# Usage of the three-sigma limits in calculating minimum force

I have carried out 10 cantilever sample tests that provide a braking force as follows (in kN):

22.1, 20.98, 21.03, 20.7, 21.03, 21.09, 20.98, 21.08, 21.24, 21.19

The goal is to recommend the worst case breaking force

The mean is 21.14 and the STD is 0.37.

If I use the 3-sigma rule, the minimum force is 20.04kN (mean-3sigma). I feel like the 22.1kN sample is skewing the minimum data.

Whats the best way to recommending the worst case scenario for the breaking force?

Data is data. This is the capability of the design and manufacturing process. Trust the data.

If there is a need to improve the quality of the data one could take for example 3 measurements for each sample and then take the average. So if there is a concern of quality of the data collection, this method will help minimize the error, with regard to the 22.1kN sample.

Other than that the method is spot on. What the data is telling 99.73% of the samples have a have a breaking force of 20.04kN and above. If you need better reliability use 6-sigma instead of 3-sigma.

The problem with your proposed solution of using 3 times the standard deviation, is that it doesn't take into account how accurately the standard deviation has been determined. As you only have 10 test values, it has a significant error bar, which needs to be accounted for somehow.

I would recommend using the method in EN1990 annex D or whatever design code is applicable in your case. That would mean using 3.04 times the standard deviation if you had performed an infinite number of tests, but 4.51 times the standard deviation when it is only known from 10 test values. The safety index is the same in the two cases.

Keep in mind that the failure doesn't have to follow a normal distribution. Weibull distribution is a much better fit for real life situations. And to be more specific, a weibull with a shape like the following. Near one end is very similar to a normal distribution but it is very skewed. So regarding your data, the 22.1[kN] and the 20.7 [kN] seem like outliers based on the following kernel density plot. However, unless you can identify valid reasons why they should be discarded, I suggest leaving them in the sample. The points could carry significant information. For example, it might give indication of the skewness of the distribution, or it might help you identify that its a multimodal distribution.

Regarding your question what is the best way to present the breaking force I would either go for:

a) plain simple average and standard deviation or

b) (preferably) give the value that 90% of the specimens are survived.

In this case, I would sort the data based on ascending order and take the average of the two lowest observations: $$\frac{20.98+ 20.7}{2}= 20.84 [kN]$$

For 20.84 kN only one 1 in 10 specimens failed.

There are other fancy ways of doing it, but depending on who you are reporting to, it will most likely end up being "Lies, God damn lies and Statistics".