Conesville, Iowa

From Wikipedia, the free encyclopedia

Conesville is a city in Muscatine County, Iowa, United States. The population was 424 at the 2000 census.

[edit] Geography

Conesville is located at 41°22′46″N, 91°21′1″W (41.379323, -91.350211)^GR1.

According to the United States Census Bureau, the city has a total area of 1.0 km² (0.4 mi²), all land.

[edit] Demographics

As of the census^GR2 of 2000, there were 424 people, 133 households, and 100 families residing in the city. The population density was 442.5/km² (1,147.7/mi²). There were 144 housing units at an average density of 150.3/km² (389.8/mi²). The racial makeup of the city was 78.77% White, 0.24% Asian, 20.28% from other races, and 0.71% from two or more races. Hispanic or Latino of any race were 58.96% of the population.

There were 133 households out of which 42.1% had children under the age of 18 living with them, 54.9% were married couples living together, 13.5% had a female householder with no husband present, and 24.1% were non-families. 18.8% of all households were made up of individuals and 4.5% had someone living alone who was 65 years of age or older. The average household size was 3.19 and the average family size was 3.62.

In the city the population was spread out with 36.3% under the age of 18, 7.8% from 18 to 24, 33.3% from 25 to 44, 14.2% from 45 to 64, and 8.5% who were 65 years of age or older. The median age was 28 years. For every 100 females there were 126.7 males. For every 100 females age 18 and over, there were 114.3 males.

The median income for a household in the city was $29,464, and the median income for a family was $30,000. Males had a median income of $28,000 versus $19,167 for females. The per capita income for the city was $10,097. About 11.1% of families and 13.9% of the population were below the poverty line, including 19.3% of those under age 18 and 4.9% of those age 65 or over.