Five Regression Models - Assignment Example

Five Regression Models A) Use Excel to run each of these five regression models listed in Table 7.1. Create your own new version of Table 7.1 that fills in the estimated figures you have found for the same information presented in Table 7.1 Results of Regressions of Test Scores on the Student-Teacher Ratio and        Student Characteristic Control variables Using California Elementary School Districts Dependent variable: average test score in the district Regressor (1) (2) (3) (4) (5) Student-teacher ratio (X1) -2.28** (0.48) -1.10** (0.38) -1.00* (0.24) -1.31 (0.31) -1.01 (0.24) Percent English learners (X2) -0.65** (0.04) -0.12 (0.03) -0.49 (0.03) -0.13 (0.03) Percent eligible for subsidized lunch (X3) -0.55 (0.02) -0.53 (0.03) Percent on public income assistance (X4) -0.79 (0.52) -0.05 (0.06) Intercept Summary statistics 698.9** (9.47) 686.0** (7.41) 700.2** (4.69) 698.0** (6.02) 700.4** (4.70) SER 18.58 14.46 9.02 11.65 9.08 Adjusted R-Squared 0.049 0.423 0.773 0.626 0.773 n 420 420 420 420 420 B) The estimated figures that have been found to be the SAME as those presented in the original Table 7.1 are; coefficient estimate, SER, adjusted R2, n The estimated regression coefficients remain unchanged since there was no change in the number of variables regressed, same to the sample size n C) The estimated figures that have been found to be DIFFERENT from those presented in the original Table 7.1 are; estimated coefficient standard error Standard errors of regression coefficient change with introduction of new variable D) Run a regression of STR on EL_PCT along with an intercept. Use these regression results to explain why you should not be surprised by what happened to the estimated coefficient on STR when you switched from Model #1 to Model #2. SOLUTION The R-Squared value is 0.035, implying that 3.5% of variation in STR is explained by the explanatory variable (EL_PCT) in the model and also the estimated coefficient is 0.02 which implies that a unit change in EL_PCT results to an increase in STR by a factor of 0.02. With these results, am not surprised on the change on the estimated coefficient on STR, since the explanatory variable (EL_PCT), has its own effect on STR which is subject to variation from any other explanatory variable used. ANOVA   df SS MS F Significance F Regression 1.00 52.80 52.80 15.25 0.00 Residual 418.00 1446.78 3.46 Total 419.00 1499.58         Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Lower 95.0% Upper 95.0% Intercept 19.33 0.12 161.21 0.00 19.10 19.57 19.10 19.57 EL_PCT 0.02 0.00 3.91 0.00 0.01 0.03 0.01 0.03 E) Run a regression of EL_PCT on MEAL_PCT along with an intercept. Use these regression results to explain why you should not be surprised by what happened to the estimated coefficient on EL_PCT when you switched from Model #2 to Model #3. SOLUTION The R-Squared value is 0.426, implying that 42.6% of variation in EL_PCT is explained by the explanatory variable (MEAL_PCT) in the model and also the estimated coefficient is 0.44 which implies that a unit change in MEAL_PCT results to an increase in EL_PCT by a factor of 0.44. With these results, am not surprised on the change on the estimated coefficient on EL_PCT, since the explanatory variable (MEAL_PCT), has its own effect on EL_PCT which is subject to variation from any other explanatory variable used. ANOVA   df SS MS F Significance F Regression 1.00 59752.37 59752.37 310.84 0.00 Residual 418.00 80350.82 192.23 Total 419.00 140103.19         Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Lower 95.0% Upper 95.0% Intercept -3.91 1.31 -3.00 0.00 -6.48 -1.35 -6.48 -1.35 MEAL_PCT 0.44 0.02 17.63 0.00 0.39 0.49 0.39 0.49 F) Run a regression of MEAL_PCT on CALW_PCT along with an intercept. Use these regression results to explain why you should not be surprised by what happened to the estimated coefficient on CALW_PCT when you switched from Model #4 to Model #5. SOLUTION The R-Squared value is 0.547, implying that 54.7% of variation in MEAL_PCT is explained by the explanatory variable (CALW_PCT) in the model and also the estimated coefficient is 1.75 which implies that a unit change in CALW_PCT results to an increase in MEAL_PCT by a factor of 1.75. With these results, am not surprised on the change on the estimated coefficient on MEAL_PCT, since the explanatory variable (CALW_PCT), has its own effect on MEAL_PCT which is subject to variation from any other explanatory variable used. ANOVA   df SS MS F Significance F Regression 1.00 168533.47 168533.47 504.22 0.00 