We have two candidates and, based on their sales numbers, we have to decide which one should get a proper training in sales. A mock example is the following:
Individual $A$ has a $88.3\%$ completion rate with a total of $94$ cold-calls.
Individual $B$ has a $97.2\%$ completion rate with a total of $215$ cold-calls.
Obviously, individual $A$ has a lower $\%$-rate, but this doesn't necessarily mean he is the 'worst' of the two and requires management to choose him above the other to receive training.
My initial though process was to just use a simple ratio
$$\frac{\text{calls}}{\text{rate}}$$
to give me a normalized value for both individuals. The higher number indicates the amount of calls needed, mathematically, to achieve a $100\%$ success rate. This higher number tells me that the person with $97.2\%$ success is 'worse off' and is in higher need of training than the one with $88.3\%$?
Am I correct to approach this problem this way? Or is there a simpler or a proven mathematical concept that was invented for this kind of thinking?
I apologize if this question doesn't belong here, but I am trying to approach this problem mathematically. Thank you.
I don't think calls divided by success rate measures anything meaningful. If the completion rates were completely reliable, and given no other relevant data (such as feedback from both buyers and non-buyers), I would assume that the person with the lower completion rate would be the higher priority candidate for training.
The catch is that what you have are estimates of the long-term completion rates for the two salespeople. Those are basically sample means of Bernoulli distributions (where each call is a Bernoulli variable with some probability of success and the complementary probability of failure). So what you really want to do is test whether the difference in the sample means (0.883 probability of a success v. 0.972) is an actual difference or an artifact of randomness. That's a standard statistical question, and searching "test for difference in means" or something like that should get you going.