Chapter 15 Multiple Regression

Transcription

1 Multiple Regression Learning Objectives 1. Understand how multiple regression analysis can be used to develop relationships involving one dependent variable and several independent variables. 2. Be able to interpret the coefficients in a mu ltiple regression analysis. 3. Know the assumptions necessary to conduct statistical tests involving the hypothesized regression model. 4. Understand the role of computer packages in performing multiple regression analysis. 5. Be able to interpret and use computer output to develop the estimated regression equation. 6. Be able to determine how good a fit is provided by the estimated regression equation. 7. Be able to test for the significance of the regression equation. 8. Understand how multicollinearity affects multiple regression analysis. 9. Know how residual analysis can be used to make a judgement as to the appropriateness of the model, identify outliers, and determine which observations are influential. 15 1

2 Solutions: 1. a. b 1 =.5906 is an estimate of the change in y corresponding to a 1 unit change in x 1 when x 2 is held constant. b 2 =.4980 is an estimate of the change in y corresponding to a 1 unit change in x 2 when x 1 is held constant. 2. a. The estimated regression equation is ŷ = x 1 An estimate of y when x 1 = 45 is ŷ = (45) = b. The estimated regression equation is ŷ = x 2 An estimate of y when x 2 = 15 is ŷ = (15) = c. The estimated regression equation is ŷ = x x 2 An estimate of y when x 1 = 45 and x 2 = 15 is ŷ = (45) (15) = a. b 1 = 3.8 is an estimate of the change in y corresponding to a 1 unit change in x 1 when x 2, x 3, and x 4 are held constant. b 2 = 2.3 is an estimate of the change in y corresponding to a 1 unit change in x 2 when x 1, x 3, and x 4 are held constant. b 3 = 7.6 is an estimate of the change in y corresponding to a 1 unit change in x 3 when x 1, x 2, and x 4 are held constant. b 4 = 2.7 is an estimate of the change in y corresponding to a 1 unit change in x 4 when x 1, x 2, and x 3 are held constant. 4. a. ŷ = (15) + 8(10) = 255; sales estimate: $255,000 b. Sales can be expected to increase by $10 for every dollar increase in inventory investment when advertising exp enditure is held constant. Sales can be expected to increase by $8 for every dollar increase in advertising expenditure when inventory investment is held constant. 15 2

3 Multiple Regression 5. a. The Minitab output is shown below: Revenue = TVAdv Constant TVAdv S = RSq = 65.3% RSq(adj) = 59.5% Regression Residual Error Total b. The Minitab output is shown below: Revenue = TVAdv NewsAdv Constant TVAdv NewsAdv S = RSq = 91.9% RSq(adj) = 88.7% Regression Residual Error Total Source DF Seq SS TVAdv NewsAdv c. No, it is 1.60 in part 2(a) and 2.99 above. In this exercise it represents the marginal change in revenue due to an increase in television advertising with newspaper advertising held constant. d. Revenue = (3.5) (1.8) = $93.56 or $93, a. The Minitab output is shown below: Speed = Weight Constant Weight

4 S = RSq = 31.1% RSq(adj) = 26.2% Regression Error Total b. The Minitab output is shown below: Speed = Weight Horsepwr Constant Weight Horsepwr S = RSq = 88.0% RSq(adj) = 86.2% Regression Residual Error Total a. The Minitab output is shown below: Sales = Compet$ Heller$ Constant Compet$ Heller$ S = RSq = 65.3% RSq(adj) = 55.4% Regression Residual Error Total b. b 1 =.414 is an estimate of the change in the quantity sold (1000s) of the Heller mower with respect to a $1 change in price in competitor s mower with the price of the Heller mower held constant. b 2 =.270 is an estimate of the change in the quantity sold (1000s) of the Heller mower with respect to a $1 change in its price with the price of the competitor s mower held constant. 15 4

5 Multiple Regression c. ŷ = (170) 0.270(160) = or 93,680 units 8. a. The Minitab output is shown below: Return = Safety ExpRatio Constant Safety ExpRatio S = RSq = 58.2% RSq(adj) = 53.3% Regression Residual Error Total b. y ˆ = (7.5) (2) = a. The Minitab output is shown below: %College = Size SatScore Constant Size SatScore S = RSq = 38.2% RSq(adj) = 30.0% Regression Residual Error Total b. ŷ = (20) (1000) = 73.8 Estimate is 73.8% 10. a. The Minitab output is shown below: Revenue = Cars 15 5

