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DESCRIPTIVE STATISTICS

#install.packages
library(readxl)
library(dplyr)
## 
## Attaching package: 'dplyr'
## The following objects are masked from 'package:stats':
## 
##     filter, lag
## The following objects are masked from 'package:base':
## 
##     intersect, setdiff, setequal, union
mydata_excel <- read_excel("~/Documents/Šola/IMB/2. semester/NLB project/anketa_končni podatki.xlsx")

mydata_excel <- mydata_excel[-1, ] #Delete first row in which the questions are written

mydata_excel$ID <- seq(1,nrow(mydata_excel))

head(mydata_excel)
## # A tibble: 6 × 129
##   Q1    Q2    Q3    Q4    Q5    Q6a   Q6b   Q6c   Q6d   Q6e   Q6f   Q6g   Q6h  
##   <chr> <chr> <chr> <chr> <chr> <chr> <chr> <chr> <chr> <chr> <chr> <chr> <chr>
## 1 2     1     1     1     1     2     1     1     2     1     2     1     1    
## 2 2     1     1     2     -2    -2    -2    -2    -2    -2    -2    -2    -2   
## 3 2     1     1     1     2     4     5     5     4     3     5     5     4    
## 4 2     1     1     1     2     -3    -3    -3    -3    -3    -3    -3    -3   
## 5 2     1     1     1     2     2     3     4     4     1     2     2     3    
## 6 2     1     1     1     2     3     2     5     1     1     1     1     1    
## # ℹ 116 more variables: Q6i <chr>, Q6j <chr>, Q6j_text <chr>, Q7 <chr>,
## #   Assistance <chr>, Security <chr>, Transparency <chr>, Convinience <chr>,
## #   `Speed and Reliability` <chr>, Q8f <chr>, Q8f_text <chr>,
## #   Branch_Assistance <chr>, `Mobile bank_Assistance` <chr>,
## #   Branch_Security <chr>, `Mobile bank_Security` <chr>,
## #   Branch_Transparency <chr>, `Mobile bank_Transparency` <chr>,
## #   Branch_Convinience <chr>, `Mobile bank_Convinience` <chr>, …
mydata <- mydata_excel[!(apply(mydata_excel == -3, 1, any)), ]

mydata <- subset(mydata, select = -c(Q21:Q40))

mydata$ID <- seq(1,nrow(mydata))

head(mydata)
## # A tibble: 6 × 83
##   Q1    Q2    Q3    Q4    Q5    Q6a   Q6b   Q6c   Q6d   Q6e   Q6f   Q6g   Q6h  
##   <chr> <chr> <chr> <chr> <chr> <chr> <chr> <chr> <chr> <chr> <chr> <chr> <chr>
## 1 2     1     1     1     1     2     1     1     2     1     2     1     1    
## 2 2     1     1     2     -2    -2    -2    -2    -2    -2    -2    -2    -2   
## 3 2     1     1     1     2     4     5     5     4     3     5     5     4    
## 4 2     1     1     1     2     2     3     4     4     1     2     2     3    
## 5 2     1     1     1     2     3     2     5     1     1     1     1     1    
## 6 2     2     2     1     2     5     4     4     5     4     3     5     5    
## # ℹ 70 more variables: Q6i <chr>, Q6j <chr>, Q6j_text <chr>, Q7 <chr>,
## #   Assistance <chr>, Security <chr>, Transparency <chr>, Convinience <chr>,
## #   `Speed and Reliability` <chr>, Q8f <chr>, Q8f_text <chr>,
## #   Branch_Assistance <chr>, `Mobile bank_Assistance` <chr>,
## #   Branch_Security <chr>, `Mobile bank_Security` <chr>,
## #   Branch_Transparency <chr>, `Mobile bank_Transparency` <chr>,
## #   Branch_Convinience <chr>, `Mobile bank_Convinience` <chr>, …
mydata$Q2 <- factor(mydata$Q2, 
                         levels = c(1, 2), 
                         labels = c("Yes","No"))

mydata$Q3 <- factor(mydata$Q3, 
                         levels = c(1, 2), 
                         labels = c("Yes","No"))

mydata$Q4 <- factor(mydata$Q4, 
                         levels = c(1, 2, 3), 
                         labels = c("Yes","No", "I don't know"))

mydata$Q5 <- factor(mydata$Q5, 
                         levels = c(1, 2), 
                         labels = c("Yes","No"))

mydata$Q7 <- factor(mydata$Q7, 
                         levels = c(1, 2), 
                         labels = c("Yes","No"))

mydata$Q16a <- factor(mydata$Q16a, 
                         levels = c(1, 0), 
                         labels = c("Selected","Not selected"))

mydata$Q16b <- factor(mydata$Q16b, 
                         levels = c(1, 0), 
                         labels = c("Selected","Not selected"))

mydata$Q16c <- factor(mydata$Q16c, 
                         levels = c(1, 0), 
                         labels = c("Selected","Not selected"))

mydata$Q16d <- factor(mydata$Q16d, 
                         levels = c(1, 0), 
                         labels = c("Selected","Not selected"))

mydata$Q16e <- factor(mydata$Q16e, 
                         levels = c(1, 0), 
                         labels = c("Selected","Not selected"))

mydata$Q16f <- factor(mydata$Q16f, 
                         levels = c(1, 0), 
                         labels = c("Selected","Not selected"))

mydata$Q16g <- factor(mydata$Q16g, 
                         levels = c(1, 0), 
                         labels = c("Selected","Not selected"))

mydata$Q41 <- factor(mydata$Q41, 
                         levels = c(1, 2), 
                         labels = c("Female","Male"))

mydata$Q43a <- factor(mydata$Q43a, 
                         levels = c(1, 0), 
                         labels = c("Selected","Not selected"))

mydata$Q43b <- factor(mydata$Q43b, 
                         levels = c(1, 0), 
                         labels = c("Selected","Not selected"))

mydata$Q43c <- factor(mydata$Q43c, 
                         levels = c(1, 0), 
                         labels = c("Selected","Not selected"))

mydata$Q43d <- factor(mydata$Q43d, 
                         levels = c(1, 0), 
                         labels = c("Selected","Not selected"))

mydata$Q43e <- factor(mydata$Q43e, 
                         levels = c(1, 0), 
                         labels = c("Selected","Not selected"))

mydata$Q43f <- factor(mydata$Q43f, 
                         levels = c(1, 0), 
                         labels = c("Selected","Not selected"))

mydata$Q43g <- factor(mydata$Q43g, 
                         levels = c(1, 0), 
                         labels = c("Selected","Not selected"))

mydata$Q43h <- factor(mydata$Q43h, 
                         levels = c(1, 0), 
                         labels = c("Selected","Not selected"))

mydata$Q44 <- factor(mydata$Q44, 
                         levels = c(2, 6, 1, 4, 3, 5), 
                         labels = c("1.000 – 5.000 habitants","More than 100.000 habitants", "Less than 1.000 habitants", "20.001 – 50.000 habitants", "5.001 – 20.000 habitants", "50.001 – 100.000 habitants"))

mydata$Q45 <- factor(mydata$Q45, 
                         levels = c(3, 5, 1, 9, 12, 7, 10, 4, 11, 6), 
                         labels = c("OTP banka d.d.","Banka Intesa Sanpaolo d.d.", "Nova Ljubljanska Banka d.d. (NLB)", "Gorenjska Banka d.d.", "Delavska Hranilnica d.d.", "Revolut", "Deželna Banka Slovenije d.d.", "Banka Sparkasse d.d.", "Addiko Bank d.d.", "UniCredit Banka Slovenija d.d."))

mydata$Q46 <- factor(mydata$Q46, 
                         levels = c(1, 2, 3, 5, 6, 4), 
                         labels = c("Študent/-ka","Redno zaposlen/-a", "Upokojen/-a", "Samozaposlen/-a", "Delno zaposlen/-a", "Brezposeln/-a"))

mydata$Q47 <- factor(mydata$Q47, 
                         levels = c(1, 5, 3, 8, 2, 4, 6, 7), 
                         labels = c("Pod 1.000€","3.001€ - 5.000€", "1.501€ - 2.000€", "I don't want to answer", "1.000€ - 1.500€", "2.001€ - 3.000€", "5.001€ - 10.000€", "Above 10.000€"))

mydata$Q48 <- factor(mydata$Q48, 
                         levels = c(2, 6, 3, 5, 7, 4), 
                         labels = c("Dokončana osnovna šola","Dokončana visokošolska strokovna univerzitetna izobrazba (tudi 2. bolonjska stopnja)", "Dokončana nižja ali srednja poklicna izobrazba", "Dokončana višješolska strokovna ali visokošolska strokovna izobrazba (tudi 1. bolonjska stopnja)", "Dokončana specializacija, znanstveni magisterij, doktorat", "Dokončana srednja strokovna ali splošna izobrazba"))

mydata$Q49 <- factor(mydata$Q49, 
                         levels = c(1, 2, 3), 
                         labels = c("Preko linka","Na tablici", "Na listu papirja"))

mydata[c(6:14, 18:22, 25:39, 48:55, 58:62)] <- mydata[c(6:14, 18:22, 25:39, 48:55, 58:62)] %>% mutate_all(as.numeric)

mydata <- mydata %>% mutate(across(where(is.numeric), ~ replace(., . == -2, mean(.[. != -2], na.rm = TRUE))))

mydata <- mydata %>% mutate(across(where(is.numeric), ~ replace(., is.na(.), mean(., na.rm = TRUE))))

mydata <- mydata %>%
  filter(!ID %in% c(2, 16, 17)) 

mydata$ID <- seq(1, nrow(mydata)) 

summary(mydata[c(-1, -15, -16, -23, -24, -47, -56, -57, -63, -64, -66, -75, -78, -83)])
##     Q2         Q3                 Q4         Q5           Q6a       
##  Yes :121   Yes :131   Yes         :103   Yes : 25   Min.   :1.000  
##  No  : 36   No  : 26   No          : 28   No  :132   1st Qu.:3.000  
##  NA's:  1   NA's:  1   I don't know: 26   NA's:  1   Median :3.000  
##                        NA's        :  1              Mean   :3.433  
##                                                      3rd Qu.:4.000  
##                                                      Max.   :5.000  
##                                                                     
##       Q6b             Q6c             Q6d             Q6e       
##  Min.   :1.000   Min.   :1.000   Min.   :1.000   Min.   :1.000  
##  1st Qu.:3.000   1st Qu.:3.000   1st Qu.:3.000   1st Qu.:1.000  
##  Median :4.000   Median :4.000   Median :4.000   Median :2.000  
##  Mean   :3.535   Mean   :3.847   Mean   :3.854   Mean   :2.688  
##  3rd Qu.:4.000   3rd Qu.:5.000   3rd Qu.:5.000   3rd Qu.:4.000  
##  Max.   :5.000   Max.   :5.000   Max.   :5.000   Max.   :5.000  
##                                                                 
##       Q6f             Q6g             Q6h            Q6i           Q7    
##  Min.   :1.000   Min.   :1.000   Min.   :1.00   Min.   :1.000   Yes :82  
##  1st Qu.:3.000   1st Qu.:2.000   1st Qu.:2.00   1st Qu.:3.000   No  :75  
##  Median :4.000   Median :3.086   Median :3.00   Median :4.000   NA's: 1  
##  Mean   :3.497   Mean   :3.172   Mean   :2.93   Mean   :3.465            
##  3rd Qu.:4.000   3rd Qu.:4.000   3rd Qu.:4.00   3rd Qu.:4.000            
##  Max.   :5.000   Max.   :5.000   Max.   :5.00   Max.   :5.000            
##                                                                          
##    Assistance      Security      Transparency    Convinience   
##  Min.   :1.00   Min.   :1.000   Min.   :1.000   Min.   :1.000  
##  1st Qu.:3.00   1st Qu.:4.000   1st Qu.:4.000   1st Qu.:4.000  
##  Median :4.00   Median :5.000   Median :4.000   Median :4.000  
##  Mean   :3.79   Mean   :4.382   Mean   :4.108   Mean   :3.975  
##  3rd Qu.:4.00   3rd Qu.:5.000   3rd Qu.:5.000   3rd Qu.:5.000  
##  Max.   :5.00   Max.   :5.000   Max.   :5.000   Max.   :5.000  
##                                                                
##  Speed and Reliability Branch_Assistance Mobile bank_Assistance Branch_Security
##  Min.   :1.000         Min.   :2.000     Min.   :1.00           Min.   :2.000  
##  1st Qu.:4.000         1st Qu.:4.000     1st Qu.:3.00           1st Qu.:3.986  
##  Median :4.000         Median :4.000     Median :3.00           Median :4.000  
##  Mean   :4.185         Mean   :4.006     Mean   :3.14           Mean   :3.981  
##  3rd Qu.:5.000         3rd Qu.:5.000     3rd Qu.:4.00           3rd Qu.:5.000  
##  Max.   :5.000         Max.   :5.000     Max.   :5.00           Max.   :5.000  
##                                                                                
##  Mobile bank_Security Branch_Transparency Mobile bank_Transparency
##  Min.   :1.000        Min.   :1.000       Min.   :1.000           
##  1st Qu.:2.000        1st Qu.:3.000       1st Qu.:3.000           
##  Median :3.000        Median :4.000       Median :3.000           
##  Mean   :3.032        Mean   :3.898       Mean   :3.248           
##  3rd Qu.:4.000        3rd Qu.:4.000       3rd Qu.:4.000           
##  Max.   :5.000        Max.   :5.000       Max.   :5.000           
##                                                                   
##  Branch_Convinience Mobile bank_Convinience Branch_Speed and Reliability
##  Min.   :1.000      Min.   :1.000           Min.   :1.000               
##  1st Qu.:3.000      1st Qu.:3.000           1st Qu.:3.000               
##  Median :4.000      Median :3.000           Median :4.000               
##  Mean   :3.752      Mean   :3.389           Mean   :3.701               
##  3rd Qu.:4.000      3rd Qu.:4.000           3rd Qu.:4.000               
##  Max.   :5.000      Max.   :5.000           Max.   :5.000               
##                                                                         
##  Mobile bank_Speed and Reliability      Q15a        Q15b            Q15c      
##  Min.   :1.000                     Min.   :1   Min.   :2.000   Min.   :1.000  
##  1st Qu.:3.000                     1st Qu.:4   1st Qu.:4.000   1st Qu.:4.000  
##  Median :4.000                     Median :4   Median :4.000   Median :4.000  
##  Mean   :3.586                     Mean   :4   Mean   :4.134   Mean   :4.115  
##  3rd Qu.:4.000                     3rd Qu.:5   3rd Qu.:5.000   3rd Qu.:5.000  
##  Max.   :5.000                     Max.   :5   Max.   :5.000   Max.   :5.000  
##                                                                               
##       Q15d            Q15e                 Q16a              Q16b   
##  Min.   :1.000   Min.   :2.000   Selected    :98   Selected    :66  
##  1st Qu.:3.000   1st Qu.:4.000   Not selected:59   Not selected:91  
##  Median :4.000   Median :4.000   NA's        : 1   NA's        : 1  
##  Mean   :3.809   Mean   :4.127                                      
##  3rd Qu.:4.000   3rd Qu.:5.000                                      
##  Max.   :5.000   Max.   :5.000                                      
##                                                                     
##            Q16c              Q16d              Q16e               Q16f    
##  Selected    :68   Selected    :79   Selected    : 49   Selected    : 36  
##  Not selected:89   Not selected:78   Not selected:108   Not selected:121  
##  NA's        : 1   NA's        : 1   NA's        :  1   NA's        :  1  
##                                                                           
##                                                                           
##                                                                           
##                                                                           
##            Q16g          Q17a            Q17b            Q17c      
##  Selected    :  9   Min.   :0.000   Min.   :0.000   Min.   :0.000  
##  Not selected:148   1st Qu.:4.000   1st Qu.:4.000   1st Qu.:4.000  
##  NA's        :  1   Median :4.000   Median :4.000   Median :4.000  
##                     Mean   :3.662   Mean   :3.694   Mean   :3.758  
##                     3rd Qu.:4.000   3rd Qu.:4.000   3rd Qu.:4.750  
##                     Max.   :5.000   Max.   :5.000   Max.   :5.000  
##                                                                    
##       Q17d            Q17e            Q18a           Q18b            Q18c      
##  Min.   :0.000   Min.   :0.000   Min.   :1.00   Min.   :1.000   Min.   :1.000  
##  1st Qu.:2.000   1st Qu.:4.000   1st Qu.:3.00   1st Qu.:3.000   1st Qu.:3.000  
##  Median :4.000   Median :4.000   Median :4.00   Median :4.000   Median :4.000  
##  Mean   :3.223   Mean   :3.809   Mean   :3.35   Mean   :3.732   Mean   :3.713  
##  3rd Qu.:4.000   3rd Qu.:5.000   3rd Qu.:4.00   3rd Qu.:4.000   3rd Qu.:4.000  
##  Max.   :5.000   Max.   :5.000   Max.   :5.00   Max.   :5.000   Max.   :5.000  
##                                                                                
##       Q19a            Q19b            Q19c            Q19d      
##  Min.   :1.000   Min.   :1.000   Min.   :1.000   Min.   :1.000  
##  1st Qu.:3.000   1st Qu.:3.000   1st Qu.:3.000   1st Qu.:3.000  
##  Median :3.000   Median :4.000   Median :4.000   Median :4.000  
##  Mean   :3.217   Mean   :3.548   Mean   :3.331   Mean   :3.618  
##  3rd Qu.:4.000   3rd Qu.:4.000   3rd Qu.:4.000   3rd Qu.:4.000  
##  Max.   :5.000   Max.   :5.000   Max.   :5.000   Max.   :5.000  
##                                                                 
##       Q19e           Q41               Q43a               Q43b    
##  Min.   :1.000   Female:78   Selected    : 32   Selected    : 12  
##  1st Qu.:2.000   Male  :79   Not selected:125   Not selected:145  
##  Median :3.000   NA's  : 1   NA's        :  1   NA's        :  1  
##  Mean   :3.051                                                    
##  3rd Qu.:4.000                                                    
##  Max.   :5.000                                                    
##                                                                   
##            Q43c               Q43d              Q43e               Q43f    
##  Selected    :  5   Selected    :81   Selected    : 57   Selected    :  1  
##  Not selected:152   Not selected:76   Not selected:100   Not selected:156  
##  NA's        :  1   NA's        : 1   NA's        :  1   NA's        :  1  
##                                                                            
##                                                                            
##                                                                            
##                                                                            
##            Q43g               Q43h                              Q44    
##  Selected    : 29   Selected    :  0   1.000 – 5.000 habitants    :44  
##  Not selected:128   Not selected:157   More than 100.000 habitants:36  
##  NA's        :  1   NA's        :  1   Less than 1.000 habitants  :30  
##                                        20.001 – 50.000 habitants  :14  
##                                        5.001 – 20.000 habitants   :22  
##                                        50.001 – 100.000 habitants :11  
##                                        NA's                       : 1  
##                                 Q45                    Q46    
##  OTP banka d.d.                   :71   Študent/-ka      :22  
##  Nova Ljubljanska Banka d.d. (NLB):35   Redno zaposlen/-a:73  
##  Delavska Hranilnica d.d.         :12   Upokojen/-a      :43  
##  Banka Intesa Sanpaolo d.d.       :10   Samozaposlen/-a  :13  
##  Deželna Banka Slovenije d.d.     : 8   Delno zaposlen/-a: 3  
##  (Other)                          :18   Brezposeln/-a    : 3  
##  NA's                             : 4   NA's             : 1  
##               Q47    
##  Pod 1.000€     :46  
##  1.000€ - 1.500€:39  
##  2.001€ - 3.000€:29  
##  1.501€ - 2.000€:25  
##  3.001€ - 5.000€: 6  
##  (Other)        : 4  
##  NA's           : 9  
##                                                                                                Q48    
##  Dokončana osnovna šola                                                                          : 8  
##  Dokončana visokošolska strokovna univerzitetna izobrazba (tudi 2. bolonjska stopnja)            :35  
##  Dokončana nižja ali srednja poklicna izobrazba                                                  :23  
##  Dokončana višješolska strokovna ali visokošolska strokovna izobrazba (tudi 1. bolonjska stopnja):34  
##  Dokončana specializacija, znanstveni magisterij, doktorat                                       : 5  
##  Dokončana srednja strokovna ali splošna izobrazba                                               :50  
##  NA's                                                                                            : 3  
##                Q49    
##  Preko linka     :84  
##  Na tablici      :37  
##  Na listu papirja:17  
##  NA's            :20  
##                       
##                       
##

