The annual salary statistics of medical records and health information technicians in Sandusky, Ohio is shown in Table 1. The wage statistics of medical records and health information technicians in Sandusky is based on the national compensation survey conducted by the U.S. Bureau of Labor Statistics in 2020 and published in April 2021 [1].
| Percentile Bracket | Average Annual Salary |
|---|---|
| 10th Percentile Wage | $25,830 |
| 25th Percentile Wage | $27,580 |
| 50th Percentile Wage | $30,480 |
| 75th Percentile Wage | $34,960 |
| 90th Percentile Wage | $37,800 |
Table 1 shows the average annual salary for medical records and health information technicians in Sandusky, Ohio in 5 percentile scales. The average annual salary for the 90th percentile (the top 10 percent of the highest paid) is $37,800. The median (50th percentile) annual salary is $30,480. The average annual salary for the bottom 10 percent earners is $25,830.
The table and chart below show the trend of the median salary of medical records and health information technicians from 2012 to 2014. Note that there is no salary data for some years between 2012 and 2020.
| Year | Median Salary | Growth | 2-Year Growth |
|---|---|---|---|
| 2014 | $30,480 | -23.26% | -21.78% |
| 2013 | $37,570 | 1.20% | - |
| 2012 | $37,120 | - | - |
From Table 3 we note that with a median annual salary of $30,480, Sandusky is the lowest paying city for medical records and health information technicians in state of Ohio, following city of Cincinnati-Middletown (median annual salary of $47,720). In comparison, the median annual salary of medical records and health information technicians in Sandusky is 36.1 percent (36.1%) lower than that in the highest paying city Cincinnati-Middletown.