Saco, Montana

From Wikipedia, the free encyclopedia

Saco is a town in Phillips County, Montana, United States. The population was 224 at the 2000 census.

[edit] Geography

Saco is located at 48°27′27″N, 107°20′31″W (48.457557, -107.342030)^GR1.

According to the United States Census Bureau, the town has a total area of 0.9 km² (0.3 mi²), all land.

[edit] Demographics

As of the census^GR2 of 2000, there were 224 people, 109 households, and 57 families residing in the town. The population density was 262.1/km² (683.9/mi²). There were 135 housing units at an average density of 158.0/km² (412.2/mi²). The racial makeup of the town was 96.43% White, 1.79% Native American, and 1.79% from two or more races.

There were 109 households out of which 23.9% had children under the age of 18 living with them, 41.3% were married couples living together, 8.3% had a female householder with no husband present, and 46.8% were non-families. 44.0% of all households were made up of individuals and 22.9% had someone living alone who was 65 years of age or older. The average household size was 2.06 and the average family size was 2.84.

In the town the population was spread out with 23.7% under the age of 18, 3.6% from 18 to 24, 26.3% from 25 to 44, 24.6% from 45 to 64, and 21.9% who were 65 years of age or older. The median age was 43 years. For every 100 females there were 98.2 males. For every 100 females age 18 and over, there were 94.3 males.

The median income for a household in the town was $23,125, and the median income for a family was $26,667. Males had a median income of $23,125 versus $24,375 for females. The per capita income for the town was $16,421. About 14.3% of families and 14.0% of the population were below the poverty line, including 22.0% of those under the age of eighteen and 13.2% of those sixty five or over.