ANALYSIS OF
VARIANCE
(ANOVA)
BCT 2053
CHAPTER 5
CONTENT
5.1 Introduction to ANOVA
5.2 One-Way ANOVA
5.3 Two-Way ANOVA
5.1 Introduction to
ANOVA
OBJECTIVES
After completing this chapter you should be able to:
1. Explain the purpose of ANOVA
2. Identify the assumptions that underlie the ANOVA
technique
3. Describe the ANOVA hypothesis testing procedure
What is ANALYSIS OF
VARIANCE (ANOVA)

the approach that allows us to use sample data
to see if the values of three or more unknown
population means are likely to be different

Also known as factorial experiments

this name is derived from the fact that in order
to test for statistical significance between
means, we are actually comparing (i.e.,
analyzing) variances. (so F-distribution will be
used)
Example of problems
involving ANOVA

A manager want to evaluate the performance of
three (or more) employees to see if any
performance different from others.

A marketing executive want to see if there’s a
difference in sales productivity in the 5 company
region.

A teacher wants to see if there’s a difference in
student’s performance if he use 3 or more
approach to teach.
The Procedural Steps for
an ANOVA Test
1.
2.
3.
4.
5.
6.
7.
State the Null and Alternative hypothesis
Select the level of significance, α
Determine the test distribution to use - Ftest
Compute the test statistic
Define rejection or critical region – Ftest > Fcritical
State the decision rule
Make the statistical decision - conclusion
5.2 One-Way
ANOVA
OBJECTIVE
After completing this chapter you should be able to:
1. Use the one-way ANOVA technique to determine
if there is a significance difference among three
or more means
One-Way ANOVA Design

Only one classification factor (variable) is
considered
Factor
1
Treatment 2
Response/ outcome/
dependent variable
(samples)
(The level of
the factor)
i
Replicates (1,… j)
The object to a
given treatment
The resulting input grid of
factorial experiment
where,
i = 1, 2, … a is the number of levels being tested.
j = 1, 2, … ni is the number of replicates at each level.
Assumptions
To use the one-way ANOVA test, the following assumptions
must be true

The population under study have normal distribution.

The samples are drawn randomly, and each sample is
independent of the other samples.

All the populations from which the samples values are
obtained, have the same unknown population variances,
that is for k number of populations,
 12   22 
  k2
The Null and Alternative
hypothesis
H o : 1  2 
 k
(All population means are equal)
If Ho is true we have k number of normal populations with
1  2 
H1 :  i   j
 k
for i  j
and
 12   22 
  k2
(Not all population means are equal)
Or H1: At least one mean is different from others
If H1 is true we may have k number of normal populations with
1  2  3
but
 12   22   32 if k  3
The format of a general
one-way ANOVA table
Source
of
Variation
Between
sample
(Factor
Variation)
Within
samples
(Error
variation)
Degrees
of
Freedom
(Df)
k-1
T-k
Sum of Squares (SS)
1 k 2 1 2
SS (Tr )   xi.  x..
n i 1
N
T-1
2
 Treatment
/ Factor  MS (Tr )
SS (Tr )

k 1
2
SSE  SST  SS (Tr )  Error
 MSE 
k
Total
Mean of Squares (MS)
n
SST   xij2 
i 1 j 1
F Value
Ftest
2
 Treatment
/ Factor

2
 Error
SSE
N k
1 2
x..
N
Reject Ho if Ftest  F ,k 1,T k
T=kn
Example 1

The data shows the Math’s test score for 4 groups of
student with 3 different methods of study. Test the
hypothesis that there is no difference between the Math’s
score for 4 groups of student at significance level 0.05.
Score
Individually &
Group study
80
70
85
89
Group study
60
55
58
62
Individually
65
60
62
58
Example 2

An experiment was performed to determine whether the
annealing temperature of ductile iron affects its tensile strength.
Five specimens were annealed at each of four temperatures. The
tensile strength (in ksi) was measured for each temperature. The
results are presented in the following table. Can you conclude
that there are differences among the mean strengths at α = 0.01?
Temperature
(oC)
Sample Values
750
19.72
20.88
19.63
18.68
17.89
800
16.01
20.04
18.10
20.28
20.53
850
16.66
17.38
14.49
18.21
15.58
900
16.93
14.49
16.15
15.53
13.25
Example 3

Three random samples of times (in minutes) that commuters are
stuck in traffic are shown below. At α = 0.05, is there a difference
in the mean times among the three cities?
Eastern Third
Middle Third
Western Third
48
95
29
57
52
40
24
64
68
10
64
38
Solve one-way ANOVA by
EXCEL

Excel – key in data (Example 1)
Solve one-way ANOVA by
EXCEL

Tools – Add Ins – Analysis Toolpak – Data Analysis – ANOVA
single factor – enter the data range – set a value for α - ok

