According to the US Census Bureau, the county has a total area of 2,628 square miles (6,810 km2), of which 2,626 square miles (6,800 km2) is land and 1.9 square miles (4.9 km2) (0.07%) is water.[5]
In comparison to the state of Rhode Island, the county has more than twice the land area, with approximately 1/500 of the population.[6]
As of the 2000 United States Census,[10] of 2000, there were 2,407 people, 1,011 households, and 679 families in the county. The population density was 1 person per square mile (0.39 person/km2). There were 1,338 housing units at an average density of 0.5 units per square mile (0.19 units/km2). The racial makeup of the county was 98.05% White, 0.12% Black or African American, 0.50% Native American, 0.12% Asian, 0.50% from other races, and 0.71% from two or more races. 1.50% of the population were Hispanic or Latino of any race. 35.1% were of German, 18.7% English, 11.2% Irish and 5.7% American ancestry.
There were 1,011 households, out of which 27.10% had children under the age of 18 living with them, 57.60% were married couples living together, 6.00% had a female householder with no husband present, and 32.80% were non-families. 29.50% of all households were made up of individuals, and 14.10% had someone living alone who was 65 years of age or older. The average household size was 2.28 and the average family size was 2.81.
The county population contained 22.60% under the age of 18, 6.10% from 18 to 24, 26.00% from 25 to 44, 26.60% from 45 to 64, and 18.70% who were 65 years of age or older. The median age was 43 years. For every 100 females there were 95.20 males. For every 100 females age 18 and over, there were 88.80 males.
The median income for a household in the county was $29,701, and the median income for a family was $33,714. Males had a median income of $25,909 versus $17,016 for females. The per capita income for the county was $15,757. About 10.70% of families and 13.40% of the population were below the poverty line, including 15.00% of those under age 18 and 15.60% of those age 65 or over.
2010 census
As of the 2010 United States Census, there were 2,484 people, 1,069 households, and 659 families in the county.[11] The population density was 0.9 people per square mile (0.35 people/km2). There were 1,338 housing units at an average density of 0.5 units per square mile (0.19 units/km2).[12] The racial makeup of the county was 96.5% white, 0.8% American Indian, 0.4% Asian, 0.2% black or African American, 0.5% from other races, and 1.6% from two or more races. Those of Hispanic or Latino origin made up 2.1% of the population.[11] In terms of ancestry, 41.5% were German, 19.0% were English, 17.7% were Irish, 6.6% were Swedish, and 3.0% were American.[13]
Of the 1,069 households, 23.2% had children under the age of 18 living with them, 50.4% were married couples living together, 7.4% had a female householder with no husband present, 38.4% were non-families, and 34.5% of all households were made up of individuals. The average household size was 2.12 and the average family size was 2.71. The median age was 46.1 years.[11]
The median income for a household in the county was $45,813 and the median income for a family was $57,153. Males had a median income of $41,898 versus $30,323 for females. The per capita income for the county was $22,885. About 6.3% of families and 11.5% of the population were below the poverty line, including 17.9% of those under age 18 and 13.3% of those age 65 or over.[14]
Like almost all of Wyoming, Niobrara County is overwhelmingly Republican. No Democratic presidential candidate has won Niobrara County since Franklin D. Roosevelt beat Alf Landon in 1936 by thirty-eight votes, and none since Lyndon Johnson in 1964 has attained thirty percent of the county's vote.
United States presidential election results for Niobrara County, Wyoming[15]
United States
Wyoming
