Saturday, August 3, 2013

CI Selection In Statistical Hypothesis Testing

A GB shows me her result of Hypothesis testing as below.
Based on her 85% CI, she conclude that P-value < 0.15, therefore it shows significant different between two group of sample data.

When I ask her, why she choose 85% as her CI, answer given to me was, if she choose the default value of 95% CI, she is not able to get a conclusion that shows significant different.


Two-Sample T-Test and CI: C1, C2

Two-sample T for C1 vs C2

     N    Mean   StDev  SE Mean
C1  10  5.0086  0.0680    0.022
C2  10  4.9627  0.0449    0.014

Difference = mu (C1) - mu (C2)
Estimate for difference:  0.045879
85% CI for difference:  (-0.009077, 0.100834)
T-Test of difference = 0 (vs not =): T-Value = 1.78  P-Value = 0.095  DF = 15

My question is : Is this the right way to manipulate result by playing around the CI ?
My personal opinion, ethically this is not a right way to manipulate the test conclusion by playing around the CI, and she set the CI beyond the practical limit of 90% ! What is the objective of doing hypothesis testing if we can manipulating the result by changing the CI ? might as well don't do !

In my past experience, I do face many cases where test result shows no different or marginal different, P-value > 0.05. I never advice my GB or BB to adjust the CI for the sake of getting a result of significant different, this is meaningless to me. 

However, we admit the fact, the most, we ask our self, are we still want to implement the solution that shows no or minor different ? what are the benefit we can gain ? if the answer is YES and we can justify our self to proceed, then we go ahead and do it. (handle it with special case)

As a professional Six Sigma practitioner, we should aware Six Sigma is a problem solving methodology that based on Data and Fact, we shouldn't abuse the Fact to mislead the people who do not aware about the result.

 

Tuesday, July 9, 2013

A Pull System For Packing Material Inventory Management


Few days ago I receive a request from the logistic manager, to think about a pull system to stream line her packing material inventory management.

Based on her description of current material management process and problems she face, I draft out a simple VSM I believe practical for her application.

In fact, I m planning this for her with a purpose. I hope, the system I proposed works in the packing material inventory management, so that this system can be expand for others material and slowly turn the whole warehouse inventory management by pull system

Below is the statement she wrote to me:

"The list below are packing material delivered direct from supplier to prod (Ship To Line, STL); i.e. without delivery to LOG Receiving. In my opinion, candidates for Kanban with supplier. Currently prod sup will inform planner (daily). Planner will inform supplier before 10am daily & supplier deliver aro 11am. No editing done by planner. We could work on direct info from prod to supplier. And looking at the long list, currently rather huge effort for supervisor to organize the requirement.

Pls give this a though. We meet up next week to discuss how best to proceed. Thnaks"


Sunday, June 30, 2013

Six Sigma & Lean By Itself Is Not a Problem Solving Approach

http://blogs.hbr.org/davenport/2008/01/why_six_sigma_is_on_the_downsl.html

I always emphasis that, in fact, there is no such thing like Six Sigma Problem Solving Approach. We have Six Sigma process level, but Not Six Sigma Approach.

This meant, we you achieve a Six Sigma process level, technically we can estimate the process will generate 3.4ppm defect level. But how to achieve this level ? This ...is a question mark. You may apply PCDA, DMAIC or other problem solving methodology to achieve the level.

And LEAN to me is another problem. Similar to Six Sigma, LEAN is just a target, how to achieve this target ? You may apply above mentioned methodoly.

Why I say LEAN is a big problem here. Question... How LEAN is consider LEAN ? We cant measure now. So most of time we end up measure LEAN in Sigma Level as well.

Monday, June 17, 2013

A Simple System for Packing Box Management System

One day a lady from my Logistic department calls me up for help, she needs a simple, easy to implement and easy to use system for her packing box inventory management.
Based on the current process map we workout, I learnt that her supervisor is responsible to monitor the packing boxes inventory from time to time, to ensure sufficient packing boxes to sustain the operation in the Logistic department.
Her problems – there is no proper stock replenishment systematic in place, decision for stock replenishment and quantity to order is solely based on the supervisor experience. Therefore resulting of unsatisfactory of inventory level, such as, over stock of slow moving item, whereas operation line down due to shortage of necessary packing box.
To overcome her problem, I m proposing her to use the Min – Max for her packing box management, what she need to do is review the past 6 month data to draw the Min – Max line for each packing box inventory level, set it and implement.
Here the result.


Here the feedback from her after 1 month of implementation.
-          She is happy with the system as she have better visibility of stock level
-          Free up more supervisor time from inventory management, stock take at every end of shift is eliminated, supervisor just required to trigger a fix quantity of packing box ordering when the inventory level is closing or below the red zone.
-          Packing box is review in monthly basis, whereas she has to do it by daily basis before the improvement in placed.
Lesson learn
Effective system is a system that meets your objective, easy to maintain and manage. Its not necessary must be a complicated high tech system that required huge amount of capital investment.

