Statistics for Managers Using Microsoft® Excel 5th Edition Chapter 18

Transcription

Statistics for Managers
Using Microsoft® Excel
5th Edition
Chapter 18
Statistical Applications in Quality
Management
Statistics for Managers Using Microsoft Excel, 5e © 2008 Prentice-Hall, Inc.
Chap 18-1
Learning Objectives
In this chapter, you learn:
 The basic themes of quality management and
Deming’s 14 points
 The basic aspects of Six Sigma management
 How to construct various control charts
 Which control chart to use for a particular
type of data
 How to measure the capability of a process
Chap 18-2
Chapter Overview
Quality Management and
Tools for Improvement
Philosophy of
Quality
Deming’s 14
Points
Sigma®
Six
Management
Tools for Quality
Improvement
Control
Charts
Process
Capability
p chart
R chart
X chart
Chap 18-3
Total Quality Management
 Primary focus is on process improvement
 Most variation in a process is due to the
system, not the individual
 Teamwork is integral to quality management
 Customer satisfaction is a primary goal
 Organization transformation is necessary
 It is important to remove fear
 Higher quality costs less
Chap 18-4
Deming’s 14 Points
1. Create a constancy of purpose toward improvement
 become more competitive, stay in business, and provide
jobs
2. Adopt the new philosophy
 Better to improve now than to react to problems later
3. Stop depending on inspection to achieve quality -- build in
quality from the start
 Inspection to find defects at the end of production is too
late
4. Stop awarding contracts on the basis of low bids
 Better to build long-run purchaser/supplier relationships
Chap 18-5
5. Improve the system continuously to improve quality and thus
constantly reduce costs
6. Institute training on the job
 Workers and managers must know the difference between
common cause and special cause variation
7. Institute leadership
 Know the difference between leadership and supervision
8. Drive out fear so that everyone may work effectively.
9. Break down barriers between departments so that people can
work as a team.
Chap 18-6
10. Eliminate slogans and targets for the workforce
 They can create adversarial relationships
11. Eliminate quotas and management by numerical goals
12. Remove barriers to pride of workmanship
13. Institute a vigorous program of education and selfimprovement
14. Make the transformation everyone’s job
Chap 18-7
The Shewhart-Deming Cycle
Plan
Act
The
ShewhartDeming
Cycle
Study
Do
The key is a
continuous cycle
of improvement
Chap 18-8
Six Sigma® Management
A method for breaking a process into a series of steps:
 The goal is to reduce defects and produce near
perfect results
 The Six Sigma® approach allows for a shift of as
much as 1.5 standard deviations, so is essentially a
±4.5 standard deviation goal
 The mean of a normal distribution ±4.5 standard
deviations includes all but 3.4 out of a million
Chap 18-9
The Six Sigma® DMAIC Model
DMAIC represents
 Define -- define the problem to be solved; list costs,
benefits, and impact to customer
 Measure – need consistent measurements for each
Critical-to-Quality characteristic
 Analyze – find the root causes of defects
 Improve – use experiments to determine importance
of each Critical-to-Quality variable
 Control – maintain gains that have been made
Chap 18-10
Theory of Control Charts
 A process is a repeatable series of steps
leading to a specific goal
 Control Charts are used to monitor variation
in a measured value from a process
 Inherent variation refers to process variation
that exists naturally. This variation can be
reduced but not eliminated
Chap 18-11
Theory of Control Charts
 Control charts indicate when changes in data are due
to:
 Special or assignable causes
 Fluctuations not inherent to a process
 Data outside control limits or trend
 Represents problems to be corrected or improvements
to incorporate into the process
 Chance or common causes
 Inherent random variations
 Consist of numerous small causes of random variability
Chap 18-12
Process Variation
Total Process
Common Cause
Special Cause
=
+
Variation
Variation
Variation
 Variation is natural; inherent in the world
around us
 No two products or service experiences are
exactly the same
 With a fine enough gauge, all things can be
seen to differ
Chap 18-13
Process Variation
Total Process
Common Cause
Special Cause
=
+
Variation
Variation
Variation
Variation is often due to differences in:
 People
 Machines
 Materials
 Methods
 Measurement
 Environment
Chap 18-14
Process Variation
Total Process
Common Cause
Special Cause
=
+
Variation
Variation
Variation
Common cause variation
 naturally occurring and expected
 the result of normal variation in materials,
