72
D. Analysis Result
1. Path Analysis Data processing techniques in resolving the research by using path
analysis. Path analysis model was used to analyze the patterns of relationship between variable with the aim to find out directly or indirectly
influence of a set variables with each other. Here are the result of path analysis :
a. Testing Relationship Between Sub Variable In the method of path analysis, for finding causal relationship or effect
of the research variables, first calculated the correlation matrix of the research variables.
Table 4.40 Correlation Coefficient Substructure I
Correlations
ServiceQua lity
BrandImag e
CustomerSati sfaction
ServiceQuality Pearson Correlation
1 .835
.897 Sig. 1-tailed
.000 .000
N 60
60 60
BrandImage Pearson Correlation
.835 1
.957 Sig. 1-tailed
.000 .000
N 60
60 60
CustomerSatisfaction Pearson Correlation
.897 .957
1 Sig. 1-tailed
.000 .000
N 60
60 60
. Correlation is significant at the 0.01 level 1-tailed.
Source: Primary data Processed 2014 Based on the table above can be known the value of the
correlation between variables.The correlation coefficient indicates that
73
the relationship between two variables is positive. According toSarwono 2007:22 for the affinity relationship determination criteria
based on : 0.00
– 0.25 = very weak relationships deemed on 0.25
– 0.5 = good relationship 0.5
– 0.75 = strong relationship 0.75
– 1 = very strong relationship
For further testing, then submitted the following hypothesis: H
: There is no correlations between two variables are significant H
1
:There is correlations between two variables are significant. Testing based on test results will be accepted based on the probability:
If the probabilities 0.05, then H is accepted and H
1
is rejected there is no significant relationship between the two variables
If the probabilities 0.05, then H is rejected and H
1
is accepted there is significant relationship between the two variables
Table 4.41 Testing Relationship between Variable Substructure I
Relationship Correlation
Coefficient
Category Probability Conclusion
Service Quality X
1
with Brand Image X
2
0.835 Very
Strong 0.000
Significant Service Quality X
1
with Customer Satisfaction Y
1
0.897 Very
Strong 0.000
Significant Brand Image X
2
with Customer Satisfaction
Y
1
0.957 Very
Strong 0.000
Significant Source: Primary data Processed 2014
74
Based on the test results above, it can be concluded that entire relationships of those variables have significant relationship, because all
values smaller than 0.05 probability.
Table 4.42 Correlation Coefficient Substructure II
Correlations
ServiceQuali ty
BrandIm age
CustomerSati sfaction
CustomerLoy alty
ServiceQuality Pearson Correlation
1 .835
.897 .643
Sig. 1-tailed .000
.000 .000
N 60
60 60
60 BrandImage
Pearson Correlation .835
1 .957
.800 Sig. 1-tailed
.000 .000
.000 N
60 60
60 60
CustomerSatisfaction Pearson Correlation
.897 .957
1 .761
Sig. 1-tailed .000
.000 .000
N 60
60 60
60 CustomerLoyalty
Pearson Correlation .643
.800 .761
1 Sig. 1-tailed
.000 .000
.000 N
60 60
60 60
. Correlation is significant at the 0.01 level 1-tailed.
Source: Primary data Processed 2014 Based on the table above can be the value of the correlation
between variables. The correlation coefficient indicates that the relationship between variables is positive. According to Sarwono
2007:22 for the affinity relationship determination criteria based on : 0.00
– 0.25 = Very weak relationships deemed on 0.25
– 0.5 = Good relationship 0.5
– 0.75 = Strong relationship 0.75
– 1 = Very strong relationship
75
For further testing, then submitted the following hypothesis : H
: There is no correlations between two variables are significant H
1
: There is correlations between two variables are significant Testing based on test results will be accepted based on the probability :
If the probabilities 0.05, then H is accepted and H
1
is rejected there is no significant relationship between the two variables
If the probabilities 0.05, then H is rejected and H
1
is accepted there is significant relationship between the two variables
Table 4.43 Testing Relationship between Variable Substructure II
Relationship Correlation
Coefficient Category Probability Conclusion
Service Quality X
1
with Customer Loyalty Y
2
0.643 Strong
0.000 Significant
Brand Image X
2
with Customer Loyalty Y
2
0.800 Very
Strong 0.000
Significant Customer satisfaction
Y
1
with Customer Loyalty Y
2
0.761 Very
Strong 0.000
Significant Source: Primary data Processed 2014
Based on the test results above, it can be concluded that entire relationships of those variable have significant relationship, because all
values smaller than 0.05 probability.
