Matakuliah
Tahun
: A0392 – Statistik Ekonomi
: 2006
Pertemuan 10
Analisis Varians Satu Arah
1
Outline Materi :
 Model tabel ANOVA klasifikasi satu arah
 ANOVA ulangan sama
 ANOVA ulangan tidak sama
2
Analysis of Variance
• The Completely Randomized Design:
One-Way Analysis of Variance
– ANOVA Assumptions
– F Test for Difference in c Means
– The Tukey-Kramer Procedure
3
General Experimental Setting
• Investigator Controls One or More
Independent Variables
– Called treatment variables or factors
– Each treatment factor contains two or more
groups (or levels)
• Observe Effects on Dependent Variable
– Response to groups (or levels) of independent
variable
• Experimental Design: The Plan Used to
Test Hypothesis
4
Completely Randomized
Design
• Experimental Units (Subjects) are
Assigned Randomly to Groups
– Subjects are assumed to be homogeneous
• Only One Factor or Independent Variable
– With 2 or more groups (or levels)
• Analyzed by One-Way Analysis of
Variance (ANOVA)
5
Randomized Design Example
Factor (Training Method)
Factor
Levels
(Groups)
Randomly
Assigned
Units
Dependent
Variable
(Response)



21 hrs
17 hrs
31 hrs
27 hrs
25 hrs
28 hrs
29 hrs
20 hrs
22 hrs
6
One-Way Analysis of Variance
F Test
• Evaluate the Difference Among the Mean Responses of
2 or More (c ) Populations
– E.g., Several types of tires, oven temperature
settings
• Assumptions
– Samples are randomly and independently drawn
• This condition must be met
– Populations are normally distributed
• F Test is robust to moderate departure from
normality
– Populations have equal variances
• Less sensitive to this requirement when samples
are of equal size from each population
7
Why ANOVA?
• Could Compare the Means One by One using Z
or t Tests for Difference of Means
• Each Z or t Test Contains Type I Error
• The Total Type I Error with k Pairs of Means is
1- (1 - a) k
– E.g., If there are 5 means and use a = .05
• Must perform 10 comparisons
• Type I Error is 1 – (.95) 10 = .40
• 40% of the time you will reject the null
hypothesis of equal means in favor of the
alternative when the null is true!
8
Hypotheses of One-Way
ANOVA
•
H 0 : 1  2 
 c
– All population means are equal
– No treatment effect (no variation in means
among groups)
•
H1 : Not all i are the same
– At least one population mean is different
(others may be the same!)
– There is a treatment effect
– Does not mean that all population means are
different
9
One-Way ANOVA
(No Treatment Effect)
H 0 : 1  2   c
H1 : Not all i are the same
The Null
Hypothesis is
True
1   2  3
10
One-Way ANOVA
(Treatment Effect Present)
H 0 : 1  2 
 c
H1 : Not all i are the same
1   2  3
The Null
Hypothesis is
NOT True
1  2  3
11
One-Way ANOVA
(Partition of Total Variation)
Total Variation SST
=






Variation Due to
Group SSA
Commonly referred to as:
Among Group Variation
Sum of Squares Among
Sum of Squares Between
Sum of Squares Model
Sum of Squares Explained
Sum of Squares Treatment
Variation Due to Random
Sampling
SSW
Commonly
referred
to as:
+




Within Group Variation
Sum of Squares Within
Sum of Squares Error
Sum of Squares
Unexplained
12
Total Variation
nj
c
SST   ( X ij  X )
2
j 1 i 1
X ij : the i -th observation in group j
n j : the number of observations in group j
n : the total number of observations in all groups
c : the number of groups
c
X 
nj
 X
j 1 i 1
n
ij
the overall or grand mean
13
Total Variation
(continued)

SST  X 11  X
 X
2
21
X
  X
2
nc c
X
Response, X
X
Group 1
Group 2
Group 3
14

2
Among-Group Variation
c
SSA   n j ( X j  X )
j 1
2
SSA
MSA 
c 1
X j : The sample mean of group j
X : The overall or grand mean
i  j
Variation Due to Differences Among Groups
15
Among-Group Variation
(continued)

