This is very good reference guide for MANOVA

1. MANOVA
Multivariate Analysis of Variance

2. One way Analysis of Variance
(ANOVA)
Comparing k Populations

3. The F test – for comparing k means
Situation
• We have k normal populations
• Let µi and σ denote the mean and standard
deviation of population i.
• i = 1, 2, 3, … k.
• Note: we assume that the standard deviation
for each population is the same.
σ1 = σ2 = … = σk = σ

4. We want to test
H 0 : µ1 = µ 2 = µ3 =  = µ k
against
H A : µ i ≠ µ j for at least one pair i, j

5. The F statistic k
∑n (x − x)
1 2
k −1 i i
F= i =1
k nj
∑∑ ( x − xi )
1 2
N −k ij
i =1 j =1
where xij = the jth observation in the i th sample.
( i = 1,2,, k and j = 1,2,, ni )
ni
∑x ij
= mean for i th sample ( i = 1,2,, k )
j =1
xi =
ni
k
N = ∑ ni = Total sample size
i =1
k ni
∑∑ x
i =1 j =1
ij
x= = Overall mean
N

6. The ANOVA table
Source S.S d.f, M.S. F
k
MS B
SS B = ∑ ni ( xi − x )
k
∑n (x − x)
2
Between MS B = 1 2 F=
i =1
k −1 k −1 i i MSW
i =1
nj nj
SSW = ∑∑ ( xij − xi ) ∑∑ ( x ij − xi )
k k
2 2
N − k MSW =
1
Within N −k
i =1 j =1 i =1 j =1
The ANOVA table is a tool for displaying the
computations for the F test. It is very important when
the Between Sample variability is due to two or more
factors

7. Computing Formulae:
Compute
ni
1) Ti = ∑ xij = Total for sample i
j =1
k k ni
2) G = ∑ Ti = ∑∑ xij = Grand Total
i =1 i =1 j =1
k
3) N = ∑ ni = Total sample size
i =1
k ni
∑∑ x
2
4) ij
i =1 j =1
k 2
Ti
5) ∑n
i =1 i

8. The data
• Assume we have collected data from each of
k populations
• Let xi1, xi2 , xi3 , … denote the ni observations
from population i.
• i = 1, 2, 3, … k.

9. Then
k 2 2
Ti G
1) SS Between =∑ −
i =1 ni N
k ni k 2
Ti
= ∑∑ xij − ∑
2
2) SSWithin
i =1 j =1 i =1 ni
SS Between ( k − 1)
3) F=
SSWithin ( N − k )

10. Anova Table
Source d.f. Sum of Mean F-ratio
Squares Square
Between k-1 SSBetween MSBetween MSB /MSW
Within N-k SSWithin MSWithin
Total N-1 SSTotal
SS
MS =
df

11. Example
In the following example we are comparing weight
gains resulting from the following six diets
1. Diet 1 - High Protein , Beef
2. Diet 2 - High Protein , Cereal
3. Diet 3 - High Protein , Pork
4. Diet 4 - Low protein , Beef
5. Diet 5 - Low protein , Cereal
6. Diet 6 - Low protein , Pork

12. Gains in weight (grams) for rats under six diets
differing in level of protein (High or Low)
and source of protein (Beef, Cereal, or Pork)
Diet 1 2 3 4 5 6
73 98 94 90 107 49
102 74 79 76 95 82
118 56 96 90 97 73
104 111 98 64 80 86
81 95 102 86 98 81
107 88 102 51 74 97
100 82 108 72 74 106
87 77 91 90 67 70
117 86 120 95 89 61
111 92 105 78 58 82
Mean 100.0 85.9 99.5 79.2 83.9 78.7
Std. Dev. 15.14 15.02 10.92 13.89 15.71 16.55
Σx 1000 859 995 792 839 787
Σx2 102062 75819 100075 64462 72613 64401

13. Thus
k
Ti 2 G 2 5272 2
SS Between = ∑ − = 467846 − = 4612.933
i =1 ni N 60
k ni k
Ti 2
SSWithin = ∑∑ xij − ∑
2
= 479432 − 467846 = 11586
i =1 j =1 i =1 ni
SS Between ( k − 1) 4612.933 / 5 922.6
F= = = = 4.3
SSWithin ( N − k ) 11586 / 54 214.56
F0.05 = 2.386 with ν 1 = 5 and ν 2 = 54
Thus since F > 2.386 we reject H0