76
2. Coefficient of Determination R
2
a. Coefficient Determination of Substructure I To see the influence of variable service quality X
1
and brand image X
2
through customer satisfaction Y
1
indicated by the table summary, especially in figure R square below :
Table 4.44 Coefficient Determination of Substructure I
Model Summary
b
Model R
R Square Adjusted R
Square Std. Error of the
Estimate Durbin-Watson
1 .973
a
.947 .945
.64795 1.952
a. Predictors: Constant, Brand Image, Service Quality b. Dependent Variable: Customer Satisfaction
The magnitude number of R square is 0.947. This number is
used to see the magnitude of the influence that belongs to variables service quality and brand image to customer satisfaction. The way
how to calculate the coefficient of determination by using formula:
KD = r
2
x 100 KD = 0.947 x 100
KD = 94.7
Based on the table above, the result of calculation using SPSS 17 program can be seen the adjusted R square is 0.947. It means
94.7 exogenous variable service quality X1 and brand image X2 effect customer satisfaction Y1 as endogenous variable.
While the residual 100-94.7 =5.3 are affected by other
77
variables or outside factor on this model. The probability customer satisfaction can be explained by 94.7 thorough variables service
quality and brand image. b. Coefficient determination of substructure II
To see the influence of variable service quality X
1
and brand image X
2
through customer satisfaction Y
1
on customer loyalty Y
2
indicated by the table summary, especially in figure R square below :
Table 4.45 Coefficient determination of Substructure II
Model Summary
b
Model R
R Square Adjusted R
Square Std. Error of the
Estimate Durbin-Watson
1 .801
a
.642 .623
2.15041 1.918
a. Predictors: Constant, Customer Satisfaction, Service Quality, Brand Image
The magnitude number of R square is 0.642. This number is used to see the magnitude of the influence that belongs to variables service
quality, brand image and customer satisfaction to customer loyalty. The way how to calculate the coefficient of determination by using
formula:
KD = r
2
x 100 KD = 0.642 x 100
KD = 64.2
78
Based on the table above, the result of calculation using SPSS 17 program can be seen the adjusted R square is 0.642. It means
64.2exogenous variable service quality X1, brand image X2 and customer satisfaction Y1 effect customer loyalty Y2 as
endogenous variable. While the remainder 100-64.2 = 35.8 are affected by other variables or factors outside this model. The
probability customer loyalty can be explained by 64.2 by variables service quality, brand image, and customer satisfaction.
3. Simultaneously Test F
Test
a. Simultaneously Test of Substructure I
Table 4.46 Analysis of Variant ANOVA of Substructure I
ANOVA
a
Model Sum of Squares
Df Mean Square
F Sig.
1 Regression
424.919 2
212.460 506.056
.000
a
Residual 23.931
57 .420
Total 448.850
59 a. Predictors: Constant, BrandImage, ServiceQuality
b. Dependent Variable: CustomerSatisfaction
F
test
basically indicates whether all exogenous variable have influence endogenous variable simultaneously. Based on the table
above, to test whether variable service quality X
1
and brand image X
2
have significant influence to customer satisfaction Y
1
simultaneously, there are several steps to testthe hypothesis by F
test
as follows :
79
H : Service quality and brand image have no influence to customer
satisfaction simultaneously H
1
: Service quality and brand image have influence to customer satisfaction simultaneously.
