Transcript Simplified Design of Experiments

Simplified Design of
Experiments
Quality Day
Orange Empire 0701
October 28, 2010
Goals of Today’s Session
• Understand why Design of
Experiments is preferable.
• Learn how to perform basic
experiments to optimize
results and reduce
variation.
Approaches to
Experimentation
• Build-test-fix (What experimentation?)
• One-factor-at-a-time (OFAT)
• Designed experiments (DOE)
Build-Test=Fix
• The tinkerer’s approach
• Impossible to know if true optimum
achieved
– Quit when it works!
• Consistently slow
– Requires intuition, luck, rework
– Continual fire-fighting
One Factor at a Time
(OFAT)
• Approach:
– Run all factors at one condition
– Repeat, changing condition of one factor
– Continuing to hold that factor at that condition,
rerun with another factor at its second condition
– repeat until all factors at their optimum
conditions
• Slow and requires many tests
• Can miss interactions!
“One Factor at a Time” is Like
The Blind Man and the Elephant
What we conclude may be determined by where we are looking!
Design of Experiments
• A statistics-based approach to designed
experiments.
• A methodology to achieve a predictive
knowledge of a complex, multi-variable
process with the fewest trials possible.
• An optimization of the experimental
process itself.
Key Concepts
• DOE is about better understanding of our processes
INPUTS
(Factors)
X variables
OUTPUTS
(Responses)
Y variables
People
Materials
PROCESS:
Equipment
responses related
to performing a
service
Policies
responses related
to producing a
produce
Procedures
A Blending of
Inputs which
Generates
Corresponding
Outputs
Methods
Environment
Illustration of a Process
responses related
to completing a task
Injection Molding Process
INPUTS
(Factors)
X variables
OUTPUTS
(Responses)
Y variables
Type of Raw
Material
Mold
Temperature
Holding
Pressure
PROCESS:
% shrinkage from
mold size
Holding Time
Gate Size
Manufacturing
Injection
Molded Parts
Screw Speed
Moisture
Content
thickness of molded
part
Manufacturing Injection Molded
Parts
number of defective
parts
Concrete Mixing Process
INPUTS
(Factors)
X variables
OUTPUTS
(Responses)
Y variables
Type of
cement
compressive
strength
Percent water
PROCESS:
Type of
Additives
Percent
Additives
Mixing Time
modulus of elasticity
Discovering
Optimal
Concrete
Mixture
Curing
Conditions
% Plasticizer
Optimum Concrete Mixture
modulus of rupture
Poisson's ratio
Microwave Popcorn Making Process
INPUTS
(Factors)
X variables
OUTPUTS
(Responses)
Y variables
Brand:
Cheap vs Costly
PROCESS:
Taste:
Scale of 1 to 10
Time:
4 min vs 6 min
Power:
75% or 100%
M aking the
Best
M icrow av e
popcorn
Height:
On bottom or raised
M aking microw av e popcorn
Bullets:
Grams of unpopped
corns
Three Key Principles
• Replication
[DOE’s version of Sample Size]
– Replication of an experiment
– Allows an estimate of experimental error
– Allows for a more precise estimate of the sample
mean value
• Randomization
[Run experiments in random order]
– Cornerstone of all statistical methods
– “Average out” effects of extraneous factors
– Reduce bias and systematic errors
• Blocking
[What can influence my experiment?]
– Increases precision of experiment
– “Factor out” variables not studied
Steps in a Designed
Experiments Study
1. Brainstorm the problem / causes
2. Design the Experiment
3. Perform the experiment
4. Analyze the data
5. Validate your results
Determining Input Factors to Optimize Output
FACTORIAL EXPERIMENT
IMPROVEMENT OF THE MEAN
Case Study
• You have been assigned to look into a
problem
with
your
company’s
lamination process. Customers have
been complaining about separation in
your wood laminate, and you are
leading a team to determine how to
reduce the separation.
Step 1. Brainstorm
• You met with your team, and through
your investigation, you determine
that the best approach would be to
minimize the amount of CURL
Laminate
Curl
Step 1. Brainstorm
• After brainstorming and list reduction,
your team decided to look at 3 factors
that they felt influenced the amount of
curl:
A Top Roll Tension, currently set at 22
B Bottom Roll Tension, currently set at 22
C Rewind Tension, currently set at 9
Step 2. Design the Experiment
The team decides to study 2 levels
for each of the factors, and
determine the effect on the response
variable, curl:
Factor
Low (-) Setting
High (+) Setting
Top Roll Tension
16
28
Bottom Roll Tension
16
28
Rewind tension
6
12
Note: Team will make sure that their selected
values are FEASIBLE.
Step 2. Design the Experiment
• This experiment will have 2 levels
(high and low settings) and 3 factors
(top roll tension, bottom roll tension,
and rewind tension)
• Number of Experiments = 23 = 8
Step 2. Design the Experiment
Run
Top Roll Tension
Bottom Roll Tension
Rewind Tension
1
-
-
-
2
+
-
-
3
-
+
-
4
+
+
-
5
-
-
+
6
+
-
+
7
-
+
+
8
+
+
+
Alternate - -,+ +
Alternate - - - ++++
Alternate -,+
Note: If another variable is added, there would be 24=16 runs
The next design generation would be “- - - - - - - -, + + + + + + +”
Step 3. Perform the Experiment
Run Top Roll
Tension
Bottom
Roll
Tension
Rewind
Tension
16
28
16
16
16
28
7
28
16
28
16
8
28
1
2
3
4
5
6
Replication
1
Replication
2
Average
Std.
Deviation
6
6
6
87
88
87.5
0.707
76
78
77.0
1.414
90
92
91.0
1.414
28
16
16
28
6
12
12
12
83
80
81.5
2.121
101
96
98.5
3.535
92
91
91.5
0.707
100
104
102.0
2.828
28
12
92
91
91.5
0.707
Perform the runs in random order to ensure statistical
validity
Step 4. Analyze the Data
• We are going to calculate the MAIN
EFFECTS for the 3 factors.
• We are also going to calculate the
INTERACTION EFFECTS between
each of the 2 factor combinations
and the three factor combination.
• Let’s start with the main effects…
Step 4. Analyze the Data
Visualization of Main Effects
102.0
+
98.5
91.5
91.5
Rewind
Tension
91.0
-
-
81.5
87.5
Top Roll Tension
77.0
+
+
Bottom Roll
Tension
Note: lowest curl achieved when Top Roll
Tension HIGH and Bottom Roll and Rewind
Tensions set LOW
Step 4 Analyze the Data
Main Effect Calculations
• Average the “High” Settings and subtract
the average of the “Low” Settings:
 77.0  81.5  91.5  91.5   87.5  91.0  98.5  102.0 
Main Effect of Top Roll Tension= 

