iSixSigma

John Noguera

Activity

  • Because, the contrast (or contrasts) that make the AxB interaction significant may not be the set of contrasts tested by just focusing on comparisons with a control.

    As for your recommendation – if your design matrix is such that the 0 setting for either A or B really means a complete absence of either A or B and not a case of A or B being at…[Read more]

  • No, the two methods do not show different results – the issue is you are not making the proper comparisons.

    In the regression you found the AxB interaction to be significant.

    In the test for group differences you chose to run a comparison against the control – this isn’t what you want to do.

    There are 4 combinations that comprise the AxB…[Read more]

  • I was looking at your data set again this evening and it occurred to me if you look at the data for count and count2 there is major difference between the two columns.  If count and count2 represent the results of a design and a full replication (I doubt this since the total counts are the same for both count and count2 but bear with me) of the 2…[Read more]

  • To your first modification of the 2**2 factorial design.  What you said you were doing and what you did are two entirely different things.  If you are going to set C = (0,0) and D = (1,1)  where the combinations are for A and B for those particular experiments then the actual design is:

    Matrix1

    What you have done is this:

    Matrix2

    This is just a standard ma…[Read more]

  • @flamepoop1120 This looks like a homework problem. Although the iSixSigma audience can be extremely helpful, they are not here to do your work for you.

    What do YOU think? Are you having trouble getting started? Where are you getting tripped up?

    The more specific you are — and the more of your own thinking you provide — the more likely you are…[Read more]

  • @miparent Have you seen this article? PepsiCo turns to Minecraft, moving virtual training away from ‘Zoom fatigue’ | HR Dive — https://www.hrdive.com/news/pepsico-turns-to-minecraft-moving-virtual-training-away-from-zoom-fatigue/593937/ 1 month ago

  • @Jummy123

    This looks like a homework problem. Although the iSixSigma audience can be extremely helpful, they are not here to do your work for you.

    What do YOU think? Are you having trouble getting started? Where are you getting tripped up?

    The more specific you are — and the more of your own thinking you provide — the more likely you are to…[Read more]

  • @[email protected] – One of my pet gripes is the fact that when the folks were developing their terminology for six sigma the took the word “sigma” and applied it to the calculation indicated in my first post.  That formula is the formula for the sigma level which has nothing to do with the sigma of standard deviation fame.  Unfortunately, a n…[Read more]

  • @davidgford iSixSigma has a weekly Insights newsletter that goes out on Mondays. It includes our featured article of the week, highlights recent discussions, and also includes news, jobs and upcoming events. You can sign up for it at this link: http://www.isixsigma.com/newsletter 3 months, 1 week ago

  • You might want to check the news letter option for this site.  As I understand it that is what the newsletter does.  Try contacting the website for additional details. 3 months, 1 week ago

  • …one additional thought.  With measurements in parts per billion/trillion you are sure to run up against round-off error in whatever analysis program you are using.  This will be true even if you have (as most programs do today) double precision.  I would recommend you express the measures in scientific notation, drop the 10’s power and run th…[Read more]

  • Cpk = minimum (USL-mean, mean – LSL)/(3*std)

    Therefore, in order for Cpk to be negative your mean will have to either be less than the LSL or greater than the USL.

     

    Sigma = minimum (USL-mean, mean – LSL)/std

    Therefore, if the mean is greater than USL then (USL – mean) will be the smaller value and it will be negative which means Sigma will…[Read more]

  • The short answer to your question is – your concern is really of no concern.

    1. From Agresti Categorical Data Analysis 2nd Edition page 3

    ” Variables are classified as continuous or discrete, according to the number of values they can take. Actual measurements of all variables occurs in a discrete manner, due to precision limitations in…[Read more]

  • It sounds like you are taking some kind of exam or course or something and trying to match whatever is offered on a multiple choice problem – this is fine but there is something you should keep in mind when you are faced with questions like this in the real world and that is the issue of significant digits.

    It is possible to get measurements out…[Read more]

  • Obviously the problem is assuming a normal distribution of the data.

    The basic formula for Cp in this case is

    Cp = (USL-LSL/(6*std)

    So, LSL = 330, USL = 330+10.3 and you want a minimum of 1.33 for Cp – therefore plug in the numbers in the above equation, re-arrange and solve for std. 4 months ago

  • @Work_Hard

    Did you search the site before asking your first question? You should find recommendations in the archives. This is a topic that has come up many times over the years.

    You’d probably want to contact ASQ directly to confirm your calculator question. 4 months, 1 week ago

  • I agree with @Straydog – the issue is that of statistics.  If we assume you have a mathematical background that includes algebra and if, for whatever reason,you can’t take a course in basic statistics then I would recommend working your way through the following books:

    1. A Cartoon Guide to Statistics – Gonick and Smith – I’ve recommended this…[Read more]

  • Things are fine on my end @cseider – hope all is well with you too.  As for the homework – it’s pretty quiet around these parts and I was feeling generous…    :-) 4 months, 2 weeks ago

  • It’s just the sums of the squares of the deviation from the mean divided by the total count minus 1.

    So, for a single item the average is  (10 + 0)/2 = 5

    The deviations from the mean are

    10 – 5 = 5

    0-5 = -5

    The squares of those deviations are

    5^2 = 25

    (-5)^2 = 25

    there are two observations so  2 – 1 = 1

    Therefore the variance per item i…[Read more]

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