6 Constant Cars S = RSq = 92.5% RSq(adj) = 91.9% Regression Error Total b. An increase of 1000 cars in service will result in an increase in revenue of $7.98 million. c. The Minitab output is shown below: Revenue = Cars Location Constant Cars Location S = RSq = 94.2% RSq(adj) = 93.2% Regression Error Total a. SSE = SST SSR = 6, , = b. c. 2 SSR 6, R = = =.924 SST 6, n Ra = 1 (1 R ) = 1 (1.924) =.902 n p d. The estimated regression equation provided an excellent fit. 12. a. b. 2 SSR 14,052.2 R = = =.926 SST 15, n Ra = 1 (1 R ) = 1 (1.926) =.905 n p

7 Multiple Regression c. Yes; after adjusting for the number of independent variables in the model, we see that 90.5% of the variability in y has been accounted for. 2 SSR a. R = = =.975 SST 1805 b. 2 2 n Ra = 1 (1 R ) = 1 (1.975) =.971 n p c. The estimated regression equation provided an excellent fit. 14. a. b. 2 SSR 12,000 R = = =.75 SST 16, n 1 9 Ra = 1 (1 R ) = 1.25 =.68 n p 1 7 c. The adjusted coefficient of determination shows that 68% of the variability has been explained by the two independent variables; thus, we conclude that the model does not explain a large amount of variability. 2 SSR a. R = = =.919 SST n Ra = 1 (1 R ) = 1 (1.919) =.887 n p b. Multiple regression analysis is preferred since both R 2 and R 2 show an increased percentage of the a variability of y explained when both independent variables are used. 16. Note: the Minitab output is shown with the solution to Exercise 6. a. No; RSq = 31.1% b. Multiple regression analysis is preferred since both RSq and RSq(adj) show an increased percentage of the variability of y explained when both independent variables are used. 2 SSR a. R = = =.382 SST n Ra = 1 (1 R ) = 1 (1.382) =.30 n p b. The fit is not very good 18. Note: The Minitab output is shown with the solution to Exercise 10. a. RSq = 94.2% RSq(adj) = 93.2% b. The fit is very good. 15 7

8 19. a. MSR = SSR/p = 6, /2 = 3, SSE MSE = n p 1 = = b. F = MSR/MSE = 3, / = F.05 = 4.74 (2 degrees of freedom numerator and 7 denominator) Since F = > F.05 = 4.74 the overall model is significant. c. t =.5906/.0813 = 7.26 t.025 = (7 degrees of freedom) Since t = > t.025 = 2.365, β 1 is significant. d. t =.4980/.0567 = 8.78 Since t = 8.78 > t.025 = 2.365, β 2 is significant. 20. A portion of the Minitab output is shown below. Y = X X2 Constant X X S = RSq = 92.6% RSq(adj) = 90.4% Regression Residual Error Total a. Since the pvalue corresponding to F = is.000 < α =.05, we reject H 0 : β 1 = β 2 = 0; there is a significant relationship. b. Since the pvalue corresponding to t = 8.13 is.000 < α =.05, we reject H 0 : β 1 = 0; β 1 is significant. c. Since the pvalue corresponding to t = 5.00 is.002 < α =.05, we reject H 0 : β 2 = 0; β 2 is significant. 21. a. In the two independent variable case the coefficient of x 1 represents the expected change in y corresponding to a one unit increase in x 1 when x 2 is held constant. In the single independent variable case the coefficient of x 1 represents the expected change in y corresponding to a one unit increase in x 1. b. Yes. If x 1 and x 2 are correlated one would expect a change in x 1 to be accompanied by a change in x

