Topic 20: Single Factor Analysis of Variance

Outline

• Analysis of Variance–One set of treatments (i.e., single

factor)• Cell means model• Factor effects model

–Link to linear regression using indicator explanatory variables

One-Way ANOVA

• The response variable Y is continuous

• The explanatory variable is categorical

– We call it a factor

– The possible values are called levels

• This approach is a generalization of the independent two-sample pooled t-test

• In other words, it can be used when there are more than two treatments

Data for One-Way ANOVA

• Y is the response variable

• X is the factor (it is qualitative/discrete)

– r is the number of levels

– often refer to these levels as groups or treatments

• Yi,j is the jth observation in the ith group

Notation• For Yi,j we use

– i to denote the level of the factor– j to denote the jth observation at factor

level i• i = 1, . . . , r levels of factor X

• j = 1, . . . , ni observations for level i of factor

X

– ni does not need to be the same in each group

KNNL Example (p 685)• Y is the number of cases of cereal sold

• X is the design of the cereal package

– there are 4 levels for X because there are 4 different package designs

• i =1 to 4 levels

• j =1 to ni stores with design i (ni=5,5,4,5)

• Will use n if ni the same across groups

Data for one-way ANOVA

data a1; infile 'c:../data/ch16ta01.txt'; input cases design store;

proc print data=a1; run;

The data

Obs cases design store1 11 1 12 17 1 23 16 1 34 14 1 45 15 1 56 12 2 17 10 2 28 15 2 3

Plot the data

symbol1 v=circle i=none;proc gplot data=a1; plot cases*design;run;

The plot

Plot the means

proc means data=a1; var cases; by design; output out=a2 mean=avcases;proc print data=a2;symbol1 v=circle i=join;proc gplot data=a2; plot avcases*design;run;

New Data Set

Obs design _TYPE_ _FREQ_ avcases1 1 0 5 14.6

2 2 0 5 13.4

3 3 0 4 19.5

4 4 0 5 27.2

Plot of the means

The Model

• We assume that the response variable is – Normally distributed with a

1. mean that may depend on the level of the factor

2. constant variance • All observations assumed independent• NOTE: Same assumptions as linear

regression except there is no assumed linear relationship between X and E(Y|X)

Cell Means Model

• A “cell” refers to a level of the factor

• Yij = μi + εij

– where μi is the theoretical mean or expected value of all observations at level (or cell) i

– the εij are iid N(0, σ2) which means

– Yij ~N(μi, σ2) and independent

– This is called the cell means model

Parameters• The parameters of the model are

– μ1, μ2, … , μr

–σ2

• Question (Version 1) – Does our explanatory variable help explain Y?

• Question (Version 2) – Do the μi vary?

H0: μ1= μ2= … = μr = μ (a constant)

Ha: not all μ’s are the same

Estimates• Estimate μi by the mean of the

observations at level i, (sample mean)

• ûi = = ΣYi,j/ni

• For each level i, also get an estimate of the variance

• = Σ(Yij- )2/(ni-1) (sample variance)

• We combine these to get an overall estimate of σ2

• Same approach as pooled t-test

iY

iY

iY2is

2is

Pooled estimate of σ2

• If the ni were all the same we would average the – Do not average the si

• In general we pool the , giving weights proportional to the df, ni -1

• The pooled estimate is

2is

2is

)(1

112

22

rnsn

nsns

Tii

iii

Running proc glm

proc glm data=a1; class design; model cases=design; means design; lsmeans designrun;

Difference 1: Need to specify factor variables

Difference 2: Ask for mean estimates

Output

Class Level Information

Class Levels Valuesdesign 4 1 2 3 4

Number of Observations Read 19Number of Observations Used 19

Important summaries to check these summaries!!!

