To test the hypothesis that the dispersion of the systolic blood pressure in the two populations of healthy subjects (status=0) and subjects with hypertension (status=1) is the same. The dataset contains observations for status=0 and observations for status=1 (dataset in Table A.1).

SAS code

proc npar1way data=blood_pressure correct=no st;
 var mmhg;
 class status;
 exact st;
run;

                   The NPAR1WAY Procedure
           Siegel–Tukey Scores for Variable mmhg
              Classified by Variable status
                 Sum of    Expected   Std Dev     Mean
status    N      Scores    Under H0   Under H0    Score
-------------------------------------------------------
0        25      655.0      700.0     59.001584   26.20
1        30      885.0      840.0     59.001584   29.50
              Average scores were used for ties.
                 Siegel–Tukey Two-Sample Test
                 Statistic             655.0000
                 Z                      −0.7627
                 One-Sided Pr <  Z       0.2228
                 Two-Sided Pr > |Z|      0.4456

R code

# Helper functions to find even or odd numbers
is.even <- function(x) x %% 2 == 0
is.odd  <- function(x) x %% 2 == 1
# Create a sorted matrix with first column the blood
# pressure and second column the status
data<-blood_pressure[order(blood_pressure$mmhg),]
x<-c(data$mmhg)
x<-cbind(x,data$status)
# If the sample size is odd then remove the observation
# in the middle
if (is.odd(nrow(x))) x<-x[-c(nrow(x)/2+0.5),]
# Calculate the (remaining) sample size
n<-nrow(x)
# y returns the Siegel–Tukey scores
y<-rep(0,times=n)
# Assigning the scores
for (i in seq(along=x)) {
 if (1<i & i <= n/2 & is.even(i))
 {
  y[i]<-2*i
 }
 else if (n/2<i & i<=n & is.even(i))
 {
  y[i]<-2*(n-i)+2
 }
 else if (1<=i & i <=n/2 & is.odd(i))
 {
  y[i]<-2*i-1
 }
 else if (n/2<i & i < n & is.odd(i))
 {
  y[i]<-2*(n-i)+1
 }
}
# Now mean scores must be created if necessary
t<-tapply(y,x[,1],mean) # Get mean scores for tied values
v<-strsplit(names(t), “ ”) # Get mmhg values
# r
r<-rep(0,times=n)
# Assign ranks and mean ranks to r
for (i in seq(along=r))
{
 for (j in seq(along=v))
 {
 if (x[i,1]==as.numeric(v[j])) r[i]=t[j]
 }
}
# Now calculate the test statistics S_0 (status 0)
# and S_1 (status 1) for both samples
S_0<-0
S_1<-0
for (i in seq(along=r)) {
 if(x[i,2]==0) S_0=S_0+r[i]
 if(x[i,2]==1) S_1=S_1+r[i]
}
# Calculate sample sizes for status=0 and status=1
n1<-sum(x[,2]==0)
n2<-sum(x[,2]==1)
# Choose the test statistic which belongs to the smallest
# sample size
if (n1<=n2) {
  # Choose the smaller |z| value
  z1<-(2*S_0-n1*(n+1)+1)/sqrt((n1*n2*(n+1)/3))
  z2<-(2*S_0-n1*(n+1)-1)/sqrt((n1*n2*(n+1)/3))
  if (abs(z1)<=abs(z2)) z=z1 else z=z2
  pvalue_B=1-pnorm(-abs(z))
  pvalue_C=pnorm(-abs(z))
}
if (n1>n2) {
  # Choose the smaller |z| value
  z1<-(2*S_1-n2*(n+1)+1)/sqrt((n1*n2*(n+1)/3))
  z2<-(2*S_1-n2*(n+1)-1)/sqrt((n1*n2*(n+1)/3))
  if (abs(z1)<=abs(z2)) z=z1 else z=z2
  pvalue_B=pnorm(-abs(z));
  pvalue_C=1-pnorm(-abs(z));
}
pvalue_A=2*min(pnorm(-abs(z)),1-pnorm(-abs(z)));
# Output results
print(“Siegel–Tukey test”)
n
S_0
S_1
z
pvalue_A
pvalue_B
pvalue_C

[1] “Siegel–Tukey test”
> n
[1] 54
> S_0
[1] 600.5
> S_1
[1] 884.5
> z
[1] -1.027058
> pvalue_A
[1] 0.3043931
> pvalue_B
[1] 0.8478035
> pvalue_C
[1] 0.1521965

To test the hypothesis that the dispersion of the systolic blood pressure in the two populations of healthy subjects (status=0) and subjects with hypertension (status=1) is the same. The dataset contains observations for status=0 and observations for status=1 (dataset in Table A.1).

