Industrial Engineering Technician Salary in Pennsylvania (2026)
Pennsylvania · 1.0% below national average · Medium Demand
| City | Avg Salary | vs National | Demand |
|---|---|---|---|
| Philadelphia | $66,087 | +8.0% | Medium |
| Pittsburgh | $59,968 | -2.0% | Medium |
Pennsylvania is home to a strong job market for Industrial Engineering Technicians. The state's key industries — Healthcare, Education, Finance — generate significant demand for skilled professionals. At $61,192 per year, Pennsylvania's Industrial Engineering Technician salaries are 1.0% lower than the national average.
The cost of living in Pennsylvania (index: 0.99) makes salaries stretch further than in higher-cost states. Entry-level professionals in Pennsylvania can expect to earn $39,648–$51,010, while experienced Industrial Engineering Technicians can command $76,572–$86,125.
Looking ahead, the Industrial Engineering Technician profession in Pennsylvania is projected to grow at 2.5% annually. Major employers are actively hiring, and the state's investment in Healthcare continues to drive new opportunities for qualified candidates.
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What is the average Industrial Engineering Technician salary in Pennsylvania?The average Industrial Engineering Technician salary in Pennsylvania is $61,192 per year in 2026, which is 1.0% below the national average of $61,810.
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What city in Pennsylvania pays Industrial Engineering Technicians the most?Major metro areas typically pay Industrial Engineering Technicians the highest wages in Pennsylvania due to higher costs of living and competitive job markets.
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Is Pennsylvania a good state for Industrial Engineering Technicians?Pennsylvania has a strong job market with major industries in Healthcare, Education, Finance. This makes it one of the better states for Industrial Engineering Technician employment.
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How does Pennsylvania Industrial Engineering Technician salary compare to the US average?At $61,192/year, Pennsylvania Industrial Engineering Technician salaries are 1.0% lower than the national average of $61,810/year.