Ozawkie, Kansas

From Wikipedia, the free encyclopedia

Ozawkie is a city in Jefferson County, Kansas, United States. The population was 552 at the 2000 census.

[edit] Geography

Ozawkie is located at 39°13′60″N, 95°27′57″W (39.233201, -95.465709)^GR1.

According to the United States Census Bureau, the city has a total area of 0.8 km² (0.3 mi²), all land.

[edit] Demographics

As of the census^GR2 of 2000, there were 552 people, 202 households, and 167 families residing in the city. The population density was 687.5/km² (1,797.3/mi²). There were 216 housing units at an average density of 269.0/km² (703.3/mi²). The racial makeup of the city was 97.28% White, 0.72% African American, 1.27% Native American, 0.18% Asian, and 0.54% from two or more races. Hispanic or Latino of any race were 0.36% of the population.

There were 202 households out of which 36.1% had children under the age of 18 living with them, 77.2% were married couples living together, 5.0% had a female householder with no husband present, and 17.3% were non-families. 13.9% of all households were made up of individuals and 7.9% had someone living alone who was 65 years of age or older. The average household size was 2.73 and the average family size was 3.03.

In the city the population was spread out with 27.4% under the age of 18, 4.3% from 18 to 24, 25.4% from 25 to 44, 29.3% from 45 to 64, and 13.6% who were 65 years of age or older. The median age was 40 years. For every 100 females there were 95.7 males. For every 100 females age 18 and over, there were 92.8 males.

The median income for a household in the city was $62,969, and the median income for a family was $67,292. Males had a median income of $42,308 versus $27,375 for females. The per capita income for the city was $21,857. About 1.1% of families and 2.3% of the population were below the poverty line, including none of those under the age of eighteen or sixty-five or over.