Chadwick Robbert Chadwick Robbert - 6 days ago 6
R Question

When should I use aov() and when anova()?

I have referred to much of online literature but it is increasing my confusion. Much of the discussion is too technical with terms unbalanced designs and I, II or III factor ANOVA and everything.

I only know that

aov()
uses
lm()
internally and is useful for data with factors. Whereas
anova()
can be used for different models on same dataset.
Is my understanding correct?

Answer

anova is substantially different from aov. Why not read R's documentation ?aov and ?anova? In short:

  • aov fits a model (as you are already aware, internally it calls lm), so it produces regression coefficients, fitted values, residuals, etc; It produces an object of primary class "aov" but also a secondary class "lm". So, it is an augmentation of an "lm" object.
  • anova is a generic function. In your scenario you are referring to anova.lm or anova.lmlist (read ?anova.lm for more info). The former analyses a fitted model (produced by lm or aov), while the latter analyses several nested (increasingly large) fitted models (by lm or aov). They both aim at producing type I (sequential) ANOVA table.

In practice, you first use lm / aov to fit a model, then use anova to analyse the result. There is nothing better than trying a small example:

fit <- aov(sr ~ ., data = LifeCycleSavings)  ## can also use `lm`
z <- anova(fit)

Now, have a look at their structure. aov returns a large object:

str(fit)

#List of 12
# $ coefficients : Named num [1:5] 28.566087 -0.461193 -1.691498 -0.000337 0.409695
#  ..- attr(*, "names")= chr [1:5] "(Intercept)" "pop15" "pop75" "dpi" ...
# $ residuals    : Named num [1:50] 0.864 0.616 2.219 -0.698 3.553 ...
#  ..- attr(*, "names")= chr [1:50] "Australia" "Austria" "Belgium" "Bolivia" ...
# $ effects      : Named num [1:50] -68.38 -14.29 7.3 -3.52 -7.94 ...
#  ..- attr(*, "names")= chr [1:50] "(Intercept)" "pop15" "pop75" "dpi" ...
# $ rank         : int 5
# $ fitted.values: Named num [1:50] 10.57 11.45 10.95 6.45 9.33 ...
#  ..- attr(*, "names")= chr [1:50] "Australia" "Austria" "Belgium" "Bolivia" ...
# $ assign       : int [1:5] 0 1 2 3 4
# $ qr           :List of 5
#  ..$ qr   : num [1:50, 1:5] -7.071 0.141 0.141 0.141 0.141 ...
#  .. ..- attr(*, "dimnames")=List of 2
#  .. .. ..$ : chr [1:50] "Australia" "Austria" "Belgium" "Bolivia" ...
#  .. .. ..$ : chr [1:5] "(Intercept)" "pop15" "pop75" "dpi" ...
#  .. ..- attr(*, "assign")= int [1:5] 0 1 2 3 4
#  ..$ qraux: num [1:5] 1.14 1.17 1.16 1.15 1.05
#  ..$ pivot: int [1:5] 1 2 3 4 5
#  ..$ tol  : num 1e-07
#  ..$ rank : int 5
#  ..- attr(*, "class")= chr "qr"
# $ df.residual  : int 45
# $ xlevels      : Named list()
# $ call         : language aov(formula = sr ~ ., data = LifeCycleSavings)
# $ terms        :Classes 'terms', 'formula'  language sr ~ pop15 + pop75 + dpi + ddpi
#  .. ..- attr(*, "variables")= language list(sr, pop15, pop75, dpi, ddpi)
#  .. ..- attr(*, "factors")= int [1:5, 1:4] 0 1 0 0 0 0 0 1 0 0 ...
#  .. .. ..- attr(*, "dimnames")=List of 2
#  .. .. .. ..$ : chr [1:5] "sr" "pop15" "pop75" "dpi" ...
#  .. .. .. ..$ : chr [1:4] "pop15" "pop75" "dpi" "ddpi"
#  .. ..- attr(*, "term.labels")= chr [1:4] "pop15" "pop75" "dpi" "ddpi"
#  .. ..- attr(*, "order")= int [1:4] 1 1 1 1
#  .. ..- attr(*, "intercept")= int 1
#  .. ..- attr(*, "response")= int 1
#  .. ..- attr(*, ".Environment")=<environment: R_GlobalEnv> 
#  .. ..- attr(*, "predvars")= language list(sr, pop15, pop75, dpi, ddpi)
#  .. ..- attr(*, "dataClasses")= Named chr [1:5] "numeric" "numeric" "numeric" "numeric" ...
#  .. .. ..- attr(*, "names")= chr [1:5] "sr" "pop15" "pop75" "dpi" ...
# $ model        :'data.frame':	50 obs. of  5 variables:
#  ..$ sr   : num [1:50] 11.43 12.07 13.17 5.75 12.88 ...
#  ..$ pop15: num [1:50] 29.4 23.3 23.8 41.9 42.2 ...
#  ..$ pop75: num [1:50] 2.87 4.41 4.43 1.67 0.83 2.85 1.34 0.67 1.06 1.14 ...
#  ..$ dpi  : num [1:50] 2330 1508 2108 189 728 ...
#  ..$ ddpi : num [1:50] 2.87 3.93 3.82 0.22 4.56 2.43 2.67 6.51 3.08 2.8 ...
#  ..- attr(*, "terms")=Classes 'terms', 'formula'  language sr ~ pop15 + pop75 + dpi + ddpi
#  .. .. ..- attr(*, "variables")= language list(sr, pop15, pop75, dpi, ddpi)
#  .. .. ..- attr(*, "factors")= int [1:5, 1:4] 0 1 0 0 0 0 0 1 0 0 ...
#  .. .. .. ..- attr(*, "dimnames")=List of 2
#  .. .. .. .. ..$ : chr [1:5] "sr" "pop15" "pop75" "dpi" ...
#  .. .. .. .. ..$ : chr [1:4] "pop15" "pop75" "dpi" "ddpi"
#  .. .. ..- attr(*, "term.labels")= chr [1:4] "pop15" "pop75" "dpi" "ddpi"
#  .. .. ..- attr(*, "order")= int [1:4] 1 1 1 1
#  .. .. ..- attr(*, "intercept")= int 1
#  .. .. ..- attr(*, "response")= int 1
#  .. .. ..- attr(*, ".Environment")=<environment: R_GlobalEnv> 
#  .. .. ..- attr(*, "predvars")= language list(sr, pop15, pop75, dpi, ddpi)
#  .. .. ..- attr(*, "dataClasses")= Named chr [1:5] "numeric" "numeric" "numeric" "numeric" ...
#  .. .. .. ..- attr(*, "names")= chr [1:5] "sr" "pop15" "pop75" "dpi" ...
# - attr(*, "class")= chr [1:2] "aov" "lm"

While anova returns:

str(z)

#Classes ‘anova’ and 'data.frame':  5 obs. of  5 variables:
# $ Df     : int  1 1 1 1 45
# $ Sum Sq : num  204.1 53.3 12.4 63.1 650.7
# $ Mean Sq: num  204.1 53.3 12.4 63.1 14.5
# $ F value: num  14.116 3.689 0.858 4.36 NA
# $ Pr(>F) : num  0.000492 0.061125 0.359355 0.042471 NA
# - attr(*, "heading")= chr  "Analysis of Variance Table\n" "Response: sr"
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