Attrition case study in Machine Learning using Logistic Regression

Here Attrition dataset is used for logistic regression analysis. Churn is the dependent variable. Based on the independent variable we have to predict the Churn. Below code is for importing and viewing our dataset.

library(readxl)
attrition_excel <- read_excel("C:/Users/DELL/Desktop/Imarticus/Assignments excel/attrition excel.xlsx")
View(attrition_excel)

Assigning dataset to a new variable for our analysis

Structure(str) explains us the about variables whether they are numeric or character

str(att)

Classes ‘tbl_df’, ‘tbl’ and 'data.frame':   160 obs. of  9 variables:
 $ CustID    : num  101901 102056 102522 103149 103866 ...
 $ Gender    : num  1 0 1 1 0 0 1 1 0 1 ...
 $ Age       : num  30 17 54 42 30 23 28 19 48 40 ...
 $ Income    : num  0 1 1 1 1 0 1 1 0 1 ...
 $ FamilySize: num  5 1 4 2 2 4 2 5 4 5 ...
 $ Education : num  20 12 18 17 12 16 18 16 15 16 ...
 $ Calls     : num  37 25 48 51 26 18 29 28 16 31 ...
 $ Visits    : num  3 1 3 2 1 0 2 1 3 3 ...
 $ Churn     : num  1 0 1 1 0 0 1 1 1 1 ...

Summary of the dataset gives us the minimum value,maximum value, quartile values,mean,median. This gives us the basic understanding of our dataset.

summary(att)

     CustID           Gender            Age            Income         FamilySize   
 Min.   :101901   Min.   :0.0000   Min.   :17.00   Min.   :0.0000   Min.   :1.000  
 1st Qu.:126719   1st Qu.:0.0000   1st Qu.:22.00   1st Qu.:0.0000   1st Qu.:2.000  
 Median :151060   Median :1.0000   Median :31.00   Median :1.0000   Median :3.000  
 Mean   :151144   Mean   :0.5813   Mean   :35.67   Mean   :0.5062   Mean   :3.131  
 3rd Qu.:176099   3rd Qu.:1.0000   3rd Qu.:46.00   3rd Qu.:1.0000   3rd Qu.:4.000  
 Max.   :199131   Max.   :1.0000   Max.   :82.00   Max.   :1.0000   Max.   :5.000  
   Education         Calls           Visits          Churn       
 Min.   :12.00   Min.   : 3.00   Min.   :0.000   Min.   :0.0000  
 1st Qu.:12.00   1st Qu.:15.75   1st Qu.:1.000   1st Qu.:0.0000  
 Median :14.00   Median :22.00   Median :2.000   Median :1.0000  
 Mean   :14.96   Mean   :25.22   Mean   :1.906   Mean   :0.5312  
 3rd Qu.:17.00   3rd Qu.:32.00   3rd Qu.:3.000   3rd Qu.:1.0000  
 Max.   :20.00   Max.   :65.00   Max.   :5.000   Max.   :1.0000  

is.na is used to find the not available values.

is.na(att)

       CustID Gender   Age Income FamilySize Education Calls Visits Churn
  [1,]  FALSE  FALSE FALSE  FALSE      FALSE     FALSE FALSE  FALSE FALSE
  [2,]  FALSE  FALSE FALSE  FALSE      FALSE     FALSE FALSE  FALSE FALSE
  [3,]  FALSE  FALSE FALSE  FALSE      FALSE     FALSE FALSE  FALSE FALSE
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Considering our dependent variable and some of the other variables as a factor will help us to plot graph based on our dependent variable.

Correlogram gives the correlation between all the variables in our dataset

Here we split our data into 80% for train data, 20% for test data for prediction.Library catools is used to split data randomly into 80% for training and 20% for testing.

library(caTools)
sample_att=sample.split(att,SplitRatio = 0.8)
sample_att

[1]  TRUE  TRUE  TRUE  TRUE  TRUE FALSE  TRUE FALSE  TRUE

train_att=subset(att,sample_att=='TRUE')

Length of logical index must be 1 or 160, not 9

train_att

test_att=subset(att,sample_att=='FALSE')

Length of logical index must be 1 or 160, not 9

test_att

We train our model for prediction including all variables. We cannot remove variable based on p value and we should consider AIC value for removing variables.

summary(ctt_eq)

Call:
glm(formula = Churn ~ ., family = "binomial", data = train_att)

Deviance Residuals: 
    Min       1Q   Median       3Q      Max  
-2.2034  -0.8575   0.2465   0.7594   2.0255  

Coefficients:
              Estimate Std. Error z value Pr(>|z|)   
(Intercept) -5.131e+00  2.290e+00  -2.241   0.0250 * 
CustID      -1.279e-05  8.796e-06  -1.454   0.1459   
Gender       1.134e+00  4.625e-01   2.451   0.0142 * 
Age         -2.520e-02  1.565e-02  -1.610   0.1075   
Income1      1.303e+00  4.847e-01   2.687   0.0072 **
FamilySize   6.359e-01  2.559e-01   2.485   0.0130 * 
Education    2.199e-01  9.391e-02   2.341   0.0192 * 
Calls        3.612e-02  1.836e-02   1.967   0.0492 * 
Visits       3.591e-01  1.578e-01   2.276   0.0229 * 
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

(Dispersion parameter for binomial family taken to be 1)

    Null deviance: 172.32  on 124  degrees of freedom
Residual deviance: 123.78  on 116  degrees of freedom
AIC: 141.78

Number of Fisher Scoring iterations: 5

We need to remove variable seperately and check model overall AIC. 2 variables are removed one by one and AIC is checked. AIC should not decrease.After seeing AIC we remove CustID variable from our model.

ctt_final

Call:  glm(formula = Churn ~ . - CustID, family = "binomial", data = train_att)

Coefficients:
(Intercept)       Gender          Age      Income1   FamilySize    Education  
   -7.35642      1.18065     -0.02505      1.42304      0.65295      0.23682  
      Calls       Visits  
    0.02934      0.38944  

Degrees of Freedom: 124 Total (i.e. Null);  117 Residual
Null Deviance:      172.3 
Residual Deviance: 125.9    AIC: 141.9

Now we predict our model using test data based on our trained model.

att_predict=predict(ctt_final,test_att,type = 'response')
att_predict
table(av=test_att$Churn,pv=att_predict>0.43)

After our prediction we need check our accuracy of our prediction. library ROCR is used to plot out actual value and predicted value to test our perfomance. From our ROCR graph we can find out our threshold value for our accuracy prediction.

Here for 0.66 threshold value we get 84% accuracy for prediction so our predicted model is good to use.

att_accu

[1] 0.8333333