Can I use Ordinal Regression for Rank and Nominal data ?
geetha...@gmail.com
I am need to design an experiment for my research work, which involves
ranking of a product finish(plastic panels).
Each of the Panels has 6 measurable variables associated (say
thickness, shininess, color, gloss, etc., which can be MEASURED by an
instrument) In a group of 100 panels I have selected only 18 panels
which have low, medium and high value of each Variable.
----------------------------------------------------
The above 18 panels are evaluated i.e, just RANKED by people based on
what they like (Ranking based on the overall Finish ONLY, so they are
not aware of the SIX variables, which could be measured its a
numerical data)
The people who are ranking the plastic panel are focusing on the
Product FINISH only. They Rank from 1 to 18 ( 1 = BEST and 18 =
WORST )
I have Y = Rank data , X1 = Thickness, X2 = Shininess, X3 = Color
etc.,
I am trying to correlate the Rank data with the a measurable data.
*************************************************************************************************
Median
(Rank) X 1 X 2 X 3
X4 X5 X 6
Panel 1. 4 22.5 38.5 41.8
27.5 22.7 11.8
Panel 2. 13 10 25.4 35
15.8 45 37.2
Panel 3. 9 15 17.5 22.4
54 23.8 22.4
Panel 4. 6.5
Panel 5. 3.7
.
.
.
Panel 20.
***************************************************************************************************
Note : ( X1 to X6 are Normally distributed and Median Rank - Ordinal
variable)
My aim is to determine which among the six variable is more
influential/ related to RANK and develop a STANDARD for this
instrument.
The Rank data does not follow a normal distribution even after
transformation.
I did calculate the Kendall Coefficient (W = 0.102).
Can I use the ORDINAL REGRESSION for this kind of problem ?
Can you please suggest me some solution for how approach this very
Tricky problem.
geetha...@gmail.com
> *************************************************************************************************
Median
(Rank) X 1 X 2 X 3 X4 X5 X 6
>
> Panel2. 13 10 25.4 35 15.8 45 37.2
> Panel3. 9 15 17.5 22.4 54 23.8 22.4
> Panel4. 6.5
> Panel5. 3.7
RichUlrich
>Hi
>I am need to design an experiment for my research work, which involves
>ranking of a product finish(plastic panels).
>
>Each of the Panels has 6 measurable variables associated (say
>thickness, shininess, color, gloss, etc., which can be MEASURED by an
>instrument) In a group of 100 panels I have selected only 18 panels
>which have low, medium and high value of each Variable.
>----------------------------------------------------
>The above 18 panels are evaluated i.e, just RANKED by people based on
>what they like (Ranking based on the overall Finish ONLY, so they are
>not aware of the SIX variables, which could be measured its a
>numerical data)
It is too bad that you did not ask for statistical advice *before*
you carried out the experiment. Almost anyone would have
told you that you would be far better off using a scaled score,
say, 1 to 20 or 1 to 100, "terrible" to "great", and using regular
ANOVA on that. Doing that, with suitably chosen anchors would
give you interesting information about what the averages were,
compared to those *anchor* labels.
If you wanted eventually to present information about ranks,
that could have been easily obtained after the fact from discrete
scoring. But if you start with ranks, you can never regain
those subjective responses.
>
>The people who are ranking the plastic panel are focusing on the
>Product FINISH only. They Rank from 1 to 18 ( 1 = BEST and 18 =
>WORST )
>
>I have Y = Rank data , X1 = Thickness, X2 = Shininess, X3 = Color
>etc.,
>I am trying to correlate the Rank data with the a measurable data.
>
>*************************************************************************************************
> Median
[restored from corrective-note]
> (Rank) X 1 X 2 X 3 X4 X5 X 6
>
> Panel1. 4 22.5 38.5 41.8 27.5 22.7 11.8
> Panel2. 13 10 25.4 35 15.8 45 37.2
> Panel3. 9 15 17.5 22.4 54 23.8 22.4
> Panel4. 6.5
> Panel5. 3.7
> .
>
>
> Panel 20.
>***************************************************************************************************
>Note : ( X1 to X6 are Normally distributed and Median Rank - Ordinal
>variable)
>
>My aim is to determine which among the six variable is more
>influential/ related to RANK and develop a STANDARD for this
>instrument.
