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。