Montezuma, Iowa

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Montezuma is a city in Poweshiek County, Iowa, United States. The population was 1,440 at the 2000 census. It is the county seat of Poweshiek County[1].



Montezuma is located at 41°35′5″N 92°31′31″W / 41.58472°N 92.52528°W / 41.58472; -92.52528 (41.584737, -92.525258).[2]

According to the United States Census Bureau, the city has a total area of 2.5 square miles (6.4 km²), of which, 2.5 square miles (6.3 km²) of it is land and 0.40% is water.


As of the census[3] of 2000, there were 1,440 people, 601 households, and 390 families residing in the city. The population density was 586.9 people per square mile (226.9/km²). There were 641 housing units at an average density of 261.3/sq mi (101.0/km²). The racial makeup of the city was 98.75% White, 0.28% African American, 0.07% Native American, 0.35% Asian, and 0.56% from two or more races. Hispanic or Latino of any race were 0.42% of the population.

There were 601 households out of which 28.3% had children under the age of 18 living with them, 53.7% were married couples living together, 7.8% had a female householder with no husband present, and 35.1% were non-families. 31.1% of all households were made up of individuals and 16.1% had someone living alone who was 65 years of age or older. The average household size was 2.31 and the average family size was 2.91.

In the city the population was spread out with 23.6% under the age of 18, 8.9% from 18 to 24, 24.3% from 25 to 44, 20.8% from 45 to 64, and 22.4% who were 65 years of age or older. The median age was 40 years. For every 100 females there were 91.0 males. For every 100 females age 18 and over, there were 83.9 males.

The median income for a household in the city was $35,820, and the median income for a family was $43,083. Males had a median income of $31,483 versus $21,450 for females. The per capita income for the city was $17,806. About 3.8% of families and 6.4% of the population were below the poverty line, including 7.8% of those under age 18 and 5.8% of those age 65 or over.

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