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2003 Prentice-Hall, Inc. Chap 11-1
Analysis of Variance&
Post-ANOVA ANALYSIS
IE 340/440
PROCESS IMPROVEMENTTHROUGH PLANNED EXPERIMENTATION
Dr. Xueping LiUniversity of Tennessee
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2003 Prentice-Hall, Inc. Chap 11-2
What If There Are More ThanTwo 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
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2003 Prentice-Hall, Inc. Chap 11-3
Figure 3.1 (p. 61)A single-wafer plasma etching tool.
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2003 Prentice-Hall, Inc. Chap 11-4
Table 3.1 (p. 62)Etch Rate Data (in /min) from the Plasma Etching Experiment)
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2003 Prentice-Hall, Inc. Chap 11-5
The Analysis of Variance (Sec. 3-3, pg. 65)
In general, there will be alevels of the factor, or atreatments,andnreplicates of the experiment, run in randomorderacompletely randomized design (CRD)
N = antotal runs We consider the fixed effectscasethe random effects case will
be discussed later Objective is to test hypotheses about the equality of the a
treatment means
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2003 Prentice-Hall, Inc. Chap 11-6
Models for the Data
There are several ways to write a model forthe data:
is called the effects model
Let , then
is called the means model
Regression models can also be employed
ij i ij
i i
ij i ij
y
y
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The Analysis of Variance
The name analysis of variance stems from apartitioning of the total variability in the responsevariable into components that are consistent with amodel for the experiment
The basic single-factor ANOVA model is
2
1, 2,...,,
1,2,...,
an overall mean, treatment effect,
experimental error, (0, )
ij i ij
i
ij
i ay
j n
ith
NID
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The Analysis of Variance
Total variability is measured by the total sum ofsquares:
The basic ANOVA partitioning is:
2
..
1 1
( )a n
T ij
i j
SS y y
2 2
.. . .. .
1 1 1 1
2 2
. .. .
1 1 1
( ) [( ) ( )]
( ) ( )
a n a n
ij i ij i
i j i j
a a n
i ij i
i i j
T Treatments E
y y y y y y
n y y y y
SS SS SS
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The Analysis of Variance
A large value ofSSTreatmentsreflects large differences in
treatment means A small value ofSSTreatments likely indicates no differences in
treatment means Formal statistical hypotheses are:
T Treatments E SS SS SS
0 1 2
1
:
: At least one mean is different
aH
H
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The Analysis of Variance
While sums of squares cannot be directly compared to testthe hypothesis of equal means, mean squares can becompared.
A mean square is a sum of squares divided by its degrees offreedom:
If the treatment means are equal, the treatment and errormean squares will be (theoretically) equal.
If treatment means differ, the treatment mean square will be
larger than the error mean square.
1 1 ( 1)
,
1 ( 1)
Total Treatments Error
Treatments E Treatments E
df df df
an a a n
SS SS MS MS
a a n
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The Analysis of Variance isSummarized in a Table
Computingsee 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
0 , 1, ( 1)a a nF F
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Features of One-Way ANOVAF Statistic
The F Statistic is the Ratio of the AmongEstimate of Variance and the Within Estimateof Variance The ratio must always be positive df1 = a-1 will typically be small df2 = N- c will typically be large
The Ratio Should Be Close to 1 if the Null isTrue
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Features of One-Way ANOVAF Statistic
If the Null Hypothesis is False The numerator should be greater than the
denominator The ratio should be larger than 1
(continued)
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The Reference Distribution:
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Table 3.1 (p. 62)Etch Rate Data (in /min) from the Plasma Etching Experiment)
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Table 3.4 (p. 71)ANOVA for the Plasma Etching Experiment
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2003 Prentice-Hall, Inc. Chap 11-17
Table 3.5 (p. 72)Coded Etch Rate Data for Example 3.2
Coding the observations
More about manualcalculation p.70-71
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Chap 11-18
Graphical View of the ResultsDESIGN-EXPERT Plot
Strength
X = A: Cotton Weight %
Design Points
A: Cotton Weight %
Strength
One Factor Plot
15 20 25 30 35
7
11.5
16
20.5
25
22
22
22 22
22 22
22
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2003 Prentice-Hall, Inc. Chap 11-19
Model Adequacy Checking in the ANOVAText reference, Section 3-4, pg. 76
Checking assumptions is important Normality
Constant variance Independence Have we fit the right model?
