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.
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.
The Fitted Line Plot was generated and results interpreted below.
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 RSq = 99.7% RSq(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.
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.
Select respective Y and X information below and by default, click OK.
The scatterplot was generated where it shown obvious form of a straight line, indicating a very strong linear trend.
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.
Click OK to find the analysis result below.
Correlations: X, Y
Pearson correlation of X and Y = 0.999
PValue = 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.
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 => p_{A }= p_{B} = p_{C}
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.
Click OK to view the analysis result from Minitab Session window below.
ChiSquare Test: A, B, C
Expected counts are printed below observed counts
ChiSquare 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
ChiSq = 0.924, DF = 2, PValue = 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.
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 => p_{new material }= p_{existing material}
Ha : Alternate Hypothesis => p_{new material }> p_{existing material}
Next, conduct Statistical Analysis. In Minitab, click Stat>Basic Statistics>2proportion and fill up relevant information. Then, click Options and select the right field below.
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 PValue = 0.070
Fisher’s exact test: PValue = 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.
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 1proportion 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 => p_{new }= 95.8%
Ha : Alternate Hypothesis => p_{new }> 95.8%
Next, conduct Statistical Analysis. In Minitab, click Stat>Basic Statistics>1proportion and fill up relevant information. Then, click Options and select the right field below.
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 PValue
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.
In this blog, it’s continued on discussing variances test for multiple samples ( > 2 samples ). It’s called as Test for Equal Variances in Minitab.
If the multiple sample data distributions do not follow Normal distribution, Levene’s test should be adopted. Vice versa, Bartlett’s test should be adopted.
Below is a case study that illustrates the use of Test for Equal Variances.
The process engineer would like to know further if different raw material ( type A, B, C & D ) give significant impact on the standard deviation of plastic breaking strength. He run the experiment by collecting the samples data for ALL material type reference below table:
A  7.2  7.3  6.8  7.0  9.1  8.2  8.5  6.6  6.9  8.4 
B  9.0  7.9  8.7  8.9  8.9  8.1  8.2  7.6  8.3  7.5 
C  7.8  7.9  8.0  7.4  8.5  7.3  8.8  8.6  8.5  6.8 
D  7.0  7.1  8.4  6.4  6.9  6.6  7.4  6.7  7.1  7.4 
So, the practical question here was “Are the variations of breaking strength of plastic using different raw material type the same?”
It was then translated to statistical question by means of Hypothesis Statement.
Ho : Null Hypothesis => σ_{A }= σ_{B} = σ_{C }= σ_{D}
Ha : Alternate Hypothesis => σ_{A }≠ σ_{B} ≠ σ_{C }≠ σ_{D}
Next, conduct Statistical Analysis. In Minitab, click Stat>ANOVA>Test for Equal Variance and fill up relevant information.
Then, click OK to see the analysis result from Minitab.
The Pvalue of Bartlett’s test and Levene’s test are 0.522 and 0.442 respectively. Regardless of what data distribution of the sample data, both Pvalue shows more than 0.05. Fail to reject Ho (assuming Alpha risk of 5% i.e. 0.05 )
Statistically, it was to say the breaking strength standard deviation of plastic were the same for all raw material type ( A, B, C & D ).
Practically, it was concluded NO significant difference in variation between breaking strength of plastic parts using different type of raw material ( A, B, C & D ).
In previous blog, Single Variance Test was discussed and demonstrated with a case study if a sample variation/variance (σ/σ^{2} ) was to be compared with a targeted variation/variance (σ/σ^{2} ).
If one who wish to compare 2 sample variation/variances (σ/σ^{2} ), F test can be used but it is subjected to the samples data distribution that must be Normally distributed. If the samples data distribution is NOT Normally distributed, Levene’s test is the choice. In fact, Levene’s test can be applied at multiple samples ( > 2 samples ).
It is reminded also that the application of all sample t test (discussed much earlier) are based on the assumption that samples data distribution are Normally distributed.
Below is a case study that illustrates the use of 2 Variances Test.
