Another way of evaluating the fit of a given logistic regression model is via a **Classification Table**. The Real Statistics Logistic Regression data analysis tool produces this table. For Example 1 of Comparing Logistic Regression Models the table produced is displayed on the right side of Figure 1.

**Figure 1 – Classification Table**

The table shows a comparison of the number of successes (y = 1) predicted by the logistic regression model compared to the number actually observed and similarly the number of failures (y = 0) predicted by the logistic regression model compared to the number actually observed.

We have four possible outcomes:

**True Positives** (**TP**) = the number of cases which were correctly classified to be positive, i.e. were predicted to be a success and were actually observed to be a success

**False Positives** (**FP**) = the number of cases which were incorrectly classified as positive, i.e. were predicted to be a success but were actually observed to be a failure

**True Negatives** (**TN**) = the number of cases which were correctly classified to be negative, i.e. were predicted to be a failure and were actually observed to be a failure

**False Negatives** (**FN**) = the number of cases which were incorrectly classified as negative, i.e. were predicted to be a negative but were actually observed to be a success

The Classification Table takes the form

where PP = predicted positive = TP + FP, PN = predicted negative = FN + TN, OP = observed positive = TP + FN, ON = observed negative = FP + TN and Tot = the total sample size = TP + FP + FN + TN.

For the data in Figure 1, we have

**TP** = 483 (cell AE6), which can be calculated by the formula =SUMIF(L6:L17,”>=”&AE12,H6:H17)

**FP** = 201 (cell AF6), which can be calculated by the formula =SUMIF(L6:L17,”>=”&AE12,I6:I17)

**FN** = 39 (cell AE7), which can be calculated by the formula =H18-AE6

**TN** = 137 (cell AF7), which can be calculated by the formula =I18-AF6

Here cell AE12 contains the **cutoff** value of .5. Predicted values (in column L) greater than or equal to this value are classified as positive (i.e. predicted to be a success), those less than this value are classified as negative (i.e. predicted to be a failure). TP is simply the sum of all the value in column H whose predicted probabilities in column L are ≥ .5.

The cutoff value is specified in the Logistic Regression dialog box (see for example Figure 4 of Finding Logistic Regression Coefficients using Excel’s Solver).

Note that FP is the type I error and FN is the type II error described in Hypothesis Testing.

We now can define the following:

**True Positive Rate** (**TPR**), aka **Sensitivity **= TP/OP = 483/522 = .925287 (cell AE10)

**True Negative Rate **(**TNR**), aka **Specificity** = TN/ON = 137/338 = .405325 (cell AF10)

**Accuracy **(**ACC**) = (TP + TN)/Tot = (483 + 137) / 860 = .720930 (cell AG10)

**False Positive Rate** (**FPR**) = 1 – TNR = FP/ON = 201/338 = .594675

**Positive Predictive Value** (**PPV**) = TP/PP = 483/684 = .70614

**Negative Predictive Value** (**NPV**) = TN/PN = 137/176 = .77841

The overall accuracy of the logistic regression model is a measure of the fit of the model. For Example 1 this is .720930, which means that the model is estimated to give an accurate prediction 72% of the time.

Note that the accuracy of each outcome is given in column P of Figure 1. E.g. the accuracy of Temp = 21 and Water = 0 is 93.75% (cell P7), which can be calculated by the formula

=100*IF(L7>=$AE$12,H7/J7,I7/J7)

The total accuracy of the model (cell P18) can then be calculated by the formula

=SUMPRODUCT(P6:P17,J6:J17)/J18

The value of cell P18 is .720930, which is the same value we obtained in the classification table (cell AG10).

Thank you so much for your help here! I’ve performed the logistic regression, but unfortunately the accuracy of my classification table/regression is very low for my false positives and false negatives. What can I do to improve this?

Classification Table

Suc-Obs Fail-Obs

Suc-Pred 590 267 857

Fail-Pred 82 157 239

672 424 1096

Accuracy 0.87797619 0.370283019 0.681569343

Cutoff 0.5

Gilly,

The accuracy is not that bad, but in any case there really isn’t anything you can do, except use some other statistical model.

Charles