Selinsgrove is a borough located in Snyder County, Pennsylvania. Selinsgrove has a 2025 population of 5,448 . Selinsgrove is currently declining at a rate of -0.96% annually and its population has decreased by -4.59% since the most recent census, which recorded a population of 5,710 in 2020.
The average household income in Selinsgrove is $70,262 with a poverty rate of 7.74%. The median age in Selinsgrove is 24.4 years: 26.2 years for males, and 23.2 years for females. For every 100 females there are 94.9 males.
Data after 2023 is projected based on recent change
The racial composition of Selinsgrove includes 87.64% White, 4.23% Black or African American, and smaller percentages for Native Hawaiian or Pacific Islander, Asian, other race, Native American and multiracial populations.
White (87.64%)
Two or more races (4.63%)
Black or African American (4.23%)
Native Hawaiian or Pacific Islander (1.41%)
Asian (1.01%)
Other race (1.01%)
Native American (0.07%)
Race | Population | Percentage (of total) |
|---|---|---|
| White | 4,956 | 87.64% |
| Two or more races | 262 | 4.63% |
| Black or African American | 239 | 4.23% |
| Native Hawaiian or Pacific Islander | 80 | 1.41% |
| Asian | 57 | 1.01% |
| Other race | 57 | 1.01% |
| Native American | 4 | 0.07% |
Married
Widowed
Divorced
Separated
Never Married
Selinsgrove's average per capita income is $42,857. Household income levels show a median of $58,484. The poverty rate stands at 7.74%.
Name | Median | Mean |
|---|---|---|
| Married Families | $105,699 | - |
| Families | $88,828 | $95,100 |
| Households | $58,484 | $70,262 |
| Non Families | $36,836 | $50,382 |
Average Income
Median Household Income
Poverty Rate