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I was always under the impression that regression is just a more general form of ANOVA and that the results would be identical. Recently, however, I have run both a regression and an ANOVA on the same data and the results differ significantly. That is, in the regression model both main effects and the interaction are significant, while in the ANOVA one main effect is not significant. I expect this has something to do with the interaction, but it's not clear to me what is different about these two ways of modeling the same question. If it's important, one predictor is categorical and the other is continuous, as indicated in the simulation below.

Here is an example of what my data looks like and what analyses I'm running, but without the same p-values or effects being significant in the results (my actual results are outlined above):

group<-c(1,1,1,0,0,0)
moderator<-c(1,2,3,4,5,6)
score<-c(6,3,8,5,7,4)

summary(lm(score~group*moderator))
summary(aov(score~group*moderator))
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  • $\begingroup$ summary(lm()) is giving you the coefficients for the contrasts you have specified, which are treatment contrasts in the absence of specification here. While summary(aov()) is giving you the anova table. If you want the anova for the lm model you need anova(lm()) $\endgroup$ Dec 20, 2011 at 6:17
  • $\begingroup$ group is a numerical vector, is this on purpose? Normally, grouping factors should have class factor, such that the transformation to contrasts can be handled automatically by functions like lm(). This will become apparent once you have more than two groups, or use a coding other than 0/1 for your group variable. $\endgroup$
    – caracal
    Dec 20, 2011 at 12:11
  • $\begingroup$ See also stats.stackexchange.com/questions/268006/… $\endgroup$ Dec 10, 2019 at 10:52

3 Answers 3

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The summary function calls different methods depending on the class of the object. The difference isn't in the aov vs lm, but in the information presented about the models. For example, if you used anova(mod1) and anova(mod2) instead, you should get the same results.

As @Glen says, the key is whether the tests reported are based on Type 1 or Type 3 sums of squares. These will differ when the correlation between your explanatory variables is not exactly 0. When they are correlated, some SS are unique to one predictor and some to the other, but some SS could be attributed to either or both. (You can visualize this by imagining the MasterCard symbol--there's a small region of overlap in the center.) There is no unique answer in this situation, and unfortunately, this is the norm for non-experimental data. One approach is for the analyst to use their judgment and assign the overlapping SS to one of the variables. That variable goes into the model first. The other variable goes into the model second and gets the SS that looks like a cookie with a bite taken out of it. It's effect can be tested by what is sometimes called $R^2$ change or F change. This approach uses Type 1 SS. Alternatively, you could do this twice with each going in first, and report the F change test for both predictors. In this way, neither variable gets the SS due to the overlap. This approach uses Type 3 SS. (I should also tell you that the latter approach is held in low regard.)

Following the suggestion of @BrettMagill in the comment below, I can try to make this a little clearer. (Note that, in my example, I'm using just 2 predictors and no interaction, but this idea can be scaled up to include whatever you like.)

Type 1: SS(A) and SS(B|A)

Type 3: SS(A|B) and SS(B|A)

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    $\begingroup$ This is a nice description of the problem. You might clarify the text a bit with this: Type I: SS_A = SS(A) SS_B = SS(B | A) and SS_AB = SS(AB | B, A) Type III: SS_A = SS(A | B, AB) and SS_B = SS(B | A, AB) and SS_AB = SS(AB|A, B) $\endgroup$
    – Brett
    Dec 19, 2011 at 21:48
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    $\begingroup$ Thank you so much for your help. I understand now what's going on in terms of how these models are different, but I'm still not clear on when it would appropriate to use either an anova or regression model. My advisor is advising anova, but I've always been taught to use regression and I'm not sure which is more appropriate to use when the results are divergent. Do you have any examples or a resource to advise on when either would be appropriate? Thanks again for your help. $\endgroup$
    – Rebecca
    Dec 31, 2011 at 22:40
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    $\begingroup$ I'm sorry, I don't quite follow. My point is that the models aren't actually different. An ANOVA is a regression with all qualitative predictors. If you have a regression model with continuous and qualitative predictors, and you enter the continuous predictor first, then the qualitative predictors (but without an interaction term) that's ANCOVA. Either approach is fine, since 'behind the scenes' they're identical. I usually code this as a regression, but that's a matter of style. OTOH, if your adviser wants it run ANOVA style, then go that route, as there is no difference. $\endgroup$ Jan 1, 2012 at 5:55
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    $\begingroup$ A few things: (3 up) an interaction does not mean your independent variables are correlated, these are just different things; (2 up) if model 3 is significantly better than model 2, then yes, this suggests the interaction is significant (since the interaction is the only thing that differs between them); (1 up) you want to avoid just fishing for significant effects unless you are thinking of your study as a pilot that you will use to plan a subsequent confirmatory study (in this case I think you're OK); I gather you ran this study to look at all three, thus go with model 3. $\endgroup$ Jan 2, 2012 at 19:15
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    $\begingroup$ In addition, an interaction implies that you should not interpret the main effects, thus presenting only model 1 could be dangerously misleading. If you want more info on types of SS, I wrote a fairly comprehensive answer here: stats.stackexchange.com/questions/20452/… Also, you should accept one of the answers, at some point, by clicking the check mark next to one of them. $\endgroup$ Jan 2, 2012 at 19:18
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The results from the aov output are giving you probabilities based on Type 1 sum of squares. This is why the interaction result is the same and the main effects differ.

If you use probabilities based on Type 3 sum of squares then they will match the linear regression results.

library(car)
Anova(aov(score~group*moderator),type=3)
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    $\begingroup$ Linear models and ANOVA will be equivalent when the models are testing the same hypotheses and when the parameterization of the factors is equivalent. So called "Type I" and "Type III" sums are squares are simply tests of different underlying hypotheses (effects of sequential sums of squares versus marginal sums of squares). ANOVA tends to hide some of these decisions as implemented in many packages--a fact that makes me believe that actually setting up and testing the hypotheses of interest through factor parameterization and model comparison in GLM is a superior approach. $\endgroup$
    – Brett
    Dec 19, 2011 at 20:34
  • $\begingroup$ +1, I think you have a typo, though. lm is using Type 1 SS and aov is using Type 3 SS. $\endgroup$ Dec 19, 2011 at 21:23
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    $\begingroup$ Type III (Marginal) Sums of Squares is used by default in lm. AOV would use Type I (Sequential) by default. LM results are invariant to order while aov results depend on the order of the factors. $\endgroup$
    – Brett
    Dec 19, 2011 at 21:28
  • $\begingroup$ I thought both lm and aov used type I by default, hence the use of capital A Anova() for type II and III. $\endgroup$ Dec 20, 2011 at 6:14
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    $\begingroup$ In general, Anova(..., type=3) will not give you correct type III SS, unless you also switch from treatment contrasts (default in R) to effect coding for unordered factors (options(contrasts=c("contr.sum", "contr.poly"))) or some other sum-to-zero contrast codes (e.g., Helmert). This will become apparent once you have unbalanced cell sizes and more than two groups and is also mentioned in the help page for Anova(). $\endgroup$
    – caracal
    Dec 20, 2011 at 11:47
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The main difference between linear regression and ANOVA is, in ANOVA the predictor variables are discrete (that is they have different levels). Whereas in linear regression, the predictor variables are continuous.

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    $\begingroup$ This is not generally true. $\endgroup$ Mar 28, 2017 at 12:10
  • $\begingroup$ I read it somewhere on the internet. Can you please explain the key difference. I am a newbie. $\endgroup$
    – vivek
    Mar 6, 2018 at 18:49

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