Page 288 - Applied Statistics with R

288 CHAPTER 13. MODEL DIAGNOSTICS

##
## Call:
## lm(formula = y ~ ., data = lev_ex_1)
##
## Coefficients:
## (Intercept) x1 x2
## 8.875 -1.375 4.625

Notice the large changes in the coefficients. Also notice that each of the coef-
ficients has changed in some way. Note that the leverages of the points would
not have changed, as we have not modified any of the values.
Now let’s see what happens to these coefficients when we modify the y value of
the point with the lowest leverage.

which.min(hatvalues(lev_fit))

## 4
## 4

lev_ex[which.min(hatvalues(lev_fit)),]

## x1 x2 y
## 4 7 3 14

We see that the original y value is 14. We’ll again create a copy of the data,
and modify this point to have a y value of 30.

lev_ex_2 = lev_ex
lev_ex_2$y[4] = 30
lm(y ~ ., data = lev_ex_2)

##
## Call:
## lm(formula = y ~ ., data = lev_ex_2)
##
## Coefficients:
## (Intercept) x1 x2
## 5.7 -0.7 4.4

This time despite a large change in the y value, there is only small change in
the coefficients. Also, only the intercept has changed!