Statistical difference between 2 failure rates

[duongv2006]

Hello all,

Scenario:
We built 2 lots of products (say lot1 and lot2) using the same assembly process, parts etc... except one variation; in one lot (lot2) we changed the substrate on one of the parts. We tested these 2 lots and obtained the following test data:

lot1: 1 failure out of 337 tested so far (rate=0.30%)
lot2: 6 failures out of 628 tested so far (rate=0.96%)

Question: How can i go about to test the statistical significance between these 2 failure rates? Can it be done using Weibull++ 7?

Thanks in advance for your help/input!

Vua

[David]

Hi Vua,

Instead of Weibull++ 7, a common method is to do the following. Since you do not have any failure or suspension times, you can treat it like a quality test problem. Assume that the probability of failure will not change over time and that the test is a pass/fail test. Please see the attached *.zip file which includes a Word file for this type of analysis. The reference for this is "Introduction to Statistical Quality Control" by Douglas C. Montgomery.


I hope this helps.

[DrDave]

Vua,

This can be treated as a test for the equality of two proportions. If the number of failures was large (say more than 5 in each case), a test based the normal distribution could be used.

Here the number of failures is small in one lot (only 1). There is an exact test due to Fisher that applies.

I dumped your data into Minitab14, and got the following results. The important parts are in bold:

Sample X N Sample p
1 1 337 0.002967
2 6 628 0.009554

Difference = p (1) - p (2)
Estimate for difference: -0.00658678
95% CI for difference: (-0.0161580, 0.00298446)
Test for difference = 0 (vs not = 0): Z = -1.35 P-Value = 0.177

* NOTE * The normal approximation may be inaccurate for small samples.
Fisher's exact test: P-Value = 0.432


The p-value is the likelihood of getting results this extreme if the null hypothesis was true. Small values, say less than 0.05, indicate that the null hypothesis is unlikely given the data and cause us to reject the null. Here the null was that the probabilty of failure was the same for both lots. The p-value is .432, so there is NO evidence to reject the null hypothesis: the lots could very well have the same probability of failure.

The differences between the two lots could be due to sampling variability.

Hope this helps.