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SPLUS program
R program
Creating the variable ffev1a = ffev1/100.
lung$ffev1a <- lung$ffev1/100
Fitting the regression for ffev1a regressed on fheight.
lm1 <- lm(ffev1a ~ fheight, lung, na.action=na.exclude) print(lm1) Coefficients: (Intercept) fheight -4.086702 0.1181052 Degrees of freedom: 150 total; 148 residual Residual standard error: 0.5638018
Predicting the fitted values for the regression.
lung$p <- predict(lm1)
Fig. 6.1, p. 87
Scatter diagram and regression line of fev1 versus height for fathers in lung function data.
Note: We are using the built in plots function for the regression objects. To interactively see all the plots use the following code. The which.plot option in the code for the scatter plot lets us see just one specific plot.
plot(lm1, which.plot=3)
The descriptive statistics at the top of the page, p. 88.
mean(lung$fheight) mean(lung$ffev1a) stdev(lung$fheight) stdev(lung$ffev1a) > mean(lung$fheight) [1] 69.26 > mean(lung$ffev1a) [1] 4.093267 > stdev(lung$fheight) [1] 2.779189 > stdev(lung$ffev1a) [1] 0.6507523
print( c(mean(lung$fheight), mean(lung$ffev1a), stdev(lung$fheight), stdev(lung$ffev1a) ) ) [1] 69.2600000 4.0932667 2.7791892 0.6507523
The correlation in the middle of the page, p. 99
cor.test(lung$fheight, lung$ffev1a)
Pearson's product-moment correlation
data: lung$fheight and lung$ffev1a
t = 7.1065, df = 148, p-value = 0
alternative hypothesis: true coef is not equal to 0
sample estimates:
cor
0.504396
Table 6.2, p. 101
ANOVA table for regression model in fig. 6.2.
aov1 <- aov(ffev1a~fheight, lung)
summary(aov1)
Df Sum of Sq Mean Sq F Value Pr(F)
fheight 1 16.05317 16.05317 50.50192 4.677037e-011
Residuals 148 47.04513 0.31787
Fig. 6.11, p. 116
Normal probability plot of the residuals of the regression of FEV1 on height for fathers in the lung function data.
plot(lm1, which.plot=4)
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