>
>The Rank data does not follow a normal distribution even after
>transformation.
With an N of 18, that should be *really* hard to be sure
about, with or without transformation. Anyway, Normality
is nowheres required for the variables entered into a regression
(although it can avoid some usual problems). The important
assumption about normality is that the residuals should be
normal and show no important dependencies.
The logical transformation that comes to mind for me would
be to convert the ranks first to percentiles, then convert the
percentiles to Logits (probably) or Probits. What was yours?
Then do ordinary regression.
The problem that arises with performing regression with ranks
as predictors or as outcome is that the units are not "equal" and
as a consequence, it is possible for interactions to arise by
artifact. That should be rather less likely when the criterion
is the average of ranks, as in this case. It should not
be very important when the N is as small as 18, since you
should not try to enter interactions into a model with such
a small sample.
>
>I did calculate the Kendall Coefficient (W = 0.102).
>
>Can I use the ORDINAL REGRESSION for this kind of problem ?
>
>Can you please suggest me some solution for how approach this very
>Tricky problem.
--
Rich Ulrich
Ray Koopman
wrote:
> Hi
> I am need to design an experiment for my research work, which involves
> ranking of a product finish(plastic panels).
>
> Each of the Panels has 6 measurable variables associated (say
> thickness, shininess, color, gloss, etc., which can be MEASURED by an
> instrument) In a group of 100 panels I have selected only 18 panels
> which have low, medium and high value of each Variable.
> ----------------------------------------------------
> The above 18 panels are evaluated i.e, just RANKED by people based on
> what they like (Ranking based on the overall Finish ONLY, so they are
> not aware of the SIX variables, which could be measured its a
> numerical data)
>
> The people who are ranking the plastic panel are focusing on the
> Product FINISH only. They Rank from 1 to 18 ( 1 = BEST and 18 =
> WORST )
>
> I have Y = Rank data , X1 = Thickness, X2 = Shininess, X3 = Color
> etc.,
> I am trying to correlate the Rank data with the a measurable data.
>
> *****************************************************************
> (Rank) X1 X2 X3 X4 X5 X6
>
> Panel 1. 4 22.5 38.5 41.8 27.5 22.7 11.8
> Panel 2. 13 10 25.4 35 15.8 45 37.2
> Panel 3. 9 15 17.5 22.4 54 23.8 22.4
> Panel 4. 6.5
> Panel 5. 3.7
> .
> .
> .
>
> Panel 20.
> *****************************************************************
> variable)
>
> My aim is to determine which among the six variable is more
> influential/ related to RANK and develop a STANDARD for this
> instrument.
>
> The Rank data does not follow a normal distribution even after
> transformation.
>
> I did calculate the Kendall Coefficient (W = 0.102).
>
> Can I use the ORDINAL REGRESSION for this kind of problem ?
>
> Can you please suggest me some solution for how approach this very
> Tricky problem.
You asked essentially the same question before, last November:
http://groups.google.ca/group/sci.stat.math/browse_frm/thread/aa9fda68325d8e7d/,
at which time you gave more information about the 18 panels,
and my answer now is the same as it was then.
Ray Koopman
> I did calculate the Kendall Coefficient (W = 0.102).
Kendall's W is more easily interpreted if it is changed to
an average Spearman r among the n(n-1)/2 pairs of raters,
where n = the # of raters: average r = W - (1-W)/(n-1).
You don't say what your n is, but whatever it is, the average r
is small, less than .102. Evidently there is little agreement
among your raters.
There is a chi-square test of the hypothesis that the average
correlation among the raters is 0; i.e., that every panel's
true average rank equals (k+1)/2, where k = the # of panels:
chi-square(k-1) = n(k-1)W, with large values of chi-square
leading to rejection of the hypothesis. Did you do this test
(which is also known as Friedman's two-way anova on ranks)?
geetha...@gmail.com
> On Mon, 21 Jul 2008 06:06:48 -0700 (PDT), "geetha.sh...@gmail.com"
Hi Mr. Rich Ulrich,
I have a sample size of 30 people evaluating 15 panels. I have taken
the average( median, to overcome outliers) of all 30 people ranks for
each of the 15 panels.
