Relating Type I and Type II Error to C-charts

#1
Hello,

I am trying to figure out how to relate type I and type II error calcuations to c-charts.

Ive already calculated the 3 sigma control limits (UCL, LCL and CL) for my data.

My goal is to calculate the "Z" statistic similar to this" z=x-mean/std dev.

Is there a formula i am missing? Thanks for the help!
 

Miner

TS Contributor
#2
c Charts monitor the number of defects per unit area. They are used for count data, and are based on the Poisson distribution. If the counts are large enough, you could use the normal approximation for the Poisson distribution allowing you to calculate a Z statistic.
 

noetsi

Fortran must die
#3
I rarely work with six sigma, but wouldn't the normal ways of calculating alpha (essentially deciding what this is most commonly .05 which then gives you the chance of a type 1 error) and power allow you to determine type I and type II error rates regardless of the chart?
 
#4
I tried using the poisson approximation for these types of questions but am still a little unsure of their outcome.

The approximation basically uses: Z=Xa-lamda/SQRT(lamda), where:

lamda = mean or C bar
SQRT lambda = standard deviation of the c-chart
Xa=Cbar-.5 or Cbar+.5 as appropriate.

What are your thoughts on using the above?

Here is one of the questions: Using c charts, you inspect 40 units and find 85 imperfections, assuming the processing is in statistical control calculate:
3 sigma control limits
Prob of type I error, and prob of type 2 error should the mean shift to 5.2.

I calculate my C bar to be 2.125., UCL=6.50, LCL=0 (since <0).

For prob of type one should i use Z=((6.5+.5)-2.125)/SQRT(2.125)=3.34 or Z=((6.5-.5)-2.125)/SQRT(2.125)=2.65.

For prob of type two, should i use Z=((6.5+.5)-5.2)/SQRT(2.125)=1.23 or Z=((6.5-.5)-5.2)/SQRT(2.125)=.548.

Thank you!
 

Miner

TS Contributor
#5
I rarely work with six sigma, but wouldn't the normal ways of calculating alpha (essentially deciding what this is most commonly .05 which then gives you the chance of a type 1 error) and power allow you to determine type I and type II error rates regardless of the chart?
If the counts are large enough for the normal approximation to the Poisson (or to the Binomial depending on the chart), then yes, you can do exactly what you propose. Otherwise, no.
 

Miner

TS Contributor
#6
I tried using the poisson approximation for these types of questions but am still a little unsure of their outcome.

The approximation basically uses: Z=Xa-lamda/SQRT(lamda), where:

lamda = mean or C bar
SQRT lambda = standard deviation of the c-chart
Xa=Cbar-.5 or Cbar+.5 as appropriate.

What are your thoughts on using the above?

Here is one of the questions: Using c charts, you inspect 40 units and find 85 imperfections, assuming the processing is in statistical control calculate:
3 sigma control limits
Prob of type I error, and prob of type 2 error should the mean shift to 5.2.

I calculate my C bar to be 2.125., UCL=6.50, LCL=0 (since <0).

For prob of type one should i use Z=((6.5+.5)-2.125)/SQRT(2.125)=3.34 or Z=((6.5-.5)-2.125)/SQRT(2.125)=2.65.

For prob of type two, should i use Z=((6.5+.5)-5.2)/SQRT(2.125)=1.23 or Z=((6.5-.5)-5.2)/SQRT(2.125)=.548.

Thank you!
This is a link to a Six Sigma Tutorial on this topic that should help.
 

BGM

TS Contributor
#7
A side question: I have just see the some definitions of those control charts. I understand that these charts typically use a symmetric confidence/control limits, constructed based on normal approximation.

But as C-chart is used to monitor counts data, and as what Miner said, one is using Poisson distribution to model these counts, why not construct an exact confidence interval for the mean \( \lambda \)? In this way it will no longer be a symmetric pair of limits.
 

Miner

TS Contributor
#8
A side question: I have just see the some definitions of those control charts. I understand that these charts typically use a symmetric confidence/control limits, constructed based on normal approximation.

But as C-chart is used to monitor counts data, and as what Miner said, one is using Poisson distribution to model these counts, why not construct an exact confidence interval for the mean \( \lambda \)? In this way it will no longer be a symmetric pair of limits.
The history of industrial control charts is pretty interesting. One thing that can be confusing to those outside of industrial statistics is the fact that control limits were not derived based on probabilities or on the statistical concept of a confidence interval. They were derived empirically and based on an economic balance between the cost of missing a process signal and the cost of reacting to a false alarm. Based on statistics and probability, one would have expected 95% confidence (~ 2 standard deviations) limits. However, the cost of reacting to false alarms made them settle on 3 standard deviation limits.

Regarding control charts for attributes. At lower defect rates, the lower control limits are often set at zero. Once you reach a level where the lower control limit is greater than zero, you have reached a point where the normal approximation kicks in and the control limits become symmetrical.
 

noetsi

Fortran must die
#9
I often have wondered in reading six sigma if they were derived from stats, because some of the equations seem to contradict statistics outside six sigma. Now I understand better.
 

Miner

TS Contributor
#10
One of the areas that ignites some furious debate is the concept of the 1.5 sigma shift. Here are two articles that discuss the origin of the concept and a possible basis for it.

My own position on the issue, is that I know from personal experience that manufacturing processes do drift from the target value, even when under statistical process control (SPC). However, the magnitude of that drift varies greatly depending on the particular process and the use of automated process controls. 1.5 sigma is as good an estimate as any if you have little process knowledge, but if you do have process knowledge, you should use that instead.