Introduction to Analysis of Variance (ANOVA)

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Introduction to Analysis of Variance (ANOVA)The Structural Model, The

Summary Table, and the One-Way ANOVA

Limitations of the t-Test

• Although the t-Test is commonly used, it has limitations– Can only test differences between 2 groups

• High school class? College year? – Can examine ONLY the effects of 1 IV on 1 DV– Limited to single group OR repeated measures

designs

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• Testing differences between group means– IV: Gender (Male & Female)– IV: High-school class (First-year, Sophomore,

Junior, & Senior)

– Using the t-Test, we must either “collapse”categories… or not run the analysis


• 1 Independent Variable– Gender differences in depression

• IV: Gender (Male & Female)• DV: Level of depression (BDI score)

• 2 Independent Variables– Gender and social support on depression

• IV1: Gender (Male & Female)• IV2: Social support (High, Medium, & Low)• DV: Level of depression (BDI score)

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• 2 or more Independent Variables– Simultaneously examine the impact of 2 or

more IVs on a single DV– Examine how the effects of 2 or more IVs

COMBINE to affect a single DV


• Single time point OR repeated measures designs– 1 group at 2 time points = repeated measures– 2 groups at 1 time point = independent groups

• Single time point AND repeated measures designs– 2 or more groups at 2 or more time points

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The Analysis of Variance (ANOVA)

• The ANOVA can test hypotheses that the t-Test cannot

• Probably the most commonly abused statistical test

• Many varieties of ANOVA– One-Way (between subjects)– Factorial ANOVA (between or within subjects)– Repeated Measures (within subjects)– Mixed-Model (between & within subjects)

Varieties of ANOVA

• One-Way ANOVA– 1 continuous Dependent Variable – 1 Independent Variable consisting of 2 or

more “categorical” groups• The one-way ANOVA with 2 groups is “equivalent”

to the independent groups t-Test

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Varieties of ANOVA

• Factorial ANOVA– 1 continuous Dependent Variable – 2 or more Independent Variables consisting of

2 or more “categorical” groups for each IV• 2 IVs = Two-Way Factorial ANOVA• 3 IVs = Three-Way Factorial ANOVA

– We call these “factorial” designs because EACH level of each IV is paired with EVERY level of ALL other IVs

2 x 2 Contingency Table

Col 2TotalCol 1 Total

Row 2 Total

DATADATALevel 2

Row 1 TotalDATADATALevel 1

IV 1

Level 2Level 1IV 2

Note: Each Level 1 of IV 1 is paired with BOTH Level 1 and Level2 of IV 2

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2 x 2 Contingency Table

Col 2TotalCol 1 Total

Row 2 Total

DATADATAFemale

Row 1 TotalDATADATAMale

Gender

LowHighSocial Support

Note: Each Level 1 of IV 1 is paired with BOTH Level 1 and Level2 of IV 2

Varieties of ANOVA

• Repeated Measures ANOVA– Time points = IV– The DV is assessed at EACH time point

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Varieties of ANOVA

• Mixed-Model ANOVA– 1 continuous Dependent Variable – 1 or more Independent Variables consisting of

2 or more “categorical” groups (between)– 1 Independent Variable consisting of 2 or

more “categorical” time points (within)• The DV is assessed at EACH time point

ANOVA

• ANOVA models we will consider– One-Way ANOVA– Two- and Three-way Factorial ANOVA – Repeated measures ANOVA– Mixed-model ANOVA

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Choosing the Best Test

The Underlying Model

• A statistical model by example:

• Assume: the average 18 year old human being weighs approximately 138 pounds– Men, on average, weigh 12 pounds more than

the average human weight– Women, on average, weigh 10 pounds less

than the average human weight

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• For any given human being, I can break weight down into 3 components:– Average weight for all individual

• 138 lbs– Average weight for each group

• Men: +12 lbs • Women: - 10 lbs

– The individual’s unique difference


• Male weightWeight = 138 lbs + 12 lbs + uniqueness

• Female weight Weight = 138 lbs – 10 lbs + uniqueness

• If you understand this process, you understand the basic theory behind the ANOVA

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Partitioning Variance

• The idea behind the ANOVA test is to divide or separate (partition) variance observed in the data into categories of what we CAN and what we CANNOT explain

Total Variance

Explained Unexplained

The Structural Model

• Mathematically, we partition the total variance of our data using the structural form of the ANOVA model– Xij = µ + τj +εij

– The structural model translates as follows: The score for any single individual is equal to the sum of the population mean plus the mean of the group plus the individual’s unique contribution

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The Structural Model

• For our weight example:– µ = population weight = 138 lbs– τ = group difference in weight = 12 or 10 lbs– ε = “unique” contribution of an individual’s

score

– µ & τ can be explained – ε cannot be explained…

Uniqueness

• Oftentimes, we value our uniqueness…– In statistics, unique variance is BAD– Since we can’t explain unique variance, we

call it “error”– Thus, the ANOVA seeks to examine the

relative proportion of explainable variance in our data to the unexplainable variance

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Assumptions of the ANOVA

• Owing to the mathematical construction of the ANOVA, the underlying assumptions of the test are very important– Homogeneity of variance– Normality– Independence of Observations– The Null Hypothesis

Homogeneity of Variance

• Homogeneity of variance refers to the variance for each group being equal to the variance of every other group– Really, we mean that the variance of each

group is equal to the variance of the error for the total analysis

– σ12 = σ2

2 = σ32 = σj

2 = σe2

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Homogeneity of Variance

• Heterogeneity of variance is another BAD thing– Heterogeneous variances can greatly

influence the results you obtain, making it either more or less likely that you will reject H0

– Visual inspection of variances– Tests of homogeneity of variance

Normality

• The ANOVA procedure assumes that scores are normally distributed– More accurately, it assumes that ERRORS

are normally distributed– Random sampling and random assignment– Lacking normality, consider mathematical

transformations• Logarithmic & square root transformations

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Independence of Observations

• Simple: The scores for 1 group are not dependent on the scores from another group– Don’t share subjects between groups– If violated…

• What is wrong with your experimental design?• Are you using the appropriate test?