tools, machines, operators, and the
environment
Chap 18-15
Process Variation
Total Process
Common Cause
Special Cause
=
+
Variation
Variation
Variation
Special cause variation
 abnormal or unexpected variation
 has an assignable cause
 variation beyond what is considered
inherent to the process
Chap 18-16
Control Limits
Forming the Upper control limit (UCL) and the Lower
control limit (LCL):
UCL = Process Mean + 3 Standard Deviations
LCL = Process Mean – 3 Standard Deviations
UCL
+3σ
Process Average
- 3σ
LCL
time
Chap 18-17
Control Chart Basics
Special Cause Variation:
Range of unexpected variability
UCL
Common Cause
Variation: range of
expected variability
+3σ
Process Mean
- 3σ
LCL
time
Chap 18-18
Process Variability
Special Cause of Variation:
A measurement this far from the process average is
very unlikely if only expected variation is present
UCL
±3σ → 99.7% of
process values should
be in this range
Process Mean
LCL
time
Chap 18-19
Using Control Charts
Control Charts are used to check for process control
If the process is found to be out of control, steps
should be taken to find and eliminate the special
causes of variation
Chap 18-20
In-control Process
 A process is said to be in control when the
control chart does not indicate any out-ofcontrol condition
 Contains only common causes of variation
 If the common causes of variation is small, then
control chart can be used to monitor the process
 If the common causes of variation is too large, you
need to alter the process
Chap 18-21
Process In Control
 Process in control: points are randomly
distributed around the center line and all
points are within the control limits
UCL
Process Mean
LCL
time
Chap 18-22
Process Not in Control
Out of control conditions:
 One or more points outside control limits
 8 or more points in a row on one side of the
center line
 8 or more points in a row moving in the same
direction
Chap 18-23
Process Not in Control
One or more points outside
control limits
Eight or more points in a row on one
side of the center line
UCL
UCL
Process
Average
Process
Average
LCL
LCL
Eight or more points in a row
moving in the same direction
UCL
Process
Average
LCL
Chap 18-24
Out-of-control Processes
 When the control chart indicates an out-of-
control condition (a point outside the control
limits or exhibiting trend, for example)
 Contains both common causes of variation and
assignable causes of variation
 The assignable causes of variation must be
identified
 If detrimental to the quality, assignable causes of
variation must be removed
 If increases quality, assignable causes must be
incorporated into the process design
Chap 18-25
Control Chart for the
Proportion: p Chart
 Control chart for proportions
 Is an attribute chart
 Shows proportion of nonconforming items
 Example -- Computer chips: Count the number
of defective chips and divide by total chips
inspected
 Chip is either defective or not defective
 Finding a defective chip can be classified a
“success”
Chap 18-26
Control Chart for the
Proportion: p Chart
 Used with equal or unequal sample sizes
(subgroups) over time
 Unequal sizes should not differ by more than
±25% from average sample sizes
 Easier to develop with equal sample sizes
Chap 18-27
Creating a p Chart
 Calculate subgroup proportions
 Graph subgroup proportions
 Compute mean proportion
 Compute the upper and lower control limits
 Add centerline and control limits to graph
Chap 18-28
Average of Subgroup
Proportions
The average of subgroup proportions = p
If equal sample sizes:
If unequal sample sizes:
k
k
p
 pi
i1
k
p
X
i1
k
n
i 1
i
i
where:
where:
pi = sample proportion for subgroup i
Xi = the number of nonconforming
k = number of subgroups of size n
items in sample i
ni = total number of items sampled in
k samples
Chap 18-29
Computing Control Limits
 The upper and lower control limits for a p chart are
UCL = Average Proportion + 3 Standard Deviations
LCL = Average Proportion – 3 Standard Deviations
 The standard deviation for the subgroup proportions
is
(p)(1  p)
n
where:
n = mean subgroup size
Chap 18-30
 The upper and lower control limits for the p
chart are
p(1  p)
UCL  p  3
n
p(1  p)
LCL  p  3
n
Proportions are never
negative, so if the
calculated lower
control limit is
negative, set LCL = 0
Chap 18-31
p Chart Example
You are the manager of a 500-room hotel.
You want to achieve the highest level of
service. For seven days, you collect data on
the readiness of 200 rooms. Is the process in
control?
Chap 18-32
p Chart Example
Day
1
2
3
4
5
6
7
# Rooms
200
200
200
200
200
200
200
# Not
Ready
16
7
21
17
25
19
16
Proportion
0.080
0.035
0.105
0.085
0.125
0.095
0.080
Chap 18-33
p Chart Example
k
p
X
i1
k
i

