Summerville, Georgia

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Summerville is a city in Chattooga County, Georgia, United States. The population was 4,556 at the 2000 census. The city is the county seat of Chattooga County[3].

Contents

Geography

Summerville is located at 34°28′47″N 85°20′53″W / 34.47972°N 85.34806°W / 34.47972; -85.34806 (34.479803, -85.347988)[4].

According to the United States Census Bureau, the city has a total area of 4.0 square miles (10.3 km²), all of it land.

Demographics

As of the census[1] of 2000, there were 4,556 people, 1,823 households, and 1,141 families residing in the city. The population density was 1,144.3 people per square mile (442.0/km²). There were 2,092 housing units at an average density of 525.4/sq mi (202.9/km²). The racial makeup of the city was 72.06% White, 25.31% African American, 0.07% Native American, 0.15% Asian, 0.90% from other races, and 1.51% from two or more races. Hispanic or Latino of any race were 1.58% of the population.

There were 1,823 households out of which 28.7% had children under the age of 18 living with them, 39.8% were married couples living together, 17.8% had a female householder with no husband present, and 37.4% were non-families. 33.0% of all households were made up of individuals and 16.4% had someone living alone who was 65 years of age or older. The average household size was 2.39 and the average family size was 3.04.

In the city the population was spread out with 24.3% under the age of 18, 9.0% from 18 to 24, 26.5% from 25 to 44, 21.7% from 45 to 64, and 18.5% who were 65 years of age or older. The median age was 38 years. For every 100 females there were 84.7 males. For every 100 females age 18 and over, there were 78.8 males.

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