Upload
sourav-datta
View
216
Download
0
Embed Size (px)
Citation preview
7/24/2019 A Nova Presentation
1/163
ANalysis Of VAriance
Amit K Biswas
Indian Statistical Institute
at
Program for undergraduate students, Kozhikode
December 16, 2013
Amit (ISI, Chennai) ANOVA December 16, 2013 1 / 33
http://find/7/24/2019 A Nova Presentation
2/163
ANalysis Of VAriance
Is a tool to test equality of mean values of several populations.
Amit (ISI, Chennai) ANOVA December 16, 2013 2 / 33
http://find/7/24/2019 A Nova Presentation
3/163
ANalysis Of VAriance
Is a tool to test equality of mean values of several populations.
However, this is achieved through a test of equality of two variances.
Amit (ISI, Chennai) ANOVA December 16, 2013 2 / 33
http://find/7/24/2019 A Nova Presentation
4/163
ANalysis Of VAriance
Is a tool to test equality of mean values of several populations.
However, this is achieved through a test of equality of two variances.
So the original intended test is the following :
Amit (ISI, Chennai) ANOVA December 16, 2013 2 / 33
http://find/7/24/2019 A Nova Presentation
5/163
ANalysis Of VAriance
Is a tool to test equality of mean values of several populations.
However, this is achieved through a test of equality of two variances.
So the original intended test is the following :
H0 :1 =2 =3 = = k
H1 :i=j for i=j
Amit (ISI, Chennai) ANOVA December 16, 2013 2 / 33
http://find/7/24/2019 A Nova Presentation
6/163
ANalysis Of VAriance
Is a tool to test equality of mean values of several populations.
However, this is achieved through a test of equality of two variances.
So the original intended test is the following :
H0 :1 =2 =3 = = k
H1 :i=j for i=j
But the actual test carried out in ANOVA is :
Amit (ISI, Chennai) ANOVA December 16, 2013 2 / 33
http://find/7/24/2019 A Nova Presentation
7/163
ANalysis Of VAriance
Is a tool to test equality of mean values of several populations.
However, this is achieved through a test of equality of two variances.
So the original intended test is the following :
H0 :1 =2 =3 = = k
H1 :i=j for i=j
But the actual test carried out in ANOVA is :
H0 :1 =2
H1 :i=j for i=j
Amit (ISI, Chennai) ANOVA December 16, 2013 2 / 33
http://goforward/http://find/http://goback/7/24/2019 A Nova Presentation
8/163
ANalysis Of VAriance
Is a tool to test equality of mean values of several populations.
However, this is achieved through a test of equality of two variances.
So the original intended test is the following :
H0 :1 =2 =3 = = k
H1 :i=j for i=j
But the actual test carried out in ANOVA is :
H0 :1 =2
H1 :i=j for i=j
HOW is this ?
Amit (ISI, Chennai) ANOVA December 16, 2013 2 / 33
http://goforward/http://find/7/24/2019 A Nova Presentation
9/163
TOH : a look back...
To test equality of two mean values we have several tests :
Amit (ISI, Chennai) ANOVA December 16, 2013 3 / 33
http://find/http://goback/7/24/2019 A Nova Presentation
10/163
TOH : a look back...
To test equality of two mean values we have several tests :
The ZTest
Amit (ISI, Chennai) ANOVA December 16, 2013 3 / 33
http://find/7/24/2019 A Nova Presentation
11/163
TOH : a look back...
To test equality of two mean values we have several tests :
The ZTest When variances are known,
Amit (ISI, Chennai) ANOVA December 16, 2013 3 / 33
http://find/http://goback/7/24/2019 A Nova Presentation
12/163
TOH : a look back...
To test equality of two mean values we have several tests :
The ZTest When variances are known, the students t-test,
Amit (ISI, Chennai) ANOVA December 16, 2013 3 / 33
http://find/7/24/2019 A Nova Presentation
13/163
TOH : a look back...
To test equality of two mean values we have several tests :
The ZTest When variances are known, the students t-test,whenvariances are unknown.
Amit (ISI, Chennai) ANOVA December 16, 2013 3 / 33
http://find/7/24/2019 A Nova Presentation
14/163
TOH : a look back...
To test equality of two mean values we have several tests :
The ZTest When variances are known, the students t-test,whenvariances are unknown.
H0 :1 =2H1 :1 =2
Amit (ISI, Chennai) ANOVA December 16, 2013 3 / 33
http://find/http://goback/7/24/2019 A Nova Presentation
15/163
TOH : a look back...
To test equality of two mean values we have several tests :
The ZTest When variances are known, the students t-test,whenvariances are unknown.
H0 :1 =2H1 :1 =2
H1 could also have > or< instead of not equal.
Amit (ISI, Chennai) ANOVA December 16, 2013 3 / 33
http://find/7/24/2019 A Nova Presentation
16/163
TOH : a look back...
To test equality of two mean values we have several tests :
The ZTest When variances are known, the students t-test,whenvariances are unknown.
H0 :1 =2H1 :1 =2
H1 could also have > or< instead of not equal.
The test statistics being :X1 X22
1n1
+2
2n2
Amit (ISI, Chennai) ANOVA December 16, 2013 3 / 33
http://find/7/24/2019 A Nova Presentation
17/163
TOH : a look back...
To test equality of two mean values we have several tests :
The ZTest When variances are known, the students t-test,whenvariances are unknown.
H0 :1 =2H1 :1 =2
H1 could also have > or< instead of not equal.
The test statistics being :X1 X22
1n1
+2
2n2
which follows a normal distribution with mean zero and standard deviation1
Amit (ISI, Chennai) ANOVA December 16, 2013 3 / 33
http://find/7/24/2019 A Nova Presentation
18/163
TOH : a look back...
To test equality of two mean values we have several tests :
The ZTest When variances are known, the students t-test,whenvariances are unknown.
H0 :1 =2H1 :1 =2
H1 could also have > or< instead of not equal.
The test statistics being :X1 X22
1n1
+2
2n2
which follows a normal distribution with mean zero and standard deviation1
X1 X22
1n1
+2
2n2
N(0, 1)
Amit (ISI, Chennai) ANOVA December 16, 2013 3 / 33
http://find/http://goback/7/24/2019 A Nova Presentation
19/163
TOH : a look back...
and how do we test equality of two variances?
Amit (ISI, Chennai) ANOVA December 16, 2013 4 / 33
http://find/7/24/2019 A Nova Presentation
20/163
TOH : a look back...
and how do we test equality of two variances?
