Seneca, Shawano County, Wisconsin
From Wikipedia, the free encyclopedia
Seneca is a town in Shawano County, Wisconsin, in the United States. As of the 2000 census, the town population was 567.
[edit] Geography
According to the United States Census Bureau, the town has a total area of 94.9 km² (36.7 mi²). 94.7 km² (36.6 mi²) of it is land and 0.2 km² (0.1 mi²) of it (0.25%) is water.
[edit] Demographics
As of the census2 of 2000, there were 567 people, 204 households, and 156 families residing in the town. The population density was 6.0/km² (15.5/mi²). There were 255 housing units at an average density of 2.7/km² (7.0/mi²). The racial makeup of the town was 94.89% White, 2.12% Native American, 0.35% Asian, 0.18% from other races, and 2.47% from two or more races. Hispanic or Latino of any race were 0.53% of the population.
There were 204 households out of which 30.4% had children under the age of 18 living with them, 64.7% were married couples living together, 9.8% had a female householder with no husband present, and 23.5% were non-families. 17.2% of all households were made up of individuals and 7.4% had someone living alone who was 65 years of age or older. The average household size was 2.78 and the average family size was 3.16.
In the town the population was spread out with 26.3% under the age of 18, 7.1% from 18 to 24, 28.6% from 25 to 44, 21.9% from 45 to 64, and 16.2% who were 65 years of age or older. The median age was 38 years. For every 100 females there were 90.9 males. For every 100 females age 18 and over, there were 95.3 males.
The median income for a household in the town was $38,750, and the median income for a family was $43,750. Males had a median income of $26,250 versus $23,500 for females. The per capita income for the town was $15,601. About 9.3% of families and 13.2% of the population were below the poverty line, including 26.7% of those under age 18 and 10.5% of those age 65 or over.
