Question: Create a model predicting “left” using logistic regression and interpret the results.

• Perform any initial data preparations.
• Split the data into training and validation.
• Develop an initial model.
• Select appropriate variables for the model.
• Report on the accuracy of the model.
• Interpret the coefficients of the model.

Applied logistic regression to the data about human resource to explore which variables affect leaving of the employees.

Result of the model:

• The total number of employees left is 1055. The model, at the cutoff of 0.49, predicted 1057.

• Predicting the variable, ‘left’ :

  • Of the total loyal, 95.3% of the employees were accurately predicted as loyal.

  • Of the total loyal, 4.7% were falsely predicted to leave.

  • Of the total left, 84.9% of the employees were accurately predicted to leave.

  • Of the total left, 15.1% were inaccurately predicted as loyal.

  • The model is 92.9% of the times correct.

  • The model is 7.1% of the times incorrect.

• In the final predictive model, variable ‘left’ is effected by:

      - satisfaction_level + 
      last_evaluation + 
      number_project + 
      average_montly_hours + 
      time_spend_company + 
      satisfaction_level:last_evaluation + 
      satisfaction_level:number_project + 
      satisfaction_level:average_montly_hours + 
      satisfaction_level:time_spend_company + 
      last_evaluation:number_project + 
      last_evaluation:time_spend_company + 
      number_project:average_montly_hours + 
      number_project:time_spend_company + 
      number_project:sales + 
      average_montly_hours:time_spend_company + 
      time_spend_company:Work_accident

CrossTable(vallr$left , vallr$predicted_left01)

## 
##  
##    Cell Contents
## |-------------------------|
## |                       N |
## | Chi-square contribution |
## |           N / Row Total |
## |           N / Col Total |
## |         N / Table Total |
## |-------------------------|
## 
##  
## Total Observations in Table:  4496 
## 
##  
##              | vallr$predicted_left01 
##   vallr$left |         0 |         1 | Row Total | 
## -------------|-----------|-----------|-----------|
##            0 |      3280 |       161 |      3441 | 
##              |   159.522 |   519.014 |           | 
##              |     0.953 |     0.047 |     0.765 | 
##              |     0.954 |     0.152 |           | 
##              |     0.730 |     0.036 |           | 
## -------------|-----------|-----------|-----------|
##            1 |       159 |       896 |      1055 | 
##              |   520.300 |  1692.821 |           | 
##              |     0.151 |     0.849 |     0.235 | 
##              |     0.046 |     0.848 |           | 
##              |     0.035 |     0.199 |           | 
## -------------|-----------|-----------|-----------|
## Column Total |      3439 |      1057 |      4496 | 
##              |     0.765 |     0.235 |           | 
## -------------|-----------|-----------|-----------|
## 
## 

Initial data preparations.

Exploring Data

There are 8 continuous variables and 2 categorical variables.

• Satisfaction level: Most employees are highly satisfied.

• Last Evaluation: Most employees are good performers with 75% of the data set being evaluated between 56 percent - 87 percent.

• Number of Projects: Most employees do 4 projects and fewer engage in over 6.

• Average Monthly Hours: Most employees spend, fairly, a higher amount of hours at work.

• Time Spent in Company: Fewer employees stay beyond 4 years.

Summary of the variable ‘Work_accident’

• Work Accident: Most employees do not meet accidents at work.

##   var frequency percentage cumulative_perc
## 1   0     12830      85.54           85.54
## 2   1      2169      14.46          100.00

Summary of the variable ‘promotion_last_5years’

• Promotion in Last 5years: Most employees have not received promotions in the last five years.

##   var frequency percentage cumulative_perc
## 1   0     14680      97.87           97.87
## 2   1       319       2.13          100.00

Summary of the variable ‘left’

• Left: Most employees stay with the organization and do not leave.

##   var frequency percentage cumulative_perc
## 1   0     11428      76.19           76.19
## 2   1      3571      23.81          100.00

Spliting data into training and validation.

• To split data into training and test set: validation, variable ‘random’ (14999 observations with 10 variables) was mutated to create random samples:
  • trainlr (10503 observations with 10 variables) and
  • vallr (4496 observations with 10 variables).

Developing an initial model

• The regression equation was applied to the trainlr data set.

• Variable “left” was tested as a factor of all variables along with interactions between them.

• Significant P values are:

  • satisfaction_level, last_evaluation, number_project, average_montly_hours, time_spend_company;

  • satisfaction_level:last_evaluation, satisfaction_level:number_project, satisfaction_level:average_montly_hours, satisfaction_level:time_spend_company, last_evaluation:number_project, last_evaluation:time_spend_company, number_project:average_montly_hours, number_project:time_spend_company, number_project:sales, average_montly_hours:time_spend_company and time_spend_company:Work_accident.

