現在，我們將從第一張表格中得到的具有兩面規格的cp的最小需求量設定為1.33，假設測試的問題就將變為：
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，我們把高的值設定為2，其實際的質量水平為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, 就示範能力而言,供應者定會提供一個 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的最小程式能力需求。需求愈高,cp(高度)/cp(低點)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式能力在以一個大小 30 ≤ n ≤ 50個部份的樣品為基礎的 cp ≥ 1.33 的水平上。清楚地，這個程式不涉及到在σ的估算中考慮樣本的不同，同時，樣本尺寸的值不斷變大在實踐中是很必要的。