search for: bonf

...NETARY=English_United States.1252 [4] LC_NUMERIC=C [5] LC_TIME=English_United States.1252 attached base packages: [1] stats graphics grDevices utils datasets methods base cheat.fit <- function(dat, key, wrongChoice, alpha = .01, rfa = c('nr', 'uni', 'bsct'), bonf = c('yes','no'), con = 1e-12, lower = 0, upper = 50){ bonf <- tolower(bonf) bonf <- match.arg(bonf) rfa <- match.arg(rfa) rfa <- tolower(rfa) dat <- t(dat) correctStuMat...

...5 % 97.5 % t value Std.Err. strategyPrice-strategyConvenience 31.10 -40.67 102.87 1.043 29.824 strategyQuality-strategyConvenience 75.45 3.68 147.22 2.530 29.824 strategyQuality-strategyPrice 44.35 -27.42 116.12 1.487 29.824 p raw p Bonf p adj strategyPrice-strategyConvenience 0.301 0.904 0.553 strategyQuality-strategyConvenience 0.014 0.043 0.037 strategyQuality-strategyPrice 0.143 0.428 0.305 This gives correct 95% confidence intervals and adjusted p-values for the Tukey multiple comparisons procedure. Next we issu...

...'d like to know whether the differences between the group means are significant. Is a pairwise t-test (for location, and a simple t-test for job) appropriate in this case? data = read.table("data.txt", header=T, nrows=90) attach(data) res1 = pairwise.t.test(all, location, p.adj="bonf") print(res1) res2 = pairwise.t.test(M, location, p.adj="bonf") print(res2) res3 = pairwise.t.test(OA, location, p.adj="bonf") print(res3) res4 = pairwise.t.test(UE, location, p.adj="bonf") print(res4) res1 = t.test(all~job) print(res1) res2 = t.test(M~job) print(...

...10.3912598689021, 8.04934992955112, 13.3935024789408, 12.1276080143311, 11.2987640203845, 2.05386456839974, 12.8386335687123, 3.84418682233211, 17.4626505871045, 6.5947625121599, -10.6664750811034, 21.3346023277454, 4.99216977422232) Then I do the tests > pairwise.t.test(d,n,p.adjust="bonf",paired=F) Pairwise comparisons using t tests with pooled SD data: d and n x y y 0.092 - z 0.326 1.000 P value adjustment method: bonferroni Assuming the data are paired: > pairwise.t.test(d,n,p.adjust="bonf",paired=T) Pairwise comparisons u...

...ests: Dunnett contrasts Call: simtest.formula(formula = y ~ f, data = dat, type = "Dunnett") Dunnett contrasts for factor f Contrast matrix: f1 f2 f3 f2-f1 0 -1 1 0 f3-f1 0 -1 0 1 Absolute Error Tolerance: 0.001 Coefficients: Estimate t value Std.Err. p raw p Bonf p adj f2-f1 4.015 -0.677 5.934 0.499 0.997 0.722 f3-f1 2.486 -0.419 5.934 0.675 0.997 0.722 --------------------------------- -- example 2 ------------------------------- require(multcomp) Loading required package: multcomp Loading required package: mvtnorm [1] TRUE y <- as.v...

...test(y ~ h,cmatrix=cbind(0,m))) rownames(result$estimate) result I want to compare my results with the results obtained with SigmaStat Software. In my results, i retrieve a correct q' value with the simtest t value How can i say if P<0.05 like the SigmStat results? Can i use p raw value, p bonf value, p adj value or other to compare directly? Or can i use the Dunnett's tables? Thanks for your help! Laurent Houdusse Analyste Programmeur

hi list, i have to ask you again, having tried and searched for several days... i want to do a TukeyHSD after an Anova, and want to get the adjusted p-values after the Tukey Correction. i found the p.adjust function, but it can only correct for "holm", "hochberg", bonferroni", but not "Tukey". Is it not possbile to get adjusted p-values after Tukey-correction? sorry, if this is an often-answered-question, but i didn??t find it on the list archive... thx a lot, list, Chris Christoph Strehblow, MD Department of Rheumatology, Diabetes and Endoc...

...one 2 1 2 one 2 2 3 two 2 3 2 two 1 4 2 three 9 8 1 three I have no normality, so I did a kruskal test which showed significant differences in some cases. As post hoc pairwise comparisons (the idea is to make an equivalent to Tukey test after an ANOVA) I tried with a paired wilcoxon test with the Bonferroni's correction (pairwise.wilcox.test(A, factor, p.adj="bonf") and other for B, and so on) and, now, I need to modify the output to show a letter-based representation of all pairwise comparisons. Some algorithms have been published (http://www.accessmylibrary.com/coms2/summary_0286...

Hello- I have run an ANOVA on 4 treatments with unequal sample sizes (n=9,7,10 and 10). I want to determine where my sig. differences are between treatments using a Bonferroni test, and have run the code: pairwise.t.test(Wk16, Treatment, p.adf="bonf") I receive an error message stating that my arguments are of unequal length: Error in tapply(x, g, mean, na.rm = TRUE) : arguments must have same length Is there a way to run this test even with unequa...

...contrasts for factor Cond, covariable: Q1 Contrast matrix: CondA CondB CondC CondB-CondA 0 -1 1 0 0 CondC-CondA 0 -1 0 1 0 CondC-CondB 0 0 -1 1 0 Absolute Error Tolerance: 0.001 Coefficients: Estimate t value Std.Err. p raw p Bonf p adj CondB-CondA 5.555 -1.461 3.802 0.151 0.453 0.319 CondC-CondB -5.248 -1.365 3.661 0.179 0.453 0.319 CondC-CondA 0.306 -0.084 3.844 0.934 0.934 0.934 The results from two analyses seem so different that I am wondering why. I do understand that multiple compa...

