AOV Assumption Checking
and
Transformations (§8.4, 8.5)
• How do we check the Normality assumption in AOV?
• How do we check the Homogeneity of variances
assumption in AOV? (§7.4)
• What to do if these assumptions are not met?
Model Assumptions
• Homoscedasticity (common group variances).
• Normality of responses (or of residuals).
• Independence of responses (or of residuals).
(Hopefully achieved through randomization…)
• Effect additivity. (Only an issue in multi-way
AOV; later).
Checking the Equal Variance Assumption
H0 : 12   22     2t
HA: some of the variances are different from each other
Little work but little power
Hartley’s Test: A logical extension of the F test for t=2.
Requires equal replication, n, among groups. Requires normality.
Fmax
2
smax
 2
smin
Reject if Fmax > Fa,t,n-1, tabulated in Table 12.
Bartlett’s Test
More work but better power
Bartlett’s Test: Allows unequal replication, but requires normality.
T.S.
t
 t

2
2
C   (ni  1) loge s   (ni  1) loge si 
i1

 i1

If C >
c2
(t-1),a then
apply the correction term



1  t
1 
1
  t
 

CF  1 
3( t  1)  i1 (ni  1) 
(ni  1) 



i1
R.R.
Reject if C/CF > c2(t-1),a
si2
s 
i1 t
t
2
Levene’s Test
More work but powerful result.
Let
zij  yij  yi
median of
yi = sample
i-th group
t
T.S.
L
2
n
(
z

