Interpreting summary.lm for a 2 factor anova

As Petr Pikal mentioned, the difficulty in interpretation is entirely due
to the set of contrasts you chose.The default treatment contrasts are
not orthogonal and are therefore the most difficult to interpret.
The note in ?aov warns of this difficulty.

sum contrasts will give you numbers that are easiest to interpret.


options(contrasts = c("contr.sum", "contr.poly"))
warpbreakssum.aov <- aov(breaks ~ wool * tension, data = warpbreaks)
coef(warpbreakssum.aov)
model.tables(warpbreakstreatment.aov, type="effects")
model.tables(warpbreakstreatment.aov, type="means")


John Fox showed the algebra using the default treatment contrasts

For full understanding you will need to read  in a text more about
sets of linear contrasts and their algebra.
I recommend Section 10.3 in mine, of course.

Statistical Analysis and Data Display:
An Intermediate Course with Examples in R
Heiberger, Richard M., Holland, Burt

http://www.springer.com/us/book/9781493921218

On Sat, Dec 3, 2016 at 11:46 PM, Ashim Kapoor <ashimkapoor at gmail.com>
wrote:

On Sun, Dec 4, 2016 at 10:03 AM, Ashim Kapoor <ashimkapoor at gmail.com>

wrote:

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.

This should be : y1 + ... yn = n . intercept

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

This should be : y_1 + ... +y_n = 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
Canada L8S 4M4
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

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