Freedom, New York – Wikipedia, the free encyclopedia

Freedom is a town in Cattaraugus County, New York, United States. The population was 2,405 at the 2010 census.[1] The town is in the northeast corner of Cattaraugus County.

As of the census[4] of 2000, there were 2,493 people, 871 households, and 680 families residing in the town. The population density was 61.8 people per square mile (23.9/km). There were 1,033 housing units at an average density of 25.6 per square mile (9.9/km). The racial makeup of the town was 98.88% White, 0.16% African American, 0.08% Native American, 0.08% Asian, 0.08% from other races, and 0.72% from two or more races. Hispanic or Latino of any race were 0.76% of the population.

There were 871 households out of which 39.8% had children under the age of 18 living with them, 63.5% were married couples living together, 7.9% had a female householder with no husband present, and 21.9% were non-families. 16.4% of all households were made up of individuals and 6.3% had someone living alone who was 65 years of age or older. The average household size was 2.86 and the average family size was 3.20.

In the town the population was spread out with 29.5% under the age of 18, 8.0% from 18 to 24, 30.0% from 25 to 44, 23.0% from 45 to 64, and 9.5% who were 65 years of age or older. The median age was 35 years. For every 100 females there were 103.3 males. For every 100 females age 18 and over, there were 103.9 males.

The median income for a household in the town was $34,654, and the median income for a family was $36,061. Males had a median income of $27,380 versus $22,188 for females. The per capita income for the town was $14,145. About 12.2% of families and 11.3% of the population were below the poverty line, including 10.3% of those under age 18 and 18.5% of those age 65 or over.

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Freedom, New York - Wikipedia, the free encyclopedia