9 Multiple Regression 22. a. SSE = SST SSR = = SSE 4000 s = n p1 = 7 = SSR MSR = = = 6000 p 2 b. F = MSR/MSE = 6000/ = F.05 = 4.74 (2 degrees of freedom numerator and 7 denominator) Since F = > F.05 = 4.74, we reject H 0. There is a significant relationship among the variables. 23. a. F = F.01 = (2 degrees of freedom, numerator and 1 denominator) Since F > F.01 = 13.27, reject H 0. Alternatively, the pvalue of.002 leads to the same conclusion. b. t = 7.53 t.025 = Since t > t.025 = 2.571, β 1 is significant and x 1 should not be dropped from the model. c. t = 4.06 t.025 = Since t > t.025 = 2.571, β 2 is significant and x 2 should not be dropped from the model. 24. Note: The Minitab output is shown in part (b) of Exercise 6 a. F = F.05 = 3.81 (2 degrees of freedom numerator and 13 denominator) Since F = > F.05 = 3.81, we reject H 0 : β 1 = β 2 = 0. Alternatively, since the pvalue =.000 < α =.05 we can reject H 0. b. For Weight: H 0 : β 1 = 0 H a : β

10 Since the pvalue = > α = 0.05, we cannot reject H 0 For Horsepower: H 0 : β 2 = 0 H a : β 2 0 Since the pvalue = < α = 0.05, we can reject H a. The Minitab output is shown below: P/E = Profit% Sales% Constant Profit% Sales% S = RSq = 47.2% RSq(adj) = 39.0% Regression Residual Error Total b. Since the pvalue = < α = 0.05, there is a significant relationship among the variables. c. For Profit%: Since the pvalue = < α = 0.05, Profit% is significant. For Sales%: Since the pvalue = > α = 0.05, Sales% is not significant. 26. Note: The Minitab output is shown with the solution to Exercise 10. a. Since the pvalue corresponding to F = is < α =.05, there is a significant relationship among the variables. b. For Cars: Since the pvalue = < α = 0.05, Cars is significant c. For Location: Since the pvalue = > α = 0.05, Location is not significant 27. a. ŷ = (180) (310) = b. The point estimate for an individual value is ŷ = , the same as the point estimate of the mean value. 28. a. Using Minitab, the 95% confidence interval is to b. Using Minitab, the 95% prediction interval is to

11 Multiple Regression 29. a. ŷ = (3.5) (1.8) = or $93,555 Note: In Exercise 5b, the Minitab output also shows that b 0 = , b 1 = , and b 2 = ; hence, ŷ = x x 2. Using this estimated regression equation, we obtain ŷ = (3.5) (1.8) = or $93,588 The difference ($93,588 $93,555 = $33) is simply due to the fact that additional significant digits are used in the computations. From a practical point of view, however, the difference is not enough to be concerned about. In practice, a computer software package is always used to perform the computations and this will not be an issue. The Minitab output is shown below: Fit Stdev.Fit 95% C.I. 95% P.I ( , ) ( , ) Note that the value of FIT ( ŷ ) is b. Confidence interval estimate: to or $92,840 to $94,335 c. Prediction interval estimate: to or $91,774 to $95, a. Since weight is not statistically significant (see Exercise 24), we will use an estimated regression equation which uses only Horsepower to predict the speed at 1/4 mile. The Minitab output is shown below: Speed = Horsepwr Constant Horsepwr S = RSq = 87.3% RSq(adj) = 86.4% Regression Residual Error Total Unusual Observations Obs Horsepwr Speed Fit SE Fit Residual St Resid R X 15 11

12 R denotes an observation with a large standardized residual X denotes an observation whose X value gives it large influence. The output shows that the point estimate is a speed of miles per hour. b. The 95% confidence interval is to miles per hour. c. The 95% prediction interval is to miles per hour. 31. a. Using Minitab the 95% confidence interval is 58.37% to 75.03%. b. Using Minitab the 95% prediction interval is 35.24% to 90.59%. 32. a. E(y) = β 0 + β 1 x 1 + β 2 x 2 where x 2 = 0 if level 1 and 1 if level 2 b. E(y) = β 0 + β 1 x 1 + β 2 (0) = β 0 + β 1 x 1 c. E(y) = β 0 + β 1 x 1 + β 2 (1) = β 0 + β 1 x 1 + β 2 d. β 2 = E(y level 2) E(y level 1) 33. a. two β 1 is the change in E(y) for a 1 unit change in x 1 holding x 2 constant. b. E(y) = β 0 + β 1 x 1 + β 2 x 2 + β 3 x 3 where x 2 x 3 Level c. E(y level 1) = β 0 + β 1 x 1 + β 2 (0) + β 3 (0) = β 0 + β 1 x 1 E(y level 2) = β 0 + β 1 x 1 + β 2 (1) + β 3 (0) = β 0 + β 1 x 1 + β 2 E(y level 3) = β 0 + β 1 x 1 + β 2 (0) + β 3 (0) = β 0 + β 1 x 1 + β 3 β 2 = E(y level 2) E(y level 1) β 3 = E(y level 3) E(y level 1) β 1 is the change in E(y) for a 1 unit change in x 1 holding x 2 and x 3 constant. 34. a. $15,300 b. Estimate of sales = (2) + 6.8(8) (0) = 56.1 or $56,100 c. Estimate of sales = (1) + 6.8(3) (1) = 41.6 or $41,

