The near-zero p-value associated with the quadratic term suggests that it leads to an improved model. We use the anova() function to further quantify the extent to which the quadratic fit is superior to the linear fit.

> lm.fit <- lm(medv ~ lstat)
> anova(lm.fit, lm.fit2)
Analysis of Variance Table

Model 1: medv ~ lstat
Model 2: medv ~ lstat + I(lstat^2)
  Res.Df   RSS Df Sum of Sq     F    Pr(>F)    
1    504 19472                                 
2    503 15347  1    4125.1 135.2 < 2.2e-16 ***
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Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Model 1 represents the linear submodel containing only one predictor, lstat, while Model 2 corresponds to the larger quadratic model that has two predictors, lstat and lstat². The anova() function performs a hypothesis test comparing the two models. The null hypothesis is that the two models fit the data equally well, and the alternative hypothesis is that the full model is superior. Here the F-statistic is 135 and the associated p-value is virtually zero. This provides very clear evidence that the model containing the predictors lstat and lstat² is far superior to the model that only contains the predictor lstat. This is not surprising, since earlier we saw evidence for non-linearity in the relationship between medv and lstat.

Use the anova() function to compare the second order and third order model (you created in the previous exercise) and store the output in Boston.anova:

MC1: The third order term of lstat did significantly improve the model

• 1: true
• 2: false

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Assume that:

• The MASS library has been loaded
• The Boston dataset has been loaded and attached
• The second and third order model are already defined and are stored in lm.fit2 and lm.fit3 respectively