Sandyville, Iowa

From Wikipedia, the free encyclopedia

Sandyville is a city in Warren County, Iowa, United States. The population was 61 at the 2000 census.

[edit] Geography

Sandyville is located at 41°22′13″N, 93°23′8″W (41.370224, -93.385675)^GR1.

According to the United States Census Bureau, the city has a total area of 1.3 km² (0.5 mi²), all land.

[edit] Demographics

As of the census^GR2 of 2000, there were 61 people, 27 households, and 18 families residing in the city. The population density was 46.2/km² (119.6/mi²). There were 28 housing units at an average density of 21.2/km² (54.9/mi²). The racial makeup of the city was 95.08% White, and 4.92% from two or more races. Hispanic or Latino of any race were 4.92% of the population.

There were 27 households out of which 25.9% had children under the age of 18 living with them, 55.6% were married couples living together, 11.1% had a female householder with no husband present, and 33.3% were non-families. 33.3% of all households were made up of individuals and 11.1% had someone living alone who was 65 years of age or older. The average household size was 2.26 and the average family size was 2.89.

In the city the population was spread out with 16.4% under the age of 18, 16.4% from 18 to 24, 27.9% from 25 to 44, 34.4% from 45 to 64, and 4.9% who were 65 years of age or older. The median age was 40 years. For every 100 females there were 134.6 males. For every 100 females age 18 and over, there were 155.0 males.

The median income for a household in the city was $31,667, and the median income for a family was $34,375. Males had a median income of $27,500 versus $13,750 for females. The per capita income for the city was $20,028. None of the population and none of the families were below the poverty line.