Seville, Ohio

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Seville is a village in Medina County, Ohio, United States. The population was 2,160 at the 2000 census. The community was named after Sevilla, a city in Andalusia[1], Spain.



Seville is located at 41°0′56″N 81°51′52″W / 41.01556°N 81.86444°W / 41.01556; -81.86444 (41.015620, -81.864363)[3].

According to the United States Census Bureau, the village has a total area of 2.0 square miles (5.2 km²), all of it land.


As of the census[2] of 2000, there were 2,160 people, 808 households, and 611 families residing in the village. The population density was 1,070.9 people per square mile (412.9/km²). There were 847 housing units at an average density of 419.9/sq mi (161.9/km²). The racial makeup of the village was 98.5% White, 0.19% African American, 0.23% Native American, 0.19% Asian, 0.05% from other races, and 0.19% from two or more races. Hispanic or Latino of any race were 0.65% of the population.

There were 808 households out of which 36.1% had children under the age of 18 living with them, 64.7% were married couples living together, 8.2% had a female householder with no husband present, and 24.3% were non-families. 20.9% of all households were made up of individuals and 8.5% had someone living alone who was 65 years of age or older. The average household size was 2.58 and the average family size was 2.97.

In the village the population was spread out with 25.5% under the age of 18, 6.0% from 18 to 24, 31.9% from 25 to 44, 22.6% from 45 to 64, and 14.0% who were 65 years of age or older. The median age was 37 years. For every 100 females there were 89.0 males. For every 100 females age 18 and over, there were 84.0 males.

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