 9.38


4
4

 

 91.0  81.5  102.0  91.5   87.5  77.0  98.5  91.5 
Main Effect of Bottom Roll Tension= 

 2.88


4
4

 

 98.5  91.5  102.0  91.5   87.5  77.0  91.0  81.5 
Main Effect of Rewind Tension= 

 11.63


4
4

 

Step 4. Analyze the Data
Interaction Effects
Run
Top Roll
Tension
(A)
Bottom Roll
Tension
(B)
Rewind
Tension
(C)
AB
AC
BC
ABC
Avg Curl
+
-
+
-
+
-
+
+
+
+
-
+
+
87.5
+
+
+
+
+
+
+
-
+
-
+
+
-
81.5
7
+
+
-
102.0
8
+
+
+
+
+
+
+
91.5
1
2
3
4
5
6
77.0
91.0
98.5
91.5
Simply multiply the signs of the columns, i.e. “+ times + equals +”
“- times - equals +” “+ times - equals -” and “- times + equals -”
Step 4 Analyze the Data
Interaction Effect Calculations
• Average the “High” Settings and subtract
the average of the “Low” Settings:
 87.5  81.5  98.5  91.5   77.0  91.0  91.5  102.0 
Interaction AB= 

 0.63


4
4

 

 87.5  91.0  91.5  91.5   77.0  81.5  98.5  102.0 
Interaction AC= 

 0.63


4
4

 

 87.5  77.0  102.0  91.5   91.0  81.5  98.5  91.5 
Interaction BC= 

 1.13


4
4

 

 77.0  91.0  98.5  91.5   87.5  81.5  91.5  102.0 
Interaction ABC= 

 1.13


4
4

 

