Regression Analysis: Understanding Relationships in Data

Chapter 16 Exercise 1 A. Regression Statistics Multiple R 0.794631 6 R Square 0.631439 37 Adjusted R Square 0.539299 22 Standard Error 7.269276 44 Observations 6 ANOVA df SS MS F Significanc e F Regression 1 362.13048 362.1304 8 6.853031 23 0.0589334 4 Residual 4 211.36952 52.84238 Total 5 573.5 Coefficie nts Standard Error t Stat P-value Lower 95% Upper 95% Lower 95.0% Upper 95.0% Intercept -6.77453 03 14.1709475 -0.47805 77 0.657562 78 -46.119388 32.57032 75 -46.11938 8 32.570327 5 x 1.229645 09 0.46971932 2.617829 49 0.058933 44 -0.0745048 2.533795 01 -0.074504 8 2.5337950 1 Regression Equation: Y = -6.7745 + 1.2296x

B. H0: There is no significant relationship between X & Y Ha: There is a significant relationship between X & Y P-value = 0.0589 > alpha = 0.05 Do not reject H0 and conclude that that there is no significant relationship between X & Y. C. Scatterplot suggests that the data is positively correlated D. SUMMARY OUTPUT Regression Statistics Multiple R 0.97201226 R Square 0.94480783 Adjusted R Square 0.90801305 Standard Error 3.24821546 Observations 6 ANOVA df SS MS F Significance F

Regression 2 541.847289 270.923644 25.6777668 0.01296631 Residual 3 31.652711 10.5509037 Total 5 573.5 Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Lower 95.0% Upper 95.0% Intercept -168.88484 39.7861933 -4.2448103 0.02394868 -295.50227 -42.267418 -295.50227 -42.267418 x 12.1870135 2.66323606 4.57601702 0.01958678 3.71140777 20.6626193 3.71140777 20.6626193 x^2 -0.1770358 0.04289548 -4.127143 0.02579792 -0.3135483 -0.0405232 -0.3135483 -0.0405232 Regression Equation: Y = -168.8848 + 12.1870x - 0.1770x 2 E. H0: Relationship between x, x 2 , and y are insignificant Ha: Relationship between x, x 2 , and y are significant T.S: F = 25.6777 P-value = 0.0129 < alpha - 0.05 Reject H0, conclude that the relationship between variables x, x 2 , and y are significant. F. Given x = 25 Y = -168.8848 + 12.1870(25) - 0.1770(25 2 ) = 25.1652

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Exercise 27 A. The scatter plot approximately shows that the weight and price variables do not have a strong relationship, with many of the data points plotted in a curved line. As such, a simple linear regression model wouldn’t be appropriate. B. SUMMARY OUTPUT Regression Statistics Multiple R 0.87761985 R Square 0.77021661 Adjusted R Square 0.74149369 Standard Error 242.80435 Observations 19

ANOVA df SS MS F Significance F Regression 2 3161747.29 1580873.64 26.8153971 7.7723E-06 Residual 16 943263.237 58953.9523 Total 18 4105010.53 Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Lower 95.0% Upper 95.0% Intercept 11375.9988 2565.18531 4.43476687 0.00041612 5938.0489 16813.9488 5938.0489 16813.9488 Weight -728.33455 193.68095 -3.7604862 0.0017095 -1138.9198 -317.74928 -1138.9198 -317.74928 Weight^2 11.9737375 3.53910869 3.38326357 0.00379172 4.47116223 19.4763127 4.47116223 19.4763127 Regression Equation: Y = 11375.9988 - 728.33456x + 11.9737x 2 C. SUMMARY OUTPUT Regression Statistics Multiple R 0.84691921 R Square 0.71727214 Adjusted R Square 0.68193116 Standard Error 269.327964 Observations 19 ANOVA df SS MS F Significance F Regression 2 2944409.69 1472204.85 20.2957614 4.0827E-05 Residual 16 1160600.83 72537.5521 Total 18 4105010.53

Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Lower 95.0% Upper 95.0% Intercept 1283.75 95.2218148 13.4816796 3.7432E-10 1081.88877 1485.61123 1081.88877 1485.61123 Fitness -571.75 153.540563 -3.7237717 0.00184727 -897.24145 -246.25855 -897.24145 -246.25855 Comfort -907.08333 145.453725 -6.2362331 1.1897E-05 -1215.4315 -598.73521 -1215.4315 -598.73521 Regression Equation: Y = 1283.75 - 571.75(Fitness) - 907.08(Comfort) Part B. Adjusted R 2 = 0.7415 Part C. Adjusted R 2 = 0.6819 Thus, the regression equation obtained in Part B. is better than the equation obtained in Part C. D. SUMMARY OUTPUT Regression Statistics Multiple R 0.9167757 R Square 0.8404776 8 Adjusted R Square 0.7791229 4 Standard Error 224.43789 6 Observations 19 ANOVA df SS MS F Significance F Regression 5 3450169.73 690033.94 6 13.698659 8 8.5356E-05 Residual 13 654840.797 50372.369 Total 18 4105010.53

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Coefficien ts Standard Error t Stat P-value Lower 95% Upper 95% Lower 95.0% Upper 95.0% Intercept 5923.5443 1546.52226 3.8302354 1 0.0020842 1 2582.48609 9264.6025 1 2582.48609 9264.60251 Fitness -6343.250 2 2596.16097 -2.443319 3 0.0295819 3 -11951.915 -734.5854 -11951.915 -734.5854 Comfort -7232.062 8 2518.19339 -2.871925 1 0.0130946 8 -12672.289 -1791.836 8 -12672.289 -1791.8368 Weight*Comfort 266.40881 4 93.9802425 2.8347321 4 0.0140625 63.3768436 469.44078 4 63.3768436 469.440784 Weight*Fitness 261.32166 8 111.842303 2.3365190 2 0.0361219 2 19.7010622 502.94227 4 19.7010622 502.942274 Weight -214.5569 6 71.4212811 -3.004104 1 0.0101584 -368.85326 -60.26066 5 -368.85326 -60.260665 Regression Equation: Y = 5923.5443 - 6343.2502(Fitness) - 7232.0682(Comfort) + 266.4088(Weight*Comfort) + 261.3217(Weight*Fitness) - 214.5570(Weight) Part B. Adjusted R 2 = 0.7415 Part C. Adjusted R 2 = 0.6819 Part D. Adjusted R 2 = 0.7791 This Regression Equation provides the highest Adjusted R 2 compared to the equation obtained from previous parts. Therefore, this equation is the best predictor of Price. Additional Exercises 1. Backward Elimination A. Scrambling will not be in the final model as its p-value = 0.149 exceeds the alpha threshold at 0.05. B. Model 1: Adjusted R-square = 18.34% Multiple R-square: 21.64% Model 2: Adjusted R-square = 19.17% Multiple R-square: 21.77% Model 3: Adjusted R-square = 19.40% Multiple R-square: 21.40% Therefore, Model 1 has the lowest Adjusted R-square and Model 3 has the lowest Multiple R-square.

C. No, since the p-value for DrDist (at 0.448) is still higher than the new alpha = 0.32. Other predictors will still be included as they still have a p-value less than the new alpha. 2. Best Subsets Statement 2 – TRUE: Win% vs 20+ has the 2nd lowest AIC value. Statement 3 – TRUE: 40+R has a T-score > 1. 3. Categorical Predictors A. Fit = 70.09 + 0.58*35 - 3.06*1 + 0.52*1 = 87.85 B. Every time it is a Seafood/Steakhouse-type restaurant, the score decreases by 3.06 C. The higher the average meal price, the higher the score; for every increase of a dollar in average meal price, the rating goes up by 0.58. D. Slope of the variable remains the same but the coefficient reverses; slope of the Italian predictor = 3.06 4. CEO Salaries A. SUMMARY OUTPUT Regression Statistics Multiple R 0.82729329 R Square 0.68441418 Adjusted R Square 0.67126477 Standard Error 901.047754 Observations 26 ANOVA df SS MS F Signific anc e F Regression 1 42257945.3 42257945.3 52.0490444 1.8692E-07 Residual 24 19485289.3 811887.054 Total 25 61743234.6

