# Canonical Correlation Analysis

• Hubert Gatignon
Chapter

## Abstract

In canonical correlation analysis the objective is to relate a set of dependent or criterion variables to another set of independent or predictor variables. For example, we would like to establish the relationship between socioeconomic status and consumption by households. A set of characteristics determines socioeconomic status: education level, age, income, etc. Another set of variables measures consumption such as purchases of cars, luxury items, or food products.

## Keywords

Criterion Variable Canonical Correlation Canonical Correlation Analysis Canonical Variable Unit Variance
These keywords were added by machine and not by the authors. This process is experimental and the keywords may be updated as the learning algorithm improves.

In canonical correlation analysis the objective is to relate a set of dependent or criterion variables to another set of independent or predictor variables. For example, we would like to establish the relationship between socioeconomic status and consumption by households. A set of characteristics determines socioeconomic status: education level, age, income, etc. Another set of variables measures consumption such as purchases of cars, luxury items, or food products.

## 7.1 The Method

In order to establish a relationship between these two sets of variables, we find two scalars, one defined as a linear combination of the dependent variables, and the other defined as a linear combination of the independent variables. The criterion used to judge the relationship between this set of independent variables with the set of dependent variables is simply the correlation between the two scalars. Canonical correlation analysis then consists in finding the weights to apply to the linear combinations of the independent and dependent variables that will maximize the correlation coefficient between those two linear combinations. The problem can be represented graphically as in Fig. 7.1. Fig. 7.1 Graphical representation of the canonical correlation model

In the figure, z and w represent two unobserved constructs that are correlated. The Xs are indicators that determine z and the Ys are indicators that determine w.

Formally, let $$\mathop{\mathbf{X}}\limits_{{N\times p}}$$ be the matrix of p predictor variables (centered, i.e., taking the deviations from their means) on N observations and $$\mathop{\mathbf{Y}}\limits_{{N\times p}}$$ be the matrix of q criterion variables (also centered) on the same N observations.

We will call z i the scalar representing a linear combination of the independent variables for observation i. Therefore
$$\mathop{{{z_i}}}\limits_{{1\times 1}}=\mathop{{{{{{\mathbf{x}}^{\prime}}}_i}}}\limits_{{1\times p}}\mathop{\mathbf{u}}\limits_{{p\times 1}}$$
(7.1)
Similarly, w i is the scalar representing a linear combination of the dependent variables for observation i:
$$\mathop{{{w_i}}}\limits_{{1\times 1}}=\mathop{{{{{{\mathbf{y}}^{\prime}}}_i}}}\limits_{{1\times q}}\mathop{\mathbf{v}}\limits_{{q\times 1}}$$
(7.2)
The correlation between variables z and w is
$${r_{zw }}=\frac{{\sum\limits_{i=1}^N {{z_i}{w_i}} }}{{\sqrt{{\left( {\sum\limits_{i=1}^N {z_i^2} } \right)\left( {\sum\limits_{i=1}^N {w_i^2} } \right)}}}}$$
(7.3)
More compactly, for the N observations
$$\mathop{\mathbf{z}}\limits_{{N\times 1}}=\mathop{\mathbf{X}}\limits_{{N\times p}}\mathop{\mathbf{u}}\limits_{{p\times 1}}$$
(7.4)
and
$$\mathop{\mathbf{w}}\limits_{{N\times 1}}=\mathop{\mathbf{Y}}\limits_{{N\times q}}\mathop{\mathbf{v}}\limits_{{q\times 1}}$$
(7.5)
The problem consists in finding the vectors (u, v) so as to maximize the correlation between z and w. In matrix notation, the correlation in Eq. (7.3) is
$${\boldsymbol r_{zw }}=\frac{{\mathbf{z} \mathbf{^{\prime} \bf w}}}{{\sqrt{{(\mathbf{z} \mathbf{^{\prime}\bf z})(\mathbf{w} \mathbf{^{\prime}\bf w})}}}}=\frac{{\mathbf{u} \mathbf{^{\prime}{ \bf X}^{\prime} \bf Yv}}}{{\sqrt{{\left( {\mathbf{u} \mathbf{^{\prime}\mathbf{X}^{\prime}\bf Xu}} \right)\left( {\mathbf{v} \mathbf{^{\prime}\mathbf{Y}^{\prime}\bf Yv}} \right)}}}}$$
(7.6)
Let $${{\mathbf{S}}_{\mathrm{ xy}}}=\mathbf{X} \mathbf{^{\prime}\bf Y}$$, $${{\mathbf{S}}_{\mathrm{ xx}}}=\mathbf{X} \mathbf{^{\prime}\bf X}$$, and $${{\mathbf{S}}_{\mathrm{ yy}}}=\mathbf{Y} \mathbf{^{\prime}\bf Y}$$. Then
$${r_{zw }}=\frac{{{\mathbf{u}}^{\prime}{{\mathbf{S}}_{\mathrm{ xy}}}\mathbf{v}}}{{\sqrt{{\left( {{\mathbf{u}}^{\prime}{{\mathbf{S}}_{\mathrm{ xx}}}\mathbf{u}} \right)\left( {{\mathbf{v}}^{\prime}{{\mathbf{S}}_{\mathrm{ yy}}}\mathbf{v}} \right)}}}}$$
(7.7)
The latent variables z and w can be normalized without loss of generality and for determinacy, i.e.,
$${\mathbf{u}}^{\prime}{{\mathbf{S}}_{\mathrm{ xx}}}\mathbf{u}={\mathbf{v}}^{\prime}{{\mathbf{S}}_{\mathrm{ yy}}}\mathbf{v}=1$$
(7.8)

