Laceyville, Pennsylvania

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Laceyville is a borough in Wyoming County, Pennsylvania, United States. The population was 396 at the 2000 census.



Laceyville is located at 41°38′43″N 76°9′38″W / 41.64528°N 76.16056°W / 41.64528; -76.16056 (41.645337, -76.160451)[1].

According to the United States Census Bureau, the borough has a total area of 0.2 square miles (0.5 km²), all of it land.


As of the census[2] of 2000, there were 396 people, 162 households, and 108 families residing in the borough. The population density was 1,880.5 people per square mile (728.1/km²). There were 179 housing units at an average density of 850.0/sq mi (329.1/km²). The racial makeup of the borough was 98.74% White, 0.51% Native American, 0.51% Asian, and 0.25% from two or more races. Hispanic or Latino of any race were 2.02% of the population.

There were 162 households out of which 35.2% had children under the age of 18 living with them, 48.8% were married couples living together, 13.6% had a female householder with no husband present, and 33.3% were non-families. 23.5% of all households were made up of individuals and 11.1% had someone living alone who was 65 years of age or older. The average household size was 2.44 and the average family size was 2.93.

In the borough the population was spread out with 25.8% under the age of 18, 10.1% from 18 to 24, 29.8% from 25 to 44, 21.0% from 45 to 64, and 13.4% who were 65 years of age or older. The median age was 33 years. For every 100 females there were 75.2 males. For every 100 females age 18 and over, there were 75.0 males.

The median income for a household in the borough was $29,375, and the median income for a family was $37,917. Males had a median income of $30,000 versus $20,000 for females. The per capita income for the borough was $14,185. About 11.0% of families and 12.5% of the population were below the poverty line, including 18.5% of those under age 18 and 3.9% of those age 65 or over.


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