iSixSigma

Determining the samples size

Six Sigma – iSixSigma Forums Old Forums General Determining the samples size

Viewing 10 posts - 1 through 10 (of 10 total)
  • Author
    Posts
  • #35519

    Merry Jalisi
    Participant

    Hi my name is Merry,
    I am trying to seek some information and need your help with determining a right and sufficient sample size for a very important project that has been assigned to me at work, since it has been while and I no longer remember much of my statistic,  I was wondering if I could get some guidance from the pro’s.
     
     I have been searching and still have not been able to find any specific information regarding the sample size requirements for inspecting lots of  incoming raw materials at my work.The problem that I’m having is as follow:
    Recently we received  a lot size of (100,000 pieces) of a injection molding assembled parts.These assembled fitments, are make of 12 cavity molds caps and 16 cavity spouts.so the combination of the fitment is 12*16=192 possible combination.
    I need to determine the sample size for this lot which would cover the 192 possible combination of parts.
    I need to know what is the right sample size for this project,
    and how to determine that? And what formula to use to achieve the right sample size?
    Please help me.
    Thank you,
    Mery

    0
    #100136

    indresh
    Participant

    assuming the 192 possible combinations are equally spread in 100000
    the there are 520 unique samples of each type
    to have a good sample (95% confidence and 4% margin error) required sample size per unique combination comes as 279 so for 192 it would be 192*279 = 53568 odd sample
    i suppose 95 % confidence with 10% margin error would also be quite appropriate and that would give you sample per unique combination as 81 so total sample to have all unique combinations =81*192= 15552
    would be eager to know from some expert if this method is valid
    rgds
     

    0
    #100172

    Merry Jalisi
    Participant

    Thank you so much for the help you have been giving me. I just could not understand how you came up with the 279? I don’t remember the formula for Margin error? could you please advise.
    I also though that there is a direct relation between margin error and confidence level, so when you increased the margin to 10%, how could we have 95% confidence.
    Thanks
    Merry

    0
    #100249

    indresh
    Participant

    worked this way
    assuming every comination is equally spread in 100000
    unique units in 100000 (192 possible combinations) =100000/192 = 521
    now all such sets of 521 (population of unique combination) you need to get a sample so at 95% confidence sample size is 279
    since there are 192 sets of 521 to have valid sample size for each 192 combination it would be 279*192 = 53568
     

    0
    #100252

    Mohandas
    Participant

    dear indresh,
    as far as i know, the formula for determining the sample size is given by,
    n = Z^2 * p* (1-p)
          E ^2
    where Z is the Z intercept of the confidence interval and p the proportion defective whilst E is the error. using the above formula, I don seem to get how u arrived at 279 ? Did u use any other formula ?
    Deepa

    0
    #100260

    indresh
    Participant

    dear deepa,
    we are not having any defective parts here
    we are trying to get the correct sample size indicative of the population at 95% confidence. At the mentioned samples we are more than 95% confident that the required sample is a good indicative sample
    and since we want to include all possible 192 combinations, each such combination has a population of 521 in the lot of 100000
    so indicative sample of each possible combination from ITS population of 521 is 279 (refer sample size calculator at this site)
    can mail me at [email protected] for further clarifications
    rgds

    0
    #100263

    Scott
    Member

    Is the power of statistics in solving problem being put to a poor use? 
    MOST IMPORTANT:  The job of determining which cavity a part came from should have already been done.  There should be method, number of dots, a number 1 to 12, etc that identifies which cavity the parts came from.   Now, the problem is less of a statistics problem than what I believe has been addressed the responses.  After separating the parts based upon which cavity they came from, how many parts do you have to test to receive a level of confidence that there are no issues with the parts. 
    SECOND MOST IMPORTANT:  The molder, in general, should be able to provide help to solve this problem.  Rather than mixing the output from the different cavities, they should place the output from each cavity into a separate container.  Now, if you find an issue with the way the parts fit or function, you now know which cavity the part came from. 
    If the original experiment fails to match up the different parts, you may not discover the issue until it is too late.  If the original experiment finds a problem, you still have to mold parts an have them separated at the time of molding by cavity. 
     
     
       

    0
    #100281

    Merry Jalisi
    Participant

    Rick,
    No I don’t think in this case the power of statistics in solving problem being put to a poor use.
    I agree with your response we need to to find out and determining which cavity a part came from already has been done. The Cap came from 12 cavity mold which has been identified on the parts, cavity 1, cavity 2,….(1-12) and the spouts came from 16 cavity mold which is number 1-16.  The problem that I have is that our vedor assambles these parts so when I recieved them they no longer called caps or spouts they are fitment pre assambled parts in the box, if I have to go throught the box and try to figer out what cavity has been used it will take me about a month just to get a samples of all the combinations.
    That is where I think statistics can help me calculate all the posiblity of combination in the 100,000 parts and help me come up with a right sample size which will give me 95% confidence leve. If  I have to go through 100,000 assambled parts, and  After seaprating the parts based upon which cavity they came from would be impossible.
    Thank you.

    0
    #100301

    Mery
    Participant

    Hi,
    First of all I would like to thank you for replying in response to the questions I had. I am still not clear with 279, I don’t understand how to calculated? Another concern on mine is to inspect such a large sample size of in according to you I calculation I have to inspect almost 50% of my population to achieve 95% confidence level?
    This just sounds confusing and not practical. Could you please advise.
    Thanks,
    Mery

    0
    #100302

    Darth
    Participant

    Did you include the Finite Population Correction factor in your calculations.  For such a small population, it might be important and might affect the calculated sample size.

    0
Viewing 10 posts - 1 through 10 (of 10 total)

The forum ‘General’ is closed to new topics and replies.