Brownsville ranks poorest in the nation

The Census Bureau ranks Brownsville as the most impoverished city in the nation, according to the bureau’s 2006 American Community Survey released on Tuesday. More than 40 percent of the city’s 171,000 residents live below the poverty line, the bureau’s figures show.

The bureau’s poverty threshold for an individual is a $10,294 annual income. For a family of four it is $20,614.

Despite the last-place standing, there are slight gains being made in this area. In 2005, the poverty level was 42.6 percent, compared to 40.6 percent in 2006.

Median household income in Brownsville, the fourth lowest in the nation, is also inching north — $26,017 last year, compared to $24,207 in 2005.

“We’ve got to get people’s attention on this,” said Traci Wickett, president of United Way of Southern Cameron County.

United Way aims to reduce poverty in two ways, she said. Its first strategy is to improve educational attainment through initiatives such as Success by Six, which focuses on high-quality early childhood education.

The other strategy is to getting people stable financially by accessing the Earned Income Tax Credit, and other state and employer benefits available to them.

“When we talk about poverty in our community, we’re not talking about people who are unemployed. We have a lot of hard working people in our community who live below the poverty line.”

In the Brownsville-Harlingen metropolitan statistical area the unemployment rate was 6.5 percent in June, compared to 7.4 percent the year prior. That’s about 9,400 people.

Area-wide, the Census Bureau rated Hidalgo County as the poorest county in the nation, with Cameron County coming second to last.

Also on Tuesday, the bureau released another population report, titled, “Income, Poverty and Health Insurance Coverage in the United States: 2006.

According to that survey, Texas had the highest population of uninsured in the country, with 24.1 percent. That survey does not give more local data.

On the Web: www.census.gov