Therefore, the problem is to find (u, v) so as to maximize $${\mathbf{u}}^{\prime}{{\mathbf{S}}_{\mathrm{ xy}}}\mathbf{v}$$ subject to $${\mathbf{u}}^{\prime}{{\mathbf{S}}_{\mathrm{ xx}}}\mathbf{u}={\mathbf{v}}^{\prime}{{\mathbf{S}}_{\mathrm{ yy}}}\mathbf{v}=1$$.

The Lagrangian is
$$\mathbf{L}\left( {\mathbf{u},\mathbf{v}} \right)={\mathbf{u}}^{\prime}{{\mathbf{S}}_{\mathrm{ xy}}}\mathbf{v}-\frac{\lambda }{2}\left( {{\mathbf{u}}^{\prime}{{\mathbf{S}}_{\mathrm{ xx}}}\mathbf{u}-1} \right)-\frac{\mu }{2}\left( {{\mathbf{v}}^{\prime}{{\mathbf{S}}_{\mathrm{ yy}}}\mathbf{v}-1} \right)$$
(7.9)
The maximum of the Lagrangian can be obtained by setting the derivatives relative to u and v equal to zero:
$$\frac{{\partial \mathbf{L}}}{{\partial \mathbf{u}}}={{\mathbf{S}}_{\mathrm{ xy}}}\mathbf{v}-\lambda {{\mathbf{S}}_{\mathrm{ xx}}}\mathbf{u}=0$$
(7.10)
and
$$\frac{{\partial \mathbf{L}}}{{\partial \mathbf{v}}}={\mathbf{u}}^{\prime}{{\mathbf{S}}_{\mathrm{ xy}}}-\mu {\mathbf{v}}^{\prime}{{\mathbf{S}}_{\mathrm{ yy}}}=0$$
(7.11)
From Eqs. (7.10) and (7.11), it follows that
$${\mathbf{u}}^{\prime}{{\mathbf{S}}_{\mathrm{ xy}}}\mathbf{v}=\lambda {\mathbf{u}}^{\prime}{{\mathbf{S}}_{\mathrm{ xx}}}\mathbf{u}$$
(7.12)
and
$${\mathbf{u}}^{\prime}{{\mathbf{S}}_{\mathrm{ xy}}}\mathbf{v}=\mu {\mathbf{v}}^{\prime}{{\mathbf{S}}_{\mathrm{ yy}}}\mathbf{v}$$
(7.13)
Consequently,
$$\lambda {\mathbf{u}}^{\prime}{{\mathbf{S}}_{\mathrm{ xx}}}\mathbf{u}=\mu {\mathbf{v}}^{\prime}{{\mathbf{S}}_{\mathrm{ yy}}}\mathbf{v}$$
(7.14)
However, because the transformed linear combination variables are standardized with unit variance, the result is
$$\lambda =\mu$$
(7.15)
Therefore, from Eq. (7.10), replacing $$\lambda$$ by $$\mu$$
$${{\mathbf{S}}_{\mathrm{ xy}}}\mathbf{v}=\mu {{\mathbf{S}}_{\mathrm{ xx}}}\mathbf{u}$$
(7.16)
and from Eq. (7.11), by taking its transpose
$${{\mathbf{S}}_{\mathrm{ yx}}}\mathbf{u}=\mu {{\mathbf{S}}_{\mathrm{ yy}}}\mathbf{v}$$
(7.17)
Solving for v in Eq. (7.17) leads to
$$\mathbf{v}=\frac{1}{\mu}\mathbf{S}_{\mathrm{ yy}}^{-1 }{{\mathbf{S}}_{\mathrm{ yx}}}\mathbf{u}$$
(7.18)
Replacing the value of v expressed in Eq. (7.18) into Eq. (7.16):
$${{\mathbf{S}}_{\mathrm{ xy}}}\left( {\frac{1}{\mu}\mathbf{S}_{\mathrm{ yy}}^{-1 }{{\mathbf{S}}_{\mathrm{ yx}}}\mathbf{u}} \right)=\mu {{\mathbf{S}}_{\mathrm{ xx}}}\mathbf{u}$$
