z score on Wilcoxon signed ranks test? I was wondering if anyone might have some answers to a question regarding running a Wilcoxon signed rank test in R. 
When looking up how to report our scores, it tells us to report a Z score. However, running a Wilcoxon signed rank test in R produces a V, not a Z, score. 
Are these two interchangeable? Or different scores? If so, how would I calculate z?
 A: Given the number of elements in your samples (and the number of ties in them), you can transform one (the test stat, say in your case the $V$ stat though as I show below, this will also work for the $W$ stat) to the other (the corresponding z score) fairly easily. 
The formula's for the W and V stats are widely available (here and here). In fact, these computations are already done in the R code when exact = FALSE --to compute the p-val-- they are just not exported. 
To have R export these statistics, copy to the source code to an R script. 
On line 397-403 you see:
    RVAL <- list(statistic = STATISTIC,
                 parameter = NULL,
                 p.value = as.numeric(PVAL),
                 null.value = mu,
                 alternative = alternative,
                 method = METHOD,
data.name = DNAME)

Change it to:
    RVAL <- list(statistic = STATISTIC,
                 parameter = NULL,
                 p.value = as.numeric(PVAL),
                 null.value = mu,
                 alternative = alternative,
                 method = METHOD,
                 z_val = z,
data.name = DNAME)

Save the function under a different name. Say wilcox_test. Then, whenever the $z$ score is available (when exact = FALSE), wilcox_test will return it. Example;
set.seed(123);
wilcox_test(runif(4000), rnorm(100), exact = FALSE)$z_val
      W 
8.77776 

or: 
 set.seed(123);
wilcox_test(runif(4000), exact = FALSE)$z_val
       V 
54.77567 

A: See Section 7.2 of BBR which discusses a simple, accurate $z$ test statistic for the Wilcoxon signed-rank test.  $z$ equals the sum of signed ranks divided by the square root of the sum of their squares.  This handles ties well also.
A: No, they are not interchangeable. The $V$ in Wilcoxon signed test from R wilcox test is "the sum of ranks assigned to the differences with positive sign".
Let me show you by using ZeaMays data.
First we do the Wilcoxon signed test using wilcox.test funciton.
 install.packages("HistData")
 library(HistData)
 data(ZeaMays)
 head(ZeaMays)
 wilcox.test(ZeaMays$cross, ZeaMays$self, paired=TRUE)

The results are:
   # Wilcoxon signed rank test
   # data:  ZeaMays$cross and ZeaMays$self
   # V = 96, p-value = 0.04126
   # alternative hypothesis: true location shift is not equal to 0

Now let us calculate V by hand
 diff<-ZeaMays$cross-ZeaMays$self #calculate paired difference
 diff
 rank_d<-rank(abs(diff))*sign(diff) #rank the difference and assign the sign

Next we calculate "the sum of ranks assigned to the differences with positive sign". i.e $V$
p_rank<-c() #postive ranks only
for (i in 1:15){
if (rank_d[i]>0){
p_rank[i]=rank_d[i]}
}
V<-sum(p_rank,na.rm=TRUE)
V
#[1] 96

You can see $V=96$ which is exactly the same as the $V$ of the wilcox.test.
Next what is $Z$ score?
$E_{H_0}(V)=\frac{n(n+1)}{4} =\frac{15\times 16}{4}=60$
$Var_{H_0}(V)=\frac{n(n+1)(2n+1)}{24}=\frac{15\times 16\times 31}{24}$
and $Z=\frac{V-E(V)}{\sqrt{Var(V)}}\sim N(0,1)$
i.e the $Z$  can be shown converge to standard normal.
Therefore your $z$ score is:$\frac{96-60}{\sqrt{\frac{15\times 16\times31}{24}}}=2.044663$
Which correspond the two sided p value 0.04088813 
  2*(1-pnorm(z,0,1))
  #p=0.04088813$

The p value is a little different, maybe because wilcox.test used some correction.
A: I do the following to obtain the Z-score when doing a Wilcoxon signed rank test.
test<-wilcox.test(mtdata$x, mydata$y, paired=TRUE, exact=TRUE) 

print(test) # get the results of the Wilcoxon signed rank test

Zstat<-qnorm(test$p.value/2)  # obtain the Z-score 
abs(Zstat)/sqrt(20)   

