What If There Are More Than
Two Factor Levels?
• Chapter 3
• Comparing more that two factor levels…the analysis of
variance
• ANOVA decomposition of total variability
• Statistical testing & analysis
• Checking assumptions, model validity
• Post-ANOVA testing of means
ANOVA_EXAMPLE P60 Brainerd
1
What If There Are More Than
Two Factor Levels?
• The t-test does not directly apply
• There are lots of practical situations where there are either
more than two levels of interest, or there are several factors
of simultaneous interest
• The analysis of variance (ANOVA) is the appropriate
analysis “engine” for these types of experiments – Chapter
3, textbook
• The ANOVA was developed by Fisher in the early 1920s,
and initially applied to agricultural experiments
• Used extensively today for industrial experiments
ANOVA_EXAMPLE P60 Brainerd
2
An Example (See pg. 60)
• Consider an investigation into the formulation of a new
“synthetic” fiber that will be used to make cloth for shirts
• The response variable is tensile strength
• The experimenter wants to determine the “best” level of
cotton (in wt %) to combine with the synthetics
• Cotton content can vary between 10 – 40 wt %; some nonlinearity in the response is anticipated
• The experimenter chooses 5 levels of cotton “content”;
15, 20, 25, 30, and 35 wt %
• The experiment is replicated 5 times – runs made in
random order
ANOVA_EXAMPLE P60 Brainerd
3
ANOVA: Design of Experiments
Chapter 3
•A
A
productdevelopment
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isisinterested
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and 35 percent.
that are desired.
•The engineer decides to test specimens at five levels of cotton weight
percent: 15 percent, 20 percent, 25 percent, 30 percent, and 35 percent. He
also decides to test five specimens at each level of cotton content.
ANOVA_EXAMPLE P60 Brainerd
4
ANOVA: Design of Experiments
Chapter 3
H 0 : µ1 = µ 2 = µ 3 = ... = µ i
H a : At least one µ i is different
Tensile
Strength
EXAMPLE PROBLEM
A single-factor experiment with a = 5 levels
of the factor and n = 5 replicates. The 25
runs should be made in random order.
15
20
25 30
35
ANOVA_EXAMPLE P60 Brainerd
Cotton %
5
ANOVA: Design of Experiments
Chapter 3
H 0 : µ1 = µ 2 = µ 3 = ... = µ i
H a : At least one µ i is different
Tensile
Strength
ONE FROM EACH
15 20 25 30 35
Can we determine which is better?
ANOVA_EXAMPLE P60 Brainerd
Cotton %
6
ANOVA: Design of Experiments
Chapter 3
H 0 : µ1 = µ 2 = µ 3 = ... = µ i
H a : At least one µ i is different
REPLICATION
Tensile
Strength
15
20
25 30
35
Cotton %
What does replication provide?
ANOVA_EXAMPLE P60 Brainerd
7
ANOVA: Design of Experiments
Chapter 3
H 0 : µ1 = µ 2 = µ 3 = ... = µ i
H a : At least one µ i is different
REPLICATION
Effect
Tensile
Strength
If the sample mean is used
to estimate the effect of a
factor in the experiment,
then replication permits
the experimenter to obtain
a more precise estimate of
this effect.
15
20
25 30
35
What else does replication provide?
ANOVA_EXAMPLE P60 Brainerd
Cotton %
8
ANOVA: Design of Experiments
Chapter 3
H 0 : µ1 = µ 2 = µ 3 = ... = µ i
H a : At least one µ i is different
REPLICATION
Effect
Error
Tensile
Strength
Allows the experimenter to
obtain an estimate of the
2
experimental error. This
2
estimate if error becomes a
xi of measurement
basic unit
for determining whether
observed differences in the
data are really statistically
different. Cotton %
σ
σ
=
n
15 20 25 30 35
What assumption does the error estimate depend upon?
