Minersville, Utah

From Wikipedia, the free encyclopedia

Minersville is a town in Beaver County, Utah, United States. The population was 817 at the 2000 census.

[edit] Geography

Minersville is located at 38°12′48″N, 112°55′30″W (38.213242, -112.924928)^GR1.

According to the United States Census Bureau, the town has a total area of 1.6 km² (0.6 mi²), all land.

[edit] Demographics

As of the census^GR2 of 2000, there were 817 people, 256 households, and 204 families residing in the town. The population density was 500.7/km² (1,289.3/mi²). There were 274 housing units at an average density of 167.9/km² (432.4/mi²). The racial makeup of the town was 93.76% White, 0.12% Asian, 0.12% Pacific Islander, 4.16% from other races, and 1.84% from two or more races. Hispanic or Latino of any race were 6.00% of the population.

There were 256 households out of which 49.2% had children under the age of 18 living with them, 73.0% were married couples living together, 5.1% had a female householder with no husband present, and 20.3% were non-families. 18.0% of all households were made up of individuals and 11.7% had someone living alone who was 65 years of age or older. The average household size was 3.19 and the average family size was 3.68.

In the town the population was spread out with 39.0% under the age of 18, 10.2% from 18 to 24, 23.6% from 25 to 44, 14.4% from 45 to 64, and 12.7% who were 65 years of age or older. The median age was 26 years. For every 100 females there were 109.5 males. For every 100 females age 18 and over, there were 100.8 males.

The median income for a household in the town was $36,563, and the median income for a family was $39,286. Males had a median income of $31,042 versus $15,179 for females. The per capita income for the town was $11,693. About 7.3% of families and 9.0% of the population were below the poverty line, including 11.3% of those under age 18 and 3.4% of those age 65 or over.