A MindPro Lean Six Sigma Simulated Training Project

MidCo Case Analysis – Candidate Edition A MindPro Lean Six Sigma Simulated Training Project ©2015 Dr. Mikel J. Harry, Ltd. and Alan M. Leduc All Rights Reserved. No part of this book may be reproduced or utilized in any form or by any means, electronic or mechanical, including photocopying, recording, or by any information storage and retrieval system, without permission in writing from the authors.

MindPro® is a registered trademark of Dr. Mikel J. Harry, Ltd.

Six Sigma® is a registered trademark of Motorola, Inc.

MINITAB® is a registered trademark of Minitab, Inc

Alan M. Leduc

Mikel J. Harry, Ph.D.

MidCoCaseAnalysisAMindProLeanSixSigmaSimulatedTrainingProject

MidCo Case Analysis 2

A MindPro ™ Lean Six Sigma Simulated Training Project

Copyright 2015 Dr. Mikel J. Harry, Ltd. & Alan M. Leduc

3370 North Hayden Road, Suite 123‐320, Scottsdale, AZ 85251 info@ss‐mi.com 480 515‐0890 fax 480 515‐0891

Preface

Dr. Mikel J. Harry is the co‐creator of Six Sigma and developer of the MindPro™ Lean Six Sigma Training. Alan

Leduc is a protégé Dr. Mikel J. Harry, and a Lean Six Sigma Master Black Belt and Executive Master Black Belt.

Dr. Harry developed MindPro™ because of two concerns with regard to existing training: 1) the variability of

training in the field and 2) the difficulty of candidates applying the concepts to projects.

Dr. Harry believed that the variability with regard to the rigor of Lean Six Sigma training had become

unacceptable. He and his team set out to identify best practices for Lean Six Sigma training and developed the

MindPro™ training based upon those best practices. The MindPro™ training is based upon the pedagogy of

providing a full understanding of the concepts and tools prior to proceeding to application via projects. As the

candidate proceeds through the training they are building a strong understanding of Lean Six Sigma concepts

and learning about the Lean Six Sigma Toolkit (some of which are classical tools which have been around for

decades). By offering the MindPro™ training online at an affordable price, more candidates are able to receive

their training direct from the co‐creator of Six Sigma based upon research of best practices, thus reducing

training variability.

Lean Six Sigma Green and Black Belt candidates expressed concerns about assimilating Lean Six Sigma concepts

and tools with regard to applying them to their initial project. Dr. Harry responded to this training gap by

incorporating a simulated project into the MindPro™ training. The simulated project is based upon a

hypothetical company called “MidCo.” The simulated project provides the candidate the opportunity to

participate in the DMAIC methodology as it applies to a project. Secondarily, Dr. Harry and his team used the

simulated project to enhance the student’s understanding of statistical analysis by providing extensive reference

to Minitab and Excel.

Alan M. Leduc performed the original case analysis of the MidCo simulated project in 2007 using Minitab 14

prior to the development of MindPro Lean Six Sigma training exam questions.

In 2015, Dr. Harry and Alan Leduc discussed ways additional support could be provided for candidates who were having difficulty with the Midco Case Study Analysis and to support directors or satellite organizations we were providing their own support of MindPro Lean Six Sigma training. These discussions lead to:

Development of “Candidate Edition” of this document which blanks out all answers to the MindPro Lean Six Sigma Exam Questions.

Development of an Exam Workbook that would provide the candidate step by step instructions with regard to how to navigate the statistical analysis using the most recent version of Minitab (Version 17) and JMP (Version 12).





This case study while updated to ensure all MindPro Lean Six Sigma Exam Questions were covered but was not updated from the perspective of statistical analysis. The graphs and data tables were performed in Minitab 14 and will be different in appearance in later versions of Minitab or other statistical packages. The purpose of this document is not to provide detailed instruction on how to do the analysis but rather provide a rational through which a candidate might proceed as they work through an actual project.





Assumptions

In the body of the analysis, you will find sections indicating contact was made with MidCo. These sections

identify assumptions necessary to complete a full analysis of the simulated case study. The assumptions are

stated in the form of questions, and possible responses obtained from MidCo project sponsors.

It is not unusual in a project to be provided data which needs clarification. Verifying and clarifying this data is an

important part of the Lean Six Sigma process.

Candidate Analysis Matrix

Black Belt

Green Belt - Industrial

Green Belt - Commercial





Executive Summary

MidCo requested that we analyze data from their Multi‐National 5 location operation. Market share has

been increasing for the past 3 years and they wish to implement a six sigma program on their core operation

with the strategic goal of increased profitability. Tactically they desire to decrease cost and increase

customer satisfaction.

Guiding Questions:

Guiding Questions are those questions that must be answered during the RDMAIC (Recognize‐Design‐

Measure‐Analyze‐Improve‐Control) process of Six Sigma Analysis.

R What business performance metric should be used to drive Six Sigma?

D What product characteristics must be isolated and improved?

M What is the actual and potential capability of the core CTQ’s?

A What is the actual and potential capability of the CTP’s?

I What are the Vital Few CTP’s and what should their optimal settings be?

C What level of process control can be sustained over time?

Recognize

What business performance metric should be used to drive Six Sigma?

Yield in the United States facility was the primary metric that we utilized to drive Six Sigma. We have yet to

do the follow‐up analysis, but since we improved yield we should see the following effects in the follow‐up

analysis: reduced COPQ; increased productivity; more consistency with deliveries; decreased costs;

increased profitability; and improved customer satisfaction. Once the follow‐up analysis validates our gains

we should use the work in the United States plant as a standard for improving the other facilities.


A MindPro ™ Lean Six Sigma Simulated Training Project Executive Summary



Define

What product characteristics must be isolated and improved?

In the United States facility, we identified four CTQ’s (1,2,3, and 7) that should be considered for isolation

and improvement. However, since this was MidCo’s first Six Sigma project, we were asked to provide

improvement estimates based only the two CTQ’s that had the highest failure rate, CTQ’s 1 and 2.

Measure

What is the actual and potential capability of the core CTQ’s?

The initial analysis of capability is summarized as follows:

CTQ‐1 had short term capability of 3.54 sigma (Cp = 1.18) and a long term capability of 2.22 sigma

(Pp = 0.74). It was evident that the overall capability of the process would be improved by

eliminating assignable causes. CTQ‐1 was also centered below (mean = 73.4) the target (80).

CTQ‐2 had a short term capability of 0.63 sigma (Cp = 0.21) and a long term capability of 0.57 sigma

(Pp = 0.19). Since the Cp and Pp values were comparable there is little room for improvement in

process without improving the technology. CTQ‐2 was centered above the (mean = 534.6) the target

(500).

Analyze

What is the actual and potential capability of the CTP’s?





Analysis of the CTP’s is summarized below:

CTP LSL/Target/USL

Sample

Mean

Centering

Issue

Spread Issue

Potential

Capability

Overall

Capability

Potential

Capability

Overall

Capability

X1 90/(N/A)/(N/A) 100.07 Yes No No 1.04 1.04

X2 31/33/35 33.17 No Yes Yes 1.76 0.99

X3 (N/A)/(N/A)/310 300.15 Yes No No 1.56 1.41

X4 100/105/110 100.33 Yes No Yes 0.35 0.11

X5 1.15/1.20/1.25 1.25 Yes No Yes ‐0.24 ‐0.07

CTQ‐2: Tuesday and Thursday were substantially different from the other days and had positive impact on

centering.

Improve

What are the Vital Few CTP’s and what should their optimal settings be?

As discovered during the analyze phase, several processes were out of control. We consulted with MidCo

team to bring these processes under control prior to proceeding to the improve stage. The table below

compares the processes before and after addressing the control issues.





Before After

CTP/CTQ SD Within SD Overall Control SD Within SD Overall Control

X1 9.70 9.71 Yes 6.72 6.69 Yes

X2 .95 1.42 No 0.71 0.74 Yes

X3 6.29 6.99 Marginal 4.79 4.58 Yes

X4 0.934 2.99 No 2.19 2.05 Yes

X5 0.006 0.021 No 0.014 0.013 Yes

CTQ‐1 9.88 15.84 No 9.29 8.98 Yes

CTQ‐2 55.89 62.82 Marginal Unavailable

MidCo advised that they were more concerned with CTQ‐1 than CTQ‐2 and made the decision to go with the

following settings:

X1 X2 X3 X4 X5

Pre‐improve settings 100 33 300 100 1.25

Our recommendations >100* 33 300 105 1.20

Post‐improve settings 80 33 300 145 1.15

* Additional analysis needed.

MidCo made the decision not to pursue CTQ‐2. By optimizing CTQ‐1 only, the process for CTQ‐2 will likely

deteriorate The variance of CTQ‐1 was optimized by reducing X1 and X5. However, the aim of X1 was below

specifications and X5 was aimed at the lower specification. X4 was increased for purposes of moving the aim

of CTQ‐1 closer to its target. However, X4 is set far beyond its upper specification limit.





Control

What level of process control can be sustained over time?

The aim of X1 was reduced below the specification. MidCo indicated that they were going to revise the

specification limit of X1 but have not done so as yet. We recommend that the unilateral specification be

revised so that the lower specification limit is 60 for a 3.14 sigma (840 ppm defective) or 40 for a six sigma

process.

X2 is properly centered but only has a 2.30 sigma value (10692 ppm defective). The overall capability is

statistically the same as the within capability so there is little if any improvement available in the process. In

order to improve X2, we will have to have a breakthrough in the process which will likely require improved

technology.

X3 is operating at 2.55 sigma (5391 ppm defective). The overall capability is statistically the same as the

within capability so like X2 in order to improve X3, a process breakthrough likely requiring new technology

will be required.

X4 like X1 has specification issues. X4 was leveraged by moving the process upward significantly in order to

improve the variance of CTQ‐1. MidCo advised that they were going to change the specifications but have

not done that yet. If MidCo moves the target of X4 to 145 and maintains the tolerance of +/‐5, the process

will still only be at 1.86 sigma (31170 ppm defective). The overall capability of X4 is statistically the same as

within capability so a breakthrough in the process, likely requiring technology improvement, will be required

to improve the process.

X5 also has some serious centering issues. MidCo advised early on that they were suspect of the

specification limits on X5 but have yet to make any adjustments. If the specification is targeted at 1.15 and

the same tolerance of +/‐ 0.05 is maintained, the process would 3.02 sigma capable with (1281 ppm

defective).

All CTP’s are under statistical control.

CTQ‐1 is essentially at six sigma capability (zero ppm defective) and is in control. No information has been

provided by MidCo as of yet on CTQ‐2 so it remains to be analyzed.





Additional Tasks

In addition to the six sigma analysis we were ask to analyze a customer survey that was given prior to the six

sigma implementation and a risk analysis calculator.

Survey

Those that have the most need for the MidCo’s product seem to be satisfied with the quality of the product; but,

have a low overall satisfaction with the product. This deserves some investigation.

The Cost and Delivery may be contributors to the overall dissatisfaction but they are not statistically

significant.

A focus group could provide some insight as to other issues that might be degrading overall satisfaction.

Once a list of potential factors is developed in the focus group, a new survey should be done using the

demographics of the group with the most need.

Those with less income tend to be less satisfied with delivery than those with higher income.

This may indicate that those with higher income are using express delivery services that cannot be

afforded by those with less income.

In the survey that is recommended above using the demographics of those with less income, questions

should be ask regarding preferred deliver methods and cost of delivery.

Investigation should be done in Region B with regard to potential delivery issues.

Risk Calculator

The risk calculator provided by MidCo either has an error or the risk factor used in the calculator is misstated.

MidCo stated that the risk factor was the risk of an individual failure exceeding its Limiting Cost. However, the

template is designed so that the risk is actually the risk of the sum of Limiting Costs of all failures of a particular

CTQ exceeding the Monthly Limiting Cost.





Introduction

MidCo Facts:

Product Line: One product line with supporting services

Size: Multi‐National with 5 locations

o South Africa

o Europe

o Pacific Rim

o United States

o Middle East

Market Share: Increasing for the last 3 years

Strategic Goal: Increase profitability

Tactical Goal: Decrease cost and increase customer satisfaction

Operational Focus: Improve yields and decrease defect rates

Operational Target: Improve capability of core process

Operational Plan: Install Six Sigma





Guiding Questions

Guiding Questions are those questions that must be answered during the RDMAIC (Recognize‐Design‐

Measure‐Analyze‐Improve‐Control) process of Six Sigma Analysis.

R What business performance metric should be used to drive Six Sigma?

D What product characteristics must be isolated and improved?

M What is the actual and potential capability of the core CTQ’s?

A What is the actual and potential capability of the CTP’s?

I What are the Vital Few CTP’s and what should their optimal settings be?

C What level of process control can be sustained over time?





Recognize

What business performance metric should be used to drive Six Sigma?

Summary

Sales

o The Middle East, United States, Pacific Rim, and South Africa seasonal patterns

are such that data collection and improvement projects requiring major effort

should be implemented at the end of the third quarter due to a downturn in

sales in the fourth quarter. Priority should be given to South Africa due to

higher sales. Projects in Europe should be implemented at the end of the first

quarter.

o The Europe market should be considered as a test location where problems

could be worked out and then later implemented at the remaining sites due to

the difference in seasonal patterns versus the other locations.

Profit

MidCo sponsors were contacted regarding the conflict between Profit and

Calculated Profit. They advised Cost $ are actually Operating Cost $. Based upon

this communication, Profit $ are more accurately described as

Operating Profit $ = Sales $ ‐ Operating Cost $

Profit $ = Operating Profit $

Profit = Net Profit Margin

o Europe should be the benchmark in terms of Overhead Costs

Candidate Conclusions

What is your conclusion regarding whether or not there is a correlation

between Sales $ and Units sold? Is there a need to analyze Sales $ and

Units Sold separately or not?


A MindPro ™ Lean Six Sigma Simulated Training Project Recognize



o The Middle East appears to be the most balanced location in terms of Profit

Analysis and has the second highest Net Profit Margin at 7.76%.

o The Pacific Rim, Europe, and the Middle East have reasonable Net Profit

Margins and an adequate overhead structure.

o Overhead Costs should be evaluated in the South Africa facility.

o The Middle East, South Africa, and Pacific Rim facilities all have Operating

Profits as a percent of total that outpace their sales as a percent of total. We

should look to these locations for benchmarking the operation side of our

facilities.

o The United States facility appears to need the most attention.

o There is a correlation between Sales and Operating Profit. This may mean that

we have utilization issues in some of the facilities.

COPQ

o Although Cost of Poor Quality is generally difficult to measure, it appears that

MidCo is being consistent in the way that it measures COPQ given the relative

stability in the Europe, Pacific Rim, and Middle East facilities which are hovering

around 5% of Sales regardless of sales levels.

o We need to investigate Quality Issues in both South Africa and the United

States.

o There is an inverse correlation between yield and COPQ, as yield increases,

COPQ decreases.


What are revenues, operating profit, net profit, and profit margin for the

United States location?





Yield

o The United States has by far the most significant variation and should be the

focus of yield analysis.

o Yield has a positive correlation to Net Profit Margin; but not to Operating Profit.

Cost

o There is no indication of a difference in the variances or means of Cost$ in the

first half of the year versus the last half of the year.

While in the long term we might want to look at the overheard structure of South Africa, we do

not currently have sufficient data to do this analysis.

As we proceed to the Design Stage the focus should be on Yield as the primary metric to drive

Six Sigma. Analysis has shown that as yield is improved, COPQ will be reduced resulting in an

increase in productivity and Net Profit Margin. More consistency in deliveries should be

attained. Focusing on yield should allow achievement of the strategic objective of “Increasing

profitability” and accomplish the tactical objectives of “decreasing costs and increasing

customer satisfaction.

The yield of the United States facility had a significant negative impact on the overall yield of

the corporation and thus should be the focus of our analysis in the Design Stage.


Is there a difference in the variances of means of Cost $ in the first half of

the year versus the last half of the year?





The data in Figure R1 was provided by MidCo for FY 2004

Case Location Quarter Units Profit Yield COPQ Sales $ Cost $ Profit $ COPQ $

1 SA 1 562 0.075 0.962 0.085 3,935,765 2,493,328 1,442,437 211,768

2 SA 2 604 0.045 0.930 0.128 4,228,535 3,194,458 1,034,077 409,337

3 SA 3 589 0.026 0.942 0.224 4,121,290 2,866,397 1,254,892 643,361

4 SA 4 271 0.058 0.956 0.094 1,894,870 1,372,529 522,341 129,387

5 EU 1 542 0.096 0.930 0.074 3,797,170 3,497,154 300,016 258,494

6 EU 2 112 0.095 0.976 0.060 784,231 730,832 53,398 43,486

7 EU 3 285 0.093 0.990 0.076 1,991,867 1,353,535 638,332 102,560

8 EU 4 578 0.065 0.967 0.084 4,047,975 2,289,659 1,758,317 192,708

9 PR 1 186 0.082 0.930 0.075 1,304,342 1,241,424 62,918 92,771

10 PR 2 194 0.134 0.961 0.039 1,357,699 892,116 465,584 34,875

11 PR 3 437 0.054 0.939 0.102 3,058,485 1,737,041 1,321,444 177,536

12 PR 4 205 0.055 0.977 0.105 1,434,454 933,015 501,439 98,217

13 US 1 141 0.031 0.748 0.132 987,624 780,857 206,767 103,344

14 US 2 146 0.064 0.940 0.091 1,020,456 803,900 216,556 90,718

15 US 3 451 0.059 0.914 0.081 3,154,245 2,310,022 844,223 227,648

16 US 4 245 0.023 0.706 0.120 1,718,113 1,438,424 279,689 398,718

17 ME 1 177 0.057 0.927 0.096 1,240,260 802,150 438,110 76,739

18 ME 2 206 0.099 0.944 0.077 1,439,847 1,038,168 401,679 79,562

19 ME 3 645 0.084 0.950 0.063 4,512,958 2,686,042 1,826,916 169,123

20 ME 4 251 0.058 0.936 0.095 1,755,401 1,227,392 528,009 116,602

Total = 6,827 47,785,587 33,688,443 14,097,144 3,656,954

MidCo Corporate Dashboard for FY 2004

FIGURE R 1





The next several pages provide a thorough analysis of the variables provided in the data from MidCo. The

purpose of the analysis is to determine which of the performance variables should be used to drive Six Sigma.

The Recognize Phase analysis was dissected in to analysis of five sections:

1. Sales Analysis

2. Profit (Operating Profit and Net Profit Margin)

3. Cost of Poor Quality

4. Yield

5. Cost

Sales Analysis

A one‐way analysis of variance was performed to determine if Sales $ were significantly different by location.

The normal probability plot of the residuals is shown in Figure R 2 and is analyzed by “the fat pencil test.” If

points in the middle 75% of the plot can be covered by a “fat pencil,” the general conclusion is that the residuals

follow a normal distribution. If the residuals are normal, it can be concluded that the data itself is normal;

validating the acceptability of using ANOVA as an analytical tool.

Candidate Analysis Required for

Black Belt, Green Belt ‐ Industrial, and Green Belt ‐

Commercial

See MidCo Exam Workbook: Recognize Task 8


p‐value? Location statistically significant with regard to

Sales $? Accept the null or alternate hypothesis?

What it the confidence level? R 2 value?


Are the residuals normal? Is the data normal?





FIGURE R 2

FIGURE R 3

While location is not a statistically significant factor, empirically, over the one year evaluation period, South

Africa and Europe contribute slightly more than half of the total sales with Middle East, Pacific Rim, and United

States sharing the remaining half approximately equally as can be seen in Figure R 3.


Black Belt, Green Belt ‐ Industrial, and

Green Belt ‐ Commercial






The Multi‐Vari charts in Figure R 4, and Quarterly Sales chart Figure R 5 indicates that the sales patterns in the

Middle East, Pacific Rim and United States follow the same seasonal cycles with a peak in the third quarter.

South Africa seems also to have only two seasons but has an extended Peak season over quarters 1‐3 and a

Down season for quarter 4. Europe’s Peak season is Quarters 1 and 4 with moderate sales in quarter 3 and a

down season in Quarter 2.

Location

Sale

s $

USSAPRMEEU

5000000

4000000

3000000

2000000

1000000

Quarter1234

Multi-Vari Chart for Sales $ by Quarter - Location

FIGURE R 4

FIGURE R 5





It is recommended that for the Middle East, United States, Pacific Rim, and South Africa that data collection and

improvement projects requiring major effort be implemented following the decline of sales beginning at the end

of the third quarter. Priority should be given to South Africa due to their short down season. Projects in Europe

should be implemented after their decline in sales at the end of the first quarter. Since the Europe market is out

of sync with the other locations this market might be considered as a test location where problems could be

worked out and then later implemented at the remaining sites.

A scatterplot was created to determine whether or not there was correlation between Units Sold and Sales

Dollars and a regression analysis was performed to verify the graphical conclusion.

FIGURE R 6






Linear Correlation between Units Sold and Sales Dollars?

If so, positive or negative correlation?





PROFIT ANALYSIS

A bar chart was developed to show Profit $ by Location.

MEUSPREUSA

4000000

3000000

2000000

1000000

0

Location

Prof

it $

Chart of Profit $

FIGURE R 7

Analysis of the data indicates that Sales $ = Profit $ + Cost $. (There is one unit variation in four data points that

appear to be simply rounding errors, as the total is correct.) However, there appears to be an issue between

Profit and the calculated Profit$/Sales$.




See MidCo Exam Workbook:

Recognize Task 2


Location with most profit? Least profit?

Ratio of most profit versus least profit by location?





Case Profit Calc Profit

(Profit$/Sales$)

1 0.075 0.366

2 0.045 0.245

3 0.026 0.304

4 0.058 0.276

5 0.096 0.079

6 0.095 0.068

7 0.093 0.320

8 0.065 0.434

9 0.082 0.048

10 0.134 0.343

Case Profit Calc Profit

(Profit$/Sales$)

11 0.054 0.432

12 0.055 0.350

13 0.031 0.209

14 0.064 0.212

15 0.059 0.268

16 0.023 0.163

17 0.057 0.353

18 0.099 0.279

19 0.084 0.405

20 0.058 0.301

FIGURE R 8

Regression analysis comparing Profit (X) to Calculated Profit (Y) results in an invalid model (p = .817). Therefore

it appears that there is no relationship between the two variables and that some of the provided data must be

erroneous. Sales and costing figures are generally accurate. Therefore it appears that there must be either an

error in the Profit $ or the Profit (proportion) data provided. Note: Profit is substantially lower than the

Calculated Profit (Profit/Sales). The following hypothesis regarding the data is likely:

As described above the relationship of Sales $ = Profit $ + Cost $ is depicted in the data. It is possible

that somebody simply gathered the Sales $ and the Cost $ and then made a subtraction to derive

the Profit $. MidCo sponsors were contacted regarding the conflict between Profit and Calculated

Profit. They advised Cost $ are actually Operating Cost $. Based upon this communication, Profit $

would be more accurately described as:

Operating Profit $ = Sales $ ‐ Operating Cost $





The Profit data provided seem reasonable as Net Profit Margin numbers.

Therefore it is possible both sets of numbers are accurate but mislabeled.

o Profit $ = Operating Profit $

o Profit = Net Profit Margin

The lack of correlation between the two could be attributable to differences in Overhead Costs.

Summary of analysis based upon these assumptions is provided in

Figure R 9 ‐ Figure R 111 and Error! Bookmark not defined.; however, until the assumptions can be

verified the analysis should be considered tentative.

SA ME EU PR US Total

Annual Net Profit $ 702,522 694,141 887,392 532,941 321,543 3,138,540

% Total 22.38% 22.12% 28.27% 16.98% 10.24% 100.00%

Annual Operating Profit $ 4,253,747 3,194,714 2,750,063 2,351,385 1,547,235 14,097,144

% Total 30.17% 22.66% 19.51% 16.68% 10.98% 100.00%

FIGURE R 9

Sales

(% Total)

Operating Profit

(% Total)

Net Profit

(% Total)

Net Profit

Margin

SA 29.68% 30.17% 22.38% 4.95%

ME 18.73% 22.66% 22.12% 7.76%

EU 22.23% 19.51% 28.27% 8.35%

PR 14.97% 16.68% 16.98% 7.45%

US 14.40% 10.98% 10.24% 4.67%

FIGURE R 10





A regression analysis was performed to determine if there was a relationship between Profit $ and

Sales $. Figure R10 reflects a strong visual correlation which coincides with the analysis indicating a statistical

significance of the equation Operating Profit $ = ‐ 146537 + 0.356 Sales $ with a p‐value of 0.000 and an R2 value

of 71.2%. A scatterplot was created to show the visual effect of the correlation. Error! Bookmark not defined.

FIGURE R 11





A regression analysis was done to determine if there is a correlation between Profit$ and the COPQ$.

