Download - Section 07 manova
MANOVA
Multivariate Analysis of Variance
One way Analysis of Variance (ANOVA)
Comparing k Populations
The F test – for comparing k means
Situation
• We have k normal populations
• Let i and 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.
1 = 2 = … = k =
We want to test
kH 3210 :
against
jiH jiA ,pair oneleast at for :
The F statistic
k
i
n
jiijkN
k
iiik
j
xx
xxnF
1 1
21
1
2
11
where xij = the jth observation in the i th sample.
injki ,,2,1 and ,,2,1
kiin
x
x th
i
n
jij
i
i
,,2,1 sample for mean 1
size sample Total 1
k
iinN
mean Overall 1 1
N
x
x
k
i
n
jij
i
The ANOVA table
k
iiiB xxnSS
1
2
W
B
MS
MSF
k
iiikB xxnMS
1
2
11
k
i
n
jiijW
j
xxSS1 1
2
k
i
n
jiijkNW
j
xxMS1 1
21
1k
kN
Source S.S d.f, M.S. F
Between
Within
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
Computing Formulae:
k
i
n
jij
i
x1 1
2
Compute
ixTin
jiji samplefor Total
1
Total Grand 1 11
k
i
n
jij
k
ii
i
xTG
size sample Total1
k
iinN
k
i i
i
n
T
1
2
1)
2)
3)
4)
5)
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.
Then
1)
2)
k
i i
ik
i
n
jijWithin n
TxSS
i
1
2
1 1
2
BetweenSS
k
i i
i
N
G
n
T
1
22
3)
kNSS
kSSF
Within
Between
1
Source d.f. Sum of Squares
Mean Square
F-ratio
Between k - 1 SSBetween MSBetween MSB /MSW
Within N - k SSWithin MSWithin
Total N - 1 SSTotal
Anova Table
SSMS
df
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
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
x 1000 859 995 792 839 787 x2 102062 75819 100075 64462 72613 64401
Thus
115864678464794321
2
1 1
2
k
i i
ik
i
n
jijWithin n
TxSS
i
BetweenSS 933.461260
5272467846
2
1
22
k
i i
i
N
G
n
T
3.4
56.214
6.922
54/11586
5/933.46121
kNSS
kSSF
Within
Between
54 and 5 with 386.2 2105.0 F
Thus since F > 2.386 we reject H0
Source d.f. Sum of Squares
Mean Square
F-ratio
Between 5 4612.933 922.587 4.3**
(p = 0.0023)
Within 54 11586.000 214.556
Total 59 16198.933
Anova Table
* - Significant at 0.05 (not 0.01)
SSSSSS
** - Significant at 0.01
Equivalence of the F-test and the t-test when k = 2
mns
yxt
Pooled
11
2
11 22
mn
smsns yx
Pooled
the t-test
the F-test
knsn
kxxn
s
sF
k
ii
k
iii
k
iii
Pooled
Between
11
2
1
2
2
2
1
1
211 21
211
211
212
211
nnsnsn
xxnxxn
212
211numerator xxnxxn
2r denominato pooleds
2
21
221122
222
nn
xnxnxnxxn
2
21
221111
211
nn
xnxnxnxxn
2
21221
221 xxnn
nn
2
21221
221 xx
nn
nn
2212
21
212
2212
222
11 xxnn
nnnnxxnxxn
221
21
21 xxnn
nn
221
21
11
1xx
nn
22
221
21
11
1t
s
xx
nn
FPooled
Hence
Factorial Experiments
Analysis of Variance
• 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.
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..
The treatment combinations can thought to be arranged in a k-dimensional rectangular block
A
1
2
a
B1 2 b
A
B
C
• 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.)
Example
In this example we are examining the effect of
We have n = 10 test animals randomly assigned to k = 6 diets
The level of protein A (High or Low) and the source of protein B (Beef, Cereal, or Pork) on weight gains (grams) in rats.