Final Model:

• The final model is:

      - satisfaction_level + last_evaluation + number_project + average_montly_hours + time_spend_company + satisfaction_level:last_evaluation + satisfaction_level:number_project + satisfaction_level:average_montly_hours + satisfaction_level:time_spend_company + last_evaluation:number_project + last_evaluation:time_spend_company + number_project:average_montly_hours + number_project:time_spend_company +   number_project:sales + average_montly_hours:time_spend_company + time_spend_company:Work_accident.

Interpreting the coefficients of the final model:

• Variables with negative coefficients are inversely proportional to the dependent variable ‘left’ , which are:

  - Satisfaction_level, 
  - last_evaluation,
  - number_project, 
  - average_montly_hours,
  - time_spend_company

For instance, lower satisfaction level would mean greater number of employees leaving, lower number of projects would mean greater number of employees leaving etcetera.

• The AIC of the initial model is 8989 whereas the AIC for the final model (fitlr2) is 4591.2

• Keeping fitlr2 (final model) due to lower AIC.

Summary of the initial model:

## 
## Call:
## glm(formula = left ~ ., family = binomial(link = "logit"), data = trainlr)
## 
## Deviance Residuals: 
##     Min       1Q   Median       3Q      Max  
## -2.2206  -0.6613  -0.3960  -0.1138   3.0280  
## 
## Coefficients:
##                         Estimate Std. Error z value Pr(>|z|)    
## (Intercept)           -1.1269779  0.2285857  -4.930 8.21e-07 ***
## satisfaction_level    -4.3165425  0.1190481 -36.259  < 2e-16 ***
## last_evaluation        0.6208105  0.1798334   3.452 0.000556 ***
## number_project        -0.3113181  0.0256478 -12.138  < 2e-16 ***
## average_montly_hours   0.0040751  0.0006196   6.577 4.81e-11 ***
## time_spend_company     0.2575157  0.0184363  13.968  < 2e-16 ***
## Work_accident         -1.5543248  0.1094542 -14.201  < 2e-16 ***
## promotion_last_5years -1.3142357  0.2833255  -4.639 3.51e-06 ***
## saleshr                0.1345257  0.1552961   0.866 0.386352    
## salesIT               -0.3283652  0.1445313  -2.272 0.023091 *  
## salesmanagement       -0.3219024  0.1824519  -1.764 0.077679 .  
## salesmarketing        -0.0082122  0.1549095  -0.053 0.957721    
## salesproduct_mng      -0.2008565  0.1541501  -1.303 0.192577    
## salesRandD            -0.7776814  0.1731848  -4.490 7.11e-06 ***
## salessales            -0.1576817  0.1199494  -1.315 0.188655    
## salessupport          -0.0246326  0.1285122  -0.192 0.847997    
## salestechnical         0.0061095  0.1251695   0.049 0.961071    
## salarylow              1.9435242  0.1519826  12.788  < 2e-16 ***
## salarymedium           1.4551725  0.1526778   9.531  < 2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 11565  on 10502  degrees of freedom
## Residual deviance:  8951  on 10484  degrees of freedom
## AIC: 8989
## 
## Number of Fisher Scoring iterations: 5

Summary of the final model:

