Health Insurance Coverage in Dane County, WI
ACS 2024 5-Year Estimates
How Dane County Compares
National Distribution
Where Dane County falls among all 3,222 U.S. counties for health insurance coverage.
■ Dane County: 96.4%
Ranking Context
| Rank | County | Value |
|---|---|---|
| #178 | Sangamon County, IL | 96.4% |
| #179 | Mercer County, OH | 96.4% |
| #180 | Floyd County, IA | 96.4% |
| #181 | Dane County, WI | 96.4% |
| #182 | Huerfano County, CO | 96.4% |
| #183 | Cecil County, MD | 96.4% |
| #184 | Olmsted County, MN | 96.4% |
Health Insurance Coverage in Dane County, Wisconsin
Dane County, Wisconsin has a health insurance coverage of 96.4%, which ranks #181 of 3,222 out of 3,222 counties nationwide. This is 5.1% above the national value of 91.7%.
Within Wisconsin, Dane County ranks #9 of 72 out of 72 counties. This places it in the 94th percentile nationally.
Percentage of the population with health insurance coverage. All data comes from the U.S. Census Bureau's American Community Survey (ACS) 5-Year Estimates for 2024.
About this data
Source: U.S. Census Bureau, American Community Survey 5-Year Estimates (2024). All values are estimates derived from survey samples and are subject to margin of error. For more information, see our methodology.