7.8.1
7.8.2
library(splines)
library(ISLR)
attach(Wage)
agelims =range(age)
age.grid = seq(from=agelims[1],to=agelims[2])
fit = lm(wage~bs(age,knots=c(25,40,60)), data=Wage)
fit2 = lm(wage~bs(age,df=2,knots=NULL), data=Wage)## Warning in bs(age, df = 2, knots = NULL): 'df' was too small; have used 3fit3 = lm(wage~bs(age,df=6,knots=NULL), data=Wage)
pred = predict(fit, newdata=list(age=age.grid), se=T)
pred2 = predict(fit2, newdata=list(age=age.grid), se=T)
pred3 = predict(fit3, newdata=list(age=age.grid), se=T)
fit.sum = summary(fit)
fit2.sum = summary(fit2)
fit3.sum = summary(fit3)
fit.sum##
## Call:
## lm(formula = wage ~ bs(age, knots = c(25, 40, 60)), data = Wage)
##
## Residuals:
## Min 1Q Median 3Q Max
## -98.832 -24.537 -5.049 15.209 203.207
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 60.494 9.460 6.394 1.86e-10 ***
## bs(age, knots = c(25, 40, 60))1 3.980 12.538 0.317 0.750899
## bs(age, knots = c(25, 40, 60))2 44.631 9.626 4.636 3.70e-06 ***
## bs(age, knots = c(25, 40, 60))3 62.839 10.755 5.843 5.69e-09 ***
## bs(age, knots = c(25, 40, 60))4 55.991 10.706 5.230 1.81e-07 ***
## bs(age, knots = c(25, 40, 60))5 50.688 14.402 3.520 0.000439 ***
## bs(age, knots = c(25, 40, 60))6 16.606 19.126 0.868 0.385338
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 39.92 on 2993 degrees of freedom
## Multiple R-squared: 0.08642, Adjusted R-squared: 0.08459
## F-statistic: 47.19 on 6 and 2993 DF, p-value: < 2.2e-16fit2.sum##
## Call:
## lm(formula = wage ~ bs(age, df = 2, knots = NULL), data = Wage)
##
## Residuals:
## Min 1Q Median 3Q Max
## -99.693 -24.562 -5.222 15.096 206.119
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 58.689 4.013 14.625 < 2e-16 ***
## bs(age, df = 2, knots = NULL)1 102.644 11.449 8.965 < 2e-16 ***
## bs(age, df = 2, knots = NULL)2 48.762 8.625 5.654 1.72e-08 ***
## bs(age, df = 2, knots = NULL)3 40.803 12.109 3.370 0.000762 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 39.93 on 2996 degrees of freedom
## Multiple R-squared: 0.0851, Adjusted R-squared: 0.08419
## F-statistic: 92.89 on 3 and 2996 DF, p-value: < 2.2e-16fit3.sum##
## Call:
## lm(formula = wage ~ bs(age, df = 6, knots = NULL), data = Wage)
##
## Residuals:
## Min 1Q Median 3Q Max
## -99.681 -24.403 -5.202 15.441 201.413
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 56.314 7.258 7.759 1.17e-14 ***
## bs(age, df = 6, knots = NULL)1 27.824 12.435 2.238 0.0253 *
## bs(age, df = 6, knots = NULL)2 54.063 7.127 7.585 4.41e-14 ***
## bs(age, df = 6, knots = NULL)3 65.828 8.323 7.909 3.62e-15 ***
## bs(age, df = 6, knots = NULL)4 55.813 8.724 6.398 1.83e-10 ***
## bs(age, df = 6, knots = NULL)5 72.131 13.745 5.248 1.65e-07 ***
## bs(age, df = 6, knots = NULL)6 14.751 16.209 0.910 0.3629
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 39.91 on 2993 degrees of freedom
## Multiple R-squared: 0.08729, Adjusted R-squared: 0.08546
## F-statistic: 47.71 on 6 and 2993 DF, p-value: < 2.2e-16anova(fit,fit2,fit3)## Analysis of Variance Table
##
## Model 1: wage ~ bs(age, knots = c(25, 40, 60))
## Model 2: wage ~ bs(age, df = 2, knots = NULL)
## Model 3: wage ~ bs(age, df = 6, knots = NULL)
## Res.Df RSS Df Sum of Sq F Pr(>F)
## 1 2993 4770777
## 2 2996 4777674 -3 -6897.3 1.4424 0.22847
## 3 2993 4766244 3 11429.9 2.3902 0.06687 .
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1