DOE help

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    Hello all,
    I need to get your thoughts on the following…I’m running a DOE where the response is an attribute (the device made is either good or bad).  If the response was a variable, I would only do 2-3 replicates, but being an attribute I was thinking about using 5-6 reps.
    Does the amount of replicates make sense given the response to be an attribute?



    Hai PRPatel.
    It depends: Does 1 replicate conssist of only 1 product that can be good or bad? Then the answer is No; much too few; think in hunderds.
    If 1 replicate consists of a number of products (n) from which you calculate #good/n. Then it does not matter much for the total number of products you have to make in all the runs together in how many replicates you split it up (unless with the replicate action you want to detemine something extra (= use up df’s).
    There are generally 2 Strategies: (1) Fixate n=number of products made and measure # good ; (2) Fixate # good you want to see and keep the run ongoing until you have found them.
    Also the Y analysed should not be #good/n but arcsin(SQRT(#good.n)). This transformation helps making the residuals normal distributed.
    Probably Stan or Darth can help you with the exact formula’s.



    I think you are going to find it difficult to analyze a bunch of 1s and 0s and get any useful information. I suppose logistic regression is the way to handle this kind of response, but I have never used it.
    If you can rate the samples on a scale of 1 to 7, you might be better off!


    Ken Feldman

    Start with this thread and see if you get any insight.


    Bernard C

    With attribute data, the output response value Y to analyze for any design is a binomial (n, p) with n being the sample size (replicate)    of each run and p being the probability of success for each run; p is unknown and will vary at each run and n is also unknown (your question). Using a lower conservative estimate of p, say p1, then each run should use a common sample size of n = 5/p1 (or greater). This will make the output variable Y approximately normal (because np1=5) with mean np1 and variance np1(1-p1). The analysis of your response is done with the transformed variable Z=arcsin(sqrt(Y)) in order to make the variance independent of the mean. For exemple, if p1=0.05 then n= 5/0.05=100

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