© OpenSourceSixSigma, LLCOSSS LSS Green Belt v9.1 - Analyze Phase 1

Paired t-test Exercise

Exercise objective: Utilize what you have learned to conduct and analyze a paired t-test using MINITABTM.

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” to determine the answer to the quality manager’s question.

© OpenSourceSixSigma, LLCOSSS LSS Green Belt v9.1 - Analyze Phase 2

Paired t-test Exercise: Solution

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.

Calc>Calculator

© OpenSourceSixSigma, LLCOSSS LSS Green Belt v9.1 - Analyze Phase 3

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.

Paired t-test Exercise: Solution

0.750.500.250.00-0.25-0.50

Median

Mean

0.60.40.20.0

Anderson-Darling Normality Test

Variance 0.14418Skewness -0.833133Kurtosis -0.233638N 11

Minimum -0.50000

A-Squared

1st Quartile -0.10000Median 0.400003rd Quartile 0.50000Maximum 0.70000

95% Confidence I nterval for Mean

-0.02782

0.45

0.48237

95% Confidence I nterval for Median

-0.11644 0.50822

95% Confidence I nterval for StDev

0.26531 0.66637

P-Value 0.222

Mean 0.22727StDev 0.37971

95% Confidence Intervals

Summary for TX_MX-Diff

0.750.500.250.00-0.25-0.50

Median

Mean

0.60.40.20.0

Anderson-Darling Normality Test

Variance 0.14418Skewness -0.833133Kurtosis -0.233638N 11

Minimum -0.50000

A-Squared

1st Quartile -0.10000Median 0.400003rd Quartile 0.50000Maximum 0.70000

95% Confidence I nterval for Mean

-0.02782

0.45

0.48237

95% Confidence I nterval for Median

-0.11644 0.50822

95% Confidence I nterval for StDev

0.26531 0.66637

P-Value 0.222

Mean 0.22727StDev 0.37971

95% Confidence Intervals

Summary for TX_MX-Diff

© OpenSourceSixSigma, LLCOSSS LSS Green Belt v9.1 - 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

© OpenSourceSixSigma, LLCOSSS LSS Green Belt v9.1 - 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.

TX_MX-Diff

Frequency

0.750.500.250.00-0.25-0.50

5

4

3

2

1

0 _X

Ho

Histogram of TX_MX-Diff(with Ho and 95% t-confidence interval for the mean)

TX_MX-Diff

Frequency

0.750.500.250.00-0.25-0.50

5

4

3

2

1

0 _X

Ho

Histogram of TX_MX-Diff(with Ho and 95% t-confidence interval for the mean)

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.