Coeffi ci e nt s Standar d Erro r t Stat P-val u e Lower 95 % Uppe r 9 5 % Lower 95. 0% Upper 95. 0% Intercept 852.858741 200.465871 4.25438373 0.00027656 439.117518 1266.59996 439.117518 1266.59996 Sales ($Mil.) 0.0688151 0.00953844 7.21450236 1.8692E-07 0.04912873 0.08850148 0.04912873 0.08850148 Correlation = 0.8273 B. SUMMARY OUTPUT Regression Statistics Multiple R 0.92145649 R Square 0.84908206 Adjusted R Square 0.84279381 Standard Error 0.15231861 Observations 26 ANOVA df SS MS F Signific anc e F Regression 1 3.13275176 3.13275176 135.026822 2.4255E-11 Residual 24 0.55682302 0.02320096 Total 25 3.68957478 Coeffi ci e nt s Standar d Erro r t Stat P-val u e Lower 95 % Uppe r 9 5 % Lower 95. 0% Upper 95. 0% Intercept 1.65692143 0.1208026 13.7159419 7.4931E-13 1.40759712 1.90624574 1.40759712 1.90624574 Log Sales 0.41397085 0.0356254 11.6201042 2.4255E-11 0.34044364 0.48749807 0.34044364 0.48749807

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Correlation = 0.9215 C. Part A. and B. does not have equal correlation as the logarithmic transformation in Part B. does not preserve the linear characteristic of the original function. D. * Required outputs have been shown in Part A & B respectively. The regression analysis shows that the residuals are less in the logarithmic scale compared to the normal scale. This shows that there is a better linear relationship when using the logarithmic scale. E. Normal Scale: CEO Pay = 852.8587 + 0.0688(Sales) = 852.8587 + 0.0688*7billion = 481,600,852.86 Logarithmic Scale: CEO Pay = 1.6569 + 0.4140*7billion = 2,898,000,001.66 F. XYZ should choose residuals closer to 0. In such a case, XYZ should choose the logarithmic model. 5. Logistic Regression A. P(1) = (-3.623 + 0.0491*45 + 1.605) / [1+(-3.623 + 0.0491*45+1.605)] = 0.1607 B. P(1) = -3.623 + 0.0491*45 / [1+(-3.623 + 0.0491*45)] = 3.4184 C. We must multiply the odds in part B by 1.605 to get the odds in part A. D. For the income variable, for every unit increase of household income, the odds of purchasing a cereal would increase by exp(0.0491) when ViewAd remains the same. For the ViewAd variable; if the shopper watches an add, then the probability of purchasing a cereal would increase by exp(1.605) when income remains the same E. The ViewAd_Yes variable is significant because its coefficient of 1.605 lies within the C.I of (1.4855, 16.6835). Interaction Effects A. Gambling Amount($) = 30.1 - 68.2 Sex - 0.481 Status + 1.088 Sex:Status + 4.970 Income = 30.1 - 68.2(1) - 0.481(70) + 1.088(1*70) + 4.970(3.5) = 21.785 B. = 30.1 - 68.2(1) - 0.481(80) + 1.088(1*80) + 4.970(3.5) = 27.855 C. = 30.1 - 68.2(0) - 0.481(70) + 1.088(0) + 4.970(3.5) = 13.825 D. = 30.1 - 68.2(0) - 0.481(80) + 1.088(0) + 4.970(3.5) = 9.015 E. For females, through the sex:status variable, we can see that there is an increased effect of status on gambling amount. Therefore, for every 1-unit increase in status, there is an

increase of $0.607 ($1.088 - $0.481) in gambling amount. As such, we can surmise that for females, with all else constant, the higher the status level, the higher the gambling amount will be. Different from males, with all else constant, the higher the status level, the lower the gambling amount will be since the sex:status variable wouldn’t be applied. Therefore, for every 1-unit increase in status, the gambling amount will be decreased by 0.481. Trend and Lag Modeling A. SUMMARY OUTPUT Regression Statistics Multiple R 0.97176371 R Square 0.94432471 Adjusted R Square 0.94246887 Standard Error 17.3943043 Observations 32 ANOVA df SS MS F Significan ce F Regression 1 153955.165 153955.16 5 508.83870 6 2.2716E-20 Residual 30 9076.85461 302.56182 Total 31 163032.02 Coeffici en ts Standard Error t Stat P-valu e Lower 95% Upper 9 5 % Lower 95. 0% Upper 95. 0% Intercept 93.9995968 6.29684354 14.928050 2 1.9922E-15 81.1397267 106.859467 81.1397267 106.859467 Trend 7.51233504 0.33303117 22.557453 4 2.2716E-20 6.83219466 8.19247542 6.83219466 8.19247542 Regression Equation: Y = 93.9996 + 7.5123(Trend)

