4

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 / OCP / WLI
WART / Pearson Correlation / 1 / .631** / .611**
Sig. (2-tailed) / .000 / .000
N / 188 / 188 / 188
OCP / Pearson Correlation / .631** / 1 / .515**
Sig. (2-tailed) / .000 / .000
N / 188 / 188 / 188
WLI / Pearson Correlation / .611** / .515** / 1
Sig. (2-tailed) / .000 / .000
N / 188 / 188 / 188
**. Correlation is significant at the 0.01 level (2-tailed).

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 Summary
Model / R / R Square / Adjusted R Square / Std. Error of the Estimate
1 / .634a / .401 / .395 / .65634
a. Predictors: (Constant), OCP, WART
ANOVAa
Model / Sum of Squares / df / Mean Square / F / Sig.
1 / Regression / 53.442 / 2 / 26.721 / 62.030 / .000b
Residual / 79.694 / 185 / .431
Total / 133.136 / 187
a. Dependent Variable: WLI
b. Predictors: (Constant), OCP, WART
Coefficientsa
Model / Unstandardized Coefficients / Standardized Coefficients / t / Sig.
B / Std. Error / Beta
1 / (Constant) / -.336 / .263 / -1.278 / .203
WART / .846 / .130 / .476 / 6.495 / .000
OCP / 1.184 / .405 / .215 / 2.928 / .004
Coefficientsa
Model / Correlations / Collinearity Statistics
Zero-Order / Semi-Partial / Tolerance / VIF
1 / (Constant)
WART / .611 / .369 / .602 / 1.660
OCP / .515 / .167 / .602 / 1.660
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

c = redundancy (aka communality)

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: -- that is, (b+c) + (c+d) – (b+c+d). For our data, that is .6112 + .5152 - .401 = .2375. Alternatively, .

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 obtain F. Take the square root of that F to obtain t. For our data, .

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.

ANOVAa
Model / Sum of Squares / df / Mean Square / F / Sig.
1 / Regression / 49.749 / 1 / 49.749 / 110.968 / .000b
Residual / 83.387 / 186 / .448
Total / 133.136 / 187
a. Dependent Variable: WLI
b. Predictors: (Constant), WART

REGRESSION

/MISSING LISTWISE

/STATISTICS COEFF OUTS R ANOVA

/CRITERIA=PIN(.05) POUT(.10)

/NOORIGIN

/DEPENDENT WLI

/METHOD=ENTER OCP.

ANOVAa
Model / Sum of Squares / df / Mean Square / F / Sig.
1 / Regression / 35.271 / 1 / 35.271 / 67.036 / .000b
Residual / 97.864 / 186 / .526
Total / 133.136 / 187
a. Dependent Variable: WLI b. Predictors: (Constant), OCP

. 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, , p < .001.

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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