[R] difference between lrm's Model L.R. and anova's Chi-Square
I am running lrm() with a single factor. I then run anova() on the fitted model to obtain a p-value associated with having that factor in the model. I am noticing that the Model L.R. in the lrm results is almost the same as the Chi-Square in the anova results, but not quite; the latter value is always slightly smaller. anova() calculates the p-value based on Chi-Square, but I have independent evidence that Model L.R. is the actual -2*log(LR), so should I be using that? Why are the values different? prob_a - inv.logit(rnorm(1,0,1)) prob_b - inv.logit(rnorm(1,0,1)) data - data.frame( factor=c(rep(a,500),rep(b,500)), outcome=c(sample(c(1,0),100,replace=T,prob=c(prob_a,1-prob_a)), sample(c(1,0),100,replace=T,prob=c(prob_b,1-prob_b fit - lrm(outcome~factor,data) fit # gives Model L.R. e.g. 8.23, 11.76, 6.89... anova(fit)# gives Chi-Square e.g. 8.19, 11.69, 6.85... __ 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.
Re: [R] difference between lrm's Model L.R. and anova's Chi-Square
Quoting Frank E Harrell Jr [EMAIL PROTECTED]: anova (anova.Design) computes Wald statistics. When the log-likelihood is very quadratic, these statistics will be very close to log-likelihood ratio chi-square statistics. In general LR chi-square tests are better; we use Wald tests for speed. It's best to take the time and do lrtest(fit1,fit2) in Design, where one of the two fits is a subset of the other. Frank Harrell Thanks, this is great, but in my case, there's just one factor, fit1 - lrm(outcome~factor,data) and I'm having trouble constructing the subset 'null model', as e.g. fit2 - lrm(outcome~1,data) returns an error message. How do I construct a null model with lrm() so that I can use lrtest() to test a model with only one predictor? I apologize for asking what must be a very simple question but I have been unable to find the answer by searching R-help. Thanks, Dan P.S. Second point: I have another case where I use lmer(), and there the null model includes a random effect so I don't get the problem above. It looks like with lmer objects anova() uses LLR, not Wald. Is that right? __ 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.
Re: [R] difference between lrm's Model L.R. and anova's Chi-Square
[EMAIL PROTECTED] wrote: Quoting Frank E Harrell Jr [EMAIL PROTECTED]: anova (anova.Design) computes Wald statistics. When the log-likelihood is very quadratic, these statistics will be very close to log-likelihood ratio chi-square statistics. In general LR chi-square tests are better; we use Wald tests for speed. It's best to take the time and do lrtest(fit1,fit2) in Design, where one of the two fits is a subset of the other. Frank Harrell Thanks, this is great, but in my case, there's just one factor, fit1 - lrm(outcome~factor,data) and I'm having trouble constructing the subset 'null model', as e.g. fit2 - lrm(outcome~1,data) returns an error message. How do I construct a null model with lrm() so that I can use lrtest() to test a model with only one predictor? The overall LR chi-square test statistic is in the standard output of lrm (which uses print.lrm). I apologize for asking what must be a very simple question but I have been unable to find the answer by searching R-help. Thanks, Dan P.S. Second point: I have another case where I use lmer(), and there the null model includes a random effect so I don't get the problem above. It looks like with lmer objects anova() uses LLR, not Wald. Is that right? Please check the lmer documentation. Frank -- Frank E Harrell Jr Professor and Chair School of Medicine Department of Biostatistics Vanderbilt University __ 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.
[R] difference between lrm's Model L.R. and anova's Chi-Square
I am running lrm() with a single factor. I then run anova() on the fitted model to obtain a p-value associated with having that factor in the model. I am noticing that the Model L.R. in the lrm results is almost the same as the Chi-Square in the anova results, but not quite; the latter value is always slightly smaller. anova() calculates the p-value based on Chi-Square, but I have independent evidence that Model L.R. is the actual -2*log(LR), so should I be using that? Why are the values different? prob_a - inv.logit(rnorm(1,0,1)) prob_b - inv.logit(rnorm(1,0,1)) data - data.frame( factor=c(rep(a,500),rep(b,500)), outcome=c(sample(c(1,0),100,replace=T,prob=c(prob_a,1-prob_a)), sample(c(1,0),100,replace=T,prob=c(prob_b,1-prob_b fit - lrm(outcome~factor,data) fit # gives Model L.R. e.g. 8.23, 11.76, 6.89... anova(fit)# gives Chi-Square e.g. 8.19, 11.69, 6.85... PreviousNext| Save| Delete | Reply | __ 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.
Re: [R] difference between lrm's Model L.R. and anova's Chi-Square
[EMAIL PROTECTED] wrote: I am running lrm() with a single factor. I then run anova() on the fitted model to obtain a p-value associated with having that factor in the model. I am noticing that the Model L.R. in the lrm results is almost the same as the Chi-Square in the anova results, but not quite; the latter value is always slightly smaller. anova() calculates the p-value based on Chi-Square, but I have independent evidence that Model L.R. is the actual -2*log(LR), so should I be using that? Why are the values different? anova (anova.Design) computes Wald statistics. When the log-likelihood is very quadratic, these statistics will be very close to log-likelihood ratio chi-square statistics. In general LR chi-square tests are better; we use Wald tests for speed. It's best to take the time and do lrtest(fit1,fit2) in Design, where one of the two fits is a subset of the other. Frank Harrell prob_a - inv.logit(rnorm(1,0,1)) prob_b - inv.logit(rnorm(1,0,1)) data - data.frame( factor=c(rep(a,500),rep(b,500)), outcome=c(sample(c(1,0),100,replace=T,prob=c(prob_a,1-prob_a)), sample(c(1,0),100,replace=T,prob=c(prob_b,1-prob_b fit - lrm(outcome~factor,data) fit # gives Model L.R. e.g. 8.23, 11.76, 6.89... anova(fit)# gives Chi-Square e.g. 8.19, 11.69, 6.85... Previous Next| Save| Delete | Reply | __ 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. -- Frank E Harrell Jr Professor and Chair School of Medicine Department of Biostatistics Vanderbilt University __ 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.