...fidence intervals: model contrasts model contrasts for factor Contrast matrix: [,1] [,2] [,3] [1,] 1 0 0 [2,] 0 1 0 [3,] 0 0 1 Absolute Error Tolerance: 1e-04 90 % quantile: 1.8431 Coefficients: Estimate -- 90 % t value Std.Err. p raw p Bonf p adj blanketb1-b0 -2.1333 -Inf 0.8226 -1.3302 1.6038 0.0958 0.2874 0.2412 blanketb2-b0 -7.4667 -Inf -4.5108 -4.6556 1.6038 0.0000 0.0001 0.0001 blanketb3-b0 -1.6667 -Inf -0.0360 -1.8837 0.8848 0.0337 0.1012 0.0924 > ## change lm to aov > aovmod <- aov(minutes ~ blanket, data=re...

...rial5 trial2-trial1 -1 1 0 0 0 trial3-trial1 -1 0 1 0 0 trial4-trial1 -1 0 0 1 0 trial5-trial1 -1 0 0 0 1 Absolute Error Tolerance: 0.001 Coefficients: Estimate t value Std.Err. p raw p Bonf p adj trial3-trial1 0.745 -2.911 0.256 0.006 0.025 0.021 trial2-trial1 -0.688 -2.686 0.256 0.011 0.033 0.029 trial4-trial1 0.091 -0.357 0.256 0.724 1.447 0.912 trial5-trial1 0.032 -0.127 0.256 0.900 1.447 0.912 But now I want to test trial3 against all other trials a...

Hi, I'm having problems while working with large data sets with R 1.5.1 in windows 2000. Given a integer matrix size of 30 columns and 15000 rows my function should return a boolean matrix size of about 5000 rows and 15000 columns. First of all I tried to run this function on computer with 256 MB of RAM. I increased memory limit of R with memory.limit() up to 512 MB. I was inspecting

...-Al400 0 0 -1 0 0 1 Al800-Al600 0 0 0 -1 1 0 control-Al600 0 0 0 -1 0 1 control-Al800 0 0 0 0 -1 1 Absolute Error Tolerance: 0.001 Coefficients: Estimate t value Std.Err. p raw p Bonf p adj Al800-Al100 -2.253 -10.467 0.213 0.000 0.000 0.000 Al600-Al100 -2.185 -10.389 0.207 0.000 0.000 0.000 Al400-Al100 -2.036 -9.850 0.210 0.000 0.000 0.000 Al200-Al100 -1.712 -8.051 0.215 0.000 0.000 0.000 control-Al100 -1.487 -7.243 0.205 0.000 0.000 0.00...

....formula(formula = response ~ treatment * (site + time), whichf = "treatment", type = "Tukey") # # Tukey contrasts for factor treatment, covariables: site +time +treatment:site + treatment:time # #Coefficients: # Estimate t value Std.Err. p raw p Bonf p adj #treatment3-treatment1 -0.655 -2.004 0.327 0.050 0.149 0.120 #treatment3-treatment2 -0.581 -1.777 0.327 0.081 0.162 0.143 #treatment2-treatment1 -0.074 -0.227 0.327 0.821 0.821 0.821 ___ drs. René Eschen CABI Bioscience Switzerland Centre 1 Rue des Grillons CH-2800 Delém...

...ugais",6), rep("francais",6))) >tousy<-data.frame(tous_y,lg) #simtest for only one response >summary(simtest(F1_nearey~lg, type="Tukey", data=tousy)) Absolute Error Tolerance: 0.001 Coefficients: Estimate t value Std.Err. p raw p Bonf p adj lgitalien-lgfrancais -0.074 -1.898 0.039 0.066 1 0.495 lgitalien-lganglais -0.064 -1.630 0.039 0.112 1 0.658 lgitalien-lgarabe -0.055 -1.413 0.039 0.167 1 0.780 lgitalien-lgespagnol -0.046 -1.166 0.039 0.252 1 0.891 lgportugais-lgfrancai...

...g-Groupctrl 0 -1 1 0 Groupshort-Groupctrl 0 -1 0 1 Groupshort-Grouplong 0 0 -1 1 Absolute Error Tolerance: 0.001 95 % quantile: 2.519 Coefficients: Estimate 2.5 % 97.5 % t value Std.Err. p raw p Bonf Grouplong-Groupctrl -134.246 -930.286 661.794 -0.425 316.076 0.675 1.000 Groupshort-Groupctrl 978.056 182.016 1774.097 3.094 316.076 0.005 0.016 Groupshort-Grouplong 1112.303 400.303 1824.303 3.934 282.707 0.001 0.002 p adj Grouplong-Groupctrl 0.906 Groupshort-...

Dear all, I am trying to run Wilcoxon Rank-Sum Test with Bonferroni's correction. I have two lists: l0, l1: mapply(function(x,y)wilcox.test(x,y)$p.value, l0, l1) How do I run Bonferroni's correction on mapply? Any help is much apperciated. Thanks, -Raj

...A,60,6)) names(df2)<-c('t','df','p','lower','upper','mean.diff') for (i in 1:60) {df2[i,1:6]<-as.numeric( unlist(t.test(df1[,i+1]~df1$trt,paired=T))[1:6])} Now, I want to adjust the confidence intervals for multiple comparisons. For a Bonferroni-adjustment, I did the following: df2$std.error.of.diff<-df2$mean.diff/df2$t ci<-qt(p=1-(0.05/nrow(df2)),df=df2$df)*df2$std.error.of.diff ci.bonf<-data.frame(lower=df2$mean.diff-ci,upper=df2$mean.diff+ci) I hope this is the correct method. However, I think, the Bonferroni-ad...