Re: [R] longitudinal survey data
Thank you for your reply. Does that mean that in order to take in account the repeated measures I denote these as another cluster in R? Dassy Quoting Thomas Lumley [EMAIL PROTECTED]: On Thu, 26 May 2005 [EMAIL PROTECTED] wrote: Dear R-Users! Is there a possibility in R to do analyze longitudinal survey data (repeated measures in a survey)? I know that for longitudinal data I can use lme() to incorporate the correlation structure within individual and I know that there is the package survey for analyzing survey data. How can I combine both? I am trying to calculate design-based estimates. However, if I use svyglm() from the survey package I would ignore the correlation structure of the repeated measures. You *can* fit regression models to these data with svyglm(). Remember that from a design-based point of view there is no such thing as a correlation structure of repeated measures -- only the sampling is random, not the population data. If you *want* to fit mixed models (eg because you are interested in estimating variance components, or perhaps to gain efficiency) then it's quite a bit trickier. You can't just use the sampling weights in lme(). You can correct for the biased sampling if you put the variables that affect the weights in as predictors in the model. Cluster sampling could perhaps then be modelled as another level of random effect. -thomas Thomas Lumley Assoc. Professor, Biostatistics [EMAIL PROTECTED] University of Washington, Seattle __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] longitudinal survey data
Sorry, still confused. If I dont have fpc's ready in my dataset (calculate myself?) that means that R will use the weight of an individual for each of his repeated observations. But is that then still correct? The cluster individual is ignored and each observation of an individual has the same weight. Thanks a lot. Dassy Quoting Thomas Lumley [EMAIL PROTECTED]: On Fri, 27 May 2005 [EMAIL PROTECTED] wrote: Thank you for your reply. Does that mean that in order to take in account the repeated measures I denote these as another cluster in R? Yes, but unless you have multistage finite population corrections to put in the design object only the first stage of clustering affects the results, so you may not need to bother. -thomas __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] MLE: Question
Hi R users! I have a likelihood ratio statistic that depends on a parameter delta and I am trying to get confidence intervals for this delta using the fact that the likelihood ratio statistic is approx. chi-squared distributed. For this I need to maximize the two likelihoods (for the ratio statistic) one of which depends on delta too and I am trying to use the function mle. But delta is neither a parameter I would like to estimate nor a fixed value that I know the value of (to give it to the mle function with fixed=list(delta=?)). Can I still use mle to find the likelihood of the data (just denoting delta as a constant)? Thank you. Hadassa __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] survreg: fitting different location parameters
Hi R-Help! My question: I have lifetime/failure data of machines with different stress levels and i think an weibull/extreme value distribution would fit this data. So I did: model1 - survreg(Surv(lfailure)~stress,data=steel,dist=extreme) (where lfailure=log(failure)) Now I would like to do a likelihood ratio test to test the hypothesis H0: location parameters of the extreme value distribution are equal for each stress level. So in order to perform the likelihood ratio test I need the likelood of the model under HA: location parameters are not equal (i.e. each stress level has its slope and intercept). How can I do this? Thanks for any help! Hadassa __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] survreg: fitting different location parameters
Hi R-Help! My question: I have lifetime/failure data of machines with different stress levels and i think an weibull/extreme value distribution would fit this data. So I did: model1 - survreg(Surv(lfailure)~stress,data=steel,dist=extreme) (where lfailure=log(failure)) Now I would like to do a likelihood ratio test to test the hypothesis H0: location parameters of the extreme value distribution are equal for each stress level. So in order to perform the likelihood ratio test I need the likelood of the model under HA: location parameters are not equal (i.e. each stress level has its slope and intercept). How can I do this? Thanks for any help! Hadassa __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] survreg: fitting different location parameters