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
TableGains 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)
Levelof Protein High Protein Low protein
Sourceof Protein Beef Cereal Pork Beef Cereal Pork
Diet 1 2 3 4 5 6
73 98 94 90 107 49102 74 79 76 95 82118 56 96 90 97 73104 111 98 64 80 86
81 95 102 86 98 81107 88 102 51 74 97100 82 108 72 74 106
87 77 91 90 67 70117 86 120 95 89 61111 92 105 78 58 82
Mean 100.0 85.9 99.5 79.2 83.9 78.7Std. Dev. 15.14 15.02 10.92 13.89 15.71 16.55
Source of Protein
Level of Protein
Beef Cereal Pork
High
Low
Treatment combinations
Diet 1 Diet 2 Diet 3
Diet 4 Diet 5 Diet 6
Level of Protein Beef Cereal Pork Overall
Low 79.20 83.90 78.70 80.60
Source of Protein
High 100.00 85.90 99.50 95.13
Overall 89.60 84.90 89.10 87.87
Summary Table of Means
Profiles of the response relative to a factor
A graphical representation of the effect of a factor on a reponse variable (dependent variable)
Profile Y for AY
Levels of A
a1 2 3 …
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
70
80
90
100
110
Beef Cereal Pork
Wei
ght
Gai
n
High Protein
Low Protein
Overall
Profiles of Weight Gain for Source and Level of Protein
70
80
90
100
110
High Protein Low Protein
Wei
ght
Gai
nBeef
Cereal
Pork
Overall
Profiles of Weight Gain for Source and Level of Protein
Example – Four factor experiment
Four factors are studied for their effect on Y (luster of paint film). The four factors are:
Two observations of film luster (Y) are taken for each treatment combination
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)
The data is tabulated below:Regular Dry Special Dry
Minutes 92 C 100 C 92C 100 C1-mil Thickness
20 3.4 3.4 19.6 14.5 2.1 3.8 17.2 13.430 4.1 4.1 17.5 17.0 4.0 4.6 13.5 14.340 4.9 4.2 17.6 15.2 5.1 3.3 16.0 17.860 5.0 4.9 20.9 17.1 8.3 4.3 17.5 13.9
2-mil Thickness20 5.5 3.7 26.6 29.5 4.5 4.5 25.6 22.530 5.7 6.1 31.6 30.2 5.9 5.9 29.2 29.840 5.5 5.6 30.5 30.2 5.5 5.8 32.6 27.460 7.2 6.0 31.4 29.6 8.0 9.9 33.5 29.5
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:
Profile Y for A – A affects the response
Y
Levels of A
a1 2 3 …
Levels of B
Profile Y for A – no affect on the response
Y
Levels of A
a1 2 3 …
Levels of B
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.
Interacting factors A and BY
Levels of A
a1 2 3 …
Levels of B
Additive factors A and BY
Levels of A
a1 2 3 …
Levels of B
• 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 .
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.
Models for factorial Experiments
The Single Factor Experiment
Situation
• We have t = a treatment combinations
• Let i and 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.
1 = 2 = … = a =
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.
The model
Note:
ij i ij i i ijy y
i ij i ij
where ij ij iy
1
1 k
iik
i i
has N(0,2) distribution
(overall mean effect)
(Effect of Factor A)
Note:1
0a
ii
by their definition.
Model 1:
ij i ijy
yij (i = 1, … , a; j = 1, …, n) are independent Normal with mean i and variance 2.
Model 2:
where ij (i = 1, … , a; j = 1, …, n) are independent Normal with mean 0 and variance 2.
ij i ijy Model 3:
where ij (i = 1, … , a; j = 1, …, n) are independent Normal with mean 0 and variance 2 and
1
0a
ii
The Two Factor Experiment
Situation
• We have t = ab treatment combinations
• Let ij and 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.
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.
