Hanlontown, Iowa

From Wikipedia, the free encyclopedia

Hanlontown is a city in Worth County, Iowa, United States. The population was 229 at the 2000 census.

[edit] Geography

Hanlontown is located at 43°16′51″N, 93°22′46″W (43.280883, -93.379370)^GR1.

According to the United States Census Bureau, the city has a total area of 2.5 km² (1.0 mi²), all land.

[edit] Demographics

As of the census^GR2 of 2000, there were 229 people, 90 households, and 65 families residing in the city. The population density was 91.2/km² (235.9/mi²). There were 97 housing units at an average density of 38.6/km² (99.9/mi²). The racial makeup of the city was 98.25% White, 0.44% African American, 0.87% Asian, and 0.44% from two or more races. Hispanic or Latino of any race were 4.37% of the population.

There were 90 households out of which 37.8% had children under the age of 18 living with them, 66.7% were married couples living together, 4.4% had a female householder with no husband present, and 26.7% were non-families. 22.2% of all households were made up of individuals and 13.3% had someone living alone who was 65 years of age or older. The average household size was 2.54 and the average family size was 2.98.

In the city the population was spread out with 28.8% under the age of 18, 5.7% from 18 to 24, 35.8% from 25 to 44, 17.5% from 45 to 64, and 12.2% who were 65 years of age or older. The median age was 33 years. For every 100 females there were 90.8 males. For every 100 females age 18 and over, there were 96.4 males.

The median income for a household in the city was $35,000, and the median income for a family was $46,458. Males had a median income of $29,792 versus $22,778 for females. The per capita income for the city was $17,320. About 3.0% of families and 2.7% of the population were below the poverty line, including none of those under the age of eighteen and 7.4% of those sixty five or over.