Andrew, Iowa
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Andrew is a city located in Jackson County, Iowa. As of the 2000 census, the city had a total population of 460.
Geography
Andrew is located at 42°9'15" North, 90°35'30" West (42.154059, -90.591765)Template:GR.
According to the United States Census Bureau, the city has a total area of 0.7 km² (0.3 mi²). 0.7 km² (0.3 mi²) of it is land and none of it is covered by water.
Demographics
As of the censusTemplate:GR of 2000, there are 460 people, 165 households, and 120 families residing in the city. The population density is 683.1/km² (1,747.9/mi²). There are 169 housing units at an average density of 251.0/km² (642.2/mi²). The racial makeup of the city is 99.13% White, 0.00% African American, 0.22% Native American, 0.00% Asian, 0.00% Pacific Islander, 0.00% from other races, and 0.65% from two or more races. 2.83% of the population are Hispanic or Latino of any race.
There are 165 households out of which 45.5% have children under the age of 18 living with them, 53.9% are married couples living together, 13.9% have a female householder with no husband present, and 26.7% are non-families. 23.0% of all households are made up of individuals and 13.9% have someone living alone who is 65 years of age or older. The average household size is 2.79 and the average family size is 3.20.
In the city the population is spread out with 34.6% under the age of 18, 6.3% from 18 to 24, 32.6% from 25 to 44, 13.9% from 45 to 64, and 12.6% who are 65 years of age or older. The median age is 32 years. For every 100 females there are 110.0 males. For every 100 females age 18 and over, there are 94.2 males.
The median income for a household in the city is $36,563, and the median income for a family is $35,938. Males have a median income of $28,333 versus $22,222 for females. The per capita income for the city is $12,860. 9.3% of the population and 6.1% of families are below the poverty line. Out of the total population, 8.1% of those under the age of 18 and 7.3% of those 65 and older are living below the poverty line.