date()
## [1] "Wed Nov 21 01:00:11 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
consumption = read.table("http://myweb.fsu.edu/jelsner/PetrolConsumption.txt",
header = TRUE)
head(consumption)
## 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
tRee = tree(Petrol.Consumption ~ ., data = consumption)
plot(tRee)
text(tRee)
The first split is proportion of drivers w/ liscenses and the second split is Average Income.
tRee3 = prune.tree(tRee, best = 3)
plot(tRee3)
text(tRee3)
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)
shut = data.frame(temp, damage)
logrm = glm(damage ~ temp, data = shut, family = binomial)
summary(logrm)
##
## Call:
## glm(formula = damage ~ temp, family = binomial, data = shut)
##
## 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
Temperature is a significant predictor of damage.
pchisq(20.315, 21, lower.tail = FALSE)
## [1] 0.5014
the model is adequate based on large pval >0.5.
exp(-0.2322 * (75 - 60))
## [1] 0.03072
The odds of damage when launched at 60 degrees farenheit is 32.56 times the odds at 75 degrees.
predict(logrm, data.frame(temp = 55), type = "response")
## 1
## 0.9067
Damage probability at 55 degrees farenheit is approximately 91 %