Between-group variance from ANOVA
On Jul 25, 2012, at 14:56 , arun wrote:
Hi, From the ANOVA results, you could get MSE and MS of group. MSE is basically sigma^2 error. MS group of MS between group contains sigma^2 error+replication*sigma^2group (please check the formula. It can be slightly different when the model complexity increases). Once, you get sigma^2 group, I guess you know how to calculate Vg and Vp. Once, you have all the values, except sigma^2 group, you can subtract and divide it by replication to get sigma^2 group. In SAS, proc glm also shows the output with formula. A.K.
Beware, though, that this works for balanced designs only (identical group sizes). For unequal replication, you need to go the lme/lmer route. -pd
----- Original Message ----- From: Ista Zahn <istazahn at gmail.com> To: tedtoal <twtoal at ucdavis.edu> Cc: r-help at r-project.org Sent: Wednesday, July 25, 2012 6:21 AM Subject: Re: [R] Between-group variance from ANOVA There is nothing about R in your question, hence it is not appropriate for this list. Please consult with a local statistician, or post on a stats help list such as http://stats.stackexchange.com/ On Tue, Jul 24, 2012 at 8:55 PM, tedtoal <twtoal at ucdavis.edu> wrote:
I'm trying also to understand how to get the between-group variance out of a one-way ANOVA, but I'm beginning to think that in a sense, the variance does not exist. Emma said: *The model is response(i,j)= group(i)+ error(i,j)* Yes, if by group(i) you mean intercept + coefficient[i]. *we assume that group~N(0,P^2) and error~N(0,sigma^2) * Only the error is assumed to be a random variable. Group is a fixed effect, not a random variable, and therefore it has no variance associated with it. The model does not predict a variance for it. One could compute the variance of the coefficients and call this a group variance, but it seems to me that isn't the right way to think about it. I'm trying to calculate a heritability value for a trait in an organism, defined as Vg/Vp, where Vg = variance due to genotype and Vp = total variance. The model is p~g, or p[i,j] = intercept + g_coefficient[i] + error[i,j]. But to get Vg, I think it is actually necessary to use a different model, where g is modelled as a random variable (a random effect), so the model can estimate a variance associated with it. If anyone can add something to this, please do. ted -- View this message in context: http://r.789695.n4.nabble.com/Between-group-variance-from-ANOVA-tp901535p4637686.html Sent from the R help mailing list archive at Nabble.com.
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______________________________________________ R-help at 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. ______________________________________________ R-help at 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.
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