14. Anova Table
Source d.f. Sum of Mean F-ratio
Squares Square
Between 5 4612.933 922.587 4.3**
(p = 0.0023)
SS
Within 54 11586.000 214.556
Total 59 16198.933
* - Significant at 0.05 (not 0.01)
** - Significant at 0.01

15. Equivalence of the F-test and the t-test
when k = 2
the t-test
x−y
t=
1 1
s Pooled +
n m
( n − 1) sx2 + ( m − 1) s 2
sPooled = y
n+m−2

16. the F-test k
∑n ( x − x)
2
2 i i k −1
s
F= Between
= i =1
s 2 k
 k 
Pooled
∑ ( ni − 1) si2  ∑ ni − k 
i =1  i =1 
n1 ( x1 − x ) + n2 ( x1 − x )
2 2
=
[ ]
( n1 − 1) s12 + ( n1 − 1) s12 ( n1 + n2 − 2)
denominator = s 2
pooled
numerator = n1 ( x1 − x ) + n2 ( x1 − x )
2 2

17. 2
 n1 x1 + n2 x2 
n1 ( x1 − x ) = n1  x1 − 
2
 n1 + n2 
 
2
n1n2
= ( x1 − x2 ) 2
( n1 + n2 ) 2
2
 n1 x1 + n2 x2 
n2 ( x2 − x ) = n2  x2 − 
2
 n1 + n2 
 
n12 n2
= ( x1 − x2 ) 2
( n1 + n2 ) 2

18. nn +n n
2 2
n1 ( x1 − x ) + n2 ( x2 − x ) ( x1 − x2 ) 2
2 2
= 1 2 2 1
( n1 + n ) 2
2
n1n2
= ( x1 − x2 ) 2
( n1 + n2 )
1
= ( x1 − x2 ) 2
1 1
 + 
n n 
 1 2 
Hence F=
1 ( x1 − x2 ) 2
= t2
 1 1  sPooled
2
 + 
n n 
 1 2 

19. Factorial Experiments
Analysis of Variance

20. • Dependent variable Y
• k Categorical independent variables A, B, C,
… (the Factors)
• Let
– a = the number of categories of A
– b = the number of categories of B
– c = the number of categories of C
– etc.

21. The Completely Randomized Design
• We form the set of all treatment combinations
– the set of all combinations of the k factors
• Total number of treatment combinations
– t = abc….
• In the completely randomized design n
experimental units (test animals , test plots,
etc. are randomly assigned to each treatment
combination.
– Total number of experimental units N = nt=nabc..

22. The treatment combinations can thought to be
arranged in a k-dimensional rectangular block
B
1 2 b
1
2
A
a

23. C
B
A

24. • The Completely Randomized Design is called
balanced
• If the number of observations per treatment
combination is unequal the design is called
unbalanced. (resulting mathematically more
complex analysis and computations)
• If for some of the treatment combinations
there are no observations the design is called
incomplete. (In this case it may happen that
some of the parameters - main effects and
interactions - cannot be estimated.)

25. Example
In this example we are examining the effect of
The level of protein A (High or Low) and
the source of protein B (Beef, Cereal, or
Pork) on weight gains (grams) in rats.
We have n = 10 test animals randomly
assigned to k = 6 diets

26. The k = 6 diets are the 6 = 3×2 Level-Source
combinations
1. High - Beef
2. High - Cereal
3. High - Pork
4. Low - Beef
5. Low - Cereal
6. Low - Pork

27. Table
Gains in weight (grams) for rats under six diets
differing in level of protein (High or Low) and s
ource of protein (Beef, Cereal, or Pork)
Level
of Protein High Protein Low protein
Source
of Protein Beef Cereal Pork Beef Cereal Pork
Diet 1 2 3 4 5 6
73 98 94 90 107 49
102 74 79 76 95 82
118 56 96 90 97 73
104 111 98 64 80 86
81 95 102 86 98 81
107 88 102 51 74 97
100 82 108 72 74 106
87 77 91 90 67 70
117 86 120 95 89 61
111 92 105 78 58 82
Mean 100.0 85.9 99.5 79.2 83.9 78.7
Std. Dev. 15.14 15.02 10.92 13.89 15.71 16.55

