Martin, North Dakota

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Martin is a city in Sheridan County, North Dakota in the United States. The population was 96 at the 2000 census.

[edit] Geography

Martin is located at 47°49′36″N, 100°6′54″W (47.826776, -100.114868)^GR1.

According to the United States Census Bureau, the city has a total area of 0.2 km² (0.1 mi²), all land.

[edit] Demographics

As of the census^GR2 of 2000, there were 96 people, 44 households, and 26 families residing in the city. The population density was 411.8/km² (1,094.7/mi²). There were 48 housing units at an average density of 205.9/km² (547.4/mi²). The racial makeup of the city was 100.00% White. Hispanic or Latino of any race were 1.04% of the population.

There were 44 households out of which 27.3% had children under the age of 18 living with them, 45.5% were married couples living together, 11.4% had a female householder with no husband present, and 40.9% were non-families. 38.6% of all households were made up of individuals and 20.5% had someone living alone who was 65 years of age or older. The average household size was 2.18 and the average family size was 2.85.

In the city the population was spread out with 24.0% under the age of 18, 5.2% from 18 to 24, 24.0% from 25 to 44, 19.8% from 45 to 64, and 27.1% who were 65 years of age or older. The median age was 43 years. For every 100 females there were 84.6 males. For every 100 females age 18 and over, there were 87.2 males.

The median income for a household in the city was $28,438, and the median income for a family was $35,313. Males had a median income of $30,000 versus $18,750 for females. The per capita income for the city was $15,767. There were 8.0% of families and 17.4% of the population living below the poverty line, including 21.7% of under eighteens and none of those over 64.