(7.19)
Or, multiplying each side of the equation by $$\mu \mathrm{ S}_{\mathrm{ xx}}^{-1 }$$:
$$\mathbf{S}_{\mathrm{ xx}}^{-1 }{{\mathbf{S}}_{\mathrm{ xy}}}\mathbf{S}_{\mathrm{ yy}}^{-1 }{{\mathbf{S}}_{\mathrm{ yx}}}\mathbf{u}={\mu^2}\mathbf{S}_{\mathrm{ xx}}^{-1 }{{\mathbf{S}}_{\mathrm{ xx}}}\mathbf{u}$$
(7.20)
Equation (7.20) results in solving for the equation
$$\left( {\mathbf{S}_{\mathrm{ xx}}^{-1 }{{\mathbf{S}}_{\mathrm{ xy}}}\mathbf{S}_{\mathrm{ yy}}^{-1 }{{\mathbf{S}}_{\mathrm{ yx}}}-{\mu^2}\mathbf{I}} \right)\mathbf{u}=0$$
(7.21)
which is resolved by finding the eigenvalues and eigenvectors of $$\mathbf{S}_{\mathrm{ xx}}^{-1 }{{\mathbf{S}}_{\mathrm{ xy}}}\mathbf{S}_{\mathrm{ yy}}^{-1 }{{\mathbf{S}}_{\mathrm{ yx}}}$$.

The eigenvalue gives the maximum squared correlation r zw. This is the percentage of variance in w explained by z.

Two additional notions can be helpful in understanding the relationships between the set of x and the set of y variables: canonical loadings and redundancy analysis.

The canonical loadings are defined as the correlations between the original x and y variables and their corresponding canonical variate z and w. For the x variables
$$\mathop{{{\rho_{\mathrm{ xz}}}}}\limits_{{p\times 1}}=\frac{1}{N-1}\mathop{{{\mathbf{X}}^{\prime}}}\limits_{{p\times N}}\mathop{\mathbf{z}}\limits_{{N\times 1}}=\frac{1}{N-1 }{\mathbf{X}}^{\prime}\left( {\mathbf{X}\mathbf{u}} \right)=\frac{1}{N-1 }{{\mathbf{S}}_{\mathrm{ xx}}}\mathbf{u}$$
(7.22)
Similarly, for the y variables
$$\mathop{{{\rho_{\mathrm{ yw}}}}}\limits_{{q\times 1}}=\frac{1}{N-1}\mathop{{{\mathbf{Y}}^{\prime}}}\limits_{{q\times N}}\mathop{\mathbf{w}}\limits_{{N\times 1}}=\frac{1}{N-1 }{\mathbf{Y}}^{\prime}\left( {\mathbf{Y}\mathbf{v}} \right)=\frac{1}{N-1 }{{\mathbf{S}}_{\mathrm{ yy}}}\mathbf{v}$$
(7.23)

### 7.1.2 Canonical Redundancy Analysis

Canonical redundancy measures how well the original variables y can be predicted from the canonical variables. It reflects the correlation between the z and the y variables. Redundancy is the product of the percentage variance in w explained by z and the percentage variance in y explained by w. The first component is the squared correlation μ 2. The second component is the sum of squares of the canonical loadings for y.

