This document discusses variance component analysis and provides examples of its applications and methodology. It begins by defining key concepts such as fixed and random factors, effects, and mixed effect models. It then explains that variance component analysis partitions total variation in a dependent variable into components associated with random effects variables. The document provides examples of estimating variance components using ANOVA and examples analyzing agricultural and interlaboratory study data. It concludes that variance component analysis helps partition variation and determine where to focus efforts to reduce variance.

1. VARIANCE COMPONENT ANALYSIS
by using ANOVA method
By
PARAVAYYA C PUJERI
PGS14AGR6424

2. Flow Of Seminar
Basic concepts
Introduction
Applications
Assumption
Steps
Examples
Case study
Conclusion
References

3. • Factor: An independent variable defining groups of cases.
• Fixed factors : are generally thought of as variables whose values
of interest are all represented in the data file.
• Random factors : are variables whose values in the data file can be
considered a random sample from a larger population of values. They
are useful for explaining excess variability in the dependent variable.
• Fixed eflects : the effects attributable to a finite set of levels of a
factor that occur in the data and which are there because we are
interested in them
• Random eflect: These are attributable to a (usually) infinite set of levels of
a factor, of which only a random sample are deemed to occur in the data.
• Mixed effect model: Models that contain both fixed and random
effects are called mixed effect models.
Basic concepts

4. VARIANCE COMPONENT ANALYSIS
Other Names: Components of variation, sources of variance,
variance analysis, intraclass correlation, random
effects models, analysis of a nested data.
Definition :Is a technique of partitioning of total variations
into different components of which some are
known and some are completely unknown.
• Variance components : are a way to assess the amount of
variation in a dependent variable that is associated with one or
more random-effects variables.

5. The Variance Components procedure, for random-effects/
mixed-effects models , estimates the contribution of each
random effect to the variance of the dependent variable.
This procedure is particularly interesting for analysis of
mixed models such as split plot, univariate repeated
measures, and random block designs. By calculating
variance components, we can determine where to focus
attention in order to reduce the variance.
Example. At an agriculture school, weight gains for pigs in six different litters
are measured after one month. The litter variable is a random factor with six
levels. (The six litters studied are a random sample from a large population of
pig litters.) The investigator finds out that the variance in weight gain is
attributable to the difference in litters much more than to the difference in pigs
within a litter.
INTRODUCTION

6. Variance components models are a way to
assess the amount of variation in a
dependent variable that is associated with
one or more random-effects variable . the
variance components procedure estimates
only variance components, not model
regression coefficients`.

7. Variance component model/ analysis can
be traced back to the works of astronomers
Airy(1861) and Chauvenel (1863). A
modern interpretation of a one way
random model is given by R A Fisher
(1918). Later Handerson(1950), estimates
variance components by “equating mean
sum square to expected mean sum square”
these methods popularly known as
Handerson methods.
Historical development

8. APPLICATIONS
• components of variance have been used widely in
agricultural genetics and animal breeding
(1) to predict the breeding values of sires or
dams and to predict real producing abilities of
cows,
(2) to indicate sources of variation which should
be considered in analyzing production
records,
• In plant breeding,
• Epidemologies, psychometric testing
• In Engineering and
• In Environmental science and etc..

9.  The observations are normally distributed (under some
conditions this assumption can be relaxed) with each
source of variance being constant for all subgroups (this
may be true only after a transformation).
 The values of errors are independent of each other and
the variables in the model.
 The errors have a normal distribution with a mean of 0.
 The data are completely balanced; this means that all
similar subgroups have the same numbers of observations
(more complex methods allow estimation of variance
components from unbalanced data).
Variance Components assumes:

10. • Analysis of variance (ANOVA),
• maximum likelihood (ML),
• minimum norm quadratic unbiased
estimator (MINQUE),
• restricted maximum likelihood(REML).
Four different methods are available for
estimating the variance components:

11. The ANOVA method is the oldest and simplest
method of estimating variance components. first
computes sums of squares and expected mean squares for
all effects following the general linear model approach.
Then a system of linear equations is established by
equating the sums of squares of the random effects to
their expected mean squares. The variables in the
equations are the variance components and the residual
variance. Any solution, if one exists, to this system of
linear equations constitutes a set of estimates for the
variance components.
Estimating the variance components:
- by using ANOVA method

12.  it is easy to calculate.
 it is easy to understand;
 it has very few basic assumptions, e.g.,When -random variable
 the resulting estimators are unbiased
 Variance components can calculated by different software like
SPSS, SAS, STRATA and MARK .
 variance component methods is very simple – to decompose
the overall variance in a phenotype into particular sources.
ADVANTAGES