7000006000005000004000003000002000001000000

2000000

1500000

1000000

500000

0

COPQ $

Prof

it $

Scatterplot of Profit $ vs COPQ $

FIGURE R 12

A request was made by MidCo to determine how the variation in Profit $ compared from location to location

versus Quarter to Quarter. The recommended tool was the multi‐vari chart ‐ provided in

Figure R 13. The chart on the left shows the location mean in red and the quarters plotted in each grouping. The

chart on the right shows the quarterly mean in red and has locations plotted in each grouping. It is difficult to

determine from these charts, whether location to location or quarter to quarter has the most variation. Multi‐

vari charts are excellent for showing patterns; calculation and comparison of standard deviation or ANOVA are

better tools for comparing variation.






Linear Correlation between Profit $ and COPQ $?

If so, positive or negative correlation?





USSAPRMEEU

2000000

1500000

1000000

500000

0

Location

Prof

it $

1234

Quarter

Multi-Vari Chart for Profit $ by Quarter - Location

4321

2000000

1500000

1000000

500000

0

Quarter

Prof

it $

EUMEPRSAUS

Location

Multi-Vari Chart for Profit $ by Location - Quarter

FIGURE R 13

Since it is not clear from the multi‐vari charts whether location to location or quarter to quarter has the most

variation, a statistical analysis was performed to determine the standard deviation.

Location Profit $ Quarter Profit $

SA 4,253,747 1 2,450,248

EU 2,750,063 2 2,171,294

PR 2,351,385 3 5,885,807

US 1,547,235 4 3,589,795

ME 3,194,714

Standard Deviation = Standard Deviation =

FIGURE R 14

An analysis of variance was also performed analyzing with location and quarters as the factors and Profit $ as

the response.





Recognize Task 6





Recognize Task 6


Black Belt, Green Belt ‐ Industrial, and Green Belt ‐ Commercial


Recognize Task 6


Considering Location‐to‐Location versus Quarter‐to‐

Quarter, where does the most variation in Profit $ occur?





Casual Observations with regard to Profit Analysis:

South Africa accounts for nearly one‐third of the Sales and Total Operating Profit for the corporation but

only about one‐fifth of the Net Profit. South Africa is near the bottom in terms of Net Profit Margin at 4.95%.

The United States is at the bottom with 4.67% and Europe is at the top with 8.35%.

o Overhead Costs should be evaluated in the South Africa facility.

Europe accounts for slightly more than one‐fifth of corporate Sales and slightly less than one‐fifth of the

Total Operating Profit; however, Europe accounts for slightly more than one‐fourth of the Net Profit and is

the Net Profit Margin leader at 8.35%.

o It appears the Europe should be the benchmark in terms of Overhead Costs.

The Middle East accounts for about one‐sixth of corporate Sales and more than one‐fifth of the Total

Operating Profit and Net Profit. The Middle East has the second highest Net Profit Margin at 7.76%.

o The Middle East appears to be the most balanced location in terms of Profit Analysis.

The Pacific Rim like the United States only accounts for about one‐seventh of corporate Sales; but, unlike

the United States, accounts for approximately one‐sixth of the Total Operating Profit and Net Profit. The

Pacific Rim has a Net Profit Margin of 7.45% which compares favorably with Europe and the Middle East.

o The Pacific Rim appears favorable when evaluating the overall profit picture.

The United States accounts for about one‐seventh of corporate Sales. However, Operating Profit and Net

Profit are only about one‐tenth of the corporation’s total. The Net Profit Margin is the lowest of all facilities

at 4.67%.

o The United States facility is substantially below the other facilities in terms of profit analysis.





General Conclusions from Profit Analysis:

The overhead structure for South Africa should be evaluated.

o This is the leading facility in terms of Revenue and Operating Profit. If the overhead structure

can be improved this facility could be the benchmark facility.

o We should look at Europe as the Benchmark facility for evaluating the South Africa overhead

structure.

The Pacific Rim, Europe, and the United States appear to have an adequate overhead structure.

However, their overhead structures should be compared to Europe in order to gain marginal

improvements.

The Middle East, South Africa, and Pacific Rim facilities all have Operating Profits as a percent of total

which outpace their sales as a percent of total. These locations should be utilized for benchmarking the

operation side the facilities.

The Middle East and Pacific Rim facilities appear to in good operational and fiscal condition.

The United States facility appears to need the most attention.

o It has the lowest revenues at 14.40% of .total

o It has the lowest operating profit at 10.98% of total.

o It has the lowest net profit at 10.24% of total

o It has the lowest net profit margin at 4.67%

Regression analysis reveals that there is a correlation between Sales and Operating Profit. The

regression equation is Operating Profit $ = ‐0.147 + 0.356 Sales $. R2 = .712, that is 71.2% of the variation

in Profit is explained by Sales $’s which for cross‐sectional data is excellent.

o This may mean there are utilization issues in some of the facilities.





Analysis of Cost of Poor Quality (COPQ)

MidCo personnel were interested in determining whether or not COPQ$ was higher in any particular quarter. A

bar Chart was developed from the reported data and is displayed in Figure R 15.

4321

1400000

1200000

1000000

800000

600000

400000

200000

0

Quarter

COPQ

$Chart of COPQ $

FIGURE R 15

FIGURE R 16





Recognize Task 1


In which Quarter did the highest COPQ $ occur?





Recognize Task 3





COPQ data is reported as dollars (COPQ$) and as a decimal (COPQ). COPQ is often reported as a percent of sales.

However, as can be seen in

Figure R 17, this calculation does not match the reported data. Only the U.S. data is reported in

Figure R 17; however, this was true for all cases. It was determined COPQ is calculated as COPQ $ / Cost $, in all

cases except for the U.S.1 As can be seen from Figure R14, Quarter 1 COPQ as a percent of cost matches the

reported data, however, the other three quarters have reporting errors.

Location Quarter COPQ $ Sales $ COPQ

(%Sales) Cost $

COPQ

(Original)

COPQ1

(% Cost) COPQ ‐ COPQ1

US 1 103344 987,624 10.46% 780,857 0.132 0.132 0.000

US 2 90718 1,020,456 8.89% 803,900 0.091 0.113 ‐0.022

US 3 227648 3,154,245 7.22% 2,310,022 0.081 0.099 ‐0.018

US 4 398718 1,718,113 23.21% 1,438,424 0.120 0.277 ‐0.157

FIGURE R 17

Is the U.S. reporting error in the COPQ decimal or the COPQ $ in the reported data? A graph was developed and

is provided in Figure R 18.1 As can be seen from the graph, the calculated values appear to be erroneous.


The Run (Line) Chart is an appropriate for determining

time related patterns in this case?





FIGURE R 18

MidCo personnel were contacted regarding the apparent error and they confirmed, the COSQ $ was incorrectly

reported in the original data and the COPQ values shown as a decimal were in fact correct.

Figure R 19 shows the corrected COPQ $ for the U.S. and the variance from the reported data.

Location Quarter COPQ $

(Reported)

COPQ $

(Corrected)

COPQ

Variance

US 1 103,344 103,073 271

US 2 90,718 73,155 17,563

US 3 227,648 187,112 40,536

US 4 398,718 172,611 226,107

FIGURE R 19

Given the reporting error for COPQ $, the requested bar chart showing COPQ $ by quarter was regenerated and

is provided in Figure R 20 . As was originally reported to MidCo personnel, Quarter 3 has higher COPQ$ than the

other quarters; however, Quarter 4 and Quarter 1 and now nearly even with Quarter 2 the lowest.Error!

Bookmark not defined.

0.000

0.050

0.100

0.150

0.200

0.250

0.300

1 2 3 4

SA

EU

PR

ME

US

US Corrected





4321

1400000

1200000

1000000

800000

600000

400000

200000

0

Quarter

COPQ

$ C

orre

cted

Chart of COPQ $ Corrected

FIGURE R 20

Cost of Poor Quality is generally difficult to measure as most accounting systems are unable to trap it. However,

it appears that MidCo is at least being consistent in the way they measure COPQ, given the relative stability in

the Europe, Pacific Rim, U.S., and Middle East facilities which are hovering around 7.5%‐10% of Cost as can been

seen from Figure R 18 above.

When COPQ is calculated as a percent of Sales, Europe, and the Pacific Rim are at about 5‐6% with the U.S. at

about 8%.

FIGURE R 21

Sales for South Africa are at a high level for the first three quarters and then declines in the fourth quarter.

COPQ as a % of Sales is at the 5‐6% level during the first quarter, beginning of the upturn in Sales, and the fourth

quarter which is the down turn. The COPQ as a % of Sales goes to nearly 10% in the second quarter and nearly

16% in the third quarter. It may be that after an extended period of high sales that resources begin to fatigue

and poor quality ensues.

0.00%

10.00%

20.00%

1 2 3 4

Quarterly COPQ (Corrected) as % of Sales

SAEU





The COPQ as % of Sales for the United State is 10.44%, 7.17% and 5.93% for the first three quarters

respectively. The first and second quarters have stable sales and the third quarter is during the peak season. It is

interesting that COPQ improves as sales increases. In the fourth quarter, sales begin to decline and the COPQ

rises to 10.05%. It is a quite peculiar pattern to have much lower COPQ during the high sales periods as

compared to the low sales periods. This should be investigated.

Quality issues need to be investigated in both South Africa and the United States.

1.000.950.900.850.800.750.70

0.16

0.14

0.12

0.10

0.08

0.06

0.04

0.02

Yield

COPQ

Cor

rect

ed (

% S

ales

)

Scatterplot of COPQ Corrected (% Sales) vs Yield

FIGURE R 22

One would expect that there would be an inverse correlation between yield and COPQ.

Figure R 22 graphs the relationship.Error! Bookmark not defined. A regression analysis reveals a linear

correlation, p‐value = 0.041, COPQ Corrected % Sales = 0.237 ‐ 0.182 Yield as yield increases, COPQ decreases.

R2 is 21.1% which means that of the 21.2% of COPQ is explained by yield which is excellent for cross sectional

data

The analysis bears out the intuitive reasoning that a focus on improving yield, will result in a decrease in COPQ,

and thus improve our profitability which is our strategic goal. It would also decrease cost and likely improve

customer satisfaction, which are our tactical goals.

The Quarterly COPQ as % of Sales chart would lead one to believe that there is not a correlation between COPQ

and Sales. However, it is prudent to check. Figure R 23 graphs the relationship.Error! Bookmark not defined.





50000004000000300000020000001000000

0.16

0.14

0.12

0.10

0.08

0.06

0.04

0.02

Sales $

COPQ

Cor

rect

ed (

% S

ales

)

Scatterplot of COPQ Corrected (% Sales) vs Sales $

FIGURE R 23

Casual observation reveals a possible linear relationship; however, this is not statistically significant when

analyzed by regression analysis (p‐value of 0.635

ANOVA’s were performed on COPQ with location and quarter as factors. Neither location nor quarter proved to

be statistically significant.

Yield Analysis

An analysis was performed to determine if there was a correlation between yield and the number of units

produced. As can be seen from the scatterplot, there does not appear to be a correlation.Error! Bookmark not

defined. This was borne out by regression analysis which was insignificant.





700600500400300200100

1.00

0.95

0.90

0.85

0.80

0.75

0.70

Units

Yie

ld

Scatterplot of Yield vs Units

FIGURE R 24

The Multi‐Vari Chart for Yield by Location – Quarter plots the average yield for each quarter in red and the yield

for each location in groupings. The United States has the lowest or near the lowest yield in all quarters and has

a significant swing in yield in quarters 1 and 2.

Quarter

Yie

ld

4321

1.00

0.95

0.90

0.85

0.80

0.75

0.70

Location

US

EUMEPRSA

Multi-Vari Chart for Yield by Location - Quarter

FIGURE R 25

The Multi‐Vari Chart for Yield by Quarter – Location plots the average yield for each location in red and yield for

each quarter in groupings





Recognize Task 7





Location

Yie

ld

USSAPRMEEU

1.00

0.95

0.90

0.85

0.80

0.75

0.70

Quarter1234

Multi-Vari Chart for Yield by Quarter - Location

FIGURE R 26

An analysis of variance (ANOVA) using Yield as the response and Location as a factor and a similar analysis using

Yield as the response and quarter as the factor were performed.


In terms of Yield which location has the strongest influence

on the Multi‐vari Chart conclusions? Why?





Recognize Task 7





Recognize Task 7





We observed from our analysis of COPQ, that if we can improve yield we will decrease the COPQ. It seems that

there should also be a natural correlation between yield and operating profit as well as yield and net profit

margin.

1.000.950.900.850.800.750.70

0.4

0.3

0.2

0.1

0.0

Yield

Ope

rati

ng P

rofit

(%

Sal

es)

Scatterplot of Operating Profit (% Sales) vs Yield

FIGURE R 27

While casual observation may indicate a possibility of a relationship between yield and operating profit (Profit

$), primarily to the two low yield observations, regression analysis indicates that there is no linear relationship

(p‐value = 0.180).


Is there more variation from Location‐to‐Location or

Quarter‐to‐Quarter? p‐values? R2 values from the two

ANOVA’s





1.000.950.900.850.800.750.70

0.14

0.12

0.10

0.08

0.06

0.04

0.02

Yield

Net

Prof

it M

argi

n

Scatterplot of Net Profit Margin vs Yield

FIGURE R 28

From casual observation it would appear there is also no correlation between yield and net profit margin in that

the apparent trend is likely due to the low yields. However, when a regression analysis is performed, the

relationship of Profit = ‐ 0.135 + 0.218 Yield proves to be statistically significant with a p‐value of 0.010 and an R2

value of 31.8%, indicating net profit margin will increase if yield increases.





Cost Analysis

MidCo requested an analysis of Cost$ to determine if there was a difference in the variances and means of Cost

$ in the first half of the year as compared to the last half of the year. A test for equal variances was performed in

Minitab using both the F‐test and the Lavene’s Test The graph is shown in Figure R 29.

1

0

225000020000001750000150000012500001000000750000500000

Bi-

Ann

ual

95% Bonferroni Confidence Intervals for StDevs

1

0

350000030000002500000200000015000001000000

Bi-

Ann

ual

Cost $

Test Statistic 0.38P-Value 0.166

Test Statistic 0.33P-Value 0.575

F-Test

Levene's Test

Test for Equal Variances for Cost $

FIGURE R 29


Black Belt and Green Belt ‐ Commercial


Recognize Task 9


Is there a statistical difference in the variances of Cost $

during the first half of the year when compared to the

second half?





A two‐sample t‐test was performed using Minitab . The graph is show in

Figure R 30.

10

3500000

3000000

2500000

2000000

1500000

1000000

Bi-Annual

Cost

$

Individual Value Plot of Cost $ vs Bi-Annual

FIGURE R 30





Recognize Task 10


Is there a statistical difference in the means of Cost $

during the first half of the year when compared to the

second half?





DEFINE

What product characteristics must be isolated and improved?

Summary

Analysis of the CTQ data provided by MidCo indicates that the United States facility has four

CTQ’s (1, 2, 3, and 7) that should be considered for isolation and improvement. If these four

CTQ’s can be brought to a 99.60% first time yield, the rolled throughput yield will increase from

88.00% to 99.26% and the short term sigma will improve from 3.74 Sigma to 4.17 Sigma which

would be a significant improvement and take us toward our strategic goal of increased

productivity and our tactical goals of decreased cost and improved customer satisfaction.

However, this is MidCo’s first Six Sigma project and they ask that we provide them with an

improvement estimate based only upon looking at only the two CTQ’s that had the highest

failure rate, CTQ’s 1 and 2. Based upon the what‐if analysis, MidCo was advised that bringing

CTQ’s 1 and 2 to a 99.60% first time yield and leaving the remaining eight CTQ’s at status quo

would provide a rolled throughput yield of 94.34% and a short term sigma of 4.03.

MidCo advised that they would prefer to limit the Measurement Phase to CTQ’s 1 and 2.

The Recognize Phase indicated that we should concentrate on yield as the performance metric to drive Six Sigma

and that our focus should be on the United States facility. MidCo provided data in the following format for the

United States facility.


A MindPro ™ Lean Six Sigma Simulated Training Project Define



Case

Critical-to-Quality Characteristics

Frequency Table

CTQ ‐

1

CTQ‐

2

CTQ‐

3

CTQ‐

4

CTQ‐

5

CTQ‐

6

CTQ‐

7

CTQ‐

8

CTQ‐

9

CTQ‐

10

Unit

Test

CTQ Fail

1 Pass Pass Pass Pass Pass Pass Pass Pass Pass Pass Pass 1 22

. Fail Pass Pass Pass Pass Pass Pass Pass Pass Pass Fail 2 16

. Pass Pass Pass Pass Pass Pass Pass Pass Pass Pass Pass 3 7

. Pass Pass Pass Pass Pass Pass Pass Pass Pass Pass Pass 4 0

500 Pass Pass Pass Pass Pass Pass Pass Pass Pass Pass Pass 5 1

6 5

7 7

8 0

9 2

10 3

Unit 59





Additional analysis of the data is summarized in the charts below.

CTQ -1 CTQ-2 CTQ-3 CTQ-4 CTQ-5

Fail Frequency

Pass Frequency

First Time Yield

Long Term Sigma

Short Term Sigma

Throughput Yield





Define Task 2

Define Task 3





Define Task 4

Define Task 5

Define Task 7

Define Task 8





CTQ-6 CTQ-7 CTQ-8 CTQ-9 CTQ-10 Y.RT Y.norm Cp

Fail Frequency

Pass Frequency

First Time Yield

Long Term Sigma

(Benchmark)

Short Term Sigma

(Benchmark)

Throughput Yield

Unit Test

Total Defects

Observed

DPU

Avg. Non-

improved Process

Fail Frequency

Pass Frequency

First Time Yield

Long Term Sigma

Short Term Sigma





Define Task 6


Use the data from the analysis to complete the above

charts.





Glossary

CTQ Pass Frequency Number of units that pass each CTQ (Critical to Quality

Characteristic)

Unit Pass Frequency Number of units that pass all CTQ's

CTQ Fail Frequency Number of units that fail each CTQ

Unit Fail Frequency Number of units that fail at least one CTQ

CTQ First Time Yield Percentage of units passing on first pass

Unit First Time Yield Percentage of units passing all CTQ's on first pass

Long Term Sigma =normsinv(First Time Yield) = Long Term Capability

Short Term Sigma

Long Term Sigma + 1.5 ==> 1.5 is added as research shows that

there are approximately 1.5 standard deviations of shift and drift

= Short Term Capability if process in control, that is absent of

special cause variation. Note this assumes (whether true or not)

that collected data includes special causes.)

Y.RT Rolled Throughput Yield = Product of First Time Yields for all CTQ's

= Probability of Total Success

Y.norm Normalized Yield = Y.RT^(1/# of CTQ's in Y.RT) = Average Yield of a

typical CTQ in the facility

Facility Long Term Sigma =normsinv(Y.norm)

Facility Short Term Sigma Facility Long Term Sigma + 1.5





MidCo is operating at 3.741 Short Term Sigma. To put this number in perspective, 1900 to 1950 manufacturing

was about 3 Sigma (66,807 defects per million); the average company today operates at about 4 Sigma (6,210

defects per million); world class companies operate at about 5 Sigma (233 defect per million); and the ultimate

goal is to operate at 6 Sigma (3.4 defects per million). MidCo is operating at a level below average.

MidCo had specific interest in CTQ‐5 for which 499 of the 500 Unit Passed, a 98.000% First‐Time Yield.

Specifically they wanted to know the confidence interval on the Yield.


How many Units were tested? How many Units passed all

CTQ’s? How many Units failed only one CTQ? How many

Units failed more than one CTQ?





Define Task 1


What is the confidence on CTQ‐5 Yield?





Define Task 11





FIGURE D 1

Figure D 1 shows the CTQ’s ranked from lowest Short Term Sigma level to the highest. The six CTQ’s with the

highest level (4,5,6,8,9,&10), the average number of units passing is 498. Another way of identifying what Juran

called the “vital few” processes upon which to focus is to develop a Pareto Chart using the Frequency of Defects

table, Figure D 2. It can be seen from the Pareto Analysis that the bottom four processes are about 80% of the

total.

Fail 22 16 7 7 5 3 3Percent 34.9 25.4 11.1 11.1 7.9 4.8 4.8Cum % 34.9 60.3 71.4 82.5 90.5 95.2 100.0

CTQ Other1067321

70

60

50

40

30

20

10

0

100

80

60

40

20

0

Fail

Perc

ent

Pareto Chart of CTQ

FIGURE D 2





Define Task 12





An analysis was done to determine the number of cases in which CTQ‐1 “Passed” but the Unit “Failed.”


Which CTQ has the poorest performing process? Which

process provides the most leverage for improvement?

What are the vital few CTQ’s




Define Task 9

Define Task 10


In how many and what percent of the cases did the Unit

Test fail while CTQ‐1 passed? In how many and what

percent of cases did the Unit Test fail when CTQ‐1 also

failed?





A what‐if analysis was performed to determine the impact on performance, if the bottom four processes were

improved to the level of the average level of the top six processes.

What if

Analysis Critical‐to‐Quality Characteristics

Case CTQ ‐1 CTQ‐2 CTQ‐3 CTQ‐4 CTQ‐5 CTQ‐6 CTQ‐7 CTQ‐8 CTQ‐9 CTQ‐

10

Fail

Frequency 2 2 2 0 1 5 2 0 2 3

Pass

Frequency 498 498 498 500 499 495 498 500 498 497

First Time

Yield

99.600

%

99.600

%

99.600

%

100.000

%

99.800

%

99.000

%

99.600

%

100.000

%

99.600

%

99.400

%

Long Term

Sigma 2.652 2.652 2.652 6.000 2.878 2.326 2.652 6.000 2.652 2.512

Short Term

Sigma 4.152 4.152 4.152 6.000 4.378 3.826 4.152 6.000 4.152 4.012

Y.RT Y.norm

96.261% 99.621%

Long Term

Sigma 2.670

Short Term

Sigma 4.170

FIGURE D 3





If the pass frequency of the bottom four processes can be increased to 498 (99.6% first time yield), then number

of defects would be decreased from 63 to 19; the rolled throughput yield would increase from 88.003% to

96.261%; and short term sigma would be increased from 3.741 to 4.17 allowing MidCo to move from below

average to above average performance.

What if the next two CTQ’s in the Pareto Chart (6 and 10) were also increased to a 498 Pass Frequency?

Only small additional gains are achieved by focusing on the bottom six processes instead of the bottom four

processes: the number of defects would be decreased to 17 versus 19; the rolled throughput yield would

increase to 97.039% from 96.261%; and short term sigma increases from 4.17 to 4.25. The effort to focus on the

additional two processes does not reap significant additional benefits. Most processes are dynamic, so it would

be best to concentrate only on the bottom four processes and once the process is stable under the new

conditions, consider an additional project CTQ‐6 and CTQ‐10.





What if


Case CTQ ‐1 CTQ‐2 CTQ‐3 CTQ‐4 CTQ‐5 CTQ‐6 CTQ‐7 CTQ‐8 CTQ‐9 CTQ‐

10

Fail

Frequency 2 2 2 0 1 2 2 0 2 2

Pass

Frequency 498 498 498 500 499 498 498 500 498 498

First Time

Yield

99.600

%

99.600

%

99.600

%

100.000

%

99.800

%

99.600

%

99.600

%

100.000

%

99.600

%

99.600

%

Long Term

Sigma 2.652 2.652 2.652 6.000 2.878 2.652 2.652 6.000 2.652 2.652

Short Term

Sigma 4.152 4.152 4.152 6.000 4.378 4.152 4.152 6.000 4.152 4.152

Y.RT Y.norm

97.039% 99.700%

Long Term

Sigma 2.748

Short Term

Sigma 4.248

FIGURE D 4

It was recommended to MidCo that the four vital few CTQ (1, 2, 3, and 7) should be considered for more

detailed analysis in the Measurement Phase. This is MidCo’s first Six Sigma project and they ask that ask what

kind of results might be expected if they only focused on the two CTQ’s that had the highest failure rate. It was





observed from the Pareto Chart (Figure D 2) that 60.3% of the total CTQ could be attributed to CTQ’s 1 and

2.Error! Bookmark not defined. The results of the analysis are shown in Figure D 5.

What if


Case CTQ ‐1 CTQ‐2 CTQ‐3 CTQ‐4 CTQ‐5 CTQ‐6 CTQ‐7 CTQ‐8 CTQ‐9 CTQ‐10

Fail

Frequen

cy

2 2 7 0 1 5 7 0 2 3

Pass

Frequen

cy

498 498 493 500 499 495 493 500 498 497

First

Time

Yield

99.600

%

99.600

%

98.600

%

100.000

%

99.800

%

99.000

%

98.600

%

100.000

%

99.600

%

99.400

%

Long

Term

Sigma

2.652 2.652 2.197 6.000 2.878 2.326 2.197 6.000 2.652 2.512

Short

Term

Sigma

4.152 4.152 3.697 6.000 4.378 3.826 3.697 6.000 4.152 4.012

Y.RT Y.norm

94.337% 99.422%

Long Term

Sigma 2.525

Short Term

Sigma 4.025





FIGURE D 5

Based upon the what‐if analysis, MidCo was advised that bringing CTQ’s 1 and 2 to a 99.60% first time yield and

leaving the remaining eight CTQ’s at status quo would provide a rolled throughput yield of 94.337% and a short

term sigma of 4.03. MidCo advised that they would prefer to limit the Measurement Phase to CTQ’s 1 and 2.





