1. © Open Source Six Sigma, LLCOSSS LSS Green Belt v10.0 - Analyze Phase 1 Paired t-test Exercise Exercise objective: Utilize what you have learned to conduct and analyze a paired t-test using MINITAB TM. 1.A corrugated packaging company produces material which has creases to make boxes easier to fold. It is a critical to quality characteristic to have a predictable Relative Crease Strength. The quality manager is having her lab test some samples labeled 1-11. Then those same samples are being sent to her colleague at another facility who will report their measurements on those same 1-11 samples. 2.The US quality manager wants to know with 95% confidence what the average difference is between the lab located in Texas and the lab located in Mexico when measuring Relative Crease Strength. 3.Use the data in columns “Texas” & “Mexico” in “RM Suppliers.mtw” to determine the answer to the quality manager’s question.

2. © Open Source Six Sigma, LLCOSSS LSS Green Belt v10.0 - Analyze Phase 2 Paired t-test Exercise: Solution Calc>Calculator Because the two labs ensured to exactly report measurement results for the same parts and the results were put in the correct corresponding row, we are able to do a paired t- test. The first thing we must do is create a new column with the difference between the two test results.

3. © Open Source Six Sigma, LLCOSSS LSS Green Belt v10.0 - Analyze Phase 3 Paired t-test Exercise: Solution We must confirm the differences (now in a new calculated column) are from a Normal Distribution. This was confirmed with the Anderson- Darling Normality Test by doing a graphical summary under Basic Statistics.

4. © Open Source Six Sigma, LLCOSSS LSS Green Belt v10.0 - Analyze Phase 4 Paired t-test Exercise: Solution As we’ve seen before, this 1 Sample T analysis is found with: Stat>Basic Stat>1-sample T

5. © Open Source Six Sigma, LLCOSSS LSS Green Belt v10.0 - Analyze Phase 5 Paired t-test Exercise: Solution Even though the Mean difference is 0.23, we have a 95% confidence interval that includes zero so we know the 1-sample t-test’s null hypothesis was “failed to be rejected”. We cannot conclude the two labs have a difference in lab results. The P-value is greater than 0.05 so we do not have the 95% confidence we wanted to confirm a difference in the lab Means. This confidence interval could be reduced with more samples taken next time and analyzed by both labs.