ANOVA_EXAMPLE P60 Brainerd
9
ANOVA: Design of Experiments
Chapter 3
H 0 : µ1 = µ 2 = µ 3 = ... = µ i
H a : At least one µ i is different
REPLICATION
Effect
Error
RANDOMIZATION
Tensile
Strength
Cotton %
15
20
25 30
35
ANOVA_EXAMPLE P60 Brainerd
Both the allocation of the
experimental material and the
order in which the individual
runs or trials of the experiment
are to be performed are
randomly determined.
10
ANOVA: Design of Experiments
Chapter 3
Need to randomize run order
COTTON% Experimental Run #
15
20
25
30
35
1
6
11
16
21
2
7
12
17
22
3
8
13
18
23
ANOVA_EXAMPLE P60 Brainerd
4
9
14
19
24
5
10
15
20
25
11
ANOVA: Design of Experiments
Chapter 3
Test Seq Run #
%
1
8 20
2
18 30
3
10 20
4
23 35
5
17 30
*
*
*
25
3 15
RANDOMIZE RUNS
ANOVA_EXAMPLE P60 Brainerd
12
ANOVA: Design of Experiments
Chapter 3
H 0 : µ1 = µ 2 = µ 3 = ... = µ i
H a : At least one µ i is different
REPLICATION
Effect
Error
RANDOMIZATION
Tensile
Strength
Cotton %
15 20 25 30 35
What if the measurements varied widely because of human operators?
ANOVA_EXAMPLE P60 Brainerd
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ANOVA: Design of Experiments
Chapter 3
• PUSHUP EXAMPLE
– Test if one can do push ups better in the
morning or afternoon. 20 DATA POINTS
– Select 40 people at random
Is PM really
better than AM?
AM
PM
ANOVA_EXAMPLE P60 Brainerd
14
ANOVA: Design of Experiments
Chapter 3
•
•
•
What if the PM group of 20 was in better shape then the AM group of
20?
What if the test was conducted on a Monday morning?
What if different people counted the push ups between AM and PM?
Is PM really
better than AM?
AM
PM
ANOVA_EXAMPLE P60 Brainerd
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ANOVA: Design of Experiments
Chapter 3
Controllable Factors
Input
PROCESS
Output
Uncontrollable Factors
ANOVA_EXAMPLE P60 Brainerd
16
ANOVA: Design of Experiments
Chapter 3
• BLOCKING
– Used to limit the uncontrollable factors
– Therefore increase precision
• PUSH UP EXAMPLE
– Paired Data = BLOCKING
– Have same person do AM and PM
– You are investigating AM vs PM not which group can
do more pushups.
– Randomly Sort experiment by Days of the Week and
have one grader
ANOVA_EXAMPLE P60 Brainerd
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THE THREE PRINCIPLES of Experimental
Design
H 0 : µ1 = µ 2 = µ 3 = ... = µ i
H a : At least one µ i is different
Tensile
Strength
REPLICATION
RANDOMIZATION
BLOCKING
A block is a portion of the
experimental material that
should be more homogeneous
than the entire set of material.