Therefore,
$$\mathrm{ Redundancy} ={\mu^2}\frac{{{{{\rho^{\prime}}}_{\mathrm{ yw}}}{\rho_{\mathrm{ yw}}}}}{q}$$
(7.24)

## 7.2 Testing the Significance of the Canonical Correlations

It is possible to test the significance of these eigenvalues directly. However, the output in SAS shows eigenvalues that are different, albeit equivalent, from these eigenvalues or canonical correlation coefficients. These eigenvalues are related to the solution to the equation
$$\left( {{{\mathbf{W}}^{{-\mathbf{1}}}}\mathbf{B}-\lambda \mathbf{I}} \right)\mathbf{u}$$
(7.25)
Such an equation corresponds to Wilk’s lambda in MANOVA (see Chap. ) and to discriminant analysis discussed in Chap. . However, canonical correlation analysis differs from these two contexts because here we do not have the notions of between- and within-group variances due to the nonexistence of groups. These notions are generalized, however, to the concepts of total variance and error variance. Therefore, Λ is redefined as
$$\Lambda =\left| {\frac{\mathbf{E}}{\mathbf{T}}} \right|$$
(7.26)
where T is the total variance–covariance matrix and E is the residual variance–covariance matrix after removing the effects of each pair of canonical variable correlations. However, it should be noted here that the solution to Eq. (7.25) or (7.26) can be expressed as a function of the eigenvalues of Eq. (7.21):
$${\lambda_i}=\frac{{{\mu_i}^2}}{{1-{\mu_i}^2}}$$
(7.27)
where the μ i 2s are the solution to Eq. (7.21) and λi is the solution to
$$\left( {{{\mathbf{E}}^{{-\mathbf{1}}}}\mathbf{H}-\lambda \mathbf{I}} \right)\mathbf{u}=0$$
(7.28)
From the generalized definition of Wilk’s lambda $$\Lambda =\left| {\frac{\mathbf{E}}{\mathbf{T}}} \right|$$, it follows that
$$\frac{1}{\Lambda}=\left| {\frac{\mathbf{T}}{\mathbf{E}}} \right|=\left| {{{\mathbf{E}}^{{-\mathbf{1}}}}\mathbf{T}} \right|=\left| {{{\mathbf{E}}^{{-\mathbf{1}}}}\left( {\mathbf{H}+\mathbf{E}} \right)} \right|=\left| {{{\mathbf{E}}^{{-\mathbf{1}}}}\mathbf{H}+\mathbf{I}} \right|=\prod\limits_i {\left( {{\lambda_i}+1} \right)}$$
(7.29)
where T = H + E because of their independence. When we replace the λ i s by the μ i s using the equality in Eq. (7.27), Λ can be expressed as a function of the μ i s, i.e., the canonical correlations:
$$\Lambda =\prod\limits_i {\frac{1}{{{\lambda_i}+1}}=\prod\limits_i {\frac{1}{{1+\displaystyle\frac{{{\mu_i}^2}}{{1-{\mu_i}^2}}}}} } =\prod\limits_i {\left( {1-{\mu_i}^2} \right)}$$
(7.30)

Based on this expression of Λ, either as a function of the λ i s or as a function of the μ i s, it is possible to compute Bartlett’s V or Rao’s R, as discussed in Chap. . The degrees of freedom are not expressed in terms of the number of groups K, since this notion of group does not fit the canonical correlation model concerned with continuous variables. Instead, the equivalent is the parameter (q − 1), the number of variates on the left side, which corresponds to the number of dummy variables that would be required to determine K groups.