z
)
 i i  /(t  1)
i 1
ni
t
2
n
(
z

z
)
 i ij i /( nT  t )
t
nT   ni
i 1
i 1 j 1
R.R.
Reject H0 if
L  Fa,df1 ,df2
Essentially an AOV on the zij
df1 = t -1
df2 = nT - t
Test for Equal Variances
Response
Resist
Factors
Sand
ConfLvl
95.0000
Minitab
Bonferroni confidence intervals for standard
deviations
Lower
Sigma
Upper
N
Factor Levels
1.70502
3.28634
14.4467
5
15
1.89209
3.64692
16.0318
5
20
1.07585
2.07364
9.1157
5
25
1.07585
2.07364
9.1157
5
30
1.48567
2.86356
12.5882
5
35
Bartlett's Test (normal distribution)
Test Statistic: 1.890
P-Value
: 0.756
Levene's Test (any continuous distribution)
Test Statistic: 0.463
P-Value
: 0.762
Stat > ANOVA > Test for
Equal Variances
Minitab Help
Use Bartlett’s test when the data come from
normal distributions; Bartlett’s test is not
robust to departures from normality. Use
Levene’s test when the data come from
continuous, but not necessarily normal,
distributions.
The computational method for Levene’s Test
is a modification of Levene’s procedure [10]
developed by [2]. This method considers the
distances of the observations from their
sample median rather than their sample
mean. Using the sample median rather than
the sample mean makes the test more robust
for smaller samples.
Do not reject H0 since p-value >
0.05 (traditional a)
SAS Program
proc glm data=stress;
class sand;
model resistance = sand / solution;
means sand / hovtest=bartlett;
means sand / hovtest=levene(type=abs);
means sand / hovtest=levene(type=square);
means sand / hovtest=bf; /* Brown and Forsythe mod of Levene
*/
title1 'Compression resistance in concrete beams as';
title2 ' a function of percent sand in the mix';
run;
Hovtest only works when one factor in (right hand side) model.
hovtest=bartlett;
Bartlett's Test for Homogeneity of resistance Variance
Source
sand
DF
Chi-Square
Pr > ChiSq
4
1.8901
0.7560
SAS
Levene's Test for Homogeneity of resistance Variance
ANOVA of Absolute Deviations from Group Means
Source
sand
Error
hovtest=levene(type=abs);
Sum of
Mean
DF
Squares
Square
F Value
Pr > F
4
8.8320
2.2080
0.95
0.4573
20
46.6080
2.3304
Levene's Test for Homogeneity of resistance Variance
ANOVA of Squared Deviations from Group Means
Source
sand
Error
hovtest=levene(type=square);
Sum of
Mean
DF
Squares
Square
F Value
Pr > F
4
202.2
50.5504
0.85
0.5076
20
1182.8
59.1400
Brown and Forsythe's Test for Homogeneity of resistance Variance
ANOVA of Absolute Deviations from Group Medians
Source
sand
Error
hovtest=bf;
Sum of
Mean
DF
Squares
Square
F Value
Pr > F
4
7.4400
1.8600
0.46
0.7623
20
80.4000
4.0200
SPSS
RESIST
Levene
Statistic
.947
Test of Homogeneity of Variances
df1
4
df2
20
Sig.
.457
Since the p-value (0.457) is greater than our (typical)
a =0.05 Type I error risk level, we do not reject the null
hypothesis.
This is Levene’s original test in which the zij are centered
on group means and not medians.
R
Tests of Homogeneity of Variances
bartlett.test(): Bartlett’s Test.
fligner.test(): Fligner-Killeen Test (nonparametric).
Checking for Normality
Reminder: Normality of the RESIDUALS is assumed. The
original data are assumed normal also, but each group may
have a different mean if HA is true. Practice is to first fit the
model, THEN output the residuals, then test for normality of
the residuals. This APPROACH is always correct.
TOOLS
1. Histogram and/or boxplot of all residuals (eij).
2. Normal probability (Q-Q) plot.
3. Formal test for normality.
Histogram of Residuals
proc glm data=stress;
class sand;
model resistance = sand / solution;
output out=resid r=r_resis p=p_resis ;
title1 'Compression resistance in concrete
beams as';
title2 ' a function of percent sand in the
mix';
run;
proc capability data=resid;
histogram r_resis / normal;
ppplot r_resis / normal square ;
run;
Probability Plots (QQ-Plots)
A scatter plot of the percentiles of the residuals
against the percentiles of a standard normal
distribution. The basic idea is that if the residuals
came from a normal distribution, values for these
percentiles should lie on a straight line.
• Compute and sort the residuals e(1), e(2),…, e(n).
• Associate with each residual a standard normal
percentile: z(i) = F-1((i-.5)/n).
• Plot z(i) versus e(i). Compare to straight line (don’t
care so much about which line).
Percentile Residual Normal Percentile
0.02
-4.4
-2.0537
0.06
-3.8
-1.5548
0.10
-3.6
-1.2816
0.14
-3.4
-1.0803
0.18
-2.6
-0.9154
0.22
-2.6
-0.7722
0.26
-2.6
-0.6433
0.30
-1.6
-0.5244
0.34
-0.8
-0.4125
0.38
-0.6
-0.3055
0.42
0.2
-0.2019
0.46
0.2
-0.1004
0.50
0.4
0.0000
0.54
0.4
0.1004
0.58
0.4
0.2019
0.62
0.4
0.3055
0.66
1.4
0.4125
0.70
1.4
0.5244
0.74
1.4
0.6433
0.78
1.6
0.7722
0.82
2.4
0.9154
0.86
2.6
1.0803
0.90
3.6
1.2816
0.94
4.2
1.5548
0.98
5.4
2.0537
Software
EXCEL: Use AddLine option.
Percentile pi = (i-0.5)/n
Normal percentile =NORMSINV(pi)
MTB:
Graph -> Probability Plot
R: with residuals in “y”
qqnorm(y)
qqline(y)
Excel Probability Plot
Probability Plot - Percent Sand Data
2.500
2.000
Normal Percentiles
1.500
1.000
0.500
0.000
-6
-4
-2
-0.500 0
-1.000
-1.500
-2.000
-2.500
Data Percentiles
2
4
6
Probability Plot
Minitab
SAS (note axes changed)
These look normal!
Formal Tests of Normality
Many, many tests (a favorite pass-time of statisticians is
developing new tests for normality.)
• Kolmogorov-Smirnov test; Anderson-Darling test
(both based on the empirical CDF).
• Shapiro-Wilk’s test; Ryan-Joiner test (both are
correlation based tests applicable for n < 50).
• D’Agostino’s test (n>=50).
All quite conservative – they fail to reject the null
hypothesis of normality more often than they should.
Shapiro-Wilk’s W test
e1, e2, …, en represent data ranked from smallest to largest.
H0: The population has a normal distribution.
HA: The population does not have a normal distribution.