Later we will talk about what to do if someof these assumptions are violated
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2003 Prentice-Hall, Inc. Chap 11-20
Model Adequacy Checking in the ANOVA
Examination ofresiduals(see text, Sec. 3-4, pg.76)
Design-Expert generatesthe residuals
Residual plots are veryuseful
Normal probability plotof residuals
.
ij ij ij
ij i
e y y
y y
Residual
Normal%p
roba
bility
-3.8 -1.55 0.7 2.95 5.2
1
5
10
20
30
50
7080
90
95
99
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Table 3.6 (p. 76)Etch Rate Data and Residuals from Example 3.1a.
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2003 Prentice-Hall, Inc. Chap 11-22
Figure 3.4 (p. 77)Normal probability plot ofresiduals for Example 3-1.
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2003 Prentice-Hall, Inc. Chap 11-23
Figure 3.5 (p. 78)Plot of residuals versus runorder or time.
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Figure 3.6 (p. 79)Plot of residuals versus fittedvalues.
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Other Important Residual Plots
22
22
22
22
22
22
22
Predicted
Residuals
.
-3.8
-1.55
0.7
2.95
5.2
9.80 12.75 15.70 18.65 21.60
Run Number
Residu
als
-3.8
-1.55
0.7
2.95
5.2
1 4 7 10 13 16 19 22 25
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2003 Prentice-Hall, Inc. Chap 11-26
Post-ANOVA Comparison of Means
The analysis of variance tests the hypothesis of equal treatmentmeans
Assume that residual analysis is satisfactory If that hypothesis is rejected, we dont know whichspecific
means are different Determining which specific means differ following an ANOVA is
called the multiple comparisons problem There are lotsof ways to do thissee text, Section 3-5, pg. 86 We will use pairwise t-tests on meanssometimes called Fishers
Least Significant Difference (or Fishers LSD) Method
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2003 Prentice-Hall, Inc. Chap 11-27
Tukeys Test
H0: Mu_i = Mu_j ; H1: Mu_i Mu_j T statistic
Whether Where
f is the DF of MSE a is the number of groups
n
MSfaqT E),(
Tyyji
..
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2003 Prentice-Hall, Inc. Chap 11-28
The Tukey-Kramer Procedure
Tells which Population Means are SignificantlyDifferent E.g., 1=23
2 groups whose meansmay be significantlydifferent
Post Hoc (A Posteriori) Procedure Done after rejection of equal means in ANOVA
Pairwise Comparisons Compare absolute mean differences with critical
range
X
f(X)
1 = 2 3
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2003 Prentice-Hall, Inc. Chap 11-29
The Tukey-Kramer Procedure:Example
1. Compute absolute meandifferences:Machine1Machine2 Machine3
25.40 23.40 20.00
26.31 21.80 22.20
24.10 23.50 19.75
23.74 22.75 20.60
25.10 21.60 20.40
1 2
1 3
2 3
24.93 22.61 2.32
24.93 20.59 4.34
22.61 20.59 2.02
X X
X X
X X
2. Compute critical range:
3. All of the absolute mean differences are greater than thecritical range. There is a significant difference between
each pair of means at the 5% level of significance.
( , )
'
1 1Critical Range 1.6182
U c n c
j j
MSWQ n n
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2003 Prentice-Hall, Inc. Chap 11-30
Fishers LSD
H0: Mu_i = Mu_j Least Significant Difference
Whether Where
E
ji
aN MSnn
tLSD )11
(,2/
n
MStLSD EaN
2,2/
LSDyyji
..