28. Treatment combinations
Source of Protein
Beef Cereal Pork
Level High Diet 1 Diet 2 Diet 3
of
Protein
Low Diet 4 Diet 5 Diet 6

29. Summary Table of Means
Source of Protein
Level of Protein Beef Cereal Pork Overall
High 100.00 85.90 99.50 95.13
Low 79.20 83.90 78.70 80.60
Overall 89.60 84.90 89.10 87.87

30. Profiles of the response relative
to a factor
A graphical representation of the
effect of a factor on a reponse
variable (dependent variable)

31. Profile Y for A
Y This could be for an
individual case or
averaged over a group
of cases
This could be for
specific level of
another factor or
averaged levels of
another factor
1 2 3 … a
Levels of A

32. Profiles of Weight Gain for
Source and Level of Protein
110
High Protein
Low Protein
Overall
100
We ight Gain
90
80
70
Beef Cereal Pork

33. Profiles of Weight Gain for
Source and Level of Protein
110
Beef
Cereal
Pork
100
Overall
We ight Gain
90
80
70
High Protein Low Protein

34. Example – Four factor experiment
Four factors are studied for their effect on Y (luster
of paint film). The four factors are:
1) Film Thickness - (1 or 2 mils)
2) Drying conditions (Regular or Special)
3) Length of wash (10,30,40 or 60 Minutes), and
4) Temperature of wash (92 ˚C or 100 ˚C)
Two observations of film luster (Y) are taken
for each treatment combination

35. The data is tabulated below:
Regular Dry Special Dry
Minutes 92 °C 100 °C 92°C 100 °C
1-mil Thickness
20 3.4 3.4 19.6 14.5 2.1 3.8 17.2 13.4
30 4.1 4.1 17.5 17.0 4.0 4.6 13.5 14.3
40 4.9 4.2 17.6 15.2 5.1 3.3 16.0 17.8
60 5.0 4.9 20.9 17.1 8.3 4.3 17.5 13.9
2-mil Thickness
20 5.5 3.7 26.6 29.5 4.5 4.5 25.6 22.5
30 5.7 6.1 31.6 30.2 5.9 5.9 29.2 29.8
40 5.5 5.6 30.5 30.2 5.5 5.8 32.6 27.4
60 7.2 6.0 31.4 29.6 8.0 9.9 33.5 29.5

36. Definition:
A factor is said to not affect the response if
the profile of the factor is horizontal for all
combinations of levels of the other factors:
No change in the response when you change
the levels of the factor (true for all
combinations of levels of the other factors)
Otherwise the factor is said to affect the
response:

37. Profile Y for A – A affects the response
Y



 Levels of B



1 2 3 … a
Levels of A

38. Profile Y for A – no affect on the response
Y



 Levels of B



1 2 3 … a
Levels of A

39. Definition:
• Two (or more) factors are said to interact if
changes in the response when you change
the level of one factor depend on the
level(s) of the other factor(s).
• Profiles of the factor for different levels of
the other factor(s) are not parallel
• Otherwise the factors are said to be
additive .
• Profiles of the factor for different levels of
the other factor(s) are parallel.

40. Interacting factors A and B
Y



 Levels of B



1 2 3 … a
Levels of A

41. Additive factors A and B
Y



 Levels of B



1 2 3 … a
Levels of A

42. • If two (or more) factors interact each factor
effects the response.
• If two (or more) factors are additive it still
remains to be determined if the factors
affect the response
• In factorial experiments we are interested in
determining
– which factors effect the response and
– which groups of factors interact .

43. The testing in factorial experiments
1. Test first the higher order interactions.
2. If an interaction is present there is no need
to test lower order interactions or main
effects involving those factors. All factors
in the interaction affect the response and
they interact
3. The testing continues with for lower order
interactions and main effects for factors
which have not yet been determined to
affect the response.

