[R] Interpreting summary.lm for a 2 factor anova
Ashim Kapoor
ashimkapoor at gmail.com
Sun Dec 4 05:46:25 CET 2016
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
>> > > >>>>
>> > > >>>> [[alternative HTML version deleted]]
>> > > >>>>
>> > > >>>> ______________________________________________
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>> > > > David Winsemius
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>> > > >
>> > > > ______________________________________________
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