Could you please let me know the procedure of how to do the logit or
probits and then to ordinal regression. I am very naive about this
concept. Your advice would be greatly appreciated.
Thanks,
Geetha.
Ray Koopman
> On Jul 22, 3:53 am, Ray Koopman <koo...@sfu.ca> wrote:
> Thanks for giving me a lead in solving this
> "Interesting" problem. I earlier had a sample size of 60 People
> to evaluate 14 panels. Then I had to reduce the sample size to 30
> People to evaluate the 14 panels for the rest of the tests. I did
> the Wilcoxon Rank Sum Test to verify if I could use just 30 people
> instead of 60. The results were positive.
What did you do the Wilcoxon test on? Why do you think the results
justify using only 30 people instead of 60?
>
> SO the final results I do have is the Rank from 30 people
> evaluating 15 panels.
How many panels? First you said 18, then you said 14, now you say 15.
>
> I did the Chi - Square test, I got a value of 47.6. I also tested
> W for significance using Snedecors Distribution for F (F = 3.4).
>
> Now F value is greater than the Critical ( tabulated) F value. So
> the null hypothesis ( which states there is no correlation among
> the judges/ people) is rejected.
A chi-square of 47.6 with 14 degrees of freedom will let you reject
the null hypothesis. Why and how did you go from chi-square to F?
What degrees of freedom did you use for F?
>
> The value of W must be close to 1 to indicate there is an
> agreement between the evaluation.
> I got a value less than 1. which indicated disagreement.
> But on the other hand the F value contradicts this statement.
> Could you please let me know who should I go about this.
Rejecting the null leads only to the conclusion that there is some
degree of agreement. In this case, there is very little agreement
(average r = .071). Whether that is sufficient is an open question.
>
> I am not using pairs so I didnt use the Spearmans R Calculation.
The average r is a way of characterizing the degree to which the
raters agree with one another. If you pick two raters at random,
you would expect their rankings to correlate only .071.
>
> P.S. I am very naive about statistics, your help would be much
> appreciated.
I strongly urge you to get local assistance. You need closer
guidance than can be provided in a forum such as this.
>
> Thanks,
> Geetha.
RichUlrich
>On Jul 21, 9:31 pm, RichUlrich <rich.ulr...@comcast.net> wrote:
>> On Mon, 21 Jul 2008 06:06:48 -0700 (PDT), "geetha.sh...@gmail.com"
>>
>> <geetha.sh...@gmail.com> wrote:
[snip, much]
>
>Hi Mr. Rich Ulrich,
>I have a sample size of 30 people evaluating 15 panels. I have taken
>the average( median, to overcome outliers) of all 30 people ranks for
>each of the 15 panels.
>Could you please let me know the procedure of how to do the logit or
>probits and then to ordinal regression. I am very naive about this
>concept. Your advice would be greatly appreciated.
Did you miss the part where I said, "Normality is not an assumption
for the variables - only for the residuals"? And, Do ordinary
regression?
Doesn't that result make sense?
You do not know enough to publish anything with the logistic
transformation until you read a bunch about it.
But here is the starter - Use Google for details.
Convert the average ranks to percentiles. Pc= R/(K+1) is not
ideal, but it will do.
Convert the percentiles into logits: Log(Pc/(1-Pc) ) for Pc
respresented as proportions ( < 1.0, not as < 100).
Do the regression.
Also, considering Ray's posts, I suspect that you mis-computed
the concordance coefficient.
Those average ranks should be represented by something much
larger than an average correlation of 0.1 .
--
Rich Ulrich
geetha...@gmail.com
> On Jul 22, 9:34 pm, <geetha.sh...@gmail.com> wrote:
>
>
>
> > On Jul 22, 3:53 am, Ray Koopman <koop...@sfu.ca> wrote:
Hi Mr. Ray Koopman,
I calculated Coefficient of Concordance, W and calculated
F by subtracting unity from the calculated "S" ( sum of sq difference
between observed and expected rank) and increasing the divisor [i.e,
m^2(n^3-n)/12 ] by 2 units.
This method is explained in detail in the book titled "Facts from
figures" by Moroney M.J.
I had 16 , 14 and 14 panels in 3 groups ( of different colors).
Thanks,
Geetha.