We just did an inventory audit at my company. The result of which is a report of the percentage of mistakes found per team member. I need to weight the mistakes percentage with the total hours worked per team member for a certain department.
The department under review had 70% of total mistakes for the company. Team member A had 30% of total mistakes, resulting in 43% of the mistakes for that department. Team member B had 40% of total mistakes, resulting in 57% of the mistakes for that department.
Team member A worked 145 hours in the time period correlating to 63% of the hours for that department. Team member B worked 85 hours for the same period correlating to 37% of the hours for that department.
I need to weight the mistakes for team members A and B as if they both worked the same number of hours and determine what percentage of mistakes would have theoretically occurred.
Any help would be appreciated since I've been unable to get further than this.
Edit:
Currently I'm thinking the following though I'm not sure if I'm doing it right:
Difference in hours = $$\frac{145-85}{85}*100$$
(If I understand it correctly, this is basically user2661923's suggestion from the comments.)
You should work out the respective rates for each team member, in mistakes per hour. This is a simple way to remove dependence on the number of hours worked.
The actual number of mistakes that each team member found is not given, but it's not going to matter in the end, so let's assume the company made $100$ mistakes in total, $70$ of which are the responsibility of your department. This mean $A$ found $30$ mistakes, while $B$ found $40$.
Since $A$ worked $145$ hours, the rate at which $A$ finds mistakes is $\frac{30}{145} \approx 0.207$ mistakes per hour. For $B$, this rate is $\frac{40}{65} \approx 0.615$ mistakes per hour (almost $3$ times faster).
The ratio between these rates is a decent measure of their comparative efficiency. So, if we assume these observations scale, had $A$ and $B$ worked the same number of hours, I would assume that $B$ would find a little less than $3$ times the mistakes as $A$ (or a little under $75\%$ of all mistakes for the department).
More precisely, $B$ is exactly $$\frac{40}{65} \div \frac{30}{145} = \frac{116}{39} \approx 2.974$$ times faster than $A$ on average. Working in tandem for the same length of time, I would expect $B$ to find $$\frac{116}{116 + 39} \approx 74.8\%$$ of errors, while $A$ would find $25.2\%$ of errors.