SAS code

proc npar1way data=blood_pressure correct=no ab;
 var mmhg;
 class status;
 exact ab;
run;

                  The NPAR1WAY Procedure
           Ansari–Bradley Scores for Variable mmhg
               Classified by Variable status
                    Sum of      Expected   Std Dev  Mean
status  N   Scores  Under H0    Under H0   Score    Score
----------------------------------------------------------
0       25  334.0   356.363636  29.533137  13.360   13.360
1       30  450.0   427.636364  29.533137  15.000   15.000
             Average scores were used for ties.
              Ansari–Bradley Two-Sample Test
              Statistic             334.0000
              Z                      -0.7572
              One-Sided Pr < Z        0.2245
              Two-Sided Pr < |Z|      0.4489

• The parameter ab enables the Ansari–Bradley test of the procedure NPAR1WAY.
• correct=value is optional. If value is YES than a continuity correction for the normal approximation is used. The default is NO.
• exact ab is optional and applies an additional exact test. Note, the computation of an exact test can be very time consuming. This is the reason why in this example no exact p-values are given in the output.
• Besides the two-sided p-value SAS also reports a one-sided p-value; which one is printed depends on the Z-statistic. If the value of the Z-statistic is greater than zero the right-sided p-value is printed. If it is less than or equal to zero the left-sided p-value is printed.
• In this example the sum of scores for the healthy subjects is compared with for the people with hypertension. So there is evidence that the scale of healthy subjects is higher than the scale of unhealthy subjects. In fact the variance of the healthy subjects is and the variance of the unhealthy subjects is . Therefore the p-value for hypothesis (C) is and the p-value for hypothesis (B) is .

R code

x<-blood_pressure$mmhg[blood_pressure$status==0]
y<-blood_pressure$mmhg[blood_pressure$status==1]
ansari.test(x,y,exact=NULL,alternative =“two.sided”)

       Ansari–Bradley test
data:  x and y
AB = 334, p-value = 0.4489
alternative hypothesis: true ratio of scales is not
                                                equal to 1

• exact=value is optional. If value is not specified or TRUE an exact p-value is computed if the combined sample size is less than . If it is NULL or FALSE the approximative p-value is computed. In the case of ties R cannot compute an exact test.
• R tests equivalent hypotheses of the type vs for hypothesis (A), and so on.
• alternative=“value” is optional and defines the type of alternative hypothesis: “two.sided”= true ratio of scales is not equal to 1 (A); “greater”=true ratio of scales is greater than 1 (C); “lower”=true ratio of scales is less than 1 (B). Default is “two.sided”.

To test the hypothesis that the dispersion of the systolic blood pressure in the two populations of healthy subjects (status=0) and subjects with hypertension (status=1) is the same. The dataset contains observations for status=0 and observations for status=1 (dataset in Table A.1).

SAS code

proc npar1way data=blood_pressure correct=no mood;
 var mmhg;
 class status;
 exact mood;
run;

SAS output

                 The NPAR1WAY Procedure
             Mood Scores for Variable mmhg
             Classified by Variable status
              Sum of     Expected    Std Dev       Mean
status   N    Scores     Under H0    Under H0      Score
----------------------------------------------------------
0        25   6864.0     6300.0      837.786511    274.560
1        30   6996.0     7560.0      837.786511    233.200
               Average scores were used for ties.
                     Mood Two-Sample Test
               Statistic             6864.0000
               Z                        0.6732
               One-Sided Pr>  Z        0.2504
               Two-Sided Pr> |Z|       0.5008

• The parameter mood enables the Mood test of the procedure NPAR1WAY.
• correct=value is optional. If value is YES than a continuity correction for the normal approximation is used. The default is NO.
• exact mood is optional and applies an additional exact test. Note, the computation of an exact test can be very time consuming. This is the reason why in this example no exact p-values are given in the output.
• Besides the two-sided p-value SAS also reports a one-sided p-value; which one is printed depends on the Z-statistic. If the observed value of the Z-statistic is greater than zero the right-sided p-value is printed. If it is less than or equal to zero the left-sided p-value is printed.
• In this example the sum of scores for the healthy subjects is compared with for the people with hypertension. So there is evidence that the scale of healthy subjects is higher than the scale of unhealthy subjects. In fact the variance of the healthy subjects is and the variance of the unhealthy subjects is . Therefore the p-value for hypothesis (C) is and the p-value for hypothesis (B) is .

R code

x<-blood_pressure$mmhg[blood_pressure$status==0]
y<-blood_pressure$mmhg[blood_pressure$status==1]
mood.test(x,y,alternative =“two.sided”)

 Mood two-sample test of scale
data:  x and y
Z = 0.6765, p-value = 0.4987
alternative hypothesis: two.sided

• R handles ties differently to SAS. Instead of mid ranks a procedure by Mielke is used (Mielke 1967).
• alternative=“value” is optional and defines the type of alternative hypothesis: “two.sided”= true ratio of scales is not equal to 1 (A); “greater”=true ratio of scales is greater than 1 (C); “lower”=true ratio of scales is less than 1 (B). Default is “two.sided”.