The modelNote:
ijk ij ijk ij ij ijky y
i j ij i j ij
where ijk ijk ijy
1 1 1 1
1 1 1, and
a b b a
ij i ij j iji j j iab b a
, ,i i j j
has N(0,2) distribution
and
i j ijkij
ij i jij
The modelNote:
ijk ij ijk ij ij ijky y
i j ij i j ij
where ijk ijk ijy
1 1 1 1
1 1 1, and
a b b a
ij i ij j iji j j iab b a
, ,i i j j
has N(0,2) distribution
Note:1
0a
ii
by their definition.
i j ijkij
ijk i j ijkijy
Model :
where ijk (i = 1, … , a; j = 1, …, b ; k = 1, …, n) are independent Normal with mean 0 and variance 2 and
1
0a
ii
1
0b
jj
1 1
and 0a b
ij iji j
Main effectsInteraction Effect
Mean Error
ijk i j ijkijy
Maximum Likelihood Estimates
where ijk (i = 1, … , a; j = 1, …, b ; k = 1, …, n) are independent Normal with mean 0 and variance 2 and
1 1 1
ˆa b n
ijki j k
y y abn
1 1
ˆb n
i i ijkj k
y y y bn y
1 1
ˆa n
j j ijki k
y y y an y
^
ij i jijy y y y
1
n
ijk i jk
y n y y y
22
1 1 1
1ˆ
a b n
ijk iji j k
y ynab
2
1 1 1
^1 ˆˆˆa b n
ijk i j iji j k
ynab
This is not an unbiased estimator of 2 (usually the case when estimating variance.)
The unbiased estimator results when we divide by ab(n -1) instead of abn
22
1 1 1
1
1
a b n
ijk iji j k
s y yab n
2
1 1 1
^1 ˆˆˆ1
a b n
ijk i j iji j k
yab n
The unbiased estimator of 2 is
1
1 Error ErrorSS MSab n
2
1 1 1
a b n
Error ijk iji j k
SS y y
where
22
1 1 1 1
^a b a b
AB ij i jiji j i j
SS y y y y
Testing for Interaction:
1
1 1 AB
AB
Error Error
SSa bMS
FMS MS
where
We want to test:
H0: ()ij = 0 for all i and j, against
HA: ()ij ≠ 0 for at least one i and j.
The test statistic
( 1)( 1), ( 1)AB
Error
MSF F a b ab n
MS
We reject
H0: ()ij = 0 for all i and j,
If
22
1 1
ˆa a
A i ii i
SS y y
Testing for the Main Effect of A:
1
1 A
A
Error Error
SSaMS
FMS MS
where
We want to test:
H0: i = 0 for all i, against
HA: i ≠ 0 for at least one i.
The test statistic
( 1), ( 1)A
Error
MSF F a ab n
MS
We reject
H0: i = 0 for all i,
If
22
1 1
ˆb b
B j jj j
SS y y
Testing for the Main Effect of B:
1
1 B
B
Error Error
SSbMS
FMS MS
where
We want to test:
H0: j = 0 for all j, against
HA: j ≠ 0 for at least one j.
The test statistic
( 1), ( 1)B
Error
MSF F b ab n
MS
We reject
H0: j = 0 for all j,
If
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
Computing Formulae
1 1 1
Let a b n
ijki j k
T y
1 1 1 1 1
, , b n a n n
i ijk j ijk ij ijkj k i k k
T y T y T y
2
2 •••
1 1 1
Then a b n
Total ijki j k
TSS y
nab
22 2 2
• ••• ••• •••
1 1
, a a
jiA B
i i
TT T TSS SS
nb nab na nab
2 22 2• • ••• •••
1 1 1
,a a a
ij jiAB
i i i
T TT TSS
n nb na nab
and Error Total A B ABSS SS SS SS SS
MANOVA
Multivariate Analysis of Variance
One way Multivariate Analysis of Variance (MANOVA)
Comparing k p-variate Normal Populations
The F test – for comparing k means
Situation
• We have k normal populations
• Let 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.
and i
1 2 k
We want to test
0 1 2 3: kH
against
: for at least one pair ,A i jH i j
The data
• Assume we have collected data from each of k populations
• Let denote the n observations from population i.
• i = 1, 2, 3, … k.
