# [R] Interpreting summary.lm for a 2 factor anova

Ashim Kapoor ashimkapoor at gmail.com
Sun Dec 4 05:33:07 CET 2016

```Dear Sir,

Many thanks for the explanation. Prior to your email (with some help from a
friend of mine) I was able to figure this one out. If we look at the model
: -

y = intercept + B1.woolB + B2. tensionM + B3.tensionH + B4. woolB.TensionM
+ B5.woolB.TensionH + error

Here woolB, tensionM, tensionH are the dummy indicator variables similar to
how you have defined them.

Now suppose we consider y1,..,yn, all in group A.L (say).

Then y1 + ... + yn = intercept => average(y1,...,yn) = intercept + 0 + 0 +
0 + 0 + 0.

What was confusing me was how to compute the cell mean in woolB,tensionH
cell.

If we have y_1,...,y_n all in group B.H then :-

y_1+ ... + y_n = intercept + B1 + 0 + B3 + 0 +  B5

Therefore average of group B.H = intercept + B1 + B3 + B5

Many thanks and Best Regards,
Ashim

On Sat, Dec 3, 2016 at 7:15 PM, Fox, John <jfox at mcmaster.ca> wrote:

> Dear Ashim,
>
> Sorry to chime in late, and my apologies if someone has already pointed
> this out, but here's the relationship between the cell means and the model
> coefficients, using the row-basis of the model matrix:
>
> -------------------------- snip ------------------------
>
> > means <- with( warpbreaks, tapply( breaks, interaction(wool, tension),
> mean ) )
> > x.A <- rep(c(0, 1), 3)
> > x.B1 <- rep(c(0, 1, 0), each=2)
> > x.B2 <- rep(c(0, 0, 1), each=2)
> > x.AB1 <- x.A*x.B1
> > x.AB2 <- x.A*x.B2
> > X.basis <- cbind(1, x.A, x.B1, x.B2, x.AB1, x.AB2)
> > X.basis
>        x.A x.B1 x.B2 x.AB1 x.AB2
> [1,] 1   0    0    0     0     0
> [2,] 1   1    0    0     0     0
> [3,] 1   0    1    0     0     0
> [4,] 1   1    1    0     1     0
> [5,] 1   0    0    1     0     0
> [6,] 1   1    0    1     0     1
> > solve(X.basis, means)
>                 x.A      x.B1      x.B2     x.AB1     x.AB2
>  44.55556 -16.33333 -20.55556 -20.00000  21.11111  10.55556
> > coef(aov(breaks ~ wool * tension, data = warpbreaks))
>    (Intercept)          woolB       tensionM       tensionH woolB:tensionM
>       44.55556      -16.33333      -20.55556      -20.00000       21.11111
> woolB:tensionH
>       10.55556
>
> -------------------------- snip ------------------------
>
> I hope this helps,
>  John
>
> -----------------------------
> John Fox, Professor
> McMaster University
> Hamilton, Ontario
> Web: socserv.mcmaster.ca/jfox
>
>
>
> > -----Original Message-----
> > From: R-help [mailto:r-help-bounces at r-project.org] On Behalf Of Ashim
> Kapoor
> > Sent: December 3, 2016 12:19 AM
> > To: David Winsemius <dwinsemius at comcast.net>
> > Cc: r-help at r-project.org
> > Subject: Re: [R] Interpreting summary.lm for a 2 factor anova
> >
> > Please allow me to rephrase myquery.