Friday, May 10, 2013

Cronbach’s Alpha

I came across the term since few years back through a MBA classmate. However, I never use it and know nothing about the test until recently where I really need it to test the internal consistency of a survey question I prepare for my Six Sigma activity. I believe most BB from engineering background do not aware the statistical tool as well.

Cronbach’s alpha is very common in assessing internal consistency or reliability of survey / test question for social science and marketing data collection.

Rule of thumb, alpha value of 0.7 is commonly used benchmark. Alpha value > 0.7 suggested strong evidence of internal consistency exist in the survey instrument. The survey instrument is said to be reliable to provide consistent results.

Minitab 14 or earlier version did not offer feature. Anyway, it is now available in Minitab 15. You can download the trial version and try it yourself. Just follow the step below.


Monday, March 25, 2013

Regression Model Simulation Through Random Data Generation

OK, it is not my target to discuss DOE step by step in this post, my key objective today is just to talk about the next step after we got the optimum solution from DOE.

Picture below shows some of the chart that use for DOE analysis; no further discussion about the regression model and charts. Well, now we got the standard optimum solution from DOE, what next then?


Usually, as an engineer, we will just pick up the solution developed from the DOE and carry out trial run with a predefine batch size, we collect the data from small pilot run and compare against the process performance based on previous solution.

Two possible solutions could be draw from the pilot run, either the result showing significant improvement or no significant improvement compare against the previous process. Certainly, we how to have a result that showing significant improvement, on the other hand, we may need to revisit the DOE, to fine tune the process parameter in case the result is not following our wish.

Don’t forget DOE is carry out in a well controlled condition, data collected from DOE, or the regression model form from the DOE can only represent a short term process capability, there are many other influencing factor that exist in the production was not factor in, therefore, it is quite common to see that solution from a DOE model do not really fit to the actual production condition.

Actually, we can counter check or predict the long term process performance with the regression model developed with a set of random data generated with consideration of predefine standard deviation.

For example, I m trying to generate a set of random data with introduction of 5% variation (recommended by the process engineer) and plug it into the regression model developed from the DOE.

Here is the respond distribution with 500 random data of each factor with 5% process variation. Now, I m able to estimate the process capability before the pilot run, performance necessary corrective action before pilot run, therefore lower down the risk of wasting unnecessary resource.


OK, that's all for today...

Friday, January 18, 2013

Calculating Sample Probabilities ( Poisson )

Ref : Design for Six Sigma Statistic (3.1.4.4)

When counting defects in a sample from a contineuous medium

The probability of observing exactly "X" defects in the sample is



Example
Let failure rate of a control systme : 30 failure per million hour

Assuming that if any one failure happen, it will be detected and repaired within 24hour.
If a second failure happpens before the 1st failure is repaired, the system will shut down unexpectedly causing great damage.



Assuming the spare part is replanished only once per year. How many spare should be keep if we want to have a 99% confidence we have enough spare to up the system in case failure



Friday, January 11, 2013

Calculating Sample Probability ( Binomail )

Ref : Design For Six Sigma Statistics (3.1.4.3)

A process engineer is performing a quality check on a large batch of incoming material. He is asuming the proable defective rate is 4% ( p=0.04) based on the past record.

Since every time he draw 5 sample ( n = 5 ) for his inspection, what would be the probability of selecting at least 1 defective part ?

Solution

If the selection has at least 1 defective part, then, it could be also containing 2, 3, 4 or 5 defective part as well.

Therefore, the probability is the sum of 5 Binomial probability expression, one for each of possibilities.




Wednesday, January 9, 2013

Bernoulli Trial

Ref : Design for Six Sigma Statistics (3.1.4.3)

A Bernoulli Trial is a simple experiement with following characteristic :
  1. Each Bernoulli trial has only two outcome :
    • Yes / No
    • Pass / Fail
  2. In an experiment with multi Bernoulli trials, the probability of each of the outcomes is the same in every trial
  3. In an experiment with multiple Bernoulli trials, all trials are mutually independent.
Suppose :
  • p = probability of selecting a defective part
  • n = part are selected from a infinite population
What is the probability of selecting exactly x defective parts ?

This situation has 'n' Bernaoulli trial with a constand probability of selecting a defect 'p'. Since the trials are indenpendent, the joint probabilities are products of individual probabilities.

Each possible sample with 'x' defective parts and (n-x) nondefective parts has a probability of

Also there arecombination of samples of size 'n'with 'x' defects.
Therefore, the probability that a 'n' sample size from a population with a constand probability fo defects 'p' with have exactly 'x' defects is


This probability model is known as Binomial Distribution