University South Florida of: QMB 3200-Project 2-Answers,100% CORRECT

University South Florida of: QMB 3200-Project 2-Answers PART I - Data Description – 1. Describe the dependent variable that you are trying to predict in your project. Identify the units that it will be measured in. (2 points) - The dependent variable that we are trying to predict is the price of homes. The units they will be measured in is US dollars. 2. Describe the quantitative independent variable that you are using in your project. Identify the units that it will be measured in. (2 points) - The quantitative independent variable that we are using is size. The units that it will be measured in is square feet. 3. Describe the qualitative independent variable that you are using in your project. Identify how you coded the various levels in your regression model. (2 points) - The qualitative independent variable that we are using is location. “0” is Riverview and “1” is Orlando. 4. Give your experimental unit (the item you measured to get the values of your variables) for this data. (2 points) - A single home PART II – Model Building - use both the QN and QL independent variable to perform a multiple regression analysis of the data. Answer the following questions. 5. Model Building: For the first three tests that you conducted (the Global F- test, the quadratics test, and the interaction test), provide the information that I ask for in the space below. In addition, for each test, include the printout used in the appropriate space. a. Global F-test (6 points) Complete 2nd-Order Model: E(y) = β0 + β1x1 + β2x12 + β3x2 + β4x1x2 + β5x12x2 Fill in the following information for your test: Test: Ho: β1= β2= β3 = β4 = β5= 0 Ha: At least one βi ≠0 Test Statistic: 16.35 P-value: _0.0000 Conclusion: At α = .01 we reject Ho. There is sufficient evidence to indicate that something works. Predictor Variables Coefficient Std Error T P VIF Constant -41274.5 -0.36 0.7203 0.0 Size 140.201 110.954 1.26 0.2118 62.8 Location - -1.16 0.2492 74.9 SxS -9.732E-03 0.02539 -0.38 0.7030 51.6 SxL 248.437 158.477 1.57 0.1228 297.6 SxSxL -0.06098 0.04204 -1.45 0.1527 100.8 Source DF SS MS F P Regression 5 3.496E+11 6.993E+10 16.35 0.0000 Residual 54 2.310E+11 4.278E+09 Total 59 5.806E+11 b. Quadratics Test (8 points) – Fill in the following information for your test. Hint: You can copy and paste the model from the previous page and make the appropriate changes as an easy way of writing the reduced model that you are testing. Full Model: E(y) = β0 + β1x1 + β2x12 + β3x2 + β4x1x2 + β5x12x2 Reduced Model: E(y) = β0 + β1x1 + β3x2 + β4x1x2 Test: Ho: β2= β5 = 0 Ha: At least one βi ≠0 Test Statistic: 2.30 P-value: 0.1100 Conclusion: At α = .10 we fail to reject Ho. There is insufficient evidence to indicate that the quadratics are useful to our model. The reduced model is better at predicting the prices of homes than the full model. Adjusted AICc - P Cp R Square Min AICc Resid SS F P(F) Model Variables 4 6.6 0.5452 1340.30 2.507E+11 A B C 5 4.1 0.5721 1338.03 2.316E+11 4.52 0.0379 A B C E 5 6.1 0.5566 1340.16 2.400E+11 2.45 0.1235 A B C D 6 6.0 0.5653 1340.43 2.310E+11 2.30 0.1100 A B C D E c. Interaction Test (8 points) - Fill in the following information for your test. Full Model: E(y) = β0 + β1x1 + β3x2 + β4x1x2 Reduced Model: E(y) = β0 + β1x1 + β3x2 Test: Ho: β4= 0 Ha: β4 ≠ 0 Test Statistic: 1.87 P-value: 0.0666 Conclusion: At : At α = .10 we reject Ho. The interaction term is useful at predicting the price of the homes. Predictor Variables Coefficient Std Error T P VIF Constant -158.927 41427.5 0.00 0.9970 0.0 Size 98.3285 19.8538 4.95 0.0000 1.9 Location -38269.4 57166.0 -0.67 0.5060 11.0 X1X2 56.0554 29.9583 1.87 0.0666 10.2 R² 0.5683 Mean Square Error (MSE) 4.476E+09 Adjusted R² 0.5452 Standard Deviation 66905.2 AICc 1340.3 PRESS 2.94E+11 Source DF SS MS F P Regression 3 3.300E+11 1.100E+11 24.57 0.0000 Residual 56 2.507E+11 4.476E+09 Total 59 5.806E+11 6. Would your model building be finished after the interaction test of question 5? Why or why not? (2 points) Our model building is finished because our interaction works, and we do not need to test the lower terms individually. PART III – Final Model Interpretations - Answer the following questions about your best model. Predictor Variables Coefficient Std Error T P VIF Constant -158.927 41427.5 0.00 0.9970 0.0 Size 98.3285 19.8538 4.95 0.0000 1.9 Location -38269.4 57166.0 -0.67 0.5060 11.0 X1X2 56.0554 29.9583 1.87 0.0666 10.2 R² 0.5683 Mean Square Error (MSE) 4.476E+09 Adjusted R² 0.5452 Standard Deviation 66905.2 AICc 1340.3 PRESS 2.94E+11 Source DF SS MS F P Regression 3 3.300E+11 1.100E+11 24.57 0.0000 Residual 56 2.507E+11 4.476E+09 Total 59 5.806E+11 7. Identify the least squares prediction equation (use #’s) for your best model after all your testing was completed (you do not need to show the printouts of any additional tests conducted, just the results of your best model). Use the values from the printout and write the prediction equation below. (3 points) ŷ = -158.927 + 98.3285x1 + -38269.4x2 + 56.0554x1x2 8. Interpret the key statistics (s and R2) associated with your best model. (4 points) - 56.83% of the variation in the sampled prices of the homes around their mean can be explained by the regression model - We expect most of our sampled prices of the homes to fall within $133,810.4 of their least squares predicted value 9. Would you use your model in practice? Why or why not? (3 points) - We would not use our model in practice because our standard deviation is large. 