Type III sum of squares for ANOVA are readily available through the `Anova()`

function from the car package.

Contrast coding can be done in several ways, using `C()`

, the `contr.*`

family (as indicated by @nico), or directly the `contrasts()`

function/argument. This is detailed in §6.2 (pp. 144-151) of *Modern Applied Statistics with S* (Springer, 2002, 4th ed.). Note that `aov()`

is just a wrapper function for the `lm()`

function. It is interesting when one wants to control the error term of the model (like in a within-subject design), but otherwise they both yield the same results (and whatever the way you fit your model, you still can output ANOVA or LM-like summaries with `summary.aov`

or `summary.lm`

).

I don't have SPSS to compare the two outputs, but something like

```
> library(car)
> sample.data <- data.frame(IV=factor(rep(1:4,each=20)),
DV=rep(c(-3,-3,1,3),each=20)+rnorm(80))
> Anova(lm1 <- lm(DV ~ IV, data=sample.data,
contrasts=list(IV=contr.poly)), type="III")
Anova Table (Type III tests)
Response: DV
Sum Sq Df F value Pr(>F)
(Intercept) 18.08 1 21.815 1.27e-05 ***
IV 567.05 3 228.046 < 2.2e-16 ***
Residuals 62.99 76
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
```

is worth to try in first instance.

About factor coding in R vs. SAS: R considers the baseline or reference level as the first level in lexicographic order, whereas SAS considers the last one. So, to get comparable results, either you have to use `contr.SAS()`

or to `relevel()`

your R factor.

What you are calling type II SS, I would call type III SS. Lets imagine that there are just two factors A and B (and we'll throw in the A*B interaction later to distinguish type II SS). Further, lets imagine that there are different $n$s in the four cells (e.g., $n_{11}$=11, $n_{12}$=9, $n_{21}$=9, and $n_{22}$=11). Now your two factors are correlated with each other. (Try this yourself, make 2 columns of 1's and 0's and correlate them, $r=.1$; n.b. it doesn't matter if $r$ is 'significant', this is the whole population that you care about). The problem with your factors being correlated is that there are sums of squares that are associated with *both* A and B. When computing an ANOVA (or any other linear regression), we want to *partition* the sums of squares. A partition puts all sums of squares into *one and only one* of several subsets. (For example, we might want to divide the SS up into A, B and error.) However, since your factors (still only A and B here) are not orthogonal there is no unique partition of these SS. In fact, there can be very many partitions, and if you are willing to slice your SS up into fractions (e.g., "I'll put .5 into this bin and .5 into that one"), there are infinite partitions. A way to visualize this is to imagine the MasterCard symbol: The rectangle represents the total SS, and each of the circles represents the SS that are attributable to that factor, but notice the overlap between the circles in the center, those SS could be given to either circle.

The question is: How are we to choose the 'right' partition out of all of these possibilities? Let's bring the interaction back in and discuss some possibilities:

## Type I SS:

- SS(A)
- SS(B|A)
- SS(A*B|A,B)

## Type II SS:

- SS(A|B)
- SS(B|A)
- SS(A*B|A,B)

## Type III SS:

- SS(A|B,A*B)
- SS(B|A,A*B)
- SS(A*B|A,B)

Notice how these different possibilities work. Only type I SS actually uses those SS in the overlapping portion between the circles in the MasterCard symbol. That is, the SS that could be attributed to either A or B, *are* actually attributed to one of them when you use type I SS (specifically, the one you entered into the model first). In both of the other approaches, the overlapping SS are not used *at all*. Thus, type I SS gives to A all the SS attributable to A (including those that could also have been attributed elsewhere), then gives to B all of the *remaining* SS that are attributable to B, then gives to the A*B interaction all of the *remaining* SS that are attributable to A*B, and leaves the left-overs that couldn't be attributed to anything to the error term.

Type III SS only gives A those SS that are *uniquely* attributable to A, likewise it only gives to B and the interaction those SS that are *uniquely* attributable to them. The error term only gets those SS that couldn't be attributed to any of the factors. Thus, those 'ambiguous' SS that could be attributed to 2 or more possibilities are not used. If you sum the type III SS in an ANOVA table, you will notice that they do not equal the total SS. In other words, this analysis *must* be wrong, but errs in a kind of epistemically conservative way. Many statisticians find this approach egregious, however government funding agencies (I believe the FDA) requires their use.

The type II approach is intended to capture what might be worthwhile about the idea behind type III, but mitigate against its excesses. Specifically, it only adjusts the SS for A and B for each other, not the interaction. However, in practice type II SS is essentially never used. You would need to know about all of this and be savvy enough with your software to get these estimates, and the analysts who are typically think this is bunk.

There are more types of SS (I believe IV and V). They were suggested in the late 60's to deal with certain situations, but it was later shown that they do not do what was thought. Thus, at this point they are just a historical footnote.

As for what questions these are answering, you basically have that right already in your question:

- Estimates using type I SS tell you how much of the variability in Y can be explained by A, how much of the residual variability can be explained by B, how much of the remaining residual variability can be explained by the interaction, and so on,
*in order*.
- Estimates based on type III SS tell you how much of the residual variability in Y can be accounted for by A
*after* having accounted for everything else, and how much of the residual variability in Y can be accounted for by B *after* having accounted for everything else as well, and so on. (Note that both go both first and last simultaneously; if this makes sense to you, and accurately reflects your research question, then use type III SS.)

## Best Answer

I realize this is a year old post but its likely to come up again.

There are many factors that play into this, I'd argue the most important is your hypothesis. So there is no clear answer but I generally follow these rules-of-thumb:

1) Type II is only when you don't have an interaction term.

2) Type I vs Type III to test the interaction term...I go Type I all the time reason is this

Type I SS for dv ~ A + B + A*B depends on order so...its sequential SS(A) SS(B|A) SS(A*B|A B)

This is great to test your interaction...but not great to test the main effects since the effect of B is dependent on A.

Type II gets around this SS(A|B) SS(B|A)

Which looks great to test your main effects IF there is no interaction term.

Type III: SS(A| A*B B) SS(B| A*B A)

Which given the most common hypotheses...doesn't seem very useful since most people are interested in the interaction term, not the main effects when an interaction term is present.

So in this case I'd use Type-I to test the interaction term. If not significant I'd refit without the interaction term and use Type-II to test the main effects.

warning: anova() in R is Type-I, to get Type-II (or III) use the Anova() in the car package.