13 Multiple Regression 35. a. Let Type = 0 if a mechanical repair Type = 1 if an electrical repair The Minitab output is shown below: Time = Type Constant Type S = RSq = 8.7% RSq(adj) = 0.0% Regression Residual Error Total b. The estimated regression equation did not provide a good fit. In fact, the pvalue of.408 shows that the relationship is not significant for any reasonable value of α. c. Person = 0 if Bob Jones performed the service and Person = 1 if Dave Newton performed the service. The Minitab output is shown below: Time = Person Constant Person S = RSq = 61.1% RSq(adj) = 56.2% Regression Residual Error Total d. We see that 61.1% of the variability in repair time has been explained by the repair person that performed the service; an acceptable, but not good, fit. 36. a. The Minitab output is shown below: Time = Months Type Person 15 13

14 Constant Months Type Person S = RSq = 90.0% RSq(adj) = 85.0% Regression Residual Error Total b. Since the pvalue corresponding to F = is.002 < α =.05, the overall model is statistically significant. c. The pvalue corresponding to t = 1.57 is.167 > α =.05; thus, the addition of Person is not statistically significant. Person is highly correlated with Months (the sample correlation coefficient is.691); thus, once the effect of Months has been accounted for, Person will not add much to the model. 37. a. Let Position = 0 if a guard Position = 1 if an offensive tackle. b. The Minitab output is shown below: Rating = Position Weight 2.28 Speed Constant Position Weight Speed S = RSq = 47.5% RSq(adj) = 40.1% Regression Residual Error Total c. Since the pvalue corresponding to F = 6.35 is.003 < α =.05, there is a significant relationship between rating and the independent variables. d. The value of RSq (adj) is 40.1%; the estimated regression equation did not provide a very good fit. e. Since the pvalue for Position is t = 2.53 < α =.05, position is a significant factor in the player s rating

15 Multiple Regression f. y ˆ = (1) (300) 2.28(5.1) = a. The Minitab output is shown below: Risk = Age Pressure Smoker Constant Age Pressure Smoker S = RSq = 87.3% RSq(adj) = 85.0% Regression Residual Error Total b. Since the pvalue corresponding to t = 2.91 is.010 < α =.05, smoking is a significant factor. c. Using Minitab, the point estimate is 34.27; the 95% prediction interval is to Thus, the probability of a stroke (.2135 to.4718 at the 95% confidence level) appears to be quite high. The physician would probably recommend that Art quit smoking and begin some type of treatment designed to reduce his blood pressure. 39. a. The Minitab output is shown below: Y = X Constant X S = RSq = 84.5% RSq(adj) = 79.3% Regression

16 Residual Error Total b. Using Minitab we obtained the following values: x i y i y ˆi Standardized Residual The point (3,5) does not appear to follow the trend of remaining data; however, the value of the standardized residual for this point, 1.65, is not large enough for us to conclude that (3, 5) is an outlier. c. Using Minitab, we obtained the following values: x i y i Studentized Deleted Residual t.025 = (n p 2 = = 2 degrees of freedom) Since the studentized deleted residual for (3, 5) is 4.42 < 4.303, we conclude that the 3rd observation is an outlier. 40. a. The Minitab output is shown below: Y = X Predicator Coef Stdev tratio p Constant X s = Rsq = 98.8% Rsq (adj) = 98.3% SOURCE DF SS MS F p Regression Error Total b. Using the Minitab we obtained the following values: x i y i Studentized Deleted Residual

