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Risk of being wrong v sample size

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  • #46203

    Schuette
    Participant

    Hello,
    We designed 2 jigs that are used in a press that has 3 entry bays. The jig makes 16 parts at the same time. We would like to assess PRIOR to carrying out the trial the confidence we could have in the jig delivering results within specification, DEPENDING of the number of samples measured.
    I wonder if anyone could guide me towards relevant information that could assess the two following scenarios:
    Scenario 1. Keep entry bay the same, only the jig to change. Make 16 parts for each of the jig (total 32 parts). How do I calculate the required number of samples I need to measure so that the risk of being wrong from the number of sample measured is only say less than 10%? Would I need to take the same amount of samples from jig 1 experiment and jig 2 experiement. I presume yes and also that I would need to take random samples as well?
    Scenario 2. Run full DOE. Entry bays 1,2 and 3 and both jigs used. Total 96 parts. Same question as above.
    I would really appreciate your input.
    Many thanks
    Jim

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    #152364

    BTDT
    Participant

    Jim:You are asking for a calculation of beta. The calculation is unique for each statistical test and depends on alpha, the size of the difference you are looking for (delta) and the sample size. Research “Power of a test” or “beta” for the test you are performing.Cheers, BTDT

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    #152374

    Schuette
    Participant

    Hello BTDT and many thanks for your answer.
    I have read a bit regarding the power of a test as you suggested and would need a bit of clarification. Below detailed what I understand. Please correct me if I said anything not right.
    Say I want to assess that the thicknesses produced on the 16 parts made with the jig 1 are compliant to specifications (say target 200 microns +/- 50 microns). I should consider the null hypothesis that the parts made are compliant (what I want) and evaluate the risk of being wrong beta based on the number of samples tested out of a maximum of 16. I should want beta to be as little as possible less than 0.20 so that power is greater than 0.80. How would I proceed next to calculate the power (the chance that the mean obtained doesn’t deviate from Ho hypothesis) depending of the number of the parts’ thickness measured?
    So for 15 parts measured what would be the chance that the thickness average deviates more than 50 microns? What about 10, 8, 6?
    Any comments or links to relevant information would be greatly appreciated.
    Many thanks
    Jim
     
     

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    #152379

    Jim Shelor
    Participant

    Jim,
    When you say “16 parts at the same time” do you mean:
    1.  16 parts in sequence from one piece of input material placed in the inlet?
    2.  16 parts in parallel from one piece of input material placed in the inlet?
    Jim Shelor

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    #152390

    Schuette
    Participant

    Hello Jim,
    Answer 2. 16 parts in parallel. The jig can accommodate 16 discrete parts which are pressed at the same time. The 16 parts are independant from each other.
    By measuring 16/16 or only 14, 12, 10, 8 etc… parts what would be the risk of being wrong compared to target specifications. Shall I evaluate the power using a double sided test of hypothese on the mean with an unknown variance (assuming a normal distribution)? Say target thickness is 200 microns and spec are +/- 50microns.
    Jim

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    #152394

    Kevin S
    Participant

    I’m curious how you worked out the parts were independent because I have a similar problem on a drive shaft. What test did you use to prove independence?

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    #152395

    Schuette
    Participant

    Good point. I didn’t prove they are independent.
    I assumed they would be independent based on the fact that the pressure applied on each part for different thicknesses was pretty consistent within the jig and therefore made the assumption that for a variability in the thicknesses of the parts, it wouldn’t affect the other parts.
    Jim

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    #152397

    Jim Shelor
    Participant

    Jim,
    In this case, what you have is 16 different dies operated by the same press.  You cannot determine what you want to know from a single set of 16 parts.  What you have on a single operation is the first sample of parts made from 16 dies.  Each die could be (and probably is) slightly different from each of the other dies and will probably make parts with a slightly different mean and a slightly different variation.  In addition, the alignment/flex of the press and the exact length of your dies can invalidate your assumption that all dies get the same force from the press.
    One method of accomplishing what you really want to know (is each die capable of making parts meeting the specifications) is to run a set of 30 groups of 16 parts, making sure you carefully separate the parts from each die, and plot the parts from each die on an I, mR chart.  Then run a capability analysis of each of the 30 parts from each die.
    30 parts is the recommended standard for an I, mR chart.  You can run fewer parts but your risk of the answer being inaccurate increases with each part you delete.  Less than 30 parts will also affect the accuracy of your capability analysis.
    I do not remember the relationship of how many parts causes how much degradation in the answers from an I, mR chart and a capability analysis.  I am researching to try to find that information again.
    You also need to make sure you run a well controlled gage study on your measuring system just prior to the test.  You need to make sure the acccurcy, linearity, and R&R of your measurement are sufficient to provide you with accurate answers and an R&R that is fully capable of providing acceptable discrimination between readings to support a good analysis of the variation in the parts.
    I know you are thinking this is a long and expensive process.  It is.  But comparing the 16 parts from a single punch from your jig is like comparing the output of 16 different machines using 1 sample from each machine.  It is comparing apples and grapefruit.
    Jim Shelor

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    #152398

    Jim Shelor
    Participant

    Jim,
    Not to change the subject, but when you run your DOE, you are going to need to block by die.
    Jim Shelor

