Mgt 6203 final exam - Study guides, Class notes & Summaries

Week 1 Q1) What would be the null hypothesis for the regression F-test for the following equation Y = 0 + 1 x1 + 2 x2 + u A. 1 = 0 and 2 = 0 B. 1 = 0 or 2 = 0 C. 0 = 0 and 1 = 0 and 2 = 0 D. 0 = 0 or 1 = 0 or 2 = 0 Ans: A, the regression F-statistics test the joint null hypothesis that the values of all the coefficients of explanatory variables except the constant are 0. The intercept is not included in the hypothesis. (Week 1, slide 24 and 32) W...

Summer MGT 6203 FINAL EXAM Updated version 2023 with Correct solutions. Week 1 Q1) What would be the null hypothesis for the regression F-test for the following equation Y = 0 + 1 x1 + 2 x2 + u A. 1 = 0 and 2 = 0 B. 1 = 0 or 2 = 0 C. 0 = 0 and 1 = 0 and 2 = 0 D. 0 = 0 or 1 = 0 or 2 = 0 Ans: A, the regression F-statistics test the joint null hypothesis that the values of all the coefficients of explanatory variables except the constant are 0. The...

MGT 6203 FINAL EXAM PART 2 – CODING 1. Using the function lm, create a linear regression that regresses “Rings” onto “Diameter” and “Height” (i.e., “Rings” is the response variable and “Diameter” ,“Height” are the independent variables). Which one of the following statements is FALSE? a. Diameter is significant at a 5% Confidence Interval b. R-Squared is 0.35 c. The intercept is 11.71 d. One unit of change in height causes the number of rings to increase by 19.81 on ...

MGT 6203 FINAL EXAM PART 1 Q1. If the explained sum of squares is 35 and the total sum of squares is 49, what is the residual sum of squares? A. 10 B. 14 C. 12 D. 18 Solution: B Explanation: Total Sum of Squares = Explained Sum of Squares + Residual Sum of Squares SST = SSE + SSR SSR = SST – SSE = 49 – 35 = 14 Q2. Consider a regression model predicting the amount someone has spent (on some arbitrary product) based on their salary and age. To improve the model we have created an interacti...

Week 1 Q1) What would be the null hypothesis for the regression F-test for the following equation Y = β0 + β1 x1 + β2 x2 + u A. β1 = 0 and β2 = 0 B. β1 = 0 or β2 = 0 C. β0 = 0 and β1 = 0 and β2 = 0 D. β0 = 0 or β1 = 0 or β2 = 0 Answer: A. Explanation: the regression F-statistics test the joint null hypothesis that the values of all the coefficients of explanatory variables except the constant are 0. The intercept is not included in the hypothesis. (Week 1, slide 24 and 32...

Georgia Institute Of Technology - MGT 6203 /MGT 6203 FINAL EXAM QUESTIONS AND ANSWERS.Theory - 20 questions Week 1 Questions Q1. If the explained sum of squares is 35 and the total sum of squares is 49, what is the residual sum of squares? A. 10 B. 14 C. 12 D. 18 Solution: B Explanation: Total Sum of Squares = Explained Sum of Squares + Residual Sum of Squares SST = SSE + SSR SSR = SST – SSE = 49 – 35 = 14 Week 2 Questions Q2. Consider a regression model predicting the amount someone has spe...

Week 4 Use the dataset to answer questions from 1 to 3: 1. Using the function lm, create a linear regression that regresses “Rings” onto “Diameter” and “Height” (i.e., “Rings” is the response variable and “Diameter” ,“Height” are the independent variables). Which one of the following statements is FALSE? a. Diameter is significant at a 5% Confidence Interval b. R-Squared is 0.35 c. The intercept is 11.71 d. One unit of change in height causes the number of rings ...

MGT 6203 FINAL EXAM PART 2 ,GEORGIA INSTITUTE OF TECHNOLOGY MGT 6203 FINAL EXAM PART 2 ,GEORGIA INSTITUTE OF TECHNOLOGY MGT 6203 FINAL EXAM PART 2 ,GEORGIA INSTITUTE OF TECHNOLOGY

MGT 6203 FINAL EXAM PART 1 THEORY GEORGIA INSTITUTE OF TECHNOLOGY MGT 6203 FINAL EXAM PART 1 THEORY GEORGIA INSTITUTE OF TECHNOLOGY MGT 6203 FINAL EXAM PART 1 THEORY GEORGIA INSTITUTE OF TECHNOLOGY