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2003 Prentice-Hall, Inc. Chap 11-31
Design-Expert Output
Treatment Means (Adjusted, If Necessary)Estimated Standard
Mean Error
1-15 9.80 1.27
2-20 15.40 1.27
3-25 17.60 1.27
4-30 21.60 1.27
5-35 10.80 1.27
Mean Standard t for H0
Treatment Difference DF Error Coeff=0 Prob > |t|
1 vs 2 -5.60 1 1.80 -3.12 0.0054
1 vs 3 -7.80 1 1.80 -4.34 0.0003
1 vs 4 -11.80 1 1.80 -6.57 < 0.0001
1 vs 5 -1.00 1 1.80 -0.56 0.5838
2 vs 3 -2.20 1 1.80 -1.23 0.23472 vs 4 -6.20 1 1.80 -3.45 0.0025
2 vs 5 4.60 1 1.80 2.56 0.0186
3 vs 4 -4.00 1 1.80 -2.23 0.0375
3 vs 5 6.80 1 1.80 3.79 0.0012
4 vs 5 10.80 1 1.80 6.01 < 0.0001
Fi 3 12 ( 99)
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Figure 3.12 (p. 99)Design-Expert computer output forExample 3-1.
Figure 3 13 (p 100)
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2003 Prentice-Hall, Inc. Chap 11-33
Figure 3.13 (p. 100)Minitab computer output forExample 3-1.
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Graphical Comparison of MeansText, pg. 89
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For the Case ofQuantitative Factors, aRegression Model is often Useful
Response:StrengthANOVA 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 < 0.0001A3 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 Error161.20 20 8.06
Cor Total 636.96 24
Coefficient Standard 95% CI 95% CI
Factor Estimate DF Error Low High VIF
Intercept 19.47 1 0.95 17.49 21.44
A-Cotton % 8.10 1 2.59 2.71 13.49 9.03
A2 -8.86 1 1.46 -11.89 -5.83 1.00
A3 -7.60 1 2.87 -13.58 -1.62 9.03
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Chap 11-37
The Regression Model
Final Equation in Termsof Actual Factors:
Strength = +62.61143
-9.01143* Cotton Weight% +0.48143 * CottonWeight %^2 -7.60000E-003 * Cotton Weight %^3
This is an empirical modelof the experimental results
15.00 20.00 25.00 30.00 35.00
7
11.5
16
20.5
25
A: Cotton Weight %
Strength
22
22
22 22
22 22
22
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Figure 3.7 (p. 83)Plot of residuals versus ij for Example 3-5.
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Table 3.9 (p. 83)Variance-Stabilizing Transformations
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2003 Prentice-Hall, Inc. Chap 11-40
Figure 3.8 (p. 84)Plot of log Siversus logfor the peak discharge datafrom Example 3.5.
.iy
Figure 3 12 (p 99)
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2003 Prentice-Hall, Inc. Chap 11-41
Figure 3.12 (p. 99)Design-Expert computer output forExample 3-1.
Figure 3.13 (p. 100)
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2003 Prentice-Hall, Inc. Chap 11-42
g (p )Minitab computer output forExample 3-1.
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Display on page 103 y
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Example 3-1 p70 EX3-1 p112
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One Way ANOVA F Test
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One-Way ANOVA FTestExample
As production manager, youwant to see if 3 fillingmachines have different mean
filling times. You assign 15similarly trained &experienced workers, 5 permachine, to the machines. At
the .05 significance level, isthere a difference in meanfilling times?
Machine1Machine2
Machine3
25.40 23.40 20.00
26.31 21.80 22.2024.10 23.50 19.75
23.74 22.75 20.60
25.10 21.60 20.40
One Wa ANOVA E ample
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One-Way ANOVA Example:Scatter Diagram
27
26
25
24
23
22
21
20
19
Machine1Machine2
Machine3
25.40 23.40 20.00
26.31 21.80 22.20
24.10 23.50 19.75
23.74 22.75 20.60
25.10 21.60 20.40
1 2
3
24.93 22.61
20.59 22.71
X X
X X
1X
2X
3X
X
One Way ANOVA Example
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One-Way ANOVA ExampleComputations
Machine1Machine2 Machine3
25.40 23.40 20.00
26.31 21.80 22.20
24.10 23.50 19.75
23.74 22.75 20.6025.10 21.60 20.40
2 2 2
5 24.93 22.71 22.61 22.71 20.59 22.71
47.164
4.2592 3.112 3.682 11.0532
/( -1) 47.16 / 2 23.5820
/( - ) 11.0532 /12 .9211
SSA
SSW
MSA SSA c
MSW SSW n c
1
2
3
24.93
22.61
20.59
22.71
X
X
X
X
5
3
15
jn
c
n
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Summary Table
Source ofVariation
Degreesof
Freedom
Sum ofSquares
MeanSquares
(Variance)
FStatistic
Among(Factor)
3-1=2 47.1640 23.5820MSA/MSW
=25.60
Within
(Error) 15-3=12 11.0532 .9211Total 15-1=14 58.2172
One Way ANOVA Example
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One-Way ANOVA ExampleSolution
F0 3.89
H0: 1=2=3H1: Not All Equal= .05
df1= 2 df2 = 12
Critical Value(s):
Test Statistic:
Decision:
Conclusion:
Reject at = 0.05.
