On Sun, Dec 4, 2016 at 10:03 AM, Ashim Kapoor <ashimkap...@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 <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/posti >> ng- >> > > >>>> 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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