date()
## [1] "Tue Nov 20 16:16:58 2012"
Due Date: November 20, 2012
Total Points: 30
1 Use the petrol consumption data set from Lecture 16 and build a regression tree to predict petrol consumption based on petrol tax, average income, amount of pavement and the proportion of the population with drivers licences. Plot the tree. Which variables are split first and second? Prune the tree leaving only three terminal nodes. Plot the final tree. (10)
require(tree)
## Loading required package: tree
## Warning: package 'tree' was built under R version 2.15.2
require(spgwr)
## Loading required package: spgwr
## Loading required package: sp
## Loading required package: maptools
## Loading required package: foreign
## Loading required package: lattice
## Checking rgeos availability: FALSE Note: when rgeos is not available,
## polygon geometry computations in maptools depend on gpclib, which has a
## restricted licence. It is disabled by default; to enable gpclib, type
## gpclibPermit()
## NOTE: This package does not constitute approval of GWR as a method of
## spatial analysis
PC = read.table("http://myweb.fsu.edu/jelsner/PetrolConsumption.txt", header = TRUE)
head(PC)
## Petrol.Tax Avg.Inc Pavement Prop.DL Petrol.Consumption
## 1 9.0 3571 1976 0.525 541
## 2 9.0 4092 1250 0.572 524
## 3 9.0 3865 1586 0.580 561
## 4 7.5 4870 2351 0.529 414
## 5 8.0 4399 431 0.544 410
## 6 10.0 5342 1333 0.571 457
PC = tree(Petrol.Consumption ~ ., data = PC)
plot(PC)
text(PC)
Prop.DL and Avg.Inc are the variables split first and second.
PC2 = prune.tree(PC, best = 3)
plot(PC2)
text(PC2)
PC2
## node), split, n, deviance, yval
## * denotes terminal node
##
## 1) root 48 6e+05 600
## 2) Prop.DL < 0.646 42 3e+05 600
## 4) Avg.Inc < 4395 27 9e+04 600 *
## 5) Avg.Inc > 4395 15 5e+04 500 *
## 3) Prop.DL > 0.646 6 9e+04 800 *
2 Use the data from Lecture 18 to model the probability of O-ring damage as a logistic regression using launch temperature as the explanatory variable. Is the temperature a significant predictor of damage? Is it adequate? What are the odds of damage when launch temperature is 60F relative to the odds of damage when the temperature is 75F? Use the model to predict the probability of damage given a launch temperature of 55F. (20)
temp = c(66, 70, 69, 68, 67, 72, 73, 70, 57, 63, 70, 78, 67, 53, 67, 75, 70,
81, 76, 79, 75, 76, 58)
damage = c(0, 1, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0,
1)
logrm = glm(damage ~ temp, family = binomial)
summary(logrm)
##
## Call:
## glm(formula = damage ~ temp, family = binomial)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -1.061 -0.761 -0.378 0.452 2.217
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 15.043 7.379 2.04 0.041 *
## temp -0.232 0.108 -2.14 0.032 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 28.267 on 22 degrees of freedom
## Residual deviance: 20.315 on 21 degrees of freedom
## AIC: 24.32
##
## Number of Fisher Scoring iterations: 5
Yes, temperature is a significant predictor of damage because the p-value is very small.
28.267-20.315 = 7.952
pchisq(7.952, 1, lower.tail = F)
## [1] 0.004803
The p-value for the model is very small, which indicates that the model is adequate in explaining damage.
predict(logrm, data.frame(temp = c(55)), type = "response")
## 1
## 0.9067
There is a probability of 90.7% that there will be damage if the launch temperature is 55F.