10. Use the computer to create a confidence interval for the E(Y) for specified values (your choice) of the X’s. Interpret this confidence interval. (4 points) - We are 95% confident that the mean prices of all the homes in Riverview with the size of 2061 square feet fall between $ and $. 11. Copy and paste a plot to check the equal variance assumption of the random error component. Based on your plot, does the assumption appear satisfied? Why or why not? (4 points) - Our assumption is not satisfied because the shape of the dots appears to be in a cone shape. 12. Identify any outliers that you have in your data set. Copy and paste any printouts that you might need below. If no outliers exist, tell me why. (3 points) - There are two y-outliers and one suspect y-outlier and two x-outliers. There are two y-outliers because, looking at the stem and leaf plot, there are two isolated values that are located at opposite ends (top and bottom) of the plot. Looking at the stem and leaf plot, we can also see a suspect y-outlier located between the bottommost value and the majority of the values that are located around the median. Looking at the scatter plot, we can see two x-outliers. They are the only two values that are beyond the value of “2900” on the x axis, and they are away from the majority of the values. Stem and Leaf Plot of zresidual Leaf Digit Unit = 0.1 Minimum -3.3716 --3 3 represents -3.3 Median -0.0334 Maximum 3.8995 Depth Stem Leaves 1 -3 3 1 -2 1 -2 2 -1 7 4 -1 32 15 -0 55 (17) -0 28 0 11 0 56689 6 1 34 4 1 57 2 2 2 2 7 1 3 1 3 8 60 cases included 0 missing cases 13. Is the normal assumption for the random errors satisfied in your project? Why or why not? You may attach a printout or refer me to one that you have used elsewhere in this project to support your answer. (3 points) - From the stem-and-leaf plot above, we conclude that the normal assumption for the random errors is satisfied. The plot shows a normal curve. 14. What does the plot of your final model look like? You can attach a scatterplot from Statistix to show me your model if you would like. (3 points) - 4.0 Scatter Plot of yhat vs Size 3.1 Location 2.2 0 1 1.3 0.4 500 1300 2100 2900 3700 Size PART IV – Mystery Regression: Use STATISTIX to build the following regression model with your collected data: E(y) = β0 + β1x1 + β3x2 + β4x1x2 Predictor Variables Coefficient Std Error T P VIF Constant -158.927 41427.5 0.00 0.9970 0.0 Size 98.3285 19.8538 4.95 0.0000 1.9 Location -38269.4 57166.0 -0.67 0.5060 11.0 X1X2 56.0554 29.9583 1.87 0.0666 10.2 R² 0.5683 Mean Square Error (MSE) 4.476E+09 Adjusted R² 0.5452 Standard Deviation 66905.2 AICc PRESS 1340.3 2.94E+11 Source DF SS MS F P Regression 3 3.300E+11 1.100E+11 24.57 0.0000 Residual 56 2.507E+11 4.476E+09 Total 59 5.806E+11 15. Note that this model is the one that produces two non-parallel lines. Use this printout to find the estimates for the y-intercepts for the x2=0 and x2=1 lines. (3 points) x2=0: y-intercept = -158.927 x2=1: y-intercept = -38428.327 ŷ = -158.927 + 98.3285x1 + -38269.4x2 + 56.0554x1x2 16. Consider the model stated above as the first model that we have built and that no testing has been performed on it (please ignore any previous testing that you conducted back in question 5). State the null and alternative hypotheses that you would use to determine whether the quantitative variable is necessary to predict Y. (3 points) Ho: β3 = 0 = β4 Ha: At least one βi ≠0 17. Use the computer to find the p-value for the desired test above. P- Value=_0.0000_. (3 points) Adjusted AICc - P Cp R Square Min AICc Resid SS F P(F) Model Variables 2 71.9 -0.0029 1385.16 5.724E+11 A 3 5.5 0.5252 1341.55 2.663E+11 65.51 0.0000 A B 3 26.5 0.3574 1359.71 3.605E+11 33.52 0.0000 A C 4 4.0 0.5452 1340.30 2.507E+11 35.94 0.0000 A B C 18. State the appropriate conclusion for the test above. (3 points) At α = .01, we reject Ho. There is sufficient evidence to indicate that the quantitative term is useful to the model. 19. State the assumptions that are made for the random error component of the regression model. (4 points) The ϵ’s are normally distributed The mean of the ϵ’s is 0 The variance of the ϵ’s is constant for all settings of the x’s The ϵ’s are independent’