How to proceed Analyze phase ? Hypothesis Testing – part 8 ( Test for Equal Variances )

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 => σA = σB = σC = σD

Ha : Alternate Hypothesis => σA ≠ σB ≠ σC ≠ σD

Next, conduct Statistical Analysis. In Minitab, click Stat>ANOVA>Test for Equal Variance and fill up relevant information.

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Then, click OK to see the analysis result from Minitab.

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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 ).

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How to proceed Analyze phase ? Hypothesis Testing – Part 7 ( 2 Variances Test )

In previous blog, Single Variance Test was discussed and demonstrated with a case study if a sample variation/variance (σ/σ2 ) was to be compared with a targeted variation/variance (σ/σ2 ).

If one who wish to compare 2 sample variation/variances (σ/σ2 ), 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 => σcurrent = σA

Ha : Alternate Hypothesis => σcurrent ≠ σA

Next, conduct Statistical Analysis. In Minitab, click Stat>Basic Statistics>2 Variance and fill up following section as highlighted.

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By default, no change on Option. Thus, click OK to see the analysis result from Minitab.

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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.

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How to proceed Analyze phase ? Hypothesis Testing – Part 6 (Test for Single Variance)

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/σ2 ) 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 => σmodified = 0.9

Ha : Alternate Hypothesis => σmodified < 0.9

Next, conduct Statistical Analysis. In Minitab, click Stat>Basic Statistics>1 Variance and fill up following section as highlighted.

image  image

Then, click Option and fill up as highlighted.

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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% One-Sided 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    Chi-Square     DF  P-Value

Breaking Strength  Standard       13.17  29.00    0.005

                   Adjusted       10.56  23.25    0.011

 

For both methods ( Standard & Adjusted ), the P-value = 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.

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How to proceed Analyze phase? Hypothesis Testing – Part 5 (ANOVA)

If one wish to compare more than 2 samples, ANOVA is the right candidate.

In short, ANOVA stands for ANalysis Of VAriance.

Let’s look at following case study to understand the application of ANOVA.

A researcher would like to know if coating material would give significant effect on the performance of the gadget he/she designed. 4 coating materials were identified ( A, B, C & D ) in which each coating material was applied on 15 identical samples. All the samples were then treated at same operating condition and performance of each sample was recorded as follows:

A 2.15 2.13 2.06 2.07 2.07 2.07 2.10 2.16 2.12 2.11 2.16 2.12 2.06 2.21 2.08
B 1.95 1.95 2.06 1.95 1.99 2.01 1.97 1.98 2.02 2.07 1.97 2.09 2.03 2.07 2.09
C 2.21 2.12 2.04 2.08 1.94 2.16 2.00 2.16 1.96 1.98 1.95 2.03 2.23 2.14 2.00
D 2.15 2.11 1.99 1.93 2.08 1.86 2.32 2.09 2.13 1.95 1.94 2.07 2.08 2.08 2.20

 

So, the practical question here was “Is coating material type significantly affect performance of the gadget?”

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

Ho : Null Hypothesis => µA = µB = µC = µD

Ha : Alternate Hypothesis => µA ≠ µB ≠ µC ≠ µD ( at least 1 pair is unequal )

Next, conduct Statistical Analysis. In Minitab, click Stat>ANOVA>One-Way(unstacked) and fill up 1following section and click OK.

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Statistical Analysis result from Minitab session window as shown below:

ANOVA_1

The P-value = 0.02. Reject Ho and Accept Ha as it’s < 0.05 (assuming Alpha risk of 5% i.e. 0.05 )

Statistically, it was to said at least 1 pair of means is unequal. It is necessary to find which pair/pairs is/are unequal by using Tukey’s test.

Click Ctrl E or Stat>ANOVA>One-Way(unstacked), click icon Comparison.. and select Tukey’s, family error rate ( 5 by default )

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

ANOVA_2The above matrix displays the pair-wise confidence intervals. If 0.000 is not contained within the confidence interval, then it’s considered this pair is statistically different. It is obvious that pair A & B is the only pair that is unequal.

Practically, it was concluded that coating material type used would give significant effect on the performance of the gadget and the pair of coating material type that shown significant difference was pair A & B.

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How to proceed Analyze phase ? Hypothesis Testing – Part 4 (Paired t test)

In previous blog on “How to proceed Analyze phase ? Hypothesis Testing – Part 2”, a case study was used to illustrate the application of 2 sample t test.

Another comparison method which uses 2 samples is Paired t test.

