Seward township is located in Kosciusko County, Indiana. Seward township has a 2026 population of 2,321. Seward township is currently growing at a rate of 0.43% annually and its population has increased by 1.22% since the most recent census, which recorded a population of 2,293 in 2020.
The median household income in Seward township is $77,267 with a poverty rate of 6.89%. The median age in Seward township is 49.2 years: 41.9 years for males, and 54.1 years for females. For every 100 females there are 113.1 males.
Data after 2023 is projected based on recent change
Overall: 49.2 years
Female: 54.1 years
Male: 41.9 years
There are 2,163 adults, (628 of whom are seniors) in Seward township.
Female: 1,180 (46.9%)
Male: 1,334 (53.1%)
The racial composition of Seward township includes 86.28% White, 8.67% other race, and smaller percentages for Black or African American, Asian and multiracial populations.
| Race | Population ↓ | Percentage (of total) |
|---|---|---|
| White | 2,169 | 86.28% |
| Other race | 218 | 8.67% |
| Two or more races | 97 | 3.86% |
| Black or African American | 16 | 0.64% |
| Asian | 14 | 0.56% |
Seward township 's average per capita income is $55,845. Household income levels show a median of $77,267. The poverty rate stands at 6.89%.
| Name | Median ↓ | Mean |
|---|---|---|
| Married Families | $102,619 | - |
| Families | $92,857 | $122,024 |
| Households | $77,267 | $105,398 |
| Non Families | $36,587 | $53,884 |
Average Income
Median Household Income
Poverty Rate