17 Multiple Regression t.025 = (n p 2 = = 2 degrees of freedom) Since none of the studentized deleted residuals are less than or greater than 4.303, none of the observations can be classified as an outlier. c. Using Minitab we obtained the following values: The critical value is 3( p + 1) 3(1+ 1) = = 1.2 n 5 x i y i h i Since none of the values exceed 1.2, we conclude that there are no influential observations in the data. d. Using Minitab we obtained the following values: x i y i D i Since D 5 = > 1 (rule of thumb critical value), we conclude that the fifth observation is influential. 41. a. The Minitab output appears in the solution to part (b) of Exercise 5; the estimated regression equation is: Revenue = TVAdv NewsAdv b. Using Minitab we obtained the following values: y ˆi Standardized Residual

18 With the relatively few observations, it is difficult to determine if any of the assumptions regarding the error term have been violated. For instance, an argument could be made that there does not appear to be any pattern in the plot; alternatively an argument could be made that there is a curvilinear pattern in the plot. c. The values of the standardized residuals are greater than 2 and less than +2; thus, using test, there are no outliers. As a further check for outliers, we used Minitab to compute the following studentized deleted residuals: Studentized Observation Deleted Residual t.025 = (n p 2 = = 4 degrees of freedom) Since none of the studentized deleted residuals is less tan or greater than 2.776, we conclude that there are no outliers in the data. d. Using Minitab we obtained the following values: Observation h i D i The critical average value is 3( p + 1) 3(2+ 1) = = n 8 Since none of the values exceed 1.125, we conclude that there are no influential observations

19 Multiple Regression However, using Cook s distance measure, we see that D 1 > 1 (rule of thumb critical value); thus, we conclude the first observation is influential. Final Conclusion: observations 1 is an influential observation. 42. a. The Minitab output is shown below: Speed = Price Horsepwr Constant Price Horsepwr S = RSq = 91.9% RSq(adj) = 90.7% Regression Residual Error Total Source DF Seq SS Price Horsepwr Unusual Observations Obs Price Speed Fit SE Fit Residual St Resid X X denotes an observation whose X value gives it large influence. b. The standardized residual plot is shown below. There appears to be a very unusual trend in the standardized residuals. xx x 1.2+ x SRES1 x x x 0.0+ x x x x x x 1.2+ x x x 15 19

20 +++++FITS c. The Minitab output shown in part (a) did not identify any observations with a large standardized residual; thus, there does not appear to be any outliers in the data. d. The Minitab output shown in part (a) identifies observation 2 as an influential observation. 43. a. The Minitab output is shown below: %College = SatScore Constant SatScore S = RSq = 29.7% RSq(adj) = 25.3% Regression Residual Error Total Unusual Observations Obs SatScore %College Fit SE Fit Residual St Resid X X denotes an observation whose X value gives it large influence. b. The Minitab output shown in part a identifies observation 3 as an influential observation. c. The Minitab output appears in the solution to Exercise 9; the estimates regression equation is %College = Size SATScore d. The following Minitab output was also provided as part of the regression output for part c. Unusual Observations Obs. Size %College Fit Stdev.Fit Residual St.Resid X X denotes an obs. whose X value gives it large influence. Observation 3 is still identified as an influential observation. 44. a. The expected increase in final college grade point average corresponding to a one point increase in high school grade point average is.0235 when SAT mathematics score does not change. Similarly, the expected increase in final college grade point average corresponding to a one point increase in the SAT mathematics score is when the high school grade point average does not change. b. ŷ = (84) (540) =

21 Multiple Regression 45. a. Job satisfaction can be expected to decrease by 8.69 units with a one unit increase in length of service if the wage rate does not change. A dollar increase in the wage rate is associated with a 13.5 point increase in the job satisfaction score when the length of service does not change. b. ŷ = (4) (6.5) = a. The computer output with the missing values filled in is as follows: Y = X X2 Predicator Coef Stdev tratio Constant X X s = 3.35 Rsq = 92.3% Rsq (adj) = 91.0% SOURCE DF SS MS F Regression Error , Total b. t.025 = (12 DF) for β 1 : 3.61 > 2.179; reject H 0 : β 1 = 0 for β 2 : 5.08 > 2.179; reject H 0 : β 2 = 0 c. See computer output. d R a = 1 (1.923) = a. Y = X X2 Predictor Coef Stdev tratio Constant X X s = Rsq = 93.7% Rsq (adj) = 91.9% SOURCE DF SS MS F Regression Error