H0 :21 =
22
H1 :21 =
22
Amit (ISI, Chennai) ANOVA December 16, 2013 4 / 33
http://find/7/24/2019 A Nova Presentation
21/163
TOH : a look back...
and how do we test equality of two variances?
H0 :21 =
22
H1 :21 =
22
Here too, H1 could also have > or< instead of not equal.
Amit (ISI, Chennai) ANOVA December 16, 2013 4 / 33
http://find/7/24/2019 A Nova Presentation
22/163
TOH : a look back...
and how do we test equality of two variances?
H0 :21 =
22
H1 :21 =
22
Here too, H1 could also have > or< instead of not equal.
The test statistics...s21s22
which follows Fishers F distribution.
Amit (ISI, Chennai) ANOVA December 16, 2013 4 / 33
http://find/7/24/2019 A Nova Presentation
23/163
TOH : a look back...
and how do we test equality of two variances?
H0 :21 =
22
H1 :21 =
22
Here too, H1 could also have > or< instead of not equal.
The test statistics...s21s22
which follows Fishers F distribution.
Lets have a look at the TOH scenario in the tables next.
Amit (ISI, Chennai) ANOVA December 16, 2013 4 / 33
http://find/7/24/2019 A Nova Presentation
24/163
Tests of Means
Tests for means with variance Unknown
Hypothesis Test Statistic Criteria for rejection
H0 : = 0 |t0| > t/2,n1H1 : = 0
H0 : = 0 t0 = y0
s/
n
H1 : < 0
Amit (ISI, Chennai) ANOVA December 16, 2013 5 / 33
http://find/7/24/2019 A Nova Presentation
25/163
Tests of Means
Tests for means with variance Unknown
Hypothesis Test Statistic Criteria for rejection
H0 : = 0 |t0| > t/2,n1H1 : = 0
H0 : = 0 t0 = y0
s/
n
H1 : < 0 t0 < t,n1
H0 : = 0H1 : > 0
Amit (ISI, Chennai) ANOVA December 16, 2013 5 / 33
http://find/7/24/2019 A Nova Presentation
26/163
Tests of Means
Tests for means with variance Unknown
Hypothesis Test Statistic Criteria for rejection
H0 : = 0 |t0| > t/2,n1H1 : = 0
H0 : = 0 t0 = y0
s/
n
H1 : < 0 t0 < t,n1
H0 : = 0H1 : > 0 t0 > t,n1H0 : 1 = 2H1 : 1 = 2
Amit (ISI, Chennai) ANOVA December 16, 2013 5 / 33
http://find/7/24/2019 A Nova Presentation
27/163
Tests of Means
Tests for means with variance Unknown
Hypothesis Test Statistic Criteria for rejection
H0 : = 0 |t0| > t/2,n1H1 : = 0
H0 : = 0 t0 = y0
s/
n
H1 : < 0 t0 < t,n1
H0 : = 0H1 : > 0 t0 > t,n1H0 : 1 = 2H1 : 1 = 2 |t0| > t/2,
H0 : 1 = 2 t0 = y1y2
Sp
1n1
+ 1n2
H1 : 1 < 2 = n1+n2 2
Amit (ISI, Chennai) ANOVA December 16, 2013 5 / 33
http://find/7/24/2019 A Nova Presentation
28/163
Tests of Means
Tests for means with variance Unknown
Hypothesis Test Statistic Criteria for rejection
H0 : = 0 |t0| > t/2,n1H1 : = 0
H0 : = 0 t0 = y0
s/
n
H1 : < 0 t0 < t,n1
H0 : = 0H1 : > 0 t0 > t,n1H0 : 1 = 2H1 : 1 = 2 |t0| > t/2,
H0 : 1 = 2 t0 = y1y2
Sp
1n1
+ 1n2
H1 : 1 < 2 = n1+n2 2 t0 < t,
if1 = 2H0 : 1 = 2
H1 : 1 > 2 =
S21n1
+S22n2
2
(S21/n1)2
n1+1 +
(S22/n2 )2
n2+1
2
Amit (ISI, Chennai) ANOVA December 16, 2013 5 / 33
http://find/7/24/2019 A Nova Presentation
29/163
Tests of Means
Tests for means with variance Unknown
Hypothesis Test Statistic Criteria for rejection
H0 : = 0 |t0| > t/2,n1H1 : = 0
H0 : = 0 t0 = y0
s/
n
H1 : < 0 t0 < t,n1
H0 : = 0H1 : > 0 t0 > t,n1H0 : 1 = 2H1 : 1 = 2 |t0| > t/2,
H0 : 1 = 2 t0 = y1y2
Sp
1n1
+ 1n2
H1 : 1 < 2 = n1+n2 2 t0 < t,
if1 = 2H0 : 1 = 2
H1 : 1 > 2 =
S21n1
+S22n2
2
(S21/n1)2
n1+1 +
(S22/n2 )2
n2+1
2 t0 > t,
if1 = 2Amit (ISI, Chennai) ANOVA December 16, 2013 5 / 33
http://find/7/24/2019 A Nova Presentation
30/163
Tests on variances of Normal Distributions
Hypothesis Test Statistic Criteria for rejection
H0 : 2 = 20
20 >
2/2,n
1 or
H1 : 2 = 20
Amit (ISI, Chennai) ANOVA December 16, 2013 6 / 33
http://find/7/24/2019 A Nova Presentation
31/163
Tests on variances of Normal Distributions
Hypothesis Test Statistic Criteria for rejection
H0 : 2 = 20
20 >
2/2,n
1 or
H1 : 2 = 20 20 <
21/2,n1
H0 : 2 = 20
2 =
(n1)S2
20
H1 : 2
< 20
Amit (ISI, Chennai) ANOVA December 16, 2013 6 / 33
http://find/7/24/2019 A Nova Presentation
32/163
Tests on variances of Normal Distributions
Hypothesis Test Statistic Criteria for rejection
H0 : 2 = 20
20 >
2/2,n
1 or
H1 : 2 = 20 20 <
21/2,n1
H0 : 2 = 20
2 =
(n1)S2
20
H1 : 2
< 20
20 <
21,n1
H0 : 2 = 20
H1 : 2
> 20
Amit (ISI, Chennai) ANOVA December 16, 2013 6 / 33
http://find/7/24/2019 A Nova Presentation
33/163
Tests on variances of Normal Distributions
Hypothesis Test Statistic Criteria for rejection
H0 : 2 = 20
20 >
2/2,n
1 or
H1 : 2 = 20 20 <
21/2,n1
H0 : 2 = 20
2 =
(n1)S2
20
H1 : 2
< 20
20 <
21,n1
H0 : 2 = 20
H1 : 2
> 20
20 >
2,n1
H0 : 21 =
22
F0 = S21
S22
H1 : 21 =
22
Amit (ISI, Chennai) ANOVA December 16, 2013 6 / 33
http://find/7/24/2019 A Nova Presentation
34/163
Tests on variances of Normal Distributions
Hypothesis Test Statistic Criteria for rejection
H0 : 2 = 20
20 >
2/2,n
1 or
H1 : 2 = 20 20 < 21/2,n1H0 :
