【www.rzshzz.com--工作总结】
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。