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Apologies if this is a duplicate. I have a website with a large number of URLs, and I also have data on the number of unique page-views and online feedback generated for each URL. I would like to find the URLs with the highest relative number of tickets, so I can prioritise the for attention.

On the surface this is straightforward - I can just divide the number of feedback tickets by the number of page-views to get the relative ticketing rate.

However, the number of page-views varies greatly, and I want to find a way to account for the absolute number of tickets as well as the relative ticketing rate.

For example (in reality I have about 500,000 URLs):

URL         page-views   tickets
/popular    1000000      500
/rare       1000         5

On the surface of it, /rare has a higher ticket rate (5000 per million versus 500 per million). But in fact I would want to look at /popular at the same time.

My first thought is that I should multiply the relative and absolute values together to get a normalised value. But is this statistically respectable?

My second thought was that maybe I should use a funnel plot.

Any suggestions very welcome.

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    $\begingroup$ Yes you can, why not? The question is: what is the aim of your analysis? $\endgroup$
    – Tim
    Dec 17, 2014 at 12:55
  • $\begingroup$ As stated: "I would like to find the URLs with the highest relative number of tickets, so I can prioritise them for attention". The underlying aim is to find the URLs most in need of attention. I was wondering if there was a standard statistical way of approaching this, or whether I should just experiment with the multiple of absolute + relative values that seems to work best. $\endgroup$
    – Richard
    Dec 17, 2014 at 15:19
  • $\begingroup$ Most of the threads with the valuation tag are essentially duplicates, but so far without any standard objective answer. $\endgroup$
    – whuber
    Dec 17, 2014 at 17:12

2 Answers 2

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Compute the rank as $r=\frac{tickets^2}{pageviews}$. You'll get 0.25 and 0.005. This is essentially $r=tickets\frac{tickets}{pageviews}$, i.e. weighted relative rate, where the weight is the number of tickets.

One more thing to be aware of: the page views could be caused by defects on them. Let's say an order confirm page is broken, and people keep clicking on it generating elevated page views. Alternatively, a page is broken (maybe too slow due to a bug), so people don't go there a lot. Of course, the tickets could be dependent on page views too. A popular page may tend to have more tickets, because more people spend more time on it and notice more things about it.

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  • $\begingroup$ Thanks. This feels intuitively like the right way of doing things, but just curious - is there a formal statistical justification for doing it this way? $\endgroup$
    – Richard
    Dec 17, 2014 at 16:22
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    $\begingroup$ @Richard, not really. This relates not to statistics but to your preferences. You have a preference function $L(tickets,pageviews)$, and it may have any form you wish. I suggested one form, if it fits your preference, use it. You could use something like $\frac{tickets}{\ln pageviews}$. $\endgroup$
    – Aksakal
    Dec 17, 2014 at 16:34
  • $\begingroup$ Statistics would come into a play if you asked questions like: does low popularity cause low ticket count? $\endgroup$
    – Aksakal
    Dec 17, 2014 at 17:58
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Assuming that you do want to look only at the pages with top rank in either of the 2 nominations (absolute, relative), I would do it it in the following way:

  • create a boolean variable the denotes that the page is in top X% on all pages in absolute values (i.e. 1 if in top 5%, 0 otherwise);
  • another boolean variable for relative value;
  • add booleans together, if result is > 0 then the page is of interest;

In that way at least you have a clear understanding of what you are looking at, as opposed to multiplying the relative and absolute values and wondering around "squared tickets per view".

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  • $\begingroup$ Thanks, this is also a good suggestion. I may use this to narrow the field and then @Aksakai's suggestion to rank the results. $\endgroup$
    – Richard
    Dec 17, 2014 at 16:23

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