Measure

What is the actual and potential capability of the core CTQ’s?

Conclusions

The sample data provided by MidCo for the Measure Phase for CTQ‐1 and CTQ‐2

proved to be different from the sample data provided in the Design Phase. MidCo was

contacted and they advised that they are confident in their data collection process and

that the analysis should be performed based upon the most recent data, which is the

data from the measure phase. CTQ1 was originally defined as the most significant CTQ

with regard to poor yield followed by CTQ2. Based upon this summary, it appears that

CTQ2 is the most significant contributor to poor yield. It is also clear that Unit Yield is

significantly deteriorated from that of the define stage with first time yield dropping

from 88.20% to 84.00% and the short term sigma dropping from 2.69 to 2.49. It is

apparent that it is important to continue with the six sigma analysis.

There is a correlation between CTQ‐1 and CTQ‐2.

CTQ‐1 Value = 172 ‐ 0.185 CTQ‐2 Value (p‐value = 0.000; R2 = 53.6%)

The between sample variation of CTQ‐1 is out of control and has a sinusoidal pattern.

An autocorrelation analysis revealed 15 lags of significance indicating the likelihood of

an assignable cause.

CTQ‐2 exhibited control

The process for CTQ‐1 Value has a Cp of 1.18 (3.54 sigma) which is average short term

capability. However, Pp is only 0.74 (2.22 sigma). If we eliminate the assignable causes

we will improve the overall capability of the process.

The process for CTQ‐1 centered below (mean = 73.4) the target (80).

The process for CTQ‐2 Value has a Cp of 0.21 (0.63 sigma) and a Pp value of 0.19 (0.57

sigma). Since the Cp and Pp values are comparable there is little improvement to be

gained within the process. In order to improve CTQ‐2 it is likely that we will have to

improve the technology of the process.

The process for CTQ‐2 is centered above the (mean = 534.6) the target (500).


A MindPro ™ Lean Six Sigma Simulated Training Project Measure



The following data was provided by MidCo for FY 2004:

Unit CTQ-1 CTQ-2

USL NA 115.0 NA 600.0 NA

T NA 80.0 NA 500.0 NA

LSL NA 45.0 NA 400.0 NA

Case Unit Test CTQ-1 Value

CTQ-1 Test

CTQ-2 Value

CTQ-2 Test

CTQ-1 Time

CTQ-2 Time

1 Pass 65.3 Pass 583 Pass 60.8 122.9

. Pass 77.4 Pass 509 Pass 61.3 116.2

. Pass 68.0 Pass 505 Pass 62.0 122.9

. Fail 51.0 Pass 601 Fail 52.9 118.9

500 Pass 92.4 Pass 508 Pass 57.3 123.9





Comparison of Data provided during the Measure Phase to the data provided in the

Design Phase

CTQ-1 Measure

Data

CTQ-1 Define Data

CTQ-2 Measure

Data

CTQ-2 Define Data

Unit Measure

Data

Unit Define Data

Pass Frequency 481 478 429 484 420 441

First Time Yield 96.20% 95.60% 85.80% 96.80% 84.00% 88.20%

Long Term Sigma 1.77 1.71 1.07 1.85 0.99 1.19

Short Term Sigma 3.27 3.21 2.57 3.35 2.49 2.69

Fail Frequency 19 22 71 16 80 59

Fail % 3.80% 4.40% 14.20% 3.20% 16.00% 11.80%

Close examination of the data provided in the Measure Phase with the data provided in the Define Phase reveals

a difference in the data sets. The above chart compares the summaries of the data. It is apparent that new data

was collected for use in the Measure Phase. It is not clear as to why new data might have been collected or

whether there may be an issue with the data collection process.

MidCo was contacted to point out the differences in the data. MidCo advised that they are confident in their data

collection process and that the analysis should be performed based upon the most recent data, which is the data

from the Measure Phase.

CTQ1 was originally defined as the most significant CTQ with regard to poor yield followed by CTQ2. Based upon

this summary, CTQ2 is now the most significant contributor to poor yield. It is also clear that Unit Yield is

significantly deteriorated from that of the Define Phase with first time yield dropping from 88.20% to 84.00%

and the short term sigma dropping from 2.69 to 2.49. It is apparent that it is important to continue with the Six

Sigma analysis and it might be necessary at some point to do a measurement systems analysis (MSA) to

determine the validity of the data collection process.





Analyze Data for Normality

A histogram was developed for CTQ‐1 and CTQ‐2 Values and Time with a normal curve overlay.

1201059075604530

80

70

60

50

40

30

20

10

0

CTQ-1 Value

Freq

uenc

y

Mean 73.37StDev 15.84N 500

Histogram of CTQ-1 ValueNormal

780720660600540480420

70

60

50

40

30

20

10

0

CTQ-2 Value

Freq

uenc

y

Mean 534.6StDev 62.79N 500

Histogram of CTQ-2 ValueNormal

Appears Normal Appears Non-Normal

CTQ‐1 Time appears to be normal, while CTQ‐2 Time does not fit the normal curve as well.

7266605448

90

80

70

60

50

40

30

20

10

0

CTQ-1 Time

Freq

uenc

y

Mean 59.87StDev 5.175N 500

Histogram of CTQ-1 TimeNormal

132126120114108102

80

70

60

50

40

30

20

10

0

CTQ-2 Time

Freq

uenc

y

Mean 119.8StDev 5.122N 500

Histogram of CTQ-2 TimeNormal

Appears Normal Appears Non‐Normal

FIGURE M 1


Do the histograms indicate that the data is normal?




Measure Task 1

Analyze Task 1





Another graphical method of evaluating normality is the normal probability plot. Historically, these plots were

evaluated by what was called a “fat pencil test.” You would place a flat pencil on the straight line and if it

covered all of the points, then it was concluded that the data was normally distributed. The plot shows a center

line which indicates a perfectly normal distribution and limit lines on each side which spread as they move away

from the middle. It is common for distributions to have diverging tails. One common method for evaluating

normal probability plots visually is to give significant weight to the middle 75% of the curve and accept the

divergence at the tails. These plots provided in Error! Reference source not found.‐Error! Reference source not

found. are a large scale as the sample size is large (500), makes it difficult to see the center and limit lines on

smaller graphs.

14012010080604020

99.9

99

9590

80706050403020

10

5

1

0.1

CTQ-1 Value

Perc

ent

Mean 73.37StDev 15.84N 500AD 1.032P-Value 0.010

Probability Plot of CTQ-1 ValueNormal - 95% CI

FIGURE M 2

The normal probability plot provides a graphical representation to determine normality. The normal probability

plot for CTQ‐1 Value has some divergence at the tails; but would seem to pass a “fat pencil test” in the middle

75% of the plot. It is difficult to tell from the plot; but, it does appear that even the middle 75% of the plot is

pushing the limits. We know from the p‐value of 0.010 that the data is not statistically normal. The low p‐value is

in part due to the fact that we have a large number of samples.





Measure Task 2





900800700600500400300

99.9

99

9590

80706050403020

10

5

1

0.1

CTQ-2 Value

Perc

ent


Probability Plot of CTQ-2 ValueNormal - 95% CI

FIGURE M 3

The normal probability plot for CTQ‐2 Value has extreme divergence at the tails. Even through the middle 75% of

the plot might pass the “fat pencil,” the extreme divergence at the tails and the apparent deviation in the

histogram indicates CTQ‐2 Value is likely non‐normal.


Just looking at the normal probability plot, i.e. without

consideration of the p‐value for the AD test, does the data

for CTQ 1 pass the “Fat Pencil” test and thus normal for all

practical purposes or not?





The plots for CTQ‐1 Time and CTQ‐2 Time

8070605040

99.9

99

9590

80706050403020

10

5

1

0.1

CTQ-1 Time

Perc

ent


Probability Plot of CTQ-1 TimeNormal - 95% CI

FIGURE M 4





140130120110100

99.9

99

9590

80706050403020

10

5

1

0.1

CTQ-2 Time

Perc

ent


Probability Plot of CTQ-2 TimeNormal - 95% CI

FIGURE M 5

While it is important to create the histogram and probability plots to discern visually the distribution of the data

and its conformance to normality, modern statistical software such as Minitab provides more sophisticated tests

for normality. Minitab was used to test for normality on each of the variables resulting in the summary table of

information and probability plots shown below.Error! Bookmark not defined.

Variable Mean Standard

Deviation

AD

P‐Value* Comment

CTQ‐1 Value 73.4 15.8 0.010 Data do not follow a normal distribution

CTQ‐1 Time 59.9 5.2 0.848 Data appear to follow a normal distribution

CTQ‐2 Value 534.6 62.8 0.006 Data do not follow a normal distribution

CTQ‐2 Time 119.8 5.1 0.188 Data appear to follow a normal distribution

* H0: data follows a normal distribution vs. H1: data do not follow a normal distribution

Candidate Analysis Required for Black Belt


Measure Task 3





Small deviations in large sample sizes can have a significant impact on the p‐value for normality. Large data sets

provide a lot of power for the test and with certain tests like the Anderson‐Darling test which applies a large

penalty to data points at the extremes, the test may be more sensitive than is important for practical means.

Therefore, it is usually a good idea to look at a histogram of the data when the normality test indicates that the

data is non‐normal. The histograms for CTQ‐1 Value and CTQ‐2 Value are provided below.

The Anderson‐Darling (AD) test is a powerful and extremely sensitive statistical test for determining departures

from a stated distribution. A p‐value below the critical value for the Anderson‐Darling test indicates the data

does not conform to specified distribution. For example if the critical p‐value is 0.05, and the AD p‐value is 0.01,

then the data would be statistically non‐normal.

The Anderson‐Darling test is a powerful test which heavily weights the tails, so it is often acceptable to assume

normality for practical purposes based upon the probability plot passing the “fat pencil test.”

An Anderson‐Darling (AD) test statistic of zero indicates a perfectly normal fit. While the actual AD test statistic

is difficult to interpret, the statistics can be compared. The AD statistic for CTQ‐? Value is _____ and for CTQ‐?

Value is _____. This indicates the CTQ‐? Value is “more (or less) normal” than CTQ‐? Value as is indicated visually

by the probability plot.

Search for Patterns and Correlation

A line chart plots the observations in time sequence and allows the analyst the opportunity to inspect the data

for possible patterns. Line charts created in Excel are shown below:

0.0

50.0

100.0

150.0

1 15 29 43 57 71 85 99 113

127

141

155

169

183

197

211

225

239

253

267

281

295

309

323

337

351

365

379

393

407

421

435

449

463

477

491


Complete the Chart above and comment as to whether or

not the AD test indicates whether or not the data is

statistically normal.




Measure Task 4





FIGURE M 6 LINE CHART FOR CTQ‐1 VALUE

Figure M 7 Line Chart for CTQ‐2 Value

300

400

500

600

700

800

900

1 15 29 43 57 71 85 99 113

127

141

155

169

183

197

211

225

239

253

267

281

295

309

323

337

351

365

379

393

407

421

435

449

463

477

491


Do the line charts indicate any type of patterns? If so

describe the pattern. Is one process more stable than the

other?





Minitab can be used to perform a run chart. The run chart in Minitab is not as easily configured (i.e. stretched

and data series formatted); however, it provides information on runs not available on the line chart. The Minitab

run charts are provided below.

500450400350300250200150100501

120

100

80

60

40

20

Observation

CTQ

-1 V

alue

Number of runs about median: 144Expected number of runs: 251.0Longest run about median: 41Approx P-Value for Clustering: 0.000Approx P-Value for Mixtures: 1.000

Number of runs up or down: 307Expected number of runs: 333.0Longest run up or down: 6Approx P-Value for Trends: 0.003Approx P-Value for Oscillation: 0.997

Run Chart of CTQ-1 Value

500450400350300250200150100501

800

700

600

500

400

Observation

CTQ

-2 V

alue



Run Chart of CTQ-2 Value

500450400350300250200150100501

80

70

60

50

40

Observation

CTQ

-1 T

ime



Run Chart of CTQ-1 Time

500450400350300250200150100501

135

130

125

120

115

110

105

100

Observation

CTQ

-2 T

ime



Run Chart of CTQ-2 Time

Figure M 8

We would consider any p‐value less than 0.05 to be significant. The two most important observations from the

run chart of CTQ‐1 value are the “Longest run about the mean: 41, and the significant p‐value of __ for trends.


Black Belt, Green Belt ‐ Industrial,

and Green Belt ‐ Commercial


Measure Task 4


What is the significant p‐value for CTQ‐1 value?

Does CTQ‐1 Value appear to have any non‐random

patterns, i.e. trends, runs, cycles?





The Run Chart for CTQ‐2 Value indicates significance for clustering, but no trends. Both the CTQ‐1 Time and CTQ‐

2 Time indicate no significance for the presence of patterns. The statistics from the run chart for CTQ‐2 Value

validate the casual observation from the line chart, the data does not have cycles and the runs are close to what

would be expected.

An X‐bar and S Chart provides further indication of whether the non‐normality of the data is causing an out of

control situation in the process. Charts were developed on all four variables using a subgroup size of 5. The

control charts for CTQ‐2 Value, CTQ‐1 Time, and CTQ‐2 exhibited process control.

Sample

Sa

mp

le M

ea

n

1009080706050403020101

100

80

60

40

__X=73.37

UC L=86.63

LC L=60.11

Sample

Sa

mp

le S

tDe

v

1009080706050403020101

20

15

10

5

0

_S=9.29

UC L=19.41

LC L=0

1

11

11

1

1

1

111

11

111

11

1

11

1

11

11

11

1

1

Xbar-S Chart of CTQ-1 Value

Figure M 9

The between sample variation is out of control as represented by the X‐bar graph. The X‐bar graph has a

sinusoidal pattern.


Black Belt and Green Belt ‐ Industrial


Measure Task 8


Does the within sample variation of CTQ‐1 Value exhibit a

state of control?





We need to investigate to determine if the sinusoidal pattern in the X‐bar graph is by chance or likely caused by

a true event. We do this by analyzing autocorrelation. Minitab states, “Autocorrelation is the correlation

between observations of a time series separated by k time units.”

Lag

Aut

ocor

rela

tion

65605550454035302520151051

1.0

0.8

0.6

0.4

0.2

0.0

-0.2

-0.4

-0.6

-0.8

-1.0

Autocorrelation Function for CTQ-1 Value(with 5% significance limits for the autocorrelations)

Figure M 10

The autocorrelation chart is analyzed by looking at the lags that extend beyond the red zone. For CTQ‐1 Value

there are approximately 15 lags. Generally, the more lags outside of the red zone, the more chance that the

event is non‐random; in other words, it is likely that the sinusoidal pattern exhibited in the X‐bar chart for CTQ‐1

can be attributed to one or more assignable causes.

A matrix plot was performed on the four variables: CTQ‐1 Value, CTQ‐2 Value, CTQ‐1 Time, and CTQ‐2 Time.

Obviously there in an unlikely correlation between CTQ‐1 Value and CTQ‐2 Time; however, the matrix plot of all

variables is a good way to get a quick look at any obvious correlation between variables. The matrix plot is

simply a series of scatterplots for all variables of consideration. One must be careful when looking at a matrix

plot to keep in mind with any correlation there is a dependent variable and an independent variable.

Additionally, the analyst must keep in mind correlation does not mean causation.





800600400 135120105

120

80

40

800

600

400

70

60

50

1208040

135

120

105

706050

CTQ-1 Value

CTQ-2 Value

CTQ-1 Time

CTQ-2 Time

Matrix Plot of CTQ-1 Value, CTQ-2 Value, CTQ-1 Time, CTQ-2 Time

Figure M 11

A scatter plot of CTQ‐1/CTQ‐2 is shown below.





CTQ-2 Value

CTQ

-1 V

alue

800700600500400

120

100

80

60

40

20

Scatterplot of CTQ-1 Value vs CTQ-2 Value

Figure M 12

A regression analysis was performed using CTQ‐1 as the response variable (Y) and CTQ‐2 as the predictor

variable (X). The regression analysis was significant with a p‐value of 0.000 for the regression equation CTQ‐1

Value = 172 ‐ 0.185 CTQ‐2 Value. The R2 value of 53.6% indicates that 53.6% of the variation in CTQ‐1 is

explained by CTQ‐2. This is an exceptional correlation for cross‐sectional data. It should be noted that the

significance of the regression equation and the high R2 for cross‐sectional data is a prediction and deserves

additional investigation but does not necessarily mean the variation in CTQ‐1 is caused by CTQ‐2.





Measure Task 5


Does there appear to be correlation between CTQ‐1 value

and CTQ‐2 value?





Capability Analysis

Capability Analysis generally is only valid on “in‐control processes.” The process for CTQ‐2 Value was in control

but the process for CTQ‐1 Value was clearly out of control. However, in the Measure Phase it is good to provide

a capability analysis on the current state for all variables that will be considered in the Analyze Phase. Therefore,

a thorough capability analysis was performed on both CTQ‐1 Value and CTQ‐2 Value. A subgroup size of 5 was

used for the capability analysis performed in Minitab. Minitab also provides the option of doing what they call a

“Capability Sixpack.” The “Capability Sixpack” was also performed and follows the chart for capability analysis.

Minitab's Capability Sixpack allows the analyst assess the capability of a process at a glance by combining the

following quality tools into a single display:

Xbar or Individuals charts to determine whether the process is in control. If the process in out of

control, the capability indices are invalid. While the X‐bar and R Charts are included in the Capability

Sixpack, it might be desirable to create them individually.

R, S, or MR chart to track the variation in the data and assesses whether the process variation is

acceptable.

A run chart to look for evidence of patterns in the data.

Probability plot to verify that a chosen distribution fits the data.

Capability histogram and capability plot to visually compare the distribution of data from the

process to the specification spread. It also includes capability statistics to assess capability of the

process quantitatively.





1201059075604530

LSL USLProcess Data

Sample N 500StDev (Within) 9.88251StDev (O v erall) 15.84360

LSL 45.00000Target *USL 115.00000Sample Mean 73.37280

Potential (Within) C apability

C C pk 1.18

O v erall C apability

Pp 0.74PPL 0.60PPU 0.88Ppk

C p

0.60C pm *

1.18C PL 0.96C PU 1.40C pk 0.96

O bserv ed PerformancePPM < LSL 28000.00PPM > USL 10000.00PPM Total 38000.00

Exp. Within PerformancePPM < LSL 2045.80PPM > USL 12.64PPM Total 2058.45

Exp. O v erall PerformancePPM < LSL 36662.30PPM > USL 4302.23PPM Total 40964.53

WithinOverall

Process Capability of CTQ-1 Value

Figure M 13




Measure Task 6


What is the short‐term (within) capability (Cp)? What is

the short term (within) Standard Deviation? What is the

long term (overall) capability (Ppk)? What is the long term

(overall) standard deviation? What is the process mean?

What is the process target? What is the offset between

the mean and the target? What portion of the semi‐

tolerance zone id consumed by the offset? What is the

value of the constant K. (Hint: See equations below)? How

does the short term standard deviation compare to the

long term standard deviation? What do you think causes

this difference? Is there a centering problem? Is there a

spread or variation problem? Is spread or centering the

biggest problem?





9181716151413121111

100

75

50Sam

ple

Mea

n

__X=73.37

UCL=86.63

LCL=60.11

9181716151413121111

40

20

0

Sam

ple

Ran

ge

_R=22.99

UCL=48.61

LCL=0

10095908580

100

75

50

Sample

Valu

es

1201059075604530

LSL USL

LSL 45USL 115

Specifications

120906030

Within

O v erall

Specs

StDev 9.88251C p 1.18C pk 0.96

WithinStDev 15.8357Pp 0.74Ppk 0.6C pm *

O v erall

1

11

11

11

1

111

11

111

11

1

11

1

111

1

11

1

Process Capability Sixpack of CTQ-1 ValueXbar Chart

R Chart

Last 25 Subgroups

Capability Histogram

Normal Prob PlotA D: 1.032, P: 0.010

Capability Plot

Figure M 14

To help understand the data it will be helpful to clarify some terms:

Short Term Capability may also be called within capability or potential capability and represents the

capability of the process if no assignable causes were present.

Long Term Capability may also be called overall capability or sustainable capability.

3Cp = Z = sigma short term




Measure Task 7


Is the process mean in statistical control?





The Cp’s are calculated using short term (within) standard deviation

o Cp is referred to as the inherent capability or instantaneous reproducibility of the process. This

in an indicator of the capability that the process should be able to achieve with its current

technology. The calculation of the standard deviation used to calculate Cp is 2d

R . Since this

calculation is based upon R , it only considers the variation “within samples” and thus excludes

the influence of assignable causes. In order to improve Cp, generally there needs to be an

improvement in the technology of the process.

o If Cpk is lower than Cp, then there is an issue with centering of the mean. If the mean is

centered, Cpk will be improved.

The Pp’s are calculated using the long term (overall) standard deviation based upon each individual

observation. Thus, the Pp’s are affected by assignable causes causing the Pp’s to be smaller than the

Cp’s.

CP, Cpk, and K are calculated using the following equations; CTQ‐1 Values are used as an example:

; = Min

| |

/

Cpk = (1 – K) Cp





780720660600540480420

LSL USLProcess Data




C C pk 0.60


Pp 0.53PPL 0.71PPU 0.35Ppk

C p

0.35C pm *

0.60C PL 0.80C PU 0.39C pk 0.39




WithinOverall


Figure M 15





9181716151413121111

640

560

480Sam

ple

Mea

n

__X=534.6

UCL=609.5

LCL=459.6

9181716151413121111

300

150

0

Sam

ple

Ran

ge

_R=130.0

UCL=274.9

LCL=0

10095908580

700

600

500

Sample

Valu

es

780720660600540480420

LSL USL

LSL 400USL 600

Specifications

800600400

Within

O v erall

Specs

StDev 55.8916C p 0.6C pk 0.39

WithinStDev 62.7922Pp 0.53Ppk 0.35C pm *

O v erall

11

1

11

1

Process Capability Sixpack of CTQ-2 ValueXbar Chart

R Chart

Last 25 Subgroups

Capability Histogram

Normal Prob PlotA D: 1.139, P: 0.006

Capability Plot

Figure M 16

The process for CTQ‐2 Value has a Cp of 0.60 (1.80 sigma) and a Pp value of 0.53 (1.59 sigma). Since the Cp and

Pp values are comparable there is little improvement to be gained within the process. In order to improve CTQ‐2

it is likely that we will have to improve the technology of the process. It appears that the process is centered

above the (mean = 534.6) the target (500).





ANOVA of Subgroups

The X‐bar chart plotted subgroups of 5 with several of these subgroups plotting outside the control limits.

Another tool for analyzing the difference in subgroups in Analysis of Variance (ANOVA). A one‐way ANOVA was

performed with CTQ‐1 Value as the response and subgroups as the factor to be analyzed with the results

provided below:

One‐way ANOVA: CTQ‐1 Value versus Subgroup

Source DF SS MS F P

Subgroup 99 86116.6 869.9 8.92 0.000

Error 400 39016.8 97.5

Total 499 125133.4




Measure Task 9


What is the F‐value? At what level would the F‐value be

statistically significant? What percent of the total variation

can be explained by subgroup‐to‐subgroup variation? Is

there a need to focus on long term capability (non‐random

pattern issues; black noise)? Is there a need to focus on

short term capability (within variation issues)? What is

more important overall variation or within variation

issues? How does total (overall) standard deviation

compare to short term (within) standard deviation?





One‐way ANOVA: CTQ‐2 Value versus Subgroup

Source DF SS MS F P

Subgroup 99 719500 7268 2.33 0.000

Error 400 1247986 3120

Total 499 1967486

S = 55.86 R‐Sq = 36.57% R‐Sq(adj) = 20.87%

The p‐value for subgroups is 0.000 indicating subgroups are statistically significant – at least one subgroup is

different from other subgroups – for both CTQ‐1 Value and CTQ‐2 Value. This should not be a surprise based

upon a casual observation of the X‐bar chart which revealed temporal points significant outside the limits. This

might leave one to believe that the assignable causes are the major problem. Elimination of the assignable

causes is the first problem that should be addressed. Secondarily, the centering issue – needing to move the

sample mean closer to the target should be addressed. However, the bigger issue is that even with the out of

control points resolved and the centering issue resolved, the short‐term capability is not acceptable indicating

that variation is the larger issue. We explained a large portion of the variation but even when the black noise is

removed, the process is not capable.

Recall that the Cp value for CTQ‐1 Value was 1.18, indicating a 3.54 Sigma process. We had a Ppk of 0.60,

indicating that we had some long term stability issues. The p‐value of 0.000 for Subgroups validated that for the

immediate situation, the sustained reproducibility (long‐term capability) is more problematic than the

instantaneous reproducibility (short‐term capability). A 3.54 Sigma process may still not meet our desired needs

but we can make a large improvement by getting the process under control.

