Chapter 15 Introduction to the Analysis of Variance IThe Omnibus Null Hypothesis
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Transcript of Chapter 15 Introduction to the Analysis of Variance IThe Omnibus Null Hypothesis
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Chapter 15
Introduction to the Analysis of Variance
I The Omnibus Null Hypothesis
H0: 1 = 2 = . . . = p
H1: j = j’
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A. Answering General Versus Specific ResearchQuestions
1. Population contrast, i, and sample contrast
1 =1 − 2
1 =Y.1 −Y.22. Pairwise and nonpairwise contrasts
1 =1 − 2
7 =1 −
2 + 3
2 2 =1 − 3
3 =1 − 4
4 = 2 − 3
5 = 2 − 4
6 = 3 − 4
8 =1 −
2 + 3 + 4
3
9 =
1 + 2
2−
3 + 4
2
, i
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B. Analysis of Variance Versus Multiple t Tests
1. Number of pairwise contrasts among p means
is given by p(p – 1)/2
p = 3 3(3 – 1)/2 = 3
p = 4 4(4 – 1)/2 = 6
p = 5 5(5 – 1)/2 = 10
2. If C = 3 contrasts among p = 3 means are tested
using a t statistic at = .05, the probability of
one or more type I errors is less than
1−(1−)C =1−(1−.05)3 =.14
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3. As C increases, the probability of making one or
more Type I errors using a t statistic increases
dramatically.
Prob. of one or more Type I errors
C =4 < [1−(1−.05)4 ] =.19
C =5 < [1−(1−.05)5] =.23
C =6 < [1−(1−.05)6 ] =.26
C =7 < [1−(1−.05)7 ] =.37
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4. Analysis of variance tests the omnibus null
hypothesis, H0: 1 = 2 = . . . = p , and controls
probability of making a Type I error at, say,
= .05 for any number of means.
5. Rejection of the null hypothesis makes the
alternative hypothesis, H1: j ≠ j’, tenable.
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II Basic Concepts In ANOVA
A. Notation
1. Two subscripts are used to denote a score, Xij.
The i subscript denotes one of the i = 1, . . . , n
participants in a treatment level. The j subscript
denotes one of the j = 1, . . . , p treatment levels.
2. The jth level of treatment A is denoted by aj.
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a1 a2 a3 a4
X11 X12 X13 X14
X21 X22 X23 X24
Xn1 Xn2 Xn3 Xn4
M M M M
Treatment Levels
X.1
X.2 X.3
X.4 X. .
X.1 =
X11 + X21 +L + Xn1
n
X . . =
X.1 + X.2 +L + X.4
p
X.4 =
X14 + X24 +L + Xn4
n
M
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B. Composite Nature of a Score
1. A score reflects the effects of four variables:
independent variable
characteristics of the participants in the
experiment
chance fluctuations in the participant’s
performance
environmental and other uncontrolled
variables
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2. Sample model equation for a score
X ij = X. . + (X. j −X. .) + (Xij −X. j )
Score Grand Treatment Error
Mean Effect Effect
3. The statistics estimate parameters of the model
equation as follows
X ij = + ( j −) + (Xij − j )
Score Grand Treatment Error
Mean Effect Effect
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4. Illustration of the sample model equation using the
weight-loss data in Table 1.
Table 1. One-Month Weight Losses for Three Diets
a1 a2 a3
7 10 12
9 13 11
8 9 15
6 7 14
Treatment Levels (Diets)
M M M
X.1 =8
X.2 =9 X.3 =12
X. . =9.67
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5. Let X11 = 7 denote Joan’s weight loss. She used
diet a1. Her score is a composite that tells a story.
X ij = X. . + (X. j −X. .) + (Xij −X. j )
7 = 9.67 + (8−9.67) + (7 −8)
Score Grand Treatment Error
Mean Effect = –1.67 Effect = –1
6. Joan used a less effective diet than other girls
(8 – 9.67 = –1.67), and she lost less weight than
other girls on the same diet (8 – 9 = –1).
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C. Partition of the Total Sum of Squares (SSTO)
1. The total variability among scores in the diet
experiment
also is a composite that can be decomposed into
between-groups sum of squares (SSBG)
within-groups sum of squares (SSWG)
SSTO = (Xij −X. .
i=1
n∑
j=1
p∑ )2
SSBG =n (X. j −X. .)
