Seymour is a city located in Wayne County, Iowa. As of the 2000 census, the city had a total population of 810.

Geography

\nSeymour is located at 40°40'58" North, 93°7'15" West (40.682854, -93.120732)
1. According to the United States Census Bureau, the city has a total area of 6.1 km² (2.4 mi²). 6.1 km² (2.3 mi²) of it is land and none of it is covered by water.

Demographics

\nAs of the
census of 2000, there are 810 people, 336 households, and 219 families residing in the city. The population density is 133.1/km² (344.5/mi²). There are 393 housing units at an average density of 64.6/km² (167.1/mi²). The racial makeup of the city is 97.53% White, 0.00% African American, 0.37% Native American, 0.00% Asian, 0.00% Pacific Islander, 0.99% from other races, and 1.11% from two or more races. 2.10% of the population are Hispanic or Latino of any race. There are 336 households out of which 27.1% have children under the age of 18 living with them, 53.3% are married couples living together, 8.9% have a female householder with no husband present, and 34.8% are non-families. 31.5% of all households are made up of individuals and 17.3% have someone living alone who is 65 years of age or older. The average household size is 2.30 and the average family size is 2.89. In the city the population is 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 are 65 years of age or older. The median age is 42 years. For every 100 females there are 91.0 males. For every 100 females age 18 and over, there are 90.8 males. The median income for a household in the city is $26,172, and the median income for a family is $32,692. Males have a median income of $24,531 versus $20,833 for females. The per capita income for the city is $13,581. 22.0% of the population and 12.7% of families are below the poverty line. Out of the total people living in poverty, 37.3% are under the age of 18 and 12.3% are 65 or older.