# Linear Regression issue with model evaluation

Of these below 3 models output , though model 3 has a insignificant value in it , it has higher r square value and lesser F stat compare to model 2. In Model 2 though both variables are significant, the F statistic and RMSE value(1.638) is slightly higher than my model 3 (rms3 is 1.636). Which model should I consider here and why ? Can someone shed some light on this pleaes

 - model 1

Call:
lm(formula = tr$Sales ~ TV + Facebook, data = tr) Residuals: Min 1Q Median 3Q Max -8.6942 -1.4420 -0.1152 1.4358 7.8306 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 5.757673 0.571084 10.082 < 2e-16 *** TV 0.049902 0.002789 17.893 < 2e-16 *** Facebook 0.036963 0.010863 3.403 0.00086 *** Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 2.963 on 147 degrees of freedom Multiple R-squared: 0.6983, Adjusted R-squared: 0.6942 F-statistic: 170.1 on 2 and 147 DF, p-value: < 2.2e-16  model 2 summary(lfit2) Call: lm(formula = tr$Sales ~ TV + Adwords, data = tr)
Residuals:
Min      1Q  Median      3Q     Max
-8.5129 -0.5817  0.1659  1.1445  2.7496
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 2.926869   0.334627   8.747 4.68e-15 ***
TV          0.047292   0.001564  30.246  < 2e-16 ***
Adwords     0.180630   0.009500  19.015  < 2e-16 ***
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 1.655 on 147 degrees of freedom
Multiple R-squared:  0.9059,    Adjusted R-squared:  0.9046
F-statistic: 707.7 on 2 and 147 DF,  p-value: < 2.2e-16


model 3

summary(lfit3)

Call:
lm(formula = tr$Sales ~ TV + Facebook + Adwords, data = tr) Residuals: Min 1Q Median 3Q Max -8.6240 -0.6014 0.1477 1.1314 2.8010 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 2.991331 0.354757 8.432 3e-14 *** TV 0.047312 0.001568 30.180 <2e-16 *** Facebook -0.003618 0.006486 -0.558 0.578 Adwords 0.182602 0.010157 17.978 <2e-16 *** Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 1.659 on 146 degrees of freedom Multiple R-squared: 0.9061, Adjusted R-squared: 0.9042 F-statistic: 469.7 on 3 and 146 DF, p-value: < 2.2e-16  ## 1 Answer The$R^2$values of model 2 and 3 are essentially identical. Adding a variable always increases$R^2\$ (technically, it could leave it exactly unchanged, but that is nearly impossible). In model 3, Facebook is adding almost nothing to the model

It looks like Adwords and Facebook are colinear, as well.

Given this, the simple solution is to use model 2. A more complicated solution is to try to combine Adwords and Facebook into one variable or to try ridge regression.

• thanks for your suggestion.. model 2 rmse is (slightly) more than model 3. Is it something like in conjunction with the good R2 values the rmse should also be lesser ? In the meanwhile, let me try the other options suggested by you. – raja ram Mar 14 '16 at 15:20
• Yes. Adding even a nonsense variable will decrease RMSE slightly. – Peter Flom - Reinstate Monica Mar 15 '16 at 10:45