44. Models for factorial
Experiments

45. The Single Factor Experiment
Situation
• We have t = a treatment combinations
• Let µi and σ denote the mean and standard
deviation of observations from treatment i.
• i = 1, 2, 3, … a.
• Note: we assume that the standard deviation
for each population is the same.
σ1 = σ2 = … = σa = σ

46. The data
• Assume we have collected data for each of
the a treatments
• Let yi1, yi2 , yi3 , … , yin denote the n observations
for treatment i.
• i = 1, 2, 3, … a.

47. The model
Note:
yij = µi + ( yij − µi ) = µi + ε ij
= µ + ( µi − µ ) + ε ij = µ + α i + ε ij
where ε ij = yij − µi has N(0,σ2) distribution
1 k
µ = ∑ µi (overall mean effect)
k i =1
α i = µi − µ (Effect of Factor A)
a
Note: ∑α
i =1
i =0 by their definition.

48. Model 1:
yij (i = 1, … , a; j = 1, …, n) are independent
Normal with mean µi and variance σ2.
Model 2: yij = µi + ε ij
where εij (i = 1, … , a; j = 1, …, n) are independent
Normal with mean 0 and variance σ2.
Model 3: yij = µ + α i + ε ij
where εij (i = 1, … , a; j = 1, …, n) are independent
Normal with mean 0 and variance σ2 and
a
∑α
i =1
i =0

49. The Two Factor Experiment
Situation
• We have t = ab treatment combinations
• Let µij and σ denote the mean and standard
deviation of observations from the
treatment combination when A = i and B =
j.
• i = 1, 2, 3, … a, j = 1, 2, 3, … b.

50. The data
• Assume we have collected data (n observations)
for each of the t = ab treatment combinations.
• Let yij1, yij2 , yij3 , … , yijn denote the n observations for
treatment combination - A = i, B = j.
• i = 1, 2, 3, … a, j = 1, 2, 3, … b.

51. The model
Note:
yijk = µij + ( yijk − µij ) = µij + ε ijk
= µ + ( µi• − µ ) + ( µ• j − µ ) + ( µij − µi• − µ• j + µ ) + ε ij
= µ + α i + β j + ( αβ ) ij + ε ijk
where ε ijk = yijk − µij has N(0,σ2) distribution
1 a b 1 b 1 a
µ = ∑∑ µij , µi• = ∑ µij and µ• j = ∑ µij
ab i =1 j =1 b j =1 a i =1
α i = µi • − µ , β j = µ• j − µ ,
and ( αβ ) ij = µij − µi• − µ• j + µ

52. The model
Note:
yijk = µij + ( yijk − µij ) = µij + ε ijk
= µ + ( µi• − µ ) + ( µ• j − µ ) + ( µij − µi• − µ• j + µ ) + ε ij
= µ + α i + β j + ( αβ ) ij + ε ijk
where ε ijk = yijk − µij has N(0,σ2) distribution
1 a b 1 b 1 a
µ = ∑∑ µij , µi• = ∑ µij and µ• j = ∑ µij
ab i =1 j =1 b j =1 a i =1
α i = µi • − µ , β j = µ• j − µ ,
a
Note: ∑α
i =1
i =0 by their definition.

53. Main effects Error
Interaction
Model : Mean
Effect
yijk = µ + α i + β j + ( αβ ) ij + ε ijk
where εijk (i = 1, … , a; j = 1, …, b ; k = 1, …, n) are
independent Normal with mean 0 and variance σ2 and
a b
∑α
i =1
i =0 ∑β
j =1
j =0
a b
and ∑ ( αβ )
i =1
ij
= ∑ ( αβ ) ij = 0
j =1

54. Maximum Likelihood Estimates
yijk = µ + α i + β j + ( αβ ) ij + ε ijk
where εijk (i = 1, … , a; j = 1, …, b ; k = 1, …, n) are
independent Normal with mean 0 and variance σ2 and
a b n
µ = y••• = ∑∑∑ yijk abn
ˆ
i =1 j =1 k =1
b n
α i = yi •• − y••• = ∑∑ yijk bn − y•••
ˆ
j =1 k =1
a n
β j = y• j • − y••• = ∑∑ yijk an − y•••
ˆ
i =1 k =1

