Seymour, Iowa

From Wikipedia, the free encyclopedia

Seymour is a city in Wayne County, Iowa, United States. The population was 810 at the 2000 census.

[edit] Geography

Seymour is located at 40°40′58″N, 93°7′15″W (40.682854, -93.120732)^GR1.

According to the United States Census Bureau, the city has a total area of 6.1 km² (2.4 mi²), all land.

[edit] Demographics

As of the census^GR2 of 2000, there were 810 people, 336 households, and 219 families residing in the city. The population density was 133.1/km² (344.5/mi²). There were 393 housing units at an average density of 64.6/km² (167.1/mi²). The racial makeup of the city was 97.53% White, 0.37% Native American, 0.99% from other races, and 1.11% from two or more races. Hispanic or Latino of any race were 2.10% of the population.

There were 336 households out of which 27.1% had children under the age of 18 living with them, 53.3% were married couples living together, 8.9% had a female householder with no husband present, and 34.8% were non-families. 31.5% of all households were made up of individuals and 17.3% had someone living alone who was 65 years of age or older. The average household size was 2.30 and the average family size was 2.89.

In the city the population was spread out with 23.0% under the age of 18, 6.4% from 18 to 24, 24.9% from 25 to 44, 20.9% from 45 to 64, and 24.8% who were 65 years of age or older. The median age was 42 years. For every 100 females there were 91.0 males. For every 100 females age 18 and over, there were 90.8 males.

The median income for a household in the city was $26,172, and the median income for a family was $32,692. Males had a median income of $24,531 versus $20,833 for females. The per capita income for the city was $13,581. About 12.7% of families and 22.0% of the population were below the poverty line, including 37.3% of those under age 18 and 12.3% of those age 65 or over.