[求助]求SPC高手進來看看 [复制链接]

我不記得了，我的資料好像也是在這論壇下的。如果是SPC新手下載,可以學習普通概念。
現在我上傳上來是求高手幫忙看看這份資料，里面有我的紅字問題,希望高手指點迷津。
大概問題是sigma有几種的求法？不同sigma求法各在什么情況下用？CPK的問題同上。

原帖由0楼楼主 roverxie2008 于2009-02-25 13:36发表
我不記得了，我的資料好像也是在這論壇下的。如果是SPC新手下載,可以學習普通概念。
現在我上傳上來是求高手幫忙看看這份資料，里面有我的紅字問題,希望高手指點迷津。
大概問題是sigma有几種的求法？不同sigma求法各在什么情況下用？CPK的問題同上。

P.8
n-1 > DOF(Degree of freedom) =1

P.69
S = Sigma

P.106 & P.107
XBar Bar = Average from XBar

P.109
Use n-1, because DOF = 1

原帖由1楼楼主 nestor 于2009-02-25 15:26发表
P.8
n-1 > DOF(Degree of freedom) =1
P.69
.......

CP=[USL-LSL]/6Sigma, CPK= /1-Ca/*CP,CPK=Min{CPU,CPL}而
sigma1=Rbar/d2   取樣數=2時，d2=1.128
sigma2的平方=[(X1-xbar)的平方+(X2-Xbar)的平方+...（Xn-Xbar）的平方]/n
sigma3的平方=[(X1-xbar)的平方+(X2-Xbar)的平方+...（Xn-Xbar）的平方]/n-1
以上可以得到3個Sigma
  1.我想知道在哪種情況下用哪個Sigma.
  2.CPK有單邊的和雙邊的，什么情況下用單邊，什么情況下用雙邊？
3.I am so sorry for i have no idea about DOF.
  4.我上傳資料P106&P107中提到的問題是CPU=[USL-Xbar的平均數]/3sigma ,還是CPU=[USL-Xbar]/3sigma?
  5.P109提到的問題，你說Use n-1, because DOF = 1。我不清楚什么時候DOF=1，什么時候在P109問題中use n？

anyway ,nestor,great thank you for helping ！我夠傻的吧，i am looking forward to your reply.

原帖由2楼楼主 roverxie2008 于2009-02-25 18:07发表
CPK=[USL-LSL]/6Sigma, CPK=Min{CPU,CPL}而
.......

CPK=[USL-LSL]/6Sigma, CPK=Min{CPU,CPL}而
sigma1=Rbar/d2   取樣數=2時，d2=1.128
sigma2的平方=[(X1-xbar)的平方+(X2-Xbar)的平方+...（Xn-Xbar）的平方]/n
sigma3的平方=[(X1-xbar)的平方+(X2-Xbar)的平方+...（Xn-Xbar）的平方]/n-1
以上可以得到3個Sigma
1.我想知道在哪種情況下用哪個Sigma.

For Averages:

UCL Xbar = Xbarbar +A2 Rbar

LCL Xbar = Xbarbar - A2 Rbar



Standard deviation or Sigma for the process:

Sigma = Rbar / d2

For Range:

UCL R = D4 Rbar

LCL R = D3 Rbar


For indivduals::


UCLx = Xbarbar + 3(Rbar/d2)

LCLx = Xbarbar - 3(Rbar/d2)




2.CPK有單邊的和雙邊的，什么情況下用單邊，什么情況下用雙邊？

The preceding analysis of the process capability and the specification was made utilizing an upper and a lower specification. Many times there is only one specification and it may be either the upper or lower. A similar and much simpler analysis would be for a single specification limit. (上述分析的过程能力及规范了利用上层和较低的规范。很多时候只有一个规格，它可以是在上限或下限。类似的和更简单的分析将是一个单一的规格限制。)

The true prcoess capabiility can't be determined until the Xbar and R charts have achieved the optimal quality improvement without a substantial investment for new equipment or equipment modification. CPk = 6 Sigma0 when the process is in statistical control. (真正的加工能力尚不能确定，直到均值和R图表都取得了最优质量的改善没有大量投资新设备或设备改造。CPk= 6西格玛0时，在统计过程控制规范，它可能是在上限或下限。类似的和更简单的分析将是一个单一的规格限制。)

Case I
6 Sigma< USL - LSL
This situation (CP) is less than the tolerance (USL - LSL).

Case II
6 Sigma = USL -LSL
This case, the CP is equal to the tolerance.

Case III
6 Sigma >USL - LSL
When the CP is greater than the tolerance, an undesirable situation exists.



3.I am so sorry for i have no idea about DOF？
- If you have learned T-test / Student Test , you will have a better picture about Degree of Freedom.
e.g.
When DOF =1, thus n-1

**Whereas, Single sample Mean. Standard Deviation of population Unknown.
- Or, 2 Mean Equal Variance t Test
e.g.
When DOF =2, thus n1 + n2 -2

***Advance level will be DOE stage for DOF

4.我上傳資料P106&P107中提到的問題是CPU=[USL-Xbar的平均數]/3sigma ,還是CPU=[USL-Xbar]/3sigma？
***Provided the process is centered:
Cpk = Min {USL- X bar) or (Xbar - LSL)} / 3 Sigma


5.P109提到的問題，你說Use n-1, because DOF = 1。我不清楚什么時候DOF=1，什么時候在P109問題中use n？

Use "n" = Standard Deviation of population is known

Use "n-1" = Standard Deviation of population is unknown

Dear nestor:
  I have received your patient reply,you are appreciated,but I still can't folow you for my stupidity,so I have to interrupt you again.I have comunicated with you via email ,looking forward your advice.

Please check your email..cheers!