The Null Hypothesis

• Less an assumption and more a theoretical point:– H0: µ1 = µ2 = µ3 = µ4= µ5

• This is almost ALWAYS the basic form of your null hypothesis…

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Calculating the One-Way ANOVA

• In order to calculate the One-Way ANOVA statistic, we need to complete a number of intermediate steps

• Because there are several intermediate steps, we keep track of our progress with something called a summary table

The Summary Table

Total

Error

Treatment

FMean Square (MS)

Sum of Squares (SS)

dfSource

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One-Way ANOVA: Partitioning Variance

• The idea behind the ANOVA test is to divide or separate (partition) variance observed in the data into categories of what we CAN and what we CANNOT explain

Total Variance

Treatment Error

The Summary Table

(N-1)Total

k(n-1)Error

(k-1)Treatment

FMean Square (MS)

Sum of Squares (SS)

dfSource

Note: dfTreatment + dfError =dfTotal

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Sum of Squares

• Note:– = The treatment group mean– = The grand mean (mean of all scores)– Xij = Each individual score

jxx..

The Summary Table

(N-1)Total

k(n-1)Error

(k-1)Treatment

FMean Square (MS)

Sum of Squares (SS)

dfSource

2.. )( XXnSS jTreatment ∑ −=

TreatmentTotalError SSSSSS −=

2.. )( XXSS ijTotal ∑ −=

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The Summary Table

(N-1)Total

k(n-1)Error

(k-1)Treatment

FMean Square (MS)

Sum of Squares (SS)

dfSource




Treatment

TreatmentTreatment df

SSMS =

Error

ErrorError df

SSMS =

The Summary Table

(N-1)Total

k(n-1)Error

(k-1)Treatment

FMean Square (MS)

Sum of Squares (SS)

dfSource




Treatment

TreatmentTreatment df

SSMS =

Error

ErrorError df

SSMS =

Error

Treatment

MSMS

F =

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Example:Anorexia Nervosa 3 Group Tx

2

3

6-4

0-1.5-10

24-3

921

6-1-8

3.50-5

302

42-2

0-2-1.5

4-10

61-6

40-5

CBTIPTControl


Descriptives

Change in Weight

12 -3.4583 3.60214 1.03985 -5.7470 -1.1696

11 .3182 1.79266 .54051 -.8861 1.5225

14 3.7500 2.45537 .65623 2.3323 5.1677

37 .3919 4.04512 .66501 -.9568 1.7406

Control

IPT

CBT

Total

N Mean Std. Deviation Std. Error Lower Bound Upper Bound

95% Confidence Interval forMean

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Test of Homogeneity of Variances

Change in Weight

2.666 2 34 .084Levene Statistic df1 df2 Sig.

H0: σ1 = σ2 = σ3

H1: σ1 ≠ σ2 ≠ σ3

Fail to reject H0 . . .


ANOVA

Change in Weight

335.827 2 167.914 22.544 .000

253.241 34 7.448

589.068 36

Between Groups

Within Groups

Total

Sum of Squares df Mean Square F Sig.

F(2,34) = 22.54, p < .05

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2

3

6-4

0-10

2-3

91

6-8

3.5-5

32

4-2

0-1.5

40

6-6

4-5

CBTControl


Descriptives

Change in Weight

12 -3.4583 3.60214 1.03985 -5.7470 -1.1696 -10.00 2.00

14 3.7500 2.45537 .65623 2.3323 5.1677 .00 9.00

26 .4231 4.71952 .92557 -1.4832 2.3293 -10.00 9.00

Control

CBT

Total



Minimum Maximum

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Test of Homogeneity of Variances

Change in Weight

2.281 1 24 .144Levene Statistic df1 df2 Sig.


H0: σ1 = σ2 = σ3

H1: σ1 ≠ σ2 ≠ σ3

Fail to reject H0

ANOVA

Change in Weight

335.742 1 335.742 36.443 .000

221.104 24 9.213

556.846 25

Between Groups

Within Groups

Total



F(1,24) = 36.44, p < .05

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• Independent Groups t-test

Group Statistics

12 -3.4583 3.60214 1.03985

14 3.7500 2.45537 .65623

Experimental GroupControl

CBT

Change in WeightN Mean Std. Deviation Std. Error Mean

• One-way ANOVADescriptives

Change in Weight

12 -3.4583 3.60214 1.03985 -5.7470 -1.1696

14 3.7500 2.45537 .65623 2.3323 5.1677

26 .4231 4.71952 .92557 -1.4832 2.3293

Control

CBT

Total




• Independent groups t-testIndependent Samples Test

2.281 .144 -6.037 24 .000 -7.20833 1.19406

-5.862 18.962 .000 -7.20833 1.22960

Equal variances assumed

Equal variances not assumed

Change in WeightF Sig.

Levene's Test for Equalityof Variances

t df Sig. (2-tailed) Mean DifferenceStd. ErrorDifference

t-test for Equality of Means

• One-way ANOVAANOVA

Change in Weight

335.742 1 335.742 36.443 .000

221.104 24 9.213

556.846 25

Between Groups

Within Groups

Total


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• Shouldn’t the results of the F- and t-tests be equal if the tests are equivalent?– F(1,24) = 36.44, p < .05– t(24) = -6.04, p < .05

• t2 = F

Documents

Introduction to Analysis of Variance (ANOVA)