However, before jumping to conclusions about Wiggins we should ask whether dopers really show significantly improved results and, if so, whether such improvements also occur for non-dopers. This can be answered with some fairly straightforward statistical testing to compare a rider's current results with their previous results. I previously defined two parameters to quantify the improvement and significance of a given rider's results during a given year:
- R: The difference between a rider’s mean placings in a given year and mean placings in previous years. Larger numbers mean greater improvement.
- P: The likelihood that this difference is real and not simply a result of random fluctuations (statistical significance). Smaller numbers mean greater significance.
I computed R and P for 934 riders over the years 2003-2009. For each rider, I only considered years in which 10 or more results were in the Cycling Quotient database. Forty-five rider/year pairings showed statistically significant improvements (see technical note for the definition of "significant"). Here they are, with 2009 cases in red:
Having improved an average of 50 places per race, Wiggins's 2009 is on this list. Columbia's Tony Martin is here as well, and has actually gained more than Wiggins this year. But there are very few convicted dopers here; where are our naughty friends? Adding Di Luca's 2009 to the list I showed on my previous post, the results for recent doping positives look like this:
Although a few of these riders show large gains in average results and fairly low P values, none of these riders appear on the above list of significant cases. So I'd have to say there isn't much support to the idea that a big improvement in results is a sign of doping.
Technical Notes: I defined significance as having a P less than 2e-5. This might sound overly conservative -- it means the chances that the rider's current and previous results are the same is only 0.002%. The problem is that I've done 2400 tests, so a cutoff of 5% would give me over 100 false positives. Dividing 0.05 by the number of tests, I get a P cutoff of 2e-5 and don't need to worry about false positives. Additional technical notes of possible relevance here and here (scroll down to the fine print).