I am still trying to find a common intercept but a different slopes for each group within my lifetime data. By stratifying the variable stress (the groups) I get different scale parameters which is not my goal. So I did this: survreg(Surv(lfailure)~as.factor(stress),data=steel,dist=extreme) and I did get different slopes but common intercepts and scales. Is this correct? I am a bit unsure as I did it with a different software and got different estimates... Thanks for any help Hadassa Quoting Göran Broström [EMAIL PROTECTED]: On Sun, Jan 23, 2005 at 03:10:18PM -0500, [EMAIL PROTECTED] wrote: Hi R-Help! My question: I have lifetime/failure data of machines with different stress levels and i think an weibull/extreme value distribution would fit this data. So I did: model1 - survreg(Surv(lfailure)~stress,data=steel,dist=extreme) (where lfailure=log(failure)) Now I would like to do a likelihood ratio test to test the hypothesis H0: location parameters of the extreme value distribution are equal for each stress level. So in order to perform the likelihood ratio test I need the likelood of the model under HA: location parameters are not equal (i.e. each stress level has its slope and intercept). How can I do this? Stratify on stress level. -- Göran Broströmtel: +46 90 786 5223 Department of Statistics fax: +46 90 786 6614 Umeå University http://www.stat.umu.se/egna/gb/ SE-90187 Umeå, Sweden e-mail: [EMAIL PROTECTED] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] lme: Variances
Hi R users! I will try to state my question again. I have longitudinal data and fitted the following model with lme: Y = X*beta + U + W(t) + Z where U ~ N(0, nu*I) I is the identity matrix, so this is the random intercept W(t)~ N(0, sigma*H) and H is a matrix which incorporates a Gaussian serial correlation (covariance) in the offdiagonal elements Z ~ N(0, tau*J) the (measurement) error So lme must have estimated three variance parameters plus the parameter for the Gaussian correlations. From the output I get, I dont know which is which. The output was: nepal.lme-lme(ht~sex+died+alive+mage+lit+bf+age+I(age^2), data=nepal,random=~1|id,correlation=corGaus(corstruct,form=~age|id) ,na.action=na.exclude) summary(nepal.lme) Linear mixed-effects model fit by REML Data: nepal AICBIClogLik 3363.547 3420.7 -1669.774 Random effects: Formula: ~1 | id (Intercept) Residual StdDev:4.240752 1.240242 Correlation Structure: Gaussian spatial correlation Formula: ~age | id Parameter estimate(s): range 4.662006 Fixed effects: ht ~ sex + died + alive + mage + lit + bf + age + I(age^2) Value Std.Error DF t-value p-value (Intercept) 51.07075 2.0274503 674 25.18964 0. sex -0.54780 0.6210172 191 -0.88210 0.3788 died-0.01686 0.3915936 191 -0.04306 0.9657 alive -0.46979 0.2388267 191 -1.96706 0.0506 mage 0.34289 0.0859849 191 3.98785 0.0001 lit 2.62584 1.5123958 191 1.73621 0.0841 bf 0.31873 0.0870432 674 3.66173 0.0003 age 0.86269 0.0198487 674 43.46318 0. I(age^2)-0.00334 0.0002347 674 -14.24842 0. Correlation: (Intr) sexdied alive mage litbf age sex -0.419 died -0.063 -0.095 alive 0.475 0.041 -0.484 mage -0.814 -0.035 0.188 -0.780 lit -0.072 -0.099 0.092 0.003 0.059 bf -0.098 0.000 0.002 -0.001 0.008 0.008 age -0.168 0.016 0.003 -0.008 -0.018 -0.008 0.250 I(age^2) 0.146 -0.010 -0.010 0.009 -0.007 0.005 -0.156 -0.917 Standardized Within-Group Residuals: Min Q1 Med Q3 Max -4.22819684 -0.45649041 -0.02996715 0.43379089 2.70363057 Thanks for any help. Hadassa __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] lme: Confusion about Variances
Dear R users! I used lme to fit a mixed model with random intercept and spatial Gaussian correlation i.e. I fitted a model of the following form: Y = X*beta + error and error = U + W(t) + Z where U is the random intercept (normally distributed), W(t) the stationary Gaussian process and Z also a normally distributed (the residual) rv. Each of these three random variables have a variance which I am not sure to which output in lme they belong to. VarCorr gives the intercept and residual variance which I assume belong to U and Z respectively. The output of lme gives another estimate called range which I assume belongs to the parameter estimate needed for the Gaussian correlation. Are my assumptions correct? And where can I get the variance for the W(t) from? Thanks for any answers...and happy new year... Hadassa __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
