Displaying 5 results from an estimated 5 matches for "tost".
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host
2006 Feb 08
3
print formula on plot
..."l", main="My nice plot of the
estimated function")
zs<-format(z,digits=4,scientific=FALSE,trim=TRUE)
text(-0.9,7,expression(1.54*x^2)) # is what I want,
but DYNAMIC
text(-0.9,6,expression(paste(zf[1],x^3))) # not really
text(-0.9,5,expression(paste(toSting(zf[1]),x^3))) # not really
#using z (double) instead of zf (string) does not help.
So my question is:
How do I evaluate zf[1] from the variable to it's (String) value? Here
it is used within an expression: this makes all the trouble.
Thanks for help, I tried 2 hrs now...
Thomas
2004 Feb 06
2
Normality Test on several groups
Hi,
I use ks.test or lillie.test to verify a normal distribution. It's performed
for a group
My users use SigmaStat software and a One Way ANOVA on several groups
In the result page there is a probability value to determine if Normality
test is failed or passed
So, how can i retrieve this probability value on several groups?
Is there another function in R to verify normality on several
2004 Feb 17
4
normality test
Hello,
I am analysing several samples whose sizes are from 9 to 110.
I would like to test their distribution with R,
whether they are normal or not.
I wonder which test for normality from R should I use .
Thank you for help.
Samuel
Samuel BERTRAND
Doctorant
Laboratoire de Biomecanique
LBM - ENSAM - CNRS UMR 8005
151, bd de l'Hopital
75013 PARIS
Tel. +33 (0) 1 44 24 64 53
Fax +33 (0) 1
2006 Apr 17
1
Equivalence test and factors
...or more levels? Can the package, or the
package plus another function, be used to test the similarity of the
predictions produced by two or more factor levels from nested or
otherwise correlated data? Which of the equivalence functions
accomodates unbalanced designs? Using functions such as tost.data to
compare observations associated with the 2 predictors in my data is
complicated both by the unbalanced design and the lack of accomodation
of the nested design.
Note that I am running R 2.0.1 which predates the version in
which "equivalence" was developed.
Thank you in advan...
2011 Sep 14
0
Confidence interval or p-value for difference in two c-statistics
...0.017 4.94 0e+00 1500
neurodat2 0.481 -0.038 0.038 0.011 3.44 6e-04 1500
Is there a way to determine if the c-statistics are significantly different/similar to one another across data sets (i.e. I want to compare the c-statistic for gendat1 with that for gendat2). I was considering using the 'tost' function within the equivalence package but that doesn't seem appropriate either.
Many thanks,
Laura
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