Hamburg, Iowa

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Hamburg is a city in Fremont County, Iowa, United States, along the Nishnabotna River. The population was 1,240 at the 2000 census.



Hamburg is located at 40°36′21″N 95°39′18″W / 40.605816°N 95.655023°W / 40.605816; -95.655023 (40.605816, -95.655023).[1]

According to the United States Census Bureau, the city has a total area of 1.1 square miles (2.9 km2), all of it land.


As of the census[2] of 2000, there were 1,240 people, 544 households, and 343 families residing in the city. The population density was 1,107.1 people per square mile (427.5/km2). There were 604 housing units at an average density of 539.3/sq mi (208.2/km2). The racial makeup of the city was 96.21% White, 0.08% African American, 0.24% Native American, 0.32% Asian, 2.66% from other races, and 0.48% from two or more races. Hispanic or Latino of any race were 5.48% of the population.

There were 544 households out of which 27.6% had children under the age of 18 living with them, 46.1% were married couples living together, 12.5% had a female householder with no husband present, and 36.8% were non-families. 32.4% of all households were made up of individuals and 18.4% had someone living alone who was 65 years of age or older. The average household size was 2.27 and the average family size was 2.83.

In the city the population was spread out with 22.9% under the age of 18, 6.9% from 18 to 24, 24.7% from 25 to 44, 24.3% from 45 to 64, and 21.2% who were 65 years of age or older. The median age was 42 years. For every 100 females there were 88.7 males. For every 100 females age 18 and over, there were 87.5 males.

The median income for a household in the city was $29,479, and the median income for a family was $42,935. Males had a median income of $28,162 versus $20,781 for females. The per capita income for the city was $16,050. About 11.7% of families and 13.9% of the population were below the poverty line, including 21.3% of those under age 18 and 11.0% of those age 65 or over.

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