Clustering
Introduction
The aim of clustering is to classify or group homogeneous objects. With the results obtained from the questionnaire, we have been able to differentiate our respondents into groups with the goal to personalize the recommendations and develop the most efficient marketing strategies.
We have performed statistical tests on various variables which we could describe our clusters with. We have tested the following demographic variables:
Age (in years)
Location (either urban or rural)
Gender (either male or female)
Education (either up to high school or undergraduate degree and more)
Employment status (either employed or self-employed)
Bank (either NLB or other)
Job (either white- or blue-collar)
Additionally, we have tested the following variables that the respondents evaluated on the Likert scale, where 1 = strongly disagree, 2 = disagree, 3 = somewhat disagree, 4 = neutral, 5 = somewhat agree, 6 = agree, and 7 = strongly agree. The following variables were tested:
Raje plačujem z gotovino kot z digitalnimi plačilnimi sredstvi. (angl. I prefer to pay with cash rather than digital means of payment.)
Zlahka najdem ponudnike, ki sprejemajo digitalna plačila za vsakodnevne nakupe. (angl. I can easily find providers that accept digital payments for everyday purchases.)
Pri plačilih manjše vrednosti (do 10€) se mi zdi govotina hitrejši način plačila od digitalnih plačil. (angl. For smaller payments (up to €10), I find cash a faster way to pay than digital payments.)
Uporaba digitalnih plačil za transkacije z majhno vrednostjo (do 10€) je priročna in enostavna, z minimalnimi dodatnimi koraki (npr. vnos PIN-a, skeniranje QR kode). (angl. Using digital payments for small value transactions (up to €10) is convenient and easy, with minimal additional steps (e.g., PIN entry, QR code scanning.)
Rad/a uporabljam mobilne plačilne platforme (kot so Apple Pay, Revolut, PayPal, Flik), ker so priročne. (angl. I like to use mobile payment platforms (such as Apple Pay, Revolut, PayPal, Flik) because they are convenient.)
Verjetno bom uporabljal/a mobilne plačilne platforme (kot so Apple Pay, Revolut, PayPal, Flik), če mi jih priporočijo prijatelji, družina ali vrstniki. (angl. I am likely to use mobile payment platforms (such as Apple Pay, Revolut, PayPal, Flik) if they are recommended to be by friends, family, or peers.)
library(readxl)
mydata <- read_xlsx("./REAL.xlsx")
head(mydata)## # A tibble: 6 × 40
## ID Q2a_1 Q2b_1 Q2c_1 Q3a_1 Q3b_1 Q3c_1 Q4a_1 Q4b_1 Q4c_1 Q5a_1 Q5b_1 Q5c_1
## <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
## 1 1 5 6 5 7 7 5 7 7 5 7 7 6
## 2 2 7 7 6 2 6 6 4 7 7 3 6 6
## 3 3 7 6 6 6 6 6 7 7 5 6 7 5
## 4 4 7 3 5 6 6 3 3 6 5 3 6 6
## 5 5 6 5 5 5 5 6 6 6 7 4 6 7
## 6 6 6 6 6 5 7 6 6 7 5 7 7 5
## # ℹ 27 more variables: Q6a_1 <dbl>, Q6b_1 <dbl>, Q6c_1 <dbl>, Q7a_1 <dbl>,
## # Q7b_1 <dbl>, Q7c_1 <dbl>, Age <dbl>, Gender <dbl>, Location <dbl>,
## # Education <dbl>, Job <dbl>, Bank <dbl>, EmplStatus <dbl>, EducFixed <dbl>,
## # BankFixed <dbl>, EmplFixed <dbl>, JobFixed <dbl>, LocationFixed <dbl>,
## # PreferCash <dbl>, KeepCash <dbl>, FindSeller <dbl>, SmallCash <dbl>,
## # DigitalEasy <dbl>, ConvenientFlik <dbl>, FriendsFlik <dbl>,
## # Hypothetical <dbl>, Income <dbl>colnames(mydata)## [1] "ID" "Q2a_1" "Q2b_1" "Q2c_1"
## [5] "Q3a_1" "Q3b_1" "Q3c_1" "Q4a_1"
## [9] "Q4b_1" "Q4c_1" "Q5a_1" "Q5b_1"
## [13] "Q5c_1" "Q6a_1" "Q6b_1" "Q6c_1"
## [17] "Q7a_1" "Q7b_1" "Q7c_1" "Age"
## [21] "Gender" "Location" "Education" "Job"
## [25] "Bank" "EmplStatus" "EducFixed" "BankFixed"
## [29] "EmplFixed" "JobFixed" "LocationFixed" "PreferCash"
## [33] "KeepCash" "FindSeller" "SmallCash" "DigitalEasy"
## [37] "ConvenientFlik" "FriendsFlik" "Hypothetical" "Income"Making Factors
First, we have made factors.
