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UCL and LCL !!
Message posted by Poileak Le (via 199.128.53.57) on February 8, 2002 at 11:11 AM (ET)
How do I calculate the UCL and LCL of the following value 5, 15, 7, 10, 2 with a subgroup size of 1 ???
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Re: UCL and LCL !!
Message posted by Darius (via 200.23.217.10) on February 8, 2002 at 12:29 PM (ET)
With
value 5, 15, 7, 10, 2 subgroup=1Calculate the average
IXA = [5+15+7+10+2]/5
calculate the moving range average
MRA= [abs(15-5)+abs(7-15)+abs(10-7)+abs(2-10)]/4
UCLmr = 3.267*MRA
LCLmr = 0
UCLix = IXA + 2.66 * MRA
LCLix = IXA - 2.66 * MRA
If no coorelation between samples (autocorrelation) exists.
Check
http://www.qualitymag.com/articles/jan98/0198wh.html
Re: UCL and LCL !!
Message posted by JG (via 152.163.213.59) on February 9, 2002 at 5:01 AM (ET)
I think this answer is too complicated - I think the standard deviation is being estimated from the range, etc. Assuming that you want 99% control limits - first calculate the mean and SD for your data - then UCL=mean+2.58*SD, LCL=mean-2.58*SD, then do either a cumualtive average and SD or moving average and SD - dropping the first observation and adding a new observation, etc.
Re: UCL and LCL !!
Message posted by Darius (via 200.23.217.10) on February 11, 2002 at 12:32 PM (ET)
"Either a cumualtive average and SD or moving average and SD - dropping the first observation and adding a new observation, etc."For SPC,It愀 wrong.
The only way is with the within_sample variation, not the total_variation nor the between_sample_variation. The within_sample_variation is less affected by the outliers than the other two, and it was also recommended by Shewhart himself.
Re: UCL and LCL !!
Message posted by JG (via 64.12.106.34) on February 11, 2002 at 10:27 PM (ET)
You may be right, but I feel that the cumulative average should always be calculated as a way of studying what the system if doing over time and to help decide when enough data has been obtained.
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