2
j=1
p∑
SSWG = (Xij −X. j
i=1
n∑
j=1
p∑ )2
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D. Degrees of Freedom for SSTO, SSBG, and SSWG
1. dfTO = np – 1
2. dfBG = p – 1
3. dfWG = p(n – 1)
E. Mean Squares, MS, and F Statistic 1. SSTO / (np −1) =MSTO
2. SSBG / ( p −1) =MSBG
3. SSWG / p(n −1) =MSWG
4. F =MSBG / MSWG
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F. Nature of MSBG and MSWG
1. Expected value of MSBG and MSWG when the
null hypothesis is true.
E( MSBG) =E(MSWG) =σε
2
2. Expected value of MSBG and MSWG when the
null hypothesis is false.
E( MSWG) =σε
2
E( MSBG) =σε
2 + n ( j −)2 / (p−1)∑
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3. MSBG represents variation among participants
who have been treated differently—received
different treatment levels.
4. MSWG represents variation among participants
who have been treated the same—received
the same treatment level.
5. F = MSBG/MSWG values close to 1 suggest that
the treatment levels did not affect the dependent
variable; large values suggest that the treatment
levels had an effect.
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III Completely Randomized Design (CR-p Design)
A. Characteristics of a CR-p Design
1. Design has one treatment, treatment A, with p
levels.
2. N = n1 + n2 + . . . + np participants are randomly
assigned to the p treatment levels.
3. It is desirable, but not necessary, to have the same
number of participants in each treatment level.
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B. Comparison of layouts for a t-test design for independent samples and a CR-3 design
Participant1 a1 Participant1 a1
Participant2 a1 Participant2 a1
Participant10 a1 Participant10 a1
Participant11 a2 Participant11 a2
Participant12 a2 Participant12 a2
Participant20 a2 Participant20 a2
Participant21 a3
Participant22 a3
Participant30 a3
Treat. Treat.
level level
M M
M
M
M
M
X.1
X.2
X.1
X.2
X.3
M
M M
M
M
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C. Descriptive Statistics for Weight-Loss Data In Table 1
Table 2. Means and Standard Deviations for Weight-Loss Data
Diet
a1 a2 a3
8.00 9.00 12.00
2.21 2.21 2.31
X. j
σ j
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4 6 8 1 0 1 2 1 4 1 6
a2
a1
O n e - M o n t h W e i g h t L o s s
a3
Figure 1. Stacked box plots for the weight-loss data. The distributions are relatively symmetrical and have similar dispersions.
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Table 3. Computational Procedures for CR-3 Design
a1 a2 a3
7 10 12
9 13 11
8 9 15
6 7 14
X ij =80∑ 90 120
X ij =7 +9 +8+L +14 =290.000∑∑
X ij
2 =[ AS] =(7)2 + (9)2 + (8)2 +L + (14)2 =3026.000∑∑
( X ij )
2 / np =[ X] =(290)2 / (10)(3) =2803.333∑∑
M M M
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X iji=1
n∑
⎛
⎝⎜⎞
⎠⎟
2
nj=1
p∑ =[ A] =
(80)2
10+
(90)2
10+
(120)2
10=2890.000
SSTO =[ AS] −[ X] =3026.000 −2803.333=222.667
SSBG =[ A] −[ X] =2890.000 −2803.333=86.667
SSWG =[ AS] −[ A] =3026.000 −2890.000 =136.000
D. Sum of Squares Formulas for CR-3 Design
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Table 4. ANOVA Table for Weight-Loss Data
Source SS df MS F
1. Between 86.667 p – 1 = 2 43.334 8.60*groups (BG)Three diets
2. Within 136.000 p(n – 1) = 27 5.037groups (WG)
3. Total 222.667 np – 1 = 29
*p < .002
1
2
⎡
⎣⎢
⎤
⎦⎥
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E. Assumptions for CR-p Design
1. The model equation,
reflects all of the sources of variation that affect
Xij.
2. Random sampling or random assignment
3. The j = 1, . . . , p populations are normally
distributed.
4. Variances of the j = 1, . . . , p populations are
equal.
X ij = + ( j −) + (Xij − j ),