Senecaville, Ohio

From Wikipedia, the free encyclopedia

Senecaville is a village in Guernsey County, Ohio, United States. The population was 453 at the 2000 census.

[edit] Geography

Senecaville is located at 39°56′7″N, 81°27′40″W (39.935164, -81.461166)^GR1.

According to the United States Census Bureau, the village has a total area of 1.3 km² (0.5 mi²), all land.

[edit] Demographics

As of the census^GR2 of 2000, there were 453 people, 176 households, and 121 families residing in the village. The population density was 356.9/km² (931.2/mi²). There were 204 housing units at an average density of 160.7/km² (419.4/mi²). The racial makeup of the village was 100.00% White.

There were 176 households out of which 34.7% had children under the age of 18 living with them, 52.8% were married couples living together, 10.2% had a female householder with no husband present, and 30.7% were non-families. 26.7% of all households were made up of individuals and 11.4% had someone living alone who was 65 years of age or older. The average household size was 2.57 and the average family size was 3.14.

In the village the population was spread out with 27.2% under the age of 18, 7.3% from 18 to 24, 31.1% from 25 to 44, 21.4% from 45 to 64, and 13.0% who were 65 years of age or older. The median age was 36 years. For every 100 females there were 85.7 males. For every 100 females age 18 and over, there were 89.7 males.

The median income for a household in the village was $27,813, and the median income for a family was $31,583. Males had a median income of $25,000 versus $17,361 for females. The per capita income for the village was $12,559. About 12.0% of families and 17.2% of the population were below the poverty line, including 26.6% of those under age 18 and 13.4% of those age 65 or over.