The CORR Procedure

Example 2.4 Applications of Fisher’s z Transformation

Subsections:

This example illustrates some applications of Fisher’s z transformation. For details, see the section Fisher’s z Transformation.

The following statements simulate independent samples of variables X and Y from a bivariate normal distribution. The first batch of 150 observations is sampled using a known correlation of 0.3, the second batch of 150 observations is sampled using a known correlation of 0.25, and the third batch of 100 observations is sampled using a known correlation of 0.3. 

data Sim (drop=i);
do i=1 to 400;
  X = rannor(135791);
  Batch = 1 + (i>150) + (i>300);
  if Batch = 1 then Y = 0.3*X + 0.9*rannor(246791);
  if Batch = 2 then Y = 0.25*X + sqrt(.8375)*rannor(246791);
  if Batch = 3 then Y = 0.3*X + 0.9*rannor(246791);
  output;
end;
run;

This data set will be used to illustrate the following applications of Fisher’s z transformation:

  • testing whether a population correlation is equal to a given value

  • testing for equality of two population correlations

  • combining correlation estimates from different samples

Testing Whether a Population Correlation Is Equal to a Given Value

You can use the following statements to test the null hypothesis against a two-sided alternative . The test is requested with the option FISHER(RHO0=0.5). 

title 'Analysis for Batch 1';
proc corr data=Sim (where=(Batch=1)) fisher(rho0=.5);
   var X Y;
run;

Output 2.4.1 displays the results based on Fisher’s transformation. The null hypothesis is rejected since the p-value is less than 0.0001.

Output 2.4.1: Fisher’s Test for

Analysis for Batch 1

The CORR Procedure

Pearson Correlation Statistics (Fisher's z Transformation)
Variable With Variable N Sample Correlation Fisher's z Bias Adjustment Correlation Estimate   H0:Rho=Rho0
95% Confidence Limits Rho0 p Value
X Y 150 0.22081 0.22451 0.0007410 0.22011 0.062034 0.367409 0.50000 <.0001


Testing for Equality of Two Population Correlations

You can use the following statements to test for equality of two population correlations, and . Here, the null hypothesis is tested against the alternative . 

ods output FisherPearsonCorr=SimCorr;
title 'Testing Equality of Population Correlations';
proc corr data=Sim (where=(Batch=1 or Batch=2)) fisher;
   var X Y;
   by Batch;
run;

The ODS OUTPUT statement saves the "FisherPearsonCorr" table into an output data set in the CORR procedure. The output data set SimCorr contains Fisher’s z statistics for both batches.

The following statements display (in Output 2.4.2) the output data set SimCorr: 

proc print data=SimCorr;
run;

Output 2.4.2: Fisher’s Correlation Statistics

Obs Batch Var WithVar NObs Corr ZVal BiasAdj CorrEst Lcl Ucl pValue
1 1 X Y 150 0.22081 0.22451 0.0007410 0.22011 0.062034 0.367409 0.0065
2 2 X Y 150 0.33694 0.35064 0.00113 0.33594 0.185676 0.470853 <.0001


The p-value for testing is derived by treating the difference as a normal random variable with mean zero and variance , where and are Fisher’s z transformation of the sample correlations and , respectively, and where and are the corresponding sample sizes.

The following statements compute the p-value in Output 2.4.3: 

data SimTest (drop=Batch);
   merge SimCorr (where=(Batch=1) keep=Nobs ZVal Batch
                  rename=(Nobs=n1 ZVal=z1))
         SimCorr (where=(Batch=2) keep=Nobs ZVal Batch
                  rename=(Nobs=n2 ZVal=z2));
   variance = 1/(n1-3) + 1/(n2-3);
   z = (z1 - z2) / sqrt( variance );
   pval = probnorm(z);
   if (pval > 0.5) then pval = 1 - pval;
   pval = 2*pval;
run;

proc print data=SimTest noobs;
run;

Output 2.4.3: Test of Equality of Observed Correlations

n1 z1 n2 z2 variance z pval
150 0.22451 150 0.35064 0.013605 -1.08135 0.27954


In Output 2.4.3, the p-value of 0.2795 does not provide evidence to reject the null hypothesis that . The sample sizes and are not large enough to detect the difference at a significance level of .

Combining Correlation Estimates from Different Samples

Assume that sample correlations and are computed from two independent samples of and observations, respectively. A combined correlation estimate is given by , where is the weighted average of the z transformations of and :

The following statements compute a combined estimate of by using Batch 1 and Batch 3: 

ods output FisherPearsonCorr=SimCorr2;
proc corr data=Sim (where=(Batch=1 or Batch=3)) fisher;
   var X Y;
   by Batch;
run;

data SimComb (drop=Batch);
   merge SimCorr2 (where=(Batch=1) keep=Nobs ZVal Batch
                   rename=(Nobs=n1 ZVal=z1))
         SimCorr2 (where=(Batch=3) keep=Nobs ZVal Batch
                   rename=(Nobs=n2 ZVal=z2));
   z = ((n1-3)*z1 + (n2-3)*z2) / (n1+n2-6);
   corr = tanh(z);
   var = 1/(n1+n2-6);
   zlcl = z - probit(0.975)*sqrt(var);
   zucl = z + probit(0.975)*sqrt(var);
   lcl= tanh(zlcl);
   ucl= tanh(zucl);
   pval= probnorm( z/sqrt(var));
   if (pval > .5)  then pval= 1 - pval;
   pval= 2*pval;
run;

proc print data=SimComb noobs;
   var n1 z1 n2 z2 corr lcl ucl pval;
run;

Output 2.4.4 displays the combined estimate of . The table shows that a correlation estimate from the combined samples is r=0.2264. The confidence interval is (0.10453,0.34156), using the variance of the combined estimate. Note that this interval contains the population correlation 0.3.

Output 2.4.4: Combined Correlation Estimate

n1 z1 n2 z2 corr lcl ucl pval
150 0.22451 100 0.23929 0.22640 0.10453 0.34156 .000319748