Measures of Dispersion
How much scores are spread out
Let’s assume that you are thinking
about the age of two groups of
subjects.
• Group A: 34.5 years
• Group B - 34.5 years
Range:
• The distance between the lowest and
highest score.
• Only identifies the minimum and maximum
values.
Range of each group
• Group A: 22 – 50
• Group B: 22 - 50
Standard Deviation
The average amount that scores vary from the
mean.
As the standard deviation gets larger there is a
greater amount of variation between scores.
Standard deviation of the groups
(Remember the mean = 34.5 years)
• Group A: standard deviation - 5 years
– In general the subjects’ ages fall between 29.5
years and 39.5 years
• Group B: standard deviation – 10 years
– In general the subjects’ ages fall between 24.5
years and 45.5 years.
Group A
Group B
S : s y m b o l fo r s ta n d a rd
d e v ia tio n o f a s a m p le .
 : s y m b o l fo r s ta n d a rd
d e v ia tio n o f th e p o p u la tio n .
For a sample:
 x  x 
2
s
n
Average of individual
scores
X 
X

n
A v e ra g e o f d iffe re n c e s
b e tw e e n s c o re s a n d th e
m ean.


X

X

 ( X – X)
N
n
H o w e v e r:
 ( X – X) = 0
So…
B y s q u a rin g th e d iffe re n c e s ,
th is p ro b le m is a v o id e d .
S
2 ( X – XX ) X 
 N

2
2
n
T h is is k n o w n a s th e
v a ria n c e .
T h e s y m b o ls fo r v a ria n c e
a re :
S
2
fo r th e s a m p le
 2 fo r th e p o p u la tio n .
X  X 


2
S
2
n
S
S
2


X



2

N
For a sample:
 x  x 
2
s
n
X
30
28
25
23
22
18
X
X = ------N
s

N= 6
 x  x 
2
n
X
30
28
25
23
22
18
146
X

X 
n
 X = 146
 X  146
n6
146
X
6
X  24.66
For a sample:
 x  x 
2
s
n
X
30
28
25
23
22
18
X–X
5.67
3.67
.67
-1.33
-2.33
-6.33
(30 – 24 .33 )
For a sample:
 x  x 
2
s
n
X
30
28
25
23
22
18
X–X
5.67
3.67
.67
-1.33
-2.33
-6.33
(X – X)
32.15
13.47
.45
1.76
5.43
40.07
93.33
2
 X  X 
2
 93.33
X  X 


2
S
2
n
93.33

6
S
2
S
 15.55
S
2
S  3.94
C o m p u ta tio n a l F o rm u la
S =S 

 
( X )2
X
2
2
(  XX ) - N
n
------------------n
N

2
X
30
28
25
23
22
18
N=6
X = 146
(X)2 = 1462
(X)2 = 21316
C o m p u ta tio n a l F o rm u la
S =S 

 
( X )2
X
2
2
(  XX ) - N
n
------------------n
N

2
X
30
28
25
23
22
18
X2
900
784
625
529
484
324
3646 = X2
C o m p u ta tio n a l F o rm u la
S =S 

 
( X )2
X
2
2
(  XX ) - N
n
------------------n
N

2
21316
3646 
2
6
S 
6
3646  3552.67
S 
6
2
93.33
S 
6
2
S  15.55
2