Transcript Chapter 11 Multifactor Analysis of Variance

Chapter 11
Multifactor Analysis
of Variance
Two-Factor ANOVA Design
(KIJ > 1)
1. Test for Interaction.
If no interaction exists:
2. Test for Main Effects.
If interaction does exist:
3. Test for Differences among
Treatment Combinations.
ANOVA Two-Factor Definitions
SSA =2 T2I* / JK – 2 T2** / IJK
SSB = 2 T2*J / IK – 2 T2** / IJK
SSTR= 2 T2IJ / K – 2 T2** / IJK
SSAB= SSTR – SSA – SSB
SSTOT= 3 x2IJK – 2 T2** / IJK
SSE = SSTOT – SSTR
Fundamental Identity K>1
SSTOT = SSA+SSB+SSAB+SSE
Note: 2
I J
=

ANOVA Two-Factor Definitions
Mean Squares Due to A:
MSA = SSA / I – 1
Mean Squares Due to B:
MSB = SSB / J – 1
Mean Square Due to Interaction:
MSAB = SSAB / (I – 1)(J – 1)
Mean Square Error:
MSE = SSE / IJ(K – 1)
Two-Factor ANOVA (KIJ > 1 & equal K)
Test for Interaction
Null Hypothesis H0AB: Interaction
Effects = 0
Test Statistic:  = MSAB / MSE
Alternative Hypothesis:
HaAB: Interaction Exists
Reject Region (upper tailed)
  F, (I-1)(J-1), IJ(K-1)
Two-Factor ANOVA (KIJ > 1 & equal K)
Test for Main Effects (Factor A)
Null Hypothesis H0A: 1A = 2A = …= IA
Test Statistic:  = MSA / MSE
Alternative Hypothesis:
HaA: Differences Exist in Factor A
Reject Region (upper tailed)
  F, I-1, IJ(K-1)
Two-Factor ANOVA (KIJ > 1 & equal K)
Test for Main Effects (Factor B)
Null Hypothesis H0B: 1B = 2B = …= JB
Test Statistic:  = MSB / MSE
Alternative Hypothesis:
HaB: Differences Exist in Factor B
Reject Region (upper tailed)
  F, J-1, IJ(K-1)
Example ANOVA Two-Factor (equal K)
A study is conducted to investigate the effect of
temperature and humidity on the force required
to separate an adhesive product from a plastic
laminate. The experimenter is interested in (4)
temperature levels and (2) humidity levels. Three
measurements are taken at each of the IJ
treatment combinations. Are there real
differences in the average responses for these
levels of factors at a level of significance of .05?
The summary data of the experiment follows:
T11 = 119 T21 = 108 T31 = 95 T41 = 85 (lbs)
T12 = 99 T22 = 83 T32 = 73 T42 = 60 (lbs)
 x2IJK = 22,722
Tukey’s Multiple Comparisons with No Interaction
on significant Main Effects A
1. Find Q, I, IJ(K-1)
2. Find w = QMSE/(JK)
3. Order the I-means (A) from smallest to largest.
Underscore all pairs that differ by less than w.
Pairs not underscored correspond to
significantly different levels of factor A.
For significant Main Effects B
1. Replace I with J in Q’s 2nd subscript.
2. Replace JK by IK in w.
3. Order the J-means (factor B).
Example: Tukey’s Multiple Comparisons with
No Interaction on significant Main Effects
The study to investigate the effect of
temperature and humidity on the force required
to separate an adhesive product from a plastic
found no statistical evidence of interaction
between temperature and humidity. Differences
in mean values among temperature levels &
differences in mean values among humidity
levels were found to exist. Identify significant
differences in levels for both factors at a level of
significance of .05.
Example: ANOVA 2-Factor (equal K) with Interaction
A study is conducted to investigate the effect of
temperature and humidity on the force required to
separate an adhesive product from a plastic laminate.
The experimenter is interested in (4) temperature levels
and (2) humidity levels. Three measurements are taken
at each of the IJ treatment combinations. Are there real
differences in the average responses for these levels of
factors at a level of significance of .05? Conduct this
analysis under the assumption that the test for
interaction found a significant interaction (difference in
response) between levels of temperature and levels of
humidity. The summary data of the experiment follows:
T11 = 119 T21 = 108 T31 = 95 T41 = 85 (lbs)
T12 = 99 T22 = 83 T32 = 73 T42 = 60 (lbs)
 x2IJK = 22,722
Example: ANOVA 2-Factor (equal K) with Interaction
You are designing a battery for use in a device that will be subjected
to extreme variations in temperature. The only design parameter
that you can select is the plate material. Three choices are available.
You decide to test all three plate materials at (3) temperature levels
that are consistent with the product end-use environment. Four
batteries are tested at each combination of plate material and
temperature. The resulting observed battery life data follows:
Temperature (0F)
Material Type
1
2
3
15
T=539
T=623
T=576
70
T=229
T=479
T=583
125
T=230 (hrs)
T=198 (hrs)
T=342 (hrs)
Test at =.05 to determine if there is evidence to conclude that there
are real differences in the battery life for these levels of temperature
and material. Is there a choice of material that would give uniformly
long life regardless of temperature? x2IJK =478,546.97
ANOVA Three-Factor Definitions
SSA =3 T2I**/ JKL – 3 T2***/ IJKL
SSAB=3 T2IJ*/ KL – 3 T2I**/ JKL –
3 T2*J*/ IKL + 3 T2***/ IJKL
SSABC=3 T2IJK/L – 3 T2IJ*/KL – 3 T2I*K/ JL
– 3 T2*JK/ IL + 3 T2*J*/IKL + 3 T2I**/JKL
+ 3 T2**K/ IJL – 3 T2***/IJKL
SSTOT= 4 x2IJKL – 3 T2***/ IJKL
SSTOT= SSA+ SSB+ SSC+ SSAB + SSAC+
SSBC+ SSABC+ SSE
Note: 3
I J K
=

