Rose Hill, Iowa

From Wikipedia, the free encyclopedia

Rose Hill is a city in Mahaska County, Iowa, United States. The population was 205 at the 2000 census.

[edit] Geography

Rose Hill is located at 41°19′17″N, 92°27′47″W (41.321389, -92.462952)^GR1.

According to the United States Census Bureau, the city has a total area of 0.4 km² (0.1 mi²), all land.

[edit] Demographics

As of the census^GR2 of 2000, there were 205 people, 82 households, and 56 families residing in the city. The population density was 565.4/km² (1,501.9/mi²). There were 91 housing units at an average density of 251.0/km² (666.7/mi²). The racial makeup of the city was 96.10% White, 0.49% Native American, 0.49% from other races, and 2.93% from two or more races. Hispanic or Latino of any race were 1.95% of the population.

There were 82 households out of which 32.9% had children under the age of 18 living with them, 52.4% were married couples living together, 9.8% had a female householder with no husband present, and 31.7% were non-families. 25.6% of all households were made up of individuals and 3.7% had someone living alone who was 65 years of age or older. The average household size was 2.50 and the average family size was 3.00.

In the city the population was spread out with 26.3% under the age of 18, 9.8% from 18 to 24, 34.6% from 25 to 44, 19.5% from 45 to 64, and 9.8% who were 65 years of age or older. The median age was 34 years. For every 100 females there were 111.3 males. For every 100 females age 18 and over, there were 104.1 males.

The median income for a household in the city was $30,469, and the median income for a family was $32,143. Males had a median income of $30,357 versus $18,125 for females. The per capita income for the city was $21,298. About 16.0% of families and 14.0% of the population were below the poverty line, including 6.5% of those under the age of eighteen and 17.2% of those sixty five or over.