Seymour is a town located in New Haven County, in the state of Connecticut (CT), United States. Nestled in the New England division of the North-east region, it covers an area of 14.6 square miles (38.0 km²) and sits at an elevation of 175 ft. With a population of approximately 15,454 residents, Seymour offers a vibrant community life.
The town's population is composed of about 48.52% males and 51.48% females, and it has a population density of 58.49 people per square mile (408.69/km²). Seymour lies at a latitude of 41.39 and a longitude of -73.08, operating in the UTC-5.
Locals reflect a blend of American ancestry, with roots primarily in United States (0%), followed by English (12.2%), German (10.4%), and Scotch-Irish (0%) heritage. The town is identified by the ZIP code 06483.