FMEA ( Failure Mode & Effect Analysis )

In Six Sigma DMAIC methodology, most traditional Six Sigma consultants / training instructors advocate the use of FMEA ( typically Process FMEA ) in Measure / Analyze phase. The project leaders were normally taught on how to use FMEA to identify and prioritize the critical potential failures.

In fact, a few advanced Six Sigma consultants / training instructors proposed FMEA to be used in Improve phase where a new process been introduced. As a result, it is necessary to assess and mitigate the risk of potential failure ( if any ) on new process.

It is nothing wrong or right to which project phases ( Measure vs Analyze vs Improve ) be used. Personally, I strongly agree that FMEA should be used to tackle potential failure in new / improved process area as risk level of potential failures have not been explored.

In this blog, one common overlooked area to be discussed here.

The area where most project leaders overlooked was the FMEA process, where it was NOT completed as it intended.

The FMEA process was started with Process steps>Potential Failure Mode>Effects>Potential Cause>Current Control> Scoring for Severity/Occurrence/Detection>RPN and stopped. There was no “close loop” where further actions i.e. corrective actions needed to be taken in response to “high risk” potential failure items and final assessment of effectiveness on action taken by evaluating the RPN score again.

Below is illustration showing on how FMEA process be completed ( close loop ) until significant improvement taken to address the high risk item.

FMEA

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Correlation & Regression Part 2

In previous blog, it was shown how to make use of scatterplot to detect linear trend and correlation analysis to quantify Pearson correlation coefficient value.

The Pearson correlation coefficient value was determined as 0.999 ( positive correlated ) and it was very strong linear relationship between the Y and X. ( Note that Pearson correlation value lie in the range -1 ≤ r ≤ +1 )

It can then be proceeded Regression analysis for constructing an estimating equation that relates Y and X.

In Minitab, click Stat>Regression>Fitted Line Plot. Fill up relevant Y and X field below and by default, click OK.

image image

The Fitted Line Plot was generated and results interpreted below.

image

The analysis result from Session window interpreted as follows:

Regression Analysis: Y versus X

 

The regression equation is

Y = 4.739 + 1.251 X

 

 

S = 0.0776200   R-Sq = 99.7%   R-Sq(adj) = 99.7%

 

 

Analysis of Variance

 

Source      DF       SS       MS        F      P

Regression   1  27.3417  27.3417  4538.14  0.000

Error       13   0.0783   0.0060

Total       14  27.4200

 

The Null Hypothesis and Alternate Hypothesis are identified below.

Ho : Null Hypothesis => The prediction equation is not statistically significant – no relationship exists between X and Y

Ha : Alternate Hypothesis => The prediction equation is statistically significant – a relationship does exists between X and Y

Since Pvalue = 0.000, reject Ho and accept Ha.

Hence, the estimating equation is a significant model that can be used for prediction.

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Correlation & Regression Part 1

Correlation analysis is to quantify the degree to which the variables are related, i.e. how well the estimating equation fits.

Regression analysis is about construction of an estimating equation that relates the predictor/s to the response.

In fact, correlation analysis is part of Regression analysis.

Typically in correlation analysis, scatterplot is a graphical analysis while we can use Pearson product moment correlation coefficient to measure the degree of linear relationship between 2 variables.

The rule of thumb is always conduct scatterplot to look for the form of relationship ( is it a line or curve? ) first. If there is linear trend, it is then necessary to proceed to Pearson correlation coefficient test to check the significance and value.

A case study below to illustrate it’s application.

A project leader wish to assess the relationship between Project Y and a parameter (X) he found which he believe it’s a significant factor. The data is as follows:

Y

7.5

7.6

8.1

8.3

8.6

9.1

9.3

9.5

9.9

10.2

10.6

10.9

11.2

11.5

11.7

X

2.2

2.4

2.6

2.8

3.1

3.4

3.7

3.9

4.1

4.4

4.7

4.9

5.1

5.4

5.6

So, the practical question here was “Was X strongly correlated with Project Y?”

In Minitab, click Graph>Scatterplot> and click OK.

image image

Select respective Y and X information below and by default, click OK.

image

The scatterplot was generated where it shown obvious form of a straight line, indicating a very strong linear trend.

image

Next, conduct Pearson correlation coefficient test with following Hypothesis Statement.

Ho : Null Hypothesis => r = 0

Ha : Alternate Hypothesis => r ≠ 0

In Minitab, click Stat>Basic Statistics>Correlation and select relevant information.

image

Click OK to find the analysis result below.

Correlations: X, Y

Pearson correlation of X and Y = 0.999

P-Value = 0.000

The Pvalue was 0.000. Reject Ho (assuming Alpha risk of 5% i.e. 0.05 )

Statistically, it was to say Pearson correlation coefficient value between Y and X  was not equal to 0.

Practically, it was concluded that there was strong linear relationship between Y and X with Pearson correlation coefficient value (r) measured as 0.999.

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How to proceed Analyze phase? Hypothesis Testing – Part 11 ( Chi Square Test )

Chi Square Test is used to test proportions of > 2 sample data.