Step 4 Analyze the Data
Pareto of Effect Size
Effect Size
14
12
10
8
6
4
2
0
Rewind
Tension (C)
Top Roll
Tension (A)
Bottom Roll
Tension (B)
BC
Interaction
3-Way
Interaction
AB
Interaction
Interaction
Effect Size
Pareto shows the absolute value of the effects for
comparability of significance.
Step 4. Analyze the Data
Visualization of Interaction Effects
Interaction – Top and Bottom Roll Tension
Curl
100
95
90
85
Slope= -10.0
Slope= -8.75
Bottom Roll Low
Bottom Roll High
80
75
Top Roll Tension
Low
Top Roll Tension
High
Note: parallel lines indicate LACK OF INTERACTION
Step 4. Analyze the Data
Visualization of Interaction Effects
Interaction – Top Roll and Rewind Tension
Curl
120
Slope= -8.75
100
80
60
Slope=-10.0
Rewind Low
Rewind High
40
20
0
Top Roll Tension Low Top Roll Tension High
Note: parallel lines indicate LACK OF INTERACTION
Step 4. Analyze the Data
Visualization of Interaction Effects
Interaction –Bottom Roll and Rewind Tension
Curl
100
95
90
85
80
Slope = 1.75
Rewind Low
Slope = 4.0
Rewind High
75
Bottom Roll Tension Bottom Roll Tension
Low
High
Note: parallel lines indicate LACK OF INTERACTION
Step 4. Analyze the Data
Test of Significance
• We can test for the significance of
the effects using the t-distribution
and confidence intervals.
• The first step is to calculate the
POOLED STANDARD DEVIATION
for all of the observations….
Step 4 Analyze the Data
Pooled Standard Deviation
We are studying 7 factors/ interactions
(A, B, C, AB, AC, BC, ABC) in 8 runs
We did 2 replications of each run.
n1S12  n2 S 22  ...  nv Sv2
Sp 
n1  n2  ...  nv
for v runs
Note: if the number of replicates is the same for all
runs, we can simply calculate the pooled standard
deviation as the square root of the average of the
variances
Step 4 Analyze the Data
Pooled Standard Deviation
Run
Replication
1
Replication
2
Average
Std.
Deviation
Variance
1
87
88
87.5
0.707
0.4998
2
76
78
77.0
1.414
1.9994
3
90
92
91.0
1.414
1.9994
4
83
80
81.5
2.121
4.4986
5
101
96
98.5
3.535
12.4962
6
92
91
91.5
0.707
0.4998
7
100
104
102.0
2.828
7.9976
8
92
91
91.5
0.707
0.4998
Simplified Calculation:
Avg Variance=3.8113
Std Dev = 1.952
Step 4. Analyze the Data
t Statistic for Significance
• Key Information for calculation:
=risk (Confidence = 1- ) [5%, 95%]
p=number of “+” per effect column [4]
r=number of replicates [2]
Sp=Pooled Standard Deviation [1.952]
f=degree of fractionalization (in our case 0,
since we are doing a full factorial)
k=number of factors [3]
Degrees of Freedom=(r-1)2k-f [8]
Step 4. Analyze the Data
t Statistic for Significance
The t statistic for 95% confidence
(5% risk), 2 tailed test, 8 degrees
of freedom is 2.306
The Error is calculated as:

Error  t.025,8  S p


2 
2 
  2.306 1.952
  2.25
pr 
(4)(2)


The confidence interval for each
of the effects is:
– Effect +/- Error
t table for 2 tailed test
Step 4. Analyze the Data
t Statistic for Significance
• Since the Error = 2.25, any effect that is
contained in the limits of: 0 +/- 2.25
• Is considered NOT STATISTICALLY
SIGNIFICANT
Factor
Effect
Rewind Tension m(C)
11.63
Top Roll Tension (A)
-9.38
Bottom Roll Tension (B)
2.88
BC Interaction
-1.13
ABC Interaction
-1.13
AB Interaction
-0.63
AC Interaction
0.63
Statistically
Significant
Not
Statistically
Significant
Step 4. Analyze the Data
Conclusion
• Based on this outcome, I will conclude that
Top Roll tension, Bottom Roll Tension, and
Rewind Tension are significant at the 95%
significance level
• I will conclude that I should set:
– Top Roll Tension HIGH (28)
– Bottom Roll Tension LOW (16)
– Rewind Tension LOW (6)
Step 4. Analyze the Data
Conclusion
Run Top Roll
Tension
Bottom
Roll
Tension
Rewind
Tension
Replication
1
Replication
2
Average
Std.
Deviation
1
16
16
6
87
88
87.5
0.707
2
28
16
6
76
78
77.0
1.414
3
16
28
6
90
92
91.0
1.414
4
28
28
6
83
80
81.5
2.121
5
16
16
12
101
96
98.5
3.535
6
28
16
12
92
91
91.5
0.707
7
16
28
12
100
104
102.0
2.828
8
28
28
12
92
91
91.5
0.707
Grand Average=90.0625
Analysis of Data - Response Model
• The factor effects can be used to
establish a model to predict responses.
 EffectA 
 EffectB 
 EffectC 
Expected Response=Grand Average+ 
A

