NEED HELP FINDING THE UCL LCL
Six Sigma – iSixSigma › Forums › Old Forums › Healthcare › NEED HELP FINDING THE UCL LCL
 This topic has 8 replies, 3 voices, and was last updated 13 years, 6 months ago by Robert Butler.

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February 2, 2009 at 5:05 am #25724
colettaParticipant@coletta Include @coletta in your post and this person will
be notified via email.Wait time (minutes) in order of occurrence
1
102
173
294
395
556
647
288
69
510
311
3912
4613
3514
3015
616
3217
3318
1119
2020
1321
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7328
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9
am hitting a brick wall i need to find the UCL FOR THE RANGE AND THE INDIVIDUAL & LCL FOR THE INDIVIDUAL. I got the average of 28.33 and the range of 700February 11, 2009 at 2:13 pm #62256It looks like we are doing the same health care project. Did you get an answer?
0February 12, 2009 at 7:03 pm #62267
colettaParticipant@coletta Include @coletta in your post and this person will
be notified via email.yes wow really
0February 12, 2009 at 7:14 pm #62269would you mind helping me out?
0February 12, 2009 at 7:19 pm #62270
colettaParticipant@coletta Include @coletta in your post and this person will
be notified via email.we can talk off this board if that is ok with you is your professor name starts with K
0February 12, 2009 at 7:19 pm #62271
Robert ButlerParticipant@rbutler Include @rbutler in your post and this person will
be notified via email.Tony, coletta – check this post.
http://healthcare.isixsigma.com/forum/showmessage.asp?messageID=44400February 12, 2009 at 7:26 pm #62272here is my email: t[email protected] shoot me an email and we can take it from there. No we have different instructors. most likely a different school as well. thank you so much. I am at my zero hour..
0February 12, 2009 at 7:30 pm #62273
colettaParticipant@coletta Include @coletta in your post and this person will
be notified via email.i got the answer already thank u so much you are really good right now am trying to work out another problem finding the lowest and upper expectation variation for a mean of 21.193548 and standard deviation of 6.901301 can u help in the steps
0February 12, 2009 at 8:15 pm #62274
Robert ButlerParticipant@rbutler Include @rbutler in your post and this person will
be notified via email.Since it sounds like the same set of data then everything I said in the message I referred to applies and the steps are exactly the same except now you are looking at the spread, not of individuals, but of samples of means from that same population. The question is how many data points would go into another sample mean from the population from which you drew the 31 measures you do have?
Usually, in problems of this type, it is assumed you would go out and get a new sample of 31 measures. Therefore, instead of 1 under the square root you would now have 31. Put that number in the equation and you will have your estimates.0 
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