13. Source D f Sum of
Squares
Mean
Squares
EMS
Mean 1 SSM
Between s-1 SSA
Within N-s SSE
Total N SST
Calculation steps
Consider, for example, the completely randomized design (or 1-way
classification) of a groups and n observations in each. The usual
model equation for y ij , the j'th observation in the i 'th group, is
For i = 1, 2, …,s. and j = 1, 2, .. ·, n. With µ representing an overall
mean - is a random variable
ANOVA

14. where
yij - value of j’th observation in I’th group
The mean sum of squares is therefore N times the means squared.
The sum of squares due to a particular effect is therefore the sum over all
observations of the estimated effect in each observation squared

15. Form the ANOVA table
Variance estimates
• Var(bwtn) =
MSA −E(MSE) −
E(MSA)
• Var(error) =E(MSE)
Source D f Sum of
Squares
Mean
Squares
EMS
Mean 1 SSM SSM
between s-1 SSA MSA nσs
2
+ σe
2
Error N-s SSE MSE σe
2
Total N SST

16. Three fabrications casting in the same facility were
randomly selected. Each casting was broken into individual
bars. Ten randomly selected bars from each casting were
tested . The interest is on identifying variations of tensile
strength caused by casting in the facility and by bar within
the casting, not about the mean differences among the tree
casting.
EXAMPLE:

17. 17
Row cast 1 cast 2 cast 3
1 88.0 85.9 94.2
2 88.0 88.6 91.5
3 94.8 90.0 92.0
4 90.0 87.1 96.5
5 93.0 85.6 95.6
6 89.0 86.0 93.8
7 86.0 91.0 92.5
8 92.9 89.6 93.2
9 89.0 93.0 96.2
10 93.0 87.5 92.5
The statistical model for identifying the two
sources of variation for this random effects in this
experiment is
ij
, i = 1,2,...t; j =1,2,..r.
where is the process mean, ' are
the random effects due to castings,
e ' are the random error due to bars within castings.
The distribution assumptio
ij i ij
i
y e
s
s
 
 
  
ij
2 2 2
2 2
ns are:
~ (0, ); e ~N(0, ),andbothareindependent.
The total variance of an observation
may be expressed by :
and are two variance components. .
i e
y e
e
N 


  
  
 
 
The ANOVA table and expected mean squares for the random effect model:
Source Df SS MS EMS
mean 1 SSM SSM
Among Castings t-1 SSA MSA=SSA/(t-1) e
2  r
2
Among Bars within Casting N-t SSW MSW=SSW/(N-t) e
2
Total N SST

18. • SSM=30×(90.86)2
˭247666.188
• Grand mean= y̅ ..=
𝑦𝑖𝑗
𝑁
=
2725.8
30
=90.86
• SST=882+882+………..+92.52 =247970.4
• SSA=10((90.35-90.86)2+(88.43-90.86)2+(93.8-90.86)2=148.086
• SSW=247970.4-247666.188-148.086=156.126

19. SOURS
E
DF SS MSS EMS
mean 1 247666.188 247666.19
Casting 2 148.086 74.043 10.00
error 27 156.125 5.78 5.78
total 30 247970.4
Source Est.
Value
%
Casting 6.826 54.17
Error 5.78 45.83
total 12.60
Variance Components
Is variance due to Casting, 
2
Is the Random Error due Bars. e
2
ANOVA TABLE

20. variance component analysis in SPSS

21.  Analyze
General Linear Model
Variance Components
► To run a Variance Components analysis, from the menus choose:

22. ► Select Amount spent as the
dependent variable.
► Select Who shopping for
and Use coupons as
fixed factors.
► Select Store ID as a
random factor.
► Click Model.

23. ► Select Interaction from the
Build Term(s) drop-down list and
select the interaction term to the
model.
► Click Continue.
► Click Options in the Variance
Components dialog box.

24. ► Select ANOVA as the
` method.
► Select Sums of squares and
Expected mean squares in the
Display group.
► Click Continue.
► Click OK in the Variance
Components dialog box.

