What is the median salary for User Acceptance Testing in Jacksonville?

The median salary for User Acceptance Testing in Jacksonville is USD 93,000 - USD 191,000 per year.

How does the User Acceptance Testing salary in Jacksonville compare to the national average?

The national average salary for User Acceptance Testing in United States is USD 95,000 - USD 118,800, while in Jacksonville it is USD 93,000 - USD 191,000.

User Acceptance Testing salary in United States

User Acceptance Testing salary in Jacksonville

Based on 6 job listings - Jacksonville, United States 2026

Average salary

$93,000 - $191,000

Median salary

$93,000 - $191,000

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Compare with national average

Jacksonville

Average salary

93,000 - 191,000

Median salary

93,000 - 191,000

6 job openings

United States

Average salary

95,000 - 118,800

Median salary

95,000 - 118,800

790 job openings

Other cities

New York30 job openings

Oberlin23 job openings

Washington16 job openings

Atlanta13 job openings

Chicago13 job openings

Columbus12 job openings

Boston11 job openings

San Diego11 job openings

Austin10 job openings

San Francisco10 job openings

Buena Park9 job openings

Charlotte9 job openings

Irving9 job openings

Baltimore7 job openings

Dallas6 job openings

Denver6 job openings

Houston6 job openings

Jersey City6 job openings

Sacramento6 job openings

Bethesda5 job openings

Columbia5 job openings

Irvine5 job openings

Nashville5 job openings

Plano5 job openings

Tampa5 job openings

Albany4 job openings

Arlington4 job openings

Milwaukee4 job openings

User Acceptance Testing

6 job openings in Jacksonville

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User Acceptance Testing in Jacksonville

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The average salary for a User Acceptance Testing in Jacksonville, United States is USD 93,000 - USD 191,000 per year in 2026, based on 6 job listings.

Explore 6 User Acceptance Testing job listings in United States on BeBee, or check our salary calculator for a personalized estimate.