15
20
25 30
35
ANOVA_EXAMPLE P60 Brainerd
Cotton %
18
Analysis of Variance (ANOVA)
H 0 : µ1 = µ 2 = µ 3 = ... = µ i
H a : At least one µ i is different
Variation
Between
Samples
Tensile
Strength
Variation
Within
Samples
15
20
25 30
35
ANOVA_EXAMPLE P60 Brainerd
Cotton %
19
Analysis of Variance (ANOVA)
When very different
Tensile
Strength
Between Sample
Variation Large
Within Sample
Variation Small
15
20
25 30
35
ANOVA_EXAMPLE P60 Brainerd
Cotton %
20
Analysis of Variance (ANOVA)
When near equal
Tensile
Strength
15
20
25 30
35
Cotton %
Between Sample Variation near equal to
Within Sample Variation
ANOVA_EXAMPLE P60 Brainerd
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Analysis of Variance (ANOVA)
Between Sample Variation
TEST STAT =
Within Sample Variation
F-TEST
REQUIRED ASSUMPTION
All data is normal with equal variance
EXTENSION OF TWO SAMPLE POOLED t
ANOVA_EXAMPLE P60 Brainerd
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Analysis of Variance (ANOVA)
SETUP (i = factor; j = replicate)
Replicates
Level
1
2
3
*
i
xi , j
Mean SD
1
2
3
*
j
11
21
31
*
i1
12
22
32
*
i2
13
23
33
*
i3
*
*
*
*
*
1j 1*
2j 2*
3j 3*
*
ij
**
= observed value i th level, jth measurement
ANOVA_EXAMPLE P60 Brainerd
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Analysis of Variance (ANOVA)
COTTON EXAMPLE
Replicates
Level
1
2
3
4
5
1
2
3
4
5
7
12
14
19
7
7
17
18
25
10
15
12
18
22
11
11
18
19
19
15
9
18
19
23
11
Mean
SD
9.80
3.35
3.13
2.07
2.61
2.86
15.40
17.60
21.60
10.80
15.04
xi• = sample mean i th level
x•• = Grand Mean
ANOVA_EXAMPLE P60 Brainerd
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Analysis of Variance (ANOVA)
I = # factors and J = # replicates
Mean Square Treatment ≡ Between Sample Variation
J
2
MSTr =
( X i • − X •• )
∑
I −1 i
Mean Square Error ≡ Within Sample Variation
S + S + S + ....S
MSE =
I
2
1
2
2
2
3
ANOVA_EXAMPLE P60 Brainerd
2
I
25
Analysis of Variance (ANOVA)
H 0 : µ1 = µ 2 = µ 3 = ... = µ i
H a : At least one µ i is different
Between Sample Variation
TEST STAT =
Within Sample Variation
MSTr
f =
MSE
Fα , I −1, I ( J −1)
If H 0 true
E ( MSTr ) = E ( MSE ) = σ 2
If H 0 false
E ( MSTr ) > E ( MSE ) = σ 2
ANOVA_EXAMPLE P60 Brainerd
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Anova Chapter 3 (See pg. 62)
•
•
Does changing the
cotton weight percent
change the mean
tensile strength?
Is there an optimum
level for cotton
content?
ANOVA_EXAMPLE P60 Brainerd
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The Analysis of Variance
SST = SSTreatments + SS E
• A large value of SSTreatments reflects large differences in
treatment means
• A small value of SSTreatments likely indicates no differences in
treatment means
• Formal statistical hypotheses are:
H 0 : µ1 = µ 2 = L = µ a
H1 : At least one mean is different
ANOVA_EXAMPLE P60 Brainerd
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The Analysis of Variance
• While sums of squares cannot be directly compared to test
the hypothesis of equal means, mean squares can be
compared. ( MS = Estimates of variances)
• A mean square is a sum of squares divided by its degrees
of freedom:
dfTotal = dfTreatments + df Error
an − 1 = a − 1 + a ( n − 1)
SSTreatments
SS E
MSTreatments =
, MS E =
a −1
a ( n − 1)
• If the treatment means are equal, the treatment and error
mean squares will be (theoretically) equal.
• If treatment means differ, the treatment mean square will
be larger than the error mean square.
