Hayesville, Iowa

From Wikipedia, the free encyclopedia

Hayesville is a city in Keokuk County, Iowa, United States. The population was 64 at the 2000 census.

[edit] Geography

Hayesville is located at 41°15′52″N, 92°14′56″W (41.264570, -92.248880)^GR1.

According to the United States Census Bureau, the city has a total area of 0.7 km² (0.3 mi²), all land.

[edit] Demographics

As of the census^GR2 of 2000, there were 64 people, 28 households, and 19 families residing in the city. The population density was 95.0/km² (250.5/mi²). There were 32 housing units at an average density of 47.5/km² (125.3/mi²). The racial makeup of the city was 100.00% White.

There were 28 households out of which 25.0% had children under the age of 18 living with them, 60.7% were married couples living together, 3.6% had a female householder with no husband present, and 32.1% were non-families. 21.4% of all households were made up of individuals and 14.3% had someone living alone who was 65 years of age or older. The average household size was 2.29 and the average family size was 2.68.

In the city the population was spread out with 18.8% under the age of 18, 7.8% from 18 to 24, 23.4% from 25 to 44, 29.7% from 45 to 64, and 20.3% who were 65 years of age or older. The median age was 45 years. For every 100 females there were 82.9 males. For every 100 females age 18 and over, there were 92.6 males.

The median income for a household in the city was $38,125, and the median income for a family was $36,250. Males had a median income of $33,750 versus $16,250 for females. The per capita income for the city was $15,973. None of the population and none of the families were below the poverty line.