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.

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...

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

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.



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

Calculating Sample Probabilities From a Finite Population

Ref : Design for Six Sigma Statistics (3.1.4.2)

Suppose "n" items selected randomly from a finite population of "N" items
If "D" = defective items in a population
Then, Non-defective items in the population = N-D

What is the probabilities samples will contain exactly "x" defective items and (n-x) non-defective items ?

Example

Suppose a box contain 50 parts

Based on historical data, average defective part found per box = 4% (2 defective part per box)

If each assembly required 5 part pick up randomly from the box, what would be the probability of assembly containing 1 or more defective part ?

This is a finite population sampling problem, where the population = 50pcs (including 2 defective parts)

There are 2 possible outcome from this case

1. assembly containing 1 defective part

2. assembly containing 2 defective part

Since these are mutually exclusive events, the probabilities add.

Solution as below

Alternatively, I calculate the probability this way

A Review Of My 6 Sigma Journey of 2012

How fast time flies by and you don’t even realize, it is coming to the end of year 2012! During the festive season, I would spend some time to review through the status of Thinking 6 Sigma + Lean program in the OP BE before the closing of year 2012.

2012 is the second year of Thinking 6 Sigma + Lean program being deploy in Infineon Melaka. Over the past 2 years, we conducted 4 Wave of Green Belt training that covered Melaka, Wuxi & Singapore site. Total 101 engineers being trained in Thinking 6 Sigma + Lean problem solving approach, 34 of them (3 from Wuxi, 29 from Melaka & 2 from Kulim) successfully fulfill the Infineon Green Belt certification criteria and awarded as certified Green Belt from IFAG. As today, there are 37 Green Belt project closed that contribute to about one million euro saving from the improvement solution implemented. There are 64 Green Belt project in the pipe line, projected to contribute about 1.5 million euro saving in FY2013.

Besides Green Belt training, there is also Black Belt training being organized in year 2012 in Munich, involving colleagues from Melaka & Wuxi site, as well as colleagues from Front End site such as Kulim, Regensburg and Business Unit. The main objective of Black Belt training is to further develop existing Green Belt  to handle more advance Thinking 6 Sigma + Lean tools, so that they are capable enough to support the local management in program deployment and supporting Green Belts in applying Thinking 6 Sigma + Lean problem solving approach to achieve breakthrough improvement.

23 November 2012, we are celebrating the Green Belt Graduation ceremony for Green Belts participate in the Wave II training, this is the second graduation ceremony since last year, total 20 Green Belt of Infineon Melaka and 1 from Wuxi participate the training program successfully fulfill the stringent certification criteria and awarded Green Belt certification from IFAG QM SI. At the same time, we officially announce Khirullah was certified as Thinking 6 Sigma + Lean trainer after going through assessment by our program partner from the HQ. With Khirullah, we further strengthen our local trainer team to face the new challenging year of 2013.

Experience and the sustainable result gained in the past two year proven Thinking 6 Sigma + Lean is a practical problem solving approach, it is a reliable strategy that drive us toward a High Performance company.

2013 will be a very challenging year ahead, overwhelming demand for Thinking 6 Sigma + Lean program support and Green Belt training shows the strong interest of management in this program. We appreciate your involvement and we welcome you as part of our team, as strong management support and involvement is the key success factor of this program. Let’s team up and make Thinking 6 Sigma + Lean systematic problem solving approach as an important driver to the OP BE toward a High Performance organization.

Last but not least, I would like to take the opportunity to wish all colleagues Merry Christmas & Happy New Year.