2 = 20
2 =
(n1)S2
20
H1 : 2
< 20
20 <
21,n1
H0 : 2 = 20
H1 : 2
> 20
20 >
2,n1
H0 : 21 =
22
F0 = S21
S22
H1 : 21 =
22 F0 > F/2,n11,n21
F0 < F1/2,n11,n21H0 :
21 =
22
F0 = S22
S21
H1 : 21 <
22
Amit (ISI, Chennai) ANOVA December 16, 2013 6 / 33
http://find/7/24/2019 A Nova Presentation
35/163
Tests on variances of Normal Distributions
Hypothesis Test Statistic Criteria for rejection
H0 : 2 = 20
20 >
2/2,n
1 or
H1 : 2 = 20 20 < 21/2,n1H0 :
2 = 20
2 =
(n1)S2
20
H1 : 2
< 20
20 <
21,n1
H0 : 2 = 20
H1 : 2
> 20
20 >
2,n1
H0 : 21 =
22
F0 = S21
S22
H1 : 21 =
22 F0 > F/2,n11,n21
F0 < F1/2,n11,n21H0 :
21 =
22
F0 =
S22S21
H1 : 21 <
22 F0 > F,n21,n11
H0 : 21 =
22
F0 = S21
S22H1 :
21 >
22
Amit (ISI, Chennai) ANOVA December 16, 2013 6 / 33
http://find/7/24/2019 A Nova Presentation
36/163
Tests on variances of Normal Distributions
Hypothesis Test Statistic Criteria for rejection
H0 : 2 = 20
20 >
2/2,n
1 or
H1 : 2 = 20 20 < 21/2,n1H0 :
2 = 20
2 =
(n1)S2
20
H1 : 2
< 20
20 <
21,n1
H0 : 2 = 20
H1 : 2
> 20
20 >
2,n1
H0 : 21 =
22
F0 = S21
S22
H1 : 21 =
22 F0 > F/2,n11,n21
F0 < F1/2,n11,n21H0 :
21 =
22
F0 =
S22S21
H1 : 21 <
22 F0 > F,n21,n11
H0 : 21 =
22
F0 = S21
S22H1 :
21 >
22 F0 > F,n11,n21
Amit (ISI, Chennai) ANOVA December 16, 2013 6 / 33
A k tt l
http://find/7/24/2019 A Nova Presentation
37/163
A kutty example
Let a set of observations be 5.1, 5.3, 5.0, 5.2, understandably the sd is low!
Amit (ISI, Chennai) ANOVA December 16, 2013 7 / 33
A kutty example
http://find/7/24/2019 A Nova Presentation
38/163
A kutty example
Let a set of observations be 5.1, 5.3, 5.0, 5.2, understandably the sd is low!
Similar is the sd in another data set, like 10.9, 10.6, 10.8, 10.7.
Amit (ISI, Chennai) ANOVA December 16, 2013 7 / 33
A kutty example
http://find/7/24/2019 A Nova Presentation
39/163
A kutty example
Let a set of observations be 5.1, 5.3, 5.0, 5.2, understandably the sd is low!
Similar is the sd in another data set, like 10.9, 10.6, 10.8, 10.7.
So an average sd for the two sets of data taken together is also going tobe low!
Amit (ISI, Chennai) ANOVA December 16, 2013 7 / 33
A kutty example
http://find/7/24/2019 A Nova Presentation
40/163
A kutty example
Let a set of observations be 5.1, 5.3, 5.0, 5.2, understandably the sd is low!
Similar is the sd in another data set, like 10.9, 10.6, 10.8, 10.7.
So an average sd for the two sets of data taken together is also going tobe low!
So how do we do this averaging ?
Amit (ISI, Chennai) ANOVA December 16, 2013 7 / 33
A kutty example
http://find/7/24/2019 A Nova Presentation
41/163
A kutty example
Let a set of observations be 5.1, 5.3, 5.0, 5.2, understandably the sd is low!
Similar is the sd in another data set, like 10.9, 10.6, 10.8, 10.7.
So an average sd for the two sets of data taken together is also going tobe low!
So how do we do this averaging ?
Do you know the paveraging problem ?
Amit (ISI, Chennai) ANOVA December 16, 2013 7 / 33
do and dont
http://find/7/24/2019 A Nova Presentation
42/163
do and don t
So, in one batch of production out of100 there were no defective,
Amit (ISI, Chennai) ANOVA December 16, 2013 8 / 33
do and dont
http://find/7/24/2019 A Nova Presentation
43/163
do and don t
So, in one batch of production out of100 there were no defective, and
only one was produced in the next batch and
Amit (ISI, Chennai) ANOVA December 16, 2013 8 / 33
do and dont
http://find/7/24/2019 A Nova Presentation
44/163
do and don t
So, in one batch of production out of100 there were no defective, and
only one was produced in the next batch and the piece unfortunately wasbad!
Amit (ISI, Chennai) ANOVA December 16, 2013 8 / 33
do and dont
http://find/7/24/2019 A Nova Presentation
45/163
do and don t
So, in one batch of production out of100 there were no defective, and
only one was produced in the next batch and the piece unfortunately wasbad!
So the data was as follows :
Amit (ISI, Chennai) ANOVA December 16, 2013 8 / 33
do and dont
http://find/7/24/2019 A Nova Presentation
46/163
do and don t
So, in one batch of production out of100 there were no defective, and
only one was produced in the next batch and the piece unfortunately wasbad!
So the data was as follows :
No. produced no. defective Proportion100 0 0
1 1 1average of two ps 0.5
Can you believe that??
Amit (ISI, Chennai) ANOVA December 16, 2013 8 / 33
do and dont
http://find/7/24/2019 A Nova Presentation
47/163
do a d do t
So, in one batch of production out of100 there were no defective, and
only one was produced in the next batch and the piece unfortunately wasbad!
So the data was as follows :
No. produced no. defective Proportion100 0 0
1 1 1average of two ps 0.5
Can you believe that??
