[R] Mann-Whitney U

Wed Aug 15 21:35:35 CEST 2007

R and SPSS are using different but equivalent statistics.  R is using
the rank sum of group1 adjusted for the mean rank. SPSS is using the
rank sum of group2 adjusted for the mean rank. 

Example.
> G1=group1
> G2=group2[-length(group2)] #get rid of the NA
> n1=length(G1) #n1=28
> n2=length(G2) #n2=27
# convert to ranks
> W=rank(c(G1,G2))
> R1=W[1:n1] #put the ranks back into the groups
> R2=W[n1+1:n2]
#Get the sum of the ranks for each group
> W1=sum(R1)
> W2=sum(R2)
#Adjust for mean rank for group 1
> W1-n1*(n1+1)/2
[1] 405.5
#Adjust for mean rank for group 2
> W2-n2*(n2+1)/2
[1] 350.5

W1-n1*(n1+1)/2 gives R's result; W2-n2*(n2+1)/2 gives SPSS's result.

Ties throw a wrench in the works.  R uses a continuity correction by
default, SPSS does not.
Taking out the continuity correction,
> wilcox.test(G1,G2,correct=FALSE)

        Wilcoxon rank sum test

data:  G1 and G2 
W = 405.5, p-value = 0.6433
alternative hypothesis: true location shift is not equal to 0 

Warning message:
cannot compute exact p-value with ties in: wilcox.test.default(G1, G2,
correct = FALSE) 

This p-value is the same as SPSS's.

Consult a serious non-parametrics text.  I used
Lehmann, E. L., Nonparametrics: Statistical methods based on ranks.
1975. Holden-Day. San Francisco, CA.

-----Original Message-----
From: r-help-bounces at stat.math.ethz.ch
[mailto:r-help-bounces at stat.math.ethz.ch] On Behalf Of Natalie O'Toole
Sent: Wednesday, August 15, 2007 1:07 PM
To: r-help at stat.math.ethz.ch
Subject: Re: [R] Mann-Whitney U

Hi,

I do want to use the Mann-Whitney test which ranks my data and then uses
those ranks rather than the actual data.

Here is the R code i am using:

 group1<-
c(1.34,1.47,1.48,1.49,1.62,1.67,1.7,1.7,1.7,1.73,1.81,1.84,1.9,1.96,2,2,
2.19,2.29,2.29,2.41,2.41,2.46,2.5,2.6,2.8,2.8,3.07,3.3)
> group2<-
c(0.98,1.18,1.25,1.33,1.38,1.4,1.49,1.57,1.72,1.75,1.8,1.82,1.86,1.9,1.9
7,2.04,2.14,2.18,2.49,2.5,2.55,2.57,2.64,2.73,2.77,2.9,2.94,NA)
> result <-  wilcox.test(group1, group2, paired=FALSE, conf.level = 
> 0.95,
na.action)

paired = FALSE so that the Wilcoxon rank sum test which is equivalent to
the Mann-Whitney test is used (my samples are NOT paired).
conf.level = 0.95 to specify the confidence level na.action is used
because i have a NA value (i suspect i am not using na.action in the
correct manner)

When i use this code i get the following error message:

Error in arg == choices : comparison (1) is possible only for atomic and
list types

When i use this code:

 group1<-
c(1.34,1.47,1.48,1.49,1.62,1.67,1.7,1.7,1.7,1.73,1.81,1.84,1.9,1.96,2,2,
2.19,2.29,2.29,2.41,2.41,2.46,2.5,2.6,2.8,2.8,3.07,3.3)
> group2<-
c(0.98,1.18,1.25,1.33,1.38,1.4,1.49,1.57,1.72,1.75,1.8,1.82,1.86,1.9,1.9
7,2.04,2.14,2.18,2.49,2.5,2.55,2.57,2.64,2.73,2.77,2.9,2.94,NA)
> result <-  wilcox.test(group1, group2, paired=FALSE, conf.level = 
> 0.95)

I get the following result:

  Wilcoxon rank sum test with continuity correction

data:  group1 and group2
W = 405.5, p-value = 0.6494
alternative hypothesis: true location shift is not equal to 0 

Warning message:
cannot compute exact p-value with ties in: wilcox.test.default(group1,
group2, paired = FALSE, conf.level = 0.95) 

The W value here is 405.5 with a p-value of 0.6494

in SPSS, i am ranking my data and then performing a Mann-Whitney U by
selecting analyze - non-parametric tests - 2 independent samples  and
then checking off the Mann-Whitney U test.

For the Mann-Whitney test in SPSS i am gettting the following results:

Mann-Whitney U = 350.5
 2- tailed p value = 0.643

I think maybe the descrepancy has to do with the specification of the NA
values in R, but i'm not sure.

If anyone has any suggestions, please let me know!

I hope i have provided enough information to convey my problem.

Thank-you, 

Nat
__________________

Natalie,

It's best to provide at least a sample of your data.  Your field names 
suggest 
that your data might be collected in units of mm^2 or some similar 
measurement of area.  Why do you want to use Mann-Whitney, which will
rank 

your data and then use those ranks rather than your actual data?  Unless

your 
sample is quite small, why not use a two sample t-test?  Also,are your 
samples paired?  If they aren't, did you use the "paired = FALSE"
option?

JWDougherty

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