similar to: Large p, small n. Impossible to calculate the p-value

Is Pearson's Chi-Square test for contingency tables asymptotically unbiased for large tables (large degrees of freedom) regardless of the expected values in each cell? The rule of thumb is that Pearson's Chi-square should not be used when large numbers of cells have expected values < 5. However, I compared the results on 4x4 contingency tables for R's chisq.test using chi-square

Dear list, I do apologize if these are basic questions. I am fitting some GAM models using the mgcv package and following the model selection criteria proposed by Wood and Augustin (2002, Ecol. Model. 157, p. 157-177). One criterion to decide if a term should be dropped from a model is if the estimated degrees of freedom (EDF) for the term are close to their lower limit. What would be the

Hi All, I'm fitting an proportional odds model using vglm() from VGAM. My response variable is the severity of diseases, going from 0 to 5 (the severity is actually an ordered factor). The independent variables are: 1 genetic marker, time of medical observation, age, sex. What I *need* is a p-value for the genetic marker. Because I have ~1.5 million markers I'd rather not faffing

Hello, I'm having trouble figuring out how to calculate a p-value for a 1-tailed test of beta_1 in a linear model fit using command lm. My model has only 1 continuous, predictor variable. I want to test the null hypothesis beta_1 is >= 0. I can calculate the p-value for a 2-tailed test using the code "2*pt(-abs(t-value), df=degrees.freedom)", where t-value and degrees.freedom

Hi , I know this question has been asked twice in the past but to my knowldege, it still hasn't been solved. I am doing a zero inflated binomial model using the VGAM package, I need to obtain p values for my Tvalues in the vglm output. code is as follows > mod2=vglm(dmat~Season+Diel+Tidal.phase+Tidal.cycle,zibinomial, data=mp1) > summary(mod2) Call: vglm(formula = dmat ~ Season +

Anyone know how to get p-values for the t-values from the coefficients produced in vglm? Attached is the code and output ? see comment added to output to show where I need p-values + print(paste("********** Using VGAM function gamma2 **********")) + modl2<- vglm(MidPoint~Count,gamma2,data=modl.subset,trace=TRUE,crit="c") + print(coef(modl2,matrix=TRUE))

I am doing an analysis that requires me to calculate correlations for a matrix of 15,000 rows x 50 columns. For each row I want to calculate the correlation to all other rows and then for each row, find the n (say 10) most correlated rows. If read in the 15,000 x 50 data from file and pass it to 'cor', this function quite appropriately (and very quickly) calculates all possible row by

Dear all, I'm a bit stunned by the behaviour of a gam model using cyclic P-spline smoothers. I cannot provide the data, as I have about 61.000 observations from a time series. I use the following model : testgam <- gam(NO~s(x)+s(y,bs="cs")+s(DD,bs="cs")+s(TT),data=Final) The problem lies with the cyclic smoother I use for seasonal trends. The variable Final$y is a

Hi,everyone, I am studying the generalized additive model and employ the package 'mgcv' developed by professor Wood. However,I can not understand the example listed in check.in function. For example, library(mgcv) set.seed(1) dat <- gamSim(1,n=400,scale=2) ## fit a GAM with quite low `k' b<-gam(y~s(x0,k=6)+s(x1,k=6)+s(x2,k=6)+s(x3,k=6),data=dat) plot(b,pages=1,residuals=TRUE)

Greetings, I am performing a simple Pearson's correlation test. Length of both vectors is 40, therefore the resulting df is 38. Nevertheless, a colleague is asking me for the "effective degrees of freedom". As far as I understand, those degrees of freedom have to be estimated for more complex regressions, but I was not able to find detailed information about it. Does any one of

Users are often surprised and alarmed that the summary of a linear mixed model fit by lmer provides estimates of the fixed-effects parameters, standard errors for these parameters and a t-ratio but no p-values. Similarly the output from anova applied to a single lmer model provides the sequential sums of squares for the terms in the fixed-effects specification and the corresponding numerator

Hi R user, I am a kind of an intermediate user of R. Now I am using GLM model (library MASS, VEGUS). I used a backward stepwise logistic regression, but i got a problem in removing those predictors which are above 0.05. I don't want to include those variables which were above 0.05 in final backward stepwise logetsic regression model. for example: first I run the model,

Anyone from Chicago area interested in this course? Please email XLSolutions so they can schedule it in Chicago. We ran out of travel budget in my company :( Date: Wed, 2 Aug 2006 13:20:23 -0700From: elvis@xlsolutions-corp.comSubject: [S] Course***Dr Frank Harrell's Regression Modeling Strategies in R/Splus course *** September 2006 near you (San Francisco, Washington DC, Atlanta)To:

Hi all, I computed a Wilks Lambda Test with manova: > df.man <- manova(df.mul ~ 1, na.rm=TRUE) > summary(df.man, intercep=T, test="Wilks") Df Wilks approx F num Df den Df Pr(>F) (Intercept) 1 0.0002824 393.3 9 1 0.03911 * Residuals 9 --- Signif. codes: 0 `***' 0.001 `**' 0.01 `*'

How can I get the Log - Rank p value to be output? The chi square value can be output, so I was thinking if I can also have the degrees of freedom output I could generate the p value, but can't see how to find df either. > (survtest <- survdiff(Surv(time, cens) ~ group, data = surv,rho=0)) Call: survdiff(formula = Surv(time, cens) ~ group, data = surv, rho = 0) N Observed

List: I'm helping a colleague with some Poisson regression modeling. He uses SAS proc GENMOD and I'm using glm() in R. Note on the SAS and R output below that our estimates, standard errors, and deviances are identical but what we get for likelihoods differs considerably. I'm assuming that these must differ just by some constant but it would be nice to have some confirmation

hello everybody i'm very new at using R so probably this is a very stupid question. I have a problem calculating a p-value. When i do this with excel i can use the method CHIDIST for 1.2654 with 1 freedom degree i get the answer 0.261 i just want to do the same thing in R but i can't find a method. can somebody help me friendly regards richard

[Please use the subject line!] In the help page for summary.lm, the "Value" section says that the returned object has a component called "fstatistic", which has the F-statistic and the associated numerator and denominator degrees of freedom. You can get the p-value by something like: fstat <- summary(speciallinearmodel)$fstatistic pval <- pf(fstat[1], fstat[2],

Hello. Slope estimates in lme are shrinkage estimates which pull the OLS slope estimates towards the population estimates, the degree of which depends on the group sample size and the distance between the group-based estimate and the overall population estimate. Although these shrinkage estimates as said to be more precise with respect to the true values, they are also biased. So there is a

I can get this summary of a model that I am running: summary(myprobit) Call: glm(formula = Response_Slot ~ trial_no, family = binomial(link = "probit"), data = neg_data, na.action = na.pass) Deviance Residuals: Min 1Q Median 3Q Max -0.9528 -0.8934 -0.8418 1.4420 1.6026 Coefficients: Estimate Std. Error z value Pr(>|z|)