Swepsonville, North Carolina

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Swepsonville is a town in Alamance County, North Carolina, United States. It is part of the Burlington, North Carolina Metropolitan Statistical Area. The current population estimate (2008) is 1,053.[3]



Swepsonville is located at 36°1′36″N 79°21′17″W / 36.02667°N 79.35472°W / 36.02667; -79.35472 (36.026744, -79.354849).[4]

According to the United States Census Bureau, the town has a total area of 1.1 square miles (2.7 km²), of which, 1.0 square miles (2.7 km²) of it is land and 0.04 square miles (0.1 km²) of it (2.83%) is water.


At the 2000 census[1], there were 922 people, 383 households, and 271 families residing in the town. The population density was 894.5 people per square mile (345.6/km²). There were 405 housing units at an average density of 392.9/sq mi (151.8/km²). The racial makeup of the town was 94.03% White, 3.80% African American, 0.43% Asian, 0.54% from other races, and 1.19% from two or more races. Hispanic or Latino of any race were 1.84% of the population.

There were 383 households out of which 29.2% had children under the age of 18 living with them, 60.1% were married couples living together, 8.6% had a female householder with no husband present, and 29.2% were non-families. 25.1% of all households were made up of individuals and 9.4% had someone living alone who was 65 years of age or older. The average household size was 2.41 and the average family size was 2.86.

In the town the population was spread out with 23.2% under the age of 18, 5.1% from 18 to 24, 28.5% from 25 to 44, 30.3% from 45 to 64, and 12.9% who were 65 years of age or older. The median age was 41 years. For every 100 females there were 88.5 males. For every 100 females age 18 and over, there were 91.9 males.

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