## 
## Call:
## glm(formula = left ~ satisfaction_level + last_evaluation + number_project + 
##     average_montly_hours + time_spend_company + satisfaction_level:last_evaluation + 
##     satisfaction_level:number_project + satisfaction_level:average_montly_hours + 
##     satisfaction_level:time_spend_company + last_evaluation:number_project + 
##     last_evaluation:time_spend_company + number_project:average_montly_hours + 
##     number_project:time_spend_company + number_project:sales + 
##     average_montly_hours:time_spend_company + time_spend_company:Work_accident, 
##     family = binomial(link = "logit"), data = trainlr)
## 
## Deviance Residuals: 
##     Min       1Q   Median       3Q      Max  
## -3.7288  -0.2939  -0.1518  -0.0074   5.6382  
## 
## Coefficients:
##                                           Estimate Std. Error z value Pr(>|z|)
## (Intercept)                              4.499e+01  1.034e+00  43.498  < 2e-16
## satisfaction_level                      -2.762e+01  1.312e+00 -21.054  < 2e-16
## last_evaluation                         -3.333e+01  1.399e+00 -23.823  < 2e-16
## number_project                          -7.987e+00  2.346e-01 -34.052  < 2e-16
## average_montly_hours                    -8.620e-02  4.645e-03 -18.556  < 2e-16
## time_spend_company                      -2.956e+00  2.002e-01 -14.763  < 2e-16
## satisfaction_level:last_evaluation       2.111e+01  1.371e+00  15.400  < 2e-16
## satisfaction_level:number_project       -9.495e-01  1.909e-01  -4.973 6.60e-07
## satisfaction_level:average_montly_hours  2.785e-02  4.735e-03   5.883 4.04e-09
## satisfaction_level:time_spend_company    1.770e+00  1.342e-01  13.192  < 2e-16
## last_evaluation:number_project           5.398e+00  2.502e-01  21.577  < 2e-16
## last_evaluation:time_spend_company       1.151e+00  1.945e-01   5.919 3.25e-09
## number_project:average_montly_hours      1.866e-02  8.598e-04  21.700  < 2e-16
## number_project:time_spend_company        1.485e-01  3.080e-02   4.822 1.42e-06
## number_project:saleshr                   5.219e-02  6.272e-02   0.832  0.40532
## number_project:salesIT                  -3.990e-02  5.556e-02  -0.718  0.47269
## number_project:salesmanagement          -2.019e-01  6.607e-02  -3.056  0.00224
## number_project:salesmarketing           -3.680e-02  6.173e-02  -0.596  0.55109
## number_project:salesproduct_mng         -8.445e-02  5.937e-02  -1.422  0.15489
## number_project:salesRandD               -1.392e-01  6.137e-02  -2.268  0.02330
## number_project:salessales               -3.363e-02  4.697e-02  -0.716  0.47410
## number_project:salessupport              3.609e-02  4.984e-02   0.724  0.46896
## number_project:salestechnical            2.879e-02  4.875e-02   0.590  0.55488
## average_montly_hours:time_spend_company  3.296e-03  6.977e-04   4.724 2.31e-06
## time_spend_company:Work_accident        -3.598e-01  3.664e-02  -9.817  < 2e-16
##                                            
## (Intercept)                             ***
## satisfaction_level                      ***
## last_evaluation                         ***
## number_project                          ***
## average_montly_hours                    ***
## time_spend_company                      ***
## satisfaction_level:last_evaluation      ***
## satisfaction_level:number_project       ***
## satisfaction_level:average_montly_hours ***
## satisfaction_level:time_spend_company   ***
## last_evaluation:number_project          ***
## last_evaluation:time_spend_company      ***
## number_project:average_montly_hours     ***
## number_project:time_spend_company       ***
## number_project:saleshr                     
## number_project:salesIT                     
## number_project:salesmanagement          ** 
## number_project:salesmarketing              
## number_project:salesproduct_mng            
## number_project:salesRandD               *  
## number_project:salessales                  
## number_project:salessupport                
## number_project:salestechnical              
## average_montly_hours:time_spend_company ***
## time_spend_company:Work_accident        ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 11565.1  on 10502  degrees of freedom
## Residual deviance:  4541.2  on 10478  degrees of freedom
## AIC: 4591.2
## 
## Number of Fisher Scoring iterations: 7

Report on the accuracy of the model:

• Using the validation data set to test the model.

• Created the variable ‘logit’ using the function predict.glm in the validation data set. However, this isn’t the predicted value of the dependent variable ‘left.’

• The predicted values will exist between 0 and 1 giving us a probability of ‘left’ for each observation.

• Created the variable, ‘predicted_left’ using the variable ‘logit.’

• The mean for the predicted_left is 0.23 same as that of the variable ‘left’ which is at 0.23.

• The mean value of ‘predicted_left’ is reflective of the actual number of employees who left on an average.

Summary of ‘predicted_left’

##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
## 0.00000 0.01006 0.03875 0.23818 0.39450 0.99962

Summary of ‘left’

##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##  0.0000  0.0000  0.0000  0.2347  0.0000  1.0000

How many ‘left’ observations in the validation data set?

Judging a cutoff:

• The ‘predicted_left’ gives us a probability between 0 and 1, however, does not enable us to judge if the employee will remain loyal or loyal.

• To classify an employee as the one to leave or not, we establish a cutoff.

• The cutoff resembles a threshold which will classify an observation as the one to leave or not.

• A total of 1055 employees actually left the organization, so, we are going to get 1055 highest probabilities.

• Also, 1055 forms lesser than 50 percent of the data making it reasonable to reject median value (0.04) as the cutoff.

• To judge a reasonable cutoff, I sorted the variable ‘predicted_left’ in descending order. Further, I filtered the data set up to the row_number of the total number of observations left (row number: 1055).

• The new data set ‘cutoff’ displays the row_number 1055 where the predicted_left value is 0.49.

• The threshold (cutoff) to classify a predicted value as ‘left’ is set to 0.49.

Applying the new cutoff:

• Mutated ‘predicted_left’ into a new variable ‘predicted_left01’ where the predicted value is classified as 1 if it is greater than or equal to the cutoff (0.49), else, 0.

• 1 represents predicted to leave and 0 is predicted as not to.

• There are a total of 1057 observations predicted to leave after applying the cutoff.

• The actual number of employees which left were 1055.

## [1] 1057