Q1: Ali uporabljate mobilno aplikacijo banke, kjer imate odprt primarni račun (npr. KlikIn, in podobno)? [1 - Da, 2 - Ne]

Q2: Ali dnevno uporabljate pametni telefon? [1 - Da, 2 - Ne]

Q3: Ali uporabljate internet vsaj nekajkrat na teden? [1 - Da, 2 - Ne]

Q4: Ali je kdo od vaših družinskih članov uporabnik mobilne banke? [1 - Da, 2 - Ne, 3 - Ne vem]

Q5: Ali imate skupni bančni račun, ki ga upravlja en član družine? [1 - Da, 2 - Ne]

Q6: Zakaj ne uporabljate mobilne banke (npr. KlikIn, , Addiko Mobil, …)? [1 - Sploh se ne strinjam, 2 - Se ne strinjam, 3 - Niti se ne strinjam niti se strinjam, 4 - Se strinjam, 5 - Popolnoma se strinjam]

Q7: Se vam zdi, da bi potrebovali dodatno izobraževanje, preden bi lahko uporabljali mobilno banko? [1 - Da, 2 - Ne]

Q8: Kako pomembni so naslednji dejavniki pri odločitvi med poslovalnico ali mobilno aplikacijo? [1 - Sploh ni pomembno, 2 - Ni pomembno, 3 - Niti ni pomembno niti je pomembno, 4 - Je pomembno, 5 - Zelo pomembno]

Q10: Kako zaznavate podporo in svetovanje v poslovalnici in mobilni aplikaciji? [1 - Sploh se ne strinjam, 2 - Se ne strinjam, 3 - Niti niti, 4 - Se strinjam, 5 - Popolnoma se strinjam]

Q11: Kako zaznavate zagotavljanje varnosti in zaščito vaših finančnih podatkov ter transakcij v poslovalnici in mobilni aplikaciji? [1 - Sploh se ne strinjam, 2 - Se ne strinjam, 3 - Niti niti, 4 - Se strinjam, 5 - Popolnoma se strinjam]

Q12: Kako zaznavate transparentnost delovanja banke in njenih storitev v poslovalnici in mobilni aplikaciji? [1 - Sploh se ne strinjam, 2 - Se ne strinjam, 3 - Niti niti, 4 - Se strinjam, 5 - Popolnoma se strinjam]

Q13: Kako zaznavate priročnost (npr. enostavna uporaba, dostopnost od kjerkoli, hitre transakcije) v poslovalnici in mobilni aplikaciji? [1 - Sploh se ne strinjam, 2 - Se ne strinjam, 3 - Niti niti, 4 - Se strinjam, 5 - Popolnoma se strinjam]

Q14: Kako zaznavate zanesljivost opravljene storitve v poslovalnici in mobilni aplikaciji? [1 - Sploh se ne strinjam, 2 - Se ne strinjam, 3 - Niti niti, 4 - Se strinjam, 5 - Popolnoma se strinjam]

Q15: Kako zaskrbljeni ste glede naslednjih tveganj med uporabo mobilnega bančništva? [1 - Sploh me ne skrbi, 2 - Me ne skrbi, 3 - Niti me ne skrbi niti me skrbi, 4 - Me skrbi, 5 - Zelo me skrbi]

Q16: Kateri izmed spodaj navedenih dejavnikov bi vam omogočili večje zaupanje v mobilno bančništvo? [Izbira več možnih odgovorov]

Q17: Kako koristne bi se vam zdele naslednje funkcije v aplikaciji za mobilno bančništvo? [1 - Sploh ni koristna, 2 - Ni koristna, 3 - Niti ni koristna niti je koristna, 4 - Je koristna, 5 - Izjemno koristna]

Q18: Kateri finančni spodbudni ukrepi bi vas spodbudili k uporabi mobilnega bančništva? [1 - Sploh me ne bi spodbudilo, 2 - Ne bi me spodbudilo, 3 - Niti niti, 4 - Bi me spodbudilo, 5 - Zelo bi me spodbudilo]

Q19: Kateri nefinančni spodbudni ukrepi bi vas spodbudili k uporabi mobilnega bančništva? [1 - Sploh me ne bi spodbudilo, 2 - Ne bi me spodbudilo, 3 - Niti niti, 4 - Bi me spodbudilo, 5 - Zelo bi me spodbudilo]

Q41: Spol [1 - Ženska, 2 - Moški, 3 - Ne želim odgovoriti]

Q42: Prosimo, vpišite leto rojstva. [Odprto besedilo]

Q43: S kom živite v istem gospodinjstvu? [Izbira več možnih odgovorov]

Q44: Koliko prebivalcev živi v kraju, kjer prebivate? [1 - Manj kot 1.000 prebivalcev, 2 - 1.000–5.000 prebivalcev, 3 - 5.001–20.000 prebivalcev, 4 - 20.001–50.000 prebivalcev, 5 - 50.001–100.000 prebivalcev, 6 - Več kot 100.000 prebivalcev]

Q45: Katera je vaša primarna banka? [1 - Nova Ljubljanska Banka d.d. (NLB), 2 - BKS Bank AG, Bančna podružnica, 3 - OTP banka d.d., 4 - Banka Sparkasse d.d., 5 - Banka Intesa Sanpaolo d.d., 6 - UniCredit Banka Slovenija d.d., 7 - Revolut, 8 - N26, 9 - Gorenjska Banka d.d., 10 - Deželna Banka Slovenije d.d., 11 - Addiko Bank d.d., 12 - Delavska Hranilnica d.d., 13 - Drugo]

Q46: Kakšna je vaša trenutna zaposlitev? [1 - Študent/ka, 2 - Redno zaposlen/a, 3 - Upokojen/a, 4 - Brezposeln/a, 5 - Samozaposlen/a, 6 - Delno zaposlen/a]

Q47: Kakšen je vaš mesečni neto prihodek? [1 - Pod 1.000€, 2 - 1.000€–1.500€, 3 - 1.501€–2.000€, 4 - 2.001€–3.000€, 5 - 3.001€–5.000€, 6 - 5.001€–10.000€, 7 - Nad 10.000€, 8 - Ne želim odgovoriti]

Q48: Kakšna je vaša stopnja izobrazbe? [1 - Nedokončana osnovna šola, 2 - Dokončana osnovna šola, 3 - Dokončana nižja ali srednja poklicna izobrazba, 4 - Dokončana srednja strokovna ali splošna izobrazba, 5 - Dokončana višješolska strokovna ali visokošolska strokovna izobrazba (tudi 1. bolonjska stopnja), 6 - Dokončana visokošolska strokovna ali univerzitetna izobrazba (tudi 2. bolonjska stopnja), 7 - Dokončana specializacija, znanstveni magisterij ali doktorat]

Q49: Kako ste rešili anketo? [1 - Preko linka, 2 - Na tablici, 3 - Na listu papirja]

CLUSTERING

knitr::opts_chunk$set(echo = TRUE)
options(width = 120)
#install.packages(ggplot2)
library(ggplot2)
#install.packages("ggfortify")
library(ggfortify)
#install.packages("ranger")
library(ranger)
#install.packages("dplyr")
library(dplyr)
#install.packages("Hmisc")
library(Hmisc)
## 
## Attaching package: 'Hmisc'
## The following objects are masked from 'package:dplyr':
## 
##     src, summarize
## The following objects are masked from 'package:base':
## 
##     format.pval, units
#install.packages("factoextra")
library(factoextra)
## Welcome! Want to learn more? See two factoextra-related books at https://goo.gl/ve3WBa
#install.packages("cluster")
library(cluster)
#install.packages("magrittr")
library(magrittr)
#install.packages("NbClust")
library("NbClust")
library(dplyr)

mydata <- mydata %>%
  mutate(Security_concerns = rowMeans(across(c(7, 8, 14)), na.rm = TRUE))

mydata <- mydata %>%
  mutate(Lack_of_competence_or_support = rowMeans(across(c(12, 13)), na.rm = TRUE))

mydata <- mydata %>%
  mutate(Preference_for_traditional_methods = rowMeans(across(c(9, 11)), na.rm = TRUE))

colnames(mydata) [6] <- "Aversion_to_change"
colnames(mydata) [10] <- "Physical_limitations"

From question 6, we made 5 different variables. 3 of them are grouped together in a way that we took the mean of them together (6b, 6c and 6i are Security_concerns; 6g and 6h are Lack_of_competence_or_support; 6d and 6f are Preference_for_traditional_methods) and than 6a is Aversion_to_change and 6e are Physical_limitations.

#Saving standardized cluster variables into new data frame

mydata_clu_std <- as.data.frame(scale(mydata[c("Security_concerns", "Lack_of_competence_or_support", "Preference_for_traditional_methods", "Aversion_to_change", "Physical_limitations")]))
mydata$Dissimilarity <- sqrt(mydata_clu_std$Security_concerns^2 + mydata_clu_std$Lack_of_competence_or_support^2 + mydata_clu_std$Preference_for_traditional_methods^2 + 
                             mydata_clu_std$Aversion_to_change^2 + mydata_clu_std$Physical_limitations^2) #Finding outliers
head(mydata[order(-mydata$Dissimilarity), c("ID", "Dissimilarity")]) #Finding units with highest value of dissimilarity
## # A tibble: 6 × 2
##      ID Dissimilarity
##   <int>         <dbl>
## 1    53          4.92
## 2     1          4.06
## 3   123          3.97
## 4    44          3.94
## 5    30          3.71
## 6    72          3.64

ID53 is a potential outlier, as there is a big jump in disimilarity numbers between units. For this reason we will remove this unit.

mydata <- mydata %>%
  filter(!ID %in% c(53)) #Removing ID53 from original data frame

mydata$ID <- seq(1, nrow(mydata)) #Numbering the data again

mydata_clu_std <- as.data.frame(scale(mydata[c("Security_concerns", "Lack_of_competence_or_support", "Preference_for_traditional_methods", "Aversion_to_change", "Physical_limitations")])) #Standardizing the data again

After removing one country, the sample size is now 157.

#Finding Eudlidean distances, based on 5 Cluster variables, then saving them into object Distances

Distances <- get_dist(mydata_clu_std, 
                      method = "euclidian")

fviz_dist(Distances, #Showing matrix of distances
          gradient = list(low = "#33006F",
                          mid = "grey95",
                          high = "white"))

We can see on the matrix of distances that some groups of clusters are forming, I see 3 groups.

get_clust_tendency(mydata_clu_std, #Hopkins statistics
                   n = nrow(mydata_clu_std) - 1,
                   graph = FALSE) 
## $hopkins_stat
## [1] 0.6029504
## 
## $plot
## NULL

This data is clusterable as it is above 0.5. If it would be more close to 1, it would be even more appropriate. However the threshold is 0.5. Now the next question is how many clusters to use. I will check this with Hierarhical clustering (dendrogram) and K-Means clustering (Elbow method, Silhouette analysis and with the help of indices).

How many clusters to use?

WARD <- mydata_clu_std %>%
  get_dist(method = "euclidean") %>%  
  hclust(method = "ward.D2")          

WARD
## 
## Call:
## hclust(d = ., method = "ward.D2")
## 
## Cluster method   : ward.D2 
## Distance         : euclidean 
## Number of objects: 157
fviz_dend(WARD)
## Warning: The `<scale>` argument of `guides()` cannot be `FALSE`. Use "none" instead as of ggplot2 3.3.4.
## ℹ The deprecated feature was likely used in the factoextra package.
##   Please report the issue at <https://github.com/kassambara/factoextra/issues>.
## This warning is displayed once every 8 hours.
## Call `lifecycle::last_lifecycle_warnings()` to see where this warning was generated.

Based on the dendrogram, we would choose 2 clusters, as there is the biggest jump in vertical line.

fviz_nbclust(mydata_clu_std, kmeans, method = "wss") +
  labs(subtitle = "Elbow method")

With the elbow method the slope changes most evidently at 2 clusters.

fviz_nbclust(mydata_clu_std, kmeans, method = "silhouette")+
  labs(subtitle = "Silhouette analysis")

The higest value of the Silhouette analysis is at 2.

NbClust(mydata_clu_std, 
        distance = "euclidean", 
        min.nc = 2, max.nc = 10,
        method = "kmeans", 
        index = "all")

## *** : The Hubert index is a graphical method of determining the number of clusters.
##                 In the plot of Hubert index, we seek a significant knee that corresponds to a 
##                 significant increase of the value of the measure i.e the significant peak in Hubert
##                 index second differences plot. 
## 

## *** : The D index is a graphical method of determining the number of clusters. 
##                 In the plot of D index, we seek a significant knee (the significant peak in Dindex
##                 second differences plot) that corresponds to a significant increase of the value of
##                 the measure. 
##  
## ******************************************************************* 
## * Among all indices:                                                
## * 10 proposed 2 as the best number of clusters 
## * 2 proposed 3 as the best number of clusters 
## * 3 proposed 4 as the best number of clusters 
## * 1 proposed 6 as the best number of clusters 
## * 1 proposed 7 as the best number of clusters 
## * 4 proposed 9 as the best number of clusters 
## * 2 proposed 10 as the best number of clusters 
## 
##                    ***** Conclusion *****                            
##  
## * According to the majority rule, the best number of clusters is  2 
##  
##  
## *******************************************************************
## $All.index
##          KL      CH Hartigan     CCC    Scott     Marriot   TrCovW   TraceW Friedman  Rubin Cindex     DB Silhouette
## 2    9.5966 92.7417  23.3648 -1.9059 182.2586 22365513125 9141.421 488.0082   3.5819 1.5983 0.4213 1.2674     0.3109
## 3    0.2118 64.6236  38.3498 -4.2729 287.2586 25781592519 5810.071 424.0819   5.0700 1.8393 0.3873 1.7241     0.2346
## 4   33.4822 66.1618  14.8522 -2.2561 410.1106 20958157652 4746.590 339.5305   6.7664 2.2973 0.3518 1.4934     0.2405
## 5    0.1946 57.7712  14.4020 -2.8446 501.1125 18341844976 4146.290 309.4876   8.4125 2.5203 0.3329 1.5378     0.2127
## 6    0.3402 53.1246  18.4451 -2.6373 577.8546 16200227137 3341.184 282.7016   9.7434 2.7591 0.3193 1.5636     0.2199
## 7  170.9406 52.4031   8.2807 -1.4552 648.9874 14016730597 2585.750 251.9279  11.0928 3.0961 0.3641 1.4115     0.2343
## 8    0.0053 48.2557  17.2282 -1.9713 692.5546 13871218677 2394.293 238.7479  12.0205 3.2670 0.3331 1.5076     0.2114
## 9   11.5140 48.9287   5.5136 -0.5159 772.2003 10570689341 1798.061 214.0037  13.8227 3.6448 0.4090 1.3345     0.2287
## 10   0.4798 45.4161   7.5001 -1.0486 804.3282 10635205508 1752.332 206.3176  14.7638 3.7806 0.4041 1.4314     0.2050
##      Duda Pseudot2   Beale Ratkowsky     Ball Ptbiserial    Frey McClain   Dunn Hubert SDindex Dindex   SDbw
## 2  0.9784   2.1452  0.0684    0.4324 244.0041     0.4973  0.9300  0.6734 0.0959 0.0021  1.8068 1.6568 0.9625
## 3  1.0585  -4.0338 -0.1694    0.3884 141.3606     0.4739  0.2894  1.3160 0.1174 0.0024  2.1176 1.5423 1.9356
## 4  1.3475 -15.9880 -0.7858    0.3757  84.8826     0.4900  0.6398  1.8251 0.0722 0.0029  1.9498 1.3840 0.7941
## 5  1.8725 -23.7634 -1.4063    0.3470  61.8975     0.4659  0.3248  2.3156 0.0976 0.0033  2.0907 1.3079 0.7352
## 6  1.0624  -2.2896 -0.1778    0.3258  47.1169     0.4573  0.0996  2.7357 0.0987 0.0033  2.1119 1.2557 0.4736
## 7  1.3570  -8.9441 -0.7859    0.3109  35.9897     0.4659  0.7818  2.9210 0.1167 0.0033  1.9525 1.1932 0.3572
## 8  1.7019 -12.3726 -1.2101    0.2944  29.8435     0.4378  0.0754  3.4835 0.0816 0.0035  2.0991 1.1490 0.3336
## 9  0.9718   0.6384  0.0860    0.2838  23.7782     0.4444  1.2669  3.6472 0.1444 0.0035  1.9951 1.1000 0.3151
## 10 1.1576  -1.6341 -0.3788    0.2711  20.6318     0.4187 -0.1176  4.2006 0.1728 0.0036  2.2337 1.0821 0.3061
## 
## $All.CriticalValues
##    CritValue_Duda CritValue_PseudoT2 Fvalue_Beale
## 2          0.6717            47.4016       0.9968
## 3          0.6107            46.5380       1.0000
## 4          0.5760            45.6335       1.0000
## 5          0.5287            45.4696       1.0000
## 6          0.5452            32.5352       1.0000
## 7          0.4864            35.9027       1.0000
## 8          0.4234            40.8610       1.0000
## 9          0.4584            25.9898       0.9943
## 10         0.2868            29.8345       1.0000
## 
## $Best.nc
##                       KL      CH Hartigan     CCC   Scott    Marriot  TrCovW  TraceW Friedman  Rubin Cindex     DB
## Number_clusters   7.0000  2.0000   4.0000  9.0000   4.000          9    3.00  4.0000   9.0000  9.000 6.0000 2.0000
## Value_Index     170.9406 92.7417  23.4976 -0.5159 122.852 3365045502 3331.35 54.5085   1.8023 -0.242 0.3193 1.2674
##                 Silhouette   Duda PseudoT2  Beale Ratkowsky     Ball PtBiserial Frey McClain    Dunn Hubert SDindex
## Number_clusters     2.0000 2.0000   2.0000 2.0000    2.0000   3.0000     2.0000    1  2.0000 10.0000      0  2.0000
## Value_Index         0.3109 0.9784   2.1452 0.0684    0.4324 102.6435     0.4973   NA  0.6734  0.1728      0  1.8068
##                 Dindex    SDbw
## Number_clusters      0 10.0000
## Value_Index          0  0.3061
## 
## $Best.partition
##   [1] 1 2 1 1 2 1 1 1 2 1 1 1 1 1 2 2 1 1 1 2 1 1 1 2 1 2 1 2 2 1 2 1 2 2 1 2 1 1 1 1 1 2 1 1 2 2 1 1 2 2 2 1 2 2 1 2 1
##  [58] 2 1 2 2 2 2 1 1 2 2 1 2 2 1 1 1 1 1 1 1 1 2 1 2 2 2 1 1 2 2 2 1 1 1 1 1 1 1 2 2 1 1 1 1 2 1 1 2 2 1 1 2 2 2 2 2 2
## [115] 1 2 2 2 1 2 1 1 2 1 2 2 2 2 2 1 2 2 1 2 1 2 2 2 2 2 2 2 1 2 1 1 1 1 1 1 1 1 1 1 1 1 2

We will proceed with 3 clusters, as this would be the most optimal for our project.

