case study BDUD

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, because 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.

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You are approached by the marketing director of a local company, who

believes that he has devised a foolproof way to measure customer

Yes he is in trouble, because a scenario cab be devised where we have a

(b) Is there a way to fix the marketing director's approach? More

5. Can you think of a situation in which identification numbers would

14. The following attributes are measured for members of a herd of

(a) We randomly select n*mi/m elements from each group.

16. Consider a document-term matrix, where tfij is the frequency of the

(a) What is the effect of this transformation if a term occurs in one

If it happens in every document it will be zero.

(b) What might be the purpose of this transformation?

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