22 Total b. F.05 = 4.74 (2 DF numerator, 7 DF denominator) F = > F.05 ; significant relationship. 2 SSR c. R = =.937 SST R = 1 (1.937) = a good fit d. t.025 = (7 DF) for B 1 : t = 2.71 > 2.365; reject H 0 : B 1 = 0 for B 2 : t = 4.51 > 2.365; reject H 0 : B 2 = a. Y = X X2 Predictor Coef Stdev tratio Constant X X s = Rsq = 90.1% Rsq (adj) = 86.1% SOURCE DF SS MS F Regression Error Total b. F.05 = 5.79 (5 DF) F = > F.05 ; significant relationship. 2 SSR c. R = =.901 SST R = 1 (1.901) = a good fit d. t.025 = (5 DF) for β 1 : t = 5.59 < 2.571; reject H 0 : β 1 =

23 Multiple Regression for β 2 : t = 6.48 > 2.571; reject H 0 : β 2 = a. The Minitab output is shown below: Price = BookVal Constant BookVal S = RSq = 29.4% RSq(adj) = 26.9% Regression Error Total b. The value of Rsq is 29.4%; the estimated regression equation does not provide a good fit. c. The Minitab output is shown below: Price = BookVal ReturnEq Constant BookVal ReturnEq S = RSq = 56.7% RSq(adj) = 53.5% Regression Error Total Since the pvalue corresponding to the F test is 0.000, the relationship is significant. 50. a. The Minitab output is shown below: Speed = Price Weight Horsepwr 2.48 Zero60 Constant Price Weight

24 Horsepwr Zero S = RSq = 95.0% RSq(adj) = 93.2% Regression Residual Error Total b. Since the pvalue corresponding to the F test is 0.000, the relationship is significant. c. Since the pvalues corresponding to the t test for both Horsepwr (pvalue =.003) and Zero60 (pvalue =.025) are less than.05, both of these independent variables are significant. d. The Minitab output is shown below: Speed = Horsepwr 3.19 Zero60 Constant Horsepwr Zero S = RSq = 93.1% RSq(adj) = 92.0% Regression Residual Error Total Source DF Seq SS Horsepwr Zero Unusual Observations Obs Horsepwr Speed Fit SE Fit Residual St Resid R X R denotes an observation with a large standardized residual X denotes an observation whose X value gives it large influence. e. The standardized residual plot is shown below: 15 24

25 Multiple Regression SRES x 1.5+ x x 2 x x 0.0+ x x 2 x xx 1.5+ x x FIT There is an unusual trend in the plot and one observation appears to be an outlier. f. The Minitab output indicates that observation 2 is an outlier g. The Minitab output indicates that observation 12 is an influential observation. 51. a. The Minitab output is shown below: 640+ x Exposure 480+ x x 320+ x 160+ x 3 x x TimesAir b. The Minitab output is shown below: Exposure = TimesAir 15 25

26 Constant TimesAir S = RSq = 96.6% RSq(adj) = 96.2% Regression Error Total Since the pvalue is 0.000, the relationship is significant. c. The Minitab output is shown below: Exposure = TimesAir BigAds Constant TimesAir BigAds S = RSq = 99.5% RSq(adj) = 99.3% Regression Error Total d. The pvalue corresponding to the t test for BigAds is 0.000; thus, the dummy variable is significant. e. The dummy variable enables us to fit two different lines to the data; this approach is referred to as piecewise linear approximation. 52. a. The Minitab output is shown below: Resale% = Price Constant Price S = RSq = 36.7% RSq(adj) = 34.4% 15 26

27 Multiple Regression Regression Residual Error Total Since the pvalue corresponding to F = is.000 < α =.05, there is a significant relationship between Resale% and Price. b. RSq = 36.7%; not a very good fit. c. Let Type1 = 0 and Type2 = 0 if a small pickup; Type1 = 1 and Type2 = 0 if a fullsize pickup; and Type1 = 0 and Type2 = 1 if a sport utility. The Minitab output using Type1, Type2, and Price is shown below: Resale% = Type Type Price Constant Type Type Price S = RSq = 63.1% RSq(adj) = 58.8% Regression Residual Error Total d. Since the pvalue corresponding to F = is.000 < α =.05, there is a significant relationship between Resale% and the independent variables. Note that individually, Price is not significant at the.05 level of significance. If we rerun the regression using just Type1 and Type2 the value of RSq (adj) decreases to 54.4%, a drop of only 4%. Thus, it appears that for these data, the type of vehicle is the strongest predictor of the resale value