55. ^
( αβ ) ij = yij• − yi•• − y• j • + y•••
n
= ∑ yijk n − yi •• − y• j • + y•••
k =1
a b n
1
∑∑∑ ( yijk − yij• )
2
σ =
ˆ 2
nab i =1 j =1 k =1
2
1 a b n   ^ 
= ∑∑∑  yijk −  µ + αi + βˆ j + ( αβ ) ij  ÷
nab i =1 j =1 k =1 
ˆ ˆ
 
 ÷

This is not an unbiased estimator of σ2 (usually the
case when estimating variance.)
The unbiased estimator results when we divide by
ab(n -1) instead of abn

56. The unbiased estimator of σ2 is
a b n
1
∑∑∑ ( y − yij • )
2
s =
2
ab ( n − 1)
ijk
i =1 j =1 k =1
2
1 a b n  ^ 
=
ab ( n − 1)
∑∑∑   yijk −  µ + α i + β j + ( αβ ) ij  ÷
ˆ ˆ ˆ
i =1 j =1 k =1
 
 ÷

1
= SS Error = MS Error
ab ( n − 1)
where
a b n
SS Error = ∑∑∑ ( yijk − yij • )
2
i =1 j =1 k =1

57. Testing for Interaction:
We want to test:
H0: (αβ)ij = 0 for all i and j, against
HA: (αβ)ij ≠ 0 for at least one i and j.
The test statistic 1
SS AB
F=
MS AB
=
( a − 1) ( b − 1)
MS Error MS Error
where
a ^b a b
= ∑∑ ( αβ ) ij = ∑∑ ( yij • − yi•• − y• j • + y••• )
2 2
SS AB
i =1 j =1 i =1 j =1

58. We reject
H0: (αβ)ij = 0 for all i and j,
If
MS AB
F= > Fα ( (a − 1)(b − 1), ab(n − 1) )
MS Error

59. Testing for the Main Effect of A:
We want to test:
H0: αi = 0 for all i, against
HA: αi ≠ 0 for at least one i.
The test statistic 1
SS A
F=
MS A
=
( a − 1)
MS Error MS Error
where
a a
SS A = ∑ α = ∑ ( yi•• − y••• )
2 2
ˆ i
i =1 i =1

60. We reject
H0: αi = 0 for all i,
If
MS A
F= > Fα ( (a − 1), ab(n − 1) )
MS Error

61. Testing for the Main Effect of B:
We want to test:
H0: βj = 0 for all j, against
HA: βj ≠ 0 for at least one j.
The test statistic 1
SS B
F=
MS B
=
( b − 1)
MS Error MS Error
where
b b
ˆ2 = ( y − y ) 2
SS B = ∑ β j ∑ • j • •••
j =1 j =1

62. We reject
H0: βj = 0 for all j,
If
MS B
F= > Fα ( (b − 1), ab(n − 1) )
MS Error

63. The ANOVA Table
Source S.S. d.f. MS =SS/df F
A SSA a-1 MSA MSA / MSError
B SSB b-1 MSB MSB / MSError
AB SSAB (a - 1)(b - 1) MSAB MSAB/ MSError
Error SSError ab(n - 1) MSError
Total SSTotal abn - 1

64. Computing Formulae
a b n
Let T••• = ∑∑∑ yijk
i =1 j =1 k =1
b n a n n
Ti•• = ∑∑ yijk , T• j • = ∑∑ yijk , Tij • = ∑ yijk
j =1 k =1 i =1 k =1 k =1
a b 2 n
T•••
Then SSTotal = ∑∑∑ yijk − 2
i =1 j =1 k =1 nab
2
T a
T 2 2
T a T 2
SS A = ∑ , SS B = ∑
• j•
− i •• •••
− •••
i =1 nb nab i =1 na nab
a T2 a 2 a T2 2
Ti•• T•••
SS AB = ∑ −∑ −∑
ij • • j•
+ ,
i =1 n i =1 nb i =1 na nab