The R‐Squared value for CTQ‐2 Value is only 36.7% indicating only about 1/3 of the variation is attributable to

subgroup‐to‐subgroup, while the remaining 2/3 of the variation is attributable to the process itself.

The standard deviation (S) for CTQ‐1 Value is shown as 9.876 and is calculated by taking the square root of the

Mean Square (MS) of the Error. Note the standard deviation for the Error in the ANOVA is the “Within Standard

Deviation” found in Minitab Capability Analysis. The Minitab Capability Analysis also indicated an Overall

Standard Deviation. This can be determined from the ANOVA table by dividing the Sum of Squared Errors (SS)

for Total by the Total Degrees of Freedom and then taking the square root [(125133.4/499)^.5 = 15.84]. The

within standard deviation for CTQ‐2 Value is 55.89, while the overall standard deviation is 62.79 revealing only a

slight difference.





Confidence Intervals

The mean of sample is calculated by X‐bar, the average of the sample. If another sample were taken, the value

for X‐bar would be different. The confidence interval provides the interval of likely values for X‐bar if other

samples were taken and is based upon some stated confidence level. The confidence interval on the mean of

sample is calculated in Minitab using a 1‐sample t‐test. The results are shown below:

Variable N Mean StDev SE Mean 95% CI

CTQ‐1 Value 500 73.373 15.836 0.708 (71.981, 74.764)

Variable N Mean StDev SE Mean 95% CI

CTQ‐2 Value 500 534.55 62.79 2.81 (529.03, 540.07)

Not only will the mean be different if other samples are taken but so will the variation. The confidence interval

for variance or standard deviation can found in Minitab by calculating basic statistics for 1 variance. The results

are shown below:

95% Confidence Intervals

Variable Method

CI for

StDev

CI for

Variance

CTQ‐1 Value Standard (14.9, 16.9) (222, 285)





Measure Task 10


What is the 95% Confidence Interval for CTQ‐1 Value

Mean?





Variable Method

CI for

StDev

CI for

Variance

CTQ‐2 Value Standard (59.1, 66.9) (3496, 4482)

Adjusted (58.6, 67.6) (3434, 4574)

It is important for the analyst to understand that the mean and standard deviation while calculated with a

specific equation are in fact statistics which will vary from sample to sample. Calculating confidence intervals

allows the analyst to understand the likely range of these values and their impact on the project.





Measure Task 11


What is the 95% Confidence Intervals for CTQ‐1 Value

Standard Deviation and Variance?





CTQ Time

MidCo indicated that when they provided the “original data” the values provided for CTQ Time was not

collected simultaneously with the CTQ Values. They further advised that the data for CTQ Time was in fact

collected as individual values (not subgroups) over a period of time. Based upon this information it was

important to use analytical tools for individuals to for these variables.

The most common control chart for individuals is the I‐MR chart. This chart plots individual values on the I chart

and a moving range (usually of two values, since consecutives values are most likely to be alike and any

difference will show as variation) representing variation on a second chart. The I‐MR charts for CTQ‐1 Time and

CTQ‐2 Time are shown below:

451401351301251201151101511

80

70

60

50

40

Observation

Ind

ivid

ua

l V

alu

e

_X=59.87

UC L=75.58

LC L=44.16

451401351301251201151101511

24

18

12

6

0

Observation

Mo

vin

g R

an

ge

__MR=5.91

UC L=19.30

LC L=0

1

1

1

11

1

I-MR Chart of CTQ-1 Time





Measure Task 11


Does the I‐chart for CTQ‐1 time reveal any out of control

conditions? Is CTQ‐1 in a “pragmatic” state of control?





451401351301251201151101511

130

120

110

100

Observation

Ind

ivid

ua

l V

alu

e

_X=119.77

UC L=134.77

LC L=104.77

451401351301251201151101511

24

18

12

6

0

Observation

Mo

vin

g R

an

ge

__MR=5.64

UC L=18.42

LC L=0

1

1

1

1

11

1

I-MR Chart of CTQ-2 Time

It would have been best to analyze out of control conditions at the time of occurrence; however, no record of

investigation is available. Similarly the red points in the MR‐chart should have been investigated

Autocorrelation analysis was performed on the CTQ Time data to determine if any pattern existed.

65605550454035302520151051

1.0

0.8

0.6

0.4

0.2

0.0

-0.2

-0.4

-0.6

-0.8

-1.0

Lag

Aut

ocor

rela

tion

Autocorrelation Function for CTQ-1 Time(with 5% significance limits for the autocorrelations)

65605550454035302520151051

1.0

0.8

0.6

0.4

0.2

0.0

-0.2

-0.4

-0.6

-0.8

-1.0

Lag

Aut

ocor

rela

tion

Autocorrelation Function for CTQ-2 Time(with 5% significance limits for the autocorrelations)



Measure Task 13





Generally, the more lags outside of the red zone, the more chance that the event is non‐random. This is

consistent with the limited out of control points and stability displayed in the I‐chart.

Evaluation of Test Results

Cross‐tabulation provides one‐way, two‐way, or multi‐way tables containing count data. The optional Chi‐

Square option tests dependence among characteristics in a two‐way classification. This procedure can be used

to test if the probabilities of items being classified for one variable are dependent upon the classification of the

other variable. Unit Test was cross tabulated and the Chi‐square option was employed for CTQ‐1 Test and CTQ‐2

Test with the results shown below:

Tabulated statistics: Unit Test, CTQ-1 Test

Rows: Unit Test Columns: CTQ-1 Test

Fail Pass All

Fail 19 61 80

3.04 76.96 80.00

Pass 0 420 420

15.96 404.04 420.00

All 19 481 500

19.00 481.00 500.00

Cell Contents: Count

Expected count

Pearson Chi-Square = 103.690, DF = 1, P-Value = 0.000

Likelihood Ratio Chi-Square = 73.826, DF = 1, P-Value = 0.000


Are there any autocorrelation lags outside the confidence

intervals? What is the interpretation of lags outside the

confidence limits?


Black Belt, Green Belt – Industrial, and



Measure Task 14





Tabulated statistics: Unit Test, CTQ-2 Test

Rows: Unit Test Columns: CTQ-2 Test

Fail Pass All

Fail 71 9 80

11.36 68.64 80.00

Pass 0 420 420

59.64 360.36 420.00

All 71 429 500

71.00 429.00 500.00

Cell Contents: Count

Expected count

Pearson Chi-Square = 434.441, DF = 1, P-Value = 0.000

Likelihood Ratio Chi-Square = 352.304, DF = 1, P-Value = 0.000


Are you able to identify the meaning of intersections, i.e.

the count where CTQ‐1 Test passes but the Unit Test fails?

What the expected counts? What is the meaning of

expected counts? Are there any expected counts below 5?

Why are you given a warning about low cell counts? Is the

Chi‐square value statistically significant? Does a low p‐

value mean there is an association between row and

column data or that there is not association? Is CTQ‐1 Test

correlated to Unit Test? Is CTQ‐1 Independent of Unit

Test? What are the first time yields of CTQ‐1 Test and CTQ‐

2 Test? What is the rolled throughput yield of CTQ‐1 Test

and CTQ‐2? What is the normalized yield of CTQ‐1 Test

and CTQ‐2? What is the Sigma Capability of CTQ‐1 Test and

CTQ‐2?





A cross‐tabulation was also done on the three variables Unit Test, CTQ‐1 Test, and CTQ‐2 Test with the results

shown below:

Tabulated statistics: CTQ-1 Test, CTQ-2 Test, Unit Test

Results for Unit Test = Fail Rows: CTQ-1 Test Columns: CTQ-2 Test Fail Pass All Fail 10 9 19 16.86 2.14 19.00 Pass 61 0 61 54.14 6.86 61.00 All 71 9 80 71.00 9.00 80.00 Cell Contents: Count Expected count Pearson Chi-Square = 32.557, DF = 1, P-Value = 0.000 Likelihood Ratio Chi-Square = 29.987, DF = 1, P-Value = 0.000 * NOTE * 1 cells with expected counts less than 5

Results for Unit Test = Pass Rows: CTQ-1 Test Columns: CTQ-2 Test Fail Pass All Fail 0 0 0 * * * Pass 0 420 420 * * * All 0 420 420 * * * Cell Contents: Count Expected count * ERROR * Cannot compute Chi-Square.


Black Belt and



Measure Task 15


Complete and interpret the above table.





Minitab can be used to compare two proportions. CTQ‐1 Test was compared to CTQ‐2 Test with the following

results.

Test and CI for Two Proportions: CTQ-1 Test, CTQ-2 Test Event = Pass Variable X N Sample p CTQ-1 Test 481 500 0.962000 CTQ-2 Test 429 500 0.858000 Difference = p (CTQ-1 Test) - p (CTQ-2 Test) Estimate for difference: 0.104 95% CI for difference: (0.0691157, 0.138884) Test for difference = 0 (vs not = 0): Z = 5.84 P-Value = 0.000 Fisher's exact test: P-Value = 0.000

As with the cross‐tabulation this test provides the yields of the two variables.

Similarly, a single proportion test may be done.

Test and CI for One Proportion: CTQ-1 Test Event = Pass Variable X N Sample p 95% CI CTQ-1 Test 481 500 0.962000 (0.941293, 0.976969) Variable X N Sample p 95% CI CTQ-2 Test 429 500 0.858000 (0.824289, 0.887403)


Black Belt, Green Belt – Industrial, and



Measure Task 16


Is First Time Yield of CTQ‐1 Test statistically different from

that fo CTQ‐2 Test?


Black Belt, Green Belt – Industrial, and Green Belt ‐ Commercial


Measure Task 17


What are the First Time Yield confidence intervals for CTQ‐

1 Test and CTQ‐2 Test? Is CTQ‐1 Test contained within the

confidence interval of CTQ‐2 Test?





The test for two proportions indicated the yields of CTQ‐1 Test and CTQ‐2 Test were different. This can be

verified by looking at the confidence intervals for the two variables





Analyze

What is the actual and potential capability of the CTP’s?

Conclusions

The following relationships were observed casually and confirmed by detailed analysis.

o As CTQ1 increases; CTQ2 decreases (This was observed in the Measure Phase).

o As X1, X4, and X5 increases CTQ1 increases and CTQ2 decreases (some have a stronger

influence than others)

o As X2 increases CTQ1 decreases and CTQ2 increases

o X3 does not seem to have any correlation to CTQ1 or CTQ2

o From a casual observation it appears that if we adjust X1, X4, and X5 up and X2 down that we

will improve centering on both CTQ‐1 Value and CTQ‐2 Value.

The technology as revealed by the S chart is relatively in control for all CTP’s.

X2, X4, and X5 have processes that are out of control and exhibit cyclic patterns.

X2 has marginal autocorrelation but X4 and X5 have many lags and are likely to have an assignable

cause present.


A MindPro ™ Lean Six Sigma Simulated Training Project Analyze



Capability analysis revealed:

o X1, X3, X4, and X5 have centering issues

o X2 has spread issues and may require technology improvement

o X4 and X5 have spread issues but appear to be potentially cable if the process is brought

under control and the centering issues are resolved.

o Overall capability ranged from ‐0.07 to 1.41 Z‐Benchmark; therefore, all CTP’s need

improvement.

A capability analysis was performed on all variables as summarized below:

CTP LSL/Target/USL

Sample

Mean

Centering

Issue

Spread Issue

Potential

Capability

Overall

Capability

Potential

Capability

Overall

Capability

X1 90/(N/A)/(N/A) 100.07 Yes No No 1.04 1.04

X2 31/33/35 33.17 No Yes Yes 1.76 0.99

X3 (N/A)/(N/A)/310 300.15 Yes No No 1.56 1.41

X4 100/105/110 100.33 Yes No Yes 0.35 0.11

X5 1.15/1.20/1.25 1.25 Yes No Yes -0.24 -0.07

Day to Day variations exists to some degree in both CTQ‐1 and CTQ‐2 but otherwise we should not be

concerned about the sampling demographics.

o CTQ‐1: Wednesday is substantially different from the other days and has a negative impact

on centering.

o CTQ‐2: Tuesday and Thursday are substantially different from the other days and has a

positive impact on centering.





The data in the following format was provided by MidCo for FY 2004:

Analyze Phase

Unit CTQ-1 CTQ-2 Critical-to-Process Variables

X1 X2 X3 X4 X5

USL NA 115.0 NA 600.0 NA NA NA 35.0 NA 310.0 NA 110.0 NA 1.25 NA

T NA 80.0 NA 500.0 NA NA NA 33.0 NA NA NA 105.0 NA 1.20 NA

LSL NA 45.0 NA 400.0 NA 90.0 NA 31.0 NA NA NA 100.0 NA 1.15 NA

Case Unit Test

CTQ-1 Value

CTQ-1

Test

CTQ-2

Value

CTQ-2

Test

X1 Value

X1 Check

X2 Value

X2 Check

X3 Value

X3 Check

X4 Value

X4 Check

X5 Value

X5 Check

1 Pass 65.3 Pass 583 Pass 88.7 Low 31.3 OK 304.0 OK 96.55 Low 1.256 High

. Pass 77.4 Pass 509 Pass 98.6 OK 30.7 Low 303.3 OK 98.44 Low 1.253 High

. Pass 68.0 Pass 505 Pass 99.4 OK 32.7 OK 290.5 OK 98.63 Low 1.250 OK

. Fail 51.0 Pass 601 Fail 87.4 Low 34.0 OK 290.5 OK 99.13 Low 1.238 OK

500 Fail 56.2 Pass 627 Fail 91.3 OK 33.9 OK 306.6 OK 96.81 Low 1.243 OK

Sampling demographics

Case Day Subgroup Sequence Inspector

1 Mon 1 1 Fred

. Mon 1 2 Fred

. Mon 1 3 Mary

. Mon 1 4 Fred

500 Mon 1 5 Fred





Overview

The Recognize Phase indicated that we should concentrate on yield as the performance metric to drive Six Sigma

and that our focus should be on the United States facility. In the Define Phase CTQ‐1 and CTQ‐2 were identified

as the most important of the ten CTQ’s. In the Measure Phase, CTQ‐1 and CTQ‐2 was analyzed in terms of its

performance measure including cycle time. In the Analyze Phase, the focus will be on determining the Critical to

Process (CTP) characteristics.

There is a significant amount of analysis required given the amount of data that has been provided. How do we

avoid analysis paralysis – that is analyzing to data ad infinitum? This analysis will systematically analyze

correlation, normality, control, and capability. The analysis will also look at the sampling demographics for

possible issues with inspectors or days of the week.

The Matrix Plot in Minitab provides a quick overview of the data to identify possible correlations.

120

80

40

1.301.251.20

800600400 35.032.530.0 11010090

800

600

400

120

100

80

35.0

32.5

30.0

320

300

280 110

100

90

1208040

1.30

1.25

1.20

12010080 320300280

CTQ-1 Value

CTQ-2 Value

X1 Value

X2 Value

X3 Value

X4 Value

X5 Value

Matrix Plot of CTQ-1 Value, CTQ-2 Value, X1 Value, X2 Value, ...




Analyze Task 2





The Matrix Plot reveals the following possible relationships:

As CTQ1-Value increases; CTQ2-Value decreases (This was observed in the Measure Phase).

While not extremely strong, X1, X4, and X5 appear to have a positive correlation with CTQ1-Value.

X1 appears to have a negative correlation with CTQ2-Value.)

The correlation between X4 and CTQ1-Value appears to be a moderate positive correlation that is heteroscedastic (variation is not equal) in nature.

As X2 it appears that CTQ1-Value may decrease while CTQ2-Value may increases. The correlation is not strong.

X3 does not seem to have any correlation to CTQ1-Value and CTQ2-Value




Analyze Task 2


What are your conclusions with regard to the Matrix Plot?

No relationship, moderate relationship, strong

relationship? Positive or negative correlation? Does any

heteroscedastic behavior exist?





Correlation

The Matrix Plot essentially provides a scatter plot for all variables. Since we observed what appears to be some

correlation between X1, X2, X4, and X5 and the two CTQ’s, we decided to perform a regression analysis between

all of the variables and the CTQ’s and have provided a summary below:

CTP (X)

CTQ1 (Y)

R2

p-Value Constant

p-Value

Intercept

X1

CTQ-1 Value = - 18.7 + 0.920 X1 Value

31.8% 0.002 0.000

X2

CTQ-1 Value = 191 - 3.56 X2 Value

10.1% 0.000 0.000

X3

Insignificant

N/A

0.474

0.090

X4

CTQ-1 Value = - 181 + 2.54 X4 Value

22.9% 0.000 0.000

X5

CTQ-1 Value = - 612 + 547 X5 Value

52.1% 0.000 0.000



Analyze Task 7


What is the regression equation? What proportion of the

variance is explained by the CTP (X)? What is the p‐value of

the Constant and Intercept? What is the meaning of the p‐

value? What are your final conclusions?





CTP (X)

CTQ2 (Y)

R2

p-Value Constant

p-Value

Intercept

X1

CTQ-2 Value =

1070 - 5.35 X1 Value

68.4%

0.000

0.000

X2

Insignificant

N/A

0.595

0.000

X3

Insignificant

N/A

0.133

0.000

X4

CTQ-2 Value =

1176 - 6.40 X4 Value

9.2%

0.000

0.000

X5

CTQ-2 Value =

1347 - 649 X5 Value

4.7%

0.000

0.000

The regression analysis confirms the casual observations from the matrix plot. It is premature to make a decision

on the CTP’s, but remember that from the measure phase we discovered that the average CTP‐1 Value was

below Target and the average CTP‐2 Value was above target. From a casual observation it appears that if we

adjust X1, X4, and X5 up and X2 down that we will improve centering on both CTQ‐1 Value and CTQ‐2 Value.

After viewing the results from the regression analysis Midco ask that a multiple regression analysis be

performed to evaluate the combined effect of X1 and X4 on CTQ1.



Analyze Task 8





Normality

Normality is a root assumption of all statistics, so it is generally a good practice to do a normality check on all

data that will be analyzed. The normality test was done in Minitab and is summarized below:

CTP Comment p-value Histogram

X1 Data follow a normal distribution 0.391

130120110100908070

50

40

30

20

10

0

X1 Value

Freq

uenc

y

Mean 100.1StDev 9.708N 500

Histogram of X1 ValueNormal

X2 Data follow a normal distribution 0.294

X2 Value

Freq

uenc

y

36.034.533.031.530.0

70

60

50

40

30

20

10

0

Mean 33.17StDev 1.417N 500



What is the resulting regression equation? What

proportion of variation is explained by the model? What is

the p‐value of the Constant and Intercept? What is the

meaning of the p‐value? What are your final conclusions?





X3 Data do not appear to follow a normal

distribution 0.044

X3 Value

Freq

uenc

y

318312306300294288

70

60

50

40

30

20

10

0

Mean 300.2StDev 6.988N 500



distribution < 0.005

X4 Value

Freq

uenc

y

108105102999693

80

70

60

50

40

30

20

10

0

Mean 100.3StDev 2.984N 500



distribution < 0.005

X5 Value

Freq

uenc

y

1.2901.2751.2601.2451.2301.2151.200

60

50

40

30

20

10

0

Mean 1.251StDev 0.02088N 500


While X3, X4, and X5 are not statistically normal, they follow the normal curve sufficiently to analyze the data for

control using an X‐bar and S. We will be using subgroups of 5 for the X‐bar and S charts and the central limit

theorem provides that when subgroups are utilized the subgroups will be normal if the individual observations

are not radically non‐normal.





Descriptive Statistics

It is always helpful to have descriptive statistics of the variables.

Variable Mean SE Mean St Dev Min Q1 Median Q3 Max SS

X1 100.800 0.4340 9.708 68.900 93.000 99.600 106.270 129.400 5,054,634

X2 33.172 0.0634 1.417 29.000 32.100 33.200 34.100 36.900 551,193

X3 300.150 0.3130 6.990 283.500 295.200 299.800 304.850 322.200 45,069,981

X4 100.330 0.1330 2.980 92.480 98.220 100.290 102.090 110.140 5,037,103

X5 1.2514 0.0009 0.0209 1.1900 1.2390 1.2510 1.2680 1.2990 783

Control

X‐bar (Between Sample Variation) and S (Within Sample Variation) charts were developed in Minitab with

subgroups of 5.


Fill in the above table.




Analyze Task 4





9181716151413121111

114

108

102

96

90

Sample

Sa

mp

le M

ea

n

__X=100.08

UC L=113.08

LC L=87.07

9181716151413121111

20

15

10

5

0

Sample

Sa

mp

le S

tDe

v

_S=9.11

UC L=19.04

LC L=0

1

Xbar-S Chart of X1 Value

9181716151413121111

36.0

34.5

33.0

31.5

30.0

Sample

Sa

mp

le M

ea

n

__X=33.172

UC L=34.452

LC L=31.892

9181716151413121111

2.0

1.5

1.0

0.5

0.0

Sample

Sa

mp

le S

tDe

v

_S=0.897

UC L=1.873

LC L=0

1

111

1

11

1

111

1

1

1

11

11

1

1

11

111

1

1

1

1






9181716151413121111

310

305

300

295

290

Sample

Sa

mp

le M

ea

n

__X=300.15

UC L=308.60

LC L=291.70

9181716151413121111

12

9

6

3

0

Sample

Sa

mp

le S

tDe

v

_S=5.92

UC L=12.36

LC L=0

1

1

1

1

1

1


9181716151413121111

110

105

100

95

Sample

Sa

mp

le M

ea

n

__X=100.33UC L=101.58

LC L=99.07

9181716151413121111

2.0

1.5

1.0

0.5

0.0

Sample

Sa

mp

le S

tDe

v

_S=0.878

UC L=1.835

LC L=0

11

11

1

11

11

11

1

11111

111

1

111

1

111

1

1

1

1

11

1

1

111

1

1

1

1111

1111

1

1

11

1

11

111

11111

1

1

1



Is the process in a state of control? Did any subgroup means exceed the control

limits? If so, how many? Did any subgroup standard deviations exceed the control

limits? If so, how many? Where is the largest problem, if any, between variation

of within variation? Is the performance consistent with the principles of SPC? Are

there any trends over time?


Black Belt and Green Belt – Industrial


Analyze Task 19





9181716151413121111

1.28

1.26

1.24

1.22

1.20

Sample

Sa

mp

le M

ea

n

__X=1.2514UC L=1.2591

LC L=1.2436

9181716151413121111

0.012

0.009

0.006

0.003

0.000

Sample

Sa

mp

le S

tDe

v

_S=0.00543

UC L=0.01135

LC L=0

111

1

1111

111

1

1

1

11

111

1

1

1

1

11

1

1111

1

1

11

111

1

1

1

1111

111

1111

11

11

1

1

111

1111

1

11


A summary of the X‐Bar and S‐Charts is provided below:

CTP S Chart (Within Sample Variation) X-bar Chart (Between Sample Variation)

X1 No out of control points One point out of control

X2 One point out of control Several points out of control and somewhat of a cyclic pattern.

X3 One point out of control A few out of control points at end of data

X4 Three points out of control Severely out of control with cyclic pattern

X5 Two points out of control Severely out of control with cyclic pattern

Since the S‐Charts are essentially under control, the technology for the variables are rather stable.


Is the process in a state of control? Did any subgroup means exceed the control

limits? If so, how many? Did any subgroup standard deviations exceed the control

limits? If so, how many? Where is the largest problem, if any, between variation

of within variation? Is the performance consistent with the principles of SPC? Are

there any trends over time?




Analyze Task 19





Autocorrelation

Autocorrelation for CTQ‐1 Value and CTQ‐1 and CTQ‐2 Time was performed in the measure phase. For CTQ‐1

Value there are approximately 15 lags, indicating a likely sinusoidal pattern exhibited in the X‐bar chart for CTQ‐

1 attributable to one or more assignable causes. CTQ‐1 and CTQ‐2 Time indicated that the data appeared to be

random.

No autocorrelation was performed on CTQ‐2 Value in the Measure Phase as the data appeared to be in control.

Midco ask that this analysis be performed along with the looking at the CTP variables X1 through X5.

65605550454035302520151051

1.0

0.8

0.6

0.4

0.2

0.0

-0.2

-0.4

-0.6

-0.8

-1.0

Lag

Aut

ocor

rela

tion


Generally, the more lags outside of the red zone, the more chance that the event is non‐random; with only 5

lags outside the confidence limits it is likely that the data is random as indicated by the evidence of control

determined in the Measure phase.

It is a good idea to check for autocorrelation when a pattern exists in the data as represented by the cycles in

the X‐bar charts of X2, X3, and X5 in order to determine if the pattern is by random chance or likely attributable

to an assignable cause.


Are any of the lags statistically significant? What is the

meaning of the lags in terms of forecasting?



Analyze Task 15





CTP Comment

X1 No Significant Lags. Data is likely to be random.

X2 3 Lags and some additional lags in the high 20’s to low 40’s – possible assignable cause present.

