Paired t-test Exercise - PowerPoint PPT Presentation

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Paired t-test Exercise

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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 ... – PowerPoint PPT presentation

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Title: Paired t-test Exercise


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 managers
    question.

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.

CalcgtCalculator
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
Paired t-test Exercise Solution
  • As weve seen before, this 1 Sample T analysis is
    found with
  • StatgtBasic Statgt1-sample T

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-tests 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.
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