Staleno: As always, you need to read the Help file carefully. From ?aov:
"The main difference from lm is in the way print, summary and so on handle the fit: this is expressed in the traditional language of the analysis of variance rather than that of linear models. " summary.aov() computes sequential ss. lm() uses the t-statistics for the estimated coefficients. They are not the same if non-orthogonal contrasts are used. Of course, the coefficients and fits **are** identical. If you don't know what this means, consult a statistician or linear models references. -- Bert On Mon, Jun 25, 2012 at 2:26 AM, Staleno <s...@bergen-plastics.no> wrote: > Hello. > > I'm a new user of R, and I have a question regarding the use of aov and > lm-functions. I'm doing a fractional factorial experiment at our production > site, and I need to familiarize myself with the analysis before I conduct > the experiment. I've been working my way through the examples provided at > http://www.itl.nist.gov/div898/handbook/pri/section4/pri472.htm > http://www.itl.nist.gov/div898/handbook/pri/section4/pri472.htm , but I > can't get the results provided in the trial model calculations. Why is > this. > Here is how I have tried to do it: > > > data.catapult=read.table("Fractional.txt",header=T) #Read the data in the > > table provided in the example. > > > data.catapult > Distance h s b l e > 1 28.00 3.25 0 1 0 80 > 2 99.00 4.00 10 2 2 62 > 3 126.50 4.75 20 2 4 80 > 4 126.50 4.75 0 2 4 45 > 5 45.00 3.25 20 2 4 45 > 6 35.00 4.75 0 1 0 45 > 7 45.00 4.00 10 1 2 62 > 8 28.25 4.75 20 1 0 80 > 9 85.00 4.75 0 1 4 80 > 10 8.00 3.25 20 1 0 45 > 11 36.50 4.75 20 1 4 45 > 12 33.00 3.25 0 1 4 45 > 13 84.50 4.00 10 2 2 62 > 14 28.50 4.75 20 2 0 45 > 15 33.50 3.25 0 2 0 45 > 16 36.00 3.25 20 2 0 80 > 17 84.00 4.75 0 2 0 80 > 18 45.00 3.25 20 1 4 80 > 19 37.50 4.00 10 1 2 62 > 20 106.00 3.25 0 2 4 80 > > > aov.catapult = > > > aov(Distance~h+s+b+l+e+h*s+h*b+h*l+h*e+s*b+s*l+s*e+b*l+b*e+l*e,data=data.catapult) > > summary(aov.catapult) > Df Sum Sq Mean Sq F value Pr(>F) > h 1 2909 2909 15.854 0.01638 * > s 1 1964 1964 10.701 0.03076 * > b 1 7537 7537 41.072 0.00305 ** > l 1 6490 6490 35.369 0.00401 ** > e 1 2297 2297 12.518 0.02406 * > h:s 1 122 122 0.667 0.45998 > h:b 1 345 345 1.878 0.24247 > h:l 1 354 354 1.929 0.23724 > h:e 1 0 0 0.001 0.97578 > s:b 1 161 161 0.877 0.40199 > s:l 1 20 20 0.107 0.75966 > s:e 1 114 114 0.622 0.47427 > b:l 1 926 926 5.049 0.08795 . > b:e 1 124 124 0.677 0.45689 > l:e 1 158 158 0.860 0.40623 > Residuals 4 734 184 > --- > Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1 > > This seems just about right to me. However, when I attempt to make the > linear model, based on main factors and two-factor interactions, I get a > completely different result: > > > lm.catapult = > > > lm(Distance~h+s+b+l+e+h*s+h*b+h*l+h*e+s*b+s*l+s*e+b*l+b*e+l*e,data=data.catapult) > > summary(lm.catapult) > > Call: > lm(formula = Distance ~ h + s + b + l + e + h * s + h * b + h * > l + h * e + s * b + s * l + s * e + b * l + b * e + l * e, > data = data.catapult) > > Residuals: > 1 2 3 4 5 6 7 8 9 > 10 > -0.8100 22.3875 -3.6763 -3.8925 -3.8925 -0.8576 7.0852 -0.8100 -0.8100 > -0.8576 > 11 12 13 14 15 16 17 18 19 > 20 > -0.8576 -0.8576 7.8875 -3.8925 -3.8925 -3.6763 -3.6763 -0.8100 -0.4148 > -3.6763 > > Coefficients: > Estimate Std. Error t value Pr(>|t|) > (Intercept) 25.031042 100.791955 0.248 0.8161 > h -3.687500 22.466457 -0.164 0.8776 > s 0.475446 2.446791 0.194 0.8554 > b -39.417973 44.906164 -0.878 0.4296 > l -18.938988 12.233954 -1.548 0.1965 > e -0.158449 1.230683 -0.129 0.9038 > h:s -0.368750 0.451546 -0.817 0.4600 > h:b 12.375000 9.030925 1.370 0.2425 > h:l 3.135417 2.257731 1.389 0.2372 > h:e 0.008333 0.258026 0.032 0.9758 > s:b -0.634375 0.677319 -0.937 0.4020 > s:l -0.055469 0.169330 -0.328 0.7597 > s:e 0.015268 0.019352 0.789 0.4743 > b:l 7.609375 3.386597 2.247 0.0879 . > b:e 0.318397 0.387008 0.823 0.4569 > l:e 0.089732 0.096760 0.927 0.4062 > --- > Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1 > > Residual standard error: 13.55 on 4 degrees of freedom > Multiple R-squared: 0.9697, Adjusted R-squared: 0.8563 > F-statistic: 8.545 on 15 and 4 DF, p-value: 0.02559 > > This result is nothing like the results provided in the example. Why is > this? Any help is very much appreciated. > > Regards, Ståle. > > -- > View this message in context: > http://r.789695.n4.nabble.com/Fractional-Factorial-Wrong-values-using-lm-function-tp4634400.html > Sent from the R help mailing list archive at Nabble.com. > > ______________________________________________ > R-help@r-project.org mailing list > 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. > -- Bert Gunter Genentech Nonclinical Biostatistics Internal Contact Info: Phone: 467-7374 Website: http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm [[alternative HTML version deleted]]
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