We have Z₁,...,Z_n independent random observations wich are symmetric about an unknown parameter q. To test H_o: q= q_o versus H₁: q> q_o (or H₁: q < q_o or H₁: q ¹ q_o) we use:

 

Assumptions	Test to use
independent observations	sign test
i .i .d. observations	Wilcoxon signed rank test
i.i.d. normal observations	t test

The previous tests also apply to paired data. We have independent paired observations (X₁,Y₁,),...,(X_n,Y_n). We assume that Z_i=Y_i-X_i has a symmetric distribution.

To do a t test, we use:

t.test(x, y=NULL, alternative="two.sided", mu=0, paired=F, var.equal=T, conf.level=.95)

This command does the z-test and the t-test. If y is not indicated, we get the one sample case. The argument alternative has the options: "greater", "less" or "two.sided".

To do a signed rank Wilcoxon test for paired data, we can do the following. Suppose that we want to test Ho:theta=theta_o versus H₁:theta<theta_o, from (X₁,Y₁,),...,(X_n,Y_n). We can do a sign test, a signed rank Wilcoxon test for paired data and a t-test.

*************************
> x_scan()
1: 1.83 0.50 1.62 2.48 1.68 1.88 1.55 3.06 1.30
10:
> y_scan()
1: 0.878 0.647 0.598 2.050 1.060 1.290 1.060 3.140 1.290
10:
> z_y-x
> sum(z>0)
[1] 2
>  pbinom(2,size=9,p=0.5)
[1] 0.08984375
>sum(rank(abs(z))*(z>0))
[1] 5
> wilcox.test(x, y, alternative="greater",paired=T) 
        Exact Wilcoxon signed-rank test
data:  x and y
signed-rank statistic V = 40, n = 9, p-value = 0.0195
alternative hypothesis: true mu is greater than 0
>  t.test(x, y, alternative="greater", mu=0, paired=T, conf.level=.95) 
	Paired t-Test
data:  x and y 
t = 3.0354, df = 8, p-value = 0.0081 
alternative hypothesis: true mean of differences is greater than 0 
95 percent confidence interval:
 0.1673028        NA 
sample estimates:
 mean of x - y 
     0.4318889
*************************

The p-value of the sign test is 0.08984375. This is not enough evidence to reject Ho at the level 0.05. We got 2 positive signs in 9 observations.

The p-value of the Wilcoxon signed rank test is 0.0195. There is enough evidence to reject Ho at the level 0.05. The sum of the positive ranks is small enough: it is only 5. Even, when we get 2 positive signs. These values are small compared to the rest.

The p-value of the t-test is 0.0081. There is enough evidence to reject Ho at the level 0.05.

For more information in the command wilcox.test, do

> help(wilcox.test)

The distribution of the Wilcoxon signed rank statistics is not found in Splus, you can find the Wilcoxon signed rank statistic using this program. Instead of 6, you plug, the corresponding n.

********3a**********************************
n_6
nn_2**n
v_rep(0,times=nn) 
for(i in 1:nn){
for(j in 1:n){
l_((i-1)%%(2**j))+1
m_(2**(j-1))+.5
k_0
if (l>m) k_1
v[i]_v[i]+(k*j) 
}
}
v1_sort(v)
a1_rle(v1)$values
a2_rle(v1)$lengths
a2_a2/nn
a3_cumsum(a2)
a4_1-a3
a_cbind(a1,a2,a3,a4)
dimnames(a)_list(c(NULL),c("value","probab.","cdf","1-cdf"))
******************************************

This program find the exact distribution of the Wilcoxon signed rank statistic. However, it takes a lot of computer time. You can do estimate the distribution of the Wilcoxon signed rank statistic using simulations:

*******3b*************
n_6
N_10000
alpha_.05
rm(xb)
xb_c(1:N)
for(i in 1:N){
xb[i]<-sum(c(1:n)*rbinom(n,1,.5))
}
res<- quantile(xb,probs=c(alpha/2,1-alpha/2))
names(res)_c("t(1-alpha/2)","t(alpha/2)")
xb_sort(xb)
a1_rle(xb)$values
a2_rle(xb)$lengths
a2_a2/N
a3_cumsum(a2)
a4_1-a3
a_cbind(a1,a2,a3,a4)
dimnames(a)_list(c(NULL),c("value","probab.","cdf","1-cdf"))
*********************

The following programs finds theta-hat and a 95 % confidence interval for theta using the data in Example 3.1 of the textbook.

*******3c*************
x_c(1.83,.5,1.62,2.48,1.68,1.88,1.55,3.06,1.3)
y_c(.878,.647,.598,2.05,1.06,1.29,1.06,3.14,1.29)
z_y-x
zabs_abs(z)
ra_rank(zabs)
si_(sign(z)+1)/2
v_sum(si*ra)
n_length(x)
rm(aver,con1,con2,conf)
aver_c((z[1]+z[1])/2)
for(j in 1:n){
for(i in 1:j){
aver_append(aver,(z[i]+z[j])/2)
}
}
aver_aver[-1]
aver_sort(aver)
theta_median(aver)
zalpha_qnorm(.975)
calpha1_((n*(n+1)/4)-zalpha*sqrt(n*(n+1)*(2*n+1)/24)-1/2)
calpha1_round(calpha1)
calpha2_((n*(n+1)/2)+1-calpha1)
diff_sort(differ)
con1_aver[calpha1]
con2_aver[calpha2]  
conf_c(con1,con2)
*********************

We run this program and get theta-hat=-0.46, theta-L=-0.806 and that-U=0.035

>source("3c")
> theta
[1] -0.46
> conf
[1] -0.806  0.035

Comments to: Miguel A. Arcones