Paired t test is used when taking 2 measurements on same samples under similar condition. Case study below is an example to illustrate the application of this statistical tool.

The Six Sigma instructor would like to assess the performance of the Green Belt (15 in total) if there is significant improvement in terms of performance (knowledge acquisition) in post test after the training. Prior to starting the training, all Green Belts were required to sit for a test and the scoring in percentage was recorded. After 1 week of training, all Green Belts required to sit for the same test again. The performance (score before and after the training) was shown on table below:

GB#

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

Before

55

50

50

45

65

60

55

65

70

70

65

55

35

75

60

After

70

65

70

60

85

80

75

90

95

90

80

80

65

100

80

So, the practical question here was “Was performance of Green Belts improved after the training?”

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

Ho : Null Hypothesis =>  µBefore = µAfter

Ha : Alternate Hypothesis =>  µBefore < µAfter

Next, conduct Statistical Analysis. In Minitab, click Stat>Basic Statistics>Paired t and fill up following section:

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Next, click Option and fill up accordingly reference below:

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Click OK and OK and following is Statistical Analysis result from Minitab session window.

 

Paired T-Test and CI: Before, After

Paired T for Before – After

N Mean StDev SE Mean

Before 15 58.33 10.63 2.75

After 15 79.00 11.68 3.02

Difference 15 -20.67 4.58 1.18

95% upper bound for mean difference: -18.59

T-Test of mean difference = 0 (vs < 0): T-Value = -17.49 P-Value = 0.000

 

The P-value = 0.000. Reject Ho and Accept Ha as it’s < 0.05 (assuming Alpha risk of 5% i.e. 0.05 )

Statistically, it was to said Performance in score of Green Belts (after) was significantly higher than Performance of Green Belts (before)

Practically, it was concluded as Performance (knowledge acquisition) of Green Belts in test was significantly improved after 1 week training, that meant training effectiveness by the instructor was effective.

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How to proceed Analyze phase ? Hypothesis Testing – Part 3

In previous blog, a case study example was used to illustrate the practical application 2 sample t test, which is one of the common Hypothesis Testing Tools.

In general, Hypothesis Testing is a method used for conducting comparison. The summary of the right selection of statistical techniques is as shown on table below:

Statistical Techniques Summary

All of us may already aware that the selection of Hypothesis Testing technique/tool depends on the followings:

· Statistic measured: Mean, Variance, Proportion and etc.

· Comparison type: Single Sample, Two Sample, Multiple Sample

Below table is the guideline used for selection of right Hypothesis Testing technique/tool.

Hypothesis Statistical Tools Summary

In next blogs, I will use more practical case studies to illustrate the application of above statistical tools.

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How to proceed Analyze phase? Hypothesis Testing – Part 2

It will be good if a case study example is used to illustrate the step by step to perform Hypothesis Testing reference below.

In this case study, there were 2 manufacturers ( A and B ) producing the same product. A distributor would like to know which product is more superior in terms of one quality performance i.e. power consumption. The product was considered superior if it consumed less power, which were tested at same operating condition with same loading.

So, the practical question here was “Is product A better than product B?”

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

Ho : Null Hypothesis =>  µA = µB

 

Ha : Alternate Hypothesis =>  µA < µB

 

Said, 20 measurement data collected from the Test.

Data_Hypothesis Testing_Part 2

Next, conduct Statistical Analysis. In this case, 2 sample t test was chosen as it met relevant statistical assumptions. Will discuss in next blog of the right selection of statistical analysis tools.

In Minitab, click Stat>Basic Statistics>2 Sample t and fill up following section :

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Click Options… and select “less than” in Alternative, others were set by default by Minitab.

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Click OK and OK. The Statistical Analysis result as follows :

Two-Sample T-Test and CI: Product A, Product B

Two-sample T for Product A vs Product B

N Mean StDev SE Mean

Product A 20 14.15 1.20 0.27

Product B 20 18.88 9.21 2.1

Difference = mu (Product A) – mu (Product B)

Estimate for difference: -4.73

95% upper bound for difference: -1.14

T-Test of difference = 0 (vs <): T-Value = -2.28 P-Value = 0.017 DF = 19

The P-value = 0.017. Reject Ho and Accept Ha as it’s < 0.05 ( assuming Alpha risk of 5% i.e. 0.05 )

Statistically, it was to said Product A power consumption was significantly lower than Product B power consumption.

Practically, it was concluded as Product A was better than Product B in terms of power consumption. Hence, it was obvious that the distributor should decide objectively which manufacturer to choose.

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