^2), ...))
the SS for 'x' is calculated with I(x^2) present in the model, i.e. it is no longer marginal.
Using poly (x, 2) instead of x + I(x^2), one gets a marginal SS for the total effect of x, but not for the linear and quadratic effects separately. (summary.aov() has a 'split' argument that can be used to get separate SSs, but these are not marginal.)
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