APPLIED STATISTICS (D1074)
THE ANALYSIS OF VARIANCE (ANOVA) & MANOVA
Week 7-10
Learning Outcomes
LO 1 : Apply statistical method to the real problem
LO 2 : Use proper statistical method to the real
problem
LO 3 : Use statistical software to conduct analysis
LO 4 : Interpret the results of output software and
statistics calculation
LO 5 : Explain the suitable decision from statistical
method solution
(1) One-Factor Analysis of Variance
(1) One-Factor Analysis of Variance
1.1 Analysis of Variance (ANOVA)
Analysis of variance is mainly used for tests
of hypotheses that three or more population
means are all equal against that at least one
mean is different.
is a technique for assessing how one or several
nominal independent variables (called
factors) affect a continuous dependent
variable.
Bina Nusantara University
4
(1) One-Factor Analysis of Variance
ANOVA in which only one nominal independent
variable is involved is called 1-way ANOVA.
ANOVA in which only two nominal independent
variables are involved is called 2-way ANOVA.
ANOVA is an extension to the independent-twosample t test.
Bina Nusantara University
5
(1) One-Factor Analysis of Variance
1.2 Experimental Design and ANOVA
 Statistical studies can be classified as either experimental or
observational.
 In an experimental statistical study, an experiment is
conducted to generate the data.
 An experiment begins with identifying a variable of interest.
Then one or more other variables, thought to be related, are
identified and controlled, and data are collected about how
those variables influence the variable of interest.
Bina Nusantara University
6
(1) One-Factor Analysis of Variance
Example 1
Suppose in an industrial experiment that an engineer is interested in
how the mean absorption of moisture in concrete varies among 5
different concrete aggregates.
Aggregates 1

Aggregates 2
Aggregates 3

Aggregates 4
Aggregates 5
Bina Nusantara University
may be interested
in making
individual
comparisons
among these 5
population means
How the different
aggregates effect
on absorption of
moisture
7
(1) One-Factor Analysis of Variance
Example 1
The samples are exposed to moisture for 48 hours. It is decided that 6
samples are to be tested for each aggregate, requiring a total of 30
samples to be tested.
Bina Nusantara University
8
(1) One-Factor Analysis of Variance
Example 1
Hypothesis :
Bina Nusantara University
9
(1) One-Factor Analysis of Variance
1.3 Assumption
1. For each population, the response variable is
normally distributed.
2. The variance of the response variable, denoted σ2, is
the same for all of the populations.
3. The observations must be independent.
Bina Nusantara University
10
(1) One-Factor Analysis of Variance
1.4 F-Distribution
Bina Nusantara University
11
(1) One-Factor Analysis of Variance
1.5 Hypothesis Test
A nominal independent variable with k treatment is
called a factor with k levels.
Bina Nusantara University
12
(1) One-Factor Analysis of Variance
Bina Nusantara University
13
(1) One-Factor Analysis of Variance
1.6 ANOVA Table
ANOVA table is use to calculate test statistics F
Bina Nusantara University
14
(1) One-Factor Analysis of Variance
SSTR
Bina Nusantara University
15
(1) One-Factor Analysis of Variance
Example 2
Continued from example 1
Demonstrate that the aggregates do not have the same mean absorption.
α = 5%
Bina Nusantara University
16
(1) One-Factor Analysis of Variance
Example 2
Hypothesis :
H0: μ1 = μ2 = · · · = μ5 (the aggregates have the same mean absorption)
H1: At least two of the means are not equal (the aggregates do not have
the same mean absorption)
Statistic test :
SST = 209377
SSA= 85356
SSE = 209377 − 85356 = 124020
F= 4.30
Critical region:
F > 2.76 with 4 and 25 degrees of freedom
Bina Nusantara University
17
(1) One-Factor Analysis of Variance
Example 2
ANOVA table is use to calculate test statistics F
85356
4
21339
124020
25
4961
209377
29
Bina Nusantara University
4,3
18
(1) One-Factor Analysis of Variance
1.7 Multiple Comparison Method
ANOVA
The population means are not all equal
Multiple comparison procedures
Purpose : determine where the differences
among means occur
Bina Nusantara University
19
(1) One-Factor Analysis of Variance
Bina Nusantara University
20
(2) Randomized Block Design
(2) Randomized Block Design
2.1 Definition
A completely randomized design is useful when the
experimental units are homogeneous. If the experimental
units are heterogeneous, blocking is often used to form
homogeneous groups.