SAS 9.3 default output for MEANS statement

MEANS statement output

Level ofdesign N

cases

Mean Std Dev1 5 14.6000000 2.302172892 5 13.4000000 3.646916513 4 19.5000000 2.645751314 5 27.2000000 3.96232255

Table of sample means and sample variances

SAS 9.3 default output for LSMEANS statement

LSMEANS statement output

design cases LSMEANStandard

Error Pr > |t|1 14.6000000 1.4523544 <.00012 13.4000000 1.4523544 <.00013 19.5000000 1.6237816 <.00014 27.2000000 1.4523544 <.0001

Provides estimates based on model (i.e., constant variance)

Notation

i iT

T

i j Tij

ij ij

nn

n

nYY

nY

..

i.

nsobservatio ofnumber total the is

mean) sample (grand /

mean) sample(trt /Y

ANOVA Table

Source df SS MS

Model r-1 Σij( - )2 SSR/dfR

Error nT-r Σij(Yij - )2 SSE/dfE

Total nT-1 Σij(Yij - )2 SST/dfT..Y

..Yi.Y

i.Y

ANOVA SAS Output

Source DFSum of

SquaresMean

SquareF

Value Pr > FModel 3 588.2210526 196.0736842 18.59 <.0001

Error 15 158.2000000 10.5466667    

Corrected Total

18 746.4210526      

R-Square Coeff Var Root MSE cases Mean0.788055 17.43042 3.247563 18.63158

Expected Mean Squares

• E(MSR) > E(MSE) when the group means are different

• See KNNL p 694 – 698 for more details• In more complicated models, these tell

us how to construct the F test

Ti ii

i ii

nn

rnE

E

/ where

1)MSR(

)MSE(

.

2.

2

2

F test

• F = MSR/MSE

• H0: μ1 = μ2 = … = μr

• Ha: not all of the μi are equal

• Under H0, F ~ F(r-1, nT-r)

• Reject H0 when F is large

• Report the P-value

Maximum Likelihood Approach

proc glimmix data=a1;

class design;

model cases=design / dist=normal;

lsmeans design;

run;

GLIMMIX OutputModel Information

Data Set WORK.A1

Response Variable cases

Response Distribution Gaussian

Link Function Identity

Variance Function Default

Variance Matrix Diagonal

Estimation Technique Restricted Maximum Likelihood

Degrees of Freedom Method Residual

GLIMMIX Output

Fit Statistics-2 Res Log Likelihood 84.12AIC (smaller is better) 94.12AICC (smaller is better) 100.79BIC (smaller is better) 97.66CAIC (smaller is better) 102.66HQIC (smaller is better) 94.08Pearson Chi-Square 158.20Pearson Chi-Square / DF 10.55

GLIMMIX OutputType III Tests of Fixed Effects

EffectNum

DFDen DF F Value Pr > F

design 3 15 18.59 <.0001

design Least Squares Means

design EstimateStandard

Error DF t Value Pr > |t|1 14.6000 1.4524 15 10.05 <.00012 13.4000 1.4524 15 9.23 <.00013 19.5000 1.6238 15 12.01 <.00014 27.2000 1.4524 15 18.73 <.0001

Factor Effects Model

• A reparameterization of the cell means model

• Useful way at looking at more complicated models

• Null hypotheses are easier to state

• Yij = μ + i + εij

– the εij are iid N(0, σ2)

Parameters

• The parameters of the model are

– μ, 1, 2, … , r

– σ2

• The cell means model had r + 1 parameters– r μ’s and σ2

• The factor effects model has r + 2 parameters– μ, the r ’s, and σ2

– Cannot uniquely estimate all parameters

An example

• Suppose r=3; μ1 = 10, μ2 = 20, μ3 = 30

• What is an equivalent set of parameters for the factor effects model?