However, the following is what you SHOULD do.
ratio of total def & tot produced 1101
Amit (ISI, Chennai) ANOVA December 16, 2013 8 / 33
Averaging SDs
http://find/7/24/2019 A Nova Presentation
48/163
g g
1
So we could find a sd taking all the eight observations from the twosets. And thats going to be large.
Amit (ISI, Chennai) ANOVA December 16, 2013 9 / 33
Averaging SDs
http://find/7/24/2019 A Nova Presentation
49/163
g g
1
So we could find a sd taking all the eight observations from the twosets. And thats going to be large.
2 However the average of the two sds calculated for each set separatelyis going to small.
Amit (ISI, Chennai) ANOVA December 16, 2013 9 / 33
Averaging SDs
http://find/7/24/2019 A Nova Presentation
50/163
g g
1
So we could find a sd taking all the eight observations from the twosets. And thats going to be large.
2 However the average of the two sds calculated for each set separatelyis going to small.
The two above would be very different and a comparison for equality willfail.
Amit (ISI, Chennai) ANOVA December 16, 2013 9 / 33
Averaging SDs
http://find/7/24/2019 A Nova Presentation
51/163
g g
1
So we could find a sd taking all the eight observations from the twosets. And thats going to be large.
2 However the average of the two sds calculated for each set separatelyis going to small.
The two above would be very different and a comparison for equality willfail.
Why is this?
Amit (ISI, Chennai) ANOVA December 16, 2013 9 / 33
Averaging SDs
http://find/7/24/2019 A Nova Presentation
52/163
1
So we could find a sd taking all the eight observations from the twosets. And thats going to be large.
2 However the average of the two sds calculated for each set separatelyis going to small.
The two above would be very different and a comparison for equality willfail.
Why is this?
Because the two sets are different!
Amit (ISI, Chennai) ANOVA December 16, 2013 9 / 33
Averaging SDs
http://find/7/24/2019 A Nova Presentation
53/163
1
So we could find a sd taking all the eight observations from the twosets. And thats going to be large.
2 However the average of the two sds calculated for each set separatelyis going to small.
The two above would be very different and a comparison for equality willfail.
Why is this?
Because the two sets are different!
But different in what?
Amit (ISI, Chennai) ANOVA December 16, 2013 9 / 33
http://find/7/24/2019 A Nova Presentation
54/163
different : In the average level of course.
Amit (ISI, Chennai) ANOVA December 16, 2013 10 / 33
http://find/7/24/2019 A Nova Presentation
55/163
different : In the average level of course.
This is why ANOVA works.
Amit (ISI, Chennai) ANOVA December 16, 2013 10 / 33
http://find/7/24/2019 A Nova Presentation
56/163
different : In the average level of course.
This is why ANOVA works.
ANOVA can be used to to test such hypothesis of equality of means due
to several factors simultaneously.
Amit (ISI, Chennai) ANOVA December 16, 2013 10 / 33
http://find/7/24/2019 A Nova Presentation
57/163
different : In the average level of course.
This is why ANOVA works.
ANOVA can be used to to test such hypothesis of equality of means due
to several factors simultaneously.
Typically used to analyze data in a Design of Experiment (DOE) scenario.
Amit (ISI, Chennai) ANOVA December 16, 2013 10 / 33
http://find/7/24/2019 A Nova Presentation
58/163
different : In the average level of course.
This is why ANOVA works.
ANOVA can be used to to test such hypothesis of equality of means due
to several factors simultaneously.
Typically used to analyze data in a Design of Experiment (DOE) scenario.
Though the method assumes normality of the underlying characteristics
and equality of variances of the populations are study, it is a very robustmethod, in the sense that minor deviations from the assumption do notinfluence the conclusions adversely.
Amit (ISI, Chennai) ANOVA December 16, 2013 10 / 33
Lets look at an Example now...
http://find/7/24/2019 A Nova Presentation
59/163
Lets look at an Example now...
http://find/7/24/2019 A Nova Presentation
60/163
The tensile strength of synthetic fibre used to make cloth for mens shirt isof interest to a manufacturer. It is suspected that the strength is affected
by the % of cotton in the fibre.
Amit (ISI, Chennai) ANOVA December 16, 2013 11 / 33
Lets look at an Example now...
http://find/7/24/2019 A Nova Presentation
61/163
The tensile strength of synthetic fibre used to make cloth for mens shirt isof interest to a manufacturer. It is suspected that the strength is affected
by the % of cotton in the fibre. Five levels of cotton % are of interest 15, 20, 25, 30 and 35.
Amit (ISI, Chennai) ANOVA December 16, 2013 11 / 33
Lets look at an Example now...
http://find/7/24/2019 A Nova Presentation
62/163
Lets look at an Example now...
http://find/7/24/2019 A Nova Presentation
63/163
The tensile strength of synthetic fibre used to make cloth for mens shirt isof interest to a manufacturer. It is suspected that the strength is affected
by the % of cotton in the fibre. Five levels of cotton % are of interest 15, 20, 25, 30 and 35. Five observations are taken at each level of cotton%,and the 25 observations are run in random order. The following tablegives the values of Tensile Strength of Synthetic Fibre (lb/in2).
Amit (ISI, Chennai) ANOVA December 16, 2013 11 / 33
Example 1 continued...
http://find/7/24/2019 A Nova Presentation
64/163
The data on Tensile strength of five repeat samples of yarn with differentcotton percentages are as in the table below.
% of Observationscotton 1 2 3 4 5
15 7 7 15 11 920 12 17 12 18 1825 14 18 18 19 1930 19 25 22 19 2335 7 10 11 15 11
Amit (ISI, Chennai) ANOVA December 16, 2013 12 / 33
Example 1 continued...
http://find/7/24/2019 A Nova Presentation
65/163
Some more initial computation of the data in the last table...
Amit (ISI, Chennai) ANOVA December 16, 2013 13 / 33
Example 1 continued...
http://find/7/24/2019 A Nova Presentation
66/163
Some more initial computation of the data in the last table...
% of cotton : Factor Observations(j) : yij TotalLevel(i) Value 1 2 3 4 5 yi.
1 15 7 7 15 11 9 492 20 12 17 12 18 18 773 25 14 18 18 19 19 884 30 19 25 22 19 23 1085 35 7 10 11 15 11 54
Total (G
=
i
jy
ij) 376
Amit (ISI, Chennai) ANOVA December 16, 2013 13 / 33
Example 1 continued...
http://find/7/24/2019 A Nova Presentation
67/163
In the following, we put the numbers in the table into the following
symbols.