SSA  n1 X 1  X
 n X
2
2
2
X

2

 nc X c  X
Response, X
X3
X1
Group 1
Group 2
X2
Group 3
X
16

2
Within-Group Variation
c
nj
SSW   ( X ij  X j )
2
j 1 i 1
SSW
MSW 
nc
X j : The sample mean of group j
X ij : The i -th observation in group j
Summing the variation
within each group and then
adding over all groups
j
17
Within-Group Variation
(continued)
SSW   X 11  X 1    X 21  X 1  
2
2

 X nc c  X c
Response, X
X3
X1
Group 1
Group 2
X2
Group 3
X
18

2
Within-Group Variation
(continued)
For c = 2, this is the
SSW
MSW 
pooled-variance in the
nc
t test.
2
2
2
(n1  1) S1  (n2  1) S2      (nc  1) Sc

(n1  1)  (n2  1)      (nc  1)
•If more than 2 groups,
use F Test.
•For 2 groups, use t test.
F Test more limited.
j
19
One-Way ANOVA
F Test Statistic
• Test Statistic
– F  MSA
MSW
• MSA is mean squares among
• MSW is mean squares within
• Degrees of Freedom
–
–
df1  c 1
df 2  n  c
20
One-Way ANOVA
Summary Table
Degrees
Source
of
of
Freedo
Variation
m
Among
c–1
(Factor)
Within
(Error)
Total
Sum of
Squares
SSA
n–c
SSW
n–1
SST =
SSA +
SSW
Mean
Squares
(Variance)
F
Statistic
MSA =
MSA/MS
SSA/(c – 1 )
W
MSW =
SSW/(n – c )
21
Features of One-Way ANOVA
F Statistic
• The F Statistic is the Ratio of the Among
Estimate of Variance and the Within
Estimate of Variance
– The ratio must always be positive
– df1 = c -1 will typically be small
– df2 = n - c will typically be large
• The Ratio Should Be Close to 1 if the Null
is True
22
Features of One-Way ANOVA
F Statistic
(continued)
• If the Null Hypothesis is False
– The numerator should be greater than the
denominator
– The ratio should be larger than 1
23
One-Way ANOVA F Test
Example
As production manager,
you want to see if 3 filling
machines have different
mean filling times. You
assign 15 similarly trained
& experienced workers, 5
per machine, to the
machines. At the .05
significance level, is there
a difference in mean filling
times?
Machine1 Machine2
Machine3
25.40
26.31
24.10
23.74
25.10
23.40
21.80
23.50
22.75
21.60
20.00
22.20
19.75
20.60
20.40
24
One-Way ANOVA Example:
Scatter Diagram
Machine1 Machine2
Machine3
25.40
26.31
24.10
23.74
25.10
23.40
21.80
23.50
22.75
21.60
27
20.00
22.20
19.75
20.60
20.40
X 1  24.93
X 2  22.61
X 3  20.59
X  22.71
26
25
24
23
22
21
20
•
••
•
•
X1
••
•
••
X2
•
••
••
X
X3
19
25
One-Way ANOVA Example
Computations
Machine1 Machine2 Machine3
25.40
26.31
24.10
23.74
25.10
23.40
21.80
23.50
22.75
21.60
20.00
22.20
19.75
20.60
20.40
X 1  24.93
nj  5
X 2  22.61
c3
X 3  20.59
n  15
X  22.71
2
2
2

SSA  5  24.93  22.71   22.61  22.71   20.59  22.71 


 47.164
SSW  4.2592  3.112  3.682  11.0532
MSA  SSA /(c -1)  47.16 / 2  23.5820
MSW  SSW /( n - c)  11.0532 /12  .9211
26
Summary Table
Source Degree
of
s of
Variatio Freedo
n
m
Among
(Factor)
Within
(Error)
Total
3-1=2
153=12
151=14
Mean
Squares
(Variance)
F
Statistic
47.1640
23.5820
MSA/MS
W
=25.60
11.0532
.9211
Sum of
Squares
58.2172
27
One-Way ANOVA Example
Solution
Test Statistic:
H0: 1 = 2 = 3
H1: Not All Equal
MSA
23.5820
 25.6
F

