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Homework 1
Partial Solution
Created integrating L A TEX and R using knitr
February 5, 2013
[1] "Tuesday, February 05, 2013 - 4:09:55 PM."
1. Compute the mean ¯x and the median m of the six numbers 3, 5, 8, 15, 20, 21. Apply the logarithm to the
data and then compute the mean ˜x and median ˜m of the transformed data. Is ln(¯x) = ˜x Is ln(m) = ˜m
The composition of ln and mean is not commutable.
numbers

prob2

0 200 600 1000
4-8am 4-8pm 8-Mid 8-Noon Noon-4pm
require(ggplot2)
p

Tue 535 93
Wed 488 76
P2

(c) Create side-by-side boxplots of the lengths of the flights, grouped by whether or not the flight was
delayed at least 30 min.
boxplot(FlightLength ~ Delayed30, data = FD, col = c("green", "red"))
100 150 200 250 300
No
Yes
p

FD$D30

(a) Create a table and a bar chart of the responses to the question about the death penalty.
TA

1500
count
1000
500
DeathPenalty
Favor
Oppose
0
Favor Oppose NA
DeathPenalty
(b) Use the table command and the summary command in R on the gun ownership variable.
additional information does the summary command give that the table command does not
What
table(GSS$OwnGun)
No Refused Yes
605 9 310
summary((GSS$OwnGun))
No Refused Yes NA's
605 9 310 1841
The summary command shows the number (1841) of missing values.
(c) Create a contingency table comparing responses to the death penalty to the question about gun
ownership.
T2

No 0.6533 0.3467
Refused 0.7778 0.2222
Yes 0.8046 0.1954
plot(GSS$DeathPenalty, GSS$OwnGun, col = c("red", "green", "blue"))
y
No Refused Yes
Favor
Oppose
0.0 0.2 0.4 0.6 0.8 1.0
x
with(data = GSS, plot(OwnGun, DeathPenalty, col = c("red", "green", "blue")))
y
Favor Oppose
No
Refused
0.0 0.2 0.4 0.6 0.8 1.0
x
A nice alternative for creating mosaic plots is to use the mosaic() function from the vcd package.
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equire(vcd)
mosaic(T2, shade = TRUE)
DeathPenalty
Favor
Oppose
Pearson
residuals:
2.34
2.00
OwnGun
Yes Refused No
0.00
-2.00
-3.16
p-value =
1.6e-05
mosaic(t(T2), shade = TRUE)
10

OwnGun
No Refused Yes
Pearson
residuals:
2.34
2.00
DeathPenalty
Oppose Favor
0.00
-2.00
-3.16
p-value =
1.6e-05
The proportion of gun owners who favor the death penalty is 0.8046. This is different from 0.6533,
the proportion of respondents who do not own a gun in favor of the death penalty.
6. Import the Black Spruce Case Study in Section 1.9 into R.
site

(b) Create a histogram and normal quantile plot for the height changes of the seedlings. Is the distribution
approximately normal
# Using base graphs
hist(BS$Ht.change, main = "Histogram", xlab = "", freq = FALSE, col = "pink")
lines(density(BS$Ht.change), col = "red", lwd = 2)
curve(dnorm(x, mean(BS$Ht.change), sd(BS$Ht.change)), 0, 60, add = TRUE, col = "blue",
lwd = 2)
Histogram
Density
0.00 0.02 0.04
10 20 30 40 50
qqnorm(BS$Ht.change, col = "red")
qqline(BS$Ht.change, col = "blue")
Normal Q-Q Plot
Sample Quantiles
10 20 30 40 50
-2 -1 0 1 2
Theoretical Quantiles
12

