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Determine Poplation

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  • #47918

    Don Omar
    Participant

    For years I have been using a plus or minus 3 sigma to esimate the populatoin.  (ie. get the mean of the samples and add 3 sigma to cclculate the upper limit and subtract 3 sigma for the low limit.)  Is this the correct of doing this?  Most of the statistical calculations deal with a confidence interval for the mean or the standard deviation of the population.  How can I use my data to estimate the variation and mean of the population?  Should I take the confidence interval of the mean and add three times the confidence interval for sigma?
    For example if I get thirty samples and my mean is 69 and my standard deviation is 1.6, how does this reflect what I can expect from the population?  Normally I would say 99.7% of my population will fall within 69 plus or minus 4.8.  Is this correct?
     
    Thanks,
     
     
     

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    #160312

    z calculation
    Member

    Here are your calculations:

    Standard Deviation
    SQRT (N)
    sdev/sqrt(n)
    z at 99%
    Add. Subtr.
    Mean
    69 +
    69 –

    1.6
    5.477226
    0.292118697
    2.55
    0.744902678
    69
    69.7449
    68.2551

    N
    30

    Intervale at 3 sigma caluclaitons = 69.75 (lower bound – 68.25 (upper bound.
    Formula: Mean (69) +/- z alpha/2 at 3 sigma = 2.55 times standard devaition (1.6) )=/square root of n.(30)

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    #160325

    Don Omar
    Participant

    Thanks for this doesn’t work out well.  The upper and lower limits are between some of the data points of my sample test data.  This would allow many data points in the population to fall outside the LSL and USL.  Ideally I think that the USL and LSL should be wider for my test data.  Should I account for the variation in the mean?  If I go back to the sigma ± 3sigma I can get all of my data within the LSL and USL.
     
    Thanks,
     

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    #160327

    Anand
    Participant

    Hi Don
    Probably you are trying to calculate the wrong thing. USL & LSL are specification limits which are defined by the customer. There is no way you can calculate them from data.
    What you can calculate in Control limits, which depending on the type of data – different formulas are available. But you can still say that they are at a difference of +/- 3 Sigma from the mean.
    Anand

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    #160328

    Don Omar
    Participant

    Thanks Anand,
    I am the customer trying to change the ULS and LSL based on the data.  The specification that we currently have is too tight and the supplier won’t be able to meet it.  I am trying to se what is a realistic specification based on the parts that we get, which are considered acceptable.
    Does the ± 3sigma applies here?

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    #160329

    Anand
    Participant

    Hi Don
    Unfortunately +/- 3 sigma doesnt apply over here. Your specification limit for the part id defined by the design, just check that relaxing it should not result in the failure of the entire assembly. Check with your R&D dept if the specification limits can be relaxed and if yes – by what extent. If it cannot be relaxed – You should look for another supplier, there would be plenty of them willing the parts to you at your specifications.
    Anand

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    #160342

    accrington
    Participant

    Don
    Is your current specification besed on fit and function, or customer requirements? If it’s not you could be batting on a sticky wicket. If your specs are not based on reality, you can end up with a lot of hassle, and increased costs for your company. 
    In my own organisation, I have been asked many times whether similar exercises can be carried out. My first response is “what are the specs for?” The usual reply is “so we know what to reject for?” My next response is “why do want to reject anything?” To which the reply is “because it’s out of spec”. A somewhat circular argument, I think.
    My next question is “1. What spec does the customer want?” and/or “2.What spec do you need to run the process cheaper, faster and better?”  and or “3.What spec do you need to make the product work, or not fall apart?” These questions are usually met with dazed and confused expressions. After the questioner has recovered from these bombshells of questions, I will politely decline to do what was asked, and tell them to come back when the can answer questions 1, 2 and 3, then, and only then, we can set sensible specifications.
    To think that some people think I’m a grumpy old f**t, when I’m only trying to help. (note the asterisked rude word, and apolitical/ non – religious content of this post)
     

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    #160365

    annon
    Participant

    D,
    If you are the customer determining the necessary specification for a given product characteristic, why is this not being determined by some rational means, ie tolenarnce stacking or a significant drop in customer satisfaction or perforance at a given level of product perforance?
    Sorry…just not understanding….tell us what your objective is?  What do you want to accomplish?  Determine a meaningful spec for a given characteristic for a specific part to be used in additional fabrication and or assembly?

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    #160366

    Don Omar
    Participant

    Ok,
    Is not always necessary to define the tolerance by calculations.
    I have worked in many injection molded tools and we define the specifications based on the design requirements of the critical areas.  (ie.  most critical areas have a .003-.005″ tolerance by design).  Being a complex part, there will be some areas where the dimension will not be as critical to the design.  Originally, these areas are controlled by a general tolerance.  When the parts are molded, sometimes these dimensions are not up to the original “general tolerance”.  Since the dimension is not critical, I would take some parts and measure what the parts are measuring from the tool in the actual process.  If this is OK by design, I will go back and change the print based on the capability of the tool.  This is where I adjust the USL and LSL to allow some level of control in the tolerance, but prevent quality from rejecting “good” parts.  Does that make sense?
    Thanks,
     

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