B

C





 2 
 2 
 2 
 EffectAB 
 EffectAC 
 EffectBC 
 EffectABC 
AB

AC

BC









 ABC
2
2
2
2
where A=setting for factor A(-1)
where B=setting for factor B(+1)
where C=setting for factor C(+1)
Calculating Expected Response
Factor
Effect
Effect/2
Setting
Grand Average
Value
90.0625
Top Roll Tension (A)
-9.38
-4.69
1
-4.69
Bottom Roll Tension (B)
2.88
1.44
-1
-1.44
Rewind Tension m(C)
11.63
5.815
-1
-5.815
AB Interaction
-0.63
-0.315
(1)(-1)
0.315
AC Interaction
0.63
0.315
(1)(-1)
-0.315
BC Interaction
-1.13
-0.565
(-1)(-1)
-0.565
ABC Interaction
-1.13
-0.565
(1)(-1)(-1)
-0.565
So, if we set A at low (-1), B high (+1) and C High (+1), we would predict:
90.0625 - 4.69 - 1.44 - 5.815 + 0.315 - 0.315 - 0.565 - 0.565 = 76.99
Step 5. Validation of Model
• Run the process with the new settings for
a trial period to collect data on the curl
response.
• Compare the new data to the historical
data to confirm improvement.
• A simple Test of Hypothesis of before and
after data with a t test can be used.
Step 5. Validation of Model
• We are trying to prove that the curl
with the new process settings is
significantly less than the curl with
the old process settings at a 95%
level of significance. Our historical
standard deviation has been 3.5.
– Ho: µold  µnew
– Ha: µold > µnew
Commonly Used Z-Values
1-α (or 1-β)
α (or β)
Z Value
.995
.005
2.575
.990
.010
2.387
.975
.025
1.960
.950
.050
1.645
.900
.100
1.282
.800
.200
0.842
43
Step 5. Validation of Model
• Sample Size Required:
•  = 5%
•  = 3.4
 = 10%
Change to detect = 2
2
Z

Z

  
2
n

2
(1.645  1.282) 3.4

 24.75  25
2
2
2
2
Step 5 Validation of Model
• Suppose our historical data (25 data
points) is as follows for curl:
– 100, 95, 98, 102, 97, 90, 91, 98, 101, 95
– 94, 105, 96, 104, 100, 96, 98, 96, 91, 99
– 100, 102, 98, 95, 96
• Average = 97.48
• Sample Standard Deviation = 3.8419
Step 5 Validation of Model
• Now, we run our new settings and
collect 25 additional sets of data:
– 86, 77, 85, 81, 81, 83, 78, 79, 81, 80
– 81, 78, 76, 75, 79, 83, 85, 81, 78, 78
– 85, 81, 79, 83, 72
• Average = 80.20
• Sample Standard Deviation = 3.391
t test – Assuming Equal
Variances in Excel
Old
100
95
98
102
97
90
New
86
77
85
81
81
83
91
98
101
95
94
105
96
104
100
96
98
96
91
99
100
102
98
95
96
78
79
81
80
81
78
76
75
79
83
85
81
78
78
85
81
79
83
72
t-Test: Two-Sample Assuming Equal Variances
Old
Mean
Variance
Observations
Pooled Variance
Hypothesized Mean Difference
df
t Stat
P(T<=t) one-tail
t Critical one-tail
P(T<=t) two-tail
t Critical two-tail
New
97.48
14.76
25
13.13
80.2
11.5
25
0
48
16.86034207
4.21133E-22
1.677224197
8.42266E-22
2.010634722
t critical (24 degrees of freedom) = 1.677
Calculated t value = 16.86
We can conclude with more than 95%
confidence that the new parameter
settings have significantly reduced the
amount of curl in our process.
Determining Input Factors to Reduce Variation
FACTORIAL EXPERIMENT
REDUCTION OF VARIATION
Variation Reduction
• Standard deviations are not normally
distributed, and therefore cannot be
used directly as a response variable.
• Options include:
– Use the natural or base 10 log to obtain
normality
– Use –log10(s) for normality
– Use the F Statistic on the average
variances for the high and low settings
Data – Curl Reduction
Run
1
2
3
4
5
6
7
8
Top
Roll
Tension
(A)
Bottom
Roll
Tension
(B)
Rewind
Tension
(C)
AB
AC
BC
ABC
+
-
+
-
+
-
+
+
+
+
-
+
+
87.5
0.707
.004998
77.0
1.414
1.9994
91.0
1.414
1.9994
+
+
+
-
+
+
+
+
-
+
-
+
-
81.5
2.121
4.4986
98.5
3.535
12.4962
91.5
0.707
004998
+
+
+
+
+
+
+
+
+
+
102.0
2.828
7.9976
91.5
0.707
004998
Avg Curl
Std Dev
Variance
Calculate an F Value for each Effect
For Top Roll Tension
2
1.994  4.4986  .004998  .004998
 1.6256
4
.004998  1.9994  12.4962  7.9976

