Homogeneity of Variance
Pooling the variances doesn’t make sense when we cannot assume all of the sample
Variances are estimating the same value.
For two groups:
Levene (1960): replace all of the individual scores with either
then run a t-test
t
F - test
( yij  y j )
2
E ( Sl2arg e )  E ( S small
)
Alternate Hypothesis:
( yij  y j ) 2
y1  y2
2 MSerror
n
Given: 1. Random and independent samples
2. Both samples approach normal distributions
Then: F is distributed with (n-large-1) and (n-small-1) df.
Null Hypothesis:
or
2
 larg e   small
F
Sl2arg e
S
2
small
K independent groups:
Hartley: If the two maximally different variances are NOT significantly different,
Then it is reasonable to assume that all k variances are estimating the population variance.
The average differences between pairs will be less than the difference between the smallest
And the largest variance.
 Sl2arg e 
 S A2 
E 2   E 2 
 SB 
 S small 
Thus:
F
Fmax
Sl2arg e
S
2
small
A and B are randomly selected pairs.
will NOT be distributed as a normal F.
(k groups, n-1) df
Then, use
Fmax
Table to test
Null Hypothesis:
 12   22  i2 ... 
Alternate Hypothesis:
 j2   2
Data Transformation: When Homogeneity of Variance is violated
Looking at the correlation between the variances (or standard deviations)
And the means or the squared means.
 ( x  x )( y  y )
r
n 1
Sx Sy
b) Use square root transformation
c) Use logarithmic transformation
d) Use reciprocal transformation