Wesley, Iowa

From Wikipedia, the free encyclopedia

Wesley is a city in Kossuth County, Iowa, United States. As of the 2000 census, the city population was 467.

[edit] Geography

Wesley is located at 43°5′15″N, 93°59′35″W (43.087557, -93.993140)^GR1.

According to the United States Census Bureau, the city has a total area of 1.5 km² (0.6 mi²), all land.

[edit] Demographics

As of the census^GR2 of 2000, there were 467 people, 189 households, and 130 families residing in the city. The population density was 310.9/km² (811.4/mi²). There were 200 housing units at an average density of 133.1/km² (347.5/mi²). The racial makeup of the city was 99.14% White, 0.86% from other races. Hispanic or Latino of any race were 0.86% of the population.

There were 189 households out of which 29.6% had children under the age of 18 living with them, 56.6% were married couples living together, 7.9% had a female householder with no husband present, and 31.2% were non-families. 29.6% of all households were made up of individuals and 17.5% 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 3.04.

In the city the population was spread out with 27.8% under the age of 18, 4.9% from 18 to 24, 25.7% from 25 to 44, 18.8% from 45 to 64, and 22.7% who were 65 years of age or older. The median age was 39 years. For every 100 females there were 103.9 males. For every 100 females age 18 and over, there were 92.6 males.

The median income for a household in the city was $39,688, and the median income for a family was $47,500. Males had a median income of $30,568 versus $21,250 for females. The per capita income for the city was $19,225. About 6.3% of families and 11.0% of the population were below the poverty line, including 14.6% of those under age 18 and 13.4% of those age 65 or over.