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Measures of Variability
Rick Balkin, Ph.D, LPC-S, NCC, 2008
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The Range
Simplest measure of dispersion Difference between the highest and
lowest score plus one
36 22 21 21 19
36-19+1=18
Interquartile Range
The range may be affected by extrme scores.
A more accurate representation of the range may be the interquartile range.
The interquartile range represents the middle 50% of a distribution.
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Computing the Interquartile Range
Find the 25th (Q1) and and 75th (Q3) percentiles
IQR = Q3 – Q1
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How do I find Q3 and Q1?
1. Find the median of the distribution,
2. Using the median as an endpoint, find a the median in the lower 50% of the distribution (Q1) and the upper 50% of the distribution (Q3).
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IQR Example 36, 22, 21, 21, 19 Place in order
19, 21, 21, 22, 36
Range = 18 Find the median
21 Find Q3 and Q1
22 and 21 IQR = Q3 – Q1 = 22 – 21 = 1
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Deviation score Distance from the mean 36, 22, 21, 21,19
36- ___ = 22- ___ = 21- ___ = 21- ___ = 19- ___ =
)( XXx −=
=X
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Deviation score Distance from the mean 36, 22, 21, 21,19
36-23.8 = 22-23.8 = 21-23.8 = 21-23.8 = 19-23.8 =
)( XXx −=
8.23=X
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Deviation score Distance from the mean 36, 22, 21, 21,19
36-23.8 = 12.2 22-23.8 = -1.8 21-23.8 = -2.8 21-23.8 = -2.8 19-23.8 = -4.8
0
)( XXx −=
8.23=X
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Variance and Standard Deviation
The standard deviation is the average deviation score from the mean
Problem: If we sum the deviation scores, we
get zero
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Variance We square the
deviation scores When we divide the
sum of the deviation scores by N, we have the Variance
The Variance is the average of the deviation scores squared
The formula differs slightly when calculating the variance for a sample
NXX∑ −=
2)(2σ
1)( 2
2
−−∑
=nXX
S
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Standard deviation We take the square
root of the deviation scores to calculate the standard deviation—the average of the deviation scores
Sample formula
nXX∑ −
=2)(
σ
1)( 2
−
−= ∑
nXX
s
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Variance and Standard deviation
XX − 2)( XX −X 36 12.2 22 -1.8 21 -2.8 21 -2.8 19 -4.8
0
8.23=X
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Variance and Standard deviation
XX − 2)( XX −X 36 12.2 22 -1.8 21 -2.8 21 -2.8 19 -4.8
0
8.23=X
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Variance and Standard deviation
XX − 2)( XX −X 36 12.2 148.84 22 -1.8 3.24 21 -2.8 7.84 21 -2.8 7.84 19 -4.8 23.04
0
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Variance and Standard deviation
XX − 2)( XX −X 36 12.2 148.84 22 -1.8 3.24 21 -2.8 7.84 21 -2.8 7.84
19 -4.8 23.04
0 190.80
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Variance and Standard deviation
=
=2
8.23σ
X
XX − 2)( XX −X 36 12.2 148.84 22 -1.8 3.24 21 -2.8 7.84 21 -2.8 7.84
19 -4.8 23.04
0 190.8
NXX∑ −=
2)(2σ
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Variance and Standard deviation
=
=
=
σ
σ 16.388.23
2
X
XX − 2)( XX −X 36 12.2 148.84 22 -1.8 3.24 21 -2.8 7.84 21 -2.8 7.84
19 -4.8 23.04
0 190.8
nXX∑ −
=2)(
σ
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Variance and Standard deviation
18.616.388.23
2
=
=
=
σ
σ
X
XX − 2)( XX −X 36 12.2 148.84 22 -1.8 3.24 21 -2.8 7.84 21 -2.8 7.84
19 -4.8 23.04
0 190.8
=
=
ss2
1)(
1)(
2
22
−
−Σ=
−
−Σ=
nXXs
nXXs
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Variance and Standard deviation
18.616.388.23
2
=
=
=
σ
σ
X
XX − 2)( XX −X 36 12.2 148.84 22 -1.8 3.24 21 -2.8 7.84 21 -2.8 7.84
19 -4.8 23.04
0 190.8
91.67.472
=
=
ss
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Why the separate formula for samples? Divide by n-1 to correct for the probability
that the most extreme cases will be excluded from a smaller sample
The correction of this bias makes the sample more representative of the population
For very small samples, the n-1 correction reduces the denominator to a larger extent If n=5 the we have a 20% reduction in the
denominator For large samples, the n-1 correction does
not have as large effect
Group Exercise Compute the range, sample variance
and standard deviation
98 85 80 76 67
97 85 80 76 67
95 85 80 75 64
93 84 80 73 60
90 82 78 72 57
88 82 78 70
87 82 78 70
87 80 77 70
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Group Exercise Answers are rounded to the nearest hundredth—your
answers may vary slightly (by decimals) due to rounding error
Range 42
Sample Variance 93.28
Sample Standard Deviation 9.66