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

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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 ),