1 2, , ,i i inx x x
Computing Formulae:
Compute
1
Total vector for sample n
i ijj
T x i
1
1 1 1
Grand Total vector ink k
i iji i j
p
G
G T x
G
1)
2)
11 1
1
n
ijj i
npi
pijj
xT
Tx
Total sample size N kn 3)
21 1
1 1 1 1
1 12
11 1 1 1
k n k n
ij ij piji j i j
k n
ij iji j k n k n
ij pij piji j i j
x x x
x x
x x x
21 1
1 1
12
11 1
1 1
1
1 1
k k
i i pii ik
i ii k k
i pi pii i
T T Tn n
TTn
T T Tn n
4)
5)
Let
1
1 1k
i ii
H TT GGn N
212 1
1 11 1
21 2 1
1 11 1
1 1
1 1
k kp
i i pii i
k kp
i pi ii i
G GGT T T
n N n N
G G GT T T
n N n N
2
1 1 1 11 1
2
1 11 1
k k
i i pi pi i
k k
i pi p pi pi i
n x x n x x x x
n x x x x n x x
= the Between SS and SP matrix
Let1 1 1
1k n k
ij ij i ii j i
E x x TTn
2 21 1 1 1
1 1 1 1 1 1
2 21 1
1 1 1 1 1 1
1 1
1 1
k n k k n k
ij i ij pij i pii j i i j i
k n k k n k
ij pij i pi pij pii j i i j i
x T x x T Tn n
x x T T x Tn n
2
1 1 1 11 1 1 1
2
1 11 1 1 1
k n k n
ij i ij i pij pii j i j
k n k n
ij i pij pi pij pii j i j
x x x x x x
x x x x x x
= the Within SS and SP matrix
Source SS and SP matrix
Between
Within
The Manova Table
11 1
1
p
p pp
h h
H
h h
11 1
1
p
p pp
e e
E
e e
There are several test statistics for testing
0 1 2 3: kH
against
: for at least one pair ,A i jH i j
1. Roy’s largest root1
1 largest eigenvalue of HE
This test statistic is derived using Roy’s union intersection principle
2. Wilk’s lambda ()
1
1
E
H E HE I
This test statistic is derived using the generalized Likelihood ratio principle
3. Lawley-Hotelling trace statistic
2 1 10 sum of the eigenvalues of T tr HE HE
4. Pillai trace statistic (V)
1V tr
H H E
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:
Student Math Biology English Sociology Student Math Biology English Sociology Student Math Biology English Sociology1 62 65 67 76 1 65 55 35 43 1 47 47 98 782 54 61 75 70 2 87 81 59 64 2 57 69 68 453 53 53 53 59 3 75 67 56 68 3 65 71 77 624 48 56 73 81 4 74 70 55 66 4 41 64 68 585 60 55 49 60 5 83 71 40 52 5 56 54 86 646 55 52 34 41 6 59 48 48 57 6 63 73 88 767 76 71 35 40 7 61 47 46 54 7 43 62 84 788 58 52 58 46 8 81 77 51 45 8 28 47 65 589 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 7311 72 74 64 59 11 68 64 35 44 11 50 53 89 8912 72 75 51 47 12 60 53 60 65 12 46 61 91 8213 76 69 69 57 13 94 88 51 63 13 74 78 99 8614 44 48 65 65 14 96 88 67 81 14 63 66 94 8615 89 71 59 67 15 84 75 46 67 15 69 82 78 73
Educational RegionA B C
The data
Summary Statistics
63.267 61.600 58.733 60.133
160.638 104.829 -32.638 -47.110104.829 92.543 -4.900 -22.229-32.638 -4.900 155.638 128.967-47.110 -22.229 128.967 159.552
Ax
A S
Bx
B S
Cx
C S
76.400 69.067 50.267 59.200
141.257 155.829 45.100 60.914155.829 185.924 61.767 71.05745.100 61.767 96.495 93.37160.914 71.057 93.371 123.600
52.733 61.667 83.600 72.400
156.067 116.976 53.814 35.257116.976 136.381 3.143 -0.42953.814 3.143 116.543 114.88635.257 -0.429 114.886 156.400
15 15 15
45 45 45A B Cx x x x
14 14 14
42 42 42Pooled A B C S S S S
64.133 64.111 64.200 63.911
152.654 125.878 22.092 16.354125.878 138.283 20.003 16.133
22.092 20.003 122.892 112.40816.354 16.133 112.408 146.517
Computations :
1
Total vector for sample n
i ijj
T x i
1
1 1 1