Amit (ISI, Chennai) ANOVA December 16, 2013 14 / 33
Example 1 continued...
http://find/7/24/2019 A Nova Presentation
68/163
In the following, we put the numbers in the table into the following
symbols..
yij=jth observation(Tensile Strength) on ith level of cotton %.
Amit (ISI, Chennai) ANOVA December 16, 2013 14 / 33
Example 1 continued...
http://find/7/24/2019 A Nova Presentation
69/163
In the following, we put the numbers in the table into the following
symbols..
yij=jth observation(Tensile Strength) on ith level of cotton %.
yi. = Total of the observations of ith level.
Amit (ISI, Chennai) ANOVA December 16, 2013 14 / 33
http://find/7/24/2019 A Nova Presentation
70/163
Example 1 continued...
7/24/2019 A Nova Presentation
71/163
In the following, we put the numbers in the table into the following
symbols..
yij=jth observation(Tensile Strength) on ith level of cotton %.
yi. = Total of the observations of ith level.
y.. = Total of all observations.
n= Number of observations per level
Amit (ISI, Chennai) ANOVA December 16, 2013 14 / 33
Example 1 continued...
http://find/7/24/2019 A Nova Presentation
72/163
In the following, we put the numbers in the table into the following
symbols..
yij=jth observation(Tensile Strength) on ith level of cotton %.
yi. = Total of the observations of ith level.
y.. = Total of all observations.
n= Number of observations per level
k= The number of levels we are comparing.
Amit (ISI, Chennai) ANOVA December 16, 2013 14 / 33
Example 1 continued...
http://find/7/24/2019 A Nova Presentation
73/163
In the following, we put the numbers in the table into the following
symbols..
yij=jth observation(Tensile Strength) on ith level of cotton %.
yi. = Total of the observations of ith level.
y.. = Total of all observations.
n= Number of observations per level
k= The number of levels we are comparing.
N= the total number of observations.
Amit (ISI, Chennai) ANOVA December 16, 2013 14 / 33
The ANOVA model
http://find/7/24/2019 A Nova Presentation
74/163
In a One Way Analysis of Variance the hypothesis to be tested is that theitems in the various classes come from universes, the means of which areequal.
Amit (ISI, Chennai) ANOVA December 16, 2013 15 / 33
The ANOVA model
http://find/7/24/2019 A Nova Presentation
75/163
In a One Way Analysis of Variance the hypothesis to be tested is that theitems in the various classes come from universes, the means of which areequal.
More precisely, the mathematical model for the analysis is
yij=+i+ij
where is a constant, the is are the class differentials and ij is randomnormal deviate with mean zero and variance 2, these being the same forall classes.
Amit (ISI, Chennai) ANOVA December 16, 2013 15 / 33
The ANOVA model
http://find/7/24/2019 A Nova Presentation
76/163
In a One Way Analysis of Variance the hypothesis to be tested is that theitems in the various classes come from universes, the means of which areequal.
More precisely, the mathematical model for the analysis is
yij=+i+ij
where is a constant, the is are the class differentials and ij is randomnormal deviate with mean zero and variance 2, these being the same forall classes.
The hypothesis to be tested is that the
is are zero for all j
.
Amit (ISI, Chennai) ANOVA December 16, 2013 15 / 33
The ANOVA model
O W f
http://find/7/24/2019 A Nova Presentation
77/163
In a One Way Analysis of Variance the hypothesis to be tested is that theitems in the various classes come from universes, the means of which are
equal.
More precisely, the mathematical model for the analysis is
yij=+i+ij
where is a constant, the is are the class differentials and ij is randomnormal deviate with mean zero and variance 2, these being the same forall classes.
The hypothesis to be tested is that the is are zero for all j.
With reference to the tensile strength example the null hypothesis means :the mean tensile strength of yarns with different cotton percentages are allequal.
Amit (ISI, Chennai) ANOVA December 16, 2013 15 / 33
The computations for a one way ANOVA
http://find/7/24/2019 A Nova Presentation
78/163
One important result we need to know is that : means of samples of size nfrom a single population tends to have a variance that equals the varianceof the universe divided by n.
2y=2yn
Amit (ISI, Chennai) ANOVA December 16, 2013 16 / 33
The computations for a one way ANOVA
http://find/7/24/2019 A Nova Presentation
79/163
One important result we need to know is that : means of samples of size nfrom a single population tends to have a variance that equals the varianceof the universe divided by n.
2y=2yn
Lets look at the formula for the standard deviation :
Amit (ISI, Chennai) ANOVA December 16, 2013 16 / 33
The computations for a one way ANOVA
http://find/7/24/2019 A Nova Presentation
80/163
One important result we need to know is that : means of samples of size nfrom a single population tends to have a variance that equals the varianceof the universe divided by n.
2y=2yn
Lets look at the formula for the standard deviation :
The observations are : x1, x2, . . . , xn
2x= 1n 1
ni=1
x2i [n
i=1xi]2n
Amit (ISI, Chennai) ANOVA December 16, 2013 16 / 33
The computations
http://find/7/24/2019 A Nova Presentation
81/163
1 The correction factor : CF =y2..
/N[
i
jyij]
2
N = 376
2
25 =
Amit (ISI, Chennai) ANOVA December 16, 2013 17 / 33
The computations
http://find/7/24/2019 A Nova Presentation
82/163
1 The correction factor : CF =y2..
/N[
i
jyij]
2
N = 376
2
25 =
2 Total sum of square SSTotal=
ijy2ij CF= (7)2 + (7)2 + (15)2 + + (15)2 CF= 636.96
Amit (ISI, Chennai) ANOVA December 16, 2013 17 / 33
The computations
http://find/7/24/2019 A Nova Presentation
83/163
1 The correction factor : CF =y2..
/N[
i
jyij]
2
N = 376
2
25 =
2 Total sum of square SSTotal=
ijy2ij CF= (7)2 + (7)2 + (15)2 + + (15)2 CF= 636.963 Sum of squares due to cotton percentage SScotton =
i[y2i.]/n] CF =
15 [(49)
2 + (77)2 + (88)2 + (108)2 + (54)2] CF= 475.76
Amit (ISI, Chennai) ANOVA December 16, 2013 17 / 33
The computations
http://goforward/http://find/http://goback/7/24/2019 A Nova Presentation
84/163
1 The correction factor : CF =y2..
/N[
i
jyij]
2
N = 376
2
25 =
2 Total sum of square SSTotal=
ijy2ij CF= (7)2 + (7)2 + (15)2 + + (15)2 CF= 636.963 Sum of squares due to cotton percentage SScotton =
i[y2i.]/n] CF =
15 [(49)