The formula to test F
test
with F
table
are : If F
test
F
table ,
then H is rejected and H
1
is accepted If F
test
F
table
, then H is accepted and H
1
is rejected According to table analysis of variant ANOVA, calculation value
of F
test
was obtained 506.056 with probabilities 0.000. F
table
with significant level based on 0.05, the degree of freedom df for df
1
= 2 and df
2
= 57, then the number of F
table
= 3.16. It can be compared, the result of F
test
506.056 F
table
3.16 and the significant level 0.000 0.05. Therefore it can be conclude, H
is rejected and H
1
is accepted. Means that, service quality X1 and brand image X2 have significant
influence to customer satisfaction Y1 simultaneously. The magnitude influence is 94.7. How big influence from other variable in outside
model can be calculate by formula : 1 - r
2
or 1 – 0.947 = 0.053 or 5.3.
b. Simultaneously test of Substructure II
Table 4.47 Analysis of Variant ANOVA of Substructure II
ANOVA
b
Model Sum of Squares
Df Mean Square
F Sig.
1 Regression
463.974 3
154.658 33.445
.000
a
Residual 258.959
56 4.624
Total 722.933
59 a. Predictors: Constant, Customer Satisfaction, Service Quality, Brand Image
b. Dependent Variable: Customer Loyalty
80
F
test
basically indicates whether all exogenous variable have influence endogenous variable simultaneously. Based on the table
above, to test whether variable service quality X
1
, brand image X
2
and customer satisfaction Y
1
have significant influence to customer loyalty Y
2
simultaneously, there are several steps to test the hypothesis by F
test
as follows : H
: Service quality, brand image and customer satisfaction have no influence to customer loyalty simultaneously.
H
1
: Service quality, brand image and customer satisfaction have influence to customer loyalty simultaneously.
Formula to test F
test
with F
table
are : If F
test
F
table,
then H is rejected and H
1
is accepted If F
test
F
table,
then H is accepted and H
1
is rejected According to table analysis of variant ANOVA, calculation value
of F
test
was obtained 33.445 with probabilities 0.000. F
table
with significant levels based on 0.05, the degree of freedom for df
l
= 3 and df
2
= 56, then the number of F
table
= 2,77. It can be compared the result of F
test
33.445 F
table
2,77 and the significant levels 0.000 0.05. Therefore it can be conclude, H
is rejected and H
1
is accepted. Means that, service quality X1, brand image X2, and customer
satisfaction Y1 have significant influence to customer loyalty Y2 simultaneously. The magnitude influence is 64.2. How big influence
from other variable can be calculate by formula : 1 - r
2
or 1 – 0.642 =
0.358 or 35.8.
81
4. Partial Test T
test
a. Partial test of Substructure I
Table 4.48 Result Partial Test Substructure I
Coefficients
a
Model Unstandardized
Coefficients Standardized
Coefficients T
Sig. Correlations
B Std. Error
Beta Zero-order Partial Part
1 Constant
1.298 .837
1.551 .126 Service Quality
.320 .055
.323 5.811 .000
.897 .610 .178
Brand Image .566
.046 .687 12.341 .000
.957 .853 .377
a. Dependent Variable: Customer Satisfaction
This is the result to test: 1 Test the influence of service quality X
1
to customer satisfaction Y
1
Define hypothesisare : H
:Service quality has no influence to customer satisfaction H
1
: Service quality has influence to customer satisfaction The formula to compare the value T
test
with T
table
are : If T
test
T
table ,
H is rejected and H
1
is accepted If T
test
T
table ,
H is accepted and H
1
is rejected According to table above, calculation of T
test
value was obtained 5.811 with the probabilities 0.000. T
table
with a significance level of 0.05 and degrees of freedom with df = n-2 or
60-2 = 58, then the number of T
table
2.001. It can be compared the result of T
test
5.811 T
table
2.001 and the significant levels
82
0.000 0.05. Therefore it can be conclude, H is rejected and
H
1
is accepted. Means that, service quality X1 has significant influence to customer satisfaction Y1 partially.