Bartlett’s V is
$$\begin{array}{ll} V=-\left[ {N-1-\left( {p+q-1} \right)/2} \right]Ln\;\Lambda =\left[ {N-\frac{3}{2}-\left( {p+q} \right)/2} \right]\sum\limits_{i=1}^q {Ln\left( {1+{\lambda_i}} \right)}\end{array}$$
(7.31)
or equivalently
$$\begin{array}{ll} V=-\left[ {N-1-\left( {p+q-1} \right)/2} \right]Ln\;\Lambda =\left[ {N-\frac{3}{2}-\left( {p+q} \right)/2} \right]\sum\limits_{i=1}^q {Ln\left( {1-\mu_i^2} \right)}\end{array}$$
(7.32)
Bartlett’s V is approximately distributed as a chi-square with pq degrees of freedom. Alternatively, Rao’s R can be computed as shown in Chap. for MANOVA, where K is replaced by q − 1:
$$R=\frac{{1-{\Lambda^{{\frac{1}{t}}}}}}{{{\Lambda^{{\frac{1}{t}}}}}}\frac{{wt-\frac{pq }{2}+1}}{pq }$$
(7.33)
where $$w=N-\displaystyle\frac{3}{2}-\displaystyle\frac{p+q }{2}$$ and $$t=\sqrt{{\displaystyle\frac{{{p^2}{q^2}-4}}{{{p^2}+{q^2}-5}}}}$$.

R is distributed approximately as an F distribution with pq degrees of freedom in the numerator and $$wt-\frac{pq }{2}+1$$ degrees of freedom in the denominator. This last test (Rao’s R) is the one reported in the SAS output (rather than Bartlett’s V ).

These tests are joint tests of the significance of the q canonical correlations. However, each term in the sum containing the eigenvalues in Eq. (7.31) or (7.32) is distributed approximately as a chi-square with p + q − (2i − 1) degrees of freedom where i is the ith eigenvalue from i = 1 to q.

Any subset of eigenvalues is the sum of that subset of terms in Ln (1 − μ i 2). Consequently, one can test if the residual canonical correlations are significant, after having removed the first canonical variate, then the first two, and so on. For example, the joint test of all q canonical correlations is V as in Eq. (7.32) with pq degrees of freedom. The test of the first eigenvalue is
$${V_1}=\left[ {N-\frac{3}{2}-\left( {p+q} \right)/2} \right]\;Ln\left( {1-\mu_1^2} \right)$$
(7.34)
with (p + q − 1) degrees of freedom.

Consequently, the joint test that the remaining canonical correlations μ 2 , μ 3 , μ 4 , … μ q are zero is obtained by subtracting V 1 from V. V − V 1 is approximately chi-square distributed and the number of degrees of freedom is the difference between the degrees of freedom of V and those of V 1 , i.e., pq − (p + q−1). This can be continued until the last qth eigenvalue. The same computations as those detailed above with Bartlett’s V can be performed with Rao’s R .

## 7.3 Multiple Regression as a Special Case of Canonical Correlation Analysis

In the case of multiple regression analysis, the dependent variable is a single variate represented by the vector $$\mathop{\mathbf{y}}\limits_{{N\times 1}}$$ for the N observations. Consequently, the vector v reduces to a single scalar, set to the value 1. It follows that w = y. The expression for the correlation between x and w in Eq. (7.7) becomes
$${\boldsymbol r_{zw }}=\frac{{\mathbf{u} \mathbf{^{\prime}\mathbf{X}^{\prime}\bf y}}}{{\sqrt{{\left( {\mathbf{u} \mathbf{^{\prime}\mathbf{X}^{\prime}\bf Xu}} \right)\left( {\mathbf{y} \mathbf{^{\prime}\bf y}} \right)}}}}$$
(7.35)
However, because the transformed independent variables are standardized and the single dependent variable y can be standardized to unit variance without loss of generality, the problem is to maximize the correlation coefficient r zw subject to the constraint $$\mathbf{u} \mathbf{^{\prime}\mathbf{X}^{\prime}\bf Xu}=1$$. This is solved by maximizing the Lagrangian:
$$\mathbf{L}=\mathbf{u} \mathbf{^{\prime}\mathbf{X}^{\prime}\bf y}-\frac{\lambda }{2}\left( {\mathbf{u} \mathbf{^{\prime}\mathbf{X}^{\prime}\bf Xu}-1} \right)$$
(7.36)
$$\frac{{\partial \mathbf{L}}}{{\partial \mathbf{u}}}=\mathbf{X} \mathbf{^\prime\bf y}-\lambda \mathbf{X} \mathbf{^{\prime}\bf Xu}=0$$
(7.37)
Solving for u leads to the least square estimator presented in Chap. :
$$\mathbf{u}=\frac{1}{\lambda }{{\left( {\mathbf{X} \mathbf{^{\prime}\bf X}} \right)}^{-\mathbf{1}}}\mathbf{X} \mathbf{^{\prime}\bf y}$$
(7.38)

