The overall injury rate at the 2006 FIFA World Cup was 2.3 injuries per match, which was not significantly different from the 2.7 injuries per match observed in the 2002 tournament (p = NS).
Observational
Absolute Event Rate: 2.3% vs 2.7%
p-value: p=NS
OBJECTIVE: To continue the injury surveillance of FIFA-sponsored football tournaments and report on other medical aspects of the 2006 FIFA World Cup. DESIGN: Prospective epidemiological injury surveillance and descriptive summary of additional medical aspects. SETTING: Major international football tournament. PARTICIPANTS: National team players, doctors and referees at the 2006 FIFA World Cup Germany. MAIN OUTCOME MEASURES: Injury type, location and rate. RESULTS: 145 injuries were reported for the 64 matches of the 2006 FIFA World Cup Germany-an overall injury rate of 68.7 per 1000 match hours (95% CI 57.5 to 79.9) or 2.3 injuries per match, in comparison with 2.7 injuries per match in the 2002 FIFA World Cup (p = NS). Physical examinations before participation uncovered no hidden cardiovascular problems. Once the tournament started, no referees were unable to complete their duties. There were no positive doping tests. CONCLUSIONS: The injury rate for this World Cup was below that of 2002, but consistent with the overall injury rate per match since data collection began in 1998. There continues to be no evidence of systematic doping in international football.
Building similarity graph...
Analyzing shared references across papers
Loading...
British Journal of Sports Medicine
Schulthess-Klinik
Fédération Internationale de Football Association
Add This Paper to Your Research Feed
Any time a new paper drops it will be there.
Dvořák et al. (Thu,) conducted a observational in Football injuries. 2006 FIFA World Cup vs. 2002 FIFA World Cup was evaluated on Injury type, location and rate (p=NS). The overall injury rate at the 2006 FIFA World Cup was 2.3 injuries per match, which was not significantly different from the 2.7 injuries per match observed in the 2002 tournament (p = NS).