X3 Lag 1 and Lag 40 are significant. Data is likely to be random.

X4 11 Lags – likely that assignable cause is present.

X5 18 Lags – likely that assignable cause is present.

Capability Analysis

A true capability analysis would not be performed until the process is under control and assignable causes are

eliminated. However, in Six Sigma analysis an “as is” capability analysis is often performed to analyze whether

the process has centering or spread issues or both. Individual data was provided for 500 samples (n=500). The

time between samples was not known. In order to come up with a rough idea of “Overall Capability” versus

“Potential Capability” it was decided the individual values would be compiled into subgroups of 5 (n=500;

g=100). A capability analysis was performed on all variables as summarized below:

CTP LSL/Target/USL

Sample

Mean

Centering

Issue

Spread Issue

Potential Capability Z-Bench

Overall Capability Z-Bench

Potential

Capability

Overall

Capability

X1 90/(N/A)/(N/A) 100.076 Yes Yes Yes 1.04 1.04

X2 31/33/35 33.172 No Yes Yes 1.76 .99

X3 (N/A)/(N/A)/310 300.15 Yes Yes Yes 1.56 1.41

X4 100/105/110 100.33 Yes No Yes 0.35 0.11

X5 1.15/1.20/1.25 1.25 Yes No Yes -0.24 -0.07


What is the short term (within capability) in terms of Cp,

Cpl, Cpu, Pp, Ppl, Ppu, Z‐Bench and Sigma Capability?




Analyze Task 6





Z‐Bench is considered long term Sigma. Short term Sigma is equal to Z‐Bench + 1.5. The Potential (Within)

Capability statistics (Cp, Cpl…) are based off of the estimate for within standard deviation which represents the

variation within subgroups. Overall standard deviation takes into account variation from the entire data set and

looks at the total process variation. The Overall Capability is based on the statistics (Pp, Ppl…). The one sided Cp

is calculated by dividing Z‐Bench by 3.

CTP X1 is in control; however, the process is performing at a very low level. First investigate the

feasibility of increasing the target value. Secondarily investigate decreasing variation as the short term

(within) standard deviation places the mean at only about one standard deviation from the lower

specification limit.

CTP X2 has numerous out of control points on the X‐Bar chart. This is reflected in the difference

between the Potential and Overall Capability. The process is on‐target, so centering is not an issue.

However, with a Potential Capability of only 1.76, variation needs to be reduced and should be

investigated after the process is brought under control.

CTP X3 has several out of control points on the X‐Bar chart. Getting the process in‐control should be the

first priority. Secondarily, investigate the feasibility of decreasing the target value, and thirdly investigate

decreasing variation.

CTP X4 is severely out of control with respect to the X‐Bar chart and has a mean just slightly above the

lower specification limit. The within standard deviation is 0.934 which would result in 10.7 standard

deviations from between specification limits; a little over a Five Sigma short term process, if the out of

control and centering issues can be resolved.

CTP X5 is severely out of control with respect to the X‐Bar chart and has a mean outside the upper

specification limit. The variation is very good if the other issues can be resolved.





Analysis of Sampling Demographics

Four variables are available for analysis in the sampling demographics: Day; Subgroup; Sequence; Inspector.

MaryFredBill

80

75

70

65

60

Inspector

CTQ

-1 V

alue

FriMonThrTue

Wed

Day

Multi-Vari Chart for CTQ-1 Value by Day - Inspector

Day

CTQ

-1 V

alue

WedTueThrMonFri

80

75

70

65

60

InspectorBill

FredMary

Multi-Vari Chart for CTQ-1 Value by Inspector - Day

MaryFredBill

560

550

540

530

520

Inspector

CTQ

-2 V

alue

FriMonThrTue

Wed

Day

Multi-Vari Chart for CTQ-2 Value by Day - Inspector

Day

CTQ

-2 V

alue

WedTueThrMonFri

560

550

540

530

520

InspectorBill

FredMary

Multi-Vari Chart for CTQ-2 Value by Inspector - Day




Analyze Task 3


Interpret the Multi‐vari Charts. Hows does inspector to

inspector variation compare to the day to day variation?

How does the within inspector variation compare to the

between inspector variation? How does the within day

variation compare to between day variation? What

inspectors and days have the largest influence, if any?




Analyze Task 3





Since day to day variation appears to be significant, a one‐way analysis of variance (ANOVA) was performed with

both a Tukey and Fishers comparison.

CTQ-1

The ANOVA for Days with respect to CTQ‐1 had a p‐value of 0.000 and an R2 of 15.00%. Both the Tukey and

Fisher comparisons indicated at a 95% confidence level that Tuesday had a statistically higher value than the

other days and that Wednesday had a statistically lower value than the other days. The target value for CTQ‐1 is

80. Since all days are lower than the target, and Wednesday is significantly lower than the other days and

Tuesday is significantly higher than the other days, Wednesday’s activities should be compared to Tuesday’s

activities to determine possible causes of the differences.

CTQ-2




Analyze Task 9


When visually comparing the variance of each day to the

other days, do you feel that the there is a substantial

difference in the variances or do they all appear to be

relatively equal? Is the day to day differences in the mean

statistically important? Is the day to day differences in the

mean pragmatically important?





The ANOVA for Days with respect to CTQ‐2 had a p‐value of 0.016 and an R2 of 2.44% indicating a statistical

difference in days, but not a pragmatic difference given the low R2. Neither the Tukey and Fisher comparisons

isolated any group at a 95% confidence level.

Midco was more familiar with t‐tests than ANOVA and requested that t‐test’s be used to compare the means for

each day.

CTQ-1 Mean

Comparison t-test

Monday (73.62)

Tuesday (79.99)

Wednesday (61.84)

Thursday (75.66)

Friday (75.75)

Monday p-value = 0.001

Significant Difference

p-value = 0.000 Significant Difference

p-value = 0.355 Not Significant No Difference


Tuesday p-value = 0.000




Wednesday



Thursday p-value = 0.968 Not Significant No Difference

The individual t‐tests were consistent with the Tukey and Fisher’s analysis indicating that Wednesday has a

lower mean than all of the other days; but included Tuesday in a group with Monday, Thursday, and Friday.

Tukey and Fisher’s comparisons both control the error rate (α) in different manners, resulting in the slight

difference. The conclusion is the same. Even through Tuesday is not showing as significantly different from

Monday, Thursday, and Friday, it is the highest of all days and closest to the target, so as noted above,

Wednesday should be compared to Tuesday to determine possible differences.




Analyze Task 11


Complete the table. What are your conclusions?





CTQ-2 Mean

Comparison t-test

Monday (73.62)

Tuesday (79.99)

Wednesday (61.84)

Thursday (75.66)

Friday (75.75)






Tuesday p-value = 0.008




Wednesday p-value = 0.057 Not Significant No Difference

p-value = 0.933 No Significant No Difference

Thursday p-value = 0.041


The Tukey and Fisher’s analysis did not isolate any groups; while the individual t‐tests isolate Tuesday and

Thursday as one group and Monday, Wednesday, and Friday as a second group. Regardless, the R2 value of

2.44% is so low, pragmatically it should be considered that there is no difference as it would likely be difficult to

see causes in an investigation.





Midco noted the difference in variances and ask that additional analysis be performed to determine whether or

not variances were different with respect to days. (Variances were calculated by squaring the Standard

Deviations done in the previous analysis displayed in charts above.)

CTQ-1 Variance

Comparison F-test

Monday (166.668)

Tuesday (224.400)

Wednesday (187.416)

Thursday (317.96)

Friday (178.757)


Not Significant No Difference




Tuesday p-value = 0.372 Not Significant No Difference



Wednesday

p-value = 0.009








Analyze Task 10







CTQ-2 Variance

Comparison

Monday (3659.040)

Tuesday (3123.692)

Wednesday (4523.152)

Thursday (4039.874)

Friday (178.757)

Monday p-value = 0.433 Not Significant No Difference


p-value = 0.623 Significant

No Difference


Tuesday p-value = 0.069 Not Significant No Difference



Wednesday p-value = 0.583

Significant No Difference



Significant No Difference

While the values for Standard Deviation, and thus variances appear to be different, the analysis reveals that with

respect to CTQ‐1 only Thursday is statistically different than the other days and that with respect to CTQ‐2 there

is no statistical difference. It is important to do the analysis, since by looking at just the standard deviations or

variances number, one might conclude that there is a difference and by simply visually evaluating, one might

conclude that there is no difference. A visual conclusion with respect to no difference in the variations of CTQ‐2

would have been accurate, but the difference between Thursday and the other days with respect to CTQ‐1

would have likely been lost in this analysis. Investigation should be done to determine why the variation of CTQ‐

1 is higher on Thursday, than the other days.

A one‐way analysis of variance (ANOVA) was performed to determine the effect of subgroups on CTQ1 and

CTQ2 Values. A similar analysis was performed during the measure phase to determine if variation was

attributable to subgroup‐to‐subgroup (process centering issues; black noise) variations indicating a need to

focus on long term capability versus variation being attributable to natural causes or other unknown causes

(white noise) within the process. In the case of the Measure Phase analysis, the subgroups were labeled 1

through 100. In this analysis, there is a concern that time of day is an issue, so subgroups are labeled 1 to 20 for

each day (subgroup 1 is the first subgroup of each day and subgroup 20 is the last subgroup of each day).





The figure below shows the means and confidence intervals for the twenty representative times of day. The

target value for CTQ1 Value is 80 and the upper and lower specifications are 115 and 45, respectively. While the

confidence limits for all times of the day are within specification it appears that a shift is taking place starting

with subgroup 4 and ending with subgroup 9. Subgroups 5 to 9 have a mean close to or higher than the target

while those before and after have a mean that is substantially below the target. Subgroups 5 to 9 are a better

situation than the other subgroups. Regardless of the R2 value, an investigation should be done to determine

what is being done during this time period and to determine how the results can be duplicated during the other

time periods.


What are the p‐values for CTQ‐1 Value and CTQ2 Value?

What is the interpretation of the p‐values? What

proportion of the variation is explained by the model? Are

CTQ‐1 Value and CTQ2 Value pragmatically significant?




Analyze Task 5





A one way ANOVA was also performed on X1 through X5 Values to determine the effect of subgroups.


What are the p‐values for X1 through X5? What is the

interpretation of the p‐values? What proportion of the

variation is explained by the model? Are X1 through X5

pragmatically significant? Where should we focus?




Analyze Task 5





A cross-tabulation was performed on CTQ-1 and CTQ-2 Test versus days.

Rows: CTQ-1 Test Columns: Day Fri Mon Thr Tue Wed All Fail 0 1 4 2 12 19 0.00 5.26 21.05 10.53 63.16 100.00 0.00 1.00 4.00 2.00 12.00 3.80 0.00 0.20 0.80 0.40 2.40 3.80 Pass 100 99 96 98 88 481 20.79 20.58 19.96 20.37 18.30 100.00 100.00 99.00 96.00 98.00 88.00 96.20 20.00 19.80 19.20 19.60 17.60 96.20 Cell Contents: Count % of Row % of Column Pearson Chi-Square = 25.386, DF = 4, P-Value = 0.000 Likelihood Ratio Chi-Square = 23.753, DF = 4, P-Value = 0.000

For CTQ‐1 Test, yield is lowest on Wednesdays with the other days being above 96%. A Pearson chi‐square

analysis resulted in a p‐value of 0.000 indicating that the influence of days on CTQ‐1 Test results is not random.

Note: Degrees of Freedom (DF) are calculated by the equation (C‐1)*(R‐1.

Rows: CTQ-2 Test Columns: Day Fri Mon Thr Tue Wed All Fail 17 14 13 10 17 71 23.94 19.72 18.31 14.08 23.94 100.00 17.00 14.00 13.00 10.00 17.00 14.20 3.40 2.80 2.60 2.00 3.40 14.20 Pass 83 86 87 90 83 429 19.35 20.05 20.28 20.98 19.35 100.00 83.00 86.00 87.00 90.00 83.00 85.80 16.60 17.20 17.40 18.00 16.60 85.80 Cell Contents: Count % of Row % of Column % of Total Pearson Chi-Square = 2.856, DF = 4, P-Value = 0.582 Likelihood Ratio Chi-Square = 2.936, DF = 4, P-Value = 0.569




Analyze Task 12


Complete the table. What are your conclusions? What is the number of degrees of

freedom? What are the p‐values? Is there an association or can results be deemed

random?





For CTQ‐2 Test, the lowest yield is on Wednesdays and Fridays, with the highest yield being on Tuesdays. A

Pearson chi‐square analysis was done.

A cross‐tabulation was performed on Inspectors versus days:

Tabulated statistics: Inspector, Day Rows: Inspector Columns: Day Fri Mon Thr Tue Wed All Bill 7 10 5 4 6 32 21.88 31.25 15.63 12.50 18.75 100.00 7.00 10.00 5.00 4.00 6.00 6.40 1.40 2.00 1.00 0.80 1.20 6.40 Fred 60 59 75 73 70 337 17.80 17.51 22.26 21.66 20.77 100.00 60.00 59.00 75.00 73.00 70.00 67.40 12.00 11.80 15.00 14.60 14.00 67.40 Mary 33 31 20 23 24 131 25.19 23.66 15.27 17.56 18.32 100.00 33.00 31.00 20.00 23.00 24.00 26.20 6.60 6.20 4.00 4.60 4.80 26.20 Cell Contents: Count % of Row % of Column Pearson Chi-Square = 11.281, DF = 8, P-Value = 0.186 Likelihood Ratio Chi-Square = 11.149, DF = 8, P-Value = 0.193

Fred is the lead inspector with 67.4% of the inspections followed by Mary with 26.2% and then Bill with only

6.4%. This ratio is fairly consistent throughout the week. A Pearson chi‐square analysis resulted in a p‐value of

0.186 indicating that the relationship between Inspectors and Days is random.

Midco, ask that we look deeper into the relationship between the CTP’s and their relationship with Days and

Inspectors. A Chi‐Square Correlation analysis was performed on the five CTP’s. The results are summarized in the

chart below.




Analyze Task 12

Analyze Task 16


Complete the table. What are your conclusions? What is the number of degrees of freedom?

What are the p‐values? Is there an association or can results be deemed random? What

proportion of the total performance evaluations were made by each inspector?





Check/Inspector/Day Pearson Chi-Square

Analysis

X1Check X2 Check X3 Check X4 Check X5 Check

Bill Invalid

Invalid Invalid P = 0.656 Random

P = 0.078 Random

Fred P = 0.344 Random

P = 0.032 Association

P = 0.991 Random



Mary p = 0.323 Random

Invalid P = 0.722 Random

Invalid P = 0.000

Association

In discussions with Midco some concern was expressed that the Subgroup and Day data might not be normally

distributed. The normality assumption is assumed for ANOVA, t‐tests, and the like. In order to address this

concern a Mood’s Median Test was performed. The Median Test is a nonparametric test, meaning that it does

not rely on the normality assumption. The results are summarized in the following table.

Moods Median Test Subgroup Days

CTQ 1 Value p = 0.000 Difference

p = 0.000 Difference

CTQ 2 Value p = 0.007 Difference

p = 0.267 No Difference

The Mood’s Median Test differed from the ANOVA performed on subgroups. The ANOVA indicated that there

was an association Subgroups and CTQ 1 Value, but not CTQ 2 Value, while the Median Test concluded there

was an association between Subgroups and both CTQ 1 and CTQ Values. As a reminder, in this case Subgroups

represents Time of Day. Since an investigation should be made with regard to CTQ‐1 Value, it is recommended

that the investigation be expanded to include CTQ‐2 Value.




Analyze Task 13


Complete the table. What are your conclusions? What are the p‐

values? Is there a statistical difference in the medians?




Analyze Task 14





The Mood’s Median Test also differed from the ANOVA with regard to Days.

The nonparametric analysis resulted in the same conclusion as the parametric analysis with respect to both sub‐

groups and days.

Yield Analysis

Midco requested that an analysis be performed to estimate yield. An “As is” capability analysis had been

performed earlier in the Analyze Phase, where the individual values were compiled into subgroups of five. The

capability analysis provides the estimated within and overall PPM (parts per million) defective, which can be

used to estimate yield. As noted previously, a subgroup size of 5 was used to estimate within variation.

CTP LSL/Target/USL Within PPM Defective

Total

Within Yield Estimate

Overall PPM Defective

Total

Overall Yield

Estimate

X-Check Discrete

Data Yield

X1 90/(N/A)/(N/A) 149,353.31 85.06% 149,664.78 85.03% 86.00%

X2 31/33/35 39,028.80 96.10% 161,192.08 83.88% 83.80%

X3 (N/A)/(N/A)/310 58,912.75 94.11% 79,388.80 92.06% 91.40%

X4 100/105/110 363,561.24 63.64% 457,078.73 54.29% 54.40%

X5 1.15/1.20/1.25 594,254.75 40.57% 526,305.80 47.37% 49.00%

The discrete data (X1‐X5 Check was also analyzed for yield and is summarized in the above chart. The discrete

data would be considered long term and as can be seen the estimates for yield from the discrete data is very

close to the Overall Yield Estimates from the capability analysis. The discrete data will be used to calculate rolled

throughput and normalized yield.



What is the rolled throughput yield based upon the X‐Check

discrete data yield? What is normalized yield based upon the X‐

Check discrete data yield?




Analyze Task 17

Analyze Task 18





The Normalized Yield provides the minimum Throughput Yield for each step of the process to achieve a given

Rolled Throughput Yield. While not the arithmetic average it can be considered the average yield of all of the

processes. By comparing the Discrete Yields of each variable to the Normalized Yield, it can be seen that CTP’s

X4 and X5 are below average, while CTP’s X1, X2, and X3 are above average.





Improve

What are the vital CTP’s and what should their setting be?

Conclusions:

MidCo performed both a 2^5‐1 Half Fractional Factorial and a 2^5 Full Factorial analysis

measuring values for both CTQ‐1 and CTQ‐2 for each run. There was some question about

the data but confirmed that appropriate procedures had been utilized. Normally we would

run the fractional factorial design for screening purposes and then run a full factorial on the

more critical factors. Both experiments were analyzed and provide consistent results.

As discovered during the analyze phase, several processes were out of control. We

consulted with MidCo on how they could bring these process under control prior to

proceeding to the improve stage. The table below compares the processes before and after

MidCo address the control issues.

Before After


X1 9.70 9.71 Yes 6.72 6.69 Yes

X2 .95 1.42 No 0.71 0.74 Yes

X3 6.29 6.99 Marginal 4.79 4.58 Yes

X4 0.934 2.99 No 2.19 2.05 Yes

X5 0.006 0.021 No 0.014 0.013 Yes

CTQ‐1 9.88 15.84 No 9.29 8.98 Yes

CTQ‐2 55.89 62.82 Marginal Unavailable


A MindPro ™ Lean Six Sigma Simulated Training Project Improve



MidCo advised that they were more concerned with CTQ‐1 than CTQ‐2 and made the

decision to go with the following settings:

X1 X2 X3 X4 X5

Pre‐improve settings 100 33 300 100 1.25


Post‐improve settings 80 33 300 145 1.15


It appears as MidCo made the decision to ignore the impact on CTQ‐2 that will result by

changing the CTP’s to optimize CTQ‐1.

They reduced X1 and X5 which should substantially reduce the variation on CTQ‐1.

o X1 is below the specification and X5 is set to aim at the lower specification.

o The centering of CTQ‐1 should be improved but the variance of CTQ will be

substantially deteriorated.

They increased X4 which should move CTQ‐1 upward and closer to its target.

o X4 is substantially outside the specification limits.

o This will also improve the centering of CTQ‐2

MidCo has not confirmed their rationale for stepping outside the specifications on

variable X1 and X4. We can only assume that they have re‐evaluated these

specifications and are in the processing of changing them. With the adjustments, only

the variance of CTQ‐2 will be negatively impacted.






Critical-to-Process Variables Critical-to-Process Variables

High 1.0 1.0 1.0 1.0 1.0 120.0 36.0 320.0 110.0 1.25

Mid 0.0 0.0 0.0 0.0 0.0 100.0 33.0 300.0 105.0 1.20

Low -1.0 -1.0 -1.0 -1.0 -1.0 80.0 30.0 280.0 100.0 1.15





2^5-1 Fractional Factorial Design


Case X1

Code X2

Code X3

Code X4

Code X5

Code X1

Value X2

Value X3

Value X4

Value X5

Value CTQ-1

CTQ-2

1 -1 -1 -1 -1 1 80.0 30.0 280.0 100.0 1.3 57.2 540

2 1 -1 -1 -1 -1 120.0 30.0 280.0 100.0 1.2 45.6 391

3 -1 1 -1 -1 -1 80.0 36.0 280.0 100.0 1.2 23.2 704

4 1 1 -1 -1 1 120.0 36.0 280.0 100.0 1.3 75.6 432

5 -1 -1 1 -1 -1 80.0 30.0 320.0 100.0 1.2 32.1 717

6 1 -1 1 -1 1 120.0 30.0 320.0 100.0 1.3 107.6 440

7 -1 1 1 -1 1 80.0 36.0 320.0 100.0 1.3 51.6 792

8 1 1 1 -1 -1 120.0 36.0 320.0 100.0 1.2 41.5 574

9 -1 -1 -1 1 -1 80.0 30.0 280.0 110.0 1.2 39.5 554

10 1 -1 -1 1 1 120.0 30.0 280.0 110.0 1.3 108.0 340

11 -1 1 -1 1 1 80.0 36.0 280.0 110.0 1.3 57.0 612

12 1 1 -1 1 -1 120.0 36.0 280.0 110.0 1.2 48.1 443

13 -1 -1 1 1 1 80.0 30.0 320.0 110.0 1.3 76.7 630

14 1 -1 1 1 -1 120.0 30.0 320.0 110.0 1.2 62.6 457

15 -1 1 1 1 -1 80.0 36.0 320.0 110.0 1.2 36.7 822

16 1 1 1 1 1 120.0 36.0 320.0 110.0 1.3 98.2 504





2^5 Full Factorial Design


Case X1

Code X2

Code X3

Code X4

Code X5

Code X1

Value X2

Value X3

Value X4

Value X5

Value CTQ-1

CTQ-2

1 -1 -1 -1 -1 -1 80.0 30.0 280.0 100.0 1.2 28.6 587

2 1 -1 -1 -1 -1 120.0 30.0 280.0 100.0 1.2 45.6 391

3 -1 1 -1 -1 -1 80.0 36.0 280.0 100.0 1.2 23.2 704

4 1 1 -1 -1 -1 120.0 36.0 280.0 100.0 1.2 37.0 470

5 -1 -1 1 -1 -1 80.0 30.0 320.0 100.0 1.2 32.1 717

6 1 -1 1 -1 -1 120.0 30.0 320.0 100.0 1.2 51.2 478

7 -1 1 1 -1 -1 80.0 36.0 320.0 100.0 1.2 26.0 861

8 1 1 1 -1 -1 120.0 36.0 320.0 100.0 1.2 41.5 574

9 -1 -1 -1 1 -1 80.0 30.0 280.0 110.0 1.2 39.5 554

10 1 -1 -1 1 -1 120.0 30.0 280.0 110.0 1.2 57.0 370

11 -1 1 -1 1 -1 80.0 36.0 280.0 110.0 1.2 33.9 665

12 1 1 -1 1 -1 120.0 36.0 280.0 110.0 1.2 48.1 443

13 -1 -1 1 1 -1 80.0 30.0 320.0 110.0 1.2 43.0 685

14 1 -1 1 1 -1 120.0 30.0 320.0 110.0 1.2 62.6 457

15 -1 1 1 1 -1 80.0 36.0 320.0 110.0 1.2 36.7 822

16 1 1 1 1 -1 120.0 36.0 320.0 110.0 1.2 52.6 548

17 -1 -1 -1 -1 1 80.0 30.0 280.0 100.0 1.3 57.2 540

18 1 -1 -1 -1 1 120.0 30.0 280.0 100.0 1.3 94.9 360

19 -1 1 -1 -1 1 80.0 36.0 280.0 100.0 1.3 45.5 648

20 1 1 -1 -1 1 120.0 36.0 280.0 100.0 1.3 75.6 432





21 -1 -1 1 -1 1 80.0 30.0 320.0 100.0 1.3 64.8 660

22 1 -1 1 -1 1 120.0 30.0 320.0 100.0 1.3 107.6 440

23 -1 1 1 -1 1 80.0 36.0 320.0 100.0 1.3 51.6 792

24 1 1 1 -1 1 120.0 36.0 320.0 100.0 1.3 85.6 528

25 -1 -1 -1 1 1 80.0 30.0 280.0 110.0 1.3 69.1 510

26 1 -1 -1 1 1 120.0 30.0 280.0 110.0 1.3 108.0 340

27 -1 1 -1 1 1 80.0 36.0 280.0 110.0 1.3 57.0 612

28 1 1 -1 1 1 120.0 36.0 280.0 110.0 1.3 88.1 408

29 -1 -1 1 1 1 80.0 30.0 320.0 110.0 1.3 76.7 630

30 1 -1 1 1 1 120.0 30.0 320.0 110.0 1.3 120.8 420

31 -1 1 1 1 1 80.0 36.0 320.0 110.0 1.3 63.1 756

32 1 1 1 1 1 120.0 36.0 320.0 110.0 1.3 98.2 504





Overview

MidCo decided not to wait on guidance and performed both a 2^5‐1 Half Fractional Factorial and a 2^5 Full

Factorial analysis measuring values for both CTQ‐1 and CTQ‐2 for each run. The data is presented in non‐random

order. MidCo was contacted to verify that in fact the runs had been performed randomly and we were assured

that a randomization scheme was utilized and the data had subsequently been recorded in non‐random order.