65. and SS Error = SSTotal − SS A − SS B − SS AB

66. MANOVA
Multivariate Analysis of Variance

67. One way Multivariate Analysis
of Variance (MANOVA)
Comparing k p-variate Normal
Populations

68. The F test – for comparing k means
Situation
• We have k normal populations
r
• Let µi and Σ denote the mean vector and
covariance matrix of population i.
• i = 1, 2, 3, … k.
• Note: we assume that the covariance matrix
for each population is the same.
Σ1 = Σ 2 = K = Σ k = Σ

69. We want to test
r r r r
H 0 : µ1 = µ 2 = µ3 = K = µ k
against
r r
H A : µi ≠ µ j for at least one pair i, j

70. The data
• Assume we have collected data from each of
k populations
r r r
• Let xi1 , xi 2 ,K , xin denote the n observations
from population i.
• i = 1, 2, 3, … k.

71. Computing Formulae:
 n 
Compute ∑ x1ij 
 
r n  j =1   T1i 
r
1) Ti = ∑ xij = Total vector for sample i = M = M 
 n   
j =1  x  Tpi 
 ∑ pij 
 j =1 
 G1 
r k r k ni r  
2) G = ∑ Ti = ∑∑ xij =  M  = Grand Total vector
i =1 i =1 j =1
G p 
 
3) N = kn = Total sample size

72.  k n 2 k n

 ∑∑ x1ij L ∑∑ x1ij x pij 
k n
r r  i =1 j =1 i =1 j =1

4) ∑∑ xij xij = 
′ M O M 
i =1 j =1  k n k n

 x2 
∑∑ x1ij x pij L
 i =1 j =1 ∑∑ pij 
 i =1 j =1 
 1 k 2 1 k 
 n ∑ T1i L ∑ T1iTpi 
n i =1
5) 1 k r r  i =1 
∑ TTi′ =  M
n i =1
i O M 
 k k 
 1 ∑ T1iTpi L 1 ∑ Tpi 2
 n i =1
 n i =1 


73. Let
1 k rr 1 r r
H = ∑ TTi′− GG ′
i
n i =1 N
 1 k 2 G12 1 k G1G p 
 ∑ T1i − N L ∑ T1iTpi − N 
 n i =1 n i =1 
= M O M 
 k 
 1 T T − G1G p 1 k 2 G12 
 n ∑ 1i pi ∑ T1i − N 
L
 i =1 N n i =1 
 k k

n∑ ( x1i• − x1•• ) L n ∑ ( x1i• − x1•• ) ( x pi• − x p•• ) 
2

 i =1 i =1

= M O M 
 k k 
 n∑ ( x1i• − x1•• ) ( x pi• − x p•• ) L n∑ ( x pi• − x p•• )
2

 i =1
 i =1 

= the Between SS and SP matrix

74. k n
r r 1 k rr
Let E = ∑∑ xij xij − ∑ TTi ′
′ i
i =1 j =1 n i =1
 k n 2 1 k 2 k n
1 k 
 ∑∑ x1ij − n ∑ T1i L ∑∑ x1ij x pij − ∑ T1iTpi 
n i =1
 i =1 j =1 i =1 i =1 j =1

= M O M 
 k n 
 1 k k n
1 k
x 2 − ∑ Tpi 
 ∑∑ 1ij pij n ∑ 1i pi ∑∑ pij n i=1 2 
x x − TT L
 i =1 j =1 i =1 i =1 j =1 
 k n k n

∑∑ ( x1ij − x1i• ) L ∑∑ ( x1ij − x1i• ) ( x pij − x pi • ) 
2

 i =1 j =1 i =1 j =1

= M O M 
 k n k n

 ∑∑ ( 1ij 1i• ) ( pij
 x − x pi• ) ∑∑ ( x pij − x pi• ) 
2
x −x L
 i =1 j =1 i =1 j =1


= the Within SS and SP matrix

75. The Manova Table
Source SS and SP matrix
 h11 L h1 p 
 
Between H = M O M 
 h1 p L hpp 
 
 e11 L e1 p 
 
Within E= M O M 
e1 p L e pp 
 

76. There are several test statistics for testing
r r r r
H 0 : µ1 = µ 2 = µ3 = K = µ k
against
r r
H A : µi ≠ µ j for at least one pair i, j