The result from this research was accordance with previous research Santouridis and Trivellas 2010 with title “Investigating
the impact of service quality and customer satisfaction on customer loyalty in mobile telephone on Greece”. The result of this research
was indicated service quality has significant influence to customer satisfaction.
2 Test the influence of brand image X
2
to customer satisfaction Y
1
Define hypothesis are : H
: Brand image has no influence to customer satisfaction. H
1
: Brand image has influence to customer satisfaction. Compare the value T
test
with T
table
are : If T
test
T
table,
H is rejected and H
1
is accepted. If T
test
T
table ,
H is accepted and H
1
is rejected. According to table above, calculation of T
test
value was obtained 12.341 with probabilities 0.000. T
table
with a significance level of 0.05 and degrees of freedom = n-2 or 60-2 = 58,then the
number of T
table
2.001. Itcan be compared the result of T
test
12.341 T
table
2.001 and the significant levels 0.000 0.05. Therefore it can be conclude, H
is rejected and H
1
is accepted. Means that, brand image X2 has significant influence to customer
satisfaction Y1 partially.
83
The result of this research was accordance with previous research of Thakur and Singh 2012
with title “Brand Image, Customer Satisfaction, and Loyalty Intention : A Study in Context
of Cosmetic Product among the People of Central India”. The result of this research was indicated brand image has influence to
customer satisfaction.
Table 4.49 Partial Test
The Influence of Service Quality and Brand Image to Customer Satisfaction
NO Hypothesis Path Analysis
T
test
T
table
Conclusion
1 ρx
1
y
1
X
1
≠0 ρx
1
y
1
X
1
= 0.323 5.811
2.001 H
rejected 2
ρx
2
y
1
X
2
≠0 ρx
2
y
1
X
2
= 0.687 12,341 2.001
H rejected
Source: Primary data Processed 2014 Based on the result above showsthat variable partial testing
service quality and brand image influence significantly to customer satisfaction.Thus, the equation of path analysis that form is as
follows:
Y
1
= ρx
1
y
1
X
1
+ ρx
2
y
1
X
2
+ ρЄ
1
y
1
Y
1
= 0.323X
1
+ 0.687X
2
+0.053 Є
1
The number of residues obtained from 1- R
2
1 – 0.947 = 0.053
84
b. Partial test of Substructure II
Table 4.50 Result Partial Test Substructure II
Coefficients
a
Model Unstandardized
Coefficients Standardized
Coefficients T
Sig. Correlations
B Std. Error
Beta Zero-order Partial
Part 1
Constant 12.850
2.835 4.533
.000 Service Quality
-.135- .231
-.107- -.583- .562
.643 -.078- -.047-
Brand Image .855
.292 .817 2.932
.005 .800
.365 .235
Customer Satisfaction
.095 .440
.075 .216 .830
.761 .029
.017 a. Dependent Variable: Customer Loyalty
This is the result to test: 1 Test the influence of service quality X
1
to customer loyalty Y
2
Define hypothesis are : H
: Service quality has no influence to customer loyalty. H
1
: Service quality has influence to customer loyalty. Compare the value T
test
with T
table
are
:
If T
test
T
table ,
H is rejected and H
1
is accepted. If T
test
T
table ,
H is accepted and H
1
is rejected. According to the table above, calculation of T
test
value was obtained -0.583. While T
table
with a significance level of 0.05 based on2 = 0.025 and degrees of freedom with df = n-3 or 60-3 = 57.
With the conditions obtained numbers of T
table
1.67203. It can be compared the result ofT
test
-0.583 T
table
1.67203. Therefore it can be conclude,H
is accepted and H
1
is rejected. Means that,
85
service quality X1 has no influence to customer loyaltyY2 partially.