n
i1
16  7    16
121

 .0864
200  200    200 1400
i
k
n
n
i
i1
k
200  200    200

 200
7
UCL  p  3
p(1  p)
.0864(1  .0864)
 .0864  3
 .1460
200
n
LCL  p  3
p(1  p)
.0864(1  .0864)
 .0864  3
 .0268
200
n
Chap 18-34
p Chart Example
P
0.15
UCL = .1460
_
p = .0864
0.10
0.05
0.00
LCL = .0268
1
2
3
4
5
6
7
Day
_
Individual points are distributed around p without any pattern. Any
improvement in the process must come from reduction of commoncause variation, which is the responsibility of management.
Chap 18-35
Understanding Process Variability:
Red Bead Experiment
The experiment:
 From a box with 20% red beads and 80% white
beads, have “workers” scoop out 50 beads
 Tell the workers their job is to get white beads
 Some workers will get better over time, some will
get worse
Chap 18-36
Morals of the
Red Bead Experiment
1.
2.
3.
4.
5.
Variation is an inherent part of any process.
The system is primarily responsible for worker
performance.
Only management can change the system.
Some workers will always be above average, and some will
be below.
Setting unrealistic goals is detrimental to a firm’s wellbeing.
Chap 18-37
R chart and X chart
 Used for measured numeric data from a
process
 Start with at least 20 subgroups of observed
values
 Subgroups usually contain 3 to 6
observations each
 For the process to be in control, both the R
chart and the X-bar chart must be in control
Chap 18-38
Example: Subgroups
Process measurements:
Subgroup measures
Subgroup
number
Individual measurements
(subgroup size = 4)
Mean, X
Range, R
1
15
17
15
11
14.5
6
2
12
16
9
15
13.0
7
3
17
21
18
20
19.0
4
…
…
…
…
…
…
…
Average
subgroup
Average
subgroup
mean = X
range = R
Chap 18-39
The R Chart
 Monitors variability in a process
 The characteristic of interest is measured on a
numerical scale
 Is a variables control chart
 Shows the sample range over time
 Range = difference between smallest and
largest values in the subgroup
Chap 18-40
The R Chart
Find the mean of the subgroup ranges (the
center line of the R chart)
2. Compute the upper and lower control limits
for the R chart
3. Use lines to show the center and control
limits on the R chart
4. Plot the successive subgroup ranges as a
line chart
1.
Chap 18-41
Average of Subgroup Ranges
Average of subgroup ranges:
R

R
i
k
where:
Ri = ith subgroup range
k = number of subgroups
Chap 18-42
R Chart Control Limits
The upper and lower control limits for an R chart are
UCL  D4 ( R )
LCL  D3 ( R )
where:
D4 and D3 are taken from the table
(Appendix Table E.11) for subgroup size = n
Chap 18-43
R Chart Example
You want to analyze the time it takes to
deliver luggage to the room. For 7 days, you
collect data on 5 deliveries per day. Is the
variation in the process in control?
Chap 18-44
R Chart Example
Day
1
2
3
4
5
6
7
Subgroup
Size
5
5
5
5
5
5
5
Subgroup
Average
5.32
6.59
4.89
5.70
4.07
7.34
6.79
Subgroup
Range
3.85
4.27
3.28
2.99
3.61
5.04
4.22
Chap 18-45
R Chart Example
R