The process engineer would like to know further if new raw material ( type A ) give significant impact on the standard deviation of breaking strength as compared to current material type. He run the experiment by collecting the samples data for both current material type vs type A reference below table:
Current  7.7  7.0  7.6  5.7  7.9  5.6  6.5  6.8  8.0  7.0  6.4  7.7  7.7  7.5  7.1 
Type A  6.6  6.7  7.6  9.0  5.3  6.9  7.3  8.0  6.0  8.1  9.1  6.9  7.6  7.7  6.5 
So, the practical question here was “Is the variation of breaking strength of plastic parts using type A raw material significantly different from current raw material type?”
It was then translated to statistical question by means of Hypothesis Statement.
Ho : Null Hypothesis => σ_{current }= σ_{A}
Ha : Alternate Hypothesis => σ_{current }≠ σ_{A}
Next, conduct Statistical Analysis. In Minitab, click Stat>Basic Statistics>2 Variance and fill up following section as highlighted.
By default, no change on Option. Thus, click OK to see the analysis result from Minitab.
The Pvalue of F test and Levene’s test are 0.264 and 0.287 respectively. Regardless of what data distribution of the sample data, both Pvalue shows more than 0.05. Fail to reject Ho (assuming Alpha risk of 5% i.e. 0.05 )
Statistically, it was to say the breaking strength standard deviation of plastic parts using type A raw material was the same as current raw material type.
Practically, it was concluded NO significant difference in variation between breaking strength of plastic parts using type A raw material with current raw material type.
As discussed in previous blog, t test and ANOVA were the statistical tools for those interested to find out if significant differences in the mean ( µ ) level of response produced by different methods or treatments.
However, it is sometimes the degree of variation/variance (σ/σ^{2} ) of the sample data that is of interest. As such, process improvements that reduce variation, even though the mean ( µ ) of the sample data is unchanged, can be of great importance.
Below is a case study that illustrates the application of Single Variance Test.
A process engineer modified a few injection processes to mold a plastic part. One of the key performances of the plastic part was breaking strength and the process engineer would like to know if the standard deviation of breaking strength was improved. Prior to modification, the historical standard deviation was measured as 0.9 psi.
30 sample parts were produced and the measurement result as follows:
7.8  7.3  7.6  7.0  8.2  8.5  8.4  8.7  7.6  7.7  7.7  7.8  7.2  7.8  7.6 
6.5  7.6  7.8  8.2  6.6  7.4  7.6  7.8  8.3  7.3  6.3  7.3  9.0  7.8  7.8 
So, the practical question here was “Is the variation of new plastic parts reduced significantly?”
It was then translated to statistical question by means of Hypothesis Statement.
Ho : Null Hypothesis => σ_{modified }= 0.9
Ha : Alternate Hypothesis => σ_{modified }< 0.9
Next, conduct Statistical Analysis. In Minitab, click Stat>Basic Statistics>1 Variance and fill up following section as highlighted.
Then, click Option and fill up as highlighted.
Click OK and OK and the Statistical Analysis result from Minitab session window as shown below:
Test and CI for One Standard Deviation: Breaking Strength
Method
Null hypothesis Sigma = 0.9
Alternative hypothesis Sigma = < 0.9
The standard method is only for the normal distribution.
The adjusted method is for any continuous distribution.
Statistics
Variable N StDev Variance
Breaking Strength 30 0.606 0.368
95% OneSided Confidence Intervals
Upper Bound Upper Bound
Variable Method for StDev for Variance
Breaking Strength Standard 0.776 0.602
Adjusted 0.802 0.644
Tests
Variable Method ChiSquare DF PValue
Breaking Strength Standard 13.17 29.00 0.005
Adjusted 10.56 23.25 0.011
For both methods ( Standard & Adjusted ), the Pvalue = 0.005 and 0.011 respectively. Reject Ho and Accept Ha as both < 0.05 (assuming Alpha risk of 5% i.e. 0.05 )
Statistically, it was to said the standard deviation of new plastic parts breaking strength after modification was significantly reduced as compared to historical standard deviation (before process modification)
Practically, it was concluded as variation of new plastic parts breaking strength was significantly improved after modification on injection process.