2 + (77)2 + (88)2 + (108)2 + (54)2] CF= 475.76
4
Sum of squares due to errorSS
Error =SS
total SS
Cotton
Amit (ISI, Chennai) ANOVA December 16, 2013 17 / 33
The computations
http://find/7/24/2019 A Nova Presentation
85/163
1 The correction factor : CF =y2..
/N[
i
jyij]
2
N = 376
2
25 =
2 Total sum of square SSTotal=
ijy2ij CF= (7)2 + (7)2 + (15)2 + + (15)2 CF= 636.963 Sum of squares due to cotton percentage SScotton =
i[y2i.]/n] CF =
15 [(49)
2 + (77)2 + (88)2 + (108)2 + (54)2] CF= 475.76
4 Sum of squares due to error SSError
=SStotal
SSCotton
5 The above SSErrorcan also be calculated independently.
Amit (ISI, Chennai) ANOVA December 16, 2013 17 / 33
Degree of Freedom
http://find/7/24/2019 A Nova Presentation
86/163
Degrees of freedom is the number of observations that are free to vary!
Amit (ISI, Chennai) ANOVA December 16, 2013 18 / 33
Degree of Freedom
http://find/7/24/2019 A Nova Presentation
87/163
Degrees of freedom is the number of observations that are free to vary!
Well, its not easy, unfortunately to understand degree of freedom as a
concept in its entirety with little experience!
Amit (ISI, Chennai) ANOVA December 16, 2013 18 / 33
Degree of Freedom
http://find/7/24/2019 A Nova Presentation
88/163
Degrees of freedom is the number of observations that are free to vary!
Well, its not easy, unfortunately to understand degree of freedom as a
concept in its entirety with little experience!
Do have a look athttp://www.creative-wisdom.com/computer/sas/df.html
Amit (ISI Chennai) ANOVA December 16 2013 18 / 33
Degree of Freedom
This is a deep and wide concept, not easy to get in a single go.
http://find/7/24/2019 A Nova Presentation
89/163
Amit (ISI Chennai) ANOVA December 16 2013 19 / 33
http://find/7/24/2019 A Nova Presentation
90/163
Degree of Freedom
This is a deep and wide concept, not easy to get in a single go.
7/24/2019 A Nova Presentation
91/163
However, here is an analogy! an analogy only not the whole concept.
Amit (ISI Chennai) ANOVA December 16 2013 19 / 33
Degree of Freedom
This is a deep and wide concept, not easy to get in a single go.
http://find/7/24/2019 A Nova Presentation
92/163
However, here is an analogy! an analogy only not the whole concept.
This is a piece of chocolate I want to divide in three pieces.
Amit (ISI Chennai) ANOVA December 16 2013 19 / 33
Degree of Freedom
This is a deep and wide concept, not easy to get in a single go.
http://find/7/24/2019 A Nova Presentation
93/163
However, here is an analogy! an analogy only not the whole concept.
This is a piece of chocolate I want to divide in three pieces.
For my friend!
Amit (ISI Chennai) ANOVA December 16 2013 19 / 33
Degree of Freedom
This is a deep and wide concept, not easy to get in a single go.
H h i l ! l l h h l
http://find/7/24/2019 A Nova Presentation
94/163
However, here is an analogy! an analogy only not the whole concept.
This is a piece of chocolate I want to divide in three pieces.
For my friend!
A big piece for myself!
Amit (ISI Chennai) ANOVA December 16 2013 19 / 33
Degree of Freedom
This is a deep and wide concept, not easy to get in a single go.
H h i l ! l l h h l
http://find/7/24/2019 A Nova Presentation
95/163
However, here is an analogy! an analogy only not the whole concept.
This is a piece of chocolate I want to divide in three pieces.
For my friend!
A big piece for myself!
Do I have a choice for the last?
Amit (ISI Chennai) ANOVA December 16 2013 19 / 33
Degree of Freedom
This is a deep and wide concept, not easy to get in a single go.
H h i l ! l l t th h l t
http://find/7/24/2019 A Nova Presentation
96/163
However, here is an analogy! an analogy only not the whole concept.
This is a piece of chocolate I want to divide in three pieces.
For my friend!
A big piece for myself!
Do I have a choice for the last?
Amit (ISI Chennai) ANOVA December 16 2013 19 / 33
http://find/7/24/2019 A Nova Presentation
97/163
7/24/2019 A Nova Presentation
98/163
7/24/2019 A Nova Presentation
99/163
Hypothesis Testing
7/24/2019 A Nova Presentation
100/163
H0 :1 =2 =4 =5H1 :i=j for some i=j
Conclusion1 : H0 rejected.
A it (ISI Ch i) ANOVA D b 16 2013 21 / 33
Hypothesis Testing
H
http://find/7/24/2019 A Nova Presentation
101/163
H0 :1 =2 =4 =5H1 :i=j for some i=j
Conclusion1 : H0 rejected.
Further Analysis
A it (ISI Ch i) ANOVA D b 16 2013 21 / 33
http://find/7/24/2019 A Nova Presentation
102/163
Hypothesis Testing
H
7/24/2019 A Nova Presentation
103/163
H0 :1 =2 =4 =5H1 :i=j for some i=j
Conclusion1 : H0 rejected.
Further Analysis
Level 1 2 3 4 5Avg.Response 9.8 15.4 17.6 21.6 10.8
Conclusion2
1 1 =5 < 2 =3< 4 and
A i (ISI Ch i) ANOVA D b 16 2013 21 / 33
Hypothesis Testing
H
http://find/7/24/2019 A Nova Presentation
104/163
H0 :1 =2 =4 =5H1 :i=j for some i=j
Conclusion1 : H0 rejected.
Further Analysis
Level 1 2 3 4 5Avg.Response 9.8 15.4 17.6 21.6 10.8
Conclusion2
1 1 =5 < 2 =3< 4 and
2 95% confidence interval for 4 is 18.954 24.25
Amit (ISI, Chennai) ANOVA December 16, 2013 21 / 33
Terminologies
1 Factor
http://find/7/24/2019 A Nova Presentation
105/163
Amit (ISI, Chennai) ANOVA December 16, 2013 22 / 33
http://find/7/24/2019 A Nova Presentation
106/163
Terminologies
1 Factor
7/24/2019 A Nova Presentation
107/163
2 Level
3 Response
Amit (ISI, Chennai) ANOVA December 16, 2013 22 / 33
http://find/7/24/2019 A Nova Presentation
108/163
7/24/2019 A Nova Presentation
109/163
Terminologies
1 Factor
7/24/2019 A Nova Presentation
110/163
2 Level
3 Response
4
Main effect
5 Interaction effect
6 One way ANOVA
Amit (ISI, Chennai) ANOVA December 16, 2013 22 / 33
Terminologies
1 Factor
http://find/7/24/2019 A Nova Presentation
111/163
2 Level
3 Response
4
Main effect
5 Interaction effect
6 One way ANOVA
7 Two way ANOVA
Amit (ISI, Chennai) ANOVA December 16, 2013 22 / 33
http://find/7/24/2019 A Nova Presentation
112/163
Example : Two-way
The maximum output voltage of a particular type of storage battery isthought to be influenced by the material used in the plates and thetemperature in the location at which the battery is installed.
