Chapter 2
Exercise 1
21,36,42,24,25,36,35,49,32
• a=c(21,36,42,24,25,36,35,49,32)
mean(a)
[1] 33.33333
tmean(a)
[1] 32.85714
median(a)
[1] 35
Exercise 2
a=c(21,36,42,24,25,36,35,200,32)
mean(a)
[1] 50.11111
tmean(a)
[1] 32.85714
median(a)
• [1] 35
Sample mean is not resistant as its value largely
inflated with a change of a single value.
Exercise 3
The resistance of the 20% trimmed mean is 0.2,
meaning that a change in more than 20% of the
observations are required to generate a large
change in its value.
In this case n=9, so 9×0.2=1.8, rounded down to
the nearest integer, =1. This means that a large
change in the 20% trimmed mean value requires
a change of at least 2 observations.
Exercise 4
The resistance of the Median is 0.5, meaning
that a change in more than 50% of the
observations are required to generate a large
change in its value.
In this case n=9, so 9×0.5=4.5. This means that a
large change in the median requires a change of
at least 5 observations.
Exercise 5
b=c(6,3,2,7,6,5,8,9,8,11)
a=c(21,36,42,24,25,36,35,49,32)
> tmean(a, tr=.1)
[1] 33.33333
> b=c(6,3,2,7,6,5,8,9,8,11)
> mean(b)
[1] 6.5
> tmean(b)
[1] 6.666667
> median(b)
[1] 6.5
Exercise 6
c=c(250,220,281,247,230,209,240,160,370,274,210
,204,243,251,190,200,130,150,177,475,221,350,22
4,163,272,236,200,171,98)
mean(c)
[1] 229.1724
> tmean(c)
[1] 220.7895
> median(c)
[1] 221
Exercise 7
f1 = 5, f2 = 8, f3 = 20, f4 = 32, f5 = 23
n=5+8+20+32+23=88
5´1+ 8´ 2 + 20 ´ 3+ 32 ´ 4 + 23´ 5
X=
= 3.68
88
Exercise 8
f1 =12, f2 =18, f3 =15, f4 =10, f5 = 8, f6 = 5
12 ´1+18´ 2 +15´ 3+10 ´ 4 + 8´ 5+ 5´ 6
= 2.98
68
Exercise 9
d=(21,36,42,24,25,36,35,49,32)
var(d)
[1] 81
> winvar(d)
[1] 51.36111
Exercise 10
d=(21,36,42,24,25,36,35,102,32)
winvar(d)
[1] 51.36111
Winsorized variance remained the same.
Exercise 11
Yes, because we shift the extreme values closer
to the mean. This reduces the dispersion in the
data. The mean squared distances from the
mean decreases accordingly.
Exercise 12
The variance has s sample breakdown point of
1/n, thus a single observation can render it
value arbitrarily large or small.
Exercise 13
The sample breakdown point of the 20%
Winsorized variance is 0.2. In the case of n=25,
this would be 25×0.2= 5.
Thus, we need at to change at least 6
observation to render the Winsorized variance
arbitrarily large.
Exercise 14
e=c(6,3,2,7,6,5,8,9,8,11)
var(e)
[1] 7.388889
> winvar(e)
[1] 1.822222
Exercise 15
c=c(250,220,281,247,230,209,240,160,370,274,
210,204,243,251,190,200,130,150,177,475,221,
350,224,163,272,236,200,171,98)
var(c)
[1] 5584.933
> winvar(c)
[1] 1375.606
Exercise 16
e=c(6,3,2,7,6,5,8,9,8,11)
• idealf(e)
• $ql
• [1] 4.833333
• $qu
• [1] 8.083333
IQR=8.08-4.83=3.25
Exercise 17
c=c(250,220,281,247,230,209,240,160,370,274,
210,204,243,251,190,200,130,150,177,475,221,
350,224,163,272,236,200,171,98)
out(c)
$out.val
[1] 370 475 350 98
Exercise 18
1
var ( x ) = å( xi - X )
i
2
fxi
n
Exercise 19
X:
0 1
2
3
4
5
Fx/n: 0.1 0.2 0.25 0.29 0.12 0.04
X = 0 ´ 0.1+1´ 0.2 + 2 ´ 0.25+3´ 0.29 + 4 ´ 0.12 + 5´ 0.04 = 2.25
s 2 = 2.252 ´ 0.1+1.252 ´ 0.2 + 0.252 ´ 0.25+ 0.752 ´ 0.29 +1.752 ´ 0.12 + 2.752 ´ 0.04 = 0.975
s 2 =1.452 ´ 0.2 + 0.452 ´ 0.4 + 0.552 ´ 0.2 +1.552 ´ 0.15+ 2.552 ´ 0.05 =1.2475
Exercise 20
X:
0
1
2
3
4
μ=1.45
Fx/n: 0.2 0.4 0.2 0.15 0.05
s 2 =1.452 ´ 0.2 + 0.452 ´ 0.4 + 0.552 ´ 0.2 +1.552 ´ 0.15+ 2.552 ´ 0.05 =1.2475
Exercise 21
80
70
60
50
out(f)
$out.val
[1] 51
90
f=c(90,76,90,64,86,51,72,90,95,78)
Exercise 22
print(boxplot(g,plot=F)$out)
numeric(0)
print(boxplot(g,plot=F)$out)
[1] 20 20
5
20
5
10
10
15
15
20
3 outliers (none detected)
2 outliers detected
(one is masked on graph)
Exercise 23
The boxplot has a sample break down point of
0.25%. The number of outliers it detects does
not exceed 25% of the sample. For example,
when we had 3 outliers with n=10, all outliers
disappeared.
Exercise 24
•
m=c(0,0.12,.16,.19,.33,.36,.38,.46,.47,.60,.61,.61,.66,.67,.68,.69,.75,.77,.81,.81,.82,
.87,.87,.87,.91,.96,.97,.98,.98,1.02,1.06,1.08,1.08,1.11,1.12,1.12,1.13,1.2,1.2,1.32,
1.33,1.35,1.38,1.38,1.41,1.44,1.46,1.51,1.58,1.62,1.66,1.68,1.68,1.70,1.78,1.82,1.8
9,1.93,1.94,2.05,2.09,2.16,2.25,2.76,3.05)
•
•
print(boxplot(m,plot=F)$out)
[1] 3.05
Exercise 25
The upper and lower quartiles of figure 2.2 are
125 and 50, respectively, so an outlier will be
declared when
1. x>125+1.5(125-50)
2. X<50-1.5(125-50)
Exercise 26
out(z)
$out.val
[1] 1 2 2 2 3 3 3 6 8 9 11 11 11 12 18 32 32 41
Histogram of z
15
10
5
Histogram detected far fewer outliers
than the other methods
0
Frequency
20
25
30
outbox(z)
$out.val
[1] 18 32 32 41
0
10
20
30
z
40
Exercise 1
21,36,42,24,25,36,35,49,32
• a=c(21,36,42,24,25,36,35,49,32)
mean(a)
[1] 33.33333
tmean(a)
[1] 32.85714
median(a)
[1] 35