25. This table displays variance estimates for each of the variance
components.
 we can use this table to figure out how much each component
contributes to the total variance.
In this example Var(STOREID)=665.237 and
Var(Error)=3835.388.
 Thus, the store effect explains 665.237/(665.237+3835.388) = 14.78%
of the random variation. Error accounts for the 85.22% of the random
variation.
RESULT in output of SPSS

26. Example in SPSS

27. CASE STUDIES

28. 28
(Wernimont,1985).
A study of a chromatographic method
for determining malathion
CASE STUDY-1

29. In this study ten labs participated; each lab
received a subsample of a technical grade
malathion (Tech), two wetable powders (25%
WP and 50% WP), and an emulsifiable
concentrate (58% EC), and a dust.
The statistical model is

30. 30
Row lab Rep WP25 WP50
1 1 1 26.17 50.76
2 1 2 26.22 50.67
3 1 3 25.85 50.81
4 1 4 25.80 50.72
5 2 1 26.44 50.82
6 2 2 26.57 50.90
7 2 3 25.80 51.04
8 2 4 26.06 50.96
9 3 1 26.95 52.53
10 3 2 26.91 52.54
11 3 3 26.98 52.55
12 3 4 26.91 52.47
13 5 1 26.23 50.20
14 5 2 26.00 50.47
15 5 3 26.22 50.39
16 5 4 26.18 50.43
17 6 1 25.45 51.65
18 6 2 25.62 51.67
Row lab Rep WP25 WP50
19 6 3 27.01 51.72
20 6 4 25.72 52.07
21 7 1 26.14 50.53
22 7 2 26.78 50.75
23 7 3 26.04 49.99
24 7 4 25.97 50.92
25 8 1 25.70 50.00
26 8 2 25.90 50.30
27 8 3 25.80 50.50
28 8 4 25.70 50.60
29 9 1 26.13 50.26
30 9 2 26.13 50.36
31 9 3 25.91 50.97
32 9 4 25.86 50.44
33 10 1 26.22 50.23
34 10 2 26.20 50.27
35 10 3 25.84 50.29
36 10 4 25.84 49.97
Raw Data for the Malathion Interlaboratory Study

31. 31
Analysis of Variance for WP50%_1
Source DF SS MS F P
laborato 8 19.1570 2.3946 50.958 0.000
Error 27 1.2688 0.0470
Total 35 20.4258
Variance Components
Source Var Comp. % of Total StDev
laborato 0.587 92.59 0.766
Error 0.047 7.41 0.217
Total 0.634 0.796
Expected Mean Squares
1 laborato 1.00(2) + 4.00(1)
2 Error 1.00(2)
2 2 2
+y L e  
It is clearly indicates that 92.6% of the total
variance of each observation is the between-
lab.That is the lab averages are very
different.

32. CASE STUDY -2
Estimating variance components
in Stata
Yulia Marchenko
(2006)

33. The research problem,estimates the
variability of measurements among
machines operated over several days. Four
machines (b = 4) were selected for the
study, with two measurements (r = 2)
obtained from each machine for each of the
4 days (a = 4).

34. (%) 28.82 37.23 22.39 11.54
Variance estimates
ANOVA TABLE

35. CASE STUDY -3
Methods of variance component
estimation
D. Rasch, O. Mašata(2006)

36. By random number generation , received a data set with a = 100
sires with ni daughters as given in table(1). milk yields of heifers
during the full first lactation with an assumed heritability
coefficient.
table 1.Numbers of daughters of 100 sires

38. Estimates of the variance components from the data set by different
methods

39. conclusion
From all this study we can conclude that, variance
component analyses helps to partitioning total
variation into different components. All the results
are useful, naturally. But to be applicable to real-
life situations they demand a numerical value .it
also help to know the percentage contributions of
random factors to variation of the dependent
variable.

40. Dorothy L. Robinson (1987) Estimation and Use of Variance
Components Journal of the Royal Statistical Society. Series D (The
Statistician), Vol. 36, No. 1
pp. 3-14
D. Rasch, O. Mašata (2006) .,Methods of variance component
estimation, Biometric Unit, Research Institute of Animal
Production`Czech J. Anim. Sci., 51, 2006 (6): 227–235
REFERENCES
Henderson, C.R., 1953. Estimation of variance and covariance components
Biometrics9:226-252.
http://www.jstor.org/stable/2988267 .

41. REFERENCES
Shayle R. Searle(April1994) An Overview of Variance Component
Estimation Biometrics Unit, Cornell University, Ithaca, N.Y., U.S.A.,
14853
Searle, S.R., Casella, G., and McCulloch, C.E. (1992) Variance
components. John
Wiley and Sons, NY.
Yulia Marchenko(2006)., Estimating variance components in Stata The Stata
Journal (2006)6, Number 1, pp. 1–21
P.J. SOLOMON (2005) Variance Components Volume 8, pp. 5685–5697
Encyclopedia Of Biostatistics Second Edition (ISBN 0-470-84907-X)