iSixSigma

How Should I Treat a Variable That Is Both Discrete and Continuous in Nature?

Six Sigma – iSixSigma Forums General Forums Tools & Templates How Should I Treat a Variable That Is Both Discrete and Continuous in Nature?

Viewing 2 posts - 1 through 2 (of 2 total)
  • Author
    Posts
  • #253648

    Anonymous
    Inactive

    How should I treat a variable that is both discrete and continuous in nature?

    One of my factors is the addition of an extra treatment. One setting is no extra treatment, while the other is adding the treatment, which results in a discrete variable. However, if/once added, the extra treatment has continuous variables, such as temperature and pressure. This results in, if you see it as one factor (which I highly doubt I should), an ‘off’ level with a discrete step towards the minimal ‘on’ level and a continuous step towards the maximum ‘on’ level. I don’t see how this can be 1 factor, but I also don’t see how I can separate it into multiple factors.

    How should I go about adding this in a DoE?

    Thanks in advance!

    0
    #253689

    Robert Butler
    Participant

    Given what you have posted, I think you are faced with one of two scenarios.

    (In order to provide some clarity with respect to what follows let’s pretend the treatment has as part of its application changes in temperature and pressure. Given this then there are at least two different versions of “no treatment”.)

    1. The situation for “no treatment” actually incorporates fixed settings of temperature and pressure. Since treatment also encompasses changes in temperature and pressure you would need to include the temperature and pressure settings for “no treatment” in the matrix of temperature and pressure settings for treatment. The analysis would consist of two parts.
    a. Evaluate the effects of changes in temperature and pressure under the treatment condition.
    b. Assess the effects of changes between treatment and no treatment at the temperature and pressure settings associated with no treatment.

    2. The situation of “no treatment” does not involve the variables of temperature and pressure. In this case you would treat the “no treatment” situation as an external target and test the findings using treatment for different setting of temperature and pressure against that target.
    a. Since treatment also requires the addition of other variables to the mix you won’t be able to directly compare “no treatment” and “treatment”.

    0
Viewing 2 posts - 1 through 2 (of 2 total)

You must be logged in to reply to this topic.