Today's finish to golf's British Open (or "The Open" as the hosts call it) will probably be remembered primarily for the play of Padraig Harrington and Sergio Garcia on the 18th hole in regulation and then the four-hole play-off between the two, won by Harrington.

For sheer hot streaks, though, the Sunday round of Andres Romero, the third-place finisher, would be hard to top. He made 10 birdies for the day, including a stretch of 6-out-of-7 holes in the latter half of the round.

As pointed out during the ABC television broadcast (and can be seen on Romero's scorecard), he had bested par only nine times during the first three days (54 holes) of the tournament (8 birdies and 1 eagle).

Statistical tests on one athlete in one event are always dicey because of the relatively small sample size. However, with the ready availability of online statistical calculators -- in this case, for chi-square -- let's go for it!

We start with a basic 2 X 2 contingency table, with the values referring to numbers of holes (the dashes have been inserted to make sure the spacing comes out right):

----------Below par-----Par or above
Day 1-3--------9------------45------
Day 4---------10-------------8------

The calculator site I used offers three different versions of the chi-square test. Regardless of which one is used, the obtained difference in Romero's percentage of below-par holes between the first three days and the final day would be expected to come up purely by chance less than .005 of the time (5 in 1,000 or 1 in 200). We thus conclude that he performed significantly better on Sunday than during the three previous days.

Of course, the usual cautions apply: I was drawn to doing this analysis by the unusual nature of Romero's spectacular round, I did not test a random cross-section of golfers, and in the aggregate "big picture" of all golfers in all major tournaments, a round like his may not occur any more often than would be expected by chance.