1 1 1
ANOVA Three-Factor Definitions
Mean Squares Due to A:
MSA = SSA / I – 1
Mean Square 2-Factor Interaction:
MSAB = SSAB / (I – 1)(J – 1)
Mean Square 3-Factor Interaction:
MSABC = SSABC / (I – 1)(J – 1)(K-1)
Mean Square Error:
MSE = SSE / IJK(L – 1)
ANOVA - Three Factor
An experiment is designed to investigate the effects
of three factors on productivity (measured in
thousands of dollars of items produced) per
40-hour week at a manufacturing plant. The
factors tested are:
1. Length of week (5 day-8 hrs or 4 day-10 hrs)
2. Shift (Day or Night)
3. Number of Breaks (0, 1, 2)
The experiment was conducted over a 24-week
period. The data for this completely randomized
design are shown on the next page.
Perform an ANOVA at α = .05.
DATA:
Day
0
1 2
4-Days 94 105 96
97 106 91
Night
0
1
2
90 102 103
89 97 98
5-Days 96 100 82
92 103 88
81 90
84 92
NOTE: L = 2
94
96
CF=213,948.10
2P Factorial Experiments
> P = Number of Factors
> 2 Levels per Factor (Low / High)
> Complete Design = 2P Combinations
22 Factorial Experiment
A
B > b
> (1)
ab
a
Mean
(b+ab)/2n
((1)+a)/2n
Mean (1)+b a+ab
2n
2n
(1), a, b, and ab signify cell Totals.
Contrasts using Cells Totals
A Contrast = ab + a – b – (1)
B Contrast = ab –a + b – (1)
AB Contrast = ab – a – b + (1)
Main Effects ATotals = (A contrast)
2P-1 n
Main Effects BTotals = (B contrast)
2P-1 n
Contrasts using Cell Means
A Contrast = ab + a – b – (1)
B Contrast = ab –a + b – (1)
AB Contrast = ab – a – b + (1)
Main Effects AMeans = (A contrast)
2P-1
Main Effects BMeans = (B contrast)
P-1
2
Signs for Contrasts in a 22 Factorial
Treatment
Factorial Effect
Combination
A B AB
(1)
a
b
ab
–
+
–
+
–
–
+
+
+
–
–
+
Positive values to the treatment combinations that
is at the High level.
Negative values at the Lower level. Interactions by
multiplying signs of the interacting factors.
Sum of Squares > 2P Factorial (Totals)
SSA = (A contrast)2
2P n
SSB = (B contrast)2
P
2 n
2
contrast)
SSAB = (AB
P
2 n
SSE = SST – SSA – SSB – SSAB
Sum of Squares > 2P Factorial (Means)
SSA = n(A contrast)2
2P
2
SSB = n(B contrast)
2P
SSAB = n(AB contrast)2
2P
df for Mean Square
>A=I–1=1
>B=J–1=1
>C=K–1=1
>AB = (I-1)(J-1) = 1
>ABC = (I-1)(J-1)(K-1) = 1
P
>Error = 2 (n-1)
Factorial Experiment Example
A 23 factorial experiment was conducted to estimate the
effects of three factors on the yield of a chemical reaction.
The factors were A: catalyst concentration (high or low),
B: reagent (std. or new), C: stirring rate (slow or fast).
Three replicates were obtained for each treatment
measured as a percent. Cell means are listed below.
Calculate effects & test for each main effect and
interaction at α = 0.05.
Treatment Cell Mean
1
69.8733
a
78.5500
b
77.9067
ab
78.1000
Treatment Cell Mean
c
72.4067
ac
76.2733
bc
76.1833
abc
75.8333