Sageville, Iowa

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Sageville is a city in Dubuque County, Iowa, United States. The population was 203 at the 2000 census.

[edit] Geography

Sageville is located at 42°33′10″N, 90°42′44″W (42.552850, -90.712160)^GR1.

According to the United States Census Bureau, the city has a total area of 1.6 km² (0.6 mi²), all land.

[edit] Demographics

As of the census^GR2 of 2000, there were 203 people, 85 households, and 54 families residing in the city. The population density was 124.4/km² (324.5/mi²). There were 95 housing units at an average density of 58.2/km² (151.8/mi²). The racial makeup of the city was 96.55% White, 0.99% African American, 0.49% Asian, 1.48% from other races, and 0.49% from two or more races. Hispanic or Latino of any race were 1.97% of the population.

There were 85 households out of which 36.5% had children under the age of 18 living with them, 50.6% were married couples living together, 11.8% had a female householder with no husband present, and 35.3% were non-families. 29.4% of all households were made up of individuals and 9.4% had someone living alone who was 65 years of age or older. The average household size was 2.39 and the average family size was 3.00.

In the city the population was spread out with 27.1% under the age of 18, 12.8% from 18 to 24, 34.0% from 25 to 44, 15.8% from 45 to 64, and 10.3% who were 65 years of age or older. The median age was 31 years. For every 100 females there were 99.0 males. For every 100 females age 18 and over, there were 89.7 males.

The median income for a household in the city was $29,167, and the median income for a family was $31,667. Males had a median income of $24,375 versus $19,375 for females. The per capita income for the city was $13,700. About 8.6% of families and 14.8% of the population were below the poverty line, including 26.3% of those under the age of eighteen and 18.2% of those sixty five or over.