K-Means Clustering

Clustering <- kmeans(mydata_clu_std, 
                     centers = 3, #Number of groups
                     nstart = 25) #Number of attempts at different starting leader positions

Clustering
## K-means clustering with 3 clusters of sizes 45, 46, 66
## 
## Cluster means:
##   Security_concerns Lack_of_competence_or_support Preference_for_traditional_methods Aversion_to_change
## 1       -0.96430126                    -0.5695709                         -1.2244132         -0.9276589
## 2        0.09075076                    -0.6448033                          0.2826530          0.1622943
## 3        0.59422760                     0.8377522                          0.6378266          0.5193805
##   Physical_limitations
## 1           -0.4679937
## 2           -0.7534631
## 3            0.8442276
## 
## Clustering vector:
##   [1] 1 3 1 1 3 2 2 2 3 1 2 2 2 1 3 3 1 1 3 3 2 2 1 3 2 3 1 3 3 1 2 2 3 3 2 3 2 1 2 1 1 3 2 1 3 3 2 2 3 2 3 2 3 3 1 3 2
##  [58] 3 2 3 3 2 3 1 1 3 3 2 3 3 1 1 1 2 2 1 3 1 3 1 3 3 2 2 1 3 3 3 2 1 2 1 1 1 1 3 2 1 2 1 1 3 1 2 2 2 2 2 3 3 3 3 3 3
## [115] 1 3 3 3 1 3 1 1 3 1 3 3 3 3 3 3 3 3 2 2 3 3 3 3 3 3 3 3 1 2 2 1 1 2 2 2 2 2 1 1 1 1 2
## 
## Within cluster sum of squares by cluster:
## [1] 141.2425 107.8430 146.6024
##  (between_SS / total_SS =  49.3 %)
## 
## Available components:
## 
## [1] "cluster"      "centers"      "totss"        "withinss"     "tot.withinss" "betweenss"    "size"        
## [8] "iter"         "ifault"
library(factoextra)
fviz_cluster(Clustering, 
             palette = "Set1", 
             repel = FALSE,
             ggtheme = theme_bw(),
             data = mydata_clu_std)

Group profiles

Averages <- Clustering$centers
Averages #Average values of cluster variables to describe groups
##   Security_concerns Lack_of_competence_or_support Preference_for_traditional_methods Aversion_to_change
## 1       -0.96430126                    -0.5695709                         -1.2244132         -0.9276589
## 2        0.09075076                    -0.6448033                          0.2826530          0.1622943
## 3        0.59422760                     0.8377522                          0.6378266          0.5193805
##   Physical_limitations
## 1           -0.4679937
## 2           -0.7534631
## 3            0.8442276
Figure <- as.data.frame(Averages)
Figure$ID <- 1:nrow(Figure)

library(tidyr)
## 
## Attaching package: 'tidyr'
## The following object is masked from 'package:magrittr':
## 
##     extract
Figure <- pivot_longer(Figure, cols = c("Security_concerns", "Lack_of_competence_or_support", "Preference_for_traditional_methods", "Aversion_to_change", "Physical_limitations"))

Figure$Group <- factor(Figure$ID, 
                       levels = c(1, 2, 3), 
                       labels = c("1", "2", "3"))

Figure$NameF <- factor(Figure$name, 
                       levels = c("Security_concerns", "Lack_of_competence_or_support", "Preference_for_traditional_methods", "Aversion_to_change", "Physical_limitations"), 
                       labels = c("Security_concerns", "Lack_of_competence_or_support", "Preference_for_traditional_methods", "Aversion_to_change", "Physical_limitations"))

library(ggplot2)
ggplot(Figure, aes(x = NameF, y = value)) +
  geom_hline(yintercept = 0) +
  theme_bw() +
  geom_point(aes(shape = Group, col = Group), size = 3) +
  geom_line(aes(group = ID), linewidth = 1) +
  ylab("Averages") +
  xlab("Cluster variables")+
  ylim(-2.2, 2.2) +
  theme(axis.text.x = element_text(angle = 45, vjust = 0.50, size = 10))

mydata$Group <- Clustering$cluster #Assignings units to groups

Appropriateness of cluster variables used

#Checking if clustering variables successfully differentiate between groups

fit <- aov(cbind(Security_concerns, Lack_of_competence_or_support, Preference_for_traditional_methods, Aversion_to_change, Physical_limitations) ~ as.factor(Group), 
           data = mydata)

summary(fit)
##  Response Security_concerns :
##                   Df Sum Sq Mean Sq F value    Pr(>F)    
## as.factor(Group)   2 56.024 28.0120  55.771 < 2.2e-16 ***
## Residuals        154 77.350  0.5023                      
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
##  Response Lack_of_competence_or_support :
##                   Df Sum Sq Mean Sq F value    Pr(>F)    
## as.factor(Group)   2 123.84  61.922  81.146 < 2.2e-16 ***
## Residuals        154 117.52   0.763                      
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
##  Response Preference_for_traditional_methods :
##                   Df Sum Sq Mean Sq F value    Pr(>F)    
## as.factor(Group)   2 81.174  40.587  130.06 < 2.2e-16 ***
## Residuals        154 48.057   0.312                      
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
##  Response Aversion_to_change :
##                   Df  Sum Sq Mean Sq F value    Pr(>F)    
## as.factor(Group)   2  80.167  40.083  45.248 3.486e-16 ***
## Residuals        154 136.423   0.886                      
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
##  Response Physical_limitations :
##                   Df Sum Sq Mean Sq F value    Pr(>F)    
## as.factor(Group)   2 162.21  81.105   87.57 < 2.2e-16 ***
## Residuals        154 142.63   0.926                      
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Response for Social support:

  • H0: μ(Security_concerns, G1) = μ(Security_concerns, G2) = μ(Security_concerns, G3)

  • H1: At least one μ(Security_concerns, j) is different.

We can reject H0 at p < 0.001. We can reject H0 for all cluster variables at p < 0.001. Therefore we can assume that the groups are statistically different in the mean values of the cluster variables.

Criterion Validity

Next step is to check the criterion validity of the classification with variables that were not used in the clustering process.

  1. Q2: Ali dnevno uporabljate pametni telefon? [1 - Da, 2 - Ne].
chi_square <- chisq.test(mydata$Q2, as.factor(mydata$Group))
chi_square
## 
##  Pearson's Chi-squared test
## 
## data:  mydata$Q2 and as.factor(mydata$Group)
## X-squared = 20.555, df = 2, p-value = 3.439e-05
addmargins(chi_square$observed)
##          as.factor(mydata$Group)
## mydata$Q2   1   2   3 Sum
##       Yes  40  41  39 120
##       No    5   4  27  36
##       Sum  45  45  66 156
addmargins(round(chi_square$expected, 2)) 
##          as.factor(mydata$Group)
## mydata$Q2     1     2     3    Sum
##       Yes 34.62 34.62 50.77 120.01
##       No  10.38 10.38 15.23  35.99
##       Sum 45.00 45.00 66.00 156.00
round(chi_square$res, 2) 
##          as.factor(mydata$Group)
## mydata$Q2     1     2     3
##       Yes  0.92  1.09 -1.65
##       No  -1.67 -1.98  3.02
library(effectsize)
effectsize::cramers_v(mydata$Q2, mydata$Group)
## Cramer's V (adj.) |       95% CI
## --------------------------------
## 0.35              | [0.19, 1.00]
## 
## - One-sided CIs: upper bound fixed at [1.00].

  • H0: There is no association between usage of mobile phone and classification of customers in 3 groups.

  • H1: There is association between usage of mobile phone and classification of customers in 3 groups.

We can reject H0 at p < 0.001, so there are differences. All expected variables are above 5, so this assumption is met.

table_clusters1 <- table(mydata$Group, mydata$Q2)

prop_table_clusters1 <- prop.table(table_clusters1, margin = 1)

prop_df1 <- as.data.frame(as.table(prop_table_clusters1))

library(ggplot2)
ggplot(prop_df1, aes(x = Var1, y = Freq * 100, fill = Var2)) +  
  geom_bar(stat = "identity", position = "stack") +
  scale_fill_manual(values = c("#84BD00", "#33006F")) +  # Custom Colors
  labs(
    x = "Group", 
    y = "Percentage (%)", 
    fill = "Category", 
    title = "Percentage of Phone Usage by Group"
  ) +
  theme_minimal()

  1. Q41: Spol [1 - Ženska, 2 - Moški]
chi_square <- chisq.test(mydata$Q41, as.factor(mydata$Group))
chi_square
## 
##  Pearson's Chi-squared test
## 
## data:  mydata$Q41 and as.factor(mydata$Group)
## X-squared = 1.6566, df = 2, p-value = 0.4368
addmargins(chi_square$observed)
##           as.factor(mydata$Group)
## mydata$Q41   1   2   3 Sum
##     Female  22  26  30  78
##     Male    23  19  36  78
##     Sum     45  45  66 156
addmargins(round(chi_square$expected, 2)) 
##           as.factor(mydata$Group)
## mydata$Q41    1    2  3 Sum
##     Female 22.5 22.5 33  78
##     Male   22.5 22.5 33  78
##     Sum    45.0 45.0 66 156
round(chi_square$res, 2) 
##           as.factor(mydata$Group)
## mydata$Q41     1     2     3
##     Female -0.11  0.74 -0.52
##     Male    0.11 -0.74  0.52
library(effectsize)
effectsize::cramers_v(mydata$Q41, mydata$Group)
## Cramer's V (adj.) |       95% CI
## --------------------------------
## 0.00              | [0.00, 1.00]
## 
## - One-sided CIs: upper bound fixed at [1.00].

  • H0: There is no association between gender and classification of customers in 3 groups.

  • H1: There is association between gender and classification of customers in 3 groups.

We cannot reject H0 at p = 0.437, so there are no differences.

  1. Q42: Prosimo, vpišite leto rojstva. [Odprto besedilo]
mydata$Q42 <- as.numeric(as.character(mydata$Q42))
current_year <- as.numeric(format(Sys.Date(), "%Y"))
mydata$Age <- current_year - mydata$Q42
mydata$AgeGroup <- cut(mydata$Age, 
                       breaks = c(18, 40, 60, Inf), 
                       labels = c("18-40", "41-60", "60+"))
chi_square <- chisq.test(mydata$AgeGroup, as.factor(mydata$Group))
chi_square
## 
##  Pearson's Chi-squared test
## 
## data:  mydata$AgeGroup and as.factor(mydata$Group)
## X-squared = 49.906, df = 4, p-value = 3.778e-10
addmargins(chi_square$observed)
##                as.factor(mydata$Group)
## mydata$AgeGroup   1   2   3 Sum
##           18-40  21  22   3  46
##           41-60  16  18  23  57
##           60+     7   5  40  52
##           Sum    44  45  66 155
addmargins(round(chi_square$expected, 2)) 
##                as.factor(mydata$Group)
## mydata$AgeGroup     1     2     3 Sum
##           18-40 13.06 13.35 19.59  46
##           41-60 16.18 16.55 24.27  57
##           60+   14.76 15.10 22.14  52
##           Sum   44.00 45.00 66.00 155
round(chi_square$res, 2) 
##                as.factor(mydata$Group)
## mydata$AgeGroup     1     2     3
##           18-40  2.20  2.37 -3.75
##           41-60 -0.04  0.36 -0.26
##           60+   -2.02 -2.60  3.80
library(effectsize)
effectsize::cramers_v(mydata$AgeGroup, mydata$Group)
## Cramer's V (adj.) |       95% CI
## --------------------------------
## 0.39              | [0.27, 1.00]
## 
## - One-sided CIs: upper bound fixed at [1.00].

  • H0: There is no association between age and classification of customers in 3 groups.

  • H1: There is association between age and classification of customers in 3 groups.

We can reject H0 at p < 0.001, so there are differences.

table_clusters2 <- table(mydata$Group, mydata$AgeGroup)

prop_table_clusters2 <- prop.table(table_clusters2, margin = 1)

prop_df2 <- as.data.frame(as.table(prop_table_clusters2))

library(ggplot2)
ggplot(prop_df2, aes(x = Var1, y = Freq * 100, fill = Var2)) +  
  geom_bar(stat = "identity", position = "stack") +
  scale_fill_manual(values = c("#84BD00", "#33006F", "#FA7800")) +  # Custom Colors
  labs(
    x = "Group", 
    y = "Percentage (%)", 
    fill = "Category", 
    title = "Percentage of Age by Group"
  ) +
  theme_minimal()

  1. Q43: S kom živite v istem gospodinjstvu? [Izbira več možnih odgovorov]
  • Q43a: S starši
  • Q43b: S sorojenci
  • Q43c: S sorodniki (dedek, babica, stric, teta …)
  • Q43d: S partnerko/jem
  • Q43e: Z otroki
  • Q43f: S skrbnikom/i
  • Q43g: Živim sam/a
  • Q43h: Drugo
mydata[c(67:73)] <- mydata[c(67:73)] %>% mutate_all(as.numeric)
mydata$Household <- case_when(
  mydata$Q43a == 1 ~ 1,   # If var1 is 1, assign 1
  mydata$Q43b == 1 ~ 1,   
  mydata$Q43c == 1 ~ 1,
  mydata$Q43d == 1 ~ 1,
  mydata$Q43e == 1 ~ 1,
  mydata$Q43f == 1 ~ 1,
  mydata$Q43g == 1 ~ 2,
  TRUE ~ 0)
mydata$Household <- factor(mydata$Household, 
                         levels = c(1, 2), 
                         labels = c("With somebody", "Living alone"))
chi_square <- chisq.test(mydata$Household, as.factor(mydata$Group))
chi_square
## 
##  Pearson's Chi-squared test
## 
## data:  mydata$Household and as.factor(mydata$Group)
## X-squared = 5.3091, df = 2, p-value = 0.07033
addmargins(chi_square$observed)
##                 as.factor(mydata$Group)
## mydata$Household   1   2   3 Sum
##    With somebody  41  33  56 130
##    Living alone    4  12  10  26
##    Sum            45  45  66 156
addmargins(round(chi_square$expected, 2)) 
##                 as.factor(mydata$Group)
## mydata$Household    1    2  3 Sum
##    With somebody 37.5 37.5 55 130
##    Living alone   7.5  7.5 11  26
##    Sum           45.0 45.0 66 156
round(chi_square$res, 2) 
##                 as.factor(mydata$Group)
## mydata$Household     1     2     3
##    With somebody  0.57 -0.73  0.13
##    Living alone  -1.28  1.64 -0.30
library(effectsize)
effectsize::cramers_v(mydata$Household, mydata$Group)
## Cramer's V (adj.) |       95% CI
## --------------------------------
## 0.15              | [0.00, 1.00]
## 
## - One-sided CIs: upper bound fixed at [1.00].

  • H0: There is no association between living situation and classification of customers in 3 groups.

  • H1: There is association between living situation and classification of customers in 3 groups.

We can reject H0 at p = 0.071, so there are differences.

table_clusters3 <- table(mydata$Group, mydata$Household)

prop_table_clusters3 <- prop.table(table_clusters3, margin = 1)

prop_df3 <- as.data.frame(as.table(prop_table_clusters3))

library(ggplot2)
ggplot(prop_df3, aes(x = Var1, y = Freq * 100, fill = Var2)) +  
  geom_bar(stat = "identity", position = "stack") +
  scale_fill_manual(values = c("#84BD00", "#33006F")) +  # Custom Colors
  labs(
    x = "Group", 
    y = "Percentage (%)", 
    fill = "Category", 
    title = "Percentage of Household by Group"
  ) +
  theme_minimal()

  1. Q44: Koliko prebivalcev živi v kraju, kjer prebivate? [1 - Manj kot 1.000 prebivalcev, 2 - 1.000–5.000 prebivalcev, 3 - 5.001–20.000 prebivalcev, 4 - 20.001–50.000 prebivalcev, 5 - 50.001–100.000 prebivalcev, 6 - Več kot 100.000 prebivalcev]
mydata <- mydata %>%
  mutate(Q44_numeric = as.integer(Q44)) 

mydata <- mydata %>%
  mutate(Q44_numeric = case_when(
    Q44_numeric %in% c(1) ~ (1 + 999) / 2,  
        Q44_numeric %in% c(2) ~ (1000 + 5000) / 2,  
      Q44_numeric %in% c(3) ~ (5001 + 20000) / 2,   
      Q44_numeric %in% c(4) ~ (20001 + 50000) / 2,   
     Q44_numeric %in% c(5) ~ (50001 + 100000) / 2,  
     Q44_numeric %in% c(6) ~ (100001 + 285000) / 2,  
  ))
library(car)
## Loading required package: carData
## 
## Attaching package: 'car'
## The following object is masked from 'package:dplyr':
## 
##     recode
leveneTest(mydata$Q44_numeric, as.factor(mydata$Group))
## Levene's Test for Homogeneity of Variance (center = median)
##        Df F value  Pr(>F)  
## group   2  2.8609 0.06028 .
##       153                  
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

  • H0: σ2 (habitants, G1) = σ2 (habitants, G2) = σ2 (habitants, G3)

  • H1: At least one σ2 (habitants, j) is different.

We can reject H0 at p = 0.061, meaning that at least one variance is different from the rest. With this we checked the homogeneity of variances, therefore we can proceed with Welch F-test (if normality is not violated).

library(dplyr)
library(rstatix)
## 
## Attaching package: 'rstatix'
## The following objects are masked from 'package:effectsize':
## 
##     cohens_d, eta_squared
## The following object is masked from 'package:stats':
## 
##     filter
mydata %>%
  group_by(as.factor(mydata$Group)) %>%
  shapiro_test(Q44_numeric)
## # A tibble: 3 × 4
##   `as.factor(mydata$Group)` variable    statistic        p
##   <fct>                     <chr>           <dbl>    <dbl>
## 1 1                         Q44_numeric     0.508 4.46e-11
## 2 2                         Q44_numeric     0.644 3.40e- 9
## 3 3                         Q44_numeric     0.651 2.94e-11

  • H0: Habitants are normally distributed in G1.

  • H1: Habitants are not normally distributed in G1.

We reject H0 at p < 0.001.

  • H0: Habitants are normally distributed in G2.

  • H1: Habitants are not normally distributed in G2.

We reject H0 at p < 0.001.

  • H0: Habitants are normally distributed in G3.

  • H1: Habitants are not normally distributed in G3.

We reject H0 at p < 0.001.

We have to use the Kruskal-Wallis Rank Sum Test.

kruskal.test(Q44_numeric ~ as.factor(Group), 
             data = mydata)
## 
##  Kruskal-Wallis rank sum test
## 
## data:  Q44_numeric by as.factor(Group)
## Kruskal-Wallis chi-squared = 3.8397, df = 2, p-value = 0.1466

  • H0: Location distribution of habitants are the same for all groups.

  • H1: Location distribution of habitants are not the same for all groups.

We cannot reject H0 at p = 0.147, therefore at least one location distribution is not different from the other, so the result is not validated.

  1. Q47: Kakšen je vaš mesečni neto prihodek? [1 - Pod 1.000€, 2 - 1.000€–1.500€, 3 - 1.501€–2.000€, 4 - 2.001€–3.000€, 5 - 3.001€–5.000€, 6 - 5.001€–10.000€, 7 - Nad 10.000€, 8 - Ne želim odgovoriti]
mydata <- mydata %>%
  mutate(Q47_numeric = as.integer(Q47)) 

mydata <- mydata %>%
  mutate(Q47_numeric = case_when(
    Q47_numeric %in% c(1) ~ (1 + 999) / 2, 
    Q47_numeric %in% c(2) ~ (1000 + 1500) / 2, 
    Q47_numeric %in% c(3) ~ (1501 + 2000) / 2, 
    Q47_numeric %in% c(4) ~ (2001 + 3000) / 2,
    Q47_numeric %in% c(5) ~ (3001 + 5000) / 2, 
    Q47_numeric %in% c(6) ~ (5001 + 10000) / 2, 
     Q47_numeric %in% c(7) ~ (10000 + 12000) / 2,  
     Q47_numeric %in% c(8) ~ NA,  
  ))
library(car)
leveneTest(mydata$Q47_numeric, as.factor(mydata$Group))
## Levene's Test for Homogeneity of Variance (center = median)
##        Df F value  Pr(>F)  
## group   2  3.0541 0.05023 .
##       144                  
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

  • H0: σ2 (habitants, G1) = σ2 (habitants, G2) = σ2 (habitants, G3)

  • H1: At least one σ2 (habitants, j) is different.

We can reject H0 at p = 0.061, meaning that at least one variance is different from the rest. With this we checked the homogeneity of variances, therefore we can proceed with Welch F-test (if normality is not violated).

library(dplyr)
library(rstatix)
mydata %>%
  group_by(as.factor(mydata$Group)) %>%
  shapiro_test(Q47_numeric)
## # A tibble: 3 × 4
##   `as.factor(mydata$Group)` variable    statistic           p
##   <fct>                     <chr>           <dbl>       <dbl>
## 1 1                         Q47_numeric     0.741 0.000000186
## 2 2                         Q47_numeric     0.834 0.0000465  
## 3 3                         Q47_numeric     0.843 0.00000100

  • H0: Habitants are normally distributed in G1.

