Classical Assumption Test Analysis and Discussion

102 The analysis of the data tolerance value shows there is no independent variable, which has a tolerance value less than 0.10 that means there is no correlation among independent variables that value more than 95 percent. On the other hand VIF shows similar things that there is no independent variable that has a VIF value more than 10, thus, it can be concluded that there is no multicolinearity among independent variables in regression model and feasible to use. b. Heteroscesdastisity Test Figure 4.1 Heteroscesdastisity Test Source: Processed Primary data by SPSS 20 According Imam Ghozali 2005:105 multiple linear regression free of heterocedasticity as follows: 1 There is no clear pattern 103 2 Point spread above and below zero on the Y axis Heteroscesdasticity test is aimed to examine whether in the model occur any residual variance in certain monitoring period to other monitoring period. If the characteristic is fulfilled, it means that the factors of intruder variation toward the data have the characteristic of heteroscedasticity. A good model is homocesdastisity, not heteroscedasticity. From the Scatter plot diagram in figure 4.1 above, it can be seen that the dots are spread biased. This shows that there is a heteroscesdastisity problem. Therefore the Glejser test will be conducted to make sure this content is free from heteroscesdastisity problem. And the result from Glejser test is shown in table below. Table 4.68 Glejser Test Coefficients a Model Unstandardized Coefficients Standardized Coefficients T Sig. B Std. Error Beta 1 Constant .525 1.428 .368 .714 CAE .087 .062 .338 1.414 .161 BAW .019 .104 .023 .179 .858 BAS .069 .074 .191 .922 .359 BPT -.092 .060 -.387 -1.521 .132 a. Dependent Variable: ABS_RES Source: Processed Primary data by SPSS 20 104 From the result of the Glejser test, the significancy result from that testing is more then 0.05 so, it is can be concluded there is no heteroscesdastisity problem. c. Normality Test Figure 4.2 Normality Test Result Source: Processed Primary Data by SPSS 20 Normality data test is aimed to know whether the data obtained from the research activities have a normal distribution or not. Good data is to be considered, as data that has a normal distribution. One of the ways to see whether the data in this research are normal or not, is by applying the P-P Plot graph. When the plots in the graph are distributed along the diagonal line, it can be said that the data has a normal distribution. Based on figure 4.2 this research has done normality data 105 distribution test. The result acquired from SPSS 20 statistic software. From the P-P Plots diagram above, it can be seen that the plots are distributed along the diagonal line. Thus, it can be concluded that the data used in this research has a normal distribution. Figure 4.3 Chart Source: Processed Primary data by SPSS 20 Based on chart 4.3 above, the Histogram Graphic shows normal distribution. Because, these images form the normal curve and most of the bars rods under the curve, then the variable is normally distributed. 106

4. Multiple Linear Regression

a. Coefficient Determination R 2 Table 4.69 Coefficient Determination Model R R Square Adjusted R Square Std. Error of the Estimate 1 .913 a .834 .827 3.27598 Source: Processed primary data by SPSS 20 From the coefficient determination table 4.69 above, the Adjusted R square number is 0.827 or 82.7 . This means all of the independent variables like celebrity athlete endorser, brand awareness, brand association, and brand personality towards purchase intention have significant influence 82.7 . Thus, the residual coefficient, around 17,3 it will be explained by the other factors that is not calculated in this research. b. F - Test Table 4.70 F Test ANOVA a Model Sum of Squares Df Mean Square F Sig. 1 Regression 4976.880 4 1244.220 115.935 .000 b Residual 987.347 92 10.732 Total 5964.227 96 a. Dependent Variable: PIN b. Predictors: Constant, BPT, BAW, BAS, CAE Source: Processed primary data by SPSS 20 107 The ANOVA test or F test provides a test value of 115.935 with probabilities 0.000. The degree of freedom df for the F test is df = n – k = 97 – 5 = 92. Thus the F table for F test is 2.471 with the significance level = 0.05; because the probability is smaller than the significance level 0.000 0.05, and the F test value is higher than F table value the value of F-test is 115.935 F-table 2.471. Therefore this means Ho is rejected and Ha is accepted, and can be concluded that the independent variables such as celebrity athlete endorser X 1 , brand awareness X 2 , brand association X 3 , and brand personality X 4 are simultaneously influence purchase intention as dependent variables Y. c. T - Test Table 4.71 T-Test Coefficients a Model Unstandardized Coefficients Standardize d Coefficients t Sig. Collinearity Statistics B Std. Error Beta Tolerance VIF 1 Constant -2.907 2.384 -1.220 .226 CAE .402 .103 .391 3.904 .000 .180 5.562 BAW .416 .173 .129 2.398 .019 .624 1.602 BAS .355 .124 .247 2.850 .005 .239 4.178 BPT .221 .101 .234 2.193 .031 .159 6.307 a. Dependent Variable: PIN Source: Processed primary data by SPSS 20