In the vast toolbox of Lean Six Sigma, the importance of statistical methods for improving and maintaining quality in processes cannot be overstated. Among these methods, hypothesis testing plays a pivotal role, especially when dealing with normal data distributions. A critical subset of hypothesis testing is parametric tests, which are used when the dataset is assumed to follow a normal distribution. This article delves into the Two-Sample T-Test, a fundamental parametric test, elucidating its purpose, application, and significance within the Lean Six Sigma framework.

What is the Two-Sample T-Test?

The Two-Sample T-Test, also known as the independent samples t-test or the student t-test, is a statistical method used to compare the means of two independent groups to determine if there is a statistically significant difference between them. This test assumes that the data from both groups follow a normal distribution, have similar variances, and that the observations are independent and randomly drawn from the population.

Application in Lean Six Sigma

Lean Six Sigma practitioners leverage the Two-Sample T-Test to make informed decisions about process improvements, quality control, and optimization. Specifically, this test is applied when comparing:

• The performance of two different processes or treatments.

• Outcomes before and after implementing a change in a process (assuming the groups are independent).

• Differences in quality metrics between two production lines or batches of products.

The primary goal is to determine if the changes or differences observed are due to a genuine effect rather than random variation.

Steps to Perform a Two-Sample T-Test

1. Define the Hypotheses:

   • Null Hypothesis (H0): Assumes that there is no difference between the means of the two groups.

   • Alternative Hypothesis (H1): Assumes that there is a difference.

2. Collect and Prepare Data: Ensure data is normally distributed, independent, and samples have equal variances. If variances are unequal, adjustments can be made.

3. Perform the Test: Calculate the t-statistic, which involves the difference between the sample means, the sample sizes, and the standard deviations of the groups.

4. Determine Significance: Compare the calculated t-statistic to the critical t-value from the t-distribution table at the desired significance level (commonly 0.05). If the t-statistic exceeds the critical value, the null hypothesis is rejected.

5. Interpret Results: A rejection of the null hypothesis indicates that there is a statistically significant difference between the group means, suggesting that the change or difference observed is likely not due to chance.

   
   

Significance in Lean Six Sigma

In Lean Six Sigma projects, the Two-Sample T-Test is invaluable for:

• Validating Improvements: Confirming that process improvements have statistically significant effects on outcomes.

• Comparing Processes: Identifying the more efficient or higher-quality process when evaluating alternatives.

• Data-Driven Decisions: Facilitating objective, statistical evidence-based decision-making.

  
  

Limitations and Considerations

While the Two-Sample T-Test is a powerful tool, it is essential to be mindful of its assumptions about normality, variance, and sample independence. Violations of these assumptions can lead to misleading results. Additionally, it's crucial to consider the size of the effect in addition to its statistical significance, as very small differences may be statistically significant but not practically important in some Lean Six Sigma projects.

Conclusion

The Two-Sample T-Test stands out as a cornerstone parametric test for normal data in the Lean Six Sigma methodology. It provides a rigorous, statistical means to compare two groups, helping practitioners make data-backed decisions that drive process improvements and quality enhancements. Like any statistical tool, its effective application requires a deep understanding of its assumptions, limitations, and the context of the data being analyzed. By adhering to these principles, Lean Six Sigma professionals can continue to leverage the Two-Sample T-Test to achieve excellence in quality management and process optimization.

Scenario: Comparing Machine Output Quality

A manufacturing company uses two machines to produce widgets. The quality control department wants to determine if there is a significant difference in the diameters of widgets produced by Machine A and Machine B. The company aims to ensure consistent product quality regardless of the machine used. For this test, the quality control team measures the diameters (in mm) of a random sample of widgets from each machine.

Sample Data

• Machine A (Sample 1): [10.1, 10.3, 10.2, 9.9, 10.1]

• Machine B (Sample 2): [10.4, 10.5, 10.3, 10.6, 10.4]

  
  

Step-by-Step Calculation

Step 1: Define the Hypotheses

• Null Hypothesis (H0): The mean diameter of widgets from Machine A is equal to the mean diameter of widgets from Machine B.

• Alternative Hypothesis (H1): The mean diameter of widgets from Machine A is not equal to the mean diameter of widgets from Machine B.

  
  

Step 2: Calculate Sample Means and Variances

Sample Means:

The sample mean (xˉ) is calculated as the sum of all measurements divided by the number of measurements in the sample. It represents the average value of the sample data.

xˉA=10.12 mm

xˉB=10.44 mm

Sample Standard Deviations

The sample standard deviation (s) is the square root of the sample variance and provides a measure of the dispersion or spread of the sample data points around the mean. It's calculated as follows:

For Machine A:

Calculate the mean (we just did it) = xˉA=10.12 mm

Then calculate each data point's deviation from the mean, square these deviations, and sum them up:

∑(xi​−xˉA​)^2

=(10.1−10.12)^2+(10.3−10.12)^2+(10.2−10.12)^2+(9.9−10.12)^2+(10.1−10.12)^2

=(−0.02)^2+0.18^2+0.08^2+(−0.22)^2+(−0.02)^2

=0.0004+0.0324+0.0064+0.0484+0.0004

=0.088

Calculate the variance

The variance is the sum of the squared deviations divided by the number of observations minus 1 (N - 1), which corrects for bias in the estimation of the population variance from a sample.

Calculate the standard deviation

The standard deviation is the square root of the variance.

For Machine B:

Calculate the mean (we just did before) = xˉB=10.44 mm

Then calculate each data point's deviation from the mean, square these deviations, and sum them up:

∑(xi​−xˉB​)^2

=(10.4−10.44)^2+(10.5−10.44)^2+(10.3−10.44)^2+(10.6−10.44)^2+(10.4−10.44)^2

=(−0.04)^2+0.06^2+(−0.14)^2+0.16^2+(−0.04)^2

=0.0016+0.0036+0.0196+0.0256+0.0016

=0.052

Calculate the variance

Calculate the standard deviation

Step 3: Perform the Two-Sample T-Test

We will use the formula for the t-statistic when variances are assumed equal (for simplicity in this example):

• Mean diameter from Machine A (xˉA​) = 10.12 mm

• Mean diameter from Machine B (xˉB​) = 10.44 mm

• Sample size for each machine (n) = 5

• Pooled standard deviation (sp​) = 0.132 mm

Formula and Calculation:

Step 4: Determine the t-Critical Value and Compare

Given the degrees of freedom (df = 8) and a significance level of 0.05 (for a two-tailed test), the critical t-value from a t-distribution table is approximately ±2.306.

Step 5: Conclusion Based on t-Statistic and t-Critical

The calculated t-statistic (-3.825) is beyond the critical t-value range (±2.306), which means we reject the null hypothesis.

In addition, find below the visual representation.

The chart above illustrates the T-distribution with 8 degrees of freedom, highlighting the critical region (in red) for a two-tailed test at a significance level of 0.05, where the critical t-value is ±2.306. The calculated t-statistic of -3.825 is shown as a blue dashed line, falling within the critical region. This graphical representation helps us visualize that the t-statistic lies beyond the critical t-value range, indicating that the difference in means between the two groups (Machine A and Machine B) is statistically significant, leading to the rejection of the null hypothesis.

Interpretation

There is statistically significant evidence to conclude that there is a difference in the mean diameters of widgets produced by Machine A and Machine B at the 0.05 significance level. This implies that the manufacturing process may need adjustments to ensure consistency in product quality across both machines.