There is evidence that atleast one i differs fromthe rest.
= 0.05
FMSA
MSW
23 5820
9211
25 6.
.
.
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Solution in Excel
Use Tools | Data Analysis | ANOVA: SingleFactor
Excel Worksheet that Performs the One-FactorANOVA of the Example
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2003 Prentice-Hall, Inc. Chap 11-52
The Tukey-Kramer Procedure
Tells which Population Means are SignificantlyDifferent E.g., 1=23
2 groups whose meansmay be significantlydifferent
Post Hoc (A Posteriori) Procedure
Done after rejection of equal means in ANOVA Pairwise Comparisons
Compare absolute mean differences with criticalrange
X
f(X)
1 = 2 3
The T ke K ame P oced e
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2003 Prentice-Hall, Inc. Chap 11-53
The Tukey-Kramer Procedure:Example
1. Compute absolute meandifferences:Machine1Machine2 Machine3
25.40 23.40 20.00
26.31 21.80 22.20
24.10 23.50 19.75
23.74 22.75 20.60
25.10 21.60 20.40
1 2
1 3
2 3
24.93 22.61 2.32
24.93 20.59 4.3422.61 20.59 2.02
X X
X XX X
2. Compute critical range:
3. All of the absolute mean differences are greater than thecritical range. There is a significant difference betweeneach pair of means at the 5% level of significance.
( , )
'
1 1Critical Range 1.6182
U c n c
j j
MSWQ n n
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2003 Prentice-Hall, Inc. Chap 11-54
Solution in PHStat
Use PHStat | c-Sample Tests | Tukey-KramerProcedure
Excel Worksheet that Performs the Tukey-Kramer Procedure for the Previous Example
Levenes Test for
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Levene s Test forHomogeneity of Variance
The Null Hypothesis The cpopulation variances are all equal
The Alternative Hypothesis Not all the cpopulation variances are equal
2 2 2
0 1 2: cH
2
1 : Not all are equal ( 1,2, , )jH j c
Levenes Test for Homogeneity
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2003 Prentice-Hall, Inc. Chap 11-56
Levene s Test for Homogeneityof Variance: Procedure
1. For each observation in each group, obtainthe absolute value of the difference betweeneach observation and the median of thegroup.
2. Perform a one-way analysis of variance onthese absolute differences.
Levenes Test for Homogeneity
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2003 Prentice-Hall, Inc. Chap 11-57
Levene s Test for Homogeneityof Variances: Example
As production manager, youwant to see if 3 fillingmachines have different
variance in filling times. Youassign 15 similarly trained &experienced workers, 5 permachine, to the machines. At
the .05 significance level, isthere a difference in thevariance in filling times?
