## Introduction

Sensitivity, Specificity, and Accuracy are the terms which are most commonly associated with a Binary classification test and they statistically measure the performance of the test. In a binary classification, we divide a given data set into two categories on the basis of whether they have common properties or not by identifying their significance and in a binary classification test, as the name itself conveys, we deal with two datasets. Of these two categories, in general, Sensitivity indicates, how well the test predicts one category and Specificity measures how well the test predicts the other category. Whereas Accuracy is expected to measure how well the test predicts both categories.

### A simple example to illustrate the idea

Imagine the scenario of a tough interview. The interviewer is supposed to be an experienced and sharp person so that we can blindly say that, all the candidates who are selected in the interview are exactly all the true excellent candidates who deserve the selection and no other candidates except those excellent ones are selected in the interview. I.e., an ideal or truly accurate interview will always give a positive result with excellent candidates only and a negative result with poor performers only.

This is not the case for all interviews. In practice this means that all excellent candidates may not be selected and this can be described by the term Sensitivity. Sensitivity measures the proportion of excellent candidates who were correctly identified to the total number of excellent candidates.. Similarly not all poor candidates may not be rejected and this can be described by the term Specificity. Specificity measures the proportion of poor candidates who were correctly rejected to the total number of poor candidates. Whereas Accuracy measures the proportion of Excellent and poor candidates those are selected correctly to the total number of excellent and poor candidates.

In the above figure, the box ‘Total Candidates’ contains 4 excellent candidates represented by 4 +ve squares and 4 poor candidates represented by 4 –ve circles. Suppose there is an interview, which aims at selecting excellent candidates into an output box ‘Selected’ and to rejecting poor performers into another output box ‘Rejected’.

In practice interviews are not perfect, so not all excellent applicants may be picked up by the interview. Suppose, in this case the interview gives a correct positive result in 2 out of the 4 who are excellent, therefore 2 are represented as True Positive in the ‘Correctly Selected’ portion of the output box ‘Selected’ in the figure. Also, the interview gives a correct negative result in 2 out of 4 who are not excellent; therefore 2 are represented as True Negative in the ‘Correctly Rejected’ section of the output box ‘Rejected’ in the figure.

At the same time, 2 candidates have a false positive result, even if they are not excellent. These are therefore represented as False Positive in the ‘Mistakenly Selected’ section of the output box ‘Selected’ in the figure, as the selection by the interviewer is incorrect. Similarly 2 candidates have a false negative result, even if they are excellent. These are therefore represented as False Negative in the ‘Mistakenly Rejected’ Section of the output box ‘Rejected’ in the figure, as the interviewer is mistaken. Here,

Correctly Selected ---> True Positive (Excellent candidates correctly selected in box ‘Selected’) = 2

Mistakenly Selected ---> False Positive (Poor candidates wrongly selected in box ‘Selected’) = 2

Correctly Rejected ---> True Negative (Poor candidates correctly selected in box ‘Rejected’) = 2

Mistakenly Rejected ---> False Negative (Excellent candidates wrongly selected in box ‘Rejected’) = 2

Sensitivity = True Positive / True Positive + False Negative

= Correctly Selected / Correctly Selected + Mistakenly Rejected

= Correctly Selected/ Total Excellent candidates who actually deserved Selection

I.e., 2/2+2= 2/4= 50% If the output box ‘Excellent’ bring together all the 4 excellent candidates and no excellent candidates in the output box ‘Poor’, then, the sensitivity will have its maximum value.I.e., 4/4+0 = 100%

Specificity = True Negative/ True Negative + False Positive

= Correctly Rejected / Correctly Rejected + Mistakenly Selected

= Correctly Rejected/ Total poor candidates who actually deserved Rejection

I.e., 2/2+2= 2/4= 50% If, the output box ‘Poor’ bring together all the 4 poor candidates and no poor candidates in the output box ‘Excellent’, then, the specificity will have its maximum value. I.e., 4/4+0 = 100%

Accuracy = (True Positive +True Negative)/ (True Positive+ False Positive + True Negative + False Negative)

= (Correctly Selected + Correctly Rejected)/ (Correctly Selected + Mistakenly Selected + Correctly Rejected

+ Mistakenly Rejected)

= (Correctly Selected + Correctly Rejected)/ (Total Excellent candidates who actually deserved Selection + Total poor candidates who actually deserved Rejection)

I.e., (2+2)/ (2+2+2+2) =4/8 =0.50

A comparative study is done, by considering some random outputs to identify the relationships between Accuracy, Sensitivity and Specificity, by exemplifying two datasets with, same number of data and different number of data. It is observed that, from the results, Accuracy alone is not a dependable factor, since Accuracy is derived from Sensitivity and Specificity. I.e. Accuracy= (Sensitivity + Specificity)/2. In the case where, the number of excellent candidates and poor performers are equal, if any one of the factors, Sensitivity or Specificity is high then Accuracy will bias towards that highest value. (I.e., if Sensitivity is high, Accuracy will bias towards Sensitivity, or, if Specificity if high, Accuracy will bias towards Specificity. If both are high, Accuracy will also high and if both are low, then Accuracy will be low.). But in the case where, the count of excellent candidates is low and the count of poor performers is high, Accuracy varies with Specificity with out considering Sensitivity. Similarly, in the case where the number of excellent candidates is high and the number of poor performers is low then, Accuracy tends to vary with Sensitivity with out considering Specificity also.

To nullify the Accuracy variations regarding the imbalance in the count of data in the two datasets, we can apply data normalization. Through this, the variations of Accuracy with the predominance of Sensitivity or Specificity could brought to a normal state, i.e. by considering Sensitivity and Specificity together, rather than considering either of them.

### Conclusion

Hence, a good binary classification test always results with high values for all the three factors, Sensitivity, Specificity and Accuracy, whereas a poor binary classification test results with low values for all. If Sensitivity is high and Specificity is low then, there is no need to bother about the excellent candidates but the poor candidates must be reexamined to eliminate false positives (poor candidates mistakenly selected). But If Sensitivity is low and Specificity is high then, there is no need to bother about the poor candidates but the excellent candidates must be reexamined to eliminate false negatives (excellent candidates mistakenly rejected). An average binary classification test always results with average values which are almost similar for all the three factors.