 5.6245
4
S 
S
2

2
F
S L arg er
S
2
Smaller
5.6245

 3.460
1.6256
(r  1)2k  f 1x 230
Degrees of Freedom=

4
2
2
Confidence 95%, Risk 5%
F.025,4,4  9.60
r=# replicates (2)
k=# factors (3)
f=fractionalization (0)
Any effects with a F value
greater than 9.60 are significant
at the 95% level.
Significant for reduced variation???
F Table for  = .025
Calculate an F Value for each Effect
For Bottom Roll Tension
2
1.9994  4.4986  7.9976  .004998
 3.625
4
.004998  1.9994  12.4962  .004998

 3.626
4
S 
S
2

2
F
S L arg er
S
2
Smaller
3.626

 1.000
3.625
(r  1)2k  f 1x 230
Degrees of Freedom=

4
2
2
Confidence 95%, Risk 5%
F.025,4,4  9.60
r=# replicates (2)
k=# factors (3)
f=fractionalization (0)
Any effects with a F value
greater than 9.60 are significant
at the 95% level.
Significant for reduced variation???
Calculate an F Value for each Effect
For Rewind Tension
2
12.4962  .004998  7.9976  .004998
 5.126
4
.004998  1.9994  1.9994  4.4986

 2.1256
4
S 
S
2

2
F
S L arg er
S
2
Smaller
5.126

 2.412
2.1256
(r  1)2k  f 1x 230
Degrees of Freedom=

4
2
2
Confidence 95%, Risk 5%
F.025,4,4  9.60
r=# replicates (2)
k=# factors (3)
f=fractionalization (0)
Any effects with a F value
greater than 9.60 are significant
at the 95% level.
Significant for reduced variation???
Calculate an F Value for each Effect
For Interaction of Top Roll and Bottom Roll Tension
2
.004998  4.4986  12.4962  .004998
 4.2512
4
1.9994  1.9994  .004998  7.9976

 3.0003
4
S 
S
2

2
F
S L arg er
S
2
Smaller
4.2512

 1.4169
3.0003
(r  1)2k  f 1x 230
Degrees of Freedom=

4
2
2
Confidence 95%, Risk 5%
F.025,4,4  9.60
r=# replicates (2)
k=# factors (3)
f=fractionalization (0)
Any effects with a F value
greater than 9.60 are significant
at the 95% level.
Significant for reduced variation???
Calculate an F Value for each Effect
For Interaction of Top Roll and Rewind Tension
2
.004998  1.9994  .004998  .004998
 0.5036
4
1.9994  4.4986  12.4962  7.9976

 6.74795
4
S 
S
2

2
F
S L arg er
S
2
Smaller
6.74795

 13.3994
0.5036
(r  1)2k  f 1x 230
Degrees of Freedom=

4
2
2
Confidence 95%, Risk 5%
F.025,4,4  9.60
r=# replicates (2)
k=# factors (3)
f=fractionalization (0)
Any effects with a F value
greater than 9.60 are significant
at the 95% level.
Significant for reduced variation???
Calculate an F Value for each Effect
For Interaction of Bottom Roll and Rewind Tension
2
.004998  1.9994  7.9976  .004998
 2.5017
4
1.9994  4.4986  12.4962  .004998

 4.7498
4
S 
S
2

2
F
S L arg er
S
2
Smaller
4.7498

 1.8986
2.5017
(r  1)2k  f 1x 230
Degrees of Freedom=

4
2
2
Confidence 95%, Risk 5%
F.025,4,4  9.60
r=# replicates (2)
k=# factors (3)
f=fractionalization (0)
Any effects with a F value
greater than 9.60 are significant
at the 95% level.
Significant for reduced variation???
Calculate an F Value for each Effect
For Interaction of All Factors
2
1.9994  1.9994  12.4962  .004998
 4.1250
4
.004998  4.4986  .004998  7.9976

