© 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.

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Presentation transcript:

© Open Source Six Sigma, LLCOSSS LSS Green Belt v 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 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.

© Open Source Six Sigma, LLCOSSS LSS Green Belt v 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.

© Open Source Six Sigma, LLCOSSS LSS Green Belt v 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.

© Open Source Six Sigma, LLCOSSS LSS Green Belt v 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

© Open Source Six Sigma, LLCOSSS LSS Green Belt v 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.