> >
> > > model.tables(model,"m")
> > Tables of means
> > Grand mean
> >
> > 28.14815
> >
> >  wool
> > wool
> >      A      B
> > 31.037 25.259
> >
> >  tension
> > tension
> >     L     M     H
> > 36.39 26.39 21.67
> >
> >  wool:tension
> >     tension
> > wool L     M     H
> >    A 44.56 24.00 24.56
> >    B 28.22 28.78 18.78
> > >
> >
> >
> > The above is the same as :
> >
> > with( warpbreaks, tapply( breaks, interaction(wool, tension), mean ) )
> >      A.L      B.L      A.M      B.M      A.H      B.H
> > 44.55556 28.22222 24.00000 28.77778 24.55556 18.77778
> >
> > For reference:
> >
> > > model <- aov(breaks ~ wool * tension, data = warpbreaks)
> > > summary.lm(model)
> >
> > Call:
> > aov(formula = breaks ~ wool * tension, data = warpbreaks)
> >
> > Residuals:
> >      Min       1Q   Median       3Q      Max
> > -19.5556  -6.8889  -0.6667   7.1944  25.4444
> >
> > Coefficients:
> >                Estimate Std. Error t value Pr(>|t|)
> > (Intercept)      44.556      3.647  12.218 2.43e-16 ***
> > woolB           -16.333      5.157  -3.167 0.002677 **
> > tensionM        -20.556      5.157  -3.986 0.000228 ***
> > tensionH        -20.000      5.157  -3.878 0.000320 ***
> > woolB:tensionM   21.111      7.294   2.895 0.005698 **
> > woolB:tensionH   10.556      7.294   1.447 0.154327
> > ---
> > Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
> >
> > Residual standard error: 10.94 on 48 degrees of freedom
> > Multiple R-squared:  0.3778,    Adjusted R-squared:  0.3129
> > F-statistic: 5.828 on 5 and 48 DF,  p-value: 0.0002772
> >
> >
> > Now I'll explain what is confusing me in the output of summary.lm.
> >
> > Coeff of Intercept = 44.556  = cell mean for A.L. This is the base.
> >
> > Coeff of woolB:L = -16.333 = 28.22222 - 44.556. This is the difference
> of this
> > cell mean(B:L) from the base.
> >
> > Coeff of woolA:tensionM = -20.556  = 24.000- 44.556. This is the
> difference of
> > this cell mean (A:M)  from the base.
> >
> > Coeff of woolA:tensionH = -20.000  = 24.55556 - 44.556. This is the
> difference
> > of this cell mean(A:H) from the base.
> >
> > This is where it stops being the difference from the base.
> >
> > Coeff of woolB:tensionM = 21.111 should turn out to be 28.77778 - 44.556
> but
> > this is -15.77822
> >
> > Coeff of woolB:tensionH = 10.556 should turn out to be  18.77778 -
> 44.556 but
> > this is -25.77822
> >
> > In the above 2 cases, we can't say that the coefficient = cell mean -
> base case.
> > Can you tell me what should be the statement to be made ?
> >
> >
> > Best Regards,
> > Ashim
> >
> > PS : My apologies for emailing my query to this list. Can you tell me
> the names
> > of a few (active) statistics help list ?
> >
> > On Sat, Dec 3, 2016 at 1:33 AM, David Winsemius <dwinsemius at comcast.net>
> > wrote:
> >
> > >
> > > > On Dec 2, 2016, at 9:09 AM, David Winsemius <dwinsemius at comcast.net>
> > > wrote:
> > > >
> > > >>
> > > >> On Dec 2, 2016, at 6:16 AM, Ashim Kapoor <ashimkapoor at gmail.com>
> > wrote:
> > > >>
> > > >> Dear Pikal,
> > > >>
> > > >> All levels except the interactions are compared to the Intercept.
> > > >> I'm a little confused as to what's going on in interaction terms
> > > >> eg. the cell wool B : tension M. It's mean is :
> > > >> 28.78 and 28.78 - 44.56 = -15.78 != 21.111.
> > > >>
> > > >> It's something like 44.56 (intercept) -16.333 (wool B) -.20.556
> > > >> (tension
> > > >> M)  + 21.111 (woolB:tensionM) = 28.782.
> > > >>
> > > >> I don't know how to sum up the above line in terms of differences
> > > >> succinctly.
> > > >
> > > > The aov estimate will not exactly equal the observed mean (this is
> > > _statistics_ after all). You should be comparing the mean of that cell
> > > to the estimate:
> > > >
> > > > 44.556 + (-16.33) +(-20.556) + (21.11)
> > >
> > > A respected participant advised me to look at this more closely. In
> > > this case (and I think in most such cases)  where there are the same
> > > number of parameters as there are means, the model is "saturated" and
> > > there is no
> > > difference:
> > >
> > >  with( warpbreaks, tapply( breaks, interaction(wool, tension), mean ) )
> > >      A.L      B.L      A.M      B.M      A.H      B.H
> > > 44.55556 28.22222 24.00000 28.77778 24.55556 18.77778
> > >
> > > So the B:M estimate is identical up to rounding with the observed mean:
> > >
> > >  44.556 + (-16.33) +(-20.556) + (21.11) [1] 28.78
> > >
> > >
> > >
> > > >
> > > > The difference between the observed mean and the estimated mean is
> > known
> > > as a 'residual'
> > >
> > > I've also been privately but gently chided for this misstatement.
