First, you have to make sure whether these data are the means of the subgroups or individual samples. If they are individual samples(I guess this is the case you are talking about), the standard deviation of the data are estimated by the moving range, which is related to the sequence of the data. So if you change the sequence of the data, you'll get different standard deviation and thus different Cpk given the process specifications are fixed.

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.

在诸如汽车业的一些产业中，人们将对CPK的计算称作PPK。至于人们为什么用32或者更多的数据来计算CPK，我对此做了一些研究。人们在运算中视cp1。33为普通理解与当前能力与4sigma的离差保持一致。如果你用这个数据做一些计算然后对照美国质量出版社出版的表格。（我曾尝试着邮寄那张表，但都没有成行。这简直太糟了，我猜想管理部门遗失了该邮件）。你可以取值接近32，但即使32有时候也不足以得到一个没有误差的加工能力比率

What I wanna stress again is that capability ratio is not everything, there are too many misuses in the industry, don't count all on it.我想再一次强调的是加工能力比率并不是万能的，在工业上有很多的误用，不要全部依靠它来计算。

Here is my answer to the question of 32 sample size:这里是我对样本尺寸为32的问题的回答。

A practice that is increasingly common in industry is to require a supplier to demonstrate process capability as part of the contractual agreement. Thus, it is frequently necessary to prove that the process capability ratio Cp meets or exceeds some particular target value---say, Cp0. This problem may be formulated as a hypothesis testing problem:

一个要在工业中日渐成熟的练习是需要一个供应者示范如契约的协议部份般的程序能力。 因此，有必要经常证明加工能力比率CP等于或者超过如CP0的一些特殊目标价值。这个问题可能被制定为一个假设的测试问题:

H0: Cp= Cp0 (or the process is not capable)

H1: Cp≥ Cp0 (or the process is capable)

We would like to reject H0 (recall that in statistical hypothesis testing rejection of Null hypothesis is always a strong conclusion), thereby demonstrating that the process is capable. We can formulate the statistical test in terms of Cp’, so that we will reject H0 if Cp’ exceeds a critical value C.

我们想要否定H0( 取消对统计的假设中无效力假设的测试否定一直是一个强大的结论)。因此，示范加工是有能力的。我们可以根据 Cp' 制定统计的测试, 所以如果 Cp'超过一个关键的价值 C，那么我们会否定H0 。

Kane(1986) has investigated this test, and provide a table of sample sizes and critical values for C to assist in testing process capability. We may define Cp(High) as a process capability that we would like to accept with probability (1-α) and Cp(low) as a process capability that we’d like to reject with probability (1-β). Please refer to the table created by Kane and used by American Society for Quality Control.

凯恩 (1986) 已经调查这上述测试, 而且向C提供一张有样品大小和关键值的表给来协助测试的加工能力。就如我们喜欢接受(1-α)的可能性和CP（低）作为程序能力和否定(1-β)的可能性一样，我们可以将CP（高）定义为一个加工能力。请查阅凯恩所创建的并为美国社会质量控制所用的表格。

Now we take the minimum required Cp value from the first table for two-sided specifications, which is 1.33. thus, the hypothesis testing problem then becomes:

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= 1.2*Cp(low)=1.2*1.33=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。