A completely randomized block design is an extension of
paired samples to accommodate the comparison a set of k
population means or factor levels
Bina Nusantara University
22
(2) Randomized Block Design
A typical layout for the randomized complete block design using 3
measurements in 4 blocks is as follows :
Bina Nusantara University
23
(2) Randomized Block Design
k × b Array for the RCB Design
Bina Nusantara University
24
(2) Randomized Block Design
2.2 Hypothesis Test
TEST FOR THE EQUALITY OF k TREATMENT
Hypothesis
H0 : µ1 = µ2 = µ 3 = …. = µk
H1: At least two of the means are not equal
Test Statistics
MSTR
F 
MSE
Rejection Rule
Reject H0 if F > F(α,k-1,(k-1)(b-1))
Bina Nusantara University
25
(2) Randomized Block Design
TEST FOR THE EQUALITY OF b BLOCKS
Hypothesis
H0 : β1 = β 2 = β 3 = …. = βb
H1: At least two of the means are not equal
Test Statistics
MSBL
F 
MSE
Rejection Rule
Reject H0 if F > F(α,b-1,(k-1)(b-1))
Bina Nusantara University
26
(2) Randomized Block Design
2.3 ANOVA Table
ANOVA table is use to calculate test statistics F
Bina Nusantara University
27
(2) Randomized Block Design
SSTR
SSBL
Bina Nusantara University
28
(2) Randomized Block Design
Example 3
Four different machines, M1, M2, M3, and M4, are being considered for
the assembling of a particular product. It was decided that six different
operators would be used in a randomized block experiment to
compare the machines. The machines were assigned in a random order
to each operator. The operation of the machines requires physical
dexterity, and it was anticipated that there would be a difference
among the operators in the speed with which they operated the
machines
Bina Nusantara University
29
(2) Randomized Block Design
Example 3
Bina Nusantara University
30
(2) Randomized Block Design
Example 3
ANOVA Table
Bina Nusantara University
31
(2) Multivariate ANOVA (MANOVA)
• MANOVA model for comparing g population mean vectors
parallels univariate ANOVA:
Observation Vectors
• Each component of Xlj satisfies the 1-way ANOVA model,
but now the model includes covariances among the
components.
• These covariances are assumed to be equal across
populations.
• A vector of observations can be decomposed as
Sums-of-Squares and Cross-Products (SSCP)
• First we’ll find the total corrected squares and crossproducts.
• and now sum all of this over cases and groups.
• Since addition is distributive, we’ll do this in pieces and
look just at cross-product first. . .
Sum of Squares
• Now summing the rest over j and l we get
A Closer Look at Within Groups SSCP
• where Sl is the sample covariance matrix for the lth group
(treatment, condition, etc).
• W (E) is proportional to a pooled estimated of the
common Σ
Between Groups SSCP & Test Statistic
• With respect to between groups SSCP,
• where T = W + B (i.e., the total corrected SSCP).
• is known as “Wilk’s Lambda”. It’s equivalent to likelihood
• ratio statistic.
Hypothesis Testing with
Distribution of Wilk’s Lambda
Example One Way MANOVA
An experiment was conducted for comparing 2 methods (A &
B) of teaching shorthand to 60 female seniors in a vocational
high school (a dated example). Also of interest were the
effects of distributed versus massed practice
C1: 2 hours of instruction/day for 6 weeks
C2: 3 hours of instruction/day for 4 weeks
C3: 4 hours of instruction/day for 3 weeks
So each subject received a total of 12 hours of instruction.