7/24/2019 A Nova Presentation
113/163
temperature in the location at which the battery is installed.
Amit (ISI, Chennai) ANOVA December 16, 2013 23 / 33
http://find/7/24/2019 A Nova Presentation
114/163
Example : Two-way
The maximum output voltage of a particular type of storage battery isthought to be influenced by the material used in the plates and thetemperature in the location at which the battery is installed.
7/24/2019 A Nova Presentation
115/163
p y
Four replicates of a factorial experiment are run in the laboratory for threetypes of material and three temperatures, and the results are presented asfollows.
Material Temperature(oF)Type 50 65 80
1 130 155 34 40 20 7074 180 80 75 82 58
2 150 188 136 122 25 70159 126 106 115 58 45
3 138 110 174 120 96 104168 160 150 139 82 60
Amit (ISI, Chennai) ANOVA December 16, 2013 23 / 33
http://find/7/24/2019 A Nova Presentation
116/163
7/24/2019 A Nova Presentation
117/163
7/24/2019 A Nova Presentation
118/163
7/24/2019 A Nova Presentation
119/163
The two-way Model
yijk=+i+j+ijk
7/24/2019 A Nova Presentation
120/163
where is a constant, the is are the class differentials corresponding tothe row factor (in this example material type), jare the class differentialscorresponding to the column factor(temperature) and ijk is random normaldeviate with mean zero and variance 2, these being the same for all theclasses.
The hypotheses to be tested are that the is are zero for all i, js are zerofor all j.
With reference to the output voltage example the null hypothesis means :
the mean mean output voltage of batteries with different material type areall equal and also the mean output voltages for all material types are same.
i= 1, , 3. j= 1, , 3. k= 1, , 4.
Amit (ISI, Chennai) ANOVA December 16, 2013 24 / 33
Interaction Effect
http://find/7/24/2019 A Nova Presentation
121/163
In an experiment with more than one factor there is another effect that weare concerned with, known as the interaction effect.
Amit (ISI, Chennai) ANOVA December 16, 2013 25 / 33
http://find/7/24/2019 A Nova Presentation
122/163
7/24/2019 A Nova Presentation
123/163
Interaction demonstrated
Consider two factors A and B, both at two levels 1 and 2. Let theresponses be as follows :
7/24/2019 A Nova Presentation
124/163
Amit (ISI, Chennai) ANOVA December 16, 2013 26 / 33
http://find/7/24/2019 A Nova Presentation
125/163
Interaction demonstrated
Consider two factors A and B, both at two levels 1 and 2. Let theresponses be as follows :
7/24/2019 A Nova Presentation
126/163
B1 B2A1 12 30A2 50 20
Pictorially the above scenario is as follows :
Amit (ISI, Chennai) ANOVA December 16, 2013 26 / 33
Interaction demonstrated
Consider two factors A and B, both at two levels 1 and 2. Let theresponses be as follows :
B B
http://find/7/24/2019 A Nova Presentation
127/163
B1 B2A1 12 30A2 50 20
Pictorially the above scenario is as follows :
B1 B2
Amit (ISI, Chennai) ANOVA December 16, 2013 26 / 33
Interaction demonstrated
Consider two factors A and B, both at two levels 1 and 2. Let theresponses be as follows :
B B
http://find/7/24/2019 A Nova Presentation
128/163
B1 B2A1 12 30A2 50 20
Pictorially the above scenario is as follows :
B1 B2
Amit (ISI, Chennai) ANOVA December 16, 2013 26 / 33
http://find/7/24/2019 A Nova Presentation
129/163
Interaction demonstrated
Consider two factors A and B, both at two levels 1 and 2. Let theresponses be as follows :
B B
7/24/2019 A Nova Presentation
130/163
B1 B2A1 12 30A2 50 20
Pictorially the above scenario is as follows :
B1 B2
A1
Amit (ISI, Chennai) ANOVA December 16, 2013 26 / 33
Interaction demonstrated
Consider two factors A and B, both at two levels 1 and 2. Let theresponses be as follows :
B B
http://find/7/24/2019 A Nova Presentation
131/163
B1 B2A1 12 30A2 50 20
Pictorially the above scenario is as follows :
B1 B2
A1
Amit (ISI, Chennai) ANOVA December 16, 2013 26 / 33
http://find/7/24/2019 A Nova Presentation
132/163
7/24/2019 A Nova Presentation
133/163
Interaction demonstrated
Consider two factors A and B, both at two levels 1 and 2. Let theresponses be as follows :
B1 B2
7/24/2019 A Nova Presentation
134/163
B1 B2A1 12 30A2 50 20
Pictorially the above scenario is as follows :
B1 B2
A1
A2
Amit (ISI, Chennai) ANOVA December 16, 2013 26 / 33
http://find/7/24/2019 A Nova Presentation
135/163
7/24/2019 A Nova Presentation
136/163
The Voltage Example
Correction Factor (CF) = (3799)2
36
Total Sum of Squares
7/24/2019 A Nova Presentation
137/163
Total Sum of Squares(SSTotal) = (130)
2 + (155)2 + (74)2 + + (82)2 + (60)2 CF= 77, 646.96
Sum of Squares due to Material(SSM) =
13x4
[(998)2 + (1300)2 + (1501)2] CF = 10, 683.72
Amit (ISI, Chennai) ANOVA December 16, 2013 27 / 33
The Voltage Example
Correction Factor (CF) = (3799)2
36
Total Sum of Squares
http://find/7/24/2019 A Nova Presentation
138/163
q(SSTotal) = (130)
2 + (155)2 + (74)2 + + (82)2 + (60)2 CF= 77, 646.96
Sum of Squares due to Material(SSM) =
13x4
[(998)2 + (1300)2 + (1501)2] CF = 10, 683.72
Sum of Squares due to Temperature(SST) =
13x4 [(1738)
2 + (1291)2 + (770)2] CF = 39, 118.72
Amit (ISI, Chennai) ANOVA December 16, 2013 27 / 33
The Voltage Example
Correction Factor (CF) = (3799)2
36
Total Sum of Squares
http://find/7/24/2019 A Nova Presentation
139/163
q(SSTotal) = (130)
2 + (155)2 + (74)2 + + (82)2 + (60)2 CF= 77, 646.96
Sum of Squares due to Material(SSM) =
13x4
[(998)2 + (1300)2 + (1501)2] CF = 10, 683.72
Sum of Squares due to Temperature(SST) =
13x4 [(1738)
2 + (1291)2 + (770)2] CF = 39, 118.72
Sum of Squares due to Interaction of Material and TemperatureSSMxT =
14 [(539)
2 + (229)2 + + (342)2] CF SSM SST = 9, 613.77
Amit (ISI, Chennai) ANOVA December 16, 2013 27 / 33
The Voltage Example
Correction Factor (CF) = (3799)2
36
Total Sum of Squares2 2 2 2 2
http://find/7/24/2019 A Nova Presentation
140/163
q(SSTotal) = (130)
2 + (155)2 + (74)2 + + (82)2 + (60)2 CF= 77, 646.96
Sum of Squares due to Material(SSM) =
13x4
[(998)2 + (1300)2 + (1501)2] CF = 10, 683.72
Sum of Squares due to Temperature(SST) =
13x4 [(1738)
2 + (1291)2 + (770)2] CF = 39, 118.72
Sum of Squares due to Interaction of Material and TemperatureSSMxT =
14 [(539)
2 + (229)2 + + (342)2] CF SSM SST = 9, 613.77
Sum of Squares due to Error SSError =SSTotal SSM SST SSMxT =77, 646.96 10, 683.72 39, 118.72 9, 613.77 = 18, 230.75.
