Sheffield Lake, Ohio

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Sheffield Lake is a city in Lorain County, Ohio, United States. The population was 9,371 at the 2000 census.



Sheffield Lake is located at 41°29′20″N 82°5′59″W / 41.48889°N 82.09972°W / 41.48889; -82.09972 (41.488939, -82.099694)[3].

According to the United States Census Bureau, the city has a total area of 2.5 square miles (6.5 km²), all of it land.


As of the census[2] of 2000, there were 9,371 people, 3,498 households, and 2,571 families residing in the city. The population density was 3,715.3 people per square mile (1,435.8/km²). There were 3,776 housing units at an average density of 1,497.1/sq mi (578.5/km²). The racial makeup of the city was 96.67% White, 0.97% African American, 0.41% Native American, 0.29% Asian, 0.02% Pacific Islander, 0.61% from other races, and 1.04% from two or more races. Hispanic or Latino of any race were 3.00% of the population.

There were 3,498 households out of which 35.2% had children under the age of 18 living with them, 59.3% were married couples living together, 9.9% had a female householder with no husband present, and 26.5% were non-families. 21.7% of all households were made up of individuals and 7.0% had someone living alone who was 65 years of age or older. The average household size was 2.68 and the average family size was 3.13.

In the city the population was spread out with 26.5% under the age of 18, 8.6% from 18 to 24, 31.6% from 25 to 44, 24.0% from 45 to 64, and 9.3% who were 65 years of age or older. The median age was 36 years. For every 100 females there were 95.4 males. For every 100 females age 18 and over, there were 93.6 males.

The median income for a household in the city was $48,984, and the median income for a family was $55,078. Males had a median income of $40,394 versus $26,913 for females. The per capita income for the city was $20,219. About 3.5% of families and 4.7% of the population were below the poverty line, including 4.4% of those under age 18 and 3.6% of those age 65 or over.

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