Based
Dustin Stevens-Baier
Comp
578
9-7-06
Assignment
#2
3. You are
approached by the marketing director of a local company, who believes that he
has devised a foolproof way to measure customer satisfaction. He explains his
scheme as follows: "It's so simple that I can't believe that no one has thought
of it before. I just keep track of the number of customer complaints for each
product. I read in a data mining book that counts are ratio attributes, and so,
my measure of product satisfaction must be a ratio attribute. But when i rated
the products based on my new customer satisfaction measure and showed them to my
boss, he told me that I had overlooked the obvious, and that my measure was
worthless. I think that he was just mad because our best selling product had the
worst satisfaction since it had the most complaints. Could you help me set him
straight?"
A. Who is right, the
marketing director or his boss? If you answered his boss, what would you do to
fix the measure of satisfaction?
The marketing director is wrong and
his boss is correct. The number of complaints on a particular product
cannot determine the level of its satisfaction when compared to
other products. If the best selling
product receives the most complaints, that does not mean the
worst customer satisfaction. The number of items sold plays the
most important role in getting the number of customer's
feedback. The more you sell the bigger the sample to get both
negative and positive data from.
B. What can you
say about the attribute type of the original product satisfaction attribute?
The marketing director is correct
when he talks about ratio attributes, but he does not use percent variation
allowing proportional comparison with other products. A better way to
describe the data would be through soemthing tracking the number of complaints
per items sold.
4. (a) Is the marketing
director in trouble? Will his approach work for generating an ordinal ranking of
the product variations in terms of customer preference?
Explain.
Yes he is in trouble, becuase a scenario cab be devised where
we have a contradiction. For instance if the person says 1 > 2 and 2
> 3 but then says 3 > 1 we can see
that no ordinal data can be gotten from this. The first two says 1 > 2
> 3 but then the last one contradicts that and says 3 >
1. The only time we need to test for the last one is if
no ordinal data can be gotten from 1 and 3 for instance 1 > 2 and 3
> 2 we know that 2 is lowest and then we need to find out how 1 and 3
rank.
(b) Is there a way to fix the marketing director's
approach? More generally, what can you say about trying to create an ordinal
measurement scale based on pairwise comparisons?
I don't think
that this approach makes the most since the customers should be able to compare
them all. However, sometimes there are too many to get reasonable ordinal
data. This means that you are going to need conditional questions that
only get asked if you need them. This way you can avoid the contradictions
that can arise with the marketing directors.
(c) For the
original product evaluation scheme, the overall rankings of each product
variation are found by computing its average over all test subjects.
Comment on whether you think that this is a reasonable approach. What
other approaches might you take?
I think that the original idea
would work okay as long as the value we are averaging is uniform across the
entire scope of items. For instance if every customer is rating their
experience on a scale of 1-10 then it would be fine but if some items rate on a
different scale or some customers rate on a different scale then it wouldn't
work. There would be obvious issues with this idea as well, their would be
people that rate everything a 10 or a 1. Another approach would be to take
all of the complaints made and divide it by the number of purchases, then this
could be ordered.
5. Can you think of a situation in which
identification numbers would be useful for prediction?
I can
think of a couple of ways if the id was tied into age or location. For
instance a phone number is kind of an id number and area codes give you a good
prediction of location. aAso check numbers are a form of id and give a good
prediction of time and possibly age of the person with the
account.
14. The following attributes are
measured for members of a herd of Asian elephants: weight, height, tusk length,
trunk length, and ear area. Based on these measurements, what sort of
similarity measure from Section 2.4 would you use to compare or group these
elephants? Justify your answer and explain any special
circumstances.
Based off of what I know about elephants as one of
these would increase they all would so I would use something to see what
kind of correlation they have with one another. A good idea would probably
be Pearsons's correlation coefficient. This would enable us to predict
some of the data based off of the other values. For instance if we know
the weight and height then we can probably predict the others based off the
correlation data we have.
15. You are given a set of m objects
that is divided into K groups, where the ith group is of size
mi. If the goal is to obtain a sample of size n < m, what is the
difference between the following two sampling schemes?
(a) We randomly
select n*mi/m elements from each group.
(b)We randomly select n elements
from the data set, without regard for the group to which an object
belongs.
In the first group you get a much better
proportional representation of each group something that can't be said for
the second one, because n < m. It is possible that the number of K is alot larger than each
sample. This means that you may not be able to proportionately sample
each group. Another issue is that as you the size of n grows closer to m, you loose any advantage you once had
by sampling.
16. Consider a document-term matrix, where tfij
is the frequency of the ith word in the jth document and m is the
number of documents. Consider the variable transformation that is defined
by tf'ij = tfij*log
m/df i
where dfi is
the number of documents in which the ith term appears, which is known as the
document frequency of the term. This transformation is known as the
inverse document frequency transformation.
(a) What is the
effect of this transformation if a term occurs in one
docuemnt?
In this situation it measn that the term will weighted
very heavily.
i.e. Let the frequency be 2, the documents total 6,
and the term appears in 1 document.
2*log(6/1) = 1.57
In every document?
If it happens in
every document it will be zero.
i.e. Let the frequency be 2, the
documents total 6, and the term appears in 6 documents
2*log(6/6)=
0
(b) What might be the purpose of this
transformation?
A problem with data
mining in particular for search engines and other practical purposes on the web,
common terms that are on ervy page cannot be weighted as much as the more useful
terms. An example is if someone was serching for "the baseball" They
might be interested in every document that has the word "baseball" in it but
they definately don't care about every document that has the word "the" in
it.