Machine1Machine2
Machine3
25.40 23.40 20.00
26.31 21.80 22.2024.10 23.50 19.75
23.74 22.75 20.60
25.10 21.60 20.40
Levenes Test:
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Levene s Test:Absolute Difference from the Median
Machine1 Machine2 Machine3 Machine1 Machine2 Machine3
25.4 23.4 20 0.3 0.65 0.4
26.31 21.8 22.2 1.21 0.95 1.8
24.1 23.5 19.75 1 0.75 0.65
23.74 22.75 20.6 1.36 0 0.2
25.1 21.6 20.4 0 1.15 0
median 25.1 22.75 20.4
abs(Time - median(Time))Time
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Summary Table
SUMMARY
Groups Count Sum Average Variance
Machine1 5 3.87 0.774 0.35208
Machine2 5 3.5 0.7 0.19
Machine3 5 3.05 0.61 0.5005
ANOVA
Source of Variation SS df MS F P-value F crit
Between Groups 0.067453 2 0.033727 0.097048 0.908218 3.88529Within Groups 4.17032 12 0.347527
Total 4.237773 14
Levenes Test Example:
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Levene s Test Example:Solution
F0 3.89
H0:H1: Not All Equal= .05
df1= 2 df2 = 12
Critical Value(s):
Test Statistic:
Decision:
Conclusion:
Do not reject at = 0.05.
There is no evidence that
at least one differs
from the rest.
= 0.05
2 2 2
1 2 3
0.03370.0970
0.3475
MSAF
MSW
2
j
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Randomized Blocked Design Items are Divided into Blocks
Individual items in different samples are matched,or repeated measurements are taken
Reduced within group variation (i.e., remove theeffect of block before testing)
Response of Each Treatment Group isObtained
Assumptions Same as completely randomized design No interaction effect between treatments and
blocks
Randomized Blocked Design
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Randomized Blocked Design(Example)
Factor (Training Method)
Factor Levels(Groups)
BlockedExperiment
Units
DependentVariable
(Response)
21 hrs 17 hrs 31 hrs27 hrs 25 hrs 28 hrs
29 hrs 20 hrs 22 hrs
Randomized Block Design
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Randomized Block Design(Partition of Total Variation)
Variation Dueto Group
SSA
VariationAmong
Blocks
SSBL
Variation
Among All
Observations
SST
Commonly referred to as: Sum of Squares Error Sum of Squares
Unexplained
Commonly referred to as: Sum of Squares AmongAmong Groups Variation
=
+
+Variation Dueto Random
Sampling
SSW
Commonly referred to as: Sum of Squares Among
Block
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Total Variation
2
1
the number of blocks
the number of groups or levels
the total number of observations
the value in the -th block for the -th treatment level
the mean of all val
c r
ij
j i
ij
i
SST X X
r
c
n n rc
X i j
X
ues in block
the mean of all values for treatment level
1
j
i
X j
df n
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Among-Group Variation
2
1
1 (treatment group means)
1
1
c
j
j
r
ij
ij
SSA r X X
XX
r
df cSSA
MSAc
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Among-Block Variation
2
1
1(block means)
1
1
r
i
i
c
ij
j
i
SSBL c X X
X
Xc
df rSSBL
MSBLr
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Random Error
2
1 1
1 1
1 1
c r
ij i j
j i
SSE X X X X
df r c
SSEMSE
r c
Randomized Block F Test for
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Randomized BlockF Test forDifferences in c Means
No treatment effect
Test Statistic
Degrees of Freedom
0 1 2: cH
1 : Not all are equaljH
MSAF
MSE
1 1df c
2 1 1df r c 0
UFF
Reject
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Summary Table
Source ofVariation
Degrees ofFreedom
Sum ofSquares
MeanSquares
FStatistic
AmongGroup c1 SSA MSA =SSA/(c1) MSA/MSE
AmongBlock
r1 SSBLMSBL =
SSBL/(r1)
MSBL/MSE
Error (r1)c1) SSEMSE =
SSE/[(r1)(c1)]
Total rc1 SST
Randomized Block Design:
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Randomized Block Design:Example
As production manager, youwant to see if 3 filling machineshave different mean fillingtimes. You assign 15 workers
with varied experience into 5groups of 3 based on similarityof their experience, andassigned each group of 3
workers with similar experienceto the machines. At the .05significance level, is there adifference in mean filling
times?