77. 1. Roy’s largest root
λ1 = largest eigenvalue of HE −1
This test statistic is derived using Roy’s union
intersection principle
2. Wilk’s lambda (Λ)
E 1
Λ= =
H+E HE −1 + I
This test statistic is derived using the generalized
Likelihood ratio principle

78. 3. Lawley-Hotelling trace statistic
T02 = trHE −1 = sum of the eigenvalues of HE−1
4. Pillai trace statistic (V)
V = trH ( H + E )
−1

79. Example
In the following study, n = 15 first year
university students from three different School
regions (A, B and C) who were each taking the
following four courses (Math, biology, English
and Sociology) were observed: The marks on
these courses is tabulated on the following slide:

80. The data
Educational Region
A B C
Student Math Biology English Sociology Student Math Biology English Sociology Student Math Biology English Sociology
1 62 65 67 76 1 65 55 35 43 1 47 47 98 78
2 54 61 75 70 2 87 81 59 64 2 57 69 68 45
3 53 53 53 59 3 75 67 56 68 3 65 71 77 62
4 48 56 73 81 4 74 70 55 66 4 41 64 68 58
5 60 55 49 60 5 83 71 40 52 5 56 54 86 64
6 55 52 34 41 6 59 48 48 57 6 63 73 88 76
7 76 71 35 40 7 61 47 46 54 7 43 62 84 78
8 58 52 58 46 8 81 77 51 45 8 28 47 65 58
9 75 71 60 59 9 77 68 42 49 9 47 54 90 78
10 55 51 69 75 10 82 84 63 70 10 42 44 79 73
11 72 74 64 59 11 68 64 35 44 11 50 53 89 89
12 72 75 51 47 12 60 53 60 65 12 46 61 91 82
13 76 69 69 57 13 94 88 51 63 13 74 78 99 86
14 44 48 65 65 14 96 88 67 81 14 63 66 94 86
15 89 71 59 67 15 84 75 46 67 15 69 82 78 73

81. Summary Statistics
r r 15 r 15 r 15 r
x′ =
A
63.267 61.600 58.733 60.133 ′
x• = x′ +
A
′ ′
xB + xC =
45 45 45
160.638 104.829 -32.638 -47.110
SA = 104.829 92.543 -4.900 -22.229
64.133 64.111 64.200 63.911
-32.638 -4.900 155.638 128.967
-47.110 -22.229 128.967 159.552
r 76.400 69.067 50.267 59.200
x′ =
B 14 14 14
141.257 155.829 45.100 60.914 S Pooled = S A + S B + SC =
155.829 185.924 61.767 71.057 42 42 42
SB = 45.100 61.767 96.495 93.371
60.914 71.057 93.371 123.600 152.654 125.878 22.092 16.354
125.878 138.283 20.003 16.133
22.092 20.003 122.892 112.408
r
′
xC = 52.733 61.667 83.600 72.400 16.354 16.133 112.408 146.517
156.067 116.976 53.814 35.257
SC = 116.976
53.814
136.381
3.143
3.143
116.543
-0.429
114.886
35.257 -0.429 114.886 156.400

82. Computations :
r n
r
1) Ti = ∑ xij = Total vector for sample i
j =1
 G1 
r k r k ni
r  
2) G = ∑ Ti = ∑∑ xij =  M  = Grand Total vector
i =1 i =1 j =1
G p 
 
Math Biology English Sociology
A 949 924 881 902
Totals B 1146 1036 754 888
C 791 925 1254 1086
Grand Totals G 2886 2885 2889 2876
3) N = kn = Total sample size = 45

83.  k n 2 k n

 ∑∑ x1ij L ∑∑ x1ij x pij 
k n
r r  i =1 j =1 i =1 j =1

4) ∑∑ xij xij = 
′ M O M 
i =1 j =1  k n k n

 x2 
∑∑ x1ij x pij L
 i =1 j =1 ∑∑ pij 
 i =1 j =1 
195718 191674 180399 182865
191674 191321 184516 184542
= 180399 184516 199641 193125
182865 184542 193125 191590