SUMMARY OUTPUT Regression Statistics Multiple R 0.98578672 R Square 0.97177546 Adjusted R Square 0.9708022 Standard Error 12.2841678 Observations 31 ANOVA df SS MS F Significan ce F Regression 1 150670.605 150670.60 5 998.47466 5 5.0921E-24 Residual 29 4376.12258 150.90077 9 Total 30 155046.727 Coeffici en ts Standard Error t Stat P-valu e Lower 95% Upper 9 5 % Lower 95. 0% Upper 95. 0% Intercept 3.50603991 7.22165849 0.4854895 7 0.6309769 4 -11.26391 18.2759899 -11.26391 18.2759899 Lag 1 1.01632759 0.03216364 31.598649 7 5.0921E-24 0.95054556 1.08210962 0.95054556 1.08210962 Regression Equation: Y = 3.5060 + 1.0163 (Lag1) B. Price vs Lag1 R-Square = 0.9718 Price vs Trend R-Square = 0.9443 Using the Price vs Lag1 Model is more accurate C. Lag1, 2021 = 342.9595 Lag1, 2022 = 356.0701 Trend, 2021 = 334.3943 Trend, 2022 = 341.9067

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Chapter 17 Exercise 5 A. A horizontal pattern exists in the data. B. Week Value Forecast Forecast Error Squared FE 1 18 2 13 3 16 4 11 15.6666667 -4.666666667 21.7777778 5 17 13.3333333 3.666666667 13.4444444 6 14 14.6666667 -0.666666667 0.44444444 Total 35.6666667

MSE 11.8888889 Forecast Week 7 14 C. Week Value Forecast Forecast Error Squared FE 1 18 2 13 18 -5 25 3 16 17 -1 1 4 11 16.8 -5.8 33.64 5 17 15.64 1.36 1.85 6 14 15.91 -1.91 3.65 Total 65.14 MSE 13.03 Forecast Week 7 15.53 D. The 3-week moving average provides a better forecast as its MSE is lower at 11.89 compared to the smoothing forecast using a = 0.2 which results in a higher MSE at 13.03. E. Week Value Forecast Forecast Error Squared FE 1 18 2 13 18 -5 25 3 16 16 0 0 4 11 16 -5 25 5 17 14 3 9.00 6 14 15.2 -1.2 1.44 Total 60.44 MSE 12.09

A smoothing constant of a = 0.4 results in a lower MSE of 12.09, thus providing a more accurate forecast. Exercise 11 A. A horizontal pattern exists in the data. B. Three-month Moving Average Month Percentage ForecastFE FE^2 1 80 2 82 3 84 4 83 82.00 1.00 1.00 5 83 83.00 0.00 0.00 6 84 83.33 0.67 0.44 7 85 83.33 1.67 2.78

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8 84 84.00 0.00 0.00 9 82 84.33 -2.33 5.44 10 83 83.67 -0.67 0.44 11 84 83.00 1.00 1.00 12 83 83.00 0.00 0.00 Total 11.11 MSE 1.23 Smoothing a = 0.2 Month Percentage ForecastFE FE^2 1 80 2 82 80 2.00 4.00 3 84 80 3.60 12.96 4 83 81 1.88 3.53 5 83 81 1.50 2.26 6 84 82 2.20 4.85 7 85 82 2.76 7.63 8 84 83 1.21 1.46 9 82 83 -1.03 1.06 10 83 83 0.17 0.03 11 84 83 1.14 1.30 12 83 83 -0.09 0.01 Total 39.11 MSE 3.56 The 3-month moving average provides a lower and therefore more accurate MSE at 1.23. C. Forecast for next month = (83 + 84 + 83) / 3 = 83.33

Exercise 25 A. An upward, linear pattern exists in the data. B. Linear Equation: y = 16.779x + 11.465 C. Quadratic Line Equation: y = 1.563x 2 + 5.842x + 26.048 D. Linear MSE = 15.3 Quadratic MSE = 0.11 The Quadratic Model is better due to its lower MSE. E. @period = 7 Using Linear, y = $128.9 million Using Quadratic, y = $143.5 million F. I would use the forecast using the Quadratic model as it would provide a more accurate data due to its lower MSE as calculated in part D.

Extra Credit Internship Dataset Full-time Job No Full-time Job Total No Internship 458 316 774 1 Internship 600 87 687 2 Internship 319 11 330 Total 1377 414 1791 A student who did 1 internship before graduation increases their probability of getting a full-time job before or within 6 months of graduation by (600/687 - 458/774) * 100% = 19.25%. Research question: Maximum Likelihood Estimation, used to estimate the coefficients in logistic regression.

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MGMT 305 Assignment 3 Gabriel Chang

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