Download TM 720 Lecture 05: Variation Comparisons, Process Capability

ENGM 720 - Lecture 05
Variation Comparisons,
Process Capability
5/5/2017
ENGM 720: Statistical Process Control
1
Assignment:

Reading:
•
•

Chapter 4 & 8
•
•
Finish reading through 4.3.4, 4.4 - 4.4.3, and CH 8 - 8.3
Begin reading 4.5
Chapter 5
•
Begin reading through 5.2, and 5.4
Assignments:
•
•
•
Obtain the Hypothesis Test (Chart &) Tables – Materials Page
Obtain the Exam Tables DRAFT – Materials Page
•
Verify accuracy as you work assignments
Access New Assignment and Previous Assignment Solutions:
•
Download Assignment 3 Instruction & Solutions
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ENGM 720: Statistical Process
Control
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Comparison of Variances
• The second types of comparison are those that compare the
spread of two distributions. To do this:
• Compute the ratio of the two variances, and then compare the ratio
to one of two known distributions as a check to see if the magnitude
of that ratio is sufficiently unlikely for the distribution.
Definitely
Different
Probably
Different
Probably NOT
Different
Definitely NOT
Different
• The assumption that the data come from Normal distributions is very
important. Assess how normally data are distributed prior to
conducting either test.
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Situation VII: Variance Test
With 0 Known
 Used
when:
• existing comparison process has been operating without
much change in variation for a long time
 Procedure:
• form ratio of a sample variance (t-distribution variable) to a
population variance (Normal distribution variable),
v = n - 1 degrees of freedom

2
0

n  1S 2

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
2
0
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Situation VIII: Variance Test
With 0 Unknown
 Use:
• worst case variation comparison process for when there is
not enough prior history
 Procedure:
• form ratio of the sample variances (two 2-distributions),
v1 = n1 – 1 degrees freedom for numerator, and
v2 = n2 – 1 degrees freedom for the denominator
F0 
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S 12
S 22
Note:
F1,v1,v 2 
ENGM 720: Statistical Process
Control
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F,v 2,v1
5
Table for Variance Comparisons

Decision on which test to use is based
on answering the following:
• Do we know a theoretical variance (2) or
•
should we estimate it by the sample variances
(s2) ?
What are we trying to decide (alternate
hypothesis)?
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Table for Variance Comparisons


These questions tell us:
•
•
•
•
Four primary test statistics for variance
comparisons
•
•
•

What sampling distribution to use
What test statistic(s) to use
What criteria to use
How to construct the confidence interval
Two sampling distributions
Two confidence intervals
Six alternate hypotheses
Table construction
•
Note: F1-, v1, v2 = 1/ F, v2, v1
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Grip Strength Example


True Corporate Training Example
•
How could grip strength vary among people in the SPC
training room?
Data collection to detect difference in dominant
hand mean between the left and right sides of
the training room
•
Expectations?
•
•
•
•
Significance Level?
Known parameters?
Best test?
Result?
• Direction of comparison?
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Grip Strength Data Results

R-L Side, Equal Variance
Dominant Hand Means
 Two-Sided Test at  = .05
Comparison:
• HA: There is a difference
2
• L = x1 = 129.4, S1 = 2788,
 Test: Is | t0 | > t.025, 52?
n1 = 34 people
• R = x2 = 104.0, S22 = 1225, • |1.91| > 2.009 - NO!
 Keep the Null Hypothesis:
n2 = 20 people
• There is NOT a difference btwn
• Sp = 47.1, v = 52
L&R!
t0 
x1  x 2   0
Sp
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1
1

n1 n 2
(n1  1) S12  (n2  1) S 22
Sp 
n1  n2  2
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Grip Strength Data Results

R-L Side, No Assumptions
Dom. Hand Means
Comparison:
• L = x1 = 129.4, S1
t0
2

Two-Sided Test at  = .05

Test: Is | t0 | > t.025, 51?
= 2788,
n1 = 34 people

• R = x2 = 104.0, S22 = 1225,
n2 = 20 people
• v = 51
2
 S12 S 22 


n n 
2 
 1
v

x  x2  0
2
2
 1
 S12 
 S 22 




S12 S 22
n 
n 
 1  2

n1 n 2
n1  1
n2  1
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• HA: There is a difference
• |2.12| > 2.009 - YES!
Reject the Null Hypothesis:
• There IS a difference btwn
L & R!
• Why is this wimpy test
significant when the other
wasn’t?
ANS: Check the equal
variance assumption!
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Grip Strength Data Results

Unknown σ0
Variances Comparison:
• S12 = 2788
• n1 = 34, v1 = 33
• S22 = 1225
• n2 = 20, v2 = 19

Two-Sided Test at  = .10

Test: Is F0 > F.05, 33, 19?
• HA: There is a difference
• 2.276 > 2.07 - YES!
(Should also check F1– /2, 33, 19)

Reject the Null Hypothesis:
• There IS a difference in variance!
• At  = .05, this test is just barely
F0 
S
S
2
1
2
2
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F1 ,v1 ,v2 
1
not significant
• (Should also have checked for
Normality with Normal Prob. Plot)
F ,v 2 ,v1
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Control
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Statistical Quality Improvement

