In the FIAT standard there is a section for the calculation of the Capability when the number of the sample is very low.

Cp = (USL-LSL)/k*S

The values of k are as follows:


N k
3 14,82
4 11,69
5 10,28
6 9,50
7 8,99
8 8,63
9 8,36
10 8,15
11 7,97
12 7,84
13 7,72
14 7,62
15 7,54
16 7,46
17 7,38
18 7,33
19 7,27
20 7,24


Can someone explain me how this table is obtained?
What is the formula or the rationale behind it?

The standard says that k is calculated with a confidence level of 80%.

K = coverage factor.

"Coverage Factor" - the numerical factor used as a multiplier of the combined standard uncertainty to expand the uncertainty corresponding to a specific level of confidence.

"Expanded uncertainty" - the quantity defining an interval enveloping the analytical measurement that captures a large fraction of the distribution of analyte concentrations that could be attributable to the quantity measured. The combined standard uncertainty is multiplied by the coverage factor to calculate the expanded uncertainty.


In layman's terms:

As the number of samples increases, K decreases. The greater the sample size, the higher the level of confidence that your process capability is representative of what is actually occurring.

can samebody provide me with the formula for getting the values in the table?
thank you

Not exactly, but with a little math (a good aproximation), just playing with the numbers.

k = 5.765025341 + 10.72395707* (ln(SampleSize))^-1.808731309 :mg:

thank you and sorry if I bother you again.
I needed a formula just to understand how the people who wrote the standard obtained the values in the table.
I can't explain it exacly even if the sense is pretty clear.
Has somebody ever seen a table such that somewhere or it's the first time also for you?
bye

I hadn't seen such a table befor, but it makes sense. If you know the true population standard deviation. then
Cp = (USL-LSL)/ 6*sigma
If you don't know the true pop. st. dev., then you can estimate it using the sample st. dev., S. Since you will get a different value for the sample st dev for each sample you draw, you would have to increase the multiplier from 6 to some other number to be reaonably sure you aren't overstaing you capability.

That increased number is, of course, just the number from your list. As the number of samples increases, then the value should approach 6. In fact, Darius's equation agrees pretty well -- he gets 5.76 for the constant instead of 6, but that is pretty good for an ad hoc equation.

As for the source of the equation, there is a good possibility that it is from a Monte Carlo approach. Generate a set of data and find it's standard deviation. Reppeat lots of times. Look at the distribution of standard deviations and find the limits needed to include 80% of the values.

Tim F

Agree with Tim,

I also found that the "k" factor is related to the "c4" constant to obtain the Unbiased estimate for standard deviation (not as good correlation as the equation I obtained from your data but very good).

k = -58.01114179+ 10.72485594 * (6/c4[N])

Thank you to everybody.
I don't think there is a need for using Monte Carlo simulations: the equations we need can be obtained in an analitical way.
I found an equation on Montgomery "Statistical quality control" for the distribution of Cp (function of N and of the confidence level) and I think our k should be gotten from there.
I've attached an excel file in which, in white cells, you can find the equation.
I've found out that k/6 values are very close to the ones for a confidenze level 85%.
So why the standard says that k are calculated for a conf. level 80%?

I've found out that k/6 values are very close to the ones for a confidenze level 85%.
So why the standard says that k are calculated for a conf. level 80%?

Maybe the author of the standard had low self-esteem and just wasn't as confident as he should have been. :lmao:

Tim F

Good point. :lol:

According to Excel solver; 84.612212 %
But the errors are eteroscedastic (sorry about my english), do you like a better aproach? :confused: