date()
## [1] "Thu Oct 18 14:08:12 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
attach(reaction.time)
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
detach(reaction.time)
I conclude that reaction time is affected by use of cell 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
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
detach(reaction.time)
The women's pvalue is around 0.4, so we fail to reject the null hypothesis, that there is no difference in reaction time among women. The men's pvalue is around 0.06, suggestive of a difference in reaction amongst men, but not strong enough to reject the null hypothesis.
c) Repeat the test separately for the two age groups. What do you conclude? (4)
attach(reaction.time)
t.test(time[age == "16-24"] ~ control[age == "16-24"])
##
## 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+"])
##
## 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
detach(reaction.time)
Age 16-24 pvalue is around 0.1, suggesting that the reaction time for this group is not affected by cell phone usage, but the evidence is inconclusive. For the 25+ age group the pvalue is about 0.02, suggestive that the groups reaction time is affected by cellphone use, but not strong evidence.
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(UsingR)
require(ggplot2)
## Loading required package: ggplot2
## Attaching package: 'ggplot2'
## The following object(s) are masked from 'package:UsingR':
##
## movies
attach(diamond)
d = ggplot(diamond, aes(x = carat, y = price)) + geom_point() + ylab("Price") +
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
detach(diamond)
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)
attach(trees)
t = ggplot(trees, aes(x = Girth, y = Volume)) + geom_point() + xlab("Girth(Inches)") +
ylab("Volume")
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
detach(trees)
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)
g2 = (Girth)^2
t2 = ggplot(trees, aes(x = g2, y = Volume)) + geom_point() + xlab("Girth (inches squared)") +
ylab("Volume")
t2
t2 + geom_smooth(method = lm, se = FALSE)
sum(residuals(lm(trees$Volume ~ g2))^2)
## [1] 329.3
detach(trees)
I prefer the second model as it is a better fit to the data according to the sum of squared residuals values.