> fit.all3 Call: lm(formula = occ.prestige ~ education + income + percent.female, data = prestige) Coefficients: (Intercept) education income percent.female -6.794334 4.186637 0.00131356 -0.008905157 Degrees of freedom: 102 total; 98 residual Residual standard error: 7.846467 > summary(fit.all3) Call: lm(formula = occ.prestige ~ education + income + percent.female, data = prestige) Residuals: Min 1Q Median 3Q Max -19.82 -5.333 -0.1364 5.159 17.5 Coefficients: Value Std. Error t value Pr(>|t|) (Intercept) -6.7943 3.2391 -2.0976 0.0385 education 4.1866 0.3887 10.7708 0.0000 income 0.0013 0.0003 4.7288 0.0000 percent.female -0.0089 0.0304 -0.2929 0.7702 Residual standard error: 7.846 on 98 degrees of freedom Multiple R-Squared: 0.7982 F-statistic: 129.2 on 3 and 98 degrees of freedom, the p-value is 0 Correlation of Coefficients: (Intercept) education income education -0.7774 income 0.1280 -0.6752 percent.female -0.0563 -0.4321 0.5852 > anova(fit.all3) Analysis of Variance Table Response: occ.prestige Terms added sequentially (first to last) Df Sum of Sq Mean Sq F Value Pr(F) education 1 21608.44 21608.44 350.9741 0.0000000 income 1 2248.14 2248.14 36.5153 0.0000000 percent.female 1 5.28 5.28 0.0858 0.7702447 Residuals 98 6033.57 61.57 > > > > fit.ed Call: lm(formula = occ.prestige ~ education, data = prestige) Coefficients: (Intercept) education -10.73198 5.360878 Degrees of freedom: 102 total; 100 residual Residual standard error: 9.103291 > small _ fit.ed > big _ fit.all3 > anova(big,small) Analysis of Variance Table Response: occ.prestige Terms Resid. Df RSS Test 1 education + income + percent.female 98 6033.57 2 education 100 8286.99 -income-percent.female Df Sum of Sq F Value Pr(F) 1 2 -2 -2253.42 18.30054 1.764818e-07 > q()