How to Calculate Sigma without UCL
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 This topic has 6 replies, 4 voices, and was last updated 9 months, 3 weeks ago by MBBinWI.

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November 3, 2020 at 3:44 am #250668
weasle23Participant@weasle23 Include @weasle23 in your post and this person will
be notified via email.Can someone walk me through this problem?
Suppose that:
1) The lower specification limit (tolerance) for a particular output is 330mm
2) The specification width for this output is 10.3mm
3) This output’s process mean is “centered” within the specification interval
If our goal is a process that has a Cp that is at least 1.333,
what is the largest value for σ that would be acceptable in this situation? This topic was modified 10 months, 3 weeks ago by weasle23.
 This topic was modified 10 months, 3 weeks ago by Katie Barry.
0November 3, 2020 at 9:17 am #250674
Robert ButlerParticipant@rbutler Include @rbutler in your post and this person will
be notified via email.Obviously the problem is assuming a normal distribution of the data.
The basic formula for Cp in this case is
Cp = (USLLSL/(6*std)
So, LSL = 330, USL = 330+10.3 and you want a minimum of 1.33 for Cp – therefore plug in the numbers in the above equation, rearrange and solve for std.
0November 3, 2020 at 12:10 pm #250675
weasle23Participant@weasle23 Include @weasle23 in your post and this person will
be notified via email.I was able to get that far. I wouldn’t even have to find USL because the specification width was given. Still, for the sake of learning I will fill this out and reengineer this here to the best of my ability. I have the correct answer so I know i’m wrong anyways.
(340.3330)/6SD = 1.33
10.3/6(SD) = 1.33
10.3/(SD) = 6(1.33)
10.3/(SD) = 7.98
And I do not know how to go any further than that.
 This reply was modified 10 months, 3 weeks ago by weasle23.
0November 3, 2020 at 12:26 pm #250677
weasle23Participant@weasle23 Include @weasle23 in your post and this person will
be notified via email.I see what i did wrong, I should have put in more 3’s into the CP value
1.333*6 = 7.99998
10.3/7.99998 = 1.2875
which is the correct answer
0November 3, 2020 at 2:09 pm #250679
Robert ButlerParticipant@rbutler Include @rbutler in your post and this person will
be notified via email.It sounds like you are taking some kind of exam or course or something and trying to match whatever is offered on a multiple choice problem – this is fine but there is something you should keep in mind when you are faced with questions like this in the real world and that is the issue of significant digits.
It is possible to get measurements out to the 100,000th place, or almost any other place for that matter, but when it comes to calculations the rule is the precision of your final result can be no more precise than the precision of the least precise term used in the calculation. In this case (if we assume 330 mm has actually been measured to 330.0 mm) then when we combine this information with 10.3 this means the final value for sigma should be 1.3 – anything else is just empty precision.
0November 10, 2020 at 7:21 pm #250786
Chris SeiderParticipant@cseider Include @cseider in your post and this person will
be notified via email.December 1, 2020 at 7:56 pm #251150 
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