ANOVA_EXAMPLE P60 Brainerd
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The Analysis of Variance is
Summarized in a Table
•
•
•
Computing…see text, pp 70 – 73
The reference distribution for F0 is the Fa-1, a(n-1) distribution
Reject the null hypothesis (equal treatment means) if
F0 > Fα ,a −1,a ( n −1)
ANOVA_EXAMPLE P60 Brainerd
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Analysis of Variance (ANOVA): Excel Analysis:
ANOVA Single Factor
Setup data in EXCEL Spreadsheet in columns as:
15% Cotton
20% Cotton
Tensile Strength Tensile Strength
lb/in2
lb/in2
25% Cotton
30% Cotton
35% Cotton
Tensile
Tensile
Tensile
Strength lb/in2 Strength lb/in2 Strength lb/in2
7
12
14
19
7
7
17
18
25
10
15
12
18
22
11
11
18
19
19
15
9
18
19
23
11
ANOVA_EXAMPLE P60 Brainerd
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Analysis of Variance (ANOVA): Excel Analysis:
ANOVA Single Factor
EXCEL Data Analysis: ANOVA Single Factor
ANOVA_EXAMPLE P60 Brainerd
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Analysis of Variance (ANOVA):
EXCEL Data Analysis: ANOVA Single Factor
ANOVA_EXAMPLE P60 Brainerd
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Analysis of Variance (ANOVA):
EXCEL Data Analysis: ANOVA Single Factor
Output
EXCEL
Anova: Single Factor
SUMMARY
Groups
15% Cotton
Tensile Strength
lb/in2
20% Cotton
Tensile Strength
lb/in2
25% Cotton
Tensile Strength
lb/in2
30% Cotton
Tensile Strength
lb/in2
35% Cotton
Tensile Strength
lb/in2
Count
Sum
Average Variance
5
49
9.8
11.2
5
77
15.4
9.8
5
88
17.6
4.3
5
108
21.6
6.8
5
54
10.8
8.2
ANOVA_EXAMPLE P60 Brainerd
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Analysis of Variance (ANOVA):
EXCEL Data Analysis: ANOVA Single Factor Output
H 0 : µ1 = µ 2 = µ 3 = ... = µ i
If H 0 true
H a : At least one µ i is different
E ( MSTr ) = E ( MSE ) = σ
2
TEST STAT =
If H 0 false
MSTr
f =
MSE
E ( MSTr ) > E ( MSE ) = σ 2
ANOVA
Source of
Variation
Between
Groups
Within
Groups
Total
SS
Between Sample Variation
Within Sample Variation
df
MS
F
475.76
4
118.94
161.2
20
8.06
636.96
24
P-value F crit
14.75682 9.128E-06
P-Value: Probability of wrongly
rejecting the Null
ANOVA_EXAMPLE P60 Brainerd
2.866081
Fα , I −1, I ( J −1)
35
The Reference Distribution:
ANOVA_EXAMPLE P60 Brainerd
36
Analysis of Variance (ANOVA):
EXCEL Data Analysis: ANOVA
Manual calculations
15%
20%
25%
30%
35%
Cotton
Cotton
Cotton
Cotton
Cotton
Tensile Tensile Tensile Tensile Tensile
Strength Strength Strength Strength Strength
(Σxi)
SUMj
(SUM)SQj
SUM SQj
(Σxj)
(Σxj)2
Σ(xj2)
7
12
14
19
7
59
3481
799
7
17
18
25
10
77
5929
1387
15
12
18
22
11
78
6084
1298
11
18
19
19
15
9
18
19
23
11
82
80
6724
6400
1392
1416
Σ(Σxj)
Σ((Σxj)2)
49
77
88
108
54
Σ(Σxi)
376
Σ(Σ(xij2))
Σ(Σxij)
(Σxi)2
Σ(xi2)
2401
525
5929
1225
7744
1566
11664
2360
2916
616
Σ((Σxi2)
2
Σ(Σ(xi ))
30654
376
Σ(Σ(xj2))
28618
6292
6292
ANOVA_EXAMPLE P60 Brainerd
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Analysis of Variance (ANOVA):
EXCEL Data Analysis: ANOVA
Manual calculations
Manual calculations
Source of
Variation
SS - sum of
squares
Between
Groups
=(r( ΣΣxi )-(ΣΣxi) )/n
Within
Groups
Total
2
Error
2
SS
SS - sum of
squares
Calculation
df
MS = SS/df
estimate of
sigma
475.76
as = [(5 x30654) (376^2)]/25
4
118.94
161.2
=(n( ΣΣxij2)-(ΣΣxij)2)/n 636.96
20
as = [(25 x 6292) (376^2)]/25
24
ANOVA_EXAMPLE P60 Brainerd
8.06
F statistic =
MSBG/MSwithin
G
as =
118.94/8.06 =
14.76
38
Analysis of Variance (ANOVA):
STAT EASE Design Expert: ANOVA Single Factor
General Factorial Design
STEP 1
Press
continue ...