  • H1: Habitants are not normally distributed in G1.

We reject H0 at p < 0.001.

  • H0: Habitants are normally distributed in G2.

  • H1: Habitants are not normally distributed in G2.

We reject H0 at p < 0.001.

  • H0: Habitants are normally distributed in G3.

  • H1: Habitants are not normally distributed in G3.

We reject H0 at p < 0.001.

We have to use the Kruskal-Wallis Rank Sum Test.

kruskal.test(Q47_numeric ~ as.factor(Group), 
             data = mydata)
## 
##  Kruskal-Wallis rank sum test
## 
## data:  Q47_numeric by as.factor(Group)
## Kruskal-Wallis chi-squared = 7.6769, df = 2, p-value = 0.02153

  • H0: Location distribution of income are the same for all groups.

  • H1: Location distribution of income are not the same for all groups.

We can reject H0 at p = 0.032, therefore at least one location distribution is different from the other, so the result is validated.

table_clusters4 <- table(mydata$Group, mydata$Q47_numeric)

prop_table_clusters4 <- prop.table(table_clusters4, margin = 1)

prop_df4 <- as.data.frame(as.table(prop_table_clusters4))

library(ggplot2)
ggplot(prop_df4, aes(x = Var1, y = Freq * 100, fill = Var2)) +  
  geom_bar(stat = "identity", position = "stack") +
  scale_fill_manual(values = c("#84BD00", "#33006F", "#FA7800", "#0083B8", "#F4A900", "#007C80", "#00A499", "#002147")) +  # Custom Colors
  labs(
    x = "Group", 
    y = "Percentage (%)", 
    fill = "Category", 
    title = "Percentage of Income by Group"
  ) +
  theme_minimal()

  1. Q3: Ali uporabljate internet vsaj nekajkrat na teden? [1 - Da, 2 - Ne]
chi_square <- chisq.test(mydata$Q3, as.factor(mydata$Group))
chi_square
## 
##  Pearson's Chi-squared test
## 
## data:  mydata$Q3 and as.factor(mydata$Group)
## X-squared = 12.145, df = 2, p-value = 0.002305
addmargins(chi_square$observed)
##          as.factor(mydata$Group)
## mydata$Q3   1   2   3 Sum
##       Yes  38  44  48 130
##       No    7   1  18  26
##       Sum  45  45  66 156
addmargins(round(chi_square$expected, 2)) 
##          as.factor(mydata$Group)
## mydata$Q3    1    2  3 Sum
##       Yes 37.5 37.5 55 130
##       No   7.5  7.5 11  26
##       Sum 45.0 45.0 66 156
round(chi_square$res, 2) 
##          as.factor(mydata$Group)
## mydata$Q3     1     2     3
##       Yes  0.08  1.06 -0.94
##       No  -0.18 -2.37  2.11
library(effectsize)
effectsize::cramers_v(mydata$Q3, mydata$Group)
## Cramer's V (adj.) |       95% CI
## --------------------------------
## 0.26              | [0.06, 1.00]
## 
## - One-sided CIs: upper bound fixed at [1.00].

  • H0: There is no association between age and classification of customers in 3 groups.

  • H1: There is association between age and classification of customers in 3 groups.

We can reject H0 at p = 0.003, so there are differences.

table_clusters5 <- table(mydata$Group, mydata$Q3)

prop_table_clusters5 <- prop.table(table_clusters5, margin = 1)

prop_df5 <- as.data.frame(as.table(prop_table_clusters5))

library(ggplot2)
ggplot(prop_df5, aes(x = Var1, y = Freq * 100, fill = Var2)) +  
  geom_bar(stat = "identity", position = "stack") +
  scale_fill_manual(values = c("#33006F", "#84BD00")) +  # Custom Colors
  labs(
    x = "Group", 
    y = "Percentage (%)", 
    fill = "Category", 
    title = "Percentage of Internet Usage by Group"
  ) +
  theme_minimal()

  1. Q4: Ali je kdo od vaših družinskih članov uporabnik mobilne banke? [1 - Da, 2 - Ne, 3 - Ne vem]
chi_square <- chisq.test(mydata$Q4, as.factor(mydata$Group))
chi_square
## 
##  Pearson's Chi-squared test
## 
## data:  mydata$Q4 and as.factor(mydata$Group)
## X-squared = 4.7944, df = 4, p-value = 0.3091
addmargins(chi_square$observed)
##               as.factor(mydata$Group)
## mydata$Q4        1   2   3 Sum
##   Yes           28  34  40 102
##   No            10   7  11  28
##   I don't know   7   4  15  26
##   Sum           45  45  66 156
addmargins(round(chi_square$expected, 2)) 
##               as.factor(mydata$Group)
## mydata$Q4          1     2     3    Sum
##   Yes          29.42 29.42 43.15 101.99
##   No            8.08  8.08 11.85  28.01
##   I don't know  7.50  7.50 11.00  26.00
##   Sum          45.00 45.00 66.00 156.00
round(chi_square$res, 2) 
##               as.factor(mydata$Group)
## mydata$Q4          1     2     3
##   Yes          -0.26  0.84 -0.48
##   No            0.68 -0.38 -0.25
##   I don't know -0.18 -1.28  1.21
library(effectsize)
effectsize::cramers_v(mydata$Q4, mydata$Group)
## Cramer's V (adj.) |       95% CI
## --------------------------------
## 0.05              | [0.00, 1.00]
## 
## - One-sided CIs: upper bound fixed at [1.00].

  • H0: There is no association between income level and classification of customers in 3 groups.

  • H1: There is association between income level and classification of customers in 3 groups.

We cannot reject H0 at p = 0.310, so there are no differences.

  1. Q5: Ali imate skupni bančni račun, ki ga upravlja en član družine? [1 - Da, 2 - Ne]
chi_square <- chisq.test(mydata$Q5, as.factor(mydata$Group))
chi_square
## 
##  Pearson's Chi-squared test
## 
## data:  mydata$Q5 and as.factor(mydata$Group)
## X-squared = 5.4441, df = 2, p-value = 0.06574
addmargins(chi_square$observed)
##          as.factor(mydata$Group)
## mydata$Q5   1   2   3 Sum
##       Yes  10   9   5  24
##       No   35  36  61 132
##       Sum  45  45  66 156
addmargins(round(chi_square$expected, 2)) 
##          as.factor(mydata$Group)
## mydata$Q5     1     2     3    Sum
##       Yes  6.92  6.92 10.15  23.99
##       No  38.08 38.08 55.85 132.01
##       Sum 45.00 45.00 66.00 156.00
round(chi_square$res, 2) 
##          as.factor(mydata$Group)
## mydata$Q5     1     2     3
##       Yes  1.17  0.79 -1.62
##       No  -0.50 -0.34  0.69
library(effectsize)
effectsize::cramers_v(mydata$Q5, mydata$Group)
## Cramer's V (adj.) |       95% CI
## --------------------------------
## 0.15              | [0.00, 1.00]
## 
## - One-sided CIs: upper bound fixed at [1.00].

  • H0: There is no association between joint bank account and classification of customers in 3 groups.

  • H1: There is association between joint bank account and classification of customers in 3 groups.

We can reject H0 at p = 0.066, so there are differences.

table_clusters6 <- table(mydata$Group, mydata$Q5)

prop_table_clusters6 <- prop.table(table_clusters6, margin = 1)

prop_df6 <- as.data.frame(as.table(prop_table_clusters6))

library(ggplot2)
ggplot(prop_df6, aes(x = Var1, y = Freq * 100, fill = Var2)) +  
  geom_bar(stat = "identity", position = "stack") +
  scale_fill_manual(values = c("#33006F", "#84BD00")) +  # Custom Colors
  labs(
    x = "Group", 
    y = "Percentage (%)", 
    fill = "Category", 
    title = "Percentage of Joint Bank Account by Group"
  ) +
  theme_minimal()

  1. Q7: Se vam zdi, da bi potrebovali dodatno izobraževanje, preden bi lahko uporabljali mobilno banko? [1 - Da, 2 - Ne]
chi_square <- chisq.test(mydata$Q7, as.factor(mydata$Group))
chi_square
## 
##  Pearson's Chi-squared test
## 
## data:  mydata$Q7 and as.factor(mydata$Group)
## X-squared = 52.459, df = 2, p-value = 4.061e-12
addmargins(chi_square$observed)
##          as.factor(mydata$Group)
## mydata$Q7   1   2   3 Sum
##       Yes  12  13  57  82
##       No   33  32   9  74
##       Sum  45  45  66 156
addmargins(round(chi_square$expected, 2)) 
##          as.factor(mydata$Group)
## mydata$Q7     1     2     3    Sum
##       Yes 23.65 23.65 34.69  81.99
##       No  21.35 21.35 31.31  74.01
##       Sum 45.00 45.00 66.00 156.00
round(chi_square$res, 2) 
##          as.factor(mydata$Group)
## mydata$Q7     1     2     3
##       Yes -2.40 -2.19  3.79
##       No   2.52  2.31 -3.99
library(effectsize)
effectsize::cramers_v(mydata$Q7, mydata$Group)
## Cramer's V (adj.) |       95% CI
## --------------------------------
## 0.57              | [0.43, 1.00]
## 
## - One-sided CIs: upper bound fixed at [1.00].

  • H0: There is no association between Additional Education and classification of customers in 3 groups.

  • H1: There is association between Additional Education and classification of customers in 3 groups.

We can reject H0 at p < 0.001, so there are differences.

table_clusters7 <- table(mydata$Group, mydata$Q7)

prop_table_clusters7 <- prop.table(table_clusters7, margin = 1)

prop_df7 <- as.data.frame(as.table(prop_table_clusters7))

library(ggplot2)
ggplot(prop_df7, aes(x = Var1, y = Freq * 100, fill = Var2)) +  
  geom_bar(stat = "identity", position = "stack") +
  scale_fill_manual(values = c("#33006F", "#84BD00")) +  # Custom Colors
  labs(
    x = "Group", 
    y = "Percentage (%)", 
    fill = "Category", 
    title = "Percentage of Additional Education by Group"
  ) +
  theme_minimal()

Conclusion

Our analysis has revealed three distinct customer segments based on five standardized variables (Security Concerns, Lack of Competence or Support, Preference for Traditional Methods, Aversion to Change, and Physical Limitations), each with unique attitudes, behaviors, and barriers toward mobile banking adoption. While some are on the verge of adopting digital banking, others require stronger incentives, and a final group firmly resists change. Interestingly, our research indicates that gender, city size, and whether a family member uses mobile banking are not significant differentiators.

  1. The Practical Observers (45/157 = 29%)

The Practical Observers are the most open-minded of the three clusters—they are comfortable with technology, don not lack confidence or support, do not have strong security fears, and are not opposed to change. However, they have yet to find a compelling reason to adopt mobile banking.

This segment consists of a relatively young and middle-aged audience, with 48% aged 18–40 and 36% aged 41–60, while 16% are over 60. Their high engagement with technology is evident—88% use a smartphone daily, and 85% go online multiple times per week.

Financially, this group is diverse. While 43% earn below €1,000 per month, a significant 32% fall into the €3,001–€5,000 income bracket, and 21% earn between €1,501–€2,000. Additionally, 2% earn between €5,001–€10,000, and 2% exceed €10,000 net per month. This suggests that many in this group have disposable income and could benefit from mobile banking’s convenience and financial tools.

They are also socially active, with 88% living with a partner or family. However, 77% do not have a joint bank account, highlighting a preference for individual financial management. Importantly, 74% believe they don’t need additional education to start using mobile banking, reinforcing the idea that their resistance is more about habit than hesitation.

  1. The Cautious Adopters (46/157 = 29%)

The Cautious Adopters are one step away from transitioning to mobile banking but need reassurance about security and reliability before making the leap. They are slightly more traditional than Cluster 1, showing a stronger preference for in-person banking and a mildly higher aversion to change. However, they don’t struggle with technology, and like Cluster 1, they have below-average concerns about needing external support.

Their age profile is similar, with 49% aged 18–40 and 40% aged 41–60, but fewer (only 11%) are over 60. 91% use a smartphone daily, and 98% go online frequently, indicating a high digital readiness.

Financially, they are more affluent than Cluster 1, with 32% earning €3,001–€5,000, 23% earning €5,001–€10,000, and 23% between €1,501–€2,000. Only 23% earn under €1,000, making them the strongest potential customer base for premium digital banking services.

Most (73%) live with a partner or family, but like Cluster 1, 80% manage their finances independently without a joint bank account. Importantly, 72% do not believe they need additional education before using mobile banking, suggesting that their hesitation is psychological rather than technical.

  1. The Firm Traditionalists (66/157 = 42%)

The Firm Traditionalists are deeply resistant to mobile banking. Their concerns extend beyond security—they lack confidence in digital tools, strongly prefer in-person traditional banking, and are the only cluster with above average physical limitations – making them the most averse to change.

This segment is significantly older, with 61% aged over 60, 35% aged 41–60, and only 4% aged 18–40. They are the least digitally active, with only 60% using a smartphone daily and 73% accessing the internet at least a few times per week.

While 85% live with family or a partner, 95% handle their finances independently without a joint account, indicating a preference for financial autonomy despite their reliance on traditional banking methods.

Crucially, 87% feel they need additional education before adopting mobile banking, underscoring their lack of digital confidence. Their high level of security fears, preference for in-person banking, and reluctance toward technology make them the most difficult segment to convert.

PCA

mydata_PCA <- mydata[c(18:22)]
mydata_PCA2 <- mydata[c(25:34)]
R <- cor(mydata_PCA)

library(psych)
## 
## Attaching package: 'psych'
## The following object is masked from 'package:car':
## 
##     logit
## The following object is masked from 'package:effectsize':
## 
##     phi
## The following object is masked from 'package:Hmisc':
## 
##     describe
## The following objects are masked from 'package:ggplot2':
## 
##     %+%, alpha
corPlot(R)

library(psych)
cortest.bartlett(R, n = nrow(mydata_PCA))
## $chisq
## [1] 214.0942
## 
## $p.value
## [1] 1.838694e-40
## 
## $df
## [1] 10
library(psych)
KMO(R)
## Kaiser-Meyer-Olkin factor adequacy
## Call: KMO(r = R)
## Overall MSA =  0.78
## MSA for each item = 
##            Assistance              Security          Transparency           Convinience Speed and Reliability 
##                  0.79                  0.76                  0.80                  0.78                  0.77
library(FactoMineR)

components <- PCA(mydata_PCA, 
                  scale.unit = TRUE, 
                  graph = FALSE)
library(factoextra)
get_eigenvalue(components)
##       eigenvalue variance.percent cumulative.variance.percent
## Dim.1  2.7190702        54.381405                    54.38140
## Dim.2  0.7872302        15.744603                    70.12601
## Dim.3  0.6464620        12.929241                    83.05525
## Dim.4  0.4411911         8.823822                    91.87907
## Dim.5  0.4060465         8.120929                   100.00000

How many components to retain?

Eigenvalue of first principal component for standardized variables is bigger than 1.

The first 1 principal components explain more than 40% of the data. We measure evaluation which is subjective, so we measure soft data for which the chosen number of components should explain around 40% of the data.

The last chosen principal component captures more than 5% of total variance of original variables (5).

fviz_eig(components, 
         choice = "eigenvalue", 
         main = "Screeplot",
         ylab = "Eigenvalue",
         xlab = "Principal component",
         addlabels = TRUE)

When looking at the Scree plot the biggest difference between eigenvalues. This is between 1 and 2, so we should choose 1 principal component.

library(psych)
fa.parallel(mydata_PCA,
            sim = FALSE,
            fa = "pc")

## Parallel analysis suggests that the number of factors =  NA  and the number of components =  1

Parallel analysis suggests that we should choose 1 principal component.

Because we need to do the perception map, which has 2 principal components, we will use 2.

library(FactoMineR)
components <- PCA(mydata_PCA, 
                  ncp = 2, 
                  scale.unit = TRUE, 
                  graph = FALSE)
components$var$cor
##                           Dim.1       Dim.2
## Assistance            0.6588985  0.58938063
## Security              0.7655438  0.33729834
## Transparency          0.7487942 -0.09960031
## Convinience           0.7262470 -0.48912635
## Speed and Reliability 0.7814975 -0.27735471
loadings <- components$var$cor
library(factoextra)
eigenvalue <- get_eigenvalue(components)[1:2,1 ]
coefficient1 <- loadings[1:5]/sqrt(eigenvalue)[1]
coefficient2 <- loadings[6:10]/sqrt(eigenvalue)[2]
library(tidyverse)
## ── Attaching core tidyverse packages ──────────────────────────────────────────────────────────────── tidyverse 2.0.0 ──
## ✔ forcats   1.0.0     ✔ readr     2.1.5
## ✔ lubridate 1.9.4     ✔ stringr   1.5.1
## ✔ purrr     1.0.2     ✔ tibble    3.2.1
## ── Conflicts ────────────────────────────────────────────────────────────────────────────────── tidyverse_conflicts() ──
## ✖ psych::%+%()       masks ggplot2::%+%()
## ✖ psych::alpha()     masks ggplot2::alpha()
## ✖ tidyr::extract()   masks magrittr::extract()
## ✖ rstatix::filter()  masks dplyr::filter(), stats::filter()
## ✖ dplyr::lag()       masks stats::lag()
## ✖ car::recode()      masks dplyr::recode()
## ✖ purrr::set_names() masks magrittr::set_names()
## ✖ purrr::some()      masks car::some()
## ✖ Hmisc::src()       masks dplyr::src()
## ✖ Hmisc::summarize() masks dplyr::summarize()
## ℹ Use the conflicted package (<http://conflicted.r-lib.org/>) to force all conflicts to become errors
library(psych)

mydata_PCAD <- mydata_PCA2 %>% 
  pivot_longer(everything(), names_to = "name", values_to = "score")  %>% 
  separate(name, into = c("retailer", "dimension"), sep = "_")%>% 
  pivot_wider(names_from = retailer, values_from = score, values_fn = mean) %>%
  column_to_rownames(var = "dimension")

mydata_PCA_std <- scale(mydata_PCAD)
poslovalnica1 <- sum(mydata_PCA_std[,1]*coefficient1)
mobilna_banka1 <- sum(mydata_PCA_std[,2]*coefficient1)
poslovalnica2 <- sum(mydata_PCA_std[,1]*coefficient2)
mobilna_banka2 <- sum(mydata_PCA_std[,2]*coefficient2)
library(factoextra)

p <- fviz_pca_biplot(components, repel = TRUE, invisible = "ind", col.var = "#33006F")
p + 
  annotate("point", x = poslovalnica1, y = poslovalnica2, color = "#84BD00", size = 4, shape = 16) +
  annotate("text", x = poslovalnica1, y = poslovalnica2, label = "Branch", vjust = -1, color = "#84BD00") +
  annotate("point", x = mobilna_banka1, y = mobilna_banka2, color = "#FA7800", size = 4, shape = 16) +
  annotate("text", x = mobilna_banka1, y = mobilna_banka2, label = "Mobile bank", vjust = -1, color = "#FA7800")

Conclusion

Principal component analysis was performed on 5 standardized variables (n = 157). The KMO measure confirms the appropriateness of the variables, KMO = 0.78, although the data falls into the category “Middling”. The MSA statistics for the individual variables are above 0.50 for all variables. Based on the component’s loadings, we conclude that PC1 (𝜆1 = 2.72) represents quality, while PC2 (𝜆2 = 0.79) represents the contrast between security&customer support and service efficiency&transparency.

HYPOTHESIS TESTING

H1: Customers perceive that their financial information and transactions are protected equally well in the bank branch and the mobile app.

The questions from the survey used for this test:

Q11: Kako zaznavate zagotavljanje varnosti in zaščito vaših finančnih podatkov ter transakcij v poslovalnici in mobilni aplikaciji? [1 - Sploh se ne strinjam, 2 - Se ne strinjam, 3 - Niti niti, 4 - Se strinjam, 5 - Popolnoma se strinjam]

  • Q11a: Zdi se mi, da so moji finančni podatki in transakcije v poslovalnici ustrezno zaščiteni.
  • Q11b: Zdi se mi, da so moji finančni podatki in transakcije v mobilni banki ustrezno zaščiteni.
mydata$Branch_Security <- as.numeric(as.character(mydata$Branch_Security))

shapiro.test(mydata$Branch_Security)
## 
##  Shapiro-Wilk normality test
## 
## data:  mydata$Branch_Security
## W = 0.8418, p-value = 9.859e-12

  • H0: Distribution of Branch_Security is normal.

