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Chapter 2

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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] tmean(a) [1] median(a) [1] 35

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Exercise 2 a=c(21,36,42,24,25,36,35,200,32) mean(a) [1] tmean(a) [1] median(a) [1] 35 Sample mean is not resistant as its value largely inflated with a change of a single value.

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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.

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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.

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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] > b=c(6,3,2,7,6,5,8,9,8,11) > mean(b) [1] 6.5 > tmean(b) [1] > median(b) [1] 6.5

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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] > tmean(c) [1] > median(c) [1] 221

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Exercise 7 n= =88

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Exercise 8

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Exercise 9 d=(21,36,42,24,25,36,35,49,32) var(d) [1] 81 > winvar(d) [1]

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Exercise 10 d=(21,36,42,24,25,36,35,102,32) winvar(d) [1] Winsorized variance remained the same.

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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.

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Exercise 12 The variance has s sample breakdown point of 1/n, thus a single observation can render it value arbitrarily large or small.

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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.

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Exercise 14 e=c(6,3,2,7,6,5,8,9,8,11) var(e) [1] > winvar(e) [1]

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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] > winvar(c) [1]

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Exercise 16 e=c(6,3,2,7,6,5,8,9,8,11) idealf(e) $ql [1] $qu [1] IQR= =3.25

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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]

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Exercise 18

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Exercise 19 X: Fx/n:

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Exercise 20 X: μ=1.45 Fx/n:

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Exercise 21 f=c(90,76,90,64,86,51,72,90,95,78) out(f) $out.val [1] 51

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Exercise 22 3 outliers (none detected) 2 outliers detected (one is masked on graph) print(boxplot(g,plot=F)$out) [1] print(boxplot(g,plot=F)$out) numeric(0)

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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.

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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

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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> (125-50) 2.X<50-1.5(125-50)

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Exercise 26 out(z) $out.val [1] outbox(z) $out.val [1] Histogram detected far fewer outliers than the other methods

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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] tmean(a) [1] median(a) [1] 35

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