Using Stress/Strength Analysis to Reduce Sample Size

A good resource to follow up on this is Wayne Taylor’s book Statistical Procedures for the Medical Device Industry (Taylor Enterprises, 2017), in which Taylor notes that:

Oftentimes, due to several constraints such as material availability, resource constraints, and unforeseen circumstances, one may not be able to use the required sample sizes. I’m proposing here that we can use the stress/strength relationship to appropriately justify the use of a smaller sample size, while at the same time not compromising on the desired reliability/confidence level combination.

Published: Monday, October 16, 2023 – 12:03

this equates to 59 samples. Similarly for 2x stress conditions, we estimate 2% failures, and here R = 0.98. Using C = 0.95 in the equation, we get the sample size required as 149.

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Following this logic, if we increase the testing stress, we will also increase the likelihood for failures. For example, by increasing the stress five-fold (7.5 lbf), we are also increasing the likelihood of failure by five-fold (5%) or more. Therefore, if we test 60 parts (one-fifth of 299 from the original study) at 7.5 lbf and see no failures, this would equate to 99% reliability at 95% confidence at 1.5 lbf. We can claim at least 99% reliability of performance at 95% confidence level during normal use of product. We were able to reduce the sample size needed to demonstrate the required 99% reliability at 95% confidence level by increasing the stress test condition.

“Many test methods contained in standards are in fact stress tests designed to provide a safety margin. For example, the ASTM packaging standards provide for conditioning units by repeated temperature/humidity cycles and dropping of units from heights that are more extreme and at intervals that are more frequent than most products would typically see during shipping. As a result, it is common practice to test smaller sample sizes. The ASTM packaging conditioning tests are shown… to be five-times stress tests.”

ln(1-C)/ln(R), where C = 0.95 and R = 0.95,

The exact number can be found by using the success run theorem. In our example, we estimate at least 95% reliability based on the 5% failures while using 5x stress test conditions, when compared to the original 1% failures.

A common depiction of a stress/strength relationship is shown below for a product. We can see that as long as the stress distribution doesn’t overlap with the strength distribution, the product should function with no issues. The space between the two distributions is referred to as the margin of safety. Often, the product manufacturer defines the normal operating parameters based on this. The specifications for the product are also based on this, and some value of margin of safety is incorporated in the specifications.

“An attribute stress test results in a pass/fail result. However, the unit is exposed to higher stresses than are typical under normal conditions. As a result, the stress test is expected to produce more failures than will occur under normal conditions. This allows the number of units tested to be reduced. Stress testing requires identifying the appropriate stressor, including time, temperature, force, humidity, and voltage. Examples of stress tests include dropping a product from a higher height, exposing a product to more cycles, and exposing a product to a wider range of operating conditions.

n = ln(1 – C)/ ln(R), where n is the sample size, ln is the natural logarithm, C is the confidence level, and R is reliability.

Using the equation

The formula for the success run theorem is given as:

Always keep on learning.

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Published Oct. 1, 2023, in Harish’s Notebook.

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