Franklin, Sauk County, Wisconsin


Franklin is a town in Sauk County, Wisconsin, United States. The population was 696 at the 2000 census.
The town is named after Benjamin Franklin.

Geography

According to the United States Census Bureau, the town has a total area of 49.3 square miles, of which, 49.2 square miles of it is land and 0.2 square miles of it is water.

Demographics

As of the census of 2000, there were 696 people, 250 households, and 198 families residing in the town. The population density was 14.2 people per square mile. There were 267 housing units at an average density of 5.4 per square mile. The racial makeup of the town was 98.71% White, 1.01% Native American, and 0.29% from two or more races. Hispanic or Latino of any race were 0.72% of the population.
There were 250 households, out of which 38.4% had children under the age of 18 living with them, 71.6% were married couples living together, 2.8% had a female householder with no husband present, and 20.4% were non-families. 17.2% of all households were made up of individuals, and 7.2% 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 3.18.
In the town, the population was spread out, with 29.2% under the age of 18, 5.0% from 18 to 24, 31.0% from 25 to 44, 23.6% from 45 to 64, and 11.2% who were 65 years of age or older. The median age was 37 years. For every 100 females, there were 109.0 males. For every 100 females age 18 and over, there were 110.7 males.
The median income for a household in the town was $45,982, and the median income for a family was $49,250. Males had a median income of $30,917 versus $22,500 for females. The per capita income for the town was $18,494. About 4.0% of families and 4.3% of the population were below the poverty line, including 5.6% of those under age 18 and 9.8% of those age 65 or over.