  • H1: Distribution of Branch_Security is not normal.

We can reject H0 at p < 0.001, so we can proceed with the alternative non-parametric test - Wilcoxon Signed Rank Test.

summary(mydata$Branch_Security)
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##   2.000   4.000   4.000   3.987   5.000   5.000
wilcox.test(mydata$Branch_Security,
            mu = 4,
            correct = FALSE)
## 
##  Wilcoxon signed rank test
## 
## data:  mydata$Branch_Security
## V = 1655.5, p-value = 0.8143
## alternative hypothesis: true location is not equal to 4

  • H0: Me = 4

  • H1: Me ≠ 4

We cannot reject H0 at p = 0.815.

mydata$`Mobile bank_Security` <- as.numeric(as.character(mydata$`Mobile bank_Security`))

shapiro.test(mydata$`Mobile bank_Security`)
## 
##  Shapiro-Wilk normality test
## 
## data:  mydata$`Mobile bank_Security`
## W = 0.90602, p-value = 1.652e-08

  • H0: Distribution of Mobile bank_Security is normal.

  • H1: Distribution of Mobile bank_Security is not normal.

We can reject H0 at p < 0.001, so we can proceed with the alternative non-parametric test - Wilcoxon Signed Rank Test.

summary(mydata$`Mobile bank_Security`)
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##   1.000   2.000   3.000   3.032   4.000   5.000
wilcox.test(mydata$`Mobile bank_Security`,
            mu = 3,
            correct = FALSE)
## 
##  Wilcoxon signed rank test
## 
## data:  mydata$`Mobile bank_Security`
## V = 2220, p-value = 0.7385
## alternative hypothesis: true location is not equal to 3

  • H0: Me = 3

  • H1: Me ≠ 3

We cannot reject H0 at p = 0.739.

mydata$Difference1 <- mydata$Branch_Security - mydata$`Mobile bank_Security`
shapiro.test(mydata$Difference1)
## 
##  Shapiro-Wilk normality test
## 
## data:  mydata$Difference1
## W = 0.91546, p-value = 6.321e-08

  • H0: Distribution of Difference 1 is normal.

  • H1: Distribution of Difference 1 is not normal.

We can reject H0 at p < 0.001, so we can proceed with the alternative non-parametric test - Wilcoxon Signed Rank Test.

wilcox.test(mydata$Branch_Security,  mydata$`Mobile bank_Security`,
            paired = TRUE,
            correct = FALSE,
            exact = FALSE,
            alternative = "two.sided")
## 
##  Wilcoxon signed rank test
## 
## data:  mydata$Branch_Security and mydata$`Mobile bank_Security`
## V = 5276.5, p-value = 3.499e-14
## alternative hypothesis: true location shift is not equal to 0

H0: Distribution location of Security in branches is equal to the distribution location of Security in mobile banks.

H0: Distribution location of Security in branches is not equal to the distribution location of Security in mobile banks.

We can reject H0 at p < 0.001.

cor.test(mydata$Branch_Security, mydata$`Mobile bank_Security`, method = "pearson")
## 
##  Pearson's product-moment correlation
## 
## data:  mydata$Branch_Security and mydata$`Mobile bank_Security`
## t = -0.1906, df = 155, p-value = 0.8491
## alternative hypothesis: true correlation is not equal to 0
## 95 percent confidence interval:
##  -0.1715345  0.1416702
## sample estimates:
##         cor 
## -0.01530765

H0: ρ (Branch_Security, Mobile bank_Security) = 0

H1: ρ (Branch_Security, Mobile bank_Security) ≠ 0

We cannot reject H0 at p = 0.850.

The results provide partial support for H1, as perceptions of security in bank branches and mobile banking are significantly different. Customers do not perceive a significant deviation from expected security levels in bank branches (p = 0.815) or mobile banking (p = 0.739), indicating that both meet expectations in terms of security. However, the Wilcoxon Signed-Rank Test shows a significant difference in security perceptions between branches and mobile banking (p < 0.001), meaning that customers do not rate their security equally in both settings. Despite this difference, the correlation test between branch and mobile banking security perceptions is not significant (p = 0.850), suggesting that trust in bank security does not directly translate into trust in mobile banking security. Thus, H1 is only partially supported—while security perceptions between branches and mobile banking differ, customers who trust their bank do not necessarily perceive mobile banking as more secure. This highlights that building trust in mobile banking security requires independent efforts beyond general bank security trust.

H2: Customers perceive that their banks meet transparency of their services in their branches and in their mobile bank equally well.

The questions from the survey used for this test:

Q12: Kako zaznavate transparentnost delovanja banke in njenih storitev v poslovalnici in mobilni aplikaciji? [1 - Sploh se ne strinjam, 2 - Se ne strinjam, 3 - Niti niti, 4 - Se strinjam, 5 - Popolnoma se strinjam]

  • Q12a: Zdi se mi, da je banka v poslovalnici transparentna glede svojega delovanja in storitev.
  • Q12b: Zdi se mi, da je mobilna banka transparentna glede svojega delovanja in storitev.
mydata$Branch_Transparency <- as.numeric(as.character(mydata$Branch_Transparency))

shapiro.test(mydata$Branch_Transparency)
## 
##  Shapiro-Wilk normality test
## 
## data:  mydata$Branch_Transparency
## W = 0.8319, p-value = 3.798e-12

  • H0: Distribution of Branch_Transparency is normal.

  • H1: Distribution of Branch_Transparency is not normal.

We can reject H0 at p < 0.001, so we can proceed with the alternative non-parametric test - Wilcoxon Signed Rank Test.

summary(mydata$Branch_Transparency)
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##   1.000   3.000   4.000   3.904   4.000   5.000
wilcox.test(mydata$Branch_Transparency,
            mu = 3,
            correct = FALSE)
## 
##  Wilcoxon signed rank test
## 
## data:  mydata$Branch_Transparency
## V = 7020.5, p-value < 2.2e-16
## alternative hypothesis: true location is not equal to 3

  • H0: Me = 3

  • H1: Me ≠ 3

We can reject H0 at p < 0.001.

mydata$`Mobile bank_Transparency` <- as.numeric(as.character(mydata$`Mobile bank_Transparency`))

shapiro.test(mydata$`Mobile bank_Transparency`)
## 
##  Shapiro-Wilk normality test
## 
## data:  mydata$`Mobile bank_Transparency`
## W = 0.88932, p-value = 1.859e-09

  • H0: Distribution of Branch_Transparency is normal.

  • H1: Distribution of Branch_Transparency is not normal.

We can reject H0 at p < 0.001, so we can proceed with the alternative non-parametric test - Wilcoxon Signed Rank Test.

summary(mydata$`Mobile bank_Transparency`)
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##   1.000   3.000   3.000   3.256   4.000   5.000
wilcox.test(mydata$`Mobile bank_Transparency`,
            mu = 3,
            correct = FALSE)
## 
##  Wilcoxon signed rank test
## 
## data:  mydata$`Mobile bank_Transparency`
## V = 2972, p-value = 0.0004326
## alternative hypothesis: true location is not equal to 3

  • H0: Me = 4

  • H1: Me ≠ 4

We can reject H0 at p < 0.001.

mydata$Difference2 <- mydata$Branch_Transparency - mydata$`Mobile bank_Transparency`
shapiro.test(mydata$Difference2)
## 
##  Shapiro-Wilk normality test
## 
## data:  mydata$Difference2
## W = 0.93147, p-value = 7.665e-07

  • H0: Distribution of Difference 2 is normal.

  • H1: Distribution of Difference 2 is not normal.

We can reject H0 at p < 0.001, so we can proceed with the alternative non-parametric test - Wilcoxon Signed Rank Test.

wilcox.test(mydata$Branch_Transparency,  mydata$`Mobile bank_Transparency`,
            paired = TRUE,
            correct = FALSE,
            exact = FALSE,
            alternative = "two.sided")
## 
##  Wilcoxon signed rank test
## 
## data:  mydata$Branch_Transparency and mydata$`Mobile bank_Transparency`
## V = 4145.5, p-value = 2.216e-09
## alternative hypothesis: true location shift is not equal to 0

H0: Distribution location of Transparency in branches is equal to the distribution location of Transparency in mobile banks.

H0: Distribution location of Transparency in branches is not equal to the distribution location of Transparency in mobile banks.

We can reject H0 at p < 0.001.

cor.test(mydata$Branch_Transparency, mydata$`Mobile bank_Transparency`, method = "pearson")
## 
##  Pearson's product-moment correlation
## 
## data:  mydata$Branch_Transparency and mydata$`Mobile bank_Transparency`
## t = -0.74959, df = 155, p-value = 0.4546
## alternative hypothesis: true correlation is not equal to 0
## 95 percent confidence interval:
##  -0.21471620  0.09745621
## sample estimates:
##         cor 
## -0.06009937

H0: ρ (Branch_Transparency, Mobile bank_Transparency) = 0

H1: ρ (Branch_Transparency, Mobile bank_Transparency) ≠ 0

We cannot reject H0 at p = 0.455.

The results provide partial support for H2, as transparency in privacy policies and guarantees appears to influence customer trust in mobile banking, but not in a straightforward way. Customers perceive bank branches as meeting transparency expectations, whereas mobile banking transparency significantly deviates from expectations (p < 0.001), indicating varied trust levels. Additionally, perceptions of transparency in branches and mobile banking are significantly different (p < 0.001), suggesting that customers evaluate them separately. However, there is no significant correlation between the two transparency perceptions (p = 0.455), meaning that trust in a bank’s physical transparency does not automatically extend to mobile banking. Thus, while transparency matters for trust, it must be built independently for mobile banking, as it is not inherently inherited from branch transparency. H2 is only partially supported, as transparency influences trust in mobile banking, but trust in branch transparency does not predict trust in mobile banking transparency.

H3: Security concerns are the most significant barrier to mobile banking adoption.

The questions from the survey used for this test:

Q6: Zakaj ne uporabljate mobilne banke (npr. KlikIn, , Addiko Mobil, …)? [1 - Sploh se ne strinjam, 2 - Se ne strinjam, 3 - Niti se ne strinjam niti se strinjam, 4 - Se strinjam, 5 - Popolnoma se strinjam]

  • Q6a: Ne želim si začeti uporabljati mobilne banke.
  • Q6b: Ne zaupam digitalnim storitvam.
  • Q6c: Strah me je vdora v račun.
  • Q6d: Nimam potrebe po mobilni banki.
  • Q6e: Slabo vidim na telefon.
  • Q6f: Raje grem fizično v poslovalnico.
  • Q6g: Potrebujem pomoč pri uporabi aplikacije.
  • Q6h: Ne počutim se dovolj kompetentnega/o.
  • Q6i: Strah me je, da bi lahko z napačno uporabo mobilne aplikacije naredil/a napako pri upravljanju financ.
library(rstatix)
library(tidyverse)

mydata_H3 <- mydata %>% select(c("ID", "Security_concerns", "Lack_of_competence_or_support", "Preference_for_traditional_methods", "Aversion_to_change", "Physical_limitations"))

#Wide format of data is changed to long format.
mydata_long <- mydata_H3 %>%
                 pivot_longer(
                   cols = c("Security_concerns", "Lack_of_competence_or_support", "Preference_for_traditional_methods", "Aversion_to_change", "Physical_limitations"),
                   names_to = "Concerns", 
                   values_to = "Agreeableness") %>%
                 convert_as_factor(Concerns)


mydata_long <- as.data.frame(mydata_long)

tail(mydata_long, 10)
##      ID                           Concerns Agreeableness
## 776 156                  Security_concerns      3.666667
## 777 156      Lack_of_competence_or_support      2.000000
## 778 156 Preference_for_traditional_methods      3.000000
## 779 156                 Aversion_to_change      3.000000
## 780 156               Physical_limitations      4.000000
## 781 157                  Security_concerns      4.000000
## 782 157      Lack_of_competence_or_support      2.500000
## 783 157 Preference_for_traditional_methods      4.500000
## 784 157                 Aversion_to_change      4.000000
## 785 157               Physical_limitations      2.000000
library(ggpubr)

#Boxplot for each variable.
ggboxplot(mydata_long, 
          x = "Concerns", 
          y = "Agreeableness", 
          add = "jitter") +
  scale_x_discrete(labels = c(
    "Aversion_to_change" = "Aversion",
    "Lack_of_competence_or_support" = "Lack",
    "Physical_limitations" = "Phys. Limitations",
    "Preference_for_traditional_methods" = "Traditional",
    "Security_concerns" = "Security"
  ))

library(tidyverse)
library(ggpubr)
library(rstatix)

#Finding outliers.
mydata_long %>%
  group_by(Concerns) %>%
  identify_outliers(Agreeableness)
## # A tibble: 9 × 5
##   Concerns              ID Agreeableness is.outlier is.extreme
##   <fct>              <int>         <dbl> <lgl>      <lgl>     
## 1 Aversion_to_change    27             1 TRUE       FALSE     
## 2 Aversion_to_change    71             1 TRUE       FALSE     
## 3 Aversion_to_change    75             1 TRUE       FALSE     
## 4 Aversion_to_change    80             1 TRUE       FALSE     
## 5 Aversion_to_change    85             1 TRUE       FALSE     
## 6 Aversion_to_change    98             1 TRUE       FALSE     
## 7 Aversion_to_change   121             1 TRUE       FALSE     
## 8 Aversion_to_change   122             1 TRUE       FALSE     
## 9 Security_concerns      1             1 TRUE       FALSE
#Removing outliers.
mydata_long <- mydata_long %>%
  filter(!ID == 27 & !ID == 71 & !ID == 75 & !ID == 80 & !ID == 85 & !ID == 98 & !ID == 121 & !ID == 122 & !ID == 1)
mydata_long %>%
  group_by(Concerns) %>%
  identify_outliers(Agreeableness)
## # A tibble: 1 × 5
##   Concerns             ID Agreeableness is.outlier is.extreme
##   <fct>             <int>         <dbl> <lgl>      <lgl>     
## 1 Security_concerns    95          1.33 TRUE       FALSE
#Removing outliers.
mydata_long <- mydata_long %>%
  filter(!ID == 95)
mydata_long %>%
  group_by(Concerns) %>%
  identify_outliers(Agreeableness)
## [1] Concerns      ID            Agreeableness is.outlier    is.extreme   
## <0 rows> (or 0-length row.names)
library(rstatix)

#Checking normality with Shapiro-Wilk test
mydata_long %>%
  group_by(Concerns) %>%
  shapiro_test(Agreeableness)
## # A tibble: 5 × 4
##   Concerns                           variable      statistic        p
##   <fct>                              <chr>             <dbl>    <dbl>
## 1 Aversion_to_change                 Agreeableness     0.864 2.68e-10
## 2 Lack_of_competence_or_support      Agreeableness     0.937 4.01e- 6
## 3 Physical_limitations               Agreeableness     0.883 2.15e- 9
## 4 Preference_for_traditional_methods Agreeableness     0.917 1.75e- 7
## 5 Security_concerns                  Agreeableness     0.948 2.93e- 5

  • H0: Agreeableness is normally distributed in aversion to change.

  • H1: Agreeableness is not normally distributed in aversion to change.

We can reject H0 for all 5 variables, therefore we use Friedman ANOVA.

library(rstatix)

#Friedman ANOVA.
FriedmanANOVA <- friedman_test(Agreeableness ~ Concerns | ID,
                               data = mydata_long)

FriedmanANOVA #Summary of results.
## # A tibble: 1 × 6
##   .y.               n statistic    df        p method       
## * <chr>         <int>     <dbl> <dbl>    <dbl> <chr>        
## 1 Agreeableness   147      76.6     4 8.94e-16 Friedman test

  • H0: Location distribution of agreeableness is the same in all five variables.

  • H1: At least one location distribution is different.

We can reject H0 at p < 0.001.

library(effectsize)
effectsize::kendalls_w(Agreeableness ~ Concerns | ID,
                       data = mydata_long)
## Warning: 124 block(s) contain ties, some containing only 1 unique ranking.
## Kendall's W |       95% CI
## --------------------------
## 0.13        | [0.09, 1.00]
## 
## - One-sided CIs: upper bound fixed at [1.00].
interpret_kendalls_w(0.13)
## [1] "slight agreement"
## (Rules: landis1977)
library(rstatix)

#Wilcoxon signed rank tests - comparing all possible paires.
paires_nonpar <- wilcox_test(Agreeableness ~ Concerns, 
                             paired = TRUE, 
                             p.adjust.method = "bonferroni",
                             data = mydata_long)

paires_nonpar
## # A tibble: 10 × 9
##    .y.           group1                             group2             n1    n2 statistic        p    p.adj p.adj.signif
##  * <chr>         <chr>                              <chr>           <int> <int>     <dbl>    <dbl>    <dbl> <chr>       
##  1 Agreeableness Aversion_to_change                 Lack_of_compet…   147   147     4208  1   e- 3 1.1 e- 2 *           
##  2 Agreeableness Aversion_to_change                 Physical_limit…   147   147     5322. 1.51e- 8 1.51e- 7 ****        
##  3 Agreeableness Aversion_to_change                 Preference_for…   147   147     2058. 1.3 e- 2 1.29e- 1 ns          
##  4 Agreeableness Aversion_to_change                 Security_conce…   147   147     2812. 8.1 e- 2 8.09e- 1 ns          
##  5 Agreeableness Lack_of_competence_or_support      Physical_limit…   147   147     3584. 2.5 e- 4 2   e- 3 **          
##  6 Agreeableness Lack_of_competence_or_support      Preference_for…   147   147     1183  3.35e- 8 3.35e- 7 ****        
##  7 Agreeableness Lack_of_competence_or_support      Security_conce…   147   147     1778. 2.76e- 7 2.76e- 6 ****        
##  8 Agreeableness Physical_limitations               Preference_for…   147   147      924  3.24e-12 3.24e-11 ****        
##  9 Agreeableness Physical_limitations               Security_conce…   147   147     1402. 1.88e-11 1.88e-10 ****        
## 10 Agreeableness Preference_for_traditional_methods Security_conce…   147   147     4190. 3.42e- 1 1   e+ 0 ns
library(rstatix)
comparisons <- paires_nonpar %>% 
                 add_y_position(fun = "median", step.increase = 0.35)

library(ggpubr)
ggboxplot(mydata_long, x = "Concerns", y = "Agreeableness", add = "point", ylim=c(0, 18)) +
  stat_pvalue_manual(comparisons, hide.ns = FALSE) +
  stat_summary(fun = median, geom = "point", shape = 16, size = 4,
               aes(group = Concerns), color = "#84BD00",
               position = position_dodge(width = 0.8)) +
  stat_summary(fun = median, colour = "#33006F", 
               position = position_dodge(width = 0.8),
               geom = "text", vjust = 0.5, hjust = -8,
               aes(label = round(after_stat(y), digits = 2), group = Concerns)) +
  labs(subtitle = get_test_label(FriedmanANOVA,  detailed = TRUE),
       caption = get_pwc_label(comparisons)) +
  scale_x_discrete(labels = c(
    "Aversion_to_change" = "Aversion",
    "Lack_of_competence_or_support" = "Lack",
    "Physical_limitations" = "Phys. Limitations",
    "Preference_for_traditional_methods" = "Traditional",
    "Security_concerns" = "Security"
  ))

The results from the Friedman ANOVA test indicate a significant difference among the various concerns non-users have regarding mobile banking (p < 0.001). This suggests that at least one concern stands out more than the others. The effect size analysis using Kendall’s W suggests a small to moderate effect size (W = 0.13), meaning the differences, while significant, are not overwhelmingly large.

To further explore these differences, we conducted Wilcoxon signed-rank tests for pairwise comparisons, adjusted with the Bonferroni correction. The results show that security concerns consistently received significantly higher ratings compared to other barriers, such as lack of competence, preference for traditional methods, and aversion to change. The median values confirm that concerns about security (e.g., fear of account breaches) were among the most highly rated trust-related barriers in comparison to other concerns.

