# Chapter 15 Introduction to the Analysis of Variance IThe Omnibus Null Hypothesis - PowerPoint PPT Presentation

1 / 23

Chapter 15 Introduction to the Analysis of Variance IThe Omnibus Null Hypothesis H 0 :  1 =  2 = . . . =  p H 1 :  j =  j ’. A.Answering General Versus Specific Research Questions 1. Population contrast,  i , and sample contrast.

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

Chapter 15 Introduction to the Analysis of Variance IThe Omnibus Null Hypothesis

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

#### Presentation Transcript

Chapter 15

Introduction to the Analysis of Variance

IThe Omnibus Null Hypothesis

H0: 1 = 2 = . . . = p

H1: j = j’

Questions

1. Population contrast, i, and sample contrast

2.Pairwise and nonpairwise contrasts

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

3.As C increases, the probability of making one or

more Type I errors using a t statistic increases

dramatically.

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.

IIBasic 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.

a1 a2 a3 a4

X11 X12 X13 X14

X21 X22 X23 X24

Xn1 Xn2 Xn3 Xn4

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

2.Sample model equation for a score

3.The statistics estimate parameters of the model

equation as follows

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

71012

91311

8915

6714

5.Let X11 = 7 denote Joan’s weight loss. She used

diet a1. Her score is a composite that tells a story.

6.Joan used a less effective diet than other girls

(8 – 9.67 = –1.67), and she lost less weight than

other girlson the same diet (8 – 9 = –1).

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)

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

F.Nature of MSBG and MSWG

1.Expected value of MSBG and MSWG when the

null hypothesis is true.

2.Expected value of MSBG and MSWG when the

null hypothesis is false.

3.MSBG represents variation among participants

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

IIICompletely 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.

B.Comparison of layouts for a t-test design for

independent samples and a CR-3 design

Participant1a1Participant1a1

Participant2a1Participant2a1

Participant10a1Participant10a1

Participant11a2Participant11a2

Participant12a2Participant12a2

Participant20a2Participant20a2

Participant21 a3

Participant22a3

Participant30 a3

C.Descriptive Statistics for Weight-Loss Data

In Table 1

Table 2. Means and Standard Deviations for Weight-Loss Data

Diet

a1a2a3

8.009.0012.00

2.212.21 2.31

Figure 1. Stacked box plots for the weight-loss data. The

distributions are relatively symmetrical and have similar

dispersions.

Table 3. Computational Procedures for CR-3 Design

a1 a2 a3

71012

91311

8915

6714

D.Sum of Squares Formulas for CR-3 Design

Table 4. ANOVA Table for Weight-Loss Data

SourceSS df MS F

1.Between86.667p – 1 = 243.3348.60*groups (BG)

Three diets

2.Within 136.000p(n – 1) = 275.037

groups (WG)

3. Total222.667np – 1 = 29

*p < .002

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.