## 7.4 Examples

Figure 7.2 shows the SAS commands to run a canonical correlation analysis. The data concern a number of new products that are characterized by a number of innovation characteristics, all rated on 7-point Likert scales from 1 (disagree) to 7 (agree): Fig. 7.2 Example of SAS code for canonical correlation analysis (examp7-1.sas)
• X1: This new product is hard to understand.

• X2: This new product is not really easy to use.

• X3: Using this new product is not very compatible with the way I do things.

• X14: I really like this new product.

• X15: I am favorably disposed towards this new product.

The SAS procedure “proc cancorr” runs the canonical correlation analysis. The X variables (see Fig. 7.1) are indicated in the list following the key word “VAR” and the Y variables (see Fig. 7.1) are listed after the key word “with.” Titles can be inserted for the output in single quotes after the word “title.”

The input for STATA is shown in Fig. 7.3. Fig. 7.3 Example of STATA code for canonical correlation analysis (examp7-1.do)

Figure 7.4 lists the SAS output from running the canonical correlation analysis. Fig. 7.4 Example of SAS output of canonical correlation analysis (examp7-1.out)

When the canonical correlations are listed, we see that one correlation coefficient of 0.35131 appears larger than the other two values. Therefore, we can concentrate on this larger value. These correlations correspond to the eigenvalues that give a solution to Eq. (7.21) (the canonical correlation is the square root of these eigenvalues).

The eigenvalues λ i , which are the solution to Eq. (7.28), are those shown under the column “Eigenvalue” in the SAS output. For example, the first (highest) eigenvalue of 0.1408 is related to the first canonical correlation as
$$0.1408=\frac{{{{{\left( {0.3513} \right)}}^2}}}{{\left[ {1-{{{\left( {0.3513} \right)}}^2}} \right]}}$$
(7.39)

Given the relationship between the λ i s and the μ i s, these eigenvalues provide the same information as the canonical correlations. The F test corresponding to Rao’s R (highlighted in grey in Fig. 7.4) indicates that the set of canonical correlations (or eigenvalues) are jointly significantly different from zero (F = 6.21 with 9 and 959.04 degrees of freedom). Then, the next row in that part of the table shows that after removing the first canonical correlation, the remaining canonical correlations are jointly statistically insignificant at the 0.05 level (F = 0.61 with 4 and 790 degrees of freedom). Therefore, we can concentrate on the results concerning the first canonical variable.

The raw (highlighted in grey in Fig. 7.4) and the standardized eigenvectors are then listed in the SAS output. The raw values are subject to variations due to the scale units of each variate and should be interpreted accordingly. It should be noted that the canonical variables are normalized to unit variance as per Eq. (7.8), and consequently, the magnitude of the coefficients that are the elements of the eigenvectors u and v are affected as well by the unit of the variates. The first eigenvector indicates that innovations that are not complex and that are easy to understand (x1, x2, and x3) are associated with greater positive responses (x13, x14, and x15).

Then, the correlation of each variate to the canonical variables (composite variable v and then w) is contained in the last tables of Fig. 7.4. This allows us to assess the strength of the relationships that form a composite (unobserved) canonical variable and of the relationship of a variable to the other composite canonical variable.

The STATA output is shown in Fig. 7.5. Fig. 7.5 Example of STATA output of canonical correlation analysis (examp7-1.log)

In the last section of the STATA output, the heading “Test of significance of canonical correlation 1–3” corresponds to the joint test of all the canonical correlations shown at the top of the output. The “Test of significance of canonical correlation 2–3” is the joint test of canonical correlations 2 through 3. Given that it is insignificant (F = 0.6138), we conclude that only the first canonical correlation is significant.

## 7.5 Assignment

Using the survey data described for the assignment in Chap. , associate certain types of consumer behaviors to their psychographic profiles. The sample SAS code file to read the data is shown in Fig. .

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