Following is a case study to illustrate it’s application.

The process engineer would like to know if different raw material ( type A, B & C ) give significant impact on the Yield. He run the experiment for 3 material types on similar production line. Below is the Yield data collected.

 

Material A

Material B

Material C

Passed

95

93

96

Failed

5

7

4

Total inspected

100

100

100

So, the practical question here was “Was there significant difference in Yield among the 3 material type?”

It was then translated to statistical question by means of Hypothesis Statement.

Ho : Null Hypothesis => pA = pB = pC

Ha : Alternate Hypothesis => At least one p is different than another.

Next, conduct Statistical Analysis. In Minitab, click Stat>Tables>Chi Square Test ( Two- way table in one worksheet ) and select relevant information.

image image

Click OK to view the analysis result from Minitab Session window below.

Chi-Square Test: A, B, C

 

Expected counts are printed below observed counts

Chi-Square contributions are printed below expected counts

 

           A      B      C  Total

    1     95     93     96    284

       94.67  94.67  94.67

       0.001  0.029  0.019

 

    2      5      7      4     16

        5.33   5.33   5.33

       0.021  0.521  0.333

 

Total    100    100    100    300

 

Chi-Sq = 0.924, DF = 2, P-Value = 0.630

 

The Pvalue was 0.630. Fail to reject Ho (assuming Alpha risk of 5% i.e. 0.05 )

Statistically, it was to say the material type did not have association with Yield.

Practically, it was concluded that No significant difference in Yield among the 3 material type.

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How to proceed Analyze phase? Hypothesis Testing – Part 10 (2-proportion Test)

Following is a case study related to the application of 2-proportion test.

Currently, there were 5 production lines running the same product. All the 5 production lines were using the same version of machines. A process engineer would like to assess if changes on raw material would improve the Yield(%) significantly. He ran 1 production line with new raw material for 7 days with Yield measured as 97.8% ( Total:1250, Pass:1222 ) as compared to 96.8% ( Total:1250, Pass:1210 ) using current raw material.

So, the practical question here was “Was the new raw material improve the Yield significantly?”

It was then translated to statistical question by means of Hypothesis Statement.

Ho : Null Hypothesis => pnew material = pexisting material

Ha : Alternate Hypothesis => pnew material > pexisting material

Next, conduct Statistical Analysis. In Minitab, click Stat>Basic Statistics>2-proportion and fill up relevant information. Then, click Options and select the right field below.

image image

Click OK and OK to view the analysis result from Minitab Session window below.

Test and CI for Two Proportions

Sample    X    N    Sample p

  1     1222 1250   0.977600

  2     1210 1250   0.968000

Difference = p (1) – p (2)

Estimate for difference: 0.0096

95% lower bound for difference: -0.00109780

Test for difference = 0 (vs > 0): Z = 1.48 P-Value = 0.070

Fisher’s exact test: P-Value = 0.088

The Pvalue was 0.070. Fail to reject Ho (assuming Alpha risk of 5% i.e. 0.05 )

Statistically, it was to say No significant difference between Yield% using new and existing raw materials.

Practically, it was concluded that there was no evidence to show new raw material would improve the Yield significantly.

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How to proceed Analyze phase ? Hypothesis Testing – Part 9 ( 1-proportion test )

In the previous blogs Hypothesis Testing ( Part 1 – 8 ), the statistical tools covered were related to “mean(m)” and “Variance(σ/σ2)”. In this blog onward, “Proportion(p)” will be discussed.

Basically, proportion data is derived from the ratio of binary data. E.g. of binary data are Yes/No, Pass/Fail, Go/No Go and etc. The ratio can be expressed as %.

Following is a case study related to the application of 1-proportion test.

A process engineer had set up a new production line due to expansion program. The new production line (equipped with newer version of machine) would produce the same product as in other older production lines. Historically, the average Yield(%) measured in older production lines was 95.8%. The process engineer would like to know if the Yield (%) of new production line was significantly better than older production lines. The new production line was run for 7 days and the Yield(%) was 96.2% ( total:1250, Pass:1203)

So, the practical question here was “Was the new production line Yield better than old production lines?”

It was then translated to statistical question by means of Hypothesis Statement.

Ho : Null Hypothesis => pnew = 95.8%

Ha : Alternate Hypothesis => pnew > 95.8%

Next, conduct Statistical Analysis. In Minitab, click Stat>Basic Statistics>1-proportion and fill up relevant information. Then, click Options and select the right field below.

image image

 

Click OK and OK to view the analysis result from Minitab Session window below.

Test and CI for One Proportion

 

Test of p = 0.958 vs p > 0.958

 

 

                              95% Lower    Exact

Sample     X     N  Sample p      Bound  P-Value

1       1203  1250  0.962400   0.952301    0.244

 

The Pvalue was 0.244. Fail to reject Ho (assuming Alpha risk of 5% i.e. 0.05 )

Statistically, it was to say No significant difference between Yield% of new and old production lines.

Practically, it was concluded Yield% of new production line was NOT better than old production lines.

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