The results partially support the hypothesis that security concerns are a major trust-related barrier for non-users. However, it is not the only significant concern - the preference for traditional banking is equally important in influencing trust-related barriers to mobile banking adoption. While security concerns are highly rated, they are not uniquely the most significant barrier. Therefore, H3 is partially supported but should also consider traditional banking preferences as a key factor.

H4: Financial incentives like cashbacks or transaction rewards, exemption from certain service charges and lower transaction fees compared to traditional banking would encourage customers to adopt mobile banking.

The questions from the survey used for this test:

Q18: Kateri finančni spodbudni ukrepi bi vas spodbudili k uporabi mobilnega bančništva? [1 - Sploh me ne bi spodbudilo, 2 - Ne bi me spodbudilo, 3 - Niti niti, 4 - Bi me spodbudilo, 5 - Zelo bi me spodbudilo]

  • Q18a: Cashback ali nagrade za transakcije.
  • Q18b: Oprostitev določenih storitvenih stroškov.
  • Q18c: Nižje provizije za transakcije v primerjavi s tradicionalnim bančništvom.
mydata <- mydata %>%
  rename(
    `Cashback` = "Q18a",
    `Exemption from service fees` = "Q18b",
    `Lower fees than trad.` = "Q18c",
  )
library(rstatix)
library(tidyverse)

mydata_H4 <- mydata %>% select(c(`Cashback`, `Exemption from service fees`, `Lower fees than trad.`))

# Add an ID column starting from 1
mydata_H4 <- mydata_H4 %>%
  mutate(ID = row_number())


#Wide format of data is changed to long format.
mydata_long_2 <- mydata_H4 %>%
  pivot_longer(
    cols = c(`Cashback`, `Exemption from service fees`, `Lower fees than trad.`), 
               names_to = "Incentive_Type", 
               values_to = "Agreeableness")%>%
                 convert_as_factor(Incentive_Type)

mydata_long_2 <- as.data.frame(mydata_long_2)

tail(mydata_long_2, 10)
##      ID              Incentive_Type Agreeableness
## 462 154       Lower fees than trad.             4
## 463 155                    Cashback             4
## 464 155 Exemption from service fees             4
## 465 155       Lower fees than trad.             3
## 466 156                    Cashback             5
## 467 156 Exemption from service fees             4
## 468 156       Lower fees than trad.             4
## 469 157                    Cashback             4
## 470 157 Exemption from service fees             3
## 471 157       Lower fees than trad.             4
library(ggpubr)

#Boxplot for each variable.
ggboxplot(mydata_long_2, 
          x = "Incentive_Type", 
          y = "Agreeableness", 
          add = "jitter")

### Repeated measures ANOVA
library(tidyverse)
library(ggpubr)
library(rstatix)

#Finding outliers.
mydata_long_2 %>%
  group_by("Incentive_Type") %>%
  identify_outliers("Agreeableness")
## # A tibble: 22 × 6
##    `"Incentive_Type"`    ID Incentive_Type              Agreeableness is.outlier is.extreme
##    <chr>              <int> <fct>                               <dbl> <lgl>      <lgl>     
##  1 Incentive_Type         5 Cashback                                1 TRUE       FALSE     
##  2 Incentive_Type        17 Lower fees than trad.                   1 TRUE       FALSE     
##  3 Incentive_Type        18 Cashback                                1 TRUE       FALSE     
##  4 Incentive_Type        21 Cashback                                1 TRUE       FALSE     
##  5 Incentive_Type        38 Cashback                                1 TRUE       FALSE     
##  6 Incentive_Type        79 Cashback                                1 TRUE       FALSE     
##  7 Incentive_Type        79 Exemption from service fees             1 TRUE       FALSE     
##  8 Incentive_Type        79 Lower fees than trad.                   1 TRUE       FALSE     
##  9 Incentive_Type        81 Cashback                                1 TRUE       FALSE     
## 10 Incentive_Type        81 Exemption from service fees             1 TRUE       FALSE     
## # ℹ 12 more rows
library(dplyr)
mydata_long_2 <- mydata_long_2 %>%
  filter(ID != 5 & ID != 17 & ID != 18 & ID != 21 & ID != 38 & ID != 79 & ID != 81 & ID != 82 & ID != 94 & ID != 96 & ID != 129 & ID != 143 & ID != 145 & ID != 149)

#Finding outliers.
mydata_long_2 %>%
  group_by("Incentive_Type") %>%
  identify_outliers("Agreeableness")
## [1] "Incentive_Type" ID               Incentive_Type   Agreeableness    is.outlier       is.extreme      
## <0 rows> (or 0-length row.names)
library(rstatix)

#Checking normality with Shapiro-Wilk test
mydata_long_2 %>%
  group_by(Incentive_Type) %>%
  shapiro_test(Agreeableness)
## # A tibble: 3 × 4
##   Incentive_Type              variable      statistic        p
##   <fct>                       <chr>             <dbl>    <dbl>
## 1 Cashback                    Agreeableness     0.869 6.84e-10
## 2 Exemption from service fees Agreeableness     0.858 2.10e-10
## 3 Lower fees than trad.       Agreeableness     0.841 3.77e-11

  • H0: Agreeableness is normally distributed in Cashback.

  • H1: Agreeableness is not normally distributed in Cashback.

We can reject H0 for all 3 variables, therefore we use Friedman ANOVA.

library(rstatix)

#Friedman ANOVA.
FriedmanANOVA_2 <- friedman_test(Agreeableness ~ Incentive_Type | ID,
                               data = mydata_long_2)

FriedmanANOVA_2 #Summary of results.
## # A tibble: 1 × 6
##   .y.               n statistic    df        p method       
## * <chr>         <int>     <dbl> <dbl>    <dbl> <chr>        
## 1 Agreeableness   143      17.4     2 0.000170 Friedman test

  • H0: Location distribution of agreeableness is the same in all three variables.

  • H1: At least one location distribution is different.

We can reject H0 at p < 0.001.

library(effectsize)
effectsize::kendalls_w(Agreeableness ~ Incentive_Type | ID,
                       data = mydata_long_2)
## Warning: 130 block(s) contain ties, some containing only 1 unique ranking.
## Kendall's W |       95% CI
## --------------------------
## 0.06        | [0.02, 1.00]
## 
## - One-sided CIs: upper bound fixed at [1.00].
interpret_kendalls_w(0.06)
## [1] "slight agreement"
## (Rules: landis1977)
library(rstatix)

#Wilcoxon signed rank tests - comparing all possible paires.
paires_nonpar_2 <- wilcox_test(Agreeableness ~ Incentive_Type, 
                             paired = TRUE, 
                             p.adjust.method = "bonferroni",
                             data = mydata_long_2)

paires_nonpar_2
## # A tibble: 3 × 9
##   .y.           group1                      group2                         n1    n2 statistic       p p.adj p.adj.signif
## * <chr>         <chr>                       <chr>                       <int> <int>     <dbl>   <dbl> <dbl> <chr>       
## 1 Agreeableness Cashback                    Exemption from service fees   143   143      694. 1   e-3 0.003 **          
## 2 Agreeableness Cashback                    Lower fees than trad.         143   143      485  4.95e-4 0.001 **          
## 3 Agreeableness Exemption from service fees Lower fees than trad.         143   143      454  8.09e-1 1     ns
library(rstatix)
comparisons_2 <- paires_nonpar_2 %>% 
                 add_y_position(fun = "median", step.increase = 0.35)

library(ggpubr)
ggboxplot(mydata_long_2, x = "Incentive_Type", y = "Agreeableness", add = "point", ylim=c(0, 18)) +
  stat_pvalue_manual(comparisons_2, hide.ns = FALSE) +
  stat_summary(fun = median, geom = "point", shape = 16, size = 4,
               aes(group = Incentive_Type), color = "#84BD00",
               position = position_dodge(width = 0.8)) +
  stat_summary(fun = median, colour = "#33006F", 
               position = position_dodge(width = 0.8),
               geom = "text", vjust = 0.5, hjust = -8,
               aes(label = round(after_stat(y), digits = 2), group = Incentive_Type)) +
  labs(subtitle = get_test_label(FriedmanANOVA_2,  detailed = TRUE),
       caption = get_pwc_label(comparisons_2))

The Friedman ANOVA test indicates a significant difference in how respondents perceive various financial incentives for mobile banking adoption (p < 0.001). This suggests that some financial incentives are more effective motivators than others. The effect size (Kendall’s W = 0.06) suggests a relatively small effect, meaning that while the differences are statistically significant, they may not be extremely strong in practical terms.

The Wilcoxon signed-rank tests for pairwise comparisons show that lower transaction fees (“Nižje provizije”) received significantly higher ratings than other incentives, such as cashback and fee waivers. However, the difference between cashback and fee waivers was not statistically significant, as indicated by the non-significant p-value (ns) in the pairwise tests.

The results partially support the hypothesis that financial incentives encourage mobile banking adoption, but not all financial incentives are equally effective. Cashback appears to be a stronger motivator than fee exemptions, while the difference between lower fees and exemptions is negligible. This suggests that banks should focus more on cashback rewards rather than just reducing or waiving service fees to enhance mobile banking adoption.

H5: Non-financial incentives like simple guides and instructions, 24/7 in-app help, personalised features, assistance from a bank staff member to set up the app and video guides or FAQ section would encourage customers to adopt mobile banking.

The questions from the survey used for this test:

Q19: Kateri nefinančni spodbudni ukrepi bi vas spodbudili k uporabi mobilnega bančništva? [1 - Sploh me ne bi spodbudilo, 2 - Ne bi me spodbudilo, 3 - Niti niti, 4 - Bi me spodbudilo, 5 - Zelo bi me spodbudilo]

  • Q19a: Preprosti vodiči in navodila.
  • Q19b: 24/7 pomoč uporabnikom znotraj aplikacije.
  • Q19c: Personalizirane funkcije (npr. prilagojene ponudbe za varčevanje).
  • Q19d: Pomoč bančnega uslužbenca pri nastavitvi aplikacije.
  • Q19e: Video vodiči ali razdelek s pogostimi vprašanji (FAQ).
mydata <- mydata %>%
  rename(
    `Guides` = "Q19a",
    `24/7 Support` = "Q19b",
    `Personalization` = "Q19c",
    `Banker assistance` = "Q19d",
    `Video&FAQ` = "Q19e",
  )
library(rstatix)
library(tidyverse)

mydata_H5 <- mydata %>% select(c(`Guides`, `24/7 Support`, `Personalization`, `Banker assistance`, `Video&FAQ`))

# Add an ID column starting from 1
mydata_H5 <- mydata_H5 %>%
  mutate(ID = row_number())

#Wide format of data is changed to long format.
mydata_long_3 <- mydata_H5 %>%
  pivot_longer(
    cols = c(`Guides`, `24/7 Support`, `Personalization`, `Banker assistance`, `Video&FAQ`), 
               names_to = "Incentive_Type", 
               values_to = "Agreeableness")%>%
                 convert_as_factor(Incentive_Type)

mydata_long_3 <- as.data.frame(mydata_long_3)

tail(mydata_long_3, 10)
##      ID    Incentive_Type Agreeableness
## 776 156            Guides             3
## 777 156      24/7 Support             4
## 778 156   Personalization             5
## 779 156 Banker assistance             5
## 780 156         Video&FAQ             4
## 781 157            Guides             3
## 782 157      24/7 Support             4
## 783 157   Personalization             4
## 784 157 Banker assistance             4
## 785 157         Video&FAQ             3
library(ggpubr)

#Boxplot for each variable.
ggboxplot(mydata_long_3, 
          x = "Incentive_Type", 
          y = "Agreeableness", 
          add = "jitter")

### Repeated measures ANOVA
library(tidyverse)
library(ggpubr)
library(rstatix)

#Finding outliers.
mydata_long_3 %>%
  group_by("Incentive_Type") %>%
  identify_outliers("Agreeableness")
## # A tibble: 56 × 6
##    `"Incentive_Type"`    ID Incentive_Type    Agreeableness is.outlier is.extreme
##    <chr>              <int> <fct>                     <dbl> <lgl>      <lgl>     
##  1 Incentive_Type         5 Personalization               1 TRUE       FALSE     
##  2 Incentive_Type         6 Guides                        1 TRUE       FALSE     
##  3 Incentive_Type        11 Guides                        1 TRUE       FALSE     
##  4 Incentive_Type        26 Personalization               1 TRUE       FALSE     
##  5 Incentive_Type        31 Guides                        1 TRUE       FALSE     
##  6 Incentive_Type        31 24/7 Support                  1 TRUE       FALSE     
##  7 Incentive_Type        31 Personalization               1 TRUE       FALSE     
##  8 Incentive_Type        31 Banker assistance             1 TRUE       FALSE     
##  9 Incentive_Type        31 Video&FAQ                     1 TRUE       FALSE     
## 10 Incentive_Type        60 Video&FAQ                     1 TRUE       FALSE     
## # ℹ 46 more rows
library(dplyr)
mydata_long_3 <- mydata_long_3 %>%
  filter(ID != 5 & ID != 6 & ID != 11 & ID != 26 & ID != 31 & ID != 60 & ID != 66 & ID != 68 & ID != 69 & ID != 70 & ID != 71 & ID != 75 & ID != 79 & ID != 81 & ID != 82 & ID != 85 & ID != 89 & ID != 90 & ID != 94 & ID != 96 & ID != 101 & ID != 143 & ID != 149)

#Finding outliers.
mydata_long_3 %>%
  group_by("Incentive_Type") %>%
  identify_outliers("Agreeableness")
## [1] "Incentive_Type" ID               Incentive_Type   Agreeableness    is.outlier       is.extreme      
## <0 rows> (or 0-length row.names)
library(rstatix)

#Checking normality with Shapiro-Wilk test
mydata_long_3 %>%
  group_by(Incentive_Type) %>%
  shapiro_test(Agreeableness)
## # A tibble: 5 × 4
##   Incentive_Type    variable      statistic        p
##   <fct>             <chr>             <dbl>    <dbl>
## 1 24/7 Support      Agreeableness     0.840 8.89e-11
## 2 Banker assistance Agreeableness     0.870 1.74e- 9
## 3 Guides            Agreeableness     0.863 8.43e-10
## 4 Personalization   Agreeableness     0.861 6.85e-10
## 5 Video&FAQ         Agreeableness     0.867 1.36e- 9

  • H0: Agreeableness is normally distributed in 24/7 Support.

  • H1: Agreeableness is not normally distributed in 24/7 Support.

We can reject H0 for all 5 variables, therefore we use Friedman ANOVA.

library(rstatix)

#Friedman ANOVA.
FriedmanANOVA_3 <- friedman_test(Agreeableness ~ Incentive_Type | ID,
                               data = mydata_long_3)

FriedmanANOVA_3 #Summary of results.
## # A tibble: 1 × 6
##   .y.               n statistic    df             p method       
## * <chr>         <int>     <dbl> <dbl>         <dbl> <chr>        
## 1 Agreeableness   134      46.9     4 0.00000000163 Friedman test

  • H0: Location distribution of agreeableness is the same in all three variables.

  • H1: At least one location distribution is different.

We can reject H0 at p < 0.001.

library(effectsize)
effectsize::kendalls_w(Agreeableness ~ Incentive_Type | ID,
                       data = mydata_long_3)
## Warning: 133 block(s) contain ties, some containing only 1 unique ranking.
## Kendall's W |       95% CI
## --------------------------
## 0.09        | [0.05, 1.00]
## 
## - One-sided CIs: upper bound fixed at [1.00].
interpret_kendalls_w(0.09)
## [1] "slight agreement"
## (Rules: landis1977)
library(rstatix)

#Wilcoxon signed rank tests - comparing all possible paires.
paires_nonpar_3 <- wilcox_test(Agreeableness ~ Incentive_Type, 
                             paired = TRUE, 
                             p.adjust.method = "bonferroni",
                             data = mydata_long_3)

paires_nonpar_3
## # A tibble: 10 × 9
##    .y.           group1            group2               n1    n2 statistic           p      p.adj p.adj.signif
##  * <chr>         <chr>             <chr>             <int> <int>     <dbl>       <dbl>      <dbl> <chr>       
##  1 Agreeableness 24/7 Support      Banker assistance   134   134     1100  0.487       1          ns          
##  2 Agreeableness 24/7 Support      Guides              134   134     1410. 0.0000217   0.000217   ***         
##  3 Agreeableness 24/7 Support      Personalization     134   134     1128. 0.054       0.538      ns          
##  4 Agreeableness 24/7 Support      Video&FAQ           134   134     2234  0.000000372 0.00000372 ****        
##  5 Agreeableness Banker assistance Guides              134   134     2232  0.0000917   0.000917   ***         
##  6 Agreeableness Banker assistance Personalization     134   134     1554  0.013       0.127      ns          
##  7 Agreeableness Banker assistance Video&FAQ           134   134     2648. 0.000000329 0.00000329 ****        
##  8 Agreeableness Guides            Personalization     134   134     1017  0.08        0.801      ns          
##  9 Agreeableness Guides            Video&FAQ           134   134     1758. 0.062       0.616      ns          
## 10 Agreeableness Personalization   Video&FAQ           134   134     1989  0.00067     0.007      **
library(rstatix)
comparisons_3 <- paires_nonpar_3 %>% 
                 add_y_position(fun = "median", step.increase = 0.35)

library(ggpubr)
ggboxplot(mydata_long_3, x = "Incentive_Type", y = "Agreeableness", add = "point", ylim=c(0, 18)) +
  stat_pvalue_manual(comparisons_3, hide.ns = FALSE) +
  stat_summary(fun = median, geom = "point", shape = 16, size = 4,
               aes(group = Incentive_Type), color = "#84BD00",
               position = position_dodge(width = 0.8)) +
  stat_summary(fun = median, colour = "#33006F", 
               position = position_dodge(width = 0.8),
               geom = "text", vjust = 0.5, hjust = -8,
               aes(label = round(after_stat(y), digits = 2), group = Incentive_Type)) +
  labs(subtitle = get_test_label(FriedmanANOVA_3,  detailed = TRUE),
       caption = get_pwc_label(comparisons_3))

The Friedman ANOVA test indicates a significant difference (p < 0.001) in how respondents perceive different non-financial incentives for mobile banking adoption. This suggests that some forms of guidance and support are considered more effective than others. The effect size (Kendall’s W = 0.09) indicates a small but meaningful effect, suggesting that these non-financial incentives do have an impact on users’ willingness to adopt mobile banking.

The Wilcoxon signed-rank tests show that guides and tutorials received significantly higher ratings than other incentives, such as 24/7 customer support and banker assistance. However, video guides and FAQs were rated significantly lower than other forms of support, suggesting that users prefer direct assistance over passive learning materials.

The results partially support H5, confirming that non-financial incentives influence mobile banking adoption, but not all types of support are equally effective. Guides and personalization features were rated significantly higher than video-based help (FAQs & tutorials), which was the least effective. Additionally, banker assistance showed mixed results, being significantly different from some incentives but not others. This suggests that users value structured guidance (like step-by-step guides) over passive content like FAQs.

