Example of Equivalence Test with Paired Data

An engineer for a vision care company tests a new cleaning solution for contact lenses. The engineer wants to verify that the new solution cleans lenses as well as the leading brand. The engineer has 14 participants wear contact lenses for a day, and then clean the lenses. Each participant cleans one lens in the new solution and the other lens in the leading brand. The engineer assesses the cleanliness of each lens by measuring the angle of contact for a drop of fluid on the lens. The angle of contact is affected by film or deposits on the lens. To be equivalent, the mean angle for the new solution should be within ±0.5 degrees of the mean angle of the leading brand.

The engineer performs an equivalence test with paired data to determine whether the two cleaning solutions are equivalent.

  1. Open the sample data, ContactLensCleaner.MTW.
  2. Choose Stat > Equivalence Tests > Paired.
  3. In Test sample, enter New.
  4. In Reference sample, enter Leading Brand.
  5. From Hypothesis about, select Test mean - reference mean.
  6. From What do you want to determine? (Alternative hypothesis), select Lower limit < test mean - reference mean < upper limit.
  7. In Lower limit, enter -0.5.
  8. In Upper limit, enter 0.5.
  9. Click OK.

Interpret the results

Because the confidence interval is completely within the equivalence interval, the engineer concludes that the two cleaning solutions are equivalent.

Equivalence Test with Paired Data: New, Leading Brand

Method Test mean = mean of New Reference mean = mean of Leading Brand
Descriptive Statistics Variable N Mean StDev SE Mean New 14 88.604 1.5578 0.41634 Leading Brand 14 88.724 1.5907 0.42514
Difference: Mean(New) - Mean(Leading Brand) 95% CI for Difference StDev SE Equivalence Equivalence Interval -0.11929 0.42324 0.11312 (-0.319605, 0.0810335) (-0.5, 0.5) CI is within the equivalence interval. Can claim equivalence.
Test Null hypothesis: Difference ≤ -0.5 or Difference ≥ 0.5 Alternative hypothesis: -0.5 < Difference < 0.5 α level: 0.05
Null Hypothesis DF T-Value P-Value Difference ≤ -0.5 13 3.3657 0.003 Difference ≥ 0.5 13 -5.4748 0.000 The greater of the two P-Values is 0.003. Can claim equivalence.

Equivalence Test: Mean(New) - Mean(Leading Brand)

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