Bonus Using permutation test SAS code , do the following: Trauma data Metabolic
ID: 3071185 • Letter: B
Question
Bonus
Using permutation test SAS code , do the following:
Trauma data
Metabolic Expenditure
Nontrauma patients: 20.1 22.9 18.8 20.9 20.9 22.7 21.4 20
trauma patients: 38.5 25.8 22 23 37.6 30 24.5
Part A
Build the permutation distribution (using 5,000 permutations) for the rank sum statistic for the Trauma
data used above. Use SAS to fit/overlay a normal curve to the resulting histogram. Compare the mean and
standard deviation of this normal curve that was fit to the permutation/randomization distribution to the
mu and sigma you found earlier in the homework.
Part B
Compare the one-sided p-value found in this permutation distribution with the one found in prior questions.
Explanation / Answer
We shall use R to answer this
Nontrauma<- c(20.1,22.9,18.8,20.9,20.9,22.7,21.4,20)
Trauma<-c(38.5,25.8,22,23,37.6,30,24.5)
# rank functions
rank(Nontrauma)
rank(Trauma)
# z stats is x-mean/sd
mtr<-mean(Trauma)
SDtr<-sd(Trauma)
zTrauma<- (Trauma-mtr)/SDtr
# for non trauma patients
mntr<-mean(Nontrauma)
SDntr<-sd(Nontrauma)
zNonTrauma<- (Nontrauma-mntr)/SDntr
# wilcox test for the patients
wilcox.test(Nontrauma,Trauma)
# medians for the patients
medTrauma<- median(Trauma)
medNTrauma <- median(Nontrauma)
# write a function in R for Z test
z.test = function(a, mu, var){
zeta = (mean(a) - mu) / (sqrt(var / length(a)))
return(zeta)
}
z.test(Trauma,medTrauma,SDtr^2)
z.test(Nontrauma,medNTrauma,SDntr^2)
The results of the above snippet are
> rank(Nontrauma)
[1] 3.0 8.0 1.0 4.5 4.5 7.0 6.0 2.0
> rank(Trauma)
[1] 7 4 1 2 6 5 3
> mtr
[1] 28.77143
> SDtr
[1] 6.835377
> zTrauma
[1] 1.4232677 -0.4347132 -0.9906445 -0.8443468 1.2915997 0.1797372 -0.6249002
> mntr
[1] 20.9625
> SDntr
[1] 1.379376
> zNonTrauma
[1] -0.62528267 1.40462048 -1.56773770 -0.04531034 -0.04531034 1.25962740
[7] 0.31717237 -0.69777921
> wilcox.test(Nontrauma,Trauma)
Wilcoxon rank sum test with continuity correction
data: Nontrauma and Trauma
W = 2, p-value = 0.00314
alternative hypothesis: true location shift is not equal to 0
> z.test(Trauma,medTrauma,SDtr^2)
[1] 1.150143
> z.test(Nontrauma,medNTrauma,SDntr^2)
[1] 0.128157