ANOVA_EXAMPLE P60 Brainerd
39
Analysis of Variance (ANOVA):
STAT EASE Design Expert: ANOVA Single Factor
General Factorial Design
STEP 2
Press
continue ...
ANOVA_EXAMPLE P60 Brainerd
40
Analysis of Variance (ANOVA):
STAT EASE Design Expert: ANOVA Single Factor
General Factorial Design
STEP 3
Define #
replicates
Press
continue ...
ANOVA_EXAMPLE P60 Brainerd
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Analysis of Variance (ANOVA):
STAT EASE Design Expert: ANOVA Single Factor
General Factorial Design
STEP 4
# and
Name
response
Press
continue ...
ANOVA_EXAMPLE P60 Brainerd
42
Analysis of Variance (ANOVA):
STAT EASE Design Expert: ANOVA Single Factor
STEP 5
Run experiment
in the defined
random order
and
measure/input
responses
Press
continue ...
ANOVA_EXAMPLE P60 Brainerd
43
Analysis of Variance (ANOVA):
STAT EASE Design Expert: ANOVA Single Factor
STEP 6
Design Expert
Analysis
ANOVA_EXAMPLE P60 Brainerd
44
Analysis of Variance (ANOVA):
STAT EASE Design Expert: ANOVA Single Factor
STEP 7
Design Expert
Analysis
ANOVA
ANOVA_EXAMPLE P60 Brainerd
45
Analysis of Variance (ANOVA):
STAT EASE Design Expert: ANOVA Single Factor
STEP 8
Design Expert
Analysis
ANOVA_EXAMPLE P60 Brainerd
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Analysis of Variance (ANOVA):
STAT EASE Design Expert: ANOVA Single Factor
STEP 9
Design Expert
Analysis
Effects
M for Model
e for error
ANOVA_EXAMPLE P60 Brainerd
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Analysis of Variance (ANOVA):
STAT EASE Design Expert: ANOVA Single Factor
STEP 10 Design Expert Analysis ANOVA
same table as before in EXCEL
ANOVA_EXAMPLE P60 Brainerd
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Analysis of Variance (ANOVA):
STAT EASE Design Expert: ANOVA Single Factor
STEP 10 Design Expert Analysis ANOVA Terms
Model: Terms estimating factor effects. For 2-level factorials: those that "fall off" the normal probability line
of the effects plot.
Sum of Squares: Total of the sum of squares for the terms in the model, as reported in the Effects List for
factorials and on the Model screen for RSM, MIX and Crossed designs.
DF: Degrees of freedom for the model. It is the number of model terms, including the intercept, minus one.
Mean Square: Estimate of the model variance, calculated by the model sum of squares divided by model degrees
of freedom.
F Value: Test for comparing model variance with residual (error) variance. If the variances are close to the
same, the ratio will be close to one and it is less likely that any of the factors have a significant effect on the
response. Calculated by Model Mean Square divided by Residual Mean Square.
Probe > F: Probability of seeing the observed F value if the null hypothesis is true (there is no factor effect).
Small probability values call for rejection of the null hypothesis. The probability equals the proportion of the
area under the curve of the F-distribution that lies beyond the observed F value. The F distribution itself is
determined by the degrees of freedom associated with the variances being compared.
(In "plain English", if the Probe>F value is very small (less than 0.05) then the terms in the
model have a significant effect on the response.)
