The Math Forum

Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Math Forum » Discussions » sci.math.* » sci.math

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

Topic: Is there a way to calculate an average ranking from uneven lists?
Replies: 15   Last Post: Oct 30, 2013 12:18 PM

Advanced Search

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View   Messages: [ Previous | Next ]
Jennifer Murphy

Posts: 24
Registered: 2/23/12
Re: Is there a way to calculate an average ranking from uneven lists?
Posted: Oct 28, 2013 8:30 AM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

On Mon, 28 Oct 2013 10:26:56 +0000 (UTC), JohnF
<> wrote:

>Jennifer Murphy <JenMurphy@jm.invalid> wrote:
>> James Waldby <not@valid.invalid> wrote:
>>>For the given problem, averages of ranks probably aren't a statistically
>>>sound approach. For example, see the "Qualitative description" section
>>>of article <>, which says:
>>>"User ratings are at best ordinal categorizations. While it is not
>>>uncommon to calculate averages or means for such data, doing so
>>>cannot be justified because in calculating averages, equal intervals
>>>are required to represent the same difference between levels of perceived
>>>quality. The key issues with aggregate data based on the kinds of rating
>>>scales commonly used online are as follow: Averages should not be
>>>calculated for data of the kind collected." (etc.)

>> Yes, I did feel a little uneasy about averaging numbers that are not
>> really numerical in the usual sense.

>Yes, I think that approach would be wrong, based on the following
>extreme case counterexample:
> Suppose you have 100 different lists, 99 of which identically contain
>the same two books, and only those two books, in the same ranking,
> Lists 1-99 contain: Book#1=The Bernie Madoff Story, #2=The Ken Lay Story
>Finally, the 100th list contains, say 100 different books, including
>our above two losers, but ranked
> List 100 contains: Book#1=The complete works of Shakespeare,
> #2=..., #3=..., ..., and finally,
> #99=The Bernie Madoff Story, #100=The Ken Lay Story
>Clearly, Bernie and Ken suck, but when they're the only two
>books on a list, then they have to rank #1 and #2.
>So you need a methodology that avoids a combined ranking giving
> Bernie: 99 #1-scores and one #99-score, and giving
> Ken: 99 #2-scores and one #100-score.
>That would significantly overestimate them.

This example is badly skewed. First, all of the lists I will use will be
from authoritative sources, they will all have a lot more than 2 books,
and no two lists will be even close to identical. So I have to quibble
with your point about the Bernie and Ken books being that terrible. If
these 100 lists are from reputable sources (as opposed to, say, from
some crackpot on usenet ;-)), then the Bernie and Ken books do not
"suck". If the last list is the 100 greatest books of all time, then
even making it to #99 and #100 would be well above the "suck" level. ;-)
But I get your point (I think).

I might include a list containing only science fiction books. This would
exclude 95% of all books, but, as you say, some book on that list will
have to be #1. The #1 book on the sci fi list would probably also be on
at least some of the other lists, so it's ranking there would be
factored in. And I can apply a weighting factor to each list so that the
most substantive lists have a greater influence on the results.

Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© The Math Forum at NCTM 1994-2018. All Rights Reserved.