Normally we would run the fractional factorial design for screening purposes and then run a full factorial on the

more critical factors. If a full factorial were run first it would not be appropriate to run a fractional factorial as

well. Since MidCo has provided us with both experiments, we will analyze both for consistency.

For the experiment, MidCo used fixed factors ‐ that is they set selected a high and low level for each factor

based upon what they believed was an appropriate linear range. The table below summarizes the specifications

and the levels chosen for each factor.

CTP LSL Target USL Low Level High Level

X1 90 N/A N/A 80 120

X2 31 33 35 30 36

X3 N/A N/A 310 280 320

X4 100 105 110 100 110

X5 1.15 1.20 1.25 1.2 1.3

MidCo was contacted to inquire as to why the range for X5 was held so tightly and skewed to the high side of

the spec range. They explained that this X5 was a difficult variable to control and that the original specifications

were suspect.





2^5‐1 Fractional Factorial

As stated above, fractional factorial is typically used as a screening factorial to identify the factors that have

significant influence on the CTP’s and to get a feeling for which factors will have the great leverage for improving

the CTQ’s.

Midco performed a 2^5‐1 Fractional Factorial which is known as a half‐fractional factorial. A full factorial would

have 2^5 = 32 runs. Whereas the half‐fractional factorial only has 16 runs. Minitab can be used to generate the

fractional factorial through its DOE Factorial Design Creator. When using a factorial design the higher level

interactions are aliased with the main effects. Minitab shows the alias structure when creating the design.

Fractional Factorial Design Factors: 5 Base Design: 5, 16 Resolution: V Runs: 16 Replicates: 1 Fraction: 1/2 Blocks: 1 Center pts (total): 0 Design Generators: E = ABCD Alias Structure I + ABCDE A + BCDE B + ACDE C + ABDE D + ABCE E + ABCD AB + CDE AC + BDE AD + BCE AE + BCD BC + ADE BD + ACE BE + ACD CD + ABE CE + ABD DE + ABC

CTQ‐1


What is the resolution for the 2^5‐1 Fractional Factorial? What is

the design identity? What is the design generator? What is the

alias structure? What interactions are confounded with the main

effects? What interactions are confounded with the two level

interactions?


Black Belt


Improve Task 1





The results of the ANOVA/General Linear Model from Minitab are provided below.

Analysis of Variance for CTQ-1, using Adjusted SS for Tests

Source DF Seq SS Adj SS Adj MS F P

X1 Value 1 2840.9 2840.9 2840.9 50.04 0.000

X2 Value 1 592.9 592.9 592.9 10.44 0.009

X3 Value 1 174.2 174.2 174.2 3.07 0.110

X4 Value 1 533.6 533.6 533.6 9.40 0.012

X5 Value 1 5722.9 5722.9 5722.9 100.80 0.000

Error 10 567.7 567.7 56.8

Total 15 10432.3

S = 7.53489 R-Sq = 94.56% R-Sq(adj) = 91.84%

It appears that all factors except X3 may have a significant impact on CTQ‐1. However, by looking at the Sum of

Squares (SS), we can see that the big hitters are X5 and X1 which account for 54.9% and 27.3% of the total SS’s

respectively.


Black Belt


Improve Task 2


Complete and interpret the ANOVA Table for the 2^5‐1 of

CTQ‐1. The main effects all have only one degree of freedom

and are aliased with higher level interactions? Explain the effect

this might have on the interpretation of p‐values.





The main effects plot gives us an indication of the influence that each factor has on the CTP.

Mea

n of

CTQ

-1

12080

80

70

60

50

40

3630 320280

110100

80

70

60

50

401.31.2

X1 Value X2 Value X3 Value

X4 Value X5 Value

Main Effects Plot (fitted means) for CTQ-1

The fractional factorial is consistent with the regression analysis performed in the analyze phase, which

indicated that there was a strong positive linear relationship between factors X1, X4, and X5 to CTQ‐1; that X2

had a mild negative linear relationship and that X3 was insignificant. From the measure phase we know that we

need to move the centering of CTQ‐1 to the right.

Comparing the observations of the fractional factorial to the measure and analyze phases:

X5 has the largest impact and we know from the measure phase that it a significant linear relationship to

CTQ‐1. While the analyze phase indicated that the process mean was on target, MidCo is now indicating

that the suspect that the specifications were not properly set. X5 should be adjusted to the upper level

value of 1.3.





X1 also has a large impact on the variation in CTQ‐1 and the analyze phase indicated that there was a

strong linear relationship as well. X1 only has unilateral specification limit on the low side, so adjusting it

upward should not be present any problems. X1 exhibited no spread issue in the analyze phase so we

should be able to adjust X1 so that its mean is near the high level of analyzed for the factor (120).

X2 accounts for less than 6% of the overall SS and thus will have only a marginal impact on CTQ‐1. We

know from the analyze phase that X2 was centered on target but had spread issues and only had a

marginal linear relationship with CTQ‐1. It appears that we would not want to adjust the aim of X2

without further analysis.

X4 like X2, only accounts for about 6% of the overall SS and thus will have minimal impact on CTQ‐1.

However, in the analyze phase X4 did show a strong linear relationship with CTQ‐1. The analyze phase

also revealed that X4 was aimed at the lower specification limit. If we aim X4 so that it is on target (105)

we will improve CTQ‐1.

X3 has little impact on CTQ‐1 so we might want to utilize X3 as a control variable.

CTQ‐2

The results of the ANOVA/General Linear Model from Minitab are provided below.

Analysis of Variance for CTQ-2, using Adjusted SS for Tests

Source DF Seq SS Adj SS Adj MS F P

X1 Value 1 200256 200256 200256 417.42 0.000

X2 Value 1 41412 41412 41412 86.32 0.000

X3 Value 1 52900 52900 52900 110.27 0.000

X4 Value 1 3249 3249 3249 6.77 0.026

X5 Value 1 8649 8649 8649 18.03 0.002

Error 10 4798 4798 480

Total 15 311264

S = 21.9032 R-Sq = 98.46% R-Sq(adj) = 97.69%


Black Belt


Improve Task 2





Mea

n of

CTQ

-2

12080

650

600

550

500

450

3630 320280

110100

650

600

550

500

450

1.31.2

X1 Value X2 Value X3 Value

X4 Value X5 Value

Main Effects Plot (fitted means) for CTQ-2

The fractional factorial is consistent with the regression analysis performed in the analyze phase, which

indicated that there was a strong negative linear relationship between factors X1 and CTQ‐2 and that there is

mildly positive linear relationship between X2, X4, and X5 to CTQ‐2. The analyze phase indicated that the linear

relationship between X3 and CTQ‐2 was insignificant but the main effects plot of the experiment indicates that

there may be a positive correlation. From the measure phase we know that we need to significantly move the

centering of CTQ‐2 downward. Since X1 and X5 have the strongest influence it we should look at adjusting these

two CTP’s to the right so as to influence CTQ‐1.


Complete and interpret the ANOVA Table for the 2^5‐1 of

CTQ‐2. What main effect has the most leverage? What are the

p‐values for the main effects and what do they mean in terms of

statistical significance? Which factors might be considered the

trivial many? Experimental error accounts for what percent of

the Total Sum of Squares?





Comparing the observations of the fractional factorial to the measure and analyze phases:

X1 has by far the most significant impact on CTQ‐2 and the analyze phase indicated that there was a

strong linear relationship to CTQ‐2 as well. We would want to adjust X1 upward in order to move the

mean of CTQ‐2 towards its target. This strategy would be consistent with our decision from looking at X1

and CTQ‐1, so it seems that it would be a good decision to move X1 to near its high level of 120.

X3 accounts for about 17.0% of the overall SS which cannot be ignored. While the analyze phase

indicated that there was no significant linear relationship between X3 and CTQ‐2, we can not ignore the

possibility. Since the main effects plot indicates a strong positive relationship, we might want to

consider lowering the aim of X3 in order to move CTQ‐2 to closer to the target. X3 has a unilateral

tolerance on the high side so it there should not be an issue in lowering the aim of X3. However,

lowering X3 will also have a negative impact on CTQ‐1 as it should mildly move CTQ‐1 downward and

further from the target. We should not adjust X3 at this stage

X2 accounts for 13.3% of the overall SS and has a mildly linear relationship to CTQ‐2 based upon the

analysis in the analyze phase. However, as stated above when analyzing X2 with regard to CTQ‐1, “We

know from the analyze phase that X2 was centered on target but had spread issues.” So we should not

adjust X2 at this stage.

X4 appeared to have some positive impact on CTQ‐1 if we adjusted the aim so that it was on target.

While there is a only a small relationship between X4 and CTQ‐2, moving X4 to target would have some

positive impact so it is recommended that X4 be moved to its target.

X5 specifications are suspect and moving X5 to its upper value of 1.3 should have a significant impact on

CTQ‐1 and will also have a marginal positive impact on CTQ‐2 therefore it is recommended that X5 be

moved to its upper level of 1.3

Summary of Recommendations based upon Fractional Factorial

CTP Comment

X1 Adjust to high level of 120.

X2 No adjustment recommended

X3 No Adjustment recommended

X4 Adjust to target of 105

X5 Adjust to upper level of 1.3





2^5 Full Factorial

Since MidCo performed a full factorial experiment as well as the half factorial, we will evaluate the full factorial

for consistency with the fractional factorial. The table below shows a comparison of the p‐values and % Sum of

Squares for the two experiments.

CTP CTQ-1 CTQ-2

Half Fractional Full Factorial Half Fractional Full Factorial

X1

0.000 27.2%

0.000 27.2%

0.000 64.3%

0.000 64.3%

X2

0.009 5.7%

0.000 5.7%

0.000 13.3%

0.000 13.3%

X3

0.110 1.7%

0.009 1.7%

0.000 17.0%

0.000 17.0%

X4

0.012 5.1%

0.000 5.2%

0.026 1.0%

0.000 1.1%

X5

0.000 54.9%

0.000 54.8%

0.002 2.8%

0.000 2.8%

Strictly speaking since the main effects have only one degree of freedom, they should not be statistically tested;

however, the relative effects can be meaningful and are often reported.




Improve Task 3


Complete the above table by adding the p‐values and Sum of

Squares of the Main Effects / Total Sum of Squares as a percent.

What is the statistical and pragmatic significance of the main

effects? Which main effect is the largest contributor? Given that

the Main effects only have one degree of freedom is it

appropriate to rely on the p‐values?





The table below shows a comparison of the main effects plots:

CTP CTQ-1 CTQ-2

Half Fractional Full Factorial Half Fractional Full Factorial

X1

X2

X3

X4

X5




Improve Task 3





The data from the full factorial is consistent with the half factorial, maybe too consistent. MidCo was contacted

and queried as to the process utilized for the experiment. We had a concern that the data was so consistent that

it might be a manipulation. MidCo assured us that their processes were spot on and that they had been properly

randomized. They further assured us that they had done two separate experiments. Given the consistency of the

data, the full factorial will be used for analysis from this point forward and the conclusions drawn from the half

factorial will stand.


Complete the above table by adding the Main Effects Plots.

Which main effects appear to have the most leverage? Which

main effects have positive slopes? Negative slopes? Which main

effects appear that they may simply be a result of chance

sampling variation? What is the meaning of the steepness of the

slope in terms of evaluating influence on the CTQ?





Interaction Plots

While there were insufficient degrees of freedom to analyze interaction between the main effects, MidCo

requested that interaction plots be provided for CTQ ‐1.

36.030.0 320.0280.0 110.000100.000 1.31.1100

75

50

100

75

50

100

75

50

100

75

50

X1 Value

X2 Value

X3 Value

X4 Value

X5 Value

80.0120.0

X1 Value

30.036.0

X2 Value

280.0320.0

X3 Value

100.000110.000

X4 Value

Interaction Plot for CTQ-1Data Means

For CTQ‐1, the interaction plots between X5 and X1, X2, and X3 and appear to have some interaction since the

lines are less parallel than the other plots. However, we cannot determine from the plot whether the

interactions are either statistically or pragmatically significant. The remaining Interaction plots indicate

essentially parallel lines indicating little or no interaction between the two‐level main effects.




Improve Task 3





36.030.0 320.0280.0 110.000100.000 1.31.1750

600

450750

600

450

750

600

450750

600

450

X1 Value

X2 Value

X3 Value

X4 Value

X5 Value

80.0120.0

X1 Value

30.036.0

X2 Value

280.0320.0

X3 Value

100.000110.000

X4 Value

Interaction Plot for CTQ-2Data Means


Develop interaction plots for all main effects.

Interpret when one factor is held low, how does this impact the

influence on the CTQ when comparing another factor at its low

and high setting? What is the meaning of parallel lines? What is

the meaning of non‐parallel lines? Do the interaction plots tell

the analyst whether or not the data is statistically or practically

significant? Which plots indicate a likely interaction?




Improve Task 3




Improve Task 3





MidCo continued to press on the issues of interactions, so the Full Factorial ANOVA was redone using a full

model which included all interactions. There were insufficient degrees of freedom to provide any p‐values but

at least the relative Sum of Square could be evaluated.

Folded Designs

CTQ‐1

Due to the minimal effect of X3 Value on CTQ‐1, it was decided to run an ANOVA which would fold X3 Value into

the design. The ANOVA table for the folded design is show below:

Source DF Seq SS Adj SS Adj MS F P X1 Value 1 5681.8 5681.8 5681.8 211.30 0.000 X2 Value 1 1188.3 1188.3 1188.3 44.19 0.000 X4 Value 1 1085.8 1085.8 1085.8 40.38 0.000 X5 Value 1 11445.8 11445.8 11445.8 425.65 0.000 X1 Value*X2 Value 1 69.0 69.0 69.0 2.57 0.129 X1 Value*X4 Value 1 1.3 1.3 1.3 0.05 0.830 X1 Value*X5 Value 1 812.0 812.0 812.0 30.20 0.000 X2 Value*X4 Value 1 0.3 0.3 0.3 0.01 0.920 X2 Value*X5 Value 1 170.2 170.2 170.2 6.33 0.023 X4 Value*X5 Value 1 3.1 3.1 3.1 0.12 0.738 X1 Value*X2 Value*X4 Value 1 0.0 0.0 0.0 0.00 0.984 X1 Value*X2 Value*X5 Value 1 11.8 11.8 11.8 0.44 0.518 X1 Value*X4 Value*X5 Value 1 0.2 0.2 0.2 0.01 0.925 X2 Value*X4 Value*X5 Value 1 0.0 0.0 0.0 0.00 0.973 X1 Value*X2 Value*X4 Value*X5 Value 1 0.0 0.0 0.0 0.00 0.995 Error 16 430.2 430.2 26.9 Total 31 20899.9 S = 5.18556 R-Sq = 97.94% R-Sq(adj) = 96.01%


Perform a Full Factorial with a full model which includes all

interactions. What is the influence of higher order interactions

in comparison to lower order interactions? Which two level

interactions have the most influence?




Improve Task 4





We can see that by folding X3 Value, that the other main effects are statistically significant at the below the 0.001 level. However, factors X2 Value and X4 Value have very little effect on CTQ‐1. Therefore it was decided that we would also fold X2 and X3 Value into the design. The resulting ANOVA table is show below:

Source DF Seq SS Adj SS Adj MS F P X1 Value 1 5681.8 5681.8 5681.8 53.74 0.000 X5 Value 1 11445.8 11445.8 11445.8 108.26 0.000 X1 Value*X5 Value 1 812.0 812.0 812.0 7.68 0.010 Error 28 2960.3 2960.3 105.7 Total 31 20899.9 S = 10.2822 R-Sq = 85.84% R-Sq(adj) = 84.32%

When we fold the design we simply remove the Sum of Squares from the removed factors from the model and those sum of squares then get added to the error term. The effect of folding the design is increasing the number of replications for the factors that are not removed from the design.


Complete the ANOVA table. When starting with a 2^5 Factorial

Design, what type of design do you have when you fold in a

single factor, i.e. 2^What Exponent? How many runs? How many

replicates? Which of the remaining main effects are statistically

significant in the folded design? Which of the interactions are

statistically significant in the folded design?




Improve Task 5



Design, what type of design do you have when you fold in a

three factors, i.e. 2^What Exponent? How many runs? How

many replicates? Which of the remaining main effects are

statistically significant in the folded design? Is the interaction

statistically significant in the folded design? How do the sum of

squares for the remaining effects in the folded designs and the

total sum of squares compare to the Full Factorial design before

folding?





CTQ‐2

For CTQ‐2, X4 Values and X5 Values had the least effect so it was decided that they should be folded into the

design. The resulting ANOVA table is show below:

Source DF Seq SS Adj SS Adj MS F P X1 Value 1 400513 400513 400513 379.64 0.000 X2 Value 1 82825 82825 82825 78.51 0.000 X3 Value 1 105570 105570 105570 100.07 0.000 X1 Value*X2 Value 1 3321 3321 3321 3.15 0.089 X1 Value*X3 Value 1 4232 4232 4232 4.01 0.057 X2 Value*X3 Value 1 882 882 882 0.84 0.370 X1 Value*X2 Value*X3 Value 1 36 36 36 0.03 0.855 Error 24 25319 25319 1055 Total 31 622698 S = 32.4804 R-Sq = 95.93% R-Sq(adj) = 94.75%

As with CTQ‐1, the folded design reveals that all of the remaining main effects are statistically significant below

0.001.




Improve Task 6



Design, what type of design do you have when you fold in a two

factors, i.e. 2^What Exponent? How many runs? How many

replicates? Which of the remaining main effects are statistically

significant in the folded design? Which interactions are

statistically significant in the folded design? What is the practical

significance of the main effects and interactions?





Distribution of CTQ’s

After the analysis was complete, the question came up in discussions with MidCo with regard to the distribution

of the CTQ‐1 and CTQ‐2 data. As a result a histogram and normal probability plot was performed on the CTQ’s

as show below:

120100806040200

6

5

4

3

2

1

0

CTQ-1

Freq

uenc

y

Mean 60.07StDev 25.97N 32

Histogram of CTQ-1Normal

120100806040200

99

95

90

80

70

60504030

20

10

5

1

CTQ-1Pe

rcen

t


Probability Plot of CTQ-1Normal

840720600480360240

6

5

4

3

2

1

0

CTQ-2

Freq

uenc

y

Mean 559.6StDev 141.7N 32

Histogram of CTQ-2Normal

900800700600500400300200

99

95

90

80

70

60504030

20

10

5

1

CTQ-2

Perc

ent


Probability Plot of CTQ-2Normal

Generally 100 or more observations are considered adequate when using histograms to identify distributions. In

this case we only have 32 runs so it is not surprising the histogram does not have the typical bell shaped curve. It

is best in these situations to perform a normal probability plot with and Andersen Darling Test.




Improve Task 7




Improve Task 7


Develop the above graphs.

Does the data appear to be normally distributed?

How does CTQ‐1 compare to CTQ‐2 in terms of normality?




Improve Task 7




Improve Task 7





Additional Analysis

Although MidCo’s Half‐Fractional and Full Factorial were analyzed thoroughly, a request was made for some

additional analysis:

MidCo ask that an F‐test be performed to determine if there was a difference in the variances of CTQ‐1

for the group where X1 Value was set low versus when X1 Value was set high.

X1 Value N StDev Variance 80.0 16 16.515 272.731 120.0 16 27.236 741.813

MidCo ask that an F‐test be performed to determine if there was a difference in the variances of CTQ‐1

for the group where X2 Value was set low versus when X2 Value was set high.

X2 Value N StDev Variance 30.0 16 28.334 802.821 36.0 16 22.612 511.290


Black Belt, Green Belt – Industrial and Green Belt Commercial


Improve Task 8


Is the variance of X2 Value when set low versus when set high

difference with respect to CTQ‐1




Improve Task 8


Is the variance of X1 Value when set low versus when set high






MidCo ask that an t‐test be performed to determine if there was a difference in the means of CTQ‐1 for

the group where X1 Value was set low versus when X1 Value was set high.

X1 Value N Mean StDev SE Mean 80.0 16 46.8 16.5 4.1 120.0 16 73.4 27.2 6.8

MidCo ask that a t‐test be performed to determine if there was a difference in the means of CTQ‐1 for

the group where X2 Value was set low versus when X2 Value was set high.

X2 Value N Mean StDev SE Mean 30.0 16 66.2 28.3 7.1 36.0 16 54.0 22.6 5.7


Is the Mean of X1 Value when set low versus when set high



Is the Mean of X2 Value when set low versus when set high





Improve Task 8




Improve Task 8





Midco wanted to see a Multi‐Vari Chart on CTQ‐1 with respect to Variables X1 Value, X2 Value, and X5

Value and on CTQ‐2 with respect to Variables X1 Value, X2 Value, and X3 Value.

36.030.0

110

100

90

80

70

60

50

40

30

20

36.030.0

1.150

X2 Value

CTQ

-1

1.25080.0

120.0

X1 Value

Multi-Vari Chart for CTQ-1 by X1 Value - X5 Value

Panel variable: X5 Value

36.030.0

800

700

600

500

400

300

36.030.0

280.0

X2 Value

CTQ

-2

320.080.0

120.0

X1 Value

Multi-Vari Chart for CTQ-2 by X1 Value - X3 Value

Panel variable: X3 Value


Black Belt, Green Belt – Industrial

and Green Belt Commercial


Improve Task 9


Does the dominant source of variation in CTQ‐1 appear to be

from X1 Value, X2 Value or X5 Value? Does the dominant source

of variation in CTQ‐2 appear to be from X1 Value, X2 Value or X3

Value?



and Green Belt Commercial


Improve Task 9





Midco requested that a regression analysis be performed and a scatter plot provided for CTQ‐1 with all

five X Value variables.

Variable Analysis Scatter Plot

X1 The regression equation is CTQ-1 = - 6.5 + 0.666 X1 Value Predictor Coef SE Coef T P Constant -6.55 20.30 -0.32 0.749 X1 Value 0.6662 0.1991 3.35 0.002 S = 22.5227 R-Sq = 27.2% R-Sq(adj) = 24.8% Analysis of Variance Source DF SS MS F P Regression 1 5681.8 5681.8 11.20 0.002 Residual Error 30 15218.2 507.3 Total 31 20899.9

1201101009080

120

100

80

60

40

20

X1 Value

CTQ

-1

Scatterplot of CTQ-1 vs X1 Value

X2 The regression equation is CTQ-1 = 127 - 2.03 X2 Value Predictor Coef SE Coef T P Constant 127.11 50.05 2.54 0.017 X2 Value -2.031 1.510 -1.34 0.189 S = 25.6331 R-Sq = 5.7% R-Sq(adj) = 2.5% Analysis of Variance Source DF SS MS F P Regression 1 1188.3 1188.3 1.81 0.189 Residual Error 30 19711.7 657.1 Total 31 20899.9

36353433323130

120

100

80

60

40

20

X2 Value

CTQ

-1






X3 The regression equation is CTQ-1 = 10.5 + 0.165 X3 Value Predictor Coef SE Coef T P Constant 10.48 69.55 0.15 0.881 X3 Value 0.1653 0.2313 0.71 0.480 S = 26.1726 R-Sq = 1.7% R-Sq(adj) = 0.0% Analysis of Variance Source DF SS MS F P Regression 1 349.8 349.8 0.51 0.480 Residual Error 30 20550.1 685.0 Total 31 20899.9

320310300290280

120

100

80

60

40

20

X3 Value

CTQ

-1


X4 The regression equation is CTQ-1 = - 62.2 + 1.16 X4 Value Predictor Coef SE Coef T P Constant -62.25 95.51 -0.65 0.520 X4 Value 1.1650 0.9086 1.28 0.210 S = 25.6996 R-Sq = 5.2% R-Sq(adj) = 2.0% Analysis of Variance Source DF SS MS F P Regression 1 1085.8 1085.8 1.64 0.210 Residual Error 30 19814.2 660.5 Total 31 20899.9

110108106104102100

120

100

80

60

40

20

X4 Value

CTQ

-1


X5 The regression equation is CTQ-1 = - 394 + 378 X5 Value Predictor Coef SE Coef T P Constant -393.82 75.38 -5.22 0.000 X5 Value 378.25 62.76 6.03 0.000 S = 17.7521 R-Sq = 54.8% R-Sq(adj) = 53.3% Analysis of Variance Source DF SS MS F P Regression 1 11446 11446 36.32 0.000 Residual Error 30 9454 315 Total 31 20900

1.2501.2251.2001.1751.150

120

100

80

60

40

20

X5 Value

CTQ

-1




Improve Task 10





MidCo commented that one of their engineers had performed a regression analysis on CTQ‐1 with X5

Value as the prediction variable and printed a graph called a four in one plot and requested that it be

interpreted for them.