For now, we’ll just look the effect of distributed versus
massed practice. Note: nl = 20 for l = 1, 2, 3
Two variables (dependent measures):
X1 = speed X2 = accuracy
Example
Hypothesis Test
No difference between massed versus distributed practice on
either speed or accuracy:
The within groups (residual) sums of squares and crossproducts matrix
Hypothesis Test (continued)
The between groups SSCP matrix:
Or T = (n − 1)S where S is the covariance matrix computed
over all groups and n is the total sample. Then B = T −W
Test Statistic & Distribution
For p = 2 and g = 3, we can use the exact sample distribution:
(3) Application with Minitab
ONE WAY ANOVA
ONE WAY ANOVA
ONE WAY ANOVA
ONE WAY ANOVA
One-way ANOVA: Durability versus Carpet
Source
Carpet
Error
Total
DF
SS
3 146.4
12 163.5
15 309.9
MS
48.8
13.6
F
P
3.58 0.047
S = 3.691 R-Sq = 47.24% R-Sq(adj) = 34.05%
Tukey 90% Simultaneous Confidence Intervals
All Pairwise Comparisons among Levels of Carpet
Individual confidence level = 97.50%
Carpet = 1 subtracted from:
Carpet
Lower Center Upper ---------+---------+---------+---------+
2
-11.428 -4.748 1.933
(-------*-------)
3
-8.356 -1.675 5.006
(-------*-------)
4
-3.048 3.633 10.313
(--------*-------)
---------+---------+---------+---------+
-8.0
0.0
8.0 16.0
Carpet = 2 subtracted from:
Carpet
Lower Center Upper ---------+---------+---------+---------+
3
-3.608 3.073 9.753
(--------*-------)
4
1.699 8.380 15.061
(-------*--------)
---------+---------+---------+---------+
-8.0
0.0
8.0 16.0
Carpet = 3 subtracted from:
Carpet
Lower Center Upper ---------+---------+---------+---------+
4
-1.373 5.308 11.988
(--------*-------)
---------+---------+---------+---------+
-8.0
0.0
8.0 16.0
MANOVA
ONE WAY MANOVA
ONE WAY MANOVA
General Linear Model: Tear, Gloss, Opacity versus Extrusion, Additive
MANOVA for Extrusion
s = 1 m = 0.5 n = 6.0
Test
Criterion
Statistic
Wilks'
0.38186
Lawley-Hotelling 1.61877
Pillai's
0.61814
Roy's
1.61877
F
7.554
7.554
7.554
DF
Num Denom
3
14
3
14
3
14
P
0.003
0.003
0.003
ONE WAY MANOVA
EIGEN Analysis for Extrusion
Eigenvalue 1.619 0.00000 0.00000
Proportion 1.000 0.00000 0.00000
Cumulative 1.000 1.00000 1.00000
Eigenvector
1
2
3
Tear
0.6541 0.0460 0.4333
Gloss
-0.3385 0.1241 0.5012
Opacity
0.0359 0.1246 0.0000
ONE WAY MANOVA
MANOVA for Additive
s = 1 m = 0.5 n = 6.0
Test
Criterion
Statistic
Wilks'
0.52303
Lawley-Hotelling 0.91192
Pillai's
0.47697
Roy's
0.91192
DF
F
Num Denom
4.256 3
14
4.256 3
14
4.256 3
14
P
0.025
0.025
0.025
ONE WAY MANOVA
MANOVA for Extrusion*Additive
s = 1 m = 0.5 n = 6.0
Test
Criterion
Statistic
Wilks'
0.77711
Lawley-Hotelling 0.28683
Pillai's
0.22289
Roy's
0.28683
DF
F
Num Denom
1.339 3
14
1.339 3
14
1.339 3
14
P
0.302
0.302
0.302
Exercises
(1)
Bina Nusantara University
61
Exercises
(2)
Bina Nusantara University
62
Exercises
(3)
Bina Nusantara University
63
THANK YOU
Bina Nusantara University
64
Reference
Anthony Hayter. (2013). Probability and Statistics for
Engineers and Scientists. 04. Thomson Brooks/Cole.
Australia. ISBN : 978-1133112143.
Applied Multivariate. Statistical Analysis.(2007) Richard
A. Johnson& Dean w. Wichern. Texas A&M University.
PEARSON.
Bina Nusantara University
65