ANOVA_EXAMPLE P60 Brainerd
49
Analysis of Variance (ANOVA):
STAT EASE Design Expert: ANOVA Single Factor
STEP 10 Design Expert Analysis Information in Help System
ANOVA Terms
Pure Error: Amount of variation in the response in replicated design points.
Sum of Squares: Pure error sum of squares from replicated points.
DF: The amount of information available from replicated points.
Mean Square: Estimate of pure error variance.
Cor Total: Totals of all information corrected for the mean.
Sum of Squares: Sum of the squared deviations of each point from the mean.
DF: Total degrees of freedom for the experiment, minus one for the mean.
ANOVA_EXAMPLE P60 Brainerd
50
Analysis of Variance (ANOVA):
STAT EASE Design Expert: ANOVA Single Factor
STEP 10 Design Expert Analysis ANOVA
ANOVA_EXAMPLE P60 Brainerd
51
Analysis of Variance (ANOVA):
STAT EASE Design Expert: ANOVA Single Factor
STEP 10 Design Expert Analysis ANOVA Information in Help System
ANOVA Terms
Next you see a collection of summary statistics for the model:
Std Dev: (Root MSE) Square root of the residual mean square. Consider this to be an estimate
of the standard deviation associated with the experiment.
Mean: Overall average of all the response data.
C.V.: Coefficient of Variation, the standard deviation expressed as a percentage of the mean.
Calculated by dividing the Std Dev by the Mean and multiplying by 100.
PRESS: Predicted Residual Error Sum of Squares – A measure of how the model fits each point
in the design. The PRESS is computed by first predicting where each point should be from a
model that contains all other points except the one in question. The squared residuals (difference
between actual and predicted values) are then summed.
R-Squared: A measure of the amount of variation around the mean explained by the model.
1-(SSresidual / (SSmodel + SSresidual))
ANOVA_EXAMPLE P60 Brainerd
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Analysis of Variance (ANOVA):
STAT EASE Design Expert: ANOVA Single Factor
STEP 10 Design Expert Analysis ANOVA Information in Help System
ANOVA Terms
Summary statistics for the model continued:
Adj R-Squared: A measure of the amount of variation around the mean explained by the model,
adjusted for the number of terms in the model. The adjusted R-squared decreases as the number of
terms in the model increases if those additional terms don’t add value to the model.
1-((SSresidual / DFresidual) / ((SSmodel + SSresidual) / (DFmodel + DFresidual)))
Pred R-Squared: A measure of the amount of variation in new data explained by the model.
1-(PRESS / (SStotal-SSblock)
The predicted r-squared and the adjusted r-squared should be within 0.20 of each other. Otherwise
there may be a problem with either the data or the model. Look for outliers, consider
transformations, or consider a different order polynomial.
ANOVA_EXAMPLE P60 Brainerd
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Analysis of Variance (ANOVA):
STAT EASE Design Expert: ANOVA Single Factor
STEP 10 Design Expert Analysis ANOVA Information in Help System
ANOVA Terms
Summary statistics for the model continued:
ANOVA_EXAMPLE P60 Brainerd
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Analysis of Variance (ANOVA):
STAT EASE Design Expert: ANOVA Single Factor
STEP 10 Design Expert Analysis ANOVA Scroll down
ANOVA_EXAMPLE P60 Brainerd
55
Analysis of Variance (ANOVA):
STAT EASE Design Expert: ANOVA Single Factor
STEP 10 Design Expert Analysis ANOVA Scroll down
ANOVA_EXAMPLE P60 Brainerd
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Analysis of Variance (ANOVA):
STAT EASE Design Expert: ANOVA Single Factor
STEP 10 Design Expert Analysis ANOVA Scroll down
ANOVA_EXAMPLE P60 Brainerd
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Analysis of Variance (ANOVA):
STAT EASE Design Expert: ANOVA Single Factor
STEP 10 Design Expert Analysis ANOVA Scroll down
ANOVA_EXAMPLE P60 Brainerd
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Analysis of Variance (ANOVA):
STAT EASE Design Expert: ANOVA Single Factor
STEP 10 Design Expert Analysis ANOVA Scroll down
ANOVA_EXAMPLE P60 Brainerd
59
Model Adequacy Checking in the ANOVA
Text reference, Section 3-4, pg. 76
•
•
•
•
•
Checking assumptions is important
Normality
Constant variance
Independence
Have we fit the right model?