R
i
k
3.85  4.27  ...  4.22

 3.894
7
UCL  D4 (R )  (2.114)(3.894)  8.232
LCL  D3 (R )  (0)(3.894)  0
D4 and D3 are from
Table E.11 (n = 5)
Chap 18-46
R Chart
Control Chart Solution
Minutes
UCL = 8.232
8
6
4
2
0
_
R = 3.894
LCL = 0
1
2
3
4
Day
5
6
7
Conclusion: Variation is in control
Chap 18-47
The X Chart
 Shows the means of successive subgroups
over time
 Monitors process average
 Must be preceded by examination of the R
chart to make sure that the variation in the
process is in control
Chap 18-48
The X Chart
 Compute the mean of the subgroup means (the
center line of the X chart)
 Compute the upper and lower control limits for
the X chart
 Graph the subgroup means
 Add the center line and control limits to the graph
Chap 18-49
Average of Subgroup
Means
Average of subgroup means:
X

X
i
k
where:
Xi = ith subgroup average
k = number of subgroups
Chap 18-50
 The upper and lower control limits for an X chart are
generally defined as
UCL = Process Average + 3 Standard Deviations
LCL = Process Average – 3 Standard Deviations
R
 Use d 2
to estimate the standard deviation of the process
average, where d2 is from appendix Table E.11
n
Chap 18-51
 The upper and lower control limits for an X chart are
generally defined as
UCL = Process Average + 3 Standard Deviations
LCL = Process Average – 3 Standard Deviations
 so
UCL  X  3
LCL  X  3
R
d2 n
R
d2 n
Chap 18-52
Simplify the control limit calculations by using
UCL  X  A 2 (R )
LCL  X  A 2 (R )
where A2 (from table E.11) =
3
d2 n
Chap 18-53
X Chart Example
You want to analyze the time it takes to
deliver luggage to the room. For seven days,
you collect data on five deliveries per day. Is
the process average in control?
Chap 18-54
X Chart Example
Day
1
2
3
4
5
6
7
Subgroup
Size
5
5
5
5
5
5
5
Subgroup
Average
5.32
6.59
4.89
5.70
4.07
7.34
6.79
Subgroup
Range
3.85
4.27
3.28
2.99
3.61
5.04
4.22
Chap 18-55
X Chart
Control Limits Solution
X

X
k
R

R
k
i
i
5.32  6.59    6.79

 5.814
7
3.85  4.27    4.22

 3.894
7
UCL  X  A2 ( R )  5.813  (0.577)(3.894)  8.061
LCL  X  A2 ( R )  5.813  (0.577)(3.894)  3.567
A2 is from Table E.11 (n = 5)
Chap 18-56
X Chart
Control Chart Solution
Minutes
8
6
4
2
0
1
UCL = 8.061
_
_
X = 5.814
LCL = 3.567
2
3
4
Day
5
6
7
Conclusion: Process average is in statistical control
Chap 18-57
Process Capability
 Process capability is the ability of a process to
consistently meet specified customer-driven
requirements
 Specification limits are set by management in
response to customers’ expectations
 The upper specification limit (USL) is the largest
value that can be obtained and still conform to
customers’ expectations
 The lower specification limit (LSL) is the smallest
value that is still conforming
Chap 18-58
Estimating Process Capability
 Must first have an in-control process
 Estimate the percentage of product or service
within specification
 Assume the population of X values is
approximately normally distributed with mean
estimated by X and standard deviation
estimated by R / d2
Chap 18-59
For a characteristic with a LSL and a USL
P(outcome will be within specifications)




