Joseph, Utah

From Wikipedia, the free encyclopedia

Joseph is a town in Sevier County, Utah, United States. The population was 269 at the 2000 census.

[edit] Geography

Joseph is located at 38°37′36″N, 112°13′2″W (38.626797, -112.217185)^GR1.

According to the United States Census Bureau, the town has a total area of 2.3 km² (0.9 mi²), all land.

[edit] Demographics

As of the census^GR2 of 2000, there were 269 people, 92 households, and 73 families residing in the town. The population density was 116.7/km² (301.7/mi²). There were 109 housing units at an average density of 47.3/km² (122.3/mi²). The racial makeup of the town was 94.05% White, 0.74% African American, 0.37% Asian, 2.97% from other races, and 1.86% from two or more races. Hispanic or Latino of any race were 6.69% of the population.

There were 92 households out of which 30.4% had children under the age of 18 living with them, 69.6% were married couples living together, 7.6% had a female householder with no husband present, and 19.6% were non-families. 16.3% of all households were made up of individuals and 5.4% had someone living alone who was 65 years of age or older. The average household size was 2.92 and the average family size was 3.31.

In the town the population was spread out with 30.5% under the age of 18, 10.4% from 18 to 24, 22.7% from 25 to 44, 23.4% from 45 to 64, and 13.0% who were 65 years of age or older. The median age was 36 years. For every 100 females there were 96.4 males. For every 100 females age 18 and over, there were 103.3 males.

The median income for a household in the town was $29,375, and the median income for a family was $33,125. Males had a median income of $36,563 versus $17,500 for females. The per capita income for the town was $12,456. About 17.4% of families and 21.7% of the population were below the poverty line, including 30.9% of those under the age of eighteen and 25.0% of those sixty five or over.