Martinsburg, Iowa

From Wikipedia, the free encyclopedia

Martinsburg is a city in Keokuk County, Iowa, United States. The population was 126 at the 2000 census.

[edit] Geography

Martinsburg is located at 41°10′42″N, 92°15′5″W (41.178350, -92.251326)^GR1.

According to the United States Census Bureau, the city has a total area of 1.0 km² (0.4 mi²), all land.

[edit] Demographics

As of the census^GR2 of 2000, there were 126 people, 51 households, and 39 families residing in the city. The population density was 128.0/km² (333.4/mi²). There were 60 housing units at an average density of 61.0/km² (158.8/mi²). The racial makeup of the city was 97.62% White and 2.38% Native American.

There were 51 households out of which 27.5% had children under the age of 18 living with them, 60.8% were married couples living together, 13.7% had a female householder with no husband present, and 23.5% were non-families. 19.6% of all households were made up of individuals and 9.8% had someone living alone who was 65 years of age or older. The average household size was 2.47 and the average family size was 2.82.

In the city the population was spread out with 23.8% under the age of 18, 8.7% from 18 to 24, 23.0% from 25 to 44, 24.6% from 45 to 64, and 19.8% who were 65 years of age or older. The median age was 42 years. For every 100 females there were 93.8 males. For every 100 females age 18 and over, there were 88.2 males.

The median income for a household in the city was $35,625, and the median income for a family was $41,250. Males had a median income of $26,250 versus $13,750 for females. The per capita income for the city was $17,807. There were no families and 1.6% of the population living below the poverty line, including no under eighteens and 8.0% of those over 64.