质量复习---抽样统计知识四(4)
发布于:2009-4-24 已被阅读: 次 
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。

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