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    #152403

    Jim Shelor
    Participant

    Jim,
    The reason for needing 30 I, mR points and 25 X-bar, R points was a conclusion from Shewhart that the natural tendency of humans is to improve performance simply because the process is starting to get measured.  Therefore, the first few points on the control chart are likely not to be representative of the actual process.  30 points are needed to provide a statistically valid basis on which to draw good conclusions of process operation.
    Since the human part of your equation is small, you might be able to get away with 20 or so points for you chart and analysis.  Be aware however that the fewer points you use the less stable your average, control limits, and sigmas will be and less data means you will not define your distribution as well which affects the accuracy of your capability analysis.
    Jim Shelor

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    #152405

    Schuette
    Participant

    Hello Jim and many thanks for your useful comments.
    My problem though is that I have 2 new designed jigs (each can press 16 parts at the same time) and the press has 3 bays and need to validate them and especially be able to come up with a risk that they won’t work before scale up can be started. How would you go about it from the start? I should have defined my problem from the start!
    I have the possibility of doing 1 trial for each jig and need to come up with a risk depending of the number of samples tested from the 16 parts from each trial. If I test 8, 10, 12, 14, 16. Here I suppose the risk is based on “repeatability” instead of reproducibility of the jig. At the end there will be no confidence in the jig being reproducible because the jig will be tested only once for now and no confidence in using a different bay of the press. For the repeatability by measuring say 8 out of the 16 samples what is the risk that all parts don’t conform to the specification?
    Many thanks
    Jim

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    #152406

    Jim Shelor
    Participant

    If I understand your process correctly, you have 16 dies on a single press plate that punch 16 parts from your feed material.
    Is that correct?
    Jim

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    #152411

    Schuette
    Participant

    Basscially the jig concept consists of a bottom plate, a middle plate where the 16 parts are placed and a top plate. The assembly is then put in a press that compresses the parts together under pressure and temperature. The press has 3 bays. One of the jig would potential puncture the parts at the same time as being pressed. 1 factor 2 levels, 1 factor 3 levels for full DOE total 6 experiments for no replicates, 16 parts made per run in any case. How would you go about this? My problem is to understand and estimate the risk of being wrong depending of the number of samples measured and number of trials to be carried out to assess for a certain risk what I need to do. This is to verify the design before it gets scaled up.
    1. Test one bay only. Assume all bays the same. Total run 2, 1 run for each jig. Depending of the number of samples measured how confident can I be that the results obtained are indeed good and representative of the 16, or maybe better to ask what is the risk of being wrong if I measure say 8 samples out of the 16, for each of the 2 jigs?
    2. Evaluate with the 3 bays. Run full DOE. What is the risk of being wrong if I measure say 8/16 samples for each run. I believe that increase the number of samples tested would decrease the stdev by sqrt of n but how does that translate in risk?
    Any help or direction would be appreciated. Many thanks.
    Jim

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    #152414

    Ashman
    Member

    Chapter 6 of Dr Wheeler’s “Advanced Topics in SPC” has a very detailed analysis of almost precisely this situation.

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    #152418

    Omeron
    Participant

    I hope you’re getting a cut of his sales, or are you really Dr. Wheeler :-)

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    #152434

    Jim Shelor
    Participant

    Jim,
     
    As a result of your last reply I am a little confused about when you want to do the measurement proof.  Before or during the DOE.
     
    Assuming you mean doing 1 set of 16 from jig 1 ad one set of 16 from jig 2, here is what I would do.  There are some that may disagree.
     
    Get the measuring device calibrated and make sure you know the bias and linearity for the device across the range you will use it.
     
    Make 16 parts, one operation from jig 1, and 16 parts, one operation from jig 2.
     
    Using your freshly calibrated device, conduct an MSA with all 16 parts from jig 1.  Use as many qualified operators as possible to conduct the MSA.  Ten would be a very good number of operators.
     
    Conduct the MSA by having each operator measure each of the 16 parts.
     
    Correct the measurements of the operators using the bias and linearity of the calibrated device.
     
    From the MSA you will get:
     

    Sigma Total
    Sigma R&R

    Sigma Gage
    Sigma Operator
    Sigma Parts
     
    The following relationships should exist:
     

    (Sigma Parts/Sigma gage)* 1.41 should equal at least 4 but this is barely adequate, as close to 10 as you can get is much better.  The problem is that increasing this number requires getting a different gage.
    (Tolerance/(6*sigma total) should be greater that or equal to 1.33.  This tells you that you are capable of making good parts if the mean is centered.
     
    Check the means for each of the 16 parts.  Conduct an AVOVA to check the means to make sure there are no means significantly different from the others.
     
    Take the mean that is greatest from the center and conduct the following:
     
    USL – Mean, and Mean –LSL.
     
    Using the sigma for that part, divide the lesser of the numbers calculated above by 3*sigma for that part.  The answer should be at equal to or greater than 1.33.  This tells you the part with the most off center mean will produces good parts.
     
    Find the part with the largest sigma.  Using this parts mean, conduct the calculation:
     
    USL – Mean, and Mean – LSL.  Take the lesser of these numbers and divide by 3*sigma for that part.  The result should be greater than or equal to 1.33.  This tells you that the part with the worst variation will make good parts.
     
    Just as a check on the absolute worst case, take the worst mean and the worst sigma and conduct the calculation above.  If the result of that is equal to or greater than 1 you are good to go.
     
    If you take more than one measurement for each part, this procedure should be completed for each measurement separately.
     
    Repeat this procedure for the parts from jig 2.
     
    This is mostly a reverse use of the MSA to give you assurance on the parts rather than assurance on the measuring system.  Your assurance on the measuring instrument comes from the calibration conducted prior to starting the procedure and the corrections to all the readings for bias and linearity.
     
    This is not a DOE but it would give me a very good feeling about making good parts in your situation.
     
    Jim Shelor

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