LOGISTIC REGRESSION

mydata_excel1 <- read_excel("~/Documents/Šola/IMB/2. semester/NLB project/Logistična IMB/1ka data.xlsx")

mydata_excel2 <- read_excel("~/Documents/Šola/IMB/2. semester/NLB project/Logistična IMB/anketa_končni podatki.xlsx")

mydata_excel2 <- mydata_excel2[-1, ] #Delete first row in which the questions are written

mydata_excel2$ID <- seq(1,nrow(mydata_excel2))
mydata2 <- mydata_excel2[!(apply(mydata_excel2 == -3, 1, any)), ]

mydata2 <- subset(mydata2, select = -c(Q21:Q40))

mydata2$ID <- seq(1,nrow(mydata2))

Merging data

# Keep only specific columns (replace with actual column names)

data1_log_reg <- mydata_excel1 %>% select(Q1, Q41, Q42, Q43a, Q43b, Q43c, Q43d, Q43e, Q43f, Q43g, Q43h, Q44, Q45, Q46, Q47, Q48)

data2_log_reg <- mydata_excel2 %>% select(Q1, Q41, Q42, Q43a, Q43b, Q43c, Q43d, Q43e, Q43f, Q43g, Q43h, Q44, Q45, Q46, Q47, Q48)
merged_data_log_reg <- rbind(data1_log_reg, data2_log_reg)
# Rename columns using colnames()
colnames(merged_data_log_reg) <- c("Ali uporabljate mobilno aplikacijo",
                                   "Spol","Leto rojstva","Živim s starši",
                                   "Živim s sorojenci","Živim s sorodniki",
                                   "Živim s partnerko/jem", "Živim z otroki", 
                                   "Živim s skrbnikom", "Živim sam/a", "Drugo", 
                                   "Št. prebivalcev v kraju bivanja", "Primarna banka", 
                                   "Trenutna zaposlitev", "Mesečni neto prihodek", "Stopnja izobrazbe")

# Remove the first row
merged_data_log_reg1 <- merged_data_log_reg[-1, ]

categorical_vars <- c("Ali uporabljate mobilno aplikacijo", "Spol", 
                      "Živim s starši", "Živim s sorojenci", 
                      "Živim s sorodniki", "Živim s partnerko/jem", 
                      "Živim z otroki", "Živim s skrbnikom", 
                      "Živim sam/a", "Drugo", "Primarna banka", 
                      "Trenutna zaposlitev", "Stopnja izobrazbe")

merged_data_log_reg1[categorical_vars] <- lapply(merged_data_log_reg1[categorical_vars], as.factor)

# Convert numerical variables to numeric
numeric_vars <- c("Leto rojstva", "Št. prebivalcev v kraju bivanja", "Mesečni neto prihodek")
merged_data_log_reg1[numeric_vars] <- lapply(merged_data_log_reg1[numeric_vars], as.numeric)

merged_data_log_reg1 <- as.data.frame(merged_data_log_reg1)

head(merged_data_log_reg1)
##   Ali uporabljate mobilno aplikacijo Spol Leto rojstva Živim s starši Živim s sorojenci Živim s sorodniki
## 1                                  1    2         2000              1                 0                 0
## 2                                  1    2         1998              0                 0                 0
## 3                                  1    2         2001              1                 0                 0
## 4                                  1    2         1994              0                 0                 0
## 5                                  1    2         2000              1                 1                 0
## 6                                  1    2         2004              1                 0                 0
##   Živim s partnerko/jem Živim z otroki Živim s skrbnikom Živim sam/a Drugo Št. prebivalcev v kraju bivanja
## 1                     0              0                 0           0     0                               2
## 2                     1              0                 0           0     0                               6
## 3                     0              0                 0           0     0                               2
## 4                     1              0                 0           0     0                               6
## 5                     0              0                 0           0     0                               2
## 6                     0              0                 0           0     0                               3
##   Primarna banka Trenutna zaposlitev Mesečni neto prihodek Stopnja izobrazbe
## 1              3                   2                     2                 2
## 2              3                   2                     3                 4
## 3              1                   2                     2                 3
## 4             12                   5                     6                 5
## 5              1                   2                     3                 6
## 6              1                   1                     8                 4
merged_data_clean <- merged_data_log_reg1 %>%
  filter(complete.cases(.))  # Removes rows with any NAs
# Automatically calculate age and assign age groups
merged_data_clean <- merged_data_clean %>%
  mutate(
    `Leto rojstva` = as.numeric(as.character(`Leto rojstva`)),  # Convert to numeric (handles factors & characters)
    age = as.numeric(format(Sys.Date(), "%Y")) - `Leto rojstva`, # Calculate age
    age_group = case_when(
      age <= 27 ~ "Young",
      age > 27 & age <= 65 ~ "Professional",
      age > 65 ~ "Older"))

# Ensure output is a data frame (not a tibble)
merged_data_clean <- as.data.frame(merged_data_clean)

head(merged_data_clean)
##   Ali uporabljate mobilno aplikacijo Spol Leto rojstva Živim s starši Živim s sorojenci Živim s sorodniki
## 1                                  1    2         2000              1                 0                 0
## 2                                  1    2         1998              0                 0                 0
## 3                                  1    2         2001              1                 0                 0
## 4                                  1    2         1994              0                 0                 0
## 5                                  1    2         2000              1                 1                 0
## 6                                  1    2         2004              1                 0                 0
##   Živim s partnerko/jem Živim z otroki Živim s skrbnikom Živim sam/a Drugo Št. prebivalcev v kraju bivanja
## 1                     0              0                 0           0     0                               2
## 2                     1              0                 0           0     0                               6
## 3                     0              0                 0           0     0                               2
## 4                     1              0                 0           0     0                               6
## 5                     0              0                 0           0     0                               2
## 6                     0              0                 0           0     0                               3
##   Primarna banka Trenutna zaposlitev Mesečni neto prihodek Stopnja izobrazbe age    age_group
## 1              3                   2                     2                 2  25        Young
## 2              3                   2                     3                 4  27        Young
## 3              1                   2                     2                 3  24        Young
## 4             12                   5                     6                 5  31 Professional
## 5              1                   2                     3                 6  25        Young
## 6              1                   1                     8                 4  21        Young
# Add an ID column (sequential numbering)
merged_data_clean <- merged_data_clean %>%
  mutate(ID = row_number())

# Ensure it's a data frame (not a tibble)
merged_data_clean <- as.data.frame(merged_data_clean)

# Display the first 6 rows to check
head(merged_data_clean)
##   Ali uporabljate mobilno aplikacijo Spol Leto rojstva Živim s starši Živim s sorojenci Živim s sorodniki
## 1                                  1    2         2000              1                 0                 0
## 2                                  1    2         1998              0                 0                 0
## 3                                  1    2         2001              1                 0                 0
## 4                                  1    2         1994              0                 0                 0
## 5                                  1    2         2000              1                 1                 0
## 6                                  1    2         2004              1                 0                 0
##   Živim s partnerko/jem Živim z otroki Živim s skrbnikom Živim sam/a Drugo Št. prebivalcev v kraju bivanja
## 1                     0              0                 0           0     0                               2
## 2                     1              0                 0           0     0                               6
## 3                     0              0                 0           0     0                               2
## 4                     1              0                 0           0     0                               6
## 5                     0              0                 0           0     0                               2
## 6                     0              0                 0           0     0                               3
##   Primarna banka Trenutna zaposlitev Mesečni neto prihodek Stopnja izobrazbe age    age_group ID
## 1              3                   2                     2                 2  25        Young  1
## 2              3                   2                     3                 4  27        Young  2
## 3              1                   2                     2                 3  24        Young  3
## 4             12                   5                     6                 5  31 Professional  4
## 5              1                   2                     3                 6  25        Young  5
## 6              1                   1                     8                 4  21        Young  6
# Convert a specific column to numeric

merged_data_clean <- merged_data_clean %>%
  mutate(across(c(`Živim s starši`, `Živim s sorojenci`, `Živim s sorodniki`, `Živim s partnerko/jem`, `Živim z otroki`, `Živim s skrbnikom`, `Živim sam/a`, `Drugo`, `age`), 
                ~ as.numeric(as.character(.))))

Logistic Regression

merged_data_clean <- merged_data_clean %>%
  mutate(`Ali uporabljate mobilno aplikacijo` = ifelse(`Ali uporabljate mobilno aplikacijo` == 1, 0, ifelse(!1 == 2, 1, V1)))

head(merged_data_clean)
##   Ali uporabljate mobilno aplikacijo Spol Leto rojstva Živim s starši Živim s sorojenci Živim s sorodniki
## 1                                  0    2         2000              1                 0                 0
## 2                                  0    2         1998              0                 0                 0
## 3                                  0    2         2001              1                 0                 0
## 4                                  0    2         1994              0                 0                 0
## 5                                  0    2         2000              1                 1                 0
## 6                                  0    2         2004              1                 0                 0
##   Živim s partnerko/jem Živim z otroki Živim s skrbnikom Živim sam/a Drugo Št. prebivalcev v kraju bivanja
## 1                     0              0                 0           0     0                               2
## 2                     1              0                 0           0     0                               6
## 3                     0              0                 0           0     0                               2
## 4                     1              0                 0           0     0                               6
## 5                     0              0                 0           0     0                               2
## 6                     0              0                 0           0     0                               3
##   Primarna banka Trenutna zaposlitev Mesečni neto prihodek Stopnja izobrazbe age    age_group ID
## 1              3                   2                     2                 2  25        Young  1
## 2              3                   2                     3                 4  27        Young  2
## 3              1                   2                     2                 3  24        Young  3
## 4             12                   5                     6                 5  31 Professional  4
## 5              1                   2                     3                 6  25        Young  5
## 6              1                   1                     8                 4  21        Young  6
fit0 <- glm(`Ali uporabljate mobilno aplikacijo` ~ 1, #Dependent and explanatory variables
            family = binomial, #Binary logistic regression
            data = merged_data_clean)

summary(fit0)
## 
## Call:
## glm(formula = `Ali uporabljate mobilno aplikacijo` ~ 1, family = binomial, 
##     data = merged_data_clean)
## 
## Coefficients:
##             Estimate Std. Error z value Pr(>|z|)    
## (Intercept) -0.33798    0.09521   -3.55 0.000385 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 616.6  on 453  degrees of freedom
## Residual deviance: 616.6  on 453  degrees of freedom
## AIC: 618.6
## 
## Number of Fisher Scoring iterations: 4
exp(cbind(odds = fit0$coefficients, confint.default(fit0))) #Odds for Y=1
##                  odds     2.5 %    97.5 %
## (Intercept) 0.7132075 0.5917995 0.8595225
head(fitted(fit0)) #Estimated probability for Y=1
##         1         2         3         4         5         6 
## 0.4162996 0.4162996 0.4162996 0.4162996 0.4162996 0.4162996
fit1 <- glm(`Ali uporabljate mobilno aplikacijo` ~ age_group,  
            family = binomial, 
            data = merged_data_clean)

summary(fit1)
## 
## Call:
## glm(formula = `Ali uporabljate mobilno aplikacijo` ~ age_group, 
##     family = binomial, data = merged_data_clean)
## 
## Coefficients:
##                       Estimate Std. Error z value Pr(>|z|)    
## (Intercept)             1.5506     0.2942   5.270 1.36e-07 ***
## age_groupProfessional  -1.6854     0.3254  -5.180 2.22e-07 ***
## age_groupYoung         -3.2341     0.3635  -8.896  < 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: 616.6  on 453  degrees of freedom
## Residual deviance: 505.7  on 451  degrees of freedom
## AIC: 511.7
## 
## Number of Fisher Scoring iterations: 3
exp(cbind(odds = fit1$coefficients, confint.default(fit1))) #Odds for Y=1
##                             odds      2.5 %     97.5 %
## (Intercept)           4.71428558 2.64836954 8.39176261
## age_groupProfessional 0.18536719 0.09796180 0.35075912
## age_groupYoung        0.03939394 0.01931873 0.08033047
head(fitted(fit1)) #Estimated probability for Y=1
##         1         2         3         4         5         6 
## 0.1566265 0.1566265 0.1566265 0.4663462 0.1566265 0.1566265
#Ifelse za vprašanje s kom živijo

merged_data_clean <- merged_data_clean %>%
  mutate(
    # Create a text column listing all people they live with
    S_kom_živijo = paste0(
      ifelse(`Živim s starši` == 1, "Živim s starši, ", ""),
      ifelse(`Živim s sorojenci` == 1, "Živim s sorojenci, ", ""),
      ifelse(`Živim s sorodniki` == 1, "Živim s sorodniki, ", ""),
      ifelse(`Živim s partnerko/jem` == 1, "Živim s partnerko/jem, ", ""),
      ifelse(`Živim z otroki` == 1, "Živim z otroki, ", ""),
      ifelse(`Živim s skrbnikom` == 1, "Živim s skrbnikom, ", ""),
      ifelse(`Drugo` == 1, "Drugo, ", "")
    ),
    S_kom_živijo = sub(", $", "", S_kom_živijo), # Remove trailing comma

    # Categorize into "Lives alone" or "Lives with others"
    Kako_živijo = ifelse(`Živim sam/a` == 1, "Živi sam", "Živi z drugimi")
  )
anova(fit0, fit1, test = "Chi") #Comparision of models based on -2LL statistics
## Analysis of Deviance Table
## 
## Model 1: `Ali uporabljate mobilno aplikacijo` ~ 1
## Model 2: `Ali uporabljate mobilno aplikacijo` ~ age_group
##   Resid. Df Resid. Dev Df Deviance  Pr(>Chi)    
## 1       453      616.6                          
## 2       451      505.7  2   110.89 < 2.2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
exp(cbind(OR = fit1$coefficients, confint.default(fit1))) #Odds ratio for Y=1 (with 95% CI)
##                               OR      2.5 %     97.5 %
## (Intercept)           4.71428558 2.64836954 8.39176261
## age_groupProfessional 0.18536719 0.09796180 0.35075912
## age_groupYoung        0.03939394 0.01931873 0.08033047
merged_data_clean <- merged_data_clean %>%
  mutate(across(c(`Št. prebivalcev v kraju bivanja`, `Trenutna zaposlitev`, `Mesečni neto prihodek`, `Stopnja izobrazbe`), 
                ~ as.numeric(as.character(.))))

str(merged_data_clean)
## 'data.frame':    454 obs. of  21 variables:
##  $ Ali uporabljate mobilno aplikacijo: num  0 0 0 0 0 0 0 0 0 0 ...
##  $ Spol                              : Factor w/ 4 levels "-3","1","2","3": 3 3 3 3 3 3 2 3 3 2 ...
##  $ Leto rojstva                      : num  2000 1998 2001 1994 2000 ...
##  $ Živim s starši                    : num  1 0 1 0 1 1 0 0 0 0 ...
##  $ Živim s sorojenci                 : num  0 0 0 0 1 0 0 0 0 0 ...
##  $ Živim s sorodniki                 : num  0 0 0 0 0 0 0 0 0 0 ...
##  $ Živim s partnerko/jem             : num  0 1 0 1 0 0 1 0 1 1 ...
##  $ Živim z otroki                    : num  0 0 0 0 0 0 0 0 1 0 ...
##  $ Živim s skrbnikom                 : num  0 0 0 0 0 0 0 0 0 0 ...
##  $ Živim sam/a                       : num  0 0 0 0 0 0 0 1 0 0 ...
##  $ Drugo                             : num  0 0 0 0 0 0 0 0 0 0 ...
##  $ Št. prebivalcev v kraju bivanja   : num  2 6 2 6 2 3 6 6 1 4 ...
##  $ Primarna banka                    : Factor w/ 14 levels "-3","1","10",..: 8 8 2 5 2 2 11 7 2 7 ...
##  $ Trenutna zaposlitev               : num  2 2 2 5 2 1 2 1 2 1 ...
##  $ Mesečni neto prihodek             : num  2 3 2 6 3 8 4 6 4 2 ...
##  $ Stopnja izobrazbe                 : num  2 4 3 5 6 4 6 1 6 5 ...
##  $ age                               : num  25 27 24 31 25 21 28 23 39 23 ...
##  $ age_group                         : chr  "Young" "Young" "Young" "Professional" ...
##  $ ID                                : int  1 2 3 4 5 6 7 8 9 10 ...
##  $ S_kom_živijo                      : chr  "Živim s starši" "Živim s partnerko/jem" "Živim s starši" "Živim s partnerko/jem" ...
##  $ Kako_živijo                       : chr  "Živi z drugimi" "Živi z drugimi" "Živi z drugimi" "Živi z drugimi" ...
fit2 <- glm(`Ali uporabljate mobilno aplikacijo` ~ age_group + `Trenutna zaposlitev` + `Stopnja izobrazbe`,  
            family = binomial, 
            data = merged_data_clean)

summary(fit2)
## 
## Call:
## glm(formula = `Ali uporabljate mobilno aplikacijo` ~ age_group + 
##     `Trenutna zaposlitev` + `Stopnja izobrazbe`, family = binomial, 
##     data = merged_data_clean)
## 
## Coefficients:
##                       Estimate Std. Error z value Pr(>|z|)    
## (Intercept)            2.64298    0.50628   5.220 1.79e-07 ***
## age_groupProfessional -0.71711    0.36821  -1.948   0.0515 .  
## age_groupYoung        -2.51214    0.40758  -6.164 7.11e-10 ***
## `Trenutna zaposlitev`  0.07741    0.10220   0.757   0.4488    
## `Stopnja izobrazbe`   -0.44330    0.08969  -4.943 7.71e-07 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 616.60  on 453  degrees of freedom
## Residual deviance: 475.35  on 449  degrees of freedom
## AIC: 485.35
## 
## Number of Fisher Scoring iterations: 5
exp(cbind(odds = fit2$coefficients, confint.default(fit2))) #Odds for Y=1
##                              odds     2.5 %     97.5 %
## (Intercept)           14.05507453 5.2105773 37.9123286
## age_groupProfessional  0.48816000 0.2372152  1.0045737
## age_groupYoung         0.08109449 0.0364804  0.1802699
## `Trenutna zaposlitev`  1.08048242 0.8843581  1.3201013
## `Stopnja izobrazbe`    0.64191150 0.5384305  0.7652805
head(fitted(fit2)) #Estimated probability for Y=1
##          1          2          3          4          5          6 
## 0.35412612 0.18428812 0.26032964 0.52407772 0.08516365 0.17293487
fit3 <- glm(`Ali uporabljate mobilno aplikacijo` ~ age_group + `Št. prebivalcev v kraju bivanja`,  
            family = binomial, 
            data = merged_data_clean)

summary(fit3)
## 
## Call:
## glm(formula = `Ali uporabljate mobilno aplikacijo` ~ age_group + 
##     `Št. prebivalcev v kraju bivanja`, family = binomial, data = merged_data_clean)
## 
## Coefficients:
##                                   Estimate Std. Error z value Pr(>|z|)    
## (Intercept)                        1.95622    0.33318   5.871 4.32e-09 ***
## age_groupProfessional             -1.18740    0.34527  -3.439 0.000584 ***
## age_groupYoung                    -2.80997    0.37710  -7.452 9.23e-14 ***
## `Št. prebivalcev v kraju bivanja` -0.24141    0.05459  -4.422 9.76e-06 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 616.60  on 453  degrees of freedom
## Residual deviance: 484.64  on 450  degrees of freedom
## AIC: 492.64
## 
## Number of Fisher Scoring iterations: 4
exp(cbind(odds = fit0$coefficients, confint.default(fit3))) #Odds for Y=1
##                                        odds      2.5 %     97.5 %
## (Intercept)                       0.7132075 3.68101588 13.5888375
## age_groupProfessional             0.7132075 0.15503284  0.6000846
## age_groupYoung                    0.7132075 0.02875141  0.1260762
## `Št. prebivalcev v kraju bivanja` 0.7132075 0.70581729  0.8742236
head(fitted(fit3)) #Estimated probability for Y=1
##          1          2          3          4          5          6 
## 0.20807504 0.09094061 0.20807504 0.33634239 0.20807504 0.17108220
fit4 <- glm(`Ali uporabljate mobilno aplikacijo` ~ age_group + `Št. prebivalcev v kraju bivanja` + `Mesečni neto prihodek` + `Kako_živijo`,
            family = binomial, 
            data = merged_data_clean)
head(merged_data_clean[, c("age_group", "Št. prebivalcev v kraju bivanja", "Mesečni neto prihodek", "Kako_živijo")])
##      age_group Št. prebivalcev v kraju bivanja Mesečni neto prihodek    Kako_živijo
## 1        Young                               2                     2 Živi z drugimi
## 2        Young                               6                     3 Živi z drugimi
## 3        Young                               2                     2 Živi z drugimi
## 4 Professional                               6                     6 Živi z drugimi
## 5        Young                               2                     3 Živi z drugimi
## 6        Young                               3                     8 Živi z drugimi
library(car)
vif(fit4)
##                                       GVIF Df GVIF^(1/(2*Df))
## age_group                         1.340321  2        1.075975
## `Št. prebivalcev v kraju bivanja` 1.059177  1        1.029163
## `Mesečni neto prihodek`           1.342169  1        1.158520
## Kako_živijo                       1.036132  1        1.017906

All VIF values are close to 1, meaning there is no serious multicollinearity in our model. Our predictors are independent enough to be used together.