Machine1Machine2
Machine3
25.40 23.40 20.00
26.31 21.80 22.2024.10 23.50 19.75
23.74 22.75 20.60
25.10 21.60 20.40
Randomized Block Design
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Randomized Block DesignExample Computation
Machine1Machine2 Machine3
25.40 23.40 20.00
26.31 21.80 22.20
24.10 23.50 19.75
23.74 22.75 20.6025.10 21.60 20.40
2 2 25 24.93 22.71 22.61 22.71 20.59 22.71
47.1648.4025
/( -1) 47.16 / 2 23.5820
/ ( -1) 1 8.4025 / 8 1.0503
SSA
SSE
MSA SSA c
MSE SSE r c
1
2
3
24.93
22.61
20.59
22.71
X
X
X
X
5
3
15
r
c
n
Randomized Block Design
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Randomized Block DesignExample: Summary Table
Source ofVariation
Degrees ofFreedom
Sum ofSquares
MeanSquares
FStatistic
AmongGroup2 SSA=
47.164MSA =23.582
23.582/1.0503=22.452
AmongBlock
4SSBL=2.6507
MSBL =
.6627
.6627/1.0503
=.6039
Error 8 SSE=8.4025
MSE =
1.0503
Total 14 SST=58.2172
Randomized Block Design
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Randomized Block DesignExample: Solution
F0 4.46
H0: 1=2=3H1: Not All Equal= .05
df1= 2 df2 = 8
Critical Value(s):
Test Statistic:
Decision:
Conclusion:
Reject at = 0.05.
There is evidence that atleast one i differs fromthe rest.
= 0.05
FMSA
MSE
23 582
1.0503
22.45.
Randomized Block Design
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Randomized Block Designin Excel
Tools | Data Analysis | ANOVA: Two FactorWithout Replication
Example Solution in Excel Spreadsheet
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The Tukey-Kramer Procedure
Similar to the Tukey-Kramer Procedure for theCompletely Randomized Design Case
Critical Range
( , 1 1 )Critical Range U c r cMSE
Qr
The Tukey-Kramer Procedure:
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The Tukey Kramer Procedure:Example
1. Compute absolute meandifferences:Machine1Machine2 Machine3
25.40 23.40 20.00
26.31 21.80 22.20
24.10 23.50 19.75
23.74 22.75 20.6025.10 21.60 20.40
1 2
1 3
2 3
24.93 22.61 2.32
24.93 20.59 4.34
22.61 20.59 2.02
X X
X X
X X
2. Compute critical range:
3. All of the absolute mean differences are greater. Thereis a significance difference between each pair of means at5% level of significance.
( , 1 1 )
1.0503
Critical Range 4.04 1.85165U c r c
MSE
Q r
The Tukey-Kramer Procedure
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The Tukey Kramer Procedurein PHStat
PHStat | c-Sample Tests | Tukey-KramerProcedure
Example in Excel Spreadsheet
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Two-Way ANOVA
Examines the Effect of: Two factors on the dependent variable
E.g., Percent carbonation and line speed on soft
drink bottling process Interaction between the different levels of these
two factors E.g., Does the effect of one particular percentage of
carbonation depend on which level the line speed isset?