 3.1265
4
S 
S
2

2
F
S L arg er
S
2
Smaller
4.1250

 1.31937
3.1265
(r  1)2k  f 1x 230
Degrees of Freedom=

4
2
2
Confidence 95%, Risk 5%
F.025,4,4  9.60
r=# replicates (2)
k=# factors (3)
f=fractionalization (0)
Any effects with a F value
greater than 9.60 are significant
at the 95% level.
Significant for reduced variation???
Lets Summarize…
Factor
F Value
Top Roll Tension (A)
3.460
Bottom Roll Tension (B)
1.0000
Rewind Tension (C)
2.412
AB Interaction
1.4169
AC Interaction
13.3994
BC Interaction
1.8986
ABC Interaction
1.31937
Note, the AC (Top Roll-Rewind Tension) is the only
factor or interaction that exceeds the F Critical
Value of 9.60.
Conclusion
• To minimize variation, we find that variation is
minimized when Top Roll Tension and Rewind
Tension are set to the same levels (i.e. High-High
or Low-Low).
• Our optimal settings to minimize curl were Top
Roll High, Bottom Roll Low, and Rewind Low. Top
Roll and Rewind influenced the curl in OPPOSITE
DIRECTIONS!
How can we decide?
Case Studies
Instructions:
Work in teams to read the case
study and then develop an
experimental design. You should
include:
• What factors to study
• How many and what levels to study
• How many replicates to take
• How many runs?
• Design your experiment.
Case Study 1
• You have been working late recently, and
subsisting on microwave popcorn. As a result, you
have decided to find the formula for the best
popcorn. You are down to two brands, A and B.
You also find that the time varies between 4 and
6 minutes and power between 75% and 100%.
• You judge the quality of the popcorn by taste and
the number of unpopped kernels.
Case 1 Design
•
•
•
•
•
What factors to study
How many and what levels to study
How many replicates to take
How many runs?
Design your experiment
Case 1 Design
•
•
•
•
•
What factors to study
How many and what levels to study
How many replicates to take
How many runs?
Design your experiment
Case Study 2
• Your company has a high-pressure chemical reactor which filters
impurities from your product. You have been asked to determine
the proper settings to maximize the filtration rate. After a
brainstorming session, your team decides that temperature,
pressure, chemical concentration % and stir rate all influence the
filtration rate (gallons per hour).
• Looking over the historical records, you see temperature has
varied from 24oC to 35oC. Pressure can range from 10 PSIG to 15
PSIG, and chemical concentration from 2% to 4%. The lowest
stir rate has been 15 RPM and the highest 30 RPM.
• Design an experiment to determine what the optimal levels would
be to maximize the filtration rate.
Case 2 Design
•
•
•
•
•
What factors to study
How many and what levels to study
How many replicates to take
How many runs?
Design your experiment
Case 2 Design
•
•
•
•
•
What factors to study
How many and what levels to study
How many replicates to take
How many runs?
Design your experiment
Case Study 3
• A manufacturer of ice crease has hired you to assist them
with achieving their target fill rate of 2.50 pounds. After
discussing with them the relevant factors, it is concluded that
there are two main variables impacting the final weight, fill
temperature and the overfill %. Overfill is the percentage of
air that is incorporated into the ice crease.
The fill
temperature is sensitive, and they have machines that vary
from 20oF to 25oF.
Overfill percentage is also tightly
controlled. Studies have shown that overfill percentages
greater than 120% are perceived as “lower quality” by
customers and percentages less than 90% cause the ice crease
to be difficult to scoop.
Case 3 Design
•
•
•
•
•
What factors to study
How many and what levels to study
How many replicates to take
How many runs?
Design your experiment
Case 3 Design
•
•
•
•
•
What factors to study
How many and what levels to study
How many replicates to take
How many runs?
Design your experiment
Points to Remember
1. Get a clear understanding of the problem you
intend to solve.
2. Conduct an exhaustive and detailed brainstorming
session.
3. Teamwork – involve people involved in all aspects
of the process being studied.
4. Randomize the experiment trial order
5. Replicate to understand and estimate variation.
6. Perform confirmatory runs and experiments to
test your model’s validity.
Your Assignment
• Find a problem and run a Designed
Experiment using the methods we learned
today within 1-2 weeks of completing this
session.
• No problem – Optimize microwave popcorn
or chocolate chip cookies!
Questions?