# How do I calculate sigma for attribute data of process - Coating spoliation

I have a process is coating for the production. Totally, I had 10 products, the defect is coating spoliation, Now the point is some product have 0 defect, someone have 3 defects and so on, I used Mintab calculate it, total defects are 17, opportunity is 1, unit is 10, but I can’t get result,

defect opportunity uint
1 1 1
3 1 1
1 1 1
1 1 1
0 1 1
2 1 1
1 1 1
5 1 1
0 1 1
3 1 1

it show
** Error ** Incorrect data - result contains a value of PPM in
excess of 1,000,000 (this is impossible);
Execution aborted.

I think the problem is “opportunity”? but how can I define the opportunity?, anyway the defect is same kind and have a some standard(only compare with a sample about it’s dimension)

Does anybody can give me some suggestion? thanks

It looks like you are using the p-chart to try to evaluate this. But the data don’t support that, as you appear to be able to have multiple defects per opportunity. This does commonly come up in real-life - such as defects per item (with multiple failure opportunities within the item) or defects per 100 square yards of fabric. In these cases, you are better off with the c-chart (defects per unit) or the u-chart (defects per area of opportunity).

Certain types of defect calculations don’t lend themselves well to the p-chart or the ppm type calculations. For example, with defects per square yard of fabric - what is the individual opportunity for failure? A thread? A square milimeter? There is a continuim across the piece of fabric, and are there 1,000 opportunities for defect in a square yard, a million?

I hope this helps.

You can always reduce errors without calculating a “sigma score”. The “sigma score” works well when there are well-defined “opportunities” and well-defined defects. Also there has to be nom more than one possible defect per opportunity.

A perfect product with basically no defects would have a sigma score apporaching infinity.
A perfectly bad product with all defects would have a sigma score approaching - infinity.
It is mathematically impossible to calculate the sigma score with more than 1 defect per opportunity.

There are various ways to get around this. The simplest would be to forget sigma scores and simply concentrate on reducing errors.

If you really want a sigma score, then you would have to define some small “opportunity” on the surface and then see how many defects you had in that area. But as Steve said, that choice is somewhat arbitrary. Is is 1 cm^2? 1 mm^2? 1 m^2? 1 panel? Each choice will give you a different sigma score.