抽样统计知识

首先，你必须确定这些数据是否为这个小组或者个体样品计算所提供的手段。如果它们为个体样品(我猜测这个正是你们在谈论的情形)，这些数据的标准偏差在允许的范围内被估计，其和数据的序列有关系。所以，如果你改变数据的序列，你将会得到不同的标准偏差。同时，如此不同的 CPK会给不同的规格提供修正。
 
Second, you have to understand thoroughly what Cpk is all about. It’s a process capability ratiCpk=min{Cpl,Cpu}. It shows how well the process is centered on the target comparing with Cp.
So generally people use Cp and Cpk togather trying to figure out the process capability.
Furthermore, there are cases where process capability is low but the process is in control,
and there are cases where the process is out of control but the process capability is comparative high.
These are all related to the variance of the process and how well the process is targeted.
There are lots of misleadings in the use of the process capability ratios in the industries.CP
 
其次，你必须全面理解什么是CPK。 它是一个加工能力比率Cpk=min{Cpl,Cpu}，其显示出该能力是目标中心并优于CP。因此，人们通常将CP和CPK一起使用，并尝试理解其加工能力。此外，这里有很多情形就是加工能力低但加工处于控制之中，而加工处于控制之外时加工能力则相对较高。他们与加工的方方面面联系，同时加工有很强的目的性。在工业CP中使用加工能力的过程中出现了许多误导的情形。
 
In some industries, such as auto industry, people call the calculation of Cpk as Ppk.
As to why people use 32 or more data to calculate Cpk, I did a little research about it. In the industry,
people accept Cp 1.33 as a commom sense for existing process which corresponds to 4 sigma variance level.
If you use this date to do a little calculatiuon and check the table published by Quality Society of America
( I was trying to post that table before, but it didn’t work. It was all messy. I guess the admin deleted that post), you will get the number approximately 32.
But even 32 is not enough sometimes to get a unbiased estimation of the process capability ratio.
现在，我们将从第一张表格中得到的具有两面规格的CP的最小需求量设置为1.33，假设测试的问题就将变为： H0: Cp= 1.33
H1: Cp≥ 1.33
Now we want to be sure, at the 95% confidence level, that the process capability is bigger or lower than 1.33
before we accept or reject it. And we set the high value as 2, which is actually 6-sigma quality level. Namely, Cp(high)=2, Cp(low)=1.33 , α =β=1-0.95=0.05.
 
目前，在信度为95%的水平下，我们通过加工能力值的高1。33或低1。33来确定是接受还是否定。同时，我们把高的值设定为2，其实际的质量水平为6-Σ，即为Cp(high)=2, Cp(low)=1.33 , α =β=1-0.95=0.05.
Cp(high)/Cp(low)=2/1.33=1.504
Then check the table, the corresponding sample size is about n=32. And 接下来核对该表，对应的样品大小为n=32
C/Cp(low)= 1.2
So, C= 1.2Cp(low)=1.21.33=1.6
Thus, to demonstrate the capability, the supplier must take a sample of n=32,
and the sample process capability ratio must exceed C=1.6.
This is obtained using minimum process capability requirement in the industry.
The higher the requirements, the smaller the Cp(high)/Cp(low) value will be. From the second table we know that the required sample sizes are increasing. It’s fairly common practice to accept the process as capable at the level Cp≥ 1.33 based on a sample of size
30≤n≤50 parts. Clearly, this procedure does not account for sampling variation in the estimate of sigma,
and larger values of sample size may be necessary in practice.
因此, 就示范能力而言,供应者定会提供一个 n=32 的样品，而且样品加工能力比一定超过 C=1.6。这被视为获得到使用工业的最小程序能力需求。需求愈高,Cp(高度)/Cp(低点)的比值愈小。从第二张表格中我们知道必需的样品尺寸正在逐渐增加。公平而常见的做法是接受程序能力在以一个大小 30 ≤ n ≤ 50个部份的样品为基础的 Cp ≥ 1.33 的水平上。清楚地，这个程序不涉及到在Σ的估算中考虑样本的不同，同时，样本尺寸的值不断变大在实践中是很必要的。