Amit (ISI, Chennai) ANOVA December 16, 2013 27 / 33
The ANOVA table
SV DF SS MS F i
http://find/7/24/2019 A Nova Presentation
141/163
SV DF SS MS F ratioMaterial 2 10,683.72 5,341.86 7.91
Temperature 2 39,118.72 19,55836 28.97
Interaction 4 9,613.77 2,403.44 3.56
Error 27 18,230.75 675.21Total 35 77,646.96
SV : Source of variation, DF : Degree of freedom,SS : Sum of squares, MS : Mean square.
Amit (ISI, Chennai) ANOVA December 16, 2013 28 / 33
http://find/7/24/2019 A Nova Presentation
142/163
An Exercise
The effective life of a cutting tool installed in a numerically controlledmachine is thought to be affected by the cutting speed and the tool angle.
Three speeds and three angles are selected, and a factorial experiment
7/24/2019 A Nova Presentation
143/163
with two replicates is performed.
Amit (ISI, Chennai) ANOVA December 16, 2013 29 / 33
http://find/7/24/2019 A Nova Presentation
144/163
7/24/2019 A Nova Presentation
145/163
Some questions...
How many factors?
7/24/2019 A Nova Presentation
146/163
Amit (ISI, Chennai) ANOVA December 16, 2013 31 / 33
Some questions...
How many factors?
How many levels in which factor?
http://find/7/24/2019 A Nova Presentation
147/163
y
Amit (ISI, Chennai) ANOVA December 16, 2013 31 / 33
http://find/7/24/2019 A Nova Presentation
148/163
7/24/2019 A Nova Presentation
149/163
7/24/2019 A Nova Presentation
150/163
7/24/2019 A Nova Presentation
151/163
7/24/2019 A Nova Presentation
152/163
Some questions...
How many factors?
How many levels in which factor?
7/24/2019 A Nova Presentation
153/163
Are there replications?
So what are all the sources of variations?
Do we know all the degrees of freedoms?
Can we write down the ANOVA Table?
Do we now know what the model should be?
Finally do we know what to compute and how?
Amit (ISI, Chennai) ANOVA December 16, 2013 31 / 33
Some questions. . .
What is pvalue?
http://find/7/24/2019 A Nova Presentation
154/163
Amit (ISI, Chennai) ANOVA December 16, 2013 32 / 33
http://find/7/24/2019 A Nova Presentation
155/163
Some questions. . .
What is pvalue?
7/24/2019 A Nova Presentation
156/163
What are Type 1 and Type 2 errors.
Level of significance?
Amit (ISI, Chennai) ANOVA December 16, 2013 32 / 33
Some questions. . .
What is pvalue?
http://goforward/http://find/http://goback/7/24/2019 A Nova Presentation
157/163
What are Type 1 and Type 2 errors.
Level of significance?
?
Amit (ISI, Chennai) ANOVA December 16, 2013 32 / 33
Some questions. . .
What is pvalue?
http://find/7/24/2019 A Nova Presentation
158/163
What are Type 1 and Type 2 errors.
Level of significance?
?
What are the purposes ofRandomization, Replication and Local control?
Amit (ISI, Chennai) ANOVA December 16, 2013 32 / 33
http://find/7/24/2019 A Nova Presentation
159/163
A case
Growth of a particular infection ia being studied on several differentmedically important substrates.
7/24/2019 A Nova Presentation
160/163
Several drugs are also being studied for their efficacy against the growth.
Amit (ISI, Chennai) ANOVA December 16, 2013 33 / 33
A case
Growth of a particular infection ia being studied on several differentmedically important substrates.
http://anova-output.pdf/http://find/7/24/2019 A Nova Presentation
161/163
Several drugs are also being studied for their efficacy against the growth.
It is desired to investigate if the substrates behave differently with
reference to the growth.
Amit (ISI, Chennai) ANOVA December 16, 2013 33 / 33
A case
Growth of a particular infection ia being studied on several differentmedically important substrates.
http://anova-output.pdf/http://find/7/24/2019 A Nova Presentation
162/163
Several drugs are also being studied for their efficacy against the growth.
It is desired to investigate if the substrates behave differently with
reference to the growth.
Also if some drugs are more effective than the other with reference to thetreatment.
Amit (ISI, Chennai) ANOVA December 16, 2013 33 / 33
A case
Growth of a particular infection ia being studied on several differentmedically important substrates.
http://anova-output.pdf/http://find/7/24/2019 A Nova Presentation
163/163
Several drugs are also being studied for their efficacy against the growth.
It is desired to investigate if the substrates behave differently with
reference to the growth.Also if some drugs are more effective than the other with reference to thetreatment.
Lets look at a typical output of ANOVA from MINITAB. Click Here
Amit (ISI, Chennai) ANOVA December 16, 2013 33 / 33
http://anova-output.pdf/http://anova-output.pdf/http://find/