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 <j...@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-boun...@r-project.org] On Behalf Of Ashim > Kapoor > > Sent: December 3, 2016 12:19 AM > > To: David Winsemius <dwinsem...@comcast.net> > > Cc: r-help@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 <dwinsem...@comcast.net> > > wrote: > > > > > > > > > On Dec 2, 2016, at 9:09 AM, David Winsemius <dwinsem...@comcast.net> > > > wrote: > > > > > > > >> > > > >> On Dec 2, 2016, at 6:16 AM, Ashim Kapoor <ashimkap...@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-boun...@r-project.org] On Behalf Of > Ashim > > > >>>> Kapoor > > > >>>> Sent: Thursday, December 1, 2016 2:48 PM > > > >>>> To: r-help@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]] > > > >>>> > > > >>>> ______________________________________________ > > > >>>> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see > > > >>>> https://stat.ethz.ch/mailman/listinfo/r-help > > > >>>> PLEASE do read the posting guide http://www.R-project.org/ > posting- > > > >>>> guide.html > > > >>>> and provide commented, minimal, self-contained, reproducible code. > > > >>> > > > >>> ________________________________ > > > >>> Tento e-mail a jakékoliv k němu připojené dokumenty jsou důvěrné a > jsou > > > >>> určeny pouze jeho adresátům. > > > >>> Jestliže jste obdržel(a) tento e-mail omylem, informujte laskavě > > > >>> neprodleně jeho odesílatele. 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