ANOVA_EXAMPLE P60 Brainerd
60
Model Adequacy Checking in the ANOVA
STEP 10 Design Expert Analysis ANOVA Diagnostics
Residuals
•
Examination of residuals
(see text, Sec. 3-4, pg. 76)
eij = yij − yˆij
= yij − yi.
•
•
•
Design-Expert generates
the residuals
Residual plots are very
useful
Normal probability plot
of residuals
ANOVA_EXAMPLE P60 Brainerd
61
Analysis of Variance (ANOVA):
STAT EASE Design Expert: ANOVA Single Factor
STEP 10 Design Expert Analysis ANOVA Model Graphs
ANOVA_EXAMPLE P60 Brainerd
62
Other Important Residual Plots
5.2
5.2
2.95
2.95
R es iduals
R es iduals
2
2
0.7
2
2
0.7
-1.55
-1.55
2
2
2
-3.8
-3.8
9.80
12.75
15.70
18.65
21.60
1
4
7
10
13
16
19
22
25
R un N um ber
Predicted
ANOVA_EXAMPLE P60 Brainerd
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Post-ANOVA Comparison of Means
• The analysis of variance tests the hypothesis of equal
treatment means
• Assume that residual analysis is satisfactory
• If that hypothesis is rejected, we don’t know which specific
means are different
• Determining which specific means differ following an
ANOVA is called the multiple comparisons problem
• There are lots of ways to do this…see text, Section 3-5, pg. 86
• We will use pairwise t-tests on means…sometimes called
Fisher’s Least Significant Difference (or Fisher’s LSD)
Method
ANOVA_EXAMPLE P60 Brainerd
64
Graphical Comparison of Means
Text, pg. 89
ANOVA_EXAMPLE P60 Brainerd
65
For the Case of Quantitative Factors, a
Regression Model is often Useful
Response:Strength
ANOVA for Response Surface Cubic Model
Analysis of variance table [Partial sum of squares]
Sum of
Mean
F
Source Squares
DF Square Value Prob > F
Model
441.81
3 147.27 15.85 < 0.0001
A
90.84
1 90.84
9.78 0.0051
A2
343.21
1 343.21 36.93 < 0.0001
A3
64.98
1 64.98
6.99
0.0152
Residual 195.15
21
9.29
Lack of Fit 33.95
1 33.95
4.21 0.0535
Pure Error 161.20
20
8.06
Cor Total 636.96
24
Coefficient
Factor Estimate
Intercept 19.47
A-Cotton % 8.10
A2
-8.86
A3
-7.60
Standard 95% CI 95% CI
DF Error
Low
High
1
0.95 17.49 21.44
1
2.59
2.71 13.49
1
1.46 -11.89
-5.83
1
2.87 -13.58
-1.62
ANOVA_EXAMPLE P60 Brainerd
VIF
9.03
1.00
9.03
66
The Regression Model
Strength = +62.61143
-9.01143* Cotton Weight
% +0.48143 * Cotton
Weight %^2 -7.60000E003 * Cotton Weight %^3
25
%
20.5
2
2
Strength
Final Equation in Terms of
Actual Factors:
2
16
2
11.5
This is an empirical model of
the experimental results
2
7
2
2
15.00
20.00
25.00
30.00
35.00
A: C otton Weight %
ANOVA_EXAMPLE P60 Brainerd
67