Spiceland, Indiana

From Wikipedia, the free encyclopedia

Spiceland from air, looking east.

Spiceland is a town in Henry County, Indiana, United States. The population was 807 at the 2000 census.

[edit] Geography

Spiceland is located at 39°50′16″N, 85°26′10″W (39.837743, -85.436089)^GR1.

According to the United States Census Bureau, the town has a total area of 1.1 km² (0.4 mi²), all land.

[edit] Demographics

As of the census^GR2 of 2000, there were 807 people, 327 households, and 240 families residing in the town. The population density was 708.1/km² (1,838.6/mi²). There were 340 housing units at an average density of 298.4/km² (774.6/mi²). The racial makeup of the town was 98.14% White, 0.12% African American, 0.25% Native American, 1.12% from other races, and 0.37% from two or more races. Hispanic or Latino of any race were 1.61% of the population.

There were 327 households out of which 30.3% had children under the age of 18 living with them, 64.2% were married couples living together, 7.6% had a female householder with no husband present, and 26.6% were non-families. 24.2% of all households were made up of individuals and 12.2% 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 2.92.

In the town the population was spread out with 25.0% under the age of 18, 7.7% from 18 to 24, 30.7% from 25 to 44, 23.9% from 45 to 64, and 12.6% who were 65 years of age or older. The median age was 36 years. For every 100 females there were 93.1 males. For every 100 females age 18 and over, there were 88.5 males.

The median income for a household in the town was $45,875, and the median income for a family was $54,904. Males had a median income of $37,333 versus $28,393 for females. The per capita income for the town was $20,419. About 2.3% of families and 2.5% of the population were below the poverty line, including 3.1% of those under age 18 and 6.9% of those age 65 or over.