現(xiàn)在，我們將從第一張表格中得到的具有兩面規(guī)格的CP的最小需求量設(shè)置為1.33，假設(shè)測試的問題就將變?yōu)椋篐0： 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，在信度為95%的水平下，我們通過加工能力值的高1.33或低1.33來確定是接受還是否定。同時(shí)，我們把高的值設(shè)定為2，其實(shí)際的質(zhì)量水平為6-Σ，即為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 接下來核對該表，對應(yīng)的樣品大小為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，供應(yīng)者定會提供一個(gè) n=32 的樣品，而且樣品加工能力比一定超過 C=1.6.這被視為獲得到使用工業(yè)的最小程序能力需求。需求愈高，Cp(高度)/Cp(低點(diǎn))的比值愈小。從第二張表格中我們知道必需的樣品尺寸正在逐漸增加。公平而常見的做法是接受程序能力在以一個(gè)大小 30 ≤ n ≤ 50個(gè)部份的樣品為基礎(chǔ)的 Cp ≥ 1.33 的水平上。清楚地，這個(gè)程序不涉及到在Σ的估算中考慮樣本的不同，同時(shí)，樣本尺寸的值不斷變大在實(shí)踐中是很必要的。http://www.Examw.com