Reject H0 if P-value ≤ α or F > F crit
P-value < 0.05 so Reject H0
4.2 Two-Way
ANOVA
OBJECTIVE
After completing this chapter you should be able to:
1. Use the two-way ANOVA technique to determine
if there is an effect of interaction between two
factors experiment
Two-Way ANOVA Design

Two classification factor is considered
Factor B
j=1
i=1
Factor A
2
2
b
k = 1,…n
a

Example

A researcher whishes to test the effects of two different types of
plant food and two different types of soil on the growth of certain
plant.
Some types of two way
ANOVA design
B1
B2
B1
B2
A1
A1
A2
A2
A3
B1
A1
B2
B3
B1
A1
A2
A2
A3
A3
A4
B2
B3
Assumptions

The standard two-way ANOVA tests are valid under the following
conditions:

The design must be complete
• Observations are taken on every possible treatment

The design must be balanced
• The number of replicates is the same for each treatment


The number of replicates per treatment, k must be at least 2
, xijk
Within any treatment, the observations xij1 ,
are a simple random sample from a normal population

The sample observations are independent of each other (the samples
are not matched or paired in any way)

The population variance is the same for all treatments.
Null & Alternative Hypothesis
interaction effect
H0: there is no interaction effect between factor A and factor B.
H1: there is an interaction effect between factor A and factor B.
Row effect
H0: there is no difference in means of factor A.
H1: there is a difference in means of factor A.
or H0: there is no effect from factor A.
H1: there is effect from factor A.
Column effect
H0: there is no difference in means of factor B.
H1: there is a difference in means of factor B.
or
H0: there is no effect from factor B.
H1: there is effect from factor B.
The format of a general
two-way ANOVA table
Source
A
(row effect)
(Df)
a-1
B
(Column
effect)
b- 1
Interaction
(interaction
effect)
(a-1)(b-1)
Sum of Squares (SS)
1 a 2 x...2
SSA   xi.. 
bn i 1
abn
x...2
1 b 2
SSB 
 x. j.  abn
an j 1
1 a b 2 x...2
SSAB   xij . 
n i 1 j 1
abn
 SSA  SSB
Error
ab(n-1)
SSE  SST  SSA
SSB  SSAB
x...2
SST   x 
abn
i 1 j 1 k 1
a
Total
abn-1
b
n
2
ijk
Mean of Squares
(MS)
SSA
a 1
SSB
MSB 
b 1
MSA 
SSAB
MSAB 
 a 1b 1
MSE 
SSE
ab  n  1
F Value
Reject if
Ftest 
MSA
MSE
Ftest 
Ftest 
MSB
MSE
Ftest 
Ftest 
MSAB F 
test
MSE F ,( a1)(b1),ab n1
F ,a 1,ab n1
F ,b 1,ab n 1


Procedure for Two-Way ANOVA
START
Ho: No interaction
between two factors
Test for an interaction
between the two
factors
Is there an effect due
to interaction between the
two factors?
No
Ho: No effects from the row
factor A (the row means are
equal)
Ho: No effects from the
column factor B (the column
means are equal)
Ftest 
MSAB
MSE
Yes
(Reject Ho)
Stop. Don’t consider the
effects of either factor
without considering the
effects of the other
(Accept Ho)
Test for effect from
row factor
Ftest 
MSA
MSE
Test for effect from
column factor
Ftest 
MSB
MSE
Example 1

A chemical engineer is studying the effects of various reagents and catalyst on
the yield of a certain process. Yield is expressed as a percentage of a theoretical
maximum. 2 runs of the process were made for each combination of 3 reagents
and 4 catalysts.
Reagent
Catalyst
1
2
3
A
86.8 82.4
93.4 85.2
77.9 89.6
B
71.9 72.1
74.5 87.1
87.5 82.7
C
65.5 72.4
66.7 77.1
72.7 77.8
D
63.9 70.4
73.7 81.6
79.8 75.7
a) Construct an ANOVA table.
b) Test is there an interaction effect between reagents and catalyst. Use α = 0.05.
c) Do we need to test whether there is an effect that is due to reagents or
catalyst? Why? If Yes, test is there an effect from reagents or catalyst.
Example 2

A study was done to determine the effects of two factors on the lather
ability of soap. The two factors were type of water and glycerol. The
outcome measured was the amount of foam produced in mL. The
experiment was repeated 3 times for each combination of factors. The
result are presented in the following table..
Water type
Glycerol
Foam (mL)
De-ionized
Absent
168
178
168
De-ionized
Present
160
197
200
Tap
Absent
152
142
142
Tap
Present
139
160
160
Construct an ANOVA table and test is there an interaction effect
between factors. Use α = 0.05.
Solve two-way ANOVA by
EXCEL

Excel – key in data
Solve two-way ANOVA by
EXCEL

tools – Data Analysis – ANOVA two factor with
replication – enter the data range – set a value for α - ok

Reject H0 if P-value ≤ α or F > F crit
Summary

The other name for ANOVA is
experimental design.

ANOVA help researchers to design an
experiment properly and analyzed the
data it produces in correctly way.
Thank You