Manhattan, Montana

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Manhattan is a town in Gallatin County, Montana, United States. The population was 1,396 at the 2000 census. It is part of the 'Bozeman Micropolitan Statistical Area'.



Manhattan is located at 45°51′27″N 111°19′52″W / 45.8575°N 111.33111°W / 45.8575; -111.33111 (45.857367, -111.331005)[1].

According to the United States Census Bureau, the town has a total area of 0.6 square miles (1.6 km²), all of it land.


As of the census[2] of 2000, there were 1,396 people, 553 households, and 389 families residing in the town. The population density was 2,303.9 people per square mile (883.6/km²). There were 582 housing units at an average density of 960.5/sq mi (368.4/km²). The racial makeup of the town was 97.42% White, 0.72% Native American, 0.29% Asian, 0.07% Pacific Islander, 0.43% from other races, and 1.07% from two or more races. Hispanic or Latino of any race were 0.93% of the population.

There were 553 households out of which 33.6% had children under the age of 18 living with them, 61.3% were married couples living together, 4.7% had a female householder with no husband present, and 29.5% were non-families. 24.6% of all households were made up of individuals and 10.1% had someone living alone who was 65 years of age or older. The average household size was 2.52 and the average family size was 3.03.

In the town the population was spread out with 26.6% under the age of 18, 7.7% from 18 to 24, 29.7% from 25 to 44, 22.8% from 45 to 64, and 13.1% who were 65 years of age or older. The median age was 38 years. For every 100 females there were 105.9 males. For every 100 females age 18 and over, there were 98.8 males.

The median income for a household in the town was $38,242, and the median income for a family was $45,521. Males had a median income of $31,319 versus $19,875 for females. The per capita income for the town was $17,024. About 5.2% of families and 7.1% of the population were below the poverty line, including 7.5% of those under age 18 and 12.4% of those age 65 or over.

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