Chapter 14Repeated Measures andTwo Factor Analysis of Variance

PowerPoint Lecture Slides

Essentials of Statistics for the Behavioral Sciences Seventh Edition

by Frederick J. Gravetter and Larry B. Wallnau

Chapter 14 Learning Outcomes

Concepts to review

• Independent-measures analysis of variance(Chapter 13)

• Repeated measures designs (Chapter 11)

• Individual differences

14.1 Overview

• Analysis of Variance– Evaluated mean differences of two or

groups

• Complex Analysis of Variance– Samples are related not independent

(Repeated measures ANOVA)– More than one factor is tested

(Factorial ANOVA, here Two-Factor)

14.2 Repeated Measures ANOVA

• Advantages of repeated measures designs– Individual differences among participants do

not influence outcomes– Smaller number of subjects needed to test all

the treatments• Repeated Measures ANOVA

– Compares two or more treatment conditions with the same subjects tested in all conditions

– Studies same group of subjects at two or more different times.

Hypotheses for repeated measures ANOVA

• Null hypothesis: in the population, there are no mean differences among the treatment groups

• Alternate hypothesis states that there are mean differences among the treatment groups.

...: 3210 H

H1: At least one treatment mean μ is different from another

Individual Differences in the Repeated Measures ANOVA

• F ratio based on variances– Same structure as independent measures– Variance due to individual differences is

not present

effect treatment no withexpected es)(differenc variance

treatments between es)(differenc varianceF

Individual differences

• Participant characteristics that vary from one person to another.– Not systematically present in any treatment

group or by research design

• Characteristics may influence measurements on the outcome variable– Eliminated from the numerator by the

research design– Must be removed from the denominator

statistically

Logic of repeated measures ANOVA

• Numerator of the F ratio includes– Systematic differences caused by treatments– Unsystematic differences caused by random

factors (reduced because same individuals in all treatments)

• Denominator estimates variance reasonable to expect from unsystematic factors– Effect of individual differences is removed– Residual (error) variance remains

Figure 14.1 Structure of the Repeated-Measures ANOVA

Stage One Repeated-Measures ANOVA equations

N

GXSStotal

22

treatment each insidetreatmentswithin SSSS

N

G

n

TSS treatmentsbetween

22

Two Stages of the Repeated-Measures ANOVA

• First stage– Identical to independent samples ANOVA

– Compute SSTotal, SSBetween treatments and SSWithin treatments

• Second stage– Removing the individual differences from the

denominator

– Compute SSBetween subjects and subtract it from SSWithin treatments to find SSError

Stage Two Repeated-Measures ANOVA equations

N

G

k

PSS subjectsbetween

22

subjectsbetweentreatments withinerror SSSSSS

Degrees of freedom for Repeated Measures ANOVA

dftotal = N – 1

dfwithin treatments = Σdfinside each treatment

dfbetween treatments = k – 1

dfbetween subjects = n – 1

dferror = dfwithin treatments – dfbetween subjects

Mean squares and F-ratio for Repeated-Measures ANOVA

error

errorerror df

SSMS

treatments between

treatments betweenntstreatme between df

SSMS

error

mentstreat between

MS

MSF

Effect size for the Repeated-Measures ANOVA

• Percentage of variance explained by the treatment differences

• Partial η2 is percentage of variability that has not already been explained by other factors

error

treatments between

subjectsbetweentotal

treatments between

SS

SS

SSSS

SS

2

Post hoc tests with Repeated Measures ANOVA

• Determine exactly where significant differences exist among more than two treatment means– Tukey’s HSD and Scheffé can be used

– Substitute SSerror and dferror in the formulas

Assumptions of the Repeated Measures ANOVA

• The observations within each treatment condition must be independent.

• The population distribution within each treatment must be normal.

• The variances of the population distribution for each treatment should be equivalent.

Learning Check• A researcher obtains an F-ratio with df = 2, 12

from an ANOVA for a repeated-measures research study. How many subjects participated in the study?

