Displaying 6 results from an estimated 6 matches for "mean_1".
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mean1
2009 Dec 18
1
linear contrasts for trends in an anova
...In the following example, based on the type of 2x3 ANOVA I'm trying to perform, does the linear polynomial contrast generated by contr.poly allow me to test for a linear trend across groups?
doi=data.frame(
Group=c(
rep(1, 5), rep(2, 5), rep(3, 5), rep(1, 5), rep(2, 5), rep(3, 5)),
Mean_1=c(
rnorm(5, mean=3, sd=1), rnorm(5, mean=5.7, sd=1.2), rnorm(5, mean=7, sd=1.3),
rnorm(5, mean=3.5, sd=1), rnorm(5, mean=3, sd=1.2), rnorm(5, mean=4.2, sd=1.7)),
eventType=c( rep("errors", 15), rep("stops", 15)))
doi$Group=factor(doi$Group, labels=c("Control"...
2007 Oct 22
0
anova and contratst
...t", "short", "short", "long",
"long", "long")
Subject = c("66101_2", "66112_2", "66118_2", "66119", "66121",
"66123", "66026_3", "66030_2", "66044_3")
Mean_1 = c(-0.399281, 6.951385, 5.986774, 3.485916, 7.081511,
1.705302, 7.859186, 4.573201, 3.931118)
Mean_2 = c(0.350127, 5.863118, 3.826101, 0.999821, 9.099477,
2.376836, 5.384967, 5.950012, 7.688557)
roi.errs = data.frame(Group, Subject, Mean_1, Mean_2)
# planned contrasts
cmat = rbind(
"c...
2010 Aug 05
3
eval-parse and lme in a loop
...e code is node being replaced as it is in the call to aov. Anybody got any idea what's wrong?
rm(list=ls())
library('nlme')
df=data.frame(
Group=rep(c("ctrl", "short", "long"), each=9, times=1),
Subject=paste("S", 1:27, sep=""),
Mean_1=c(-2.335482, 8.269832, 6.704734, -2.355664, 0.724432, 11.212987, 5.966652, 10.741065, 3.567522, 5.359832, 4.899951, 4.084452, 5.976326, 8.634508, 5.750942, 7.455622, 6.823843, 0.903801, 13.198157, 5.238985, 5.171453, 10.273715, 5.184785, 6.225874, 8.889997, 12.096968, 9.843028),...
2010 Mar 25
1
Expected pairwise.student.t and TukeyHSD behavior?
...esult.
As an aside, should, in cases like this, pairwise.t.test return an error or warning indicating what might be the cause for the NA's being returned? Also, should TukeyHSD similarly return a warning about this situation?
Thanks again for any insights and feedback.
length_1<-1
mean_1<-0.0
sd_1<-0.0
distribution_1<-rnorm(length_1, mean=mean_1, sd=sd_1)
length_2<-750
mean_2<-0.5
sd_2<-0.0
distribution_2<-rnorm(length_2, mean=mean_2, sd=sd_2)
length_3<-850
mean_3<-0.1
sd_3<-0.65
distribution_3<-rnorm(length_3, mean=mean_3, sd=sd_3)
length_4<...
2013 Jun 04
0
choose the lines2
...??????????????????????? ))
??????????????????????????????????? })
?? lst6<-lst5[lapply(lst5,nrow)!=0]
?? names(lst6)<- unlist(lapply(lst6,function(x) unique(x$patient_id)))
?? Mean_01<-do.call(rbind,lapply(lst6,function(x) cbind(Mean_Middle0=mean(x[seq(nrow(x))%%3==2,"basdai_d"]),Mean_1=mean(x[seq(nrow(x))%%3==0,"basdai_d"]))))
rownames(Mean_01)<- names(lst6)
? ?? lst7<-list(lst6,Mean_01)
?? lst7
?? #lapply(lst7,head,2)??
?? }?????????????????
fun2(dat.bru)
head(fun2(dat.bru)[[1]],3)
#$`2`
#??? X patient_id number responsed_at? t basdai_d evnmt_brutal
#13 13???...
2006 Sep 01
1
integration problem with gamma function
...Ni * si^2 + Di^2
Dsii <- (Nii-1) / Nii * sii^2 + Dii^2
DsD <- (Ni * Dsi + Nii * Dsii) / NN
ss <- sqrt(NN * (DsD - DD^2) / (NN-1))
### descriptive statistics
print(" \n------------- Data ---------------------------------------\n")
print(paste("N_1 = ",Ni ," : Mean_1 ? s_1 = ", Di," ? ",si, sep=""))
print(paste("N_2 = ",Nii," : Mean_2 ? s_2 = ", Dii," ? ",sii,
sep=""))
print(paste("N = ", NN ," : Mean_comb ? s_comb = ", DD," ? ",ss, sep=""))
pr...