date()
## [1] "Thu Oct 18 11:10:24 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("UsingR")
## Loading required package: UsingR
## Loading required package: MASS
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
# Because the p value is less than .05, we can reject the null hypothesis
# and accept that the result is statistically significant, proving that
# reaction time is indeed directly affected by the use of cellular phones.
b) Repeat the test separately for women and men. What do you conclude? (4)
attach(reaction.time)
t.test(time[gender == "F"] ~ control[gender == "F"])
##
## 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
# The p value is approx .4 is large, providing evidence that we cannot
# reject the null hypothesis that using a cellular phone does not affect
# the reaction time of women.
t.test(time[gender == "M"] ~ control[gender == "M"])
##
## 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
# bThe p value is approx .06, allowing us reject the null hypothesis that
# using a cellular phone does not affect the reaction time of men.
c) Repeat the test separately f or 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
# The P Value is approx. 0.12 providing evidence thast we can reject the
# null hypothesis that driving while using a cellular phone does not
# affect reaction time for drivers age 25+.
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)
require(ggplot2)
## Loading required package: ggplot2
## Attaching package: 'ggplot2'
## The following object(s) are masked from 'package:UsingR':
##
## movies
d = ggplot(diamond, aes(x = carat, y = price)) + geom_point() + ylab("Singapore Dollars") +
xlab("Carat")
d
b) Add a linear regression line to the plot. (2)
d + geom_smooth(method = lm, se = FALSE)
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))
## 1
## 980.7
# 1/3 carat diamond would cost approx. 981 Singapore Dollars
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)
t = ggplot(trees, aes(x = Girth, y = Height)) + geom_point() + xlab("Girth in Inches") +
ylab("Height in Feet")
t
b) Add a linear regression line to the plot. (2)
t + geom_smooth(method = lm, se = FALSE)
c) Determine the sum of squared residuals? (4)
sum(residuals(lm(trees$Volume ~ trees$Girth))^2)
## [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)
attach(trees)
Girth1 = (Girth)^2
t2 = ggplot(trees, aes(x = Girth1, y = Volume)) + geom_point() + ylab("Volume") +
xlab("Girth")
t2
t2 + geom_smooth(method = lm, se = FALSE)
sum(residuals(lm(trees$Volume ~ Girth1))^2)
## [1] 329.3
detach(trees)