Learning Check - Answer• A researcher obtains an F-ratio with df = 2, 12

from an ANOVA for a repeated-measures research study. How many subjects participated in the study?

Learning Check

• Decide if each of the following statements is True or False.

Answer

14.2 Two-Factor ANOVA

• Factorial designs– Consider more than one factor– Joint impact of factors is considered.

• Three hypotheses tested by three F-ratios– Each tested with same basic F-ratio structure

effect treatment no withexpected es)(differenc variance

treatments between es)(differenc varianceF

Main effects

• Mean differences among levels of one factor– Differences are tested for statistical

significance– Each factor is evaluated independently of the

other factor(s) in the study

21

21

:

:

1

0

AA

AA

H

H

21

21

:

:

1

0

BB

BB

H

H

Interactions between factors

• The mean differences between individuals treatment conditions, or cells, are different from what would be predicted from the overall main effects of the factors

• H0: There is no interaction between Factors A and B

• H1: There is an interaction between Factors A and B

Interpreting Interactions

• Dependence of factors– The effect of one factor depends on the level

or value of the other

• Non-parallel lines (cross or converge) in a graph– Indicate interaction is occurring

• Typically called the A x B interaction

Figure 14.2 Graph of group means with and without interaction

Structure of the Two-Factor Analysis

• Three distinct tests– Main effect of Factor A– Main effect of Factor B– Interaction of A and B

• A separate F test is conducted for each

Two Stages of the Two-Factor Analysis of Variance

• First stage– Identical to independent samples ANOVA

– Compute SSTotal, SSBetween treatments and SSWithin treatments

• Second stage– Partition the SSBetween treatments into three

separate components, differences attributable to Factor A, to Factor B, and to the AxB interaction

Figure 14.3 Structure of the Two- Factor Analysis of Variance

Stage One of the Two-Factor Analysis of Variance

N

GXSStotal

22

treatment each insidetreatmentswithin SSSS

N

G

n

TSS treatmentsbetween

22

Stage 2 of the Two Factor Analysis of Variance

• This stage determines the numerators for the three F-ratios by partitioning SSbetween treatments

N

G

n

TSS

row

rowA

22

N

G

n

TSS

col

colB

22

BAtreatments betweenAxB SSSSSSSS

Degrees of freedom for Two-Factor ANOVAdftotal = N – 1

dfwithin treatments = Σdfinside each treatment

dfbetween treatments = k – 1

dfA = number of rows – 1

dfB = number of columns– 1

dferror = dfwithin treatments – dfbetween subjects

Mean squares and F-ratio for the Two-Factor ANOVA

reatmentst within

reatmentst withinreatmentst within df

SSMS

AxB

AxBAxB

B

BB

A

AA df

SSMS

df

SSMS

df

SSMS

within

AxBAxB

within

BB

within

AA MS

MSF

MS

MSF

MS

MSF

Effect Size for Two-Factor ANOVA

• η2, is computed as the percentage of variability not explained by other factors.

treatments withinA

A

AxBBtotal

AA SSSS

SS

SSSSSS

SS

2

treatments withinB

B

AxBAtotal

BB SSSS

SS

SSSSSS

SS

2

treatments withinAxB

AxB

BAtotal

AxBAxB SSSS

SS

SSSSSS

SS

2

Figure 14.4 Sample means for Example 14.3

Assumptions for the Two-Factor ANOVA

• The validity of the ANOVA presented in this chapter depends on three assumptions common to other hypothesis tests1. The observations within each sample must

be independent of each other

2. The populations from which the samples are selected must be normally distributed

3. The populations from which the samples are selected must have equal variances (homogeneity of variance)

Learning Check

• If a two-factor analysis of variance produces a statistically significant interaction, then you can conclude that _____.

Learning Check - Answer

• If a two-factor analysis of variance produces a statistically significant interaction, then you can conclude that _____.

Learning Check

• Decide if each of the following statements is True or False.

Answer