MSW
.9211
a = .05
df1= 2
df2 = 12
Decision:
Reject at a = 0.05.
Critical Value(s):
a = 0.05
0
3.89
F
Conclusion:
There is evidence that at
least one  i differs from
the rest.
28
The Tukey-Kramer Procedure
• Tells which Population Means are Significantly
Different
– E.g., 1 = 2  3
f(X)
– 2 groups whose means
may be significantly
different
X
1= 2 3
• Post Hoc (A Posteriori) Procedure
– Done after rejection of equal means in ANOVA
• Pairwise Comparisons
– Compare absolute mean differences with
critical range
29
The Tukey-Kramer Procedure:
Example
Machine1 Machine2 Machine3
25.40
23.40
20.00
26.31
21.80
22.20
24.10
23.50
19.75
23.74
22.75
20.60
25.10
21.60
20.40
2. Compute critical range:
Critical Range  QU ( c,nc )
1. Compute absolute mean
differences:
X 1  X 2  24.93  22.61  2.32
X 1  X 3  24.93  20.59  4.34
X 2  X 3  22.61  20.59  2.02
MSW
2
1 1 
    1.618
 nj nj' 
3. All of the absolute mean differences are greater than the
critical range. There is a significant difference between
each pair of means at the 5% level of significance.
30
Levene’s Test for
Homogeneity of Variance
• The Null Hypothesis
2
2
2
H
:






–
0
1
2
c
– The c population variances are all equal
• The Alternative Hypothesis
2
– H1 : Not all  j are equal ( j  1, 2, , c)
– Not all the c population variances are equal
31
Levene’s Test for
Homogeneity of Variance:
Procedure
1. For each observation in each group,
obtain the absolute value of the
difference between each observation and
the median of the group.
2. Perform a one-way analysis of variance
on these absolute differences.
32
Levene’s Test for
Homogeneity of Variances:
Example
As production manager,
you want to see if 3 filling
machines have different
variance in filling times.
You assign 15 similarly
trained & experienced
workers, 5 per machine, to
the machines. At the .05
significance level, is there
a difference in the variance
in filling times?
Machine1 Machine2
Machine3
25.40
26.31
24.10
23.74
25.10
23.40
21.80
23.50
22.75
21.60
20.00
22.20
19.75
20.60
20.40
33
Levene’s Test:
Absolute Difference from the
Median
median
Machine1
25.4
26.31
24.1
23.74
25.1
25.1
Time
Machine2 Machine3
23.4
20
21.8
22.2
23.5
19.75
22.75
20.6
21.6
20.4
22.75
20.4
abs(Time - median(Time))
Machine1 Machine2 Machine3
0.3
0.65
0.4
1.21
0.95
1.8
1
0.75
0.65
1.36
0
0.2
0
1.15
0
34
Summary Table
SUMMARY
Groups
Machine1
Machine2
Machine3
Count
5
5
5
ANOVA
Source of Variation
SS
Between Groups
0.067453
Within Groups
4.17032
Total
4.237773
Sum
Average Variance
3.87
0.774 0.35208
3.5
0.7
0.19
3.05
0.61 0.5005
df
MS
F
P-value
F crit
2 0.033727 0.097048 0.908218 3.88529
12 0.347527
14
35
Levene’s Test Example:
Solution
2
2
2





H0: 1
2
3
H1: Not All Equal
Test Statistic:
MSA 0.0337
F

 0.0970
MSW 0.3475
a = .05
df1= 2
df2 = 12
Decision:
Critical Value(s):
Do not reject at a = 0.05.
a = 0.05
0
3.89
F
Conclusion:
There is no evidence that
2
at least one  j differs
from the rest.
36