Regression Analysis - Assignment Example

Regression Analysis Question UR 248615375NRU 92522993 + 0.854805 A) What percentage of the variance in UR does this regression explain? ANOVA   Df SS MS F Significance F Regression 1 89.83256073 89.83256073 105.0913 9.392E-13 Residual 40 34.19220117 0.854805029 Total 41 124.0247619       From the above table, 89.83256073 represent the sum of squares explained by the regression and 124.0247619 represent the total. Thus, 89.83256073/124.0247619 or 72.43% of variance is explained by the model. B) Test the null hypothesis that β0 = 1 at the 95% level of significance. Do you reject this hypothesis? Explain why or why not. How about at the 99% level of significance? Solution:   Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Lower 99.0% Upper 99.0% Intercept -1.92523 0.992885 -1.93903 0.059575 -3.93193 0.081465 -4.61045 0.759987 NRU 1.248615 0.121799 10.2514 9.39E-13 1.00245 1.494781 0.919214 1.578017 For Ho, Bo=1 at 95% level of significance, we do reject the null hypothesis since 2.9462> 1.6848, we accept the alternative hypothesis. Ho: Bo=1, Ha: Bo≠1 t = (Bo-1)/ std error of Bo =(-1.92522993-1)/ 0.992885 =-2.9462. Thus, TINV(0.05,39)= 1.6848 At 99% level of significance, we do not reject Ho since 2.9462>2.4258, we accept the alternative hypothesis. TINV(0.01,39)= 2.4258 C) Test the null hypothesis that β1 = 1.5 at the 95% level of significance. Do you reject this hypothesis? Explain why or why not. How about at the 99% level of significance? Solution: TINV(0.05,39)= 1.6848 Ho: B1=1.5, Ha: B1≠1.5 At 95% level of significance, t= (B1-1.5)/ std error of B1 = (1.248615 - 1) / 0.121799 = 2.041191 Therefore, we reject the null hypothesis since 2.041191< 1.6848. TINV(0.01,39)= 2.4258 For 99% significance level, we reject the null hypothesis since 2.041191 Read More