Roman Forest, Texas

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Roman Forest is a town in Montgomery County, Texas, United States. The population was 1,279 at the 2000 census.



Roman Forest is located at 30°10′40″N 95°9′22″W / 30.17778°N 95.15611°W / 30.17778; -95.15611 (30.177744, -95.156209)[3].

According to the United States Census Bureau, the town has a total area of 1.5 square miles (3.8 km²), all of it land.


As of the census[1] of 2000, there were 1,279 people, 420 households, and 385 families residing in the town. The population density was 864.6 people per square mile (333.7/km²). There were 432 housing units at an average density of 292.0/sq mi (112.7/km²). The racial makeup of the town was 93.67% White, 1.88% African American, 0.86% Native American, 1.49% Asian, 0.08% Pacific Islander, 0.94% from other races, and 1.09% from two or more races. Hispanic or Latino of any race were 4.77% of the population.

There were 420 households out of which 44.8% had children under the age of 18 living with them, 85.7% were married couples living together, 4.0% had a female householder with no husband present, and 8.3% were non-families. 6.9% of all households were made up of individuals and 3.1% had someone living alone who was 65 years of age or older. The average household size was 3.02 and the average family size was 3.15.

In the town the population was spread out with 29.2% under the age of 18, 4.9% from 18 to 24, 27.0% from 25 to 44, 31.4% from 45 to 64, and 7.5% who were 65 years of age or older. The median age was 39 years. For every 100 females there were 97.1 males. For every 100 females age 18 and over, there were 97.0 males.

The median income for a household in the town was $67,679, and the median income for a family was $70,313. Males had a median income of $50,250 versus $32,361 for females. The per capita income for the town was $25,243. About 3.3% of families and 5.5% of the population were below the poverty line, including 8.1% of those under age 18 and 6.5% of those age 65 or over.

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