• We need to have μ + i = μi

• μ = 0, 1 = 10, 2 = 20, 3 = 30

• μ = 20, 1 = -10, 2 = 0, 3 = 10

• μ = 5000, 1 = -4990, 2 = -4980, 3 = -4970

Problem with factor effects?• These parameters are not estimable

or not well defined (i.e., unique)• There are many solutions to the least

squares problem• There is an X΄X matrix for this

parameterization that does not have an inverse (perfect multicollinearity)

• The parameter estimators here are biased (SAS proc glm)

Factor effects solution

• Put a constraint on the i

• Common to assume Σi i = 0

• This effectively reduces the number of parameters by 1

• Numerous other constraints possible

Consequences• Regardless of constraint, we always have μi = μ + i

• The constraint Σi i = 0 implies

– μ = (Σi μi)/r (unweighted grand mean)

i = μi – μ (group effect)

• The “unweighted” complicates things when the ni are not all equal; see KNNL p 702-708

Hypotheses

• H0: μ1 = μ2 = … = μr

• H1: not all of the μi are equal

are translated into

• H0: 1 = 2 = … = r = 0

• H1: at least one i is not 0

Estimates of parameters

• With the constraint Σi i = 0

.i.i

..i i..

ˆYˆ

) (if YYˆ

nnr i

Solution used by SAS

• Recall, X΄X does not have an inverse

• We can use a generalized inverse in its place

• (X΄X)- is the standard notation

• There are many generalized inverses, each corresponding to a different constraint

Solution used by SAS

• (X΄X)- used in proc glm corresponds to the constraint r = 0

• Recall that μ and the i are not estimable

• But the linear combinations μ + i are estimable

• These are estimated by the cell means

Cereal package example

• Y is the number of cases of cereal sold

• X is the design of the cereal package

• i =1 to 4 levels

• j =1 to ni stores with design i

SAS coding for X•Class statement generates r explanatory variables •The ith explanatory variable is equal to 1 if the observation is from the ith group•In other words, the rows of X are 1 1 0 0 0 for design=1 1 0 1 0 0 for design=2 1 0 0 1 0 for design=3 1 0 0 0 1 for design=4

Some options

proc glm data=a1; class design; model cases=design /xpx inverse solution;run;

Output The X'X Matrix

Int d1 d2 d3 d4 casesInt 19 5 5 4 5 354d1 5 5 0 0 0 73d2 5 0 5 0 0 67d3 4 0 0 4 0 78d4 5 0 0 0 5 136cases 354 73 67 78 136 7342

Also contains X’Y

Output

X'X Generalized Inverse (g2)

Int d1 d2 d3 d4 casesInt 0.2 -0.2 -0.2 -0.2 0 27.2d1 -0.2 0.4 0.2 0.2 0 -12.6d2 -0.2 0.2 0.4 0.2 0 -13.8d3 -0.2 0.2 0.2 0.45 0 -7.7d4 0 0 0 0 0 0cases 27.2 -12.6 -13.8 -7.7 0 158.2

Output matrix•Actually, this matrix is

(X΄X)- (X΄X)- X΄Y Y΄X(X΄X)- Y΄Y-Y΄X(X΄X)- X΄Y

•Parameter estimates are in upper right corner, SSE is lower right corner (last column on previous page)

Parameter estimates

StPar Est Err t PInt 27.2 B 1.45 18.73 <.0001d1 -12.6 B 2.05 -6.13 <.0001d2 -13.8 B 2.05 -6.72 <.0001d3 -7.7 B 2.17 -3.53 0.0030d4 0.0 B . . .

Caution Message

NOTE: The X'X matrix has beenfound to be singular, and ageneralized inverse was usedto solve the normal equations.Terms whose estimates arefollowed by the letter 'B' arenot uniquely estimable.

Interpretation

• If r = 0 (in our case, 4 = 0), then the corresponding estimate should be zero

• the intercept μ is estimated by the mean of the observations in group 4

• since μ + i is the mean of group i, the i are the differences between the mean of group i and the mean of group 4

Recall the means output

Level ofdesign N Mean Std Dev

1 5 14.6 2.32 5 13.4 3.63 4 19.5 2.64 5 27.2 3.9

Parameter estimates based on means

Level ofdesign Mean = 27.2 = 27.21 14.6 = 14.6-27.2 = -12.62 13.4 = 13.4-27.2 = -13.83 19.5 = 19.5-27.2 = -7.74 27.2 = 27.2-27.2 = 0

1234

Last slide

• Read KNNL Chapter 16 up to 16.10• We used programs topic20.sas to generate the

output for today• Will focus more on the relationship between

regression and one-way ANOVA in next topic