# Using ggplot2
p

oxplot(Di.change ~ Fertilizer, data = BS, col = c("brown", "red"))
ggplot(data = BS, aes(x = Fertilizer, y = Di.change)) + geom_boxplot()
7.5
2 4 6 8
Di.change
5.0
2.5
F
NF
F
Fertilizer
NF
(d) Use the tapply command to find the numeric summaries of the diameter changes for the two levels
of fertilization.
with(data = BS, tapply(Di.change, list(Fertilizer), summary))
$F
Min. 1st Qu. Median Mean 3rd Qu. Max.
2.91 4.32 4.76 5.27 6.52 8.92
$NF
Min. 1st Qu. Median Mean 3rd Qu. Max.
1.02 1.92 2.71 2.72 3.16 5.71
with(data = BS, tapply(Di.change, list(Fertilizer), sd))
F NF
1.383 1.101
(e) Create a scatter plot of the height changes against the diameter changes and describe the relationship.
plot(Ht.change ~ Di.change, data = BS, cex = 0.5, pch = 19, col = "blue")
p

60
50
Ht.change
10 20 30 40 50
Ht.change
40
30
20
2 4 6 8
10
Di.change
2.5 5.0 7.5
Di.change
14. In this exercise, we investigate normal quantile plots using R.
(a) Draw a random sample of size n = 15 from N(0, 1) and plot both the normal quantile plot and the
histogram. Do the points on the quantile plot appear to fall on a straight line Is the histogram
symmetric, unimodal, and mound shaped Do this several times.
n = 15
Normal Q-Q Plot
Histogram of rs
Sample Quantiles
-1.0 0.0 1.0
Frequency
0 1 2 3 4 5 6
-1 0 1
Theoretical Quantiles
-1.5 -0.5 0.5 1.5
rs
n = 15
15

Normal Q-Q Plot
Histogram of rs
Sample Quantiles
-2 -1 0 1
Frequency
0 1 2 3 4 5
-1 0 1
Theoretical Quantiles
-2 -1 0 1 2
rs
n = 15
Normal Q-Q Plot
Histogram of rs
Sample Quantiles
-1.0 0.0 1.0 2.0
Frequency
0 1 2 3 4 5
-1 0 1
Theoretical Quantiles
-1 0 1 2
rs
n = 15
16

Normal Q-Q Plot
Histogram of rs
Sample Quantiles
-1.5 -0.5 0.5 1.5
Frequency
0 1 2 3 4 5 6
-1 0 1
Theoretical Quantiles
-2 -1 0 1 2
rs
n = 15
Normal Q-Q Plot
Histogram of rs
Sample Quantiles
-2 -1 0 1 2
Frequency
0 1 2 3 4 5
-1 0 1
Theoretical Quantiles
-3 -2 -1 0 1 2 3
rs
(b) Repeat part 14a for samples of size n = 30, n = 60, and n = 100.
n = 30
17

Normal Q-Q Plot
Histogram of rs
Sample Quantiles
-2 -1 0 1
Frequency
0 2 4 6 8
-2 -1 0 1 2
Theoretical Quantiles
-2 -1 0 1 2
rs
n = 30
Normal Q-Q Plot
Histogram of rs
Sample Quantiles
-1 0 1 2
Frequency
0 2 4 6 8 10
-2 -1 0 1 2
Theoretical Quantiles
-2 -1 0 1 2
rs
n = 60
18

Normal Q-Q Plot
Histogram of rs
Sample Quantiles
-2 -1 0 1 2
Frequency
0 5 10 15
-2 -1 0 1 2
Theoretical Quantiles
-2 -1 0 1 2
rs
n = 60
Normal Q-Q Plot
Histogram of rs
Sample Quantiles
-2 -1 0 1 2
Frequency
0 2 4 6 8 12
-2 -1 0 1 2
Theoretical Quantiles
-2 -1 0 1 2
rs
n = 100
19

Normal Q-Q Plot
Histogram of rs
Sample Quantiles
-2 -1 0 1 2
Frequency
0 5 10 15 20
-2 -1 0 1 2
Theoretical Quantiles
-2 -1 0 1 2
rs
n = 100
Normal Q-Q Plot
Histogram of rs
Sample Quantiles
-2 -1 0 1 2 3
Frequency
0 5 10 15 20
-2 -1 0 1 2
Theoretical Quantiles
-2 -1 0 1 2 3
rs
(c) What lesson do you draw about using graphs to asses whether or not a data set follows a normal
distribution
For small n, it is relatively difficult to assess normality. For moderate to large n, the data will
generally follow a normal distribution and the points will follow a straight line in a Q-Q plot.
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