date()
## [1] "Wed Oct 17 18:27:54 2012"
Due Date: October 18, 2012
Total Points: 38
1 The use of a cell phone while driving is hypothosized to increase the chance of an accident. The data set reaction.time (UsingR) is simulated data on the time it takes to react to an external event while driving. Subjects with control=C are not using a cell phone, and those with control=T are. The time to respond to some external event is recorded in seconds.
a) Perform a t-test on the difference in mean reaction time for groups T and C. What do you conclude. (2)
require(ggplot2)
## Loading required package: ggplot2
require(UsingR)
## Loading required package: UsingR
## Loading required package: MASS
## Attaching package: 'UsingR'
## The following object(s) are masked from 'package:ggplot2':
##
## movies
head(reaction.time)
## age gender control time
## 1 16-24 F T 1.360
## 2 16-24 M T 1.468
## 3 16-24 M T 1.512
## 4 16-24 F T 1.391
## 5 16-24 M T 1.384
## 6 16-24 M C 1.394
tail(reaction.time)
## age gender control time
## 55 25+ M T 1.445
## 56 25+ M C 1.443
## 57 25+ M C 1.355
## 58 25+ F T 1.615
## 59 25+ M T 1.528
## 60 25+ M T 1.467
t.test(time ~ control, data = reaction.time)
##
## Welch Two Sample t-test
##
## data: time by control
## t = -2.205, df = 29.83, p-value = 0.03529
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## -0.107793 -0.004122
## sample estimates:
## mean in group C mean in group T
## 1.390 1.446
b) Repeat the test separately for women and men. What do you conclude? (4)
t.test(time[gender == "M"] ~ control[gender == "M"], data = reaction.time)
##
## Welch Two Sample t-test
##
## data: time[gender == "M"] by control[gender == "M"]
## t = -1.989, df = 19.13, p-value = 0.06121
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## -0.138626 0.003504
## sample estimates:
## mean in group C mean in group T
## 1.372 1.439
t.test(time[gender == "F"] ~ control[gender == "F"], data = reaction.time)
##
## Welch Two Sample t-test
##
## data: time[gender == "F"] by control[gender == "F"]
## t = -0.875, df = 9.966, p-value = 0.4021
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## -0.12179 0.05313
## sample estimates:
## mean in group C mean in group T
## 1.416 1.451
c) Repeat the test separately for the two age groups. What do you conclude? (4)
t.test(time[age == "16-24"] ~ control[age == "16-24"], data = reaction.time)
##
## Welch Two Sample t-test
##
## data: time[age == "16-24"] by control[age == "16-24"]
## t = -1.8, df = 5.191, p-value = 0.1296
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## -0.14171 0.02422
## sample estimates:
## mean in group C mean in group T
## 1.336 1.394
t.test(time[age == "25+"] ~ control[age == "25+"], data = reaction.time)
##
## Welch Two Sample t-test
##
## data: time[age == "25+"] by control[age == "25+"]
## t = -2.627, df = 21.58, p-value = 0.01553
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## -0.13715 -0.01607
## sample estimates:
## mean in group C mean in group T
## 1.403 1.480
2 The data set diamond (UsingR) contains data about the price of 48 diamond rings. The variable price records the price in Singapore dollars and the variable carat records the size of the diamond and you are interested in predicting price from carat size.
a) Make a scatter plot of carat versus price. (2)
ggplot(diamond, aes(x = carat, y = price)) + geom_point(size = 3) + xlab("Carat") +
ylab("Price(Singapore Dollars)")
b) Add a linear regression line to the plot. (2)
ggplot(diamond, aes(x = carat, y = price)) + geom_point(size = 3) + geom_smooth(method = lm,
se = FALSE) + xlab("Carat") + ylab("Price(Singapore Dollars)")
c) Use the model to predict the amount a 1/3 carat diamond ring would cost. (4)
model = lm(price ~ carat, data = diamond)
predict(model, data.frame(carat = 1/3)) #A 1/3 carat diamond ring would cost about 981 Singaporean Dollars
## 1
## 980.7
3 The data set trees contains the girth (inches), height (feet) and volume of timber from 31 felled Black Cherry trees. Suppose you want to predict the volume of timber from a measure of girth.
a) Create a scatter plot of the data and label the axes. (4)
ggplot(trees, aes(x = Girth, y = Volume)) + geom_point(size = 3) + xlab("Girth(inches)") +
ylab("Volume")
b) Add a linear regression line to the plot. (2)
ggplot(trees, aes(x = Girth, y = Volume)) + geom_point(size = 3) + geom_smooth(method = lm,
se = FALSE) + xlab("Girth(inches)") + ylab("Volume")
c) Determine the sum of squared residuals? (4)
model2 = lm(Volume ~ Girth, data = trees)
deviance(model2)
## [1] 524.3
d) Repeat a, b, and c but use the square of the girth instead of girth as the explanatory variable. Which model do you prefer and why? (10)
attach(trees)
Girth2 = Girth^2
# Scatter Plot using Girth Squared
ggplot(trees, aes(x = Girth2, y = Volume)) + geom_point(size = 3) + xlab("Girth Squared(square inches)") +
ylab("Volume")
# Linear Regression using Girth Squared
ggplot(trees, aes(x = Girth2, y = Volume)) + geom_point(size = 3) + geom_smooth(method = lm,
se = FALSE) + xlab("Girth Squared(square inches)") + ylab("Volume")
# SSE Girth Squared
model3 = lm(Volume ~ Girth2, data = trees)
deviance(model3)
## [1] 329.3
detach(trees)
# After comparing the two models, I prefer the second model using Girth
# Squared. The model visually fits the data better, and fits better
# mathematically as seen when the two SSE's are compared. The first model
# gives us an SSE of 524.3, whereas the second model gives us a
# significantly lower SSE of 329.3.