Goal: Control and Reduction of Variation

Causes of Variation:
•
Chance Causes / Common Causes
•
Assignable Causes / Special Causes
•
• In Statistical Control
• Natural variation / background noise
• Out of Statistical Control
• Things we can do something about - IF
we act quickly!
Both can cause defects – because specifications are
often set regardless of process capabilities!
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Process Capability
 Process
Capability Analysis (PCA)
•Is only done when the process is in a state of
Statistical Control
• Meaning:
NO SPECIAL CAUSES are present
•Process does not have to be centered to do
PCA
• Yield will improve if process is centered, but the value
is in knowing what / where to improve the process
•PCA is done periodically when the process has
been operating in a state of statistical control
• Allows for measuring improvement over time
• Allows for marketing your competitive edge
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Process Capability - Timing
Process Capability Analysis is performed
when there are NO special causes of
variability present – ie. when the process is
in a state of statistical control, as
illustrated at this point.
Improving Process Capability and
Performance
Continually Improve the
System
Characterize Stable Process
Capability
Head Off Shifts in Location,
Spread
Time
Identify Special Causes - Bad
(Remove)
Identify Special Causes - Good
(Incorporate)
Reduce Variability
Center the Process
LSL
0
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USL
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Process Capability

Process Capability is INDEPENDENT of product
specifications
• Most specifications are set without regard for process
capability
• However, understanding process capability helps the engineer to
set more reasonable specifications
• PCA reflects only the Natural Tolerance Limits of the process
• PCA is done by examining the process
• Histogram
• Normal Probability Plot
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Natural Tolerance Limits
 The
natural tolerance limits assume:
• The process is well-modeled by the Normal Distribution
• Three sigma is an acceptable proportion of the process
to yield
 The
Upper and Lower Natural Tolerance Limits
are derived from:
• The process mean () and
• The process standard deviation ()
 Equations:
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UNTL    3
LNTL    3
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Natural Tolerance Limits
  1 :68.26% of the total area
  2 :95.46% of the total area
  3 :99.73% of the total area
-3
or
LNTL
-2
-

+
+2
+3
or
UNTL
The Natural Tolerance Limits cover 99.73% of the process output
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PCA: Histogram Construction

Verify rough shape and location of histogram

Quickly confirm applicability prior to statistical analysis
• Symmetric (roughly bell-shaped)
• Mean  median  mode
• Often hard to distinguish a Normal Distribution from a t-Distribution
• Sometimes even a Normal distribution doesn’t look normal
• More data and columns (bins) can make a difference

Verify location of process with respect to Specifications
• Quick inspection will show what to do to improve the process
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PCA: Normal Probability Plot
A
Normal Plot better clarifies whether the
distribution is Normal by a visual inspection for:
• Non-random patterns (non-Normal)
• Fat Pencil Test (Normal if passes)
C
u
m
C
u
m
C
u
m
F
r
e
q
F
r
e
q
F
r
e
q
X
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X
ENGM 720: Statistical Process
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X
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PCA: Parameter Estimation

The Normal Plot mid-point estimates the process mean

The slope of the “best fit” line for the Normal Plot estimates
the standard deviation
• Choose the 25th and 75th percentile points to calculate the slope

The Histogram mode should be close to the mean

The range/d2 (from Histogram) should be close to the
standard deviation
• Can also estimate
standard deviation by subtracting
th
50th percentile
from the 84 percentile of the Histogram
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Process Capability Indices
 Cp:
•Measures the potential capability of the current
process - if the process were centered within
the product specifications
•Two-sided Limits:
USL  LSL
Cp 
6
•One-sided Limit:
Cpu
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USL  

3
ENGM 720: Statistical Process
Control
  LSL
Cpl 
3
21
Cp Relation to Process Fallout
Cp
Ratio
0.50
0.60
0.80
1.00
1.20
1.40
1.50
1.60
1.80
2.00

Two-Sided Specification Fallout
(ppm)
133614
71861
16395
2700
318
27
7
2
0.06
0.0018
One-sided Specification Fallout
(ppm)
66807
35931
8198
1350
159
14
4
1
0.03
0.0009
Recommended Minimum Ratios:
(D. C. Montgomery, 2001)
• Existing Process
1.25 (1-sided)
• Existing, Safety / Critical Parameter 1.45
• New Process
1.45
• New, Safety / Critical Parameter
1.60
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ENGM 720: Statistical Process
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1.33 (2-sided)
1.50
1.50
1.67
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Process Capability Indices
 Cpk:
•Measures actual capability of current process at its’ current location with respect to product
specifications
•Formula:
pk
pu
pl
C
 min( C , C )
Where:
Cpu
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USL  

3
  LSL
Cpl 
3
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Process Capability Indices
 Regarding
Cp and Cpk:
•Both assume that the process is Normally
distributed
•Both assume that the process is in Statistical
Control
•When they are equal to each other, the
process is perfectly centered
•Both are pretty common reporting ratios
among vendors and purchasers
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Process Capability Indices
 Two
very different processes can have
identical Cpk values, though:
•because spread and location interact in Cpk!
LSL
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USL
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Process Capability Indices
 Cpm:
•Measures the current capability of the
process - using the process target center
point within the product specifications in the
calculation
•Formula: Cpm  USL  LSL
6  2  (   T )2
Where target T is:
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1
T  (USL  LSL )
2
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Process Capability Indices
 Cpkm:
•Similar to Cpm - just more sensitive to
departures from the process target center point
•Not really in very common use
•Formula:
C pkm 
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C pk
T
1 

  
2
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Questions & Issues
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