Analysis Result RESULT AND ANALYSIS

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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