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Two-Way ANOVA
Assumptions Normality
Populations are normally distributed Homogeneity of Variance
Populations have equal variances
Independence of Errors Independent random samples are drawn
(continued)
Two-Way ANOVA
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SSE
o ay OTotal Variation Partitioning
Variation Due to
Factor A
Variation Due to
Random Sampling
Variation Due to
Interaction
SSA
SSABSST
Variation Due toFactor B
SSBTotal Variation
d.f.= n-1
d.f.= r-1
=
+
+d.f.= c-1
+
d.f.= (r-1)(c-1)
d.f.= rc(n-1)
Two-Way ANOVA
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yTotal Variation Partitioning
'
the number of levels of factor A
the number of levels of factor B
the number of values (replications) for each cell
the total number of observations in the experiment
the value of the -th oijk
r
c
n
n
X k
bservation for level of
factor A and level of factor B
i
j
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Total Variation
' '
' 2
1 1 1
1 1 1 1 1 1
'
Sum of Squares Total
= total variation among all
observations around the grand mean
the overall or grand mean
r c n
ijk
i j k
r c n r c n
ijk ijk
i j k i j k
SST X X
X X
Xrcn n
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Factor A Variation
2
'
1
r
i
i
SSA cn X X
Sum of Squares Due to Factor A
= the difference among the various
levels of factor A and the grand
mean
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Factor B Variation
2
'
1
c
j
j
SSB rn X X
Sum of Squares Due to Factor B
= the difference among the various
levels of factor B and the grand mean
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Interaction Variation
2
'
1 1
r c
ij i j
i j
SSAB n X X X X
Sum of Squares Due to Interaction between A and B
= the effect of the combinations of factor A and
factor B
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Random Error
Sum of Squares Error
= the differences among the observations within
each cell and the corresponding cell means
' 2
1 1 1
r c n
ijijk
i j k
SSE X X
Two-Way ANOVA:
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yThe F Test Statistic
FTest for Factor B Main Effect
FTest for Interaction Effect
H0: 1 .= 2 . = = r .H1: Not all i . are equal
H0: ij = 0 (for all i and j)H1: ij 0
H0: 1 = . 2 = = cH1: Not all . j are equal
Reject if
F > FU
Reject if
F > FU
Reject if
F > FU
1
MSA SSAF MSA
MSE r
FTest for Factor A Main Effect
1
MSB SSBF MSB
MSE c
1 1
MSAB SSABF MSAB
MSE r c
Two-Way ANOVA
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ySummary Table
Source ofVariation
Degrees ofFreedom
Sum ofSquares
MeanSquares
FStatistic
Factor A
(Row)
r1 SSAMSA =
SSA/(r1)MSA/MSE
Factor B(Column)
c1 SSBMSB =
SSB/(c1)
MSB/MSE
AB
(Interaction)
(r1)(c1) SSABMSAB =
SSAB/ [(r1)(c1)]MSAB/MSE
Error rcn1) SSEMSE =
SSE/[rcn1)]
Total rcn1 SST
Features of Two-Way ANOVA
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Features of Two Way ANOVAF Test
Degrees of Freedom Always Add Up rcn-1=rc(n-1)+(c-1)+(r-1)+(c-1)(r-1) Total=Error+Column+Row+Interaction
The Denominator of the F Test is Always theSame but the Numerator is Different
The Sums of Squares Always Add Up Total=Error+Column+Row+Interaction
Kruskal-Wallis Rank Test
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for c Medians
Extension of Wilcoxon Rank Sum Test Tests the equality of more than 2 (c)
population medians
Distribution-Free Test Procedure
Used to Analyze Completely RandomizedExperimental Designs
Use 2 Distribution to Approximate if EachSample Group Size nj> 5
df= c 1
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Kruskal-Wallis Rank Test
Assumptions Independent random samples are drawn Continuous dependent variable Data may be ranked both within and amongsamples Populations have same variability Populations have same shape
Robust with Regard to Last 2 Conditions Use F test in completely randomized designs and
when the more stringent assumptions hold
Kruskal-Wallis Rank Test
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Procedure Obtain Ranks
In event of tie, each of the tied values gets theiraverage rank
Add the Ranks for Data from Each of the cGroups Square to obtain Tj2
2
1
123( 1)( 1)
cj
j j
TH nn n n
1 2 cn n n n
Kruskal-Wallis Rank Test
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Procedure
Compute Test Statistic
# of observation injth sample Hmay be approximated by chi-square distribution
with df = c1when each nj>5
(continued)
2
1
123( 1)
( 1)
cj
j j
TH n
n n n
1 2 cn n n n
jn
Kruskal-Wallis Rank Test
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Procedure
Critical Value for a Given Upper tail
Decision Rule Reject H0: M1= M2= = Mc if test statistic
H> Otherwise, do not reject H0
(continued)
2U
2
U
Kruskal-Wallis Rank Test:
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Machine1Machine2
Machine3
25.40 23.40 20.00
26.31 21.80 22.20
24.10 23.50 19.7523.74 22.75 20.60
25.10 21.60 20.40
Example
As production manager, youwant to see if 3 filling machineshave different median filling
times. You assign 15 similarlytrained & experienced workers,5 per machine, to themachines. At the .05
significance level, is there adifference in median fillingtimes?