50250-25-50

99

90

50

10

1

Residual

Per

cent

8070605040

40

20

0

-20

-40

Fitted Value

Res

idua

l

403020100-10-20-30

8

6

4

2

0

Residual

Freq

uenc

y

3230282624222018161412108642

40

20

0

-20

-40

Observation Order

Res

idua

l

Normal Probability Plot Versus Fits

Histogram Versus Order

Residual Plots for CTQ-1

The four in one plot is a plot of residuals (errors; what is left over after the prediction model is applied).

It is called a four in one plot because it has four panels each looking at the residuals from a different

perspective.

o The upper left panel is a normal probability plot of the residuals. If the residuals are normal then

this is a validation that the data itself is normal.

o The lower left panel is a histogram of the residuals. Again we are looking to see if the residuals

appear to be normally distributed. These residuals appear to be normal.

o The upper right panel shows the plot of Residuals Versus the Fitted Values (the fitted values are

the group means) which show whether the variation is same in each group. In this case small

fitted values clearly vary less than large values.



Improve Task 10



Improve Task 10





o The bottom right chart is a run chart in run observation order. This chart is used to see if there is

any trending in the residuals.

Comparison before and after control

As discovered during the analyze phase, several processes were out of control. We consulted with MidCo on

how they could bring these process under control prior to proceeding to the improve stage. MidCo did not

provide any detail with regard to the process improvement after bringing the processes in control. We ask that

they supply this additional data and they did as requested. (Note: The CTQ‐1 Simulator was used to generate

125 observations, subgroup of 5, with all processes set at their default mean values and this information was

used to evaluate process improvement as provided in this section.)

Before After


X1 9.70 9.71 Yes 6.72 6.69 Yes

X2 .95 1.42 No 0.71 0.74 Yes

X3 6.29 6.99 Marginal 4.79 4.58 Yes

X4 0.934 2.99 No 2.19 2.05 Yes

X5 0.006 0.021 No 0.014 0.013 Yes

CTQ-1 9.88 15.84 No 9.29 8.98 Yes

CTQ-2 55.89 62.82 Marginal Unavailable

All CTP processes now appear to be under control and the variation has improved on all process since the

original data was collected in the analyze phase.


What is the regression equation? What are the p‐values? Is the

regression model a good fit? Do the errors appear to be normal?





Optimizing the CTP Adjustments

The recommendations from above are repeated:

Summary of Recommendations based upon Fractional Factorial

CTP Comment

X1 Adjust to high level of 120.

X2 No adjustment recommended

X3 No Adjustment recommended

X4 Adjust to target of 105

X5 Adjust to upper level of 1.3

These recommendations were based strictly upon adjusting the mean values. Adjusting strictly on the means of

the CTP’s should bring the mean of the CTQ closer to its desired position but it may also result in an increase in

variation of the CTQ which could result in an actual decrease in performance. Since it is recommended that X2

and X3 not be adjusted, these two variables will be collapsed in the full factorial design, which will allow us to

calculate the effect on both average and variance.





CTQ‐1

CTQ-1

X1 Value X4 Value X5 Value X3-H X3-L X2-H X2-L Avg Var

80.0 100.0 1.2 26.0 23.2 32.1 28.6 27.5 14.4

120.0 100.0 1.2 41.5 37.0 51.2 45.6 43.8 36.5

80.0 110.0 1.2 36.7 33.9 43.0 39.5 38.3 15.1

120.0 110.0 1.2 52.6 48.1 62.6 57.0 55.1 38.4

80.0 100.0 1.3 51.6 45.5 64.8 57.2 54.8 67.5

120.0 100.0 1.3 85.6 75.6 107.6 94.9 90.9 185.7

80.0 110.0 1.3 63.1 57.0 76.7 69.1 66.5 70.9

120.0 110.0 1.3 98.2 88.1 120.8 108.0 103.8 194.8





CTP X1 X4 X5

SS% Avg 29.8% 5.7% 60.2%

Mea

n of

Avg

12080

80

70

60

50

40110100

1.31.2

80

70

60

50

40

X1 Value X4 Value

X5 Value

Main Effects Plot (data means) for Avg

SS% Var 28.2% 0.08% 58.5%

Mea

n of

Var

12080

120

90

60

30

110100

1.31.2

120

90

60

30

X1 Value X4 Value

X5 Value

Main Effects Plot (data means) for Var





We need to move the aim of CTQ‐1 to the right. We see from the inner array/outer array analysis that if we

increase X1 and X5 that the mean of CTQ‐1 will also be increased; however, the variance in CTQ‐1 will also

increase. This is not a desired effect. If we increase the mean of X4, the mean of CTQ‐1 will increase without any

significant impact on the variance of CTQ‐1. This is a positive effect; however, X4 is a small percentage of the

overall Sum of Squares so a significant increase in X4 will be needed.

CTQ‐2

CTQ-2

X1 Value X4 Value X5 Value X3-H X3-L X2-H X2-L Avg Var

80.0 100.0 1.2 861 704 717 587 717.3 12601.6

120.0 100.0 1.2 574 470 478 391 478.3 5616.3

80.0 110.0 1.2 822 665 685 554 681.5 12093.7

120.0 110.0 1.2 548 443 457 370 454.5 5340.3

80.0 100.0 1.3 792 648 660 540 660.0 10656.0

120.0 100.0 1.3 528 432 440 360 440.0 4736.0

80.0 110.0 1.3 756 612 630 510 627.0 10188.0

120.0 110.0 1.3 504 408 420 340 418.0 4528.0





CTP X1 X4 X5

SS% Avg 94.1% 1.5% 4.1%

Mea

n of

Avg

12080

650

600

550

500

450

110100

1.31.2

650

600

550

500

450

X1 Value X4 Value

X5 Value

Main Effects Plot (data means) for Avg

SS% Var 94.4% 0.3% 4.5%

Mea

n of

Var

12080

12000

10500

9000

7500

6000

110100

1.31.2

12000

10500

9000

7500

6000

X1 Value X4 Value

X5 Value

Main Effects Plot (data means) for Var





We need to move the aim of CTQ‐2 to the left. We see from the inner array/outer array analysis that if we

decrease X1 and X5 that the mean of CTQ‐1 will also be decreased; however, the variance in CTQ‐2 will increase.

This is not a desired effect. If we decrease the mean of X4, the mean of CTQ‐2 will decrease without any

significant impact on the variance of CTQ‐1. This is a positive effect; however, X4 is a small percentage of the

overall Sum of Squares so a significant decrease in X4 will be needed.

Optimization Summary

Improve CTQ CTP Adjustment

X1 X4 X5

CTQ-1 Avg Up – 29.8% Up Up – 60.2%

Var Down – 28.2% --- Down – 58.3%

CTQ-2 Avg Up – 94.1% Up Up - 4.1%

Var Up – 94.4% --- Up – 4.5%

We would generally prefer to first reduce variance in the CTQ’s. X1 and X4 when adjusted impact variance.

However, if we move the values of X1 and X5 down, we will improve the variance of CTQ‐1 but worsen the

variance of CTQ‐2. Conversely if we move the values of X1 and X5 up we will improve the variance of CTQ‐2 but

worsen the variance of CTQ‐1.





X1 X2 X3 X4 X5

Current Mean 100 33 300 100 1.25

LSL 90 31 N/A 100 1.15

Target N/A 33 N/A 105 1.20

USL N/A 35 310 110 1.25

Low Level 80 30 280 100 1.20

High Level 120 36 320 100 1.30

Recommendation Control No

Adjustment

Control No

Adjustment

Center Aim

Adjust to

105

Center Aim

Adjust to

1.20

Based upon the original data we need to move the mean of CTQ‐1 upward 0.67 standard deviations and the

mean of CTQ‐2 downward 0.62 standard deviations.

X5 is currently set at the upper specification limit. MidCo had advised that the specification for X5 was

questionable but we have been unable to validate that assumption.

Assuming that the specification for X5 is correct we should move X5 from 1.25 to 1.20 so that it is aiming

toward its target.

o Improves the variance of CTQ‐1 substantially.

o Deteriorate the centering of CTQ‐1 substantially.

In the final analysis it is hoped that the variance improvement will overwhelm the

centering issues and the Zbenchmark will improve.

o Improve the variance of CTQ‐2 slightly.

o Improve the centering of CTQ‐1 slightly.





We need to do a single factorial experiment at several levels on X1 while holding X2, X3, X4 a their

current setting and X5 at the recommended on target setting. This will allow us to determine the

optimal setting for X1 as we know raising the setting of X1 will have the following effects and lowering

X1 will have the converse effects.

o Lowers variance of CTQ‐1 somewhat.

o Improves the centering of CTQ‐1 somewhat.



X1 only has a lower bound specification so raising the level should not be detrimental

and we improve all but the variance of CTQ‐1.

X4 should be moved up to center the aim on the target.

o Improves centering of CTQ‐1 and CTQ‐2

o Has negligible effect on variance of CTQ‐1 and CTQ‐2

This variable has leverage over centering. However, there is little room for movement

without putting X4, itself, outside of specification.

MidCo advised that they were more concerned with CTQ‐1 than CTQ‐2 and made the decision to go with the

following settings:

X1 X2 X3 X4 X5

Pre-improve settings 100 33 300 100 1.25


Post-improve settings 80 33 300 145 1.15


This information was taken from the control phase data.





It appears as MidCo made the decision to ignore the impact on CTQ‐2 that will result by changing the CTP’s to

optimize CTQ‐1.

They reduced X1 and X5 which should substantially reduce the variation on CTQ‐1.

o X1 is below the specification and X5 is set to aim at the lower specification.

o The centering of CTQ‐1 should be improved but the variance of CTQ will be substantially

deteriorated.

They increased X4 which should move CTQ‐1 upward and closer to its target.

o X4 is substantially outside the specification limits.

o This will also improve the centering of CTQ‐2

MidCo has not confirmed their rationale for stepping outside the specifications on variable X1 and X4. We can

only assume that they have re‐evaluated these specifications and are in the processing of changing them. With

the adjustments, only the variance of CTQ‐2 will be negatively impacted.





Control

What level of process control can be sustained over time?

Conclusions

As stated in the improve phase, MidCo elected to reduce the aim of X1 below the

specification. MidCo indicated that they were going to revise the specification limit of

X1 but have not done that as yet. We recommend that they revise the unilateral

specification so that the lower specification limit is 60 for a 3.14 sigma (840 ppm

defective) or 40 for a six sigma process.

X2 is properly centered but only has a 2.30 sigma value (10692 ppm defective). The

overall capability is statistically the same as the within capability so there is little if any

improvement available in the process. In order to improve X2, we will have to have a

breakthrough in the process which will likely require improved technology.

X3 is operating at 2.55 sigma (5391 ppm defective). The overall capability is statistically

the same as the within capability so like X2 in order to improve X3, a process

breakthrough likely requiring new technology will be required.

X4 like X1 has specification issues. MidCo decided to leverage X4 by moving the process

upward significantly in order to improve the variance of CTQ‐1. MidCo advised that

they were going to change the specifications but have not done that yet. If MidCo

moves the target of X4 to 145 and maintains the tolerance of +/‐5, the process will still

only be at 1.86 sigma (31170 ppm defective). The overall capability of X4 is statistically

the same as within capability so a breakthrough in the process, likely requiring

technology improvement, will be required to improve the process.

X5 also has some serious centering issues. MidCo advised early on that they were

suspect of the specification limits on X5 but have yet to make any adjustments. If the

specification is targeted at 1.15 and the same tolerance of +/‐ 0.05 is maintained, the

process would 3.02 sigma capable with (1281 ppm defective).

All CTP’s are under statistical control.


A MindPro ™ Lean Six Sigma Simulated Training Project Control



CTQ‐1 is essentially at six sigma capability (zero ppm defective) and is in control.

No information has been provided by MidCo as of yet on CTQ‐2 so we cannot analyze

CTQ‐2.


Unit CTQ-1 Critical-to-Process Variables

X1 X2 X3 X4 X5

USL 115.0 NA 35.0 310.0 110.0 1.25

Sampling Demographics

T 80.0 NA 33.0 NA 105.0 1.20

LSL 45.0 90.0 31.0 NA 100.0 1.15

Case CTQ-1 Value

X1 Value X2 Value X3 Value X4 Value X5 Value Subgroup Sequence

1 83.0 92.1 32.5 303.9 143.84 1.158 1 1

. 76.4 75.5 32.2 288.1 144.15 1.167 1 2

. 71.5 75.9 33.3 292.7 142.39 1.156 1 3

. 72.3 85.8 33.4 294.7 142.39 1.139 1 4

125 75.1 77.9 33.7 297.0 145.96 1.151 1 5

Upon receiving the data from MidCo, it was noticed that the last five subgroups were labeled 1, 2, 3, 4, and 5. It

seemed odd that these subgroup labels would be repeated. MidCo was contacted and they confirmed that the

subgroups were mis‐label and should be subgroups 21, 22, 23, 24, and 25 in order.





Analysis for Control

One of the first steps in the control phase is to analyze the data to ensure that it is in statistical control. Note: No

information has been provided by MidCo as of yet on CTQ‐2 so we cannot analyze CTQ‐2 and the variables as

they relate to CTQ‐2.

Variable X‐bar / S Charts Comments

CTQ‐1

Value

252321191715131197531

85

80

75

70

Sample

Sa

mp

le M

ea

n

__X=77.22

UC L=84.86

LC L=69.58

252321191715131197531

10.0

7.5

5.0

2.5

0.0

Sample

Sa

mp

le S

tDe

v

_S=5.35

UC L=11.18

LC L=0

Xbar-S Chart of CTQ-1 Value

In statistical control. No apparent

assignable causes.

X1

Value

252321191715131197531

90

85

80

75

70

Sample

Sa

mp

le M

ea

n

__X=80.39

UC L=89.10

LC L=71.68

252321191715131197531

12

9

6

3

0

Sample

Sa

mp

le S

tDe

v

_S=6.10

UC L=12.75

LC L=0



assignable causes.


Black Belt and Green Belt ‐Industrial


Control Task 1

Control Task 2





X2

Value

252321191715131197531

34.0

33.5

33.0

32.5

32.0

Sample

Sa

mp

le M

ea

n__X=32.986

UC L=34.044

LC L=31.928

252321191715131197531

1.6

1.2

0.8

0.4

0.0

Sample

Sa

mp

le S

tDe

v

_S=0.741

UC L=1.548

LC L=0



assignable causes.

X3

Value

252321191715131197531

305.0

302.5

300.0

297.5

295.0

Sample

Sa

mp

le M

ea

n

__X=299.12

UC L=304.85

LC L=293.40

252321191715131197531

8

6

4

2

0

Sample

Sa

mp

le S

tDe

v

_S=4.009

UC L=8.374

LC L=0



assignable causes.

X4

Value

252321191715131197531

148

146

144

142

Sample

Sa

mp

le M

ea

n

__X=144.416

UC L=147.431

LC L=141.402

252321191715131197531

4

3

2

1

0

Sample

Sa

mp

le S

tDe

v

_S=2.112

UC L=4.412

LC L=0



assignable causes.





X5

Value

252321191715131197531

1.18

1.17

1.16

1.15

1.14

Sample

Sa

mp

le M

ea

n__X=1.15789

UC L=1.17664

LC L=1.13914

252321191715131197531

0.03

0.02

0.01

0.00

Sample

Sa

mp

le S

tDe

v

_S=0.01314

UC L=0.02744

LC L=0



assignable causes.

Discussions with MidCo regarding the labeling of subgroups resulted in the decision that the subgroups should

be labeled 21, 22, 23, 24, and 25. However, a comment was made by one engineer that she was not sure that

data was in fact subgroups and they might have just been individual measurement. It was decided that the data

would be analyzed for control based upon individual measurements by looking at the I and MR charts.

CTQ‐1

Value

12110997857361493725131

90

80

70

60

Observation

Ind

ivid

ua

l V

alu

e

_X=77.22

UC L=94.55

LC L=59.89

12110997857361493725131

24

18

12

6

0

Observation

Mo

vin

g R

an

ge

__MR=6.52

UC L=21.29

LC L=0

11

I-MR Chart of CTQ-1 Value

MR Chart: TEST 1. One point more than 3.00 standard deviations from center line. Test Failed at points: 111, 112


Are the sampling means in a state of control? Are the standard

deviation of the samples in a state of control? Does the control

chart reveal failure of any tests that might indicate assignable

causes?





X1

Value

12110997857361493725131

100

90

80

70

60

Observation

Ind

ivid

ua

l V

alu

e_X=80.39

UC L=100.08

LC L=60.70

12110997857361493725131

30

20

10

0

Observation

Mo

vin

g R

an

ge

__MR=7.40

UC L=24.19

LC L=0

1

I-MR Chart of X1 Value

I Chart: In statistical control.

MR Chart: TEST 1. One point more

than 3.00 standard deviations from

center line.

Test Failed at points: 124

X2

Value

12110997857361493725131

36.0

34.5

33.0

31.5

30.0

Observation

Ind

ivid

ua

l V

alu

e

_X=32.986

UC L=35.548

LC L=30.424

12110997857361493725131

3

2

1

0

Observation

Mo

vin

g R

an

ge

__MR=0.963

UC L=3.147

LC L=0



assignable causes.

X3

Value

12110997857361493725131

310

300

290

Observation

Ind

ivid

ua

l V

alu

e

_X=299.12

UC L=312.06

LC L=286.19

12110997857361493725131

16

12

8

4

0

Observation

Mo

vin

g R

an

ge

__MR=4.86

UC L=15.89

LC L=0



assignable causes.





X4

Value

12110997857361493725131

152

148

144

140

Observation

Ind

ivid

ua

l V

alu

e_X=144.42

UC L=150.94

LC L=137.89

12110997857361493725131

8

6

4

2

0

Observation

Mo

vin

g R

an

ge

__MR=2.453

UC L=8.014

LC L=0

1

1


I Chart: TEST 1. One point more


center line.




center line.


X5

Value

12110997857361493725131

1.20

1.18

1.16

1.14

1.12

Observation

Ind

ivid

ua

l V

alu

e

_X=1.15789

UC L=1.20072

LC L=1.11506

12110997857361493725131

0.060

0.045

0.030

0.015

0.000

Observation

Mo

vin

g R

an

ge

__MR=0.01610

UC L=0.05261

LC L=0

1


I Chart: In statistical control.



center line.


Remember 1 out of every 317 points is likely to fall outside of the three sigma control limits so the cases where

one point is outside the limits should send off a flag of alert but not necessarily an alarm.


Black Belt, Green Belt – Industrial and Green Belt ‐ Commercial


Control Task 4


Are the individual means in a state of control? Are the moving

ranges in a state of control? Does the control chart reveal failure

of any tests that might indicate assignable causes? Are there any

signs of patterned data? Are there any conditions that signal a

need to investigate possible assignable causes?





To further validate that there is no concern with regarding patterns, autocorrelation charts were developed.

CTQ‐1

Value

30282624222018161412108642

1.0

0.8

0.6

0.4

0.2

0.0

-0.2

-0.4

-0.6

-0.8

-1.0

Lag

Aut

ocor

rela

tion


No apparent pattern.

X1

Value

30282624222018161412108642

1.0

0.8

0.6

0.4

0.2

0.0

-0.2

-0.4

-0.6

-0.8

-1.0

Lag

Aut

ocor

rela

tion

Autocorrelation Function for X1 Value(with 5% significance limits for the autocorrelations)






X2

Value

30282624222018161412108642

1.0

0.8

0.6

0.4

0.2

0.0

-0.2

-0.4

-0.6

-0.8

-1.0

Lag

Aut

ocor

rela

tion



X3

Value

30282624222018161412108642

1.0

0.8

0.6

0.4

0.2

0.0

-0.2

-0.4

-0.6

-0.8

-1.0

Lag

Aut

ocor

rela

tion


No apparent pattern

X4

Value

30282624222018161412108642

1.0

0.8

0.6

0.4

0.2

0.0

-0.2

-0.4

-0.6

-0.8

-1.0

Lag

Aut

ocor

rela

tion


No apparent pattern





X5

Value

30282624222018161412108642

1.0

0.8

0.6

0.4

0.2

0.0

-0.2

-0.4

-0.6

-0.8

-1.0

Lag

Aut

ocor

rela

tion


No apparent pattern


Black Belt and Green Belt ‐Industrial


Control Task 5


Are there any lags that are beyond the statistical control limits?

Is there any indication that there is a patterned behavior over

time?





Analysis of Capability

Having analyzed the data for control, the data was then analyzed for capability.

CTQ‐1

1101009080706050

LSL USL

LSL 45Target *USL 115Sample Mean 77.2192Sample N 125StDev (Within) 5.69228StDev (O v erall) 5.57867

Process Data

Z.Bench 5.66Z.LSL 5.66Z.USL 6.64C pk 1.89

Z.Bench 5.78Z.LSL 5.78Z.USL 6.77Ppk 1.93C pm *



PPM < LSL 0.00PPM > USL 0.00PPM Total 0.00



Exp. O v erall Performance

WithinOverall


As a reminder Z.Bench is equal to Cp*3. Since we have elected to show Z.Bench values instead of Cp values, we

can calculate Cp by dividing Z.Bench by 3.




Control Task 3


What is the sustainable (overall) capability? What index provides

this information? What is the instantaneous (potential

capability? What index provides this information? What is the Z‐

Bench capability?





We can see from the Process Capability Analysis on CTQ‐1 that the within standard deviation is 5.69 and he

overall standard deviation is 5.57. We can take a closer look at this by developing a multi‐vari chart.

2019181716151413121110 9 8 7 6 5 4 3 2 1

90

85

80

75

70

Subgroup

CTQ

-1 V

alue

12345

Sequence

Multi-Vari Chart for CTQ-1 Value by Sequence - Subgroup

To further evaluate the overall variation an ANOVA was performed on CTQ‐1 looking at subgroups as a factor.

One-way ANOVA: CTQ-1 Value versus Subgroup Source DF SS MS F P Subgroup 24 635.0 26.5 0.82 0.703 Error 100 3224.0 32.2 Total 124 3859.1 S = 5.678 R-Sq = 16.46% R-Sq(adj) = 0.00%




Control Task 7


How does the within sample variation compare to he between

sample variation?




Control Task 9





X1

95908580757065

LSLProcess Data


LSL 90.00000Target *USL *Sample Mean 80.39120


C C pk -0.49


Pp *PPL -0.50PPU *Ppk

C p

-0.50C pm *

*C PL -0.49C PU *C pk -0.49

O bserv ed PerformancePPM < LSL 920000.00PPM > USL *PPM Total 920000.00

Exp. Within PerformancePPM < LSL 930620.19PPM > USL *PPM Total 930620.19

Exp. O v erall PerformancePPM < LSL 933650.91PPM > USL *PPM Total 933650.91

WithinOverall

Process Capability of X1 Value

As stated in the improve phase, MidCo elected to reduce the aim of X1 below the specification. MidCo indicated

that they were going to revise the specification limit of X1 but have not done that as yet. We recommend that

they revise the unilateral specification so that the lower specification limit is 60 for a 3.14 sigma (840 ppm

defective) or 40 for a six sigma process.


What is the confidence level that subgroup to subgroup

variation is statistically significant? What is the null and

alternate hypotheses? Is there sufficient evidence to accept Ha?

What percent of the total variation is explained by subgroup to

subgroup variation at the current p‐value level?





X2

34.834.233.633.032.431.831.2

LSL USLProcess Data




C C pk 0.85


Z.Bench 2.37Z.LSL 2.60Z.USL 2.64Ppk

Z.Bench

0.87C pm *

2.30Z.LSL 2.53Z.USL 2.57C pk 0.84




WithinOverall


X2 is properly centered but only has a 2.30 sigma value (10692 ppm defective). The overall capability is statistically the same as the within capability so there is little if any improvement available in the process. In order to improve X2, we will have to have a breakthrough in the process which will likely require improved technology.





X3

308304300296292288

USLProcess Data


LSL *Target *USL 310.00000Sample Mean 299.12640


C C pk 0.85


Z.Bench 2.55Z.LSL *Z.USL 2.55Ppk

Z.Bench

0.85C pm *

2.55Z.LSL *Z.USL 2.55C pk 0.85

O bserv ed PerformancePPM < LSL *PPM > USL 0.00PPM Total 0.00

Exp. Within PerformancePPM < LSL *PPM > USL 5391.13PPM Total 5391.13

Exp. O v erall PerformancePPM < LSL *PPM > USL 5357.06PPM Total 5357.06

WithinOverall


X3 is operating at 2.55 sigma (5391 ppm defective). The overall capability is statistically the same as the within

capability so like X2 in order to improve X3, a process breakthrough likely requiring new technology will be

required.





