date()
## [1] "Mon Nov 19 13:40:04 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)
suppressMessages(require(tree))
## Warning: package 'tree' was built under R version 2.15.2
PC = read.table("http://myweb.fsu.edu/jelsner/PetrolConsumption.txt", header = TRUE)
PCtree = tree(Petrol.Consumption ~ ., data = PC)
plot(PCtree)
text(PCtree, cex = 0.9)
The proportion of the population with drivers licenses is split first and the Average income is the second split.
PCtree2 = prune.tree(PCtree, best = 3)
plot(PCtree2)
text(PCtree2, cex = 0.9)
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
With a p value of .032, temperature is a significant predictor of damage.
pchisq(20.315, 21, lower.tail = FALSE)
## [1] 0.5014
With a p value of .501, we have no reason to reject the null hypothesis that the model is adequate.
exp(-0.2322 * (60 - 75))
## [1] 32.56
The odds are 32.56 times greater for damage at 60F than at 75F.
predict(logrm, data.frame(temp = 55), type = "response")
## 1
## 0.9067
At 55f, there is a 90.67% chance of damage.