Example Solution: Step 1
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Machine1Machine2
Machine314 9 2
15 6 7
12 10 1
11 8 413 5 3
Obtaining a Ranking
Raw Data Ranks
65 38 17
Machine1Machine2
Machine325.40 23.40 20.00
26.31 21.80 22.20
24.10 23.50 19.75
23.74 22.75 20.6025.10 21.60 20.40
Example Solution: Step 2
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Test Statistic Computation
212
3( 1)
( 1) 1
2 2 212 65 38 17
3(15 1)15(15 1) 5 5 5
11.58
Tc jH n
n n nj j
Kruskal-Wallis Test Example
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Solution
H0: M1 = M2 = M3H1: Not all equal= .05
df= c- 1 = 3 - 1 = 2Critical Value(s): Reject at
Test Statistic:
Decision:
Conclusion:
There is evidence that
population medians are
not all equal.
= .05
= .05.
H= 11.58
05.991
k l ll
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2003 Prentice-Hall, Inc. Chap 11-99
Kruskal-Wallis Test in PHStat
PHStat | c-Sample Tests | Kruskal-Wallis RankSum Test
Example Solution in Excel Spreadsheet
Friedman Rank Test foriff i di
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Differences in c Medians
Tests the equality of more than 2 (c)population medians
Distribution-Free Test Procedure Used to Analyze Randomized Block
Experimental Designs
Use2
Distribution to Approximate if theNumber of Blocks r> 5
df= c 1
F i d R k T
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Friedman Rank Test
Assumptions The r blocks are independent The random variable is continuous The data constitute at least an ordinal scale of
measurement No interaction between the r blocks and the c
treatment levels The c populations have the same variability The c populations have the same shape
Friedman Rank Test:P d
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2003 Prentice-Hall, Inc. Chap 11-102
Procedure
Replace the c observations by their ranks ineach of the r blocks; assign average rankfor ties
Test statistic:
R.j2 is the square of the rank total for groupj
FR can be approximated by a chi-squaredistribution with (c1) degrees of freedom The rejection region is in the right tail
2
1
12 3 11
c
R j
j
F R r crc c
F i d R k T t E l
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Friedman Rank Test: Example
As production manager, youwant to see if 3 fillingmachines have differentmedian filling times. Youassign 15 workers with variedexperience into 5 groups of 3based on similarity of theirexperience, and assigned eachgroup of 3 workers with similar
experience to the machines. Atthe .05 significance level, isthere a difference in medianfilling times?
Machine1Machine2
Machine3
25.40 23.40 20.00
26.31 21.80 22.2024.10 23.50 19.75
23.74 22.75 20.60
25.10 21.60 20.40
Friedman Rank Test:C t ti T bl
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Timing Rank
Machine 1 Machine 2 Machine 3 Machine 1 Machine 2 Machine 3
25.4 23.4 20 3 2 1
26.31 21.8 22.2 3 1 2
24.1 23.5 19.75 3 2 123.74 22.75 20.6 3 2 1
25.1 21.6 20.4 3 2 1
15 9 6
225 81 36
Computation Table
2
. jR
. jR
2
1
123 1
1
12342 3 5 4 8.4
5 3 4
c
R j
j
F R r crc c
Friedman Rank Test Example
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Solution
H0: M1 = M2 = M3H1: Not all equal= .05
df= c- 1 = 3 - 1 = 2Critical Value: Reject at
Test Statistic:
Decision:
Conclusion:
There is evidence thatpopulation medians are
not all equal.
= .05
= .05
FR= 8.4
0 5.991
Ch t S
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Chapter Summary
Described the Completely RandomizedDesign: One-Way Analysis of Variance ANOVA Assumptions F Test for Difference in c Means The Tukey-Kramer Procedure Levenes Test for Homogeneity of Variance
Discussed the Randomized Block Design F Test for the Difference in c Means The Tukey Procedure
Ch t S
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Chapter Summary
Described the Factorial Design: Two-WayAnalysis of Variance Examine effects of factors and interaction
Discussed Kruskal-Wallis Rank Test forDifferences in c Medians
Illustrated Friedman Rank Test for Differencesin c Medians
(continued)