Hobart, Indiana

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Hobart is a city in Lake County, Indiana, United States. The population was 25,363 at the 2000 census.

The city is pronounced by locals as Ho-bert instead of Ho-bart, which is how many people not from the area mistakenly pronounce it.



Hobart is located at 41°31′43″N 87°16′3″W / 41.52861°N 87.2675°W / 41.52861; -87.2675 (41.528715, -87.267420)[3].

According to the United States Census Bureau, the city has a total area of 26.7 square miles (69.2 km²), of which, 26.2 square miles (67.9 km²) of it is land and 0.5 square miles (1.3 km²) of it (1.83%) is water.

Hobart is approximately 31 miles (50 km) southeast of Chicago.


As of the census[1] of 2000, there were 25,363 people, 9,855 households, and 6,977 families residing in the city. The population density was 967.5 people per square mile (373.6/km²). There were 10,299 housing units at an average density of 392.9/sq mi (151.7/km²). The racial makeup of the city was 93.73% White, 1.39% African American, 0.21% Native American, 0.54% Asian, 0.02% Pacific Islander, 2.60% from other races, and 1.51% from two or more races. Hispanic or Latino of any race were 8.05% of the population.

There were 9,855 households out of which 30.6% had children under the age of 18 living with them, 56.3% were married couples living together, 10.4% had a female householder with no husband present, and 29.2% were non-families. 24.1% of all households were made up of individuals and 10.4% had someone living alone who was 65 years of age or older. The average household size was 2.55 and the average family size was 3.04.

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