This City Has Lowest Rate of Poverty in US

Sheboygan, Wisc., is atop the top 10
By Evann Gastaldo,  Newser Staff
Posted Oct 14, 2017 4:50 PM CDT
The 10 US Cities With Lowest Rates of Poverty
The Lighthouse is mired in the depth of winter's chill along Lake Michigan as seen Wednesday Jan. 7, 2015 in Sheboygan, Wisc.   (AP Photo/Sheboygan Press Media, Gary C. Klein)

An individual making less than $12,060 this year—$24,600 for a family of four—technically lives below the poverty line. About 14% of Americans fit that definition. Some cities have been hit harder than others, and 24/7 Wall St. looked at 382 US metro areas to compile its list of the 49 cities that are doing the best—meaning less than 10% of the population lives in poverty. More than half of the areas that qualified are in the Midwest or the Northeast, while just four are in the South. The list also looks at each city's median household income, percentage of households with income of less than $10,000, percentage of households receiving SNAP benefits, and unemployment rate for 2016. The top 10 and their associated poverty rates:

  1. Sheboygan, Wisc.: 5.4%
  2. Barnstable Town, Mass.: 6.5%
  3. Fond du Lac, Wisc.: 6.6%
  4. Anchorage, Alaska: 7.2%
  5. Billings, Mont. (tied for 5th): 7.3%
  6. Napa, Calif. (tied for 5th): 7.3%
  7. Appleton, Wisc.: 7.6%
  8. Chambersburg-Waynesboro, Penn. (tied for 8th): 8.0%
  9. Ogden-Clearfield, Utah (tied for 8th): 8.0%
  10. Winchester, Va.-W. Va. (tied for 8th): 8.0%
Click for the full list, along with the other measures 24/7 Wall St. looked at. (More poverty stories.)

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