The CORR Procedure

Example 2.4 Applications of Fisher’s z Transformation

Subsections:

Testing Whether a Population Correlation Is Equal to a Given Value $\rho _0$
Testing for Equality of Two Population Correlations
Combining Correlation Estimates from Different Samples

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 $\rho _0$

You can use the following statements to test the null hypothesis $H_0\colon \rho = 0.5$ against a two-sided alternative $H_1\colon \rho \neq 0.5$ . 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 $H_0: \rho = \rho _0$

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
Variable	With Variable	N	Sample Correlation	Fisher's z	Bias Adjustment	Correlation Estimate	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, $\rho _1$ and $\rho _2$ . Here, the null hypothesis $H_0\colon \rho _1 = \rho _2$ is tested against the alternative $H_1\colon \rho _1 \neq \rho _2$ .

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 $H_0$ is derived by treating the difference $z_1 - z_2$ as a normal random variable with mean zero and variance $1/(n_1-3) + 1/(n_2-3)$ , where $z_1$ and $z_2$ are Fisher’s z transformation of the sample correlations $r_1$ and $r_2$ , respectively, and where $n_1$ and $n_2$ 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 $\rho _1=\rho _2$ . The sample sizes $n_1=150$ and $n_2=150$ are not large enough to detect the difference $\rho _1-\rho _2=0.05$ at a significance level of $\alpha =0.05$ .

Combining Correlation Estimates from Different Samples

Assume that sample correlations $r_1$ and $r_2$ are computed from two independent samples of $n_1$ and $n_2$ observations, respectively. A combined correlation estimate is given by $\bar{r} = {\tanh } (\bar{z})$ , where $\bar{z}$ is the weighted average of the z transformations of $r_1$ and $r_2$ :

$\bar{z} = \frac{(n_1-3) z_1 + (n_2 -3) z_2}{n_1+n_2-6}$

The following statements compute a combined estimate of $\rho$ 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 $\rho$ . The table shows that a correlation estimate from the combined samples is r=0.2264. The $95\%$ 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