TALLAHASSEE, Fla. (WTXL) — A new study has named Tallahassee the drunkest city in Florida.
According to 24/7 Wall Street, 20.9 percent of adults in Tallahassee admitted that they drink excessively or binge drink on a regular basis, which is above the 17.5 percent share of statewide adults. Alcohol-related driving deaths were 32.7 percent, which is 8 percent higher than the state share of 24.7 percent.
A snapshot of the study's findings in Florida and Georgia is below:
Drunkest City in Florida: Tallahassee
> Adults binge or heavy drinking: 20.9%
> Florida adults binge or heavy drinking: 17.5% (19th lowest)
> Alcohol related driving deaths: 32.7%
> Florida Alcohol related driving deaths: 24.7% (8th lowest)
> Median household income: $50,825
> Florida median annual income: $52,594 (12th lowest)
Georgia: Savannah
> Adults binge or heavy drinking: 17.8%
> Georgia adults binge or heavy drinking: 15.1% (7th lowest)
> Alcohol related driving deaths: 24.5%
> Georgia Alcohol related driving deaths: 22.4% (6th lowest)
> Median household income: $56,610
> Georgia median annual income: $56,183 (19th lowest)
To identify the drunkest city in each state, 24/7 Wall St. reviewed the percentage of men and women over 18 who reported engaging in binge or heavy drinking in in each state’s metro areas in 2016. City data and state level data were aggregated from county level data assembled in 2019 by County Health Rankings & Roadmaps.
The CDC defines excessive alcohol use as binge drinking (five or more drinks on one occasion for men, four or more for women) or heavy drinking (15 or more drinks per week for men, eight or more for women).