84.  1 k 2 1 k 
 n ∑ T1i L ∑ T1iTpi 
n i =1
5) 1 k r r  i =1 
∑ TTi′ =  M
n i =1
i O M 
 k k 
 1 ∑ T1iTpi L 1 ∑ Tpi 2
 n i =1
 n i =1 

189306.53 186387.13 179471.13 182178.13
186387.13 185513.13 183675.87 183864.40
= 179471.13 183675.87 194479.53 188403.87
182178.13 183864.40 188403.87 185436.27

85. Now
1 k rr 1 r r
H = ∑ TTi′− GG ′
i
n i =1 N
4217.733333 1362.466667 -5810.066667 -2269.333333
1362.466667 552.5777778 -1541.133333 -519.1555556
=
-5810.066667 -1541.133333 9005.733333 3764.666667
-2269.333333 -519.1555556 3764.666667 1627.911111
= the Between SS and SP matrix

86. k n
r r 1 k rr
Let E = ∑∑ xij xij − ∑ TTi ′
′ i
i =1 j =1 n i =1
 k n 2 1 k 2 k n
1 k 
 ∑∑ x1ij − n ∑ T1i L ∑∑ x1ij x pij − ∑ T1iTpi 
n i =1
 i =1 j =1 i =1 i =1 j =1

= M O M 
 k n 
 1 k k n
1 k
x 2 − ∑ Tpi 
 ∑∑ 1ij pij n ∑ 1i pi ∑∑ pij n i=1 2 
x x − TT L
 i =1 j =1 i =1 i =1 j =1 
6411.467 5286.867 927.867 686.867
= 5286.867 5807.867 840.133 677.600
927.867 840.133 5161.467 4721.133
686.867 677.600 4721.133 6153.733
= the Within SS and SP matrix

87. Using SPSS to perform MANOVA

88. Selecting the variables and the Factors

89. The output
c
Multivariate Tests
Effect Value F Hypothesis df Error df Sig.
Intercept Pillai's Trace .984 586.890a 4.000 39.000 .000
Wilks' Lambda .016 586.890a 4.000 39.000 .000
Hotelling's Trace 60.194 586.890a 4.000 39.000 .000
Roy's Largest Root 60.194 586.890a 4.000 39.000 .000
High_School Pillai's Trace .883 7.913 8.000 80.000 .000
Wilks' Lambda .161 14.571a 8.000 78.000 .000
Hotelling's Trace 4.947 23.501 8.000 76.000 .000
Roy's Largest Root 4.891 48.913b 4.000 40.000 .000
a. Exact statistic
b. The statistic is an upper bound on F that yields a lower bound on the significance level.
c. Design: Intercept+High_School

90. Univariate Tests
Tests of Between-Subjects Effects
Type III Sum
Source Dependent Variable of Squares df Mean Square F Sig.
Corrected Model Math 4217.733a 2 2108.867 13.815 .000
Biology 552.578b 2 276.289 1.998 .148
English 9005.733c 2 4502.867 36.641 .000
Sociology 1627.911d 2 813.956 5.555 .007
Intercept Math 185088.800 1 185088.800 1212.473 .000
Biology 184960.556 1 184960.556 1337.555 .000
English 185473.800 1 185473.800 1509.241 .000
Sociology 183808.356 1 183808.356 1254.515 .000
High_School Math 4217.733 2 2108.867 13.815 .000
Biology 552.578 2 276.289 1.998 .148
English 9005.733 2 4502.867 36.641 .000
Sociology 1627.911 2 813.956 5.555 .007
Error Math 6411.467 42 152.654
Biology 5807.867 42 138.283
English 5161.467 42 122.892
Sociology 6153.733 42 146.517
Total Math 195718.000 45
Biology 191321.000 45
English 199641.000 45
Sociology 191590.000 45
Corrected Total Math 10629.200 44
Biology 6360.444 44
English 14167.200 44
Sociology 7781.644 44
a. R Squared = .397 (Adjusted R Squared = .368)
b. R Squared = .087 (Adjusted R Squared = .043)
c. R Squared = .636 (Adjusted R Squared = .618)
d. R Squared = .209 (Adjusted R Squared = .172)