The result from this research was accordance with previous research of Dwi Aryani and Febrina Rosinta 2010 with title “The
Influence of Service Quality toward Customer Satisfaction in shipping
Customer Loyalty
on PT
Human Resource
Deve lopment”. The result of this research was indicated service
quality has no influence to customer loyalty. 2 Test the influence of brand image X
2
to customer loyalty Y
2
Define hypothesis are : H
: Brand image has no influence to customer loyalty. H
1
: Brand image has influence to customer loyalty. Compare the value T
test
with T
table
are : If T
test
T
table ,
H is rejected and H
1
is accepted. If T
test
T
table ,
H is accepted and H
1
is rejected. According to the table above, calculation of T
test
value was obtained 2.932.While T
table
with a significance level of 0.05 based on2 = 0.025 and degrees of freedom with df = n-3 or 60-3 = 57.
With the conditions obtained numbers of T
table
1.67203. It can be compared the result of T
test
2.932 T
table
1.67203. Therefore it can be conclude,H
is rejected and H
1
is accepted. Means that brand imageX2 has influence to customer loyalty Y2 partially.
The result of this research was accordance with previous research of Satendra Thakur and DR. A. P. Singh 2012 with title
86
“Brand Image, Customer Satisfaction, and Loyalty Intention : A Study in Context of Cosmetic Product among the People of Central
India”. The result of this research was indicated brand image has no influence to customer loyalty.
3 Test the influence of customer satisfaction Y
1
to customer loyalty Y
2
Define hypothesis are : H
: Customer satisfaction has no influence to customer loyalty. H
1
: Customer satisfaction has influence to customer loyalty. Compare the value T
test
with T
table
are
:
If T
test
T
table ,
H is rejected and H
1
is accepted. If T
test
T
table
, H is accepted and H
1
is rejected. According to the table above, calculation of T
test
value was obtained 0.216.While T
table
with a significance level of 0.05 based on2 = 0.025 and degrees of freedom with df = n-3 or 60-3 = 57.
With the conditions obtained numbers of T
table
1.67203. It can be compared the result of T
test
0.216 T
table
1.67203. Therefore it can be conclude,H
is accepted and H
1
is rejected. Means that, customer satisfaction Y1 has no influence to customer loyalty
Y2 partially. The result of this research was different with previous research
of Ending Tjajaningsih 2013 with title “The Influence of Brand and Promotion to Customer Satisfaction that Impact on Customer
Loyalty Study Case on the Member of Carrefour Semarang”. The
result of this research was indicated customer satisfaction has influence customer loyalty.
87
Table 4.51 Partial Test
The Influence of Service Quality, Brand Image and Customer Satisfaction to Customer Loyalty
NO Hypothesis
Path Analysis T
test
T
table
Conclusion
1 ρx
1
y
2
X
1
≠0 ρx
1
y
2
X
1
= -0.107 -0.583 1.67203
H accepted
2 Ρx
2
y
2
X
2
≠0 ρx
2
y
2
X
2
= 0.817 2.932 1.67203
H rejected
3 ρy
1
y
2
Y
1
≠0 ρy
1
y
2
Y
1
= 0.075 0.216 1.67203
H accepted
Source: Primary data Processed 2014 Based on the result above shows that variable partial testing
service quality and customer satisfaction does not influence significantly to customer loyalty. While brand image influence
significantly to customer loyalty. Thus, the equation of path analysis that is from is as follows :
Y2 = ρx1y2x1 + ρx2y2x2 + ρy1y2y1 + Ρє
2
y
2
Y2 = -0.107 X
1
+ 0.817 X
2
+ 0.075 Y
1
+ 0.358 є
2
The number of residues obtained from 1 – R2
= 1 – 0.642 = 0.358
88
E. Path Analysis Diagram