Industrial Engineering Technologists Salary in New York (2026)
New York · 38.0% above national average · Medium Demand
| City | Avg Salary | vs National | Demand |
|---|---|---|---|
| New York City | $152,127 | +65.0% | Medium |
| Buffalo | $84,822 | -8.0% | Medium |
| Yonkers | $127,233 | +38.0% | Medium |
New York is home to a very strong job market for Industrial Engineering Technologistss. The state's key industries — Finance, Technology, Media — generate significant demand for skilled professionals. At $92,198 per year, New York's Industrial Engineering Technologists salaries are 38.0% higher than the national average.
The cost of living in New York (index: 1.38) justifies the higher pay scale. Entry-level professionals in New York can expect to earn $59,545–$75,934, while experienced Industrial Engineering Technologistss can command $112,810–$126,402.
Looking ahead, the Industrial Engineering Technologists profession in New York is projected to grow at 3.2% annually. Major employers are actively hiring, and the state's investment in Finance continues to drive new opportunities for qualified candidates.
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What is the average Industrial Engineering Technologists salary in New York?The average Industrial Engineering Technologists salary in New York is $92,198 per year in 2026, which is 38.0% above the national average of $66,810.
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What city in New York pays Industrial Engineering Technologistss the most?Major metro areas typically pay Industrial Engineering Technologistss the highest wages in New York due to higher costs of living and competitive job markets.
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Is New York a good state for Industrial Engineering Technologistss?New York has a very strong job market with major industries in Finance, Technology, Media. This makes it one of the better states for Industrial Engineering Technologists employment.
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How does New York Industrial Engineering Technologists salary compare to the US average?At $92,198/year, New York Industrial Engineering Technologists salaries are 38.0% higher than the national average of $66,810/year.