Stockville, Nebraska

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Stockville is a village in Frontier County, Nebraska, United States. The population was 36 at the 2000 census. It is the county seat of Frontier County[3].


Stockville is located at 40°32′00″N 100°23′04″W / 40.533250°N 100.384545°W / 40.533250; -100.384545Coordinates: 40°32′00″N 100°23′04″W / 40.533250°N 100.384545°W / 40.533250; -100.384545 (40.533250, -100.384545).[4]

According to the United States Census Bureau, the village has a total area of 0.3 square miles (0.7 km²), all of it land.


As of the census[1] of 2000, there were 36 people, 17 households, and 10 families residing in the village. The population density was 137.9 people per square mile (53.5/km²). There were 20 housing units at an average density of 76.6/sq mi (29.7/km²). The racial makeup of the village was 100.00% White. Hispanic or Latino of any race were 5.56% of the population.

There were 17 households out of which 29.4% had children under the age of 18 living with them, 47.1% were married couples living together, 17.6% had a female householder with no husband present, and 35.3% were non-families. 35.3% of all households were made up of individuals and 5.9% had someone living alone who was 65 years of age or older. The average household size was 2.12 and the average family size was 2.73.

In the village the population was spread out with 27.8% under the age of 18, 2.8% from 18 to 24, 27.8% from 25 to 44, 27.8% from 45 to 64, and 13.9% who were 65 years of age or older. The median age was 40 years. For every 100 females there were 125.0 males. For every 100 females age 18 and over, there were 116.7 males.

The median income for a household in the village was $31,250, and the median income for a family was $35,714. Males had a median income of $22,083 versus $21,250 for females. The per capita income for the village was $13,329. There were no families and 4.4% of the population living below the poverty line, including no under eighteens and none of those over 64.

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