search for: variances

I am trying to write a function that will solve a simple system of nonlinear equations for the parameters that describe the beta distribution (a,b) given the mean and variance. mean = a/(a+b) variance = (a*b)/(((a+b)^2) * (a+b+1)) Any help as to where to start would be welcome. -- Scott Story Graduate Student MSU Ecology Department 319 Lewis Hall Bozeman, Mt 59717 406.994.2670 sstory at

Hai I'm Dee. I'm trying to write var data from these codes inside excel file. My directory to store the data is *D:\FYP\image* . these are my codes, can you help give an advice or idea with my problem: l*ibrary("biOps") library("waveslim") library("xlsReadWrite") x <- readTiff("D:\\FYP\\image\\SignatureImage\\user186g1.tif") y <-

I have a data set with 50 different x values and 5 values for the sampling variance; each of the 5 sampling variances corresponds to 10 particular x values. I am trying to fit a mixed effect linear model and I'm not sure about the syntax for specifying the fixed variance structure. In Pinheiro's book my situation appears to be similar to the example used for varIdent, where there is a fixed value for vari...

Hai, I'm trying to write these var output data from these codes inside excel file. My directory to store the data is /D:\FYP\image / but receive an error message : /Error in write.xls(mydata, "D:\\FYP\\image.mydata.xls") : object 'mydata' not found/ these are my codes, can you help give an advice or idea with my problem: /library("biOps")

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

Hello, I have been looking for functions for calculating the within-groups variance and between-groups variance, for the case where you have several numerical variables describing samples from a number of groups. I didn't find such functions in R, so wrote my own versions myself (see below). I can calculate the within- and between-groups variance for the Sepal.length variable (iris[1]) in

Dear list, I'm trying to mimic the analysis of Wedderburn (1974) as cited by McCullagh and Nelder (1989) on p.328-332. This is the leaf-blotch on barley example, and the data is available in the `faraway' package. Wedderburn suggested using the variance function mu^2(1-mu)^2. This variance function isn't readily available in R's `quasi' family object, but it seems to me

Hello all, I'm running very simple regression but face a problem of non-homogeneity of variance, but with a decreasing variance with increasing mean...I do not know how to deal with that. this relationship doesn't seem to be strong, but it's my first time to see something like that, and would like to know what to do if one day it becomes stronger. I tested just for fun some

I'm currently working on writing up some documentation for some of my code, but am having the darndest time coding in equations. For example, the equation in the following: \details{ Calculated the R Squared for observed endogenous variables in a structural equation model, as well as several other useful summary statistics about the error in thoe variables. R Squared values are

Dear R users, I am a beginner to R. I generated 1000 samples with 15 data in each sample I tried finding the variance for each sample I used the code: m=1000;n=15 > r<-rnorm(15000) > for(i in 1:m){ x=data[,i] v=var(x)} what I got was just the variance for the last sample i.e. the 1000th sample but what I want is 1000 variance. Does anyone know what I did wrong? Thanks Chloe Smith

After doing a PCA using princomp, how do you view how much each component contributes to variance in the dataset. I'm still quite new to the theory of PCA - I have a little idea about eigenvectors and eigenvalues (these determine the variance explained?). Are the eigenvalues related to loadings in R? Thanks, Paul -- View this message in context:

Dear R-listers, I am working with generalized linear mixed models to quantify the variance due to two nested random factors, but have hit a snag in the interpretation of variance components. Despite my best efforts with Venables & Ripley 2002, Fahrmeir & Tutz 2001, R-help archives, Google, and other eminent sources (i.e. local R gurus), I have not been able to find a definitive answer

Hello, I have some patient data for my masters thesis with three groups (n=16, 19 & 20) I have completed compiling the results of 7 tests, for which one of these tests the variances are unequal. I wish to perform an ANOVA between the three groups but for the one test with unequal variance (<0.001 by both bartlett and levene's test) I am not sure what to do. I thought i would run ANOVA with bonferonni post-test for groups with equal variances, then for the test with u...

...estion is a bit more complicated and changes my random structure as well. I had originally built my model in nlme and used the multiple variance (varIdent) function to allow different variance for two of my terms (trt and state) in my nlme model because different levels in each term had different variances. There are two levels in each of these factors. Treatment is G or S (basically treated or control groups) and state is either Pre-treatment or Post-treatment. What is happening in my model is that the variance for the treated (G) is much smaller post-treatment than pre-treatment. Thus, overall,...

...the *proportions* as if there were only 14 PCs in > > total (but there were 32 originally). > > > > I would think that the summary should or could in addition show > > the usual "proportion of variance explained" like result which > > does involve all 32 variances or std.dev.s ... which are > > returned from the svd() anyway, even in the case when I use my > > new 'rank.' argument which only returns a "few" PCs instead of > > all. > > > > Would you think the current summary() output is good enough or > >...

...estimated regression coefficients and estimated standard errors for these pooled coefficients. In particular, I would like to estimate the model \beta_{i} = \mu + \eta_{i} + \epsilon_{i} \eta_{i} ~ iid N(0,\tau^2) and independent of the \epsilon_{i}, the latter themselves being independent with variances assumed known and equal to the squared standard errors reported in the regression output. I would like to use lme() to estimate \tau^2 by REML, and also get a sensibly weighted estimate for \mu from the fixed effects output. I am not sure how to do this. I have tried lme(fixed=beta~1,random=~1|...

...Source as random effects too to estimate the variance, i.e. lme(fixed=Trait~1,random=~(Treatment*Source)|Family, method="REML") However, in some cases, lme does not converge, hence I used lmer from the lme4 package: lmer(Trait~1+(Treatment*Source|Family),data=DATA) Where I extract the variances from the model using the summary function: model<-lmer(Trait~1+(Treatment*Source|Family),data=regrexpdat) results<-model at REmat variances<-results[,3] I get the same values as with the VarCorr function. I use then these values to calculate the actual percentage of variation taking the...

Hello, not understanding the output of prcomp, I reduce the number of components and the output continues to show cumulative 100% of the variance explained, which can't be the case dropping from 8 components to 3. How do i get the output in terms of the cumulative % of the total variance, so when i go from total solution of 8 (8 variables in the data set), to a reduced number of

Dear R users I would like to ask a question regarding to icc (intraclass correlation) or many biologists refer it to as repeatability. It is very useful to get icc for many reasons and it is easy to do so from linear mixed-effects models and many packages like psy, psychometric, aod and irr have functions to calculate icc. icc = between-group variance/(between-group variance + residual

I am fitting quantile regression models using data collected from a sample of 124 patients. When modeling cross-sectional associations, I have noticed that nonparametric bootstrap estimates of the variances of parameter estimates are much greater in magnitude than the empirical Huber estimates derived using summary.rq's "nid" option. The outcome variable is severely skewed, and I am afraid that this may be affecting the consistency of the bootstrap variance estimates. I have read that t...