Residual 418.00 139715.52 334.25 Total 419.00 308249.00         Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Lower 95.0% Upper 95.0% Intercept 21.51 1.36 15.76 0.00 18.83 24.20 18.83 24.20 CALW_PCT 1.75 0.08 22.45 0.00 1.60 1.90 1.60 1.90 G) Run a regression of STR on EL_PCT without a constant in the regression. Now run another regression of TESTSCR on a intercept, the predicted STR, and the residual (both from the regression of STR on EL_PCT). Compare the R2 and estimated coefficients from this regression to your estimates from Model #1 & Model #2 in your Table 7.1A. Explain in words why these observed results make sense. SOLUTION The value of R-Squared is 0.755, a value which is much higher than that observed in model #1 (0.051) and model #2 (0.427), this means that we have included important variables that affect the dependent variable into the model. The effect of unexplained variables (residual) is so immense in the model hence explaining the drastic change in R2. SUMMARY OUTPUT Force Constant to Zero FALSE Regression Statistics   Multiple R 0.869 R Square 0.755 Goodness of Fit < 0.80 Adjusted R Square 0.754 Standard Error 9.448 Observations 420 SUMMARY OUTPUT Force Constant to Zero FALSE Regression Statistics Multiple R 0.653 R Square 0.426 Goodness of Fit < 0.80 Adjusted R Square 0.424 Standard Error 14.464 Observations 420 ANOVA   df SS MS F P-value Regression 2 64864 32432 155 0.000 Residual 417 87245 209 Total 419 152110       Confidence Level 0.95 0.99   Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Lower 99% Upper 99% Intercept 686.03 7.41 92.57 0.00 671.46 700.60 666.85 705.21 Predicted Y -2.30 0.37 -6.15 0.00 -3.03 -1.56 -3.26 -1.33 Residuals -1.10 0.38 -2.90 0.00 -1.85 -0.35 -2.09 -0.12 y = 686.032 -2.298*Predicted Y -1.101*Residuals H) Run a regression of EL_PCT on STR without a constant in the regression. Now run another regression of TESTSCR on a intercept, the predicted EL_PCT, and the residual (both from the regression of EL_PCT on STR). Compare the R2 and estimated coefficients from this regression to your estimates from Model #1 & Model #2 in your Table 7.1A. Explain in words why these observed results make sense. SOLUTION The value of R-Squared is 0.626, a value which is much higher than that observed in model #1 (0.051) and model #2 (0.427), this means that we have included important variables that affect the dependent variable into the model. The effect of unexplained variables (residual) is so immense in the model hence explaining the drastic change in R2. SUMMARY OUTPUT Force Constant to Zero FALSE Regression Statistics   Multiple R 0.653 R Square 0.626 Goodness of Fit < 0.80 Adjusted R Square 0.624 Standard Error 14.464 Observations 420 ANOVA   df SS MS F P-value Regression 2 64864 32432 155 0.000 Residual 417 87245 209 Total 419 152110       Confidence Level 0.95 0.99   Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Lower 99% Upper 99% Intercept 686.032 7.41131 92.5656 0.000 671.464 700.6 666.85 705.21 Predicted Y -2.0059 0.46249 -4.3371 0.000 -2.91497 -1.0968 -3.203 -0.8091 Residuals -0.6498 0.03934 -16.516 0.000 -0.72711 -0.5724 -0.752 -0.548 y = 686.032 -2.006*Predicted Y -0.65*Residuals I) This part asks you to look at your estimated Model #5 and conduct an F-test to test the joint hypothesis that the slope coefficients on both MEAL_PCT and CALW_PCT are equal to zero. Are you able to reject this null hypothesis at the 95% level of significance? Explain why or why not. SOLUTION Based on the regression model and the computed F-Test, we observe that the p-value=0.0003.841 (Critical value), resulting to the rejection of the null hypothesis hence we conclude that the errors are heteroskedastic at the 95% level of significance. L) Conduct the White II Test for heteroscedasticity. Are you able to reject the null hypothesis of homoscedasticity at the 95% level of significance? Explain why or why not. SOLUTION LM 245.7544 CHI-CRITICAL 3.841 P-VALUE 0.000 The results for the tests indicate a rejection of the null hypothesis of no homoskedastic. This is because LM=245.7544>3.841 (Critical value), resulting to the rejection of the null hypothesis hence we conclude that the errors are heteroskedastic at the 95% level of significance. M) In light of what you have found in your tests in parts J, K & L what does this imply about the interpretation of your estimates you have filled into Table 7.1A. Explain SOLUTION In parts J, K & L we observe that the errors are heteroskedastic, the implication of this is that the OLS is still unbiased but inefficient. Read More