USL  X 
 LSL  X
 P(LSL  X  USL)  P
Z

R
R


d2
 d2

Where Z is a standardized normal random variable
Chap 18-60
For a characteristic with only an USL




USL  X 

 P( X  USL)  P Z 

R


d2


Chap 18-61
For a characteristic with only a LSL




 LSL  X

 P(LSL  X)  P
 Z
R


 d2

Chap 18-62
Process Capability
Example
You have instituted a policy that 99% of all
luggage deliveries must be completed within
ten minutes or less. For seven days, you
collect data on five deliveries per day. You
know from prior analysis that the process is
in control. Is the process capable?
Chap 18-63
Process Capability
Example
Day
Subgroup
Size
Subgroup
Average
Subgroup
Range
1
2
3
4
5
6
7
5
5
5
5
5
5
5
5.32
6.59
4.89
5.70
4.07
7.34
6.79
3.85
4.27
3.28
2.99
3.61
5.04
4.22
Chap 18-64
Process Capability
Example
n5
X  5.814
R  3.894
d 2  2.326



10  5.814 

 P( X  10)  P Z 
3.894 



2.326 

 P( Z  2.50)  .9938
Therefore, we estimate that 99.38% of the luggage deliveries will be
made within the ten minutes or less specification. The process is
capable of meeting the 99% goal.
Chap 18-65
Capability Indices
 A process capability index is an aggregate
measure of a process’s ability to meet
specification limits
 The larger the value, the more capable a
process is of meeting requirements
Chap 18-66
Cp Index
A measure of potential process performance is the
Cp index
USL  LSL specification spread
Cp 

process spread
6( R / d 2 )
Cp > 1 implies a process has the potential of having
more than 99.73% of outcomes within specifications
Cp > 2 implies a process has the potential of meeting
the expectations set forth in six sigma management
Chap 18-67
CPL and CPU
To measure capability in terms of actual process
performance:
X  LSL
CPL 
3(R / d2 )
CPU 
USL  X
3(R / d2 )
CPL (CPU) > 1 implies that the process mean is more
than 3 standard deviation away from the lower
(upper) specification limit
Chap 18-68
CPL and CPU
 Used for one-sided specification limits
 Use CPU when a characteristic only has a UCL
 Use CPL when a characteristic only has an
LCL
Chap 18-69
Cpk Index
 The most commonly used capability index is the Cpk index
 Measures actual process performance for characteristics with
two-sided specification limits
Cpk = min(CPL, CPU)
 Cpk = 1 indicates that the process average is 3 standard
deviation away from the closest specification limit
 Larger Cpk indicates greater capability of meeting the
requirements, e.g., Cpk > 1.5 indicates compliance with six
sigma management
Chap 18-70
Process Capability
Example
You have instituted a policy that all luggage
deliveries must be completed within ten
minutes or less. For seven days, you collect
data on five deliveries per day. You know
from prior analysis that the process is in
control. Compute an appropriate capability
index for the delivery process.
Chap 18-71
Process Capability
Example
n5
X  5.814
R  3.894
d 2  2.326
USL  X
10  5.814
CPU 

 0.8335
3( R / d 2 ) 3(3.894 / 2.326)
Since there is only the upper specification limit, we need to
only compute CPU. The capability index for the luggage
delivery process is .8337, which is less than 1. The upper
specification limit is less than 3 standard deviation above
the mean.
Chap 18-72
Chapter Summary
In this chapter, we have
 Reviewed the philosophy of quality management
 Deming’s 14 points
 Discussed Six Sigma® Management
 Reduce defects to no more than 3.4 per million
 Uses DMAIC model for process improvement
 Discussed the theory of control charts
 Common cause variation vs. special cause
variation
Chap 18-73
Chapter Summary
In this chapter, we have
 Constructed and interpreted p charts
 Constructed and interpreted X and R charts
 Obtained and interpreted process capability
measures
Chap 18-74

Statistics for Managers Using Microsoft® Excel 5th Edition Chapter 18

Transcription

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