Commonality Analysis in Multiple Regression
Suppose that you are interested in a model predicting Work-Life Imbalance (WLI) from two
predictors, obsessive/compulsive personality (OCP) and workaholism (measured with the Work
Addiction Risk Test, WART), all measures being continuous. First you obtain the zero-order
correlations.
FILE='C:\Users\Vati\Documents\Research-Misc\Aziz\Uhrich_Validity_WAQ&WART\JBAM\Uhrich_WAQ2.sav'.
DATASET NAME DataSet1 WINDOW=FRONT.
CORRELATIONS
/VARIABLES=WART OCP WLI
/PRINT=TWOTAIL NOSIG
/MISSING=PAIRWISE.
Correlations
Correlations
WART
Pearson
Correlation
OCP
WLI
.631**
.611**
Sig. (2-tailed)
.000
N
188
188
Pearson
.631**
1
Correlation
OCP
Sig. (2-tailed)
.000
N
188
188
Pearson
.611**
.515**
Correlation
WLI
Sig. (2-tailed)
.000
.000
N
188
188
**. Correlation is significant at the 0.01 level (2-tailed).
.000
188
WART
1
.515**
.000
188
1
188
Notice that the predictors are fairly well correlated with each other as well as with WLI. In this
situation we can expect that there will be considerable redundancy between the two predictors (with
respect to their association with WLI).
REGRESSION
/MISSING LISTWISE
/STATISTICS COEFF OUTS R ANOVA COLLIN TOL ZPP
/CRITERIA=PIN(.05) POUT(.10)
/NOORIGIN
/DEPENDENT WLI
/METHOD=ENTER WART OCP.
Model
1
R
.634a
R Square
.401
Model Summary
Adjusted R Square
.395
Std. Error of the Estimate
.65634
a. Predictors: (Constant), OCP, WART
Commonality.docx
2
Model
Regression
Residual
Total
1
Sum of Squares
53.442
79.694
133.136
ANOVAa
df
2
185
187
Mean Square
26.721
.431
F
62.030
Sig.
.000b
a. Dependent Variable: WLI
b. Predictors: (Constant), OCP, WART
Coefficientsa
Unstandardized Coefficients
Model
B
(Constant)
WART
OCP
1
-.336
.846
1.184
Std. Error
.263
.130
.405
Standardized
Coefficients
Beta
.476
.215
t
Sig.
-1.278
6.495
2.928
.203
.000
.004
Coefficientsa
Model
1
Correlations
Zero-Order
Semi-Partial
(Constant)
WART
OCP
.611
.515
Collinearity Statistics
Tolerance
VIF
.369
.167
.602
.602
a. Dependent Variable: WLI
Notice that for both predictors the semipartial coefficient is considerably lower than the
zero-order coefficient.
a+b+c+d=1
rY21  b  c
sr12 
b
b
(a  b  c  d )  1
rY22  d  c
r122  c  e
RY212  b  c  d
c = redundancy (aka communality)
1.660
1.660
3
Although not commonly done, it might be informative to estimate the size of area c, the
redundancy between the predictors with respect to their association with Y. For a trivariate regression,
here is how to estimate that proportion of variance: c  rY21  rY22  RY212 -- that is, (b+c) + (c+d) –
(b+c+d). For our data, that is .6112 + .5152 - .401 = .2375. Alternatively,
c  RY212  sr12  sr22  .401 .3692  .1672  .2370 .
If you wish to test whether this commonality differs significantly from zero, you can construct a t
test like this: Multiply the squared commonality coefficient by the corrected total sum of squares in Y.
For our data, .237(133.136) = 31.55. Divide this mean square (df = 1) by the residual mean square to
31.55
obtain F. Take the square root of that F to obtain t. For our data, t 
 8.556 .
.431
Because the coefficients were rounded to three digits, there may be a little rounding error here.
If you suffer from obsessive/compulsive disorder, that probably concerns you. Either take a selective
serotonin reuptake inhibitor or get the sums of squares with more precision, as shown below.
REGRESSION
/MISSING LISTWISE
/STATISTICS COEFF OUTS R ANOVA
/CRITERIA=PIN(.05) POUT(.10)
/NOORIGIN
/DEPENDENT WLI
/METHOD=ENTER WART.
Model
Sum of Squares
Regression
49.749
1
Residual
83.387
Total
133.136
a. Dependent Variable: WLI
b. Predictors: (Constant), WART
ANOVAa
df
1
186
187
Mean Square
49.749
.448
F
110.968
Sig.
.000b
Mean Square
35.271
.526
F
67.036
Sig.
.000b
REGRESSION
/MISSING LISTWISE
/STATISTICS COEFF OUTS R ANOVA
/CRITERIA=PIN(.05) POUT(.10)
/NOORIGIN
/DEPENDENT WLI
/METHOD=ENTER OCP.
Model
1
Regression
Residual
Total
Sum of Squares
35.271
97.864
133.136
a. Dependent Variable: WLI
ANOVAa
df
1
186
187
b. Predictors: (Constant), OCP
4
SSCommonalit
y
 SSY 1  SSY 2  SSY 12  49.749  35.271  53.442  31.578 . Of all the variance in
Y, 31.578/133.136 = 23.72 % is common to X1 and X2. To test the null that commonality = 0,
t (185) 
MSCommonalit y
MSError

31.578
 8.560 , p < .001.
.431
In December or 2102, a correspondent asked me “If the r2 between variable X1 and Y = a and the
sr2 between the same two variables, with variable X2 partialled out, is sr2 = b, what is the appropriate
test to determine if these two correlations are significantly different?
My response: “If you look back at the Venn Diagram above, you will see that the difference
between sr2 for X1 and r2 for X1 is area c, the commonality for the two predictors. The appropriate test
is that I have outlined above.”
Recommended Reading: Zientek, L. R. & Thompson, B. (2006). Commonality analysis:
Partitioning variance to facilitate better understanding of data. Journal of Early Intervention, 28,
299-307.
Karl L. Wuensch, December, 2012
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