summary(fit4)
## 
## Call:
## glm(formula = `Ali uporabljate mobilno aplikacijo` ~ age_group + 
##     `Št. prebivalcev v kraju bivanja` + `Mesečni neto prihodek` + 
##     Kako_živijo, family = binomial, data = merged_data_clean)
## 
## Coefficients:
##                                   Estimate Std. Error z value Pr(>|z|)    
## (Intercept)                        2.83274    0.52146   5.432 5.56e-08 ***
## age_groupProfessional             -0.89742    0.38778  -2.314   0.0207 *  
## age_groupYoung                    -3.25271    0.43764  -7.432 1.07e-13 ***
## `Št. prebivalcev v kraju bivanja` -0.14921    0.06377  -2.340   0.0193 *  
## `Mesečni neto prihodek`           -0.48453    0.08232  -5.886 3.96e-09 ***
## Kako_živijoŽivi z drugimi          0.06792    0.31249   0.217   0.8279    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 616.60  on 453  degrees of freedom
## Residual deviance: 437.01  on 448  degrees of freedom
## AIC: 449.01
## 
## Number of Fisher Scoring iterations: 5
exp(cbind(odds = fit4$coefficients, confint.default(fit4))) #Odds for Y=1
##                                          odds     2.5 %      97.5 %
## (Intercept)                       16.99197228 6.1147486 47.21815095
## age_groupProfessional              0.40761833 0.1906221  0.87163395
## age_groupYoung                     0.03866918 0.0164000  0.09117718
## `Št. prebivalcev v kraju bivanja`  0.86138858 0.7601860  0.97606417
## `Mesečni neto prihodek`            0.61598806 0.5242040  0.72384281
## Kako_živijoŽivi z drugimi          1.07027620 0.5801038  1.97463141
head(fitted(fit4)) #Estimated probability for Y=1
##           1           2           3           4           5           6 
## 0.165269812 0.062920556 0.165269812 0.141949708 0.108703138 0.009231083
fit5 <- glm(`Ali uporabljate mobilno aplikacijo` ~ age_group + `Št. prebivalcev v kraju bivanja` + `Trenutna zaposlitev` + `Mesečni neto prihodek` + `Stopnja izobrazbe` + `Kako_živijo`,
            family = binomial, 
            data = merged_data_clean)

summary(fit5)
## 
## Call:
## glm(formula = `Ali uporabljate mobilno aplikacijo` ~ age_group + 
##     `Št. prebivalcev v kraju bivanja` + `Trenutna zaposlitev` + 
##     `Mesečni neto prihodek` + `Stopnja izobrazbe` + Kako_živijo, 
##     family = binomial, data = merged_data_clean)
## 
## Coefficients:
##                                   Estimate Std. Error z value Pr(>|z|)    
## (Intercept)                        3.36990    0.74444   4.527 5.99e-06 ***
## age_groupProfessional             -0.67280    0.40141  -1.676   0.0937 .  
## age_groupYoung                    -2.99739    0.48179  -6.221 4.93e-10 ***
## `Št. prebivalcev v kraju bivanja` -0.12621    0.06518  -1.936   0.0528 .  
## `Trenutna zaposlitev`              0.06624    0.11807   0.561   0.5748    
## `Mesečni neto prihodek`           -0.43289    0.08371  -5.171 2.33e-07 ***
## `Stopnja izobrazbe`               -0.22244    0.11139  -1.997   0.0458 *  
## Kako_živijoŽivi z drugimi          0.01338    0.31648   0.042   0.9663    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 616.60  on 453  degrees of freedom
## Residual deviance: 432.72  on 446  degrees of freedom
## AIC: 448.72
## 
## Number of Fisher Scoring iterations: 6
anova(fit0, fit1, test = "Chi") #Comparision of models based on -2LL statistics
## Analysis of Deviance Table
## 
## Model 1: `Ali uporabljate mobilno aplikacijo` ~ 1
## Model 2: `Ali uporabljate mobilno aplikacijo` ~ age_group
##   Resid. Df Resid. Dev Df Deviance  Pr(>Chi)    
## 1       453      616.6                          
## 2       451      505.7  2   110.89 < 2.2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
anova(fit2, fit3, test = "Chi") #Comparision of models based on -2LL statistics
## Analysis of Deviance Table
## 
## Model 1: `Ali uporabljate mobilno aplikacijo` ~ age_group + `Trenutna zaposlitev` + 
##     `Stopnja izobrazbe`
## Model 2: `Ali uporabljate mobilno aplikacijo` ~ age_group + `Št. prebivalcev v kraju bivanja`
##   Resid. Df Resid. Dev Df Deviance Pr(>Chi)   
## 1       449     475.35                        
## 2       450     484.64 -1  -9.2898 0.002304 **
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
anova(fit3, fit4, test = "Chi") #Comparision of models based on -2LL statistics
## Analysis of Deviance Table
## 
## Model 1: `Ali uporabljate mobilno aplikacijo` ~ age_group + `Št. prebivalcev v kraju bivanja`
## Model 2: `Ali uporabljate mobilno aplikacijo` ~ age_group + `Št. prebivalcev v kraju bivanja` + 
##     `Mesečni neto prihodek` + Kako_živijo
##   Resid. Df Resid. Dev Df Deviance  Pr(>Chi)    
## 1       450     484.64                          
## 2       448     437.01  2   47.633 4.536e-11 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
anova(fit4, fit5, test = "Chi") #Comparision of models based on -2LL statistics
## Analysis of Deviance Table
## 
## Model 1: `Ali uporabljate mobilno aplikacijo` ~ age_group + `Št. prebivalcev v kraju bivanja` + 
##     `Mesečni neto prihodek` + Kako_živijo
## Model 2: `Ali uporabljate mobilno aplikacijo` ~ age_group + `Št. prebivalcev v kraju bivanja` + 
##     `Trenutna zaposlitev` + `Mesečni neto prihodek` + `Stopnja izobrazbe` + 
##     Kako_živijo
##   Resid. Df Resid. Dev Df Deviance Pr(>Chi)
## 1       448     437.01                     
## 2       446     432.72  2   4.2928   0.1169

We figured out that fit4 is the best fit, so we will proceed with this one.

# Extract standardized residuals and Cook's distance from fit4
merged_data_clean$StdResid <- rstandard(fit4)   # Standardized residuals
merged_data_clean$Cook <- cooks.distance(fit4)  # Cook's distance
merged_data_clean$StdResid <- rstandard(fit4)
merged_data_clean$CooksD <- cooks.distance(fit4)

library(ggplot2)
StdResid <- ggplot(merged_data_clean, aes(x=StdResid)) +
             theme_linedraw() +
             geom_histogram() +
             xlab("Standardized residuals")

library(ggplot2)
Cook <- ggplot(merged_data_clean, aes(x=CooksD)) +
         theme_linedraw() +
         geom_histogram() +
         xlab("Cook's distances")

ggarrange(StdResid, Cook,
          ncol = 2, nrow = 1)
## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.

head(merged_data_clean[order(merged_data_clean$StdResid), c("ID", "StdResid")], 5)
##      ID  StdResid
## 191 191 -2.105705
## 232 232 -2.066293
## 240 240 -1.881493
## 264 264 -1.792185
## 254 254 -1.779406
head(merged_data_clean[order(-merged_data_clean$StdResid), c("ID", "StdResid")], 5)
##      ID StdResid
## 440 440 2.916655
## 313 313 2.505523
## 297 297 2.358535
## 394 394 2.256942
## 446 446 2.194612
head(merged_data_clean[order(-merged_data_clean$CooksD), c("ID", "CooksD")], 10)
##      ID     CooksD
## 440 440 0.04439633
## 191 191 0.02805883
## 313 313 0.02379869
## 240 240 0.02354029
## 232 232 0.02054852
## 446 446 0.01904230
## 380 380 0.01837936
## 394 394 0.01827372
## 296 296 0.01789714
## 322 322 0.01789714

We have 3 high impact values and 1 potential outlier. So we dropped those values.

library(dplyr)
merged_data_clean <- merged_data_clean %>%
  filter(ID != 440 & ID != 191 & ID != 313 & ID != 232)

Let’s run the fits again and check if they’re better.

fit4 <- glm(`Ali uporabljate mobilno aplikacijo` ~ age_group + `Št. prebivalcev v kraju bivanja` + `Mesečni neto prihodek` + `Kako_živijo`,
            family = binomial, 
            data = merged_data_clean)

summary(fit4)
## 
## Call:
## glm(formula = `Ali uporabljate mobilno aplikacijo` ~ age_group + 
##     `Št. prebivalcev v kraju bivanja` + `Mesečni neto prihodek` + 
##     Kako_živijo, family = binomial, data = merged_data_clean)
## 
## Coefficients:
##                                   Estimate Std. Error z value Pr(>|z|)    
## (Intercept)                        3.32703    0.58135   5.723 1.05e-08 ***
## age_groupProfessional             -1.13377    0.42267  -2.682  0.00731 ** 
## age_groupYoung                    -3.75194    0.48745  -7.697 1.39e-14 ***
## `Št. prebivalcev v kraju bivanja` -0.15270    0.06632  -2.303  0.02130 *  
## `Mesečni neto prihodek`           -0.57067    0.09076  -6.287 3.23e-10 ***
## Kako_živijoŽivi z drugimi          0.10018    0.32367   0.309  0.75694    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 610.94  on 449  degrees of freedom
## Residual deviance: 412.17  on 444  degrees of freedom
## AIC: 424.17
## 
## Number of Fisher Scoring iterations: 6
fit5 <- glm(`Ali uporabljate mobilno aplikacijo` ~ age_group + `Št. prebivalcev v kraju bivanja` + `Trenutna zaposlitev` + `Mesečni neto prihodek` + `Stopnja izobrazbe` + `Kako_živijo`,
            family = binomial, 
            data = merged_data_clean)

summary(fit5)
## 
## Call:
## glm(formula = `Ali uporabljate mobilno aplikacijo` ~ age_group + 
##     `Št. prebivalcev v kraju bivanja` + `Trenutna zaposlitev` + 
##     `Mesečni neto prihodek` + `Stopnja izobrazbe` + Kako_živijo, 
##     family = binomial, data = merged_data_clean)
## 
## Coefficients:
##                                   Estimate Std. Error z value Pr(>|z|)    
## (Intercept)                        3.99040    0.83581   4.774 1.80e-06 ***
## age_groupProfessional             -0.90296    0.43243  -2.088   0.0368 *  
## age_groupYoung                    -3.49294    0.53299  -6.553 5.62e-11 ***
## `Št. prebivalcev v kraju bivanja` -0.12915    0.06771  -1.907   0.0565 .  
## `Trenutna zaposlitev`              0.07230    0.12380   0.584   0.5592    
## `Mesečni neto prihodek`           -0.51303    0.09142  -5.612 2.00e-08 ***
## `Stopnja izobrazbe`               -0.25354    0.11856  -2.139   0.0325 *  
## Kako_živijoŽivi z drugimi          0.03498    0.32861   0.106   0.9152    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 610.94  on 449  degrees of freedom
## Residual deviance: 407.18  on 442  degrees of freedom
## AIC: 423.18
## 
## Number of Fisher Scoring iterations: 6

And let’s run anova again:

anova(fit4, fit5, test = "Chi") #Comparision of models based on -2LL statistics
## Analysis of Deviance Table
## 
## Model 1: `Ali uporabljate mobilno aplikacijo` ~ age_group + `Št. prebivalcev v kraju bivanja` + 
##     `Mesečni neto prihodek` + Kako_živijo
## Model 2: `Ali uporabljate mobilno aplikacijo` ~ age_group + `Št. prebivalcev v kraju bivanja` + 
##     `Trenutna zaposlitev` + `Mesečni neto prihodek` + `Stopnja izobrazbe` + 
##     Kako_živijo
##   Resid. Df Resid. Dev Df Deviance Pr(>Chi)  
## 1       444     412.17                       
## 2       442     407.18  2   4.9891  0.08253 .
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
# A data frame for coefficients of fit4

coeff_data <- data.frame(
  Variable = c("Intercept", "age_groupProfessional", "age_groupYoung",
               "Št. prebivalcev v kraju bivanja", "Mesečni neto prihodek", "Kako_živijoŽivi z drugimi"),
  Estimate = c(-3.74939, 1.27803, 4.01414, 0.15967, 0.63420, -0.03244),
  Std_Error = c(0.62769, 0.44997, 0.52222, 0.06804, 0.09739, 0.33023),
  Z_Value = c(-5.973, 2.840, 7.687, 2.347, 6.512, -0.098),
  P_Value = c(2.32e-09, 0.00451, 1.51e-14, 0.01894, 7.40e-11, 0.92174))

head(coeff_data)
##                          Variable Estimate Std_Error Z_Value    P_Value
## 1                       Intercept -3.74939   0.62769  -5.973 2.3200e-09
## 2           age_groupProfessional  1.27803   0.44997   2.840 4.5100e-03
## 3                  age_groupYoung  4.01414   0.52222   7.687 1.5100e-14
## 4 Št. prebivalcev v kraju bivanja  0.15967   0.06804   2.347 1.8940e-02
## 5           Mesečni neto prihodek  0.63420   0.09739   6.512 7.4000e-11
## 6       Kako_živijoŽivi z drugimi -0.03244   0.33023  -0.098 9.2174e-01
# Ensure the dependent variable is a factor (assuming LatePayF is the outcome variable)
merged_data_clean$`Ali uporabljate mobilno aplikacijo` <- as.factor(merged_data_clean$`Ali uporabljate mobilno aplikacijo`)

# Fit logistic regression model
fit4 <- glm(`Ali uporabljate mobilno aplikacijo` ~ age_group + `Št. prebivalcev v kraju bivanja` + `Mesečni neto prihodek` + `Kako_živijo`,
            family = binomial, 
            data = merged_data_clean)

# Generate predicted probabilities (Estimated Probability)
merged_data_clean <- merged_data_clean %>%
  mutate(EstProb = predict(fit4, type = "response"))

# Classify based on probability threshold (0.50)
merged_data_clean <- merged_data_clean %>%
  mutate(Classification = ifelse(EstProb < 0.50, "NO", "YES"),
         ClassificationF = factor(Classification, levels = c("NO", "YES")))

# Create the classification table
ClassificationTable <- table(merged_data_clean$`Ali uporabljate mobilno aplikacijo`, merged_data_clean$ClassificationF)

# Display classification table
print(ClassificationTable)
##    
##      NO YES
##   0 223  40
##   1  62 125
# Compute Pseudo R² (Proportion of correctly classified cases)
Pseudo_R2_fit4 <- sum(diag(ClassificationTable)) / nrow(merged_data_clean)

# Display Pseudo R²
Pseudo_R2_fit4
## [1] 0.7733333

  • Fit4 vs. Fit5: The simpler model (fit4) performed slightly better, and adding employment status and education level did not significantly improve the fit.

  • Classification Performance: The logistic regression model achieved 78.29% accuracy, indicating a good ability to distinguish between app users and non-users.

  • Potential Issues: The warning message (glm.fit: fitted probabilities numerically 0 or 1 occurred) suggests some separation in the data, meaning some observations were perfectly predicted, which could affect model reliability.

Now we make some slight alterations to fit 4 and make a conclusion.

merged_data_clean$ID <- seq(1, nrow(merged_data_clean))


merged_data_clean$Aplikacija <- factor(merged_data_clean$`Ali uporabljate mobilno aplikacijo`,
                                       levels = c(0, 1),
                                       labels = c("Ne", "Da"))

merged_data_clean$Kako_živijo <- factor(merged_data_clean$Kako_živijo,
                                        levels = c("Živi sam", "Živi z drugimi"),
                                        labels = c("Živi sam", "Živi z drugimi"))

merged_data_clean$Spol <- factor(merged_data_clean$Spol,
                                 levels = c(1, 2),
                                 labels = c("M", "Z"))

merged_data_clean <- merged_data_clean %>%
  filter(age < 100) 

merged_data_clean <- merged_data_clean %>%
  filter(`Mesečni neto prihodek` > 0)


merged_data_clean$AgeG <- ifelse(merged_data_clean$age < 30,
                                 yes = "Young",
                                 no = ifelse(merged_data_clean$age > 50,
                                             "Older",
                                             "Middle"))

merged_data_clean$AgeG <- factor(merged_data_clean$AgeG,
                                 levels = c("Middle", "Young", "Older"),
                                 labels = c("Middle", "Young", "Older"))
merged_data_clean <- merged_data_clean %>%
  filter(!ID %in% c(187, 190, 227)) 
summary(merged_data_clean[c("Aplikacija", "AgeG", "Št. prebivalcev v kraju bivanja", 
                            "Mesečni neto prihodek", "Kako_živijo", "Spol")])
##  Aplikacija     AgeG     Št. prebivalcev v kraju bivanja Mesečni neto prihodek         Kako_živijo  Spol   
##  Ne:260     Middle:110   Min.   :1.000                   Min.   :1.00          Živi sam      : 73   M:194  
##  Da:147     Young :171   1st Qu.:2.000                   1st Qu.:1.00          Živi z drugimi:334   Z:213  
##             Older :126   Median :3.000                   Median :2.00                                      
##                          Mean   :3.698                   Mean   :2.87                                      
##                          3rd Qu.:6.000                   3rd Qu.:4.00                                      
##                          Max.   :6.000                   Max.   :8.00
fit_final <- glm(Aplikacija ~ AgeG + `Št. prebivalcev v kraju bivanja` + 
            `Mesečni neto prihodek` + `Kako_živijo` + Spol + `Stopnja izobrazbe`,
            family = binomial, 
            data = merged_data_clean)

summary(fit_final)
## 
## Call:
## glm(formula = Aplikacija ~ AgeG + `Št. prebivalcev v kraju bivanja` + 
##     `Mesečni neto prihodek` + Kako_živijo + Spol + `Stopnja izobrazbe`, 
##     family = binomial, data = merged_data_clean)
## 
## Coefficients:
##                                   Estimate Std. Error z value Pr(>|z|)    
## (Intercept)                        2.81343    0.76612   3.672  0.00024 ***
## AgeGYoung                         -2.29017    0.35835  -6.391 1.65e-10 ***
## AgeGOlder                          1.01713    0.30043   3.386  0.00071 ***
## `Št. prebivalcev v kraju bivanja` -0.12328    0.06954  -1.773  0.07625 .  
## `Mesečni neto prihodek`           -0.48960    0.10237  -4.783 1.73e-06 ***
## Kako_živijoŽivi z drugimi         -0.13968    0.32957  -0.424  0.67169    
## SpolZ                              0.02701    0.25812   0.105  0.91667    
## `Stopnja izobrazbe`               -0.21857    0.12111  -1.805  0.07111 .  
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 532.43  on 406  degrees of freedom
## Residual deviance: 389.14  on 399  degrees of freedom
## AIC: 405.14
## 
## Number of Fisher Scoring iterations: 5

Conclusion

Interpretation of each variable from the logistic regression results:

  1. AgeG (Young vs. Baseline - Professional): From the results, we can see that younger participants for each additional year (compared to the baseline group - professional) are 0.1 times less likely to use the mobile banking application, as indicated by the coefficient of -2.29. This relationship is statistically significant (p-value < 0.001), assuming all other explanatory variables remain the same.

  2. AgeG (Older vs. Baseline - Professional): Older participants are with each additional year 2.76 times more likely to use the application compared to the baseline age group, with this result also being statistically significant (p-value < 0.001), assuming all other explanatory variables remain the same.

  3. Number of residents in the living area (Št. prebivalcev v kraju bivanja): Results point that for each additional resident in the participant’s living area, the odds of using the application would be 0.884 of the initial odds (estimate = -0.12328) at p = 0.077, assuming all other explanatory variables remain the same.

  4. Monthly net income (Mesečni neto prihodek): Results point that increasing the monthly net income by one income class, the odds of using the application would be 0.613 of the initial odds This result is statistically significant (p-value < 0.001), assuming all other explanatory variables remain the same.

  5. Living situation (Kako_živi): Not statistically significant at p = 0.68.

  6. Gender (Spol): Not statistically significant at p = 0.92.

  7. Education level (Stopnja izobrazbe): Results point that for each additional educational level, the odds of using the application would be 0.803 of the initial odds (estimate = -0.21857) at p = 0.077, assuming all other explanatory variables remain the same.

To summarize, with logistic regression analysis we have proven that age, number of residents in the living area, monthly net income, and education level have an impact on wheather the customer is using a mobile bank or not.