X4

147140133126119112105

LSL USLProcess Data




C C pk 0.74


Z.Bench *Z.LSL 20.44Z.USL -15.84Ppk

Z.Bench

-5.28C pm *

*Z.LSL 19.77Z.USL -15.32C pk -5.11




WithinOverall


X4 like X1 has specification issues. MidCo decided to leverage X4 by moving the process upward significantly in

order to improve the variance of CTQ‐1. MidCo advised that they were going to change the specifications but

have not done that yet. If MidCo moves the target of X4 to 145 and maintains the tolerance of +/‐5, the process

will still only be at 1.86 sigma (31170 ppm defective). The overall capability of X4 is statistically the same as

within capability so a breakthrough in the process, likely requiring technology improvement, will be required to

improve the process.





X5

1.241.221.201.181.161.14

LSL USLProcess Data




C C pk 1.19


Z.Bench 0.57Z.LSL 0.57Z.USL 6.72Ppk

Z.Bench

0.19C pm *

0.56Z.LSL 0.56Z.USL 6.60C pk 0.19




WithinOverall


X5 also has some serious centering issues. MidCo advised early on that they were suspect of the specification

limits on X5 but have yet to make any adjustments. If the specification is targeted at 1.15 and the same

tolerance of +/‐ 0.05 is maintained, the process would 3.02 sigma capable with (1281 ppm defective).





Correlation Analysis

In order to determine the relationship between CTQ‐1 Value and the CTP’s (X1through X5 Vlaue) scatterplots

were developed with a regression line to get a visual look at correlation and a regression analysis was performed

to determine statistical and pragmatic significance.

Visual Analysis

Variable Scatterplot Comments

X1

Value

95908580757065

90

85

80

75

70

65

60

X1 Value

CTQ

-1 V

alue

Scatterplot of CTQ-1 Value vs X1 ValueThere appears to be a moderate

positive correlation between

CTQ‐1 Value and X1 Value.

X2

Value

3534333231

90

85

80

75

70

65

60

X2 Value

CTQ

-1 V

alue

Scatterplot of CTQ-1 Value vs X2 Value

There appears to be a mild

negative correlation between






X3

Value

310305300295290

90

85

80

75

70

65

60

X3 Value

CTQ

-1 V

alue


There appears to be a mild



X4

Value

152150148146144142140

90

85

80

75

70

65

60

X4 Value

CTQ

-1 V

alue


There appears to be a moderate



X5

Value

1.201.191.181.171.161.151.141.131.12

90

85

80

75

70

65

60

X5 Value

CTQ

-1 V

alue


There appears to be a moderate






Control Task 6





Statistical Analysis

Variable Regression Analysis Comments

X1

Value

The regression equation is CTQ-1 Value = 36.6 + 0.506 X1 Value Predictor Coef SE Coef T P Constant 36.551 5.188 7.05 0.000 X1 Value 0.50589 0.06433 7.86 0.000 S = 4.56922 R-Sq = 33.5% R-Sq(adj) = 32.9% Analysis of Variance Source DF SS MS F P Regression 1 1291.1 1291.1 61.84 0.000 Residual Error 123 2568.0 20.9 Total 124 3859.1

Regression is statistically significant and pragmatically significant.

X2

Value

The regression equation is CTQ-1 Value = 151 - 2.23 X2 Value Predictor Coef SE Coef T P Constant 150.65 20.63 7.30 0.000 X2 Value -2.2261 0.6253 -3.56 0.001 S = 5.33327 R-Sq = 9.3% R-Sq(adj) = 8.6% Analysis of Variance Source DF SS MS F P Regression 1 360.48 360.48 12.67 0.001 Residual Error 123 3498.59 28.44 Total 124 3859.07

Regression is statistically

significant but pragmatically

insignificant.


Is there a linear correlation? Is it positive or negative? Is it mild,

moderate, or substantial?

Candidate Analysis Required for Black Belts


Control Task 11





X3

Value

The regression equation is CTQ-1 Value = - 10.2 + 0.292 X3 Value Predictor Coef SE Coef T P Constant -10.22 34.51 -0.30 0.768 X3 Value 0.2923 0.1154 2.53 0.013 S = 5.46058 R-Sq = 5.0% R-Sq(adj) = 4.2% Analysis of Variance Source DF SS MS F P Regression 1 191.47 191.47 6.42 0.013 Residual Error 123 3667.61 29.82 Total 124 3859.07


significant but pragmatically

insignificant.

X4

Value

The regression equation is CTQ-1 Value = - 127 + 1.41 X4 Value Predictor Coef SE Coef T P Constant -127.03 27.99 -4.54 0.000 X4 Value 1.4143 0.1938 7.30 0.000 S = 4.67902 R-Sq = 30.2% R-Sq(adj) = 29.7% Analysis of Variance Source DF SS MS F P Regression 1 1166.2 1166.2 53.27 0.000 Residual Error 123 2692.9 21.9 Total 124 3859.


significant and pragmatically

significant.


Is the regression equation statistically significant? Is the

constant statistically significant? Is the slope statistically

significant? What is the regression equation? What percentage

of total variation is explained by the regression equation? Is the

regression equation pragmatically significant? What are the vital

few? What are the trivial many?





X5

Value

The regression equation is CTQ-1 Value = - 259 + 291 X5 Value Predictor Coef SE Coef T P Constant -259.24 29.73 -8.72 0.000 X5 Value 290.58 25.68 11.32 0.000 S = 3.92039 R-Sq = 51.0% R-Sq(adj) = 50.6% Analysis of Variance Source DF SS MS F P Regression 1 1968.6 1968.6 128.09 0.000 Residual Error 123 1890.4 15.4 Total 124 3859.1


significant and pragmatically

significant.

Normality Test

It was also decided to develop a normal probability plot and to test CTQ‐1 and X5 Value for normality using the

Anderson‐Darling Test.

9590858075706560

99.9

99

9590

80706050403020

10

5

1

0.1

CTQ-1 Value

Perc

ent


Probability Plot of CTQ-1 ValueNormal

1.201.191.181.171.161.151.141.131.121.11

99.9

99

9590

80706050403020

10

5

1

0.1

X5 Value

Perc

ent


Probability Plot of X5 ValueNormal





Control Task 8





Control Task 8


Do the normal probability plots pass the fat pencil test? Is there

divergence in the tails and if so how does it impact the

conclusion regarding normality? Does the Anderson‐Darling test

indicate normality?





Descriptive Statistics

MidCo ask that we provide descriptive statics for CTQ‐1 and all CTP’s: Mean, Median, Standard Deviation.

Variable Mean SE Mean StDev Variance CoefVar Median Range CTQ-1 Value 77.219 0.499 5.579 31.122 7.22 77.000 28.200 X1 Value 80.389 0.571 6.378 40.684 7.93 80.017 28.630 X2 Value 32.986 0.0685 0.766 0.587 2.32 32.984 3.361 X3 Value 299.12 0.380 4.25 18.07 1.42 299.22 20.24 X4 Value 144.42 0.194 2.17 4.70 1.50 144.28 11.88




Control Task 10


Complete the above table and explain the meaning of each

variable.





Moving Average of CTQ‐1 Value

MidCo made a request that we develop a moving average chart for CTQ‐1 Value with a moving average length of

5. The MA chart is a seldom used time weighted control chart.

252321191715131197531

86

84

82

80

78

76

74

72

70

Sample

Mov

ing

Ave

rage

__X=77.22

UCL=80.63

LCL=73.80

Moving Average Chart of CTQ-1 Value




Control Task 12


Is the moving average of CTQ‐1 in a state of control?





Survey

How do customers review MidCo’s Product?

Conclusions:

Those that have the most need for the product seem to be satisfied with the quality of the

product; but, have a low overall satisfaction with the product. This deserves some investigation.

o The Cost and Delivery tally appears to indicate that these two factors could be

contributors to the overall dissatisfaction but they are not statistically significant.

o A focus group could provide some insight as to other issues that might be degrading

overall satisfaction.

o Once a list of potential factors is developed in the focus group, a new survey should be

done using the demographics of the group with the most need.


o This may indicate that those with less income are using express delivery services that

cannot be afforded by those with less income.

o In the survey that is recommended above using the demographics of those with less

income, questions should be ask regarding preferred deliver methods and cost of

delivery.


MidCo provided customer survey data that was collected prior to the six sigma improvement program. They ask

that we analyze the survey and provide them a summary. This is something that should have been done in the

early stages of the six sigma program but MidCo was unaware of the data until recently and did not want to

commit funds to such a survey in the early stages of the six sigma program. Now that they have seen that results

can be achieved with six sigma they are interested in understanding more about the survey in their process of

deciding if they want to do a similar survey post six sigma.


A MindPro ™ Lean Six Sigma Simulated Training Project Survey



MidCo provided the following survey information:

Case Demographics Questions

Need Location Income Satisfied ? Quality ? Cost ? Delivery ?

1 C B A A B B A

. A B B B A C C

. B B B C B A A

. A A A C B C C

200 B B B C C B B

The questions were:

1. To what extent is your overall satisfaction of the product that MidCo you purchased?

A. High

B. Medium

C. Low

2. To what extent are you satisfied with the quality of the product that MidCo you purchased?

A. Satisfied

B. Somewhat Satisfied

C. Not Satisfied

3. To what extent were you satisfied with the price of the product MidCo your purchased?

A. Satisfied


C. Not Satisfied





4. To what extent were you satisfied with the delivery of the MidCo product you purchased?

A. Satisfied


C. Not Satisfied

Satisfaction

MidCo was most concerned about the overall satisfaction of their products, so the first step was to do a simple

tally of the “Satisfied? column” from the customer survey:

Satisfied ? Count Percent A 50 25.00 B 89 44.50 C 61 30.50 N= 200


What are the counts and percentages for each level of satisfaction?




Survey Task 1





A cross tabulation was performed to determine the count and percentage of the 50 survey participants that

were satisfied in terms of Income and Location.

Results for Satisfied ? = A Rows: Income Columns: Location A B C All A 3 11 4 18 16.67 61.11 22.22 100.00 17.65 50.00 36.36 36.00 6 22 8 36 B 10 10 5 25 40.00 40.00 20.00 100.00 58.82 45.45 45.45 50.00 20 20 10 50 C 4 1 2 7 57.14 14.29 28.57 100.00 23.53 4.55 18.18 14.00 8 2 4 14 All 17 22 11 50 34.00 44.00 22.00 100.00 100.00 100.00 100.00 100.00 34 44 22 100 Cell Contents: Count % of Row % of Column % of Total


What are the count and percentage cross tabulations for location

and income with respect to Satisfaction Level A ?


Black Belt and



Survey Task 3





MidCo ask that we contrast the survey participants that were most satisfied with those that were least satisfied

with respect to income and their perception of the quality of the product.

Results for Satisfied ? = A Rows: Location Columns: Quality ? A B C All A 0 7 10 17 0.00 41.18 58.82 100.00 * 33.33 34.48 34.00 0 14 20 34 B 0 10 12 22 0.00 45.45 54.55 100.00 * 47.62 41.38 44.00 0 20 24 44 C 0 4 7 11 0.00 36.36 63.64 100.00 * 19.05 24.14 22.00 0 8 14 22 All 0 21 29 50 0.00 42.00 58.00 100.00 * 100.00 100.00 100.00 0 42 58 100 Cell Contents: Count % of Row % of Column % of Total

Results for Satisfied ? = C Rows: Location Columns: Quality ? A B C All A 0 11 7 18 0.00 61.11 38.89 100.00 * 35.48 23.33 29.51 0.00 18.03 11.48 29.51 B 0 11 13 24 0.00 45.83 54.17 100.00 * 35.48 43.33 39.34 0.00 18.03 21.31 39.34 C 0 9 10 19 0.00 47.37 52.63 100.00 * 29.03 33.33 31.15 0.00 14.75 16.39 31.15 All 0 31 30 61 0.00 50.82 49.18 100.00 * 100.00 100.00 100.00 0.00 50.82 49.18 100.00 Cell Contents: Count % of Row % of Column % of Total

It is rather interesting that of the 50 respondents who were overall satisfied (A)., that none were satisfied with

the quality of the product (A).


What are the count and percentage cross tabulations for location

and quality with respect to Satisfaction Levels A and C ?




Survey Task 4





In order to determine if need was a predictor of satisfaction a Cross‐tabulation and Chi‐Square analysis was

performed.

Tabulated statistics: Need, Satisfied ? Rows: Need Columns: Satisfied ? A B C All A 0 39 26 65 0.00 60.00 40.00 100.00 0.00 43.82 42.62 32.50 0.00 19.50 13.00 32.50 16.25 28.93 19.82 65.00 B 0 50 35 85 0.00 58.82 41.18 100.00 0.00 56.18 57.38 42.50 0.00 25.00 17.50 42.50 21.25 37.83 25.93 85.00 C 50 0 0 50 100.00 0.00 0.00 100.00 100.00 0.00 0.00 25.00 25.00 0.00 0.00 25.00 12.50 22.25 15.25 50.00 A All 50 89 61 200 25.00 44.50 30.50 100.00 100.00 100.00 100.00 100.00 25.00 44.50 30.50 100.00 50.00 89.00 61.00 200.00 Cell Contents: Count % of Row % of Column % of Total Expected count Pearson Chi-Square = 200.028, DF = 4, P-Value = 0.000 Likelihood Ratio Chi-Square = 224.955, DF = 4, P-Value = 0.000




Survey Task 5


What are the count and percentage cross tabulations for need with

respect to Satisfaction? Is the Pearson Chi‐Square value and its

respective p‐value? How many degrees of freedom are available

for the Chi‐Square analysis? Is Satisfaction independent of Need?





In order to determine if delivery was a predictor of satisfaction a Cross‐tabulation and Chi‐Square analysis was

performed.

Tabulated statistics: Delivery ?, Satisfied ? Rows: Delivery ? Columns: Satisfied ? A B C All A 17 31 19 67 25.37 46.27 28.36 100.00 34.00 34.83 31.15 33.50 8.50 15.50 9.50 33.50 16.75 29.82 20.43 67.00 B 24 35 28 87 27.59 40.23 32.18 100.00 48.00 39.33 45.90 43.50 12.00 17.50 14.00 43.50 21.75 38.72 26.54 87.00 C 9 23 14 46 19.57 50.00 30.43 100.00 18.00 25.84 22.95 23.00 4.50 11.50 7.00 23.00 11.50 20.47 14.03 46.00 All 50 89 61 200 25.00 44.50 30.50 100.00 100.00 100.00 100.00 100.00 25.00 44.50 30.50 100.00 50.00 89.00 61.00 200.00 Cell Contents: Count % of Row % of Column % of Total Expected count Pearson Chi-Square = 1.678, DF = 4, P-Value = 0.795 Likelihood Ratio Chi-Square = 1.715, DF = 4, P-Value = 0.788




Survey Task 7


What are the count and percentage cross tabulations for delivery

with respect to Satisfaction? Is the Pearson Chi‐Square value and

its respective p‐value? How many degrees of freedom are available






In order to determine if delivery was a predictor of satisfaction a Cross‐tabulation and Chi‐Square analysis was

performed.

Tabulated statistics: Quality ?, Satisfied ? Rows: Quality ? Columns: Satisfied ? A B C All A 0 89 0 89 0.00 100.00 0.00 100.00 0.00 100.00 0.00 44.50 0.00 44.50 0.00 44.50 22.25 39.60 27.15 89.00 B 21 0 31 52 40.38 0.00 59.62 100.00 42.00 0.00 50.82 26.00 10.50 0.00 15.50 26.00 13.00 23.14 15.86 52.00 C 29 0 30 59 49.15 0.00 50.85 100.00 58.00 0.00 49.18 29.50 14.50 0.00 15.00 29.50 14.75 26.25 18.00 59.00 All 50 89 61 200 25.00 44.50 30.50 100.00 100.00 100.00 100.00 100.00 25.00 44.50 30.50 100.00 50.00 89.00 61.00 200.00 Cell Contents: Count % of Row % of Column % of Total Expected count Pearson Chi-Square = 201.547, DF = 4, P-Value = 0.000 Likelihood Ratio Chi-Square = 275.694, DF = 4, P-Value = 0.000




Survey Task 7


What are the count and percentage cross tabulations for quality

with respect to Satisfaction? Is the Pearson Chi‐Square value and

its respective p‐value? How many degrees of freedom are available






Note: As we saw above, even though quality is a predictor of satisfaction, it is not like we might perceive as the

survey seems to indicate the poorer the quality the more the customer is satisfied. This makes the data suspect

or there is something highly unusual with the survey. It is possible that the survey questions were reversed.

Summary Analysis

Minitab was utilized to calculate a cross tab and Chi Square. Green with Bold‐Italics indicates significant (p<0.05)

correlation. Yellow with Bold indicates somewhat significant (p<0.10) correlation. Magenta indicates no

significance.

Satisfied Quality Cost Delivery

A B C A B C A B C A B C

Need

A 0 39 26 39 11 15 21 28 16 21 27 17

B 0 50 35 50 20 15 33 35 17 29 36 20

C 50 0 0 0 21 29 15 23 12 17 24 9

Location

A 17 33 18 33 18 17 25 25 18 28 28 12

B 22 37 24 37 21 25 32 35 16 22 43 18

C 11 19 19 19 13 17 12 26 11 17 16 16

Income

A 18 37 24 37 20 22 25 33 21 26 35 18

B 25 36 28 36 24 29 35 36 18 33 31 25

C 7 16 9 16 8 8 9 17 6 8 21 33




Survey Task 2





If our objective is a high level of satisfaction, we might want to consider combining both the B and C and

analyzing at two levels. We have done that for the insignificant and somewhat significant variables.

Satisfied Quality Cost Delivery

High Low High Low High Low High Low

Need

A 0 65 39 11 21 44 21 44

B 0 85 50 20 33 52 29 56

C 50 0 0 21 15 35 17 33

Location

A 17 51 33 35 25 43 28 44

B 22 61 37 46 32 51 22 61

C 11 38 19 39 12 37 17 32

Income

A 18 61 37 42 25 54 26 53

B 25 64 36 53 35 54 33 56

C 7 25 16 16 9 23 8 24


What are the count and percentage cross tabulations for need with

respect to Satisfaction?





Observations

Strong observations:

o Those that have a low need tend to be satisfied overall.

o Those that have a high need tend to be dissatisfied overall.

o Only ¼ of the respondents are highly satisfied overall.

o Those that have a high or moderate need tend to be satisfied with quality.

o Those that have a low need tend to be dissatisfied with quality.

Weak observations:

o Those that have a low income tend to be less satisfied with delivery.

o Those that have a higher income tend to be more satisfied with delivery.

o Location see has a lower number of respondents than the other regions.

o Region B has more dissatisfaction with delivery than the other regions.





Recommendations:

Those that have the most need for the product seem to be satisfied with the quality of the product; but,

have a low overall satisfaction with the product. This deserves some investigation.

o The Cost and Delivery tally appears to indicate that these two factors could be contributors to

the overall dissatisfaction but they are not statistically significant.

o A focus group could provide some insight as to other issues that might be degrading overall

satisfaction.

o Once a list of potential factors is developed in the focus group, a new survey should be done

using the demographics of the group with the most need.


o This may indicate that those with less income are using express delivery services that cannot be

afforded by those with less income.

o In the survey that is recommended above using the demographics of those with less income,

questions should be ask regarding preferred deliver methods and cost of delivery.






Risk

How do we determine a confidence factor failure cost?

Conclusions:

The risk calculator provided by MidCo either has an error or the risk factor used in the calculator is

misstated. MidCo stated that the risk factor was the risk of an individual failure exceeding its Limiting

Cost. However, the template is designed so that the risk is actually the risk of the sum of Limiting

Costs of all failures of a particular CTQ exceeding the Monthly Limiting Cost.


A MindPro ™ Lean Six Sigma Simulated Training Project Risk



MidCo provided the following template that they were using for a risk calculator and ask that we evaluate it.

A B C D E F G H

1

2

3

Cost per Failure

( $ / F ) Failure Cost / Month

4 CTQ Failures

Nominal

Cost Limiting

Cost Nominal Cost

Limiting Cost

5

6 1 34 $927.31 $1,107.65 $31,529 $37,660

7 2 18 $755.84 $943.37 $13,605 $16,981

8 3 9 $912.26 $1,140.67 $8,210 $10,266

9 4 1 $626.87 $932.05 $627 $932

10 5 3 $653.71 $829.83 $1,961 $2,489

11 6 7 $837.61 $1,025.77 $5,863 $7,180

12 7 8 $676.39 $919.19 $5,411 $7,354

13 8 0 $949.40 $1,286.22 $0 $0

14 9 2 $550.77 $726.21 $1,102 $1,452

15 10 3 $823.81 $1,172.33 $2,471 $3,517

16

17 Pooled $7,713.96 $10,083.28 $70,779 $87,832

18

19 Proposed Limit $75,000

20 Confidence 81.25%





Cell D17 and E17 are sums of the numbers above but since the numbers above are per unit costs the summation has no practical value. These cells should be eliminated.

I J K L M

1

2

3

4 Risk Z.risk

Std DevMonth

Variance

5

6 0.0500 1.645 $3,727.63 13,895,214.3

7 0.0500 1.645 $2,052.15 4,211,336.0

8 0.1000 1.282 $1,604.07 2,573,056.5

9 0.1500 1.036 $294.45 86,700.2

10 0.0500 1.645 $321.22 103,183.0

11 0.0500 1.645 $800.75 641,207.1

12 0.0100 2.326 $834.97 697,171.2

13 0.0500 1.645 $0.00 0.0

14 0.0100 2.326 $150.83 22,750.1

15 0.0500 1.645 $635.66 404,062.9

16

17 0.0002 3.584 $4,757.59 22,634,681.2

18

19

20





MidCo stated that the assumed risk factor (column I) was per failure. However they have pooled as if the risk

factor was the risk of exceeding the limiting cost per month for a particular CTQ. The limiting cost per month

assumes that all individual failures are at the limiting value. This is a grossly conservative assumption. For

example, if there is only a 5% chance of each of the 34 CTQ‐1 failures exceeding the individual limit the risk of all

of the CTQ‐1’s exceeding the overall limit would be 5%^34. This number exceeds the limits of Excel an incorrect

Z.risk value of 8.294, which appears to be the limiting value of the Excel Z table.





The template could be expanded to list each failure on a separate row as follows (Note: Rows 5-84 were omitted from this snap shot).

A B C D E F G H

1

2

Cost per Failure

( $ / F ) Failure Cost / Month

3 CTQ Failures

Nominal

Cost Limiting

Cost Nominal Cost

Limiting Cost

4

85 8 1 0 $949.40 $1,286.22 $0 $0

86 9

1 1 $550.77 $726.21 $551 $726

87 2 1 $550.77 $726.21 $551 $726

88

10

1 1 $823.81 $1,172.33 $824 $1,172

89 2 1 $823.81 $1,172.33 $824 $1,172

90 3 1 $823.81 $1,172.33 $824 $1,172

91

92 Pooled $70,779 $87,832

93

94

Proposed Limit

$75,000

95 Confidence 99.98%





Cell G92 is the Sum of the nominal costs and Cell H92 is the sum of the limiting costs.

Cell H94 is an entry by the user for proposed limit.

Cell H95 is the calculated confidence level that the proposed limit will not be exceeded = =NORMSDIST((H94‐

G93)/L93) = NORMSDIST(( (Proposed Limit) – (Σ Nominal Cost / Month)) / (Pooled Standard Deviation))

Note: The template supplied by MidCo indicated an 81.25% Confidence which is actually the confidence as

opposed to 99.98%.

J K L M

1

2

3 Risk Z.risk

Std DevMonth

Variance

4

85 0.0500 1.645 $0.00 0.0

86 0.0100 2.326 $75.42 5,687.5

87 0.0100 2.326 $75.42 5,687.5

88 0.0500 1.645 $211.89 44,895.9

89 0.0500 1.645 $211.89 44,895.9

90 0.0500 1.645 $211.89 44,895.9

91

92 ######## 14.545 $1,172.37 1,374,445.7

93

94

95





Cell J85 the risk that the individual failure in that row will exceed the limiting cost.

Similarly for J86.J90.

Cell K85 = ‐NORMSINV(J85). Similarly for K85.K90.

Cell L85 =((G85‐F85)/L85)*C85 = ((Limiting Cost – Nominal Cost)/Standard Deviation)*# Failures

Similar for L86.L90.

Cell M85 = L85^2. Similarly for L86.L90.

Cell M92 = Sum(M5.M90)

Cell L92 = M92^.5

Cell K92 = =(I93‐H93)/M93 = ((Sum Limiting Costs – Sum Nominal Costs)/Standard Deviation. = Z.risk that the

cost will exceed that of the sum of the limit costs.

Cell J92 = =1‐NORMSDIST(K92) = The risk that cost will exceed that of the sum of the limit costs.

Documents

A MindPro Lean Six Sigma Simulated Training Project