Vineyard, Utah

From Wikipedia, the free encyclopedia

Vineyard is a town in Utah County, Utah, United States. The population was 150 at the 2000 census.

[edit] Geography

Vineyard is located at 40°18′14″N, 111°45′28″W (40.303835, -111.757722)^GR1.

According to the United States Census Bureau, the town has a total area of 11.4 km² (4.4 mi²). 10.2 km² (3.9 mi²) of it is land and 1.3 km² (0.5 mi²) of it (11.31%) is water.

[edit] Demographics

As of the census^GR2 of 2000, there were 150 people, 43 households, and 38 families residing in the town. The population density was 14.8/km² (38.3/mi²). There were 46 housing units at an average density of 4.5/km² (11.7/mi²). The racial makeup of the town was 92.67% White, 6.67% from other races, and 0.67% from two or more races. Hispanic or Latino of any race were 7.33% of the population.

There were 43 households out of which 30.2% had children under the age of 18 living with them, 83.7% were married couples living together, 4.7% had a female householder with no husband present, and 11.6% were non-families. 11.6% of all households were made up of individuals and 7.0% had someone living alone who was 65 years of age or older. The average household size was 3.49 and the average family size was 3.76.

In the town the population was spread out with 29.3% under the age of 18, 12.7% from 18 to 24, 28.7% from 25 to 44, 16.0% from 45 to 64, and 13.3% who were 65 years of age or older. The median age was 29 years. For every 100 females there were 111.3 males. For every 100 females age 18 and over, there were 103.8 males.

The median income for a household in the town was $55,313, and the median income for a family was $55,625. Males had a median income of $30,313 versus $22,500 for females. The per capita income for the town was $12,841. None of the population and none of the families were below the poverty line.