Displaying 2 results from an estimated 2 matches for "cntc".
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2005 Jul 04
2
Lack of independence in anova()
...ted and the 2xk design is
balanced, theory requires that the tests for interaction and row effects be
independent. In my program, appended below, this would translate to cntT
(approx)= cntR*cntI/N if all R routines were functioning correctly. They
aren't.
sim2=function(size,N,p){
cntR=0
cntC=0
cntI=0
cntT=0
cntP=0
for(i in 1:N){
#generate data
v=gendata(size)
#analyze after build(ing) design containing data
lm.out=lm(yield~c*r,build(size,v))
av.out=anova(lm.out)
#if column effect is significant, increment cntC
if (av.out[[5]][1]<=p)cntC=cntC+1...
2012 Sep 11
0
Question about logistic regression with ordered factor variable using the rms package (prev.Design)
...where cnct_o is the
corresponding value for the given concentration.
Probability <- inv.logit ( Odds ), the function inv.logit from package
¡§car¡¨.
The results of the glm are in the table below, which are first the odds
and then the probability¡¦s ( inv.logit (odds)).
#####
glm odds cntc:
test.value 25 cnct_o.L cnct_o.Q cnct_o.C
cnct_o^4
0 -0.90218 0.534169 0.318649 -2.96706
-0.34231
0.5 0.138132 1.574477 1.358957 -1.92675
0.698001
1 1.17844 2.614785...