Mgt 6203 - Study guides, Class notes & Summaries

Summer MGT 6203 MID EXAM PART2 – CODING Week 1 Use the inbuilt dataset ‘longley’ for questions 1 and 2. Q1) Fit a linear regression model with ‘Employed’ as the response variable and all other variables (except ‘Year’) as predictors. What are the significant predictors at 10% significance level? A. GNP B. GNP, Armed.Forces C. GNP, Unemployed, Population D. None of the predictors are significant at 10% significance level Solution: model1 = lm(Employed~.-Year, data = long...

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...

MGT 6203 Midterm Part 1 - Questions and Answers Q1) What is the adjusted R-squared: A. 0.681 B. 0.6772 C. 0.6778 D. 0.3228 Ans: B: 1-[((1-0.681)*(506-1))/(506-6-1)] or 1-[(13625.6/499)/(42716.3/505)] From Wk1 Page 19 Slide 1

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...

Self Assessment 1: Data Analytics Business - MGT-6203-OAN Self Assessment 1 Due Feb 5 at 11:59pm Points 30 Questions 12 Available Jan 24 at 8am - Feb 5 at 11:59pm 13 days Time Limit None Allowed Attempts Unlimited Instructions Attempt History Attempt Time Score LATEST Attempt 1 61 minutes 27.5 out of 30  Correct answers are hidden. Score for this attempt: 27.5 out of 30 Submitted Feb 4 at 4:42pm This attempt took 61 minutes. Self Assessment 1 covers the topics in Weeks 1, 2 and ...

MIDTERM – SOLUTION KEY CODING QNS Q21) Please estimate a linear regression model (using the lm function) with Personal as the dependent variable and Room.Board as the independent variable. What are the model’s Rsquared and adjusted R-squared values? a) 0.00549, 0.048 b) 0.0143, 0.022 c) 0.0398, 0.0385 d) 0.0325, 0.0336 Answer: C (Week 1 Lesson 4) library("ISLR") data("College") summary(lm(College$Personal~College$Room.Board)) ## ## Call: ## lm(formula = College$Personal ~ Co...

MGT 6203 MIDTERM PART 2– SOLUTION KEY CODING QUESTIONS with ANSWERS Week 1 Use the inbuilt dataset ‘longley’ for questions 1 and 2. Q1) Fit a linear regression model with ‘Employed’ as the response variable and all other variables (except ‘Year’) as predictors. What are the significant predictors at 10% significance level? A. GNP B. GNP, Armed.Forces C. GNP, Unemployed, Population D. None of the predictors are significant at 10% significance level Solution: model1 = lm(E...