11 compare all the above stepwise models using the mse's compare all the above models using the mse's from the cross-validations with the number of folds equal to 3. "{r } aggregate (mse ~ model, data = cv_results, FUN = mean) Description: df [6 x 2] model mse 9-poly2.inter.step 415290242 g-poly2.step 499307618 g-poly3.inter.step 393300509 g.poly3.step 479476608 g.quad.step 570556655 g.step 500902923 6 rows which model wins most of the cv's? 12 which models pass hypothesis testing? g. poly2. step vs g. poly2. inter . step Perform the ANOVA test of models g. poly2. step vs g. poly2. inter . step "{r} anova (g. poly2. step, 9. poly2. inter . step) Analysis of variance Table Model 1: salary ~ rank + discipline + poly(yrs. since. phd, 2) Model 2: salary ~ sex + poly(yrs. since. phd, 2) + rank + discipline + poly(yrs. service, 2) + sex : discipline + poly(yrs. since. phd, 2) : discipline + poly (yrs. since. phd, 2) : poly(yrs. service, 2) + discipline: poly(yrs. service, 2) + poly(yrs. since. phd, 2) : discipline: poly(yrs. service, Res . Df RSS Df Sum of Sq F Pr (>F) 391 1. 9825e+11 NP 375 1. 6488e+11 16 3. 3371e+10 4. 7438 8. 54e-09 ** * Signif. codes: 0 '* * * ' 0. 001 '* * ' 0. 01 ' * ' 0. 05 '. 'Analysis of Variance Table Model 1: salary ~ rank + discipline + poly(yrs. since. phd, 2) Model 2: salary ~ sex + poly(yrs. since. phd, 2) + rank + discipline + poly(yrs. service, 2) + sex : discipline + poly(yrs. since. phd, 2) : discipline + poly (yrs. since. phd, 2) : poly(yrs. service, 2) + discipline: poly(yrs. service, 2) + poly(yrs. since. phd, 2) : discipline: poly(yrs. service, 2 ) Res . Df RSS Df Sum of Sq F Pr (>F) 391 1. 9825e+11 NH 375 1. 6488e+11 16 3. 3371e+10 4. 7438 8. 54e-09 ** * Signif. codes: 0 '* **' 0. 001 '*#' 0.01 '*' 0.05 ' . ' 0.1 ' ' 1 what is the p-value? 8. 54e-09 Using alpha F) 387 1. 9039e+11 370 1. 5615e+11 17 3. 4237e+10 4. 772 3. 161e-09 *s* Signif. codes: 0 '* **' 0. 001 '**' 0.01 '*' 0.05 ' . ' 0.1 ' ' 1 what is the p-value? Using alpha