Grand Total vector ink k
i iji i j
p
G
G T x
G
1)
2)
Total sample size = 45N kn 3)
Math Biology English SociologyA 949 924 881 902B 1146 1036 754 888C 791 925 1254 1086
Grand Totals G 2886 2885 2889 2876
Totals
21 1
1 1 1 1
1 12
11 1 1 1
k n k n
ij ij piji j i j
k n
ij iji j k n k n
ij pij piji j i j
x x x
x x
x x x
4)
195718 191674 180399 182865191674 191321 184516 184542180399 184516 199641 193125182865 184542 193125 191590
=
21 1
1 1
12
11 1
1 1
1
1 1
k k
i i pii ik
i ii k k
i pi pii i
T T Tn n
TTn
T T Tn n
=
5)
189306.53 186387.13 179471.13 182178.13186387.13 185513.13 183675.87 183864.40179471.13 183675.87 194479.53 188403.87182178.13 183864.40 188403.87 185436.27
Now
1
1 1k
i ii
H TT GGn N
= the Between SS and SP matrix
4217.733333 1362.466667 -5810.066667 -2269.3333331362.466667 552.5777778 -1541.133333 -519.1555556
-5810.066667 -1541.133333 9005.733333 3764.666667-2269.333333 -519.1555556 3764.666667 1627.911111
=
Let1 1 1
1k n k
ij ij i ii j i
E x x TTn
2 21 1 1 1
1 1 1 1 1 1
2 21 1
1 1 1 1 1 1
1 1
1 1
k n k k n k
ij i ij pij i pii j i i j i
k n k k n k
ij pij i pi pij pii j i i j i
x T x x T Tn n
x x T T x Tn n
= the Within SS and SP matrix
6411.467 5286.867 927.867 686.8675286.867 5807.867 840.133 677.600927.867 840.133 5161.467 4721.133686.867 677.600 4721.133 6153.733
=
Using SPSS to perform MANOVA
Selecting the variables and the Factors
Multivariate Testsc
.984 586.890a 4.000 39.000 .000
.016 586.890a 4.000 39.000 .000
60.194 586.890a 4.000 39.000 .000
60.194 586.890a 4.000 39.000 .000
.883 7.913 8.000 80.000 .000
.161 14.571a 8.000 78.000 .000
4.947 23.501 8.000 76.000 .000
4.891 48.913b 4.000 40.000 .000
Pillai's Trace
Wilks' Lambda
Hotelling's Trace
Roy's Largest Root
Pillai's Trace
Wilks' Lambda
Hotelling's Trace
Roy's Largest Root
EffectIntercept
High_School
Value F Hypothesis df Error df Sig.
Exact statistica.
The statistic is an upper bound on F that yields a lower bound on the significance level.b.
Design: Intercept+High_Schoolc.
The output
Univariate TestsTests of Between-Subjects Effects
4217.733a 2 2108.867 13.815 .000
552.578b 2 276.289 1.998 .148
9005.733c 2 4502.867 36.641 .000
1627.911d 2 813.956 5.555 .007
185088.800 1 185088.800 1212.473 .000
184960.556 1 184960.556 1337.555 .000
185473.800 1 185473.800 1509.241 .000
183808.356 1 183808.356 1254.515 .000
4217.733 2 2108.867 13.815 .000
552.578 2 276.289 1.998 .148
9005.733 2 4502.867 36.641 .000
1627.911 2 813.956 5.555 .007
6411.467 42 152.654
5807.867 42 138.283
5161.467 42 122.892
6153.733 42 146.517
195718.000 45
191321.000 45
199641.000 45
191590.000 45
10629.200 44
6360.444 44
14167.200 44
7781.644 44
Dependent VariableMath
Biology
English
Sociology
Math
Biology
English
Sociology
Math
Biology
English
Sociology
Math
Biology
English
Sociology
Math
Biology
English
Sociology
Math
Biology
English
Sociology
SourceCorrected Model
Intercept
High_School
Error
Total
Corrected Total
Type III Sumof Squares df Mean Square F Sig.
R Squared = .397 (Adjusted R Squared = .368)a.
R Squared = .087 (Adjusted R Squared = .043)b.
R Squared = .636 (Adjusted R Squared = .618)c.
R Squared = .209 (Adjusted R Squared = .172)d.