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Measures of Variability

Rick Balkin, Ph.D, LPC-S, NCC, 2008

R. S. Balkin, 2008 2

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