1
W  d  a i (en  j1  e j ) 
 j1

k
T.S.
2
Coefficients ai come from a table.
R.R. Reject H0 if W < W0.05
n
d   (e i  e ) 2
i 1
n
If n is even
2
(n  1)
k
If n is odd.
2
k
Critical values of Wa come from a table.
Shapiro-Wilk
Coefficients
Shapiro-Wilk
Coefficients
Shapiro-Wilk
W Table
D’Agostino’s Test
e1, e2, …, en represent data ranked from smallest to largest.
H0: The population has a normal distribution.
HA: The population does not have a normal distribution.
T.S.
(D  0.28209479) n
Y
0.02998598
1
2
s   n  (e j  e ) 
 j1

n
n
R.R. (two sided test)
Reject H0 if
Y  Y0.025
or Y  Y0.975
D
[ j 
j1
1
2
1
2
(n  1)]e j
n 2s
Y0.025 and Y0.975 come from a
table of percentiles of the Y
statistic.
Transformations to Achieve
Homoscedasticity
What can we do if the homoscedasticity (equal variances)
assumption is rejected?
1. Declare that the AOV model is not an adequate model
for the data. Look for alternative models. (Later.)
2. Try to “cheat” by forcing the data to be homoscedastic
through a transformation of the response variable Y.
(Variance Stabilizing Transformations.)
Square Root Transformation
Response is positive and continuous.
zi 
yi
This transformation works when we
notice the variance changes as a linear
function of the mean.
  k i
2
i
• Useful for count data (Poisson Distributed).
• For small values of Y, use Y+.5.
35.00
30.00
Sample Variance
Typical use:
Counts of items when counts
are between 0 and 10.
k>0
25.00
20.00
15.00
10.00
5.00
0.00
0
10
20
Sample Mean
30
40
Logarithmic Transformation
Response is positive and continuous.
Z  ln(Y )
Typical use:
Growth over time.
Concentrations.
Counts of times when counts
are greater than 10.
 k
2
i
2
i
k>0
200
180
160
Sample Variance
This transformation tends to work
when the variance is a linear
function of the square of the mean
• Replace Y by Y+1 if zero occurs.
• Useful if effects are multiplicative (later).
• Useful If there is considerable
heterogeneity in the data.
140
120
100
80
60
40
20
0
0
10
20
Sample Mean
30
40
ARCSINE SQUARE ROOT
Response is a proportion.
1
Z  sin
Y  arcsin Y
With proportions, the variance is a
linear function of the mean times
(1-mean) where the sample mean
is the expected proportion.
 i2  k i 1   i 
• Y is a proportion (decimal between 0 and 1).
• Zero counts should be replaced by 1/4, and
N by N-1/4 before converting to percentages
Typical use:
Proportion of seeds germinating.
Proportion responding.
Reciprocal Transformation
Response is positive and continuous.
1
Z 
Y
This transformation works when the
variance is a linear function of the
fourth power of the mean.
• Use Y+1 if zero occurs.
• Useful if the reciprocal of the original
scale has meaning.
Typical use: Survival time.
 i2  k i4
Power Family of Transformations (1)
Suppose we apply the power transformation:
Suppose the true situation is that
the variance is proportional to the
th power of the mean.
In the transformed variable
we will have:
zy
p
 i  ki
2
i  
2
If p is taken as 1-, then the variance of Z
will not depend on the mean, i.e. the variance
will be constant. This is a Variance stabilizing
transformation.

p  1
i
Power Family of Transformations (2)
With replicated data,  can
sometimes be found
empirically by fitting:
Estimate:
ln( ˆ i )  C   ln( ˆ i )
2
1 ni
2
ˆi 

(
y

y
)
 ij i
ni  1 j 1
ˆ i  yi
 can be estimated by least squares (regression – Next Unit).
pˆ  1  ˆ
If p̂ is zero use the
logarithmic
transformation.
Box/Cox Transformations (advanced)
suggested
transformation
geometric
mean of the
original data.
 y i  1
 0



1
zi   
 ln y
 0
i

1 n

  exp   lny i 
 n i 1

Exponent, , is unknown. Hence the model can be viewed as
having an additional parameter which must be estimated (choose
the value of  that minimizes the residual sum of squares).
Handling Heterogeneity
no
Regression?
yes
ANOVA
Fit Effect Model
Fit linear
model
Test for
Homoscedasticity
Plot
residuals
reject
Transform
Not OK
OK
accep
t
Traditional
Box/Cox Family
Power Family
Transformed Data
OK
Transformations to Achieve Normality
Regression?
no
ANOVA
yes
Fit linear
model
Estimate
group means
Probability plot
Formal Tests
yes
Residuals Normal?
no
Transform
Different Model
OK