> > > Residuals are the difference between data and estimates.
> > >
> > > > and the squared sum of the all residuals is what this being minimized
> > > ... over all the cells including the one implicitly associated with the
> > > Intercept.
> > > >
> > > > This isn't really on-topic for Rhelp since you are not having
> difficulty
> > > in getting the R program to perform its duties, but are rather in need
> of
> > > statistical education. That not what this mailing list is set up for.
> > > >
> > > > --
> > > > David.
> > > >
> > > >>
> > > >>>
> > > >>>> -----Original Message-----
> > > >>>> From: R-help [mailto:r-help-bounces at r-project.org] On Behalf Of
> Ashim
> > > >>>> Kapoor
> > > >>>> Sent: Thursday, December 1, 2016 2:48 PM
> > > >>>> To: r-help at r-project.org
> > > >>>> Subject: [R] Interpreting summary.lm for a 2 factor anova
> > > >>>>
> > > >>>> Dear all,
> > > >>>>
> > > >>>> Here is a small example : -
> > > >>>>
> > > >>>>> model <- aov(breaks ~ wool * tension, data = warpbreaks)
> > > >>>>> summary.lm(model)
> > > >>>>
> > > >>>> Call:
> > > >>>> aov(formula = breaks ~ wool * tension, data = warpbreaks)
> > > >>>>
> > > >>>> Residuals:
> > > >>>>    Min       1Q   Median       3Q      Max
> > > >>>> -19.5556  -6.8889  -0.6667   7.1944  25.4444
> > > >>>>
> > > >>>> Coefficients:
> > > >>>>              Estimate Std. Error t value Pr(>|t|)
> > > >>>> (Intercept)      44.556      3.647  12.218 2.43e-16 ***
> > > >>>> woolB           -16.333      5.157  -3.167 0.002677 **
> > > >>>> tensionM        -20.556      5.157  -3.986 0.000228 ***
> > > >>>> tensionH        -20.000      5.157  -3.878 0.000320 ***
> > > >>>> woolB:tensionM   21.111      7.294   2.895 0.005698 **
> > > >>>> woolB:tensionH   10.556      7.294   1.447 0.154327
> > > >>>> ---
> > > >>>> Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
> > > >>>>
> > > >>>> Residual standard error: 10.94 on 48 degrees of freedom
> > > >>>> Multiple R-squared:  0.3778,    Adjusted R-squared:  0.3129
> > > >>>> F-statistic: 5.828 on 5 and 48 DF,  p-value: 0.0002772
> > > >>>>
> > > >>>>> model.tables(model,"e")
> > > >>>> Tables of effects
> > > >>>>
> > > >>>> wool
> > > >>>> wool
> > > >>>>     A       B
> > > >>>> 2.8889 -2.8889
> > > >>>>
> > > >>>> tension
> > > >>>> tension
> > > >>>>    L      M      H
> > > >>>> 8.241 -1.759 -6.481
> > > >>>>
> > > >>>> wool:tension
> > > >>>>   tension
> > > >>>> wool L      M      H
> > > >>>>  A  5.278 -5.278  0.000
> > > >>>>  B -5.278  5.278  0.000
> > > >>>>
> > > >>>>
> > > >>>>> model.tables(model,"m")
> > > >>>> Tables of means
> > > >>>> Grand mean
> > > >>>>
> > > >>>> 28.14815
> > > >>>>
> > > >>>> wool
> > > >>>> wool
> > > >>>>    A      B
> > > >>>> 31.037 25.259
> > > >>>>
> > > >>>> tension
> > > >>>> tension
> > > >>>>   L     M     H
> > > >>>> 36.39 26.39 21.67
> > > >>>>
> > > >>>> wool:tension
> > > >>>>   tension
> > > >>>> wool L     M     H
> > > >>>>  A 44.56 24.00 24.56
> > > >>>>  B 28.22 28.78 18.78
> > > >>>>>
> > > >>>>
> > > >>>> I don't follow the output of summary.lm. I understand the output
> of
> > > >>>> model.tables for effects and means. For instance what does 44.556
> > > >>>> represent ? Is it the grand average ? The grand mean is 28.14815.
> Can
> > > >>>> someone help me understand the output of summary.lm ?
> > > >>>>
> > > >>>> Best Regards,
> > > >>>> Ashim
> > > >>>>
> > > >>>>     [[alternative HTML version deleted]]
> > > >>>>
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