Rushford, Wisconsin
From Wikipedia, the free encyclopedia
Rushford is a town in Winnebago County, Wisconsin, United States. The population was 1,471 at the 2000 census.
[edit] Geography
According to the United States Census Bureau, the town has a total area of 92.1 km² (35.6 mi²). 90.8 km² (35.1 mi²) of it is land and 1.3 km² (0.5 mi²) of it (1.38%) is water.
[edit] Demographics
As of the census2 of 2000, there were 1,471 people, 549 households, and 409 families residing in the town. The population density was 16.2/km² (42.0/mi²). There were 594 housing units at an average density of 6.5/km² (16.9/mi²). The racial makeup of the town was 98.44% White, 0.07% African American, 0.14% Native American, 0.07% Asian, 0.34% from other races, and 0.95% from two or more races. Hispanic or Latino of any race were 1.36% of the population.
There were 549 households out of which 32.1% had children under the age of 18 living with them, 65.9% were married couples living together, 4.7% had a female householder with no husband present, and 25.5% were non-families. 20.0% of all households were made up of individuals and 7.7% had someone living alone who was 65 years of age or older. The average household size was 2.68 and the average family size was 3.13.
In the town the population was spread out with 25.4% under the age of 18, 7.1% from 18 to 24, 29.8% from 25 to 44, 26.0% from 45 to 64, and 11.8% who were 65 years of age or older. The median age was 38 years. For every 100 females there were 106.9 males. For every 100 females age 18 and over, there were 104.9 males.
The median income for a household in the town was $45,990, and the median income for a family was $52,614. Males had a median income of $34,038 versus $22,222 for females. The per capita income for the town was $20,768. About 1.5% of families and 2.3% of the population were below the poverty line, including 1.2% of those under age 18 and 3.1% of those age 65 or over.