mydata$GenderFactor <- factor(mydata$Gender,
levels = c(1, 2),
labels = c("Male", "Female"))
mydata$LocationFactor <- factor(mydata$Location,
levels = c(1, 2, 3),
labels = c("Urban", "Urban", "Rural"))
mydata$EducationFactor <- factor(mydata$Education,
levels = c(0,1, 2, 3, 4, 5, 6),
labels = c("Unifinished elementary", "Finished elementary", "Vocational school", "General high school", "Undergraduate degree", "Master's degree", "PhD"))
mydata$EmplStatusFactor <- factor (mydata$EmplStatus,
levels = c(1, 2, 3, 4),
labels = c("Employed", "Self-employed", "Self-employed", "Self-employed"))
#mydata$JobFactor <- factor (mydata$Job,
#levels = c(1, 2 ,3, 4, 5),
#labels = c("Manual", "Manual", "Office", "Office", "Office"))
mydata$EducFixed <- factor (mydata$EducFixed,
levels = c(0, 1),
labels = c("Up to high school", "Undergrad and more"))
mydata$BankFixed <- factor (mydata$BankFixed,
levels = c(0, 1,2,3,4,5,6,7,8,9),
labels = c("NLB", "Other banks","Other banks","Other banks","Other banks","Other banks","Other banks","Other banks","Other banks","Other banks"))
mydata$EmplFixed <- factor (mydata$EmplFixed,
levels = c(0, 1),
labels = c("Employed", "Others"))
mydata$JobFixed <- factor(mydata$JobFixed,
levels = c(0,1),
labels = c("White Collar", "Blue Collar"))Standardizing the Variables
Now we will standardize variables that will be used for clustering.
The description of the variables:
Q2a_1: perception of cash in terms of safety
Q3a_1: perception of cash in terms of speed of transaction
Q4a_1: perception of cash in terms of how easy it is to use
Q5a_1: perception of cash in terms of convenience
Q6a_1: perception of cash in terms of tracking expenses
mydata_clu_new <- as.data.frame(scale(mydata[c("Q2a_1", "Q3a_1", "Q4a_1", "Q5a_1", "Q6a_1", "Q7a_1")])) Let us find outliers.
mydata_clu_new[is.na(mydata_clu_new)] <- 0
mydata$Dissimilarity <- sqrt(mydata_clu_new$Q2a_1^2 + mydata_clu_new$Q3a_1^2 + mydata_clu_new$Q4a_1^2 + mydata_clu_new$Q5a_1^2 + mydata_clu_new$Q6a_1^2 + mydata_clu_new$Q7a_1^2) Let us find the units with the highest value of dissimilarity.
head(mydata[order(-mydata$Dissimilarity), c("ID", "Dissimilarity")]) ## # A tibble: 6 × 2
## ID Dissimilarity
## <dbl> <dbl>
## 1 40 4.64
## 2 14 4.49
## 3 120 4.12
## 4 34 3.92
## 5 71 3.89
## 6 138 3.81Let us show the units with the highest dissimilarity.
print(mydata[c(14,40,120), ])## # A tibble: 3 × 45
## ID Q2a_1 Q2b_1 Q2c_1 Q3a_1 Q3b_1 Q3c_1 Q4a_1 Q4b_1 Q4c_1 Q5a_1 Q5b_1 Q5c_1
## <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
## 1 14 7 7 5 3 6 7 6 6 7 3 6 7
## 2 40 5 3 3 2 7 7 2 7 7 2 7 7
## 3 120 1 6 5 6 6 7 7 5 6 6 6 6
## # ℹ 32 more variables: Q6a_1 <dbl>, Q6b_1 <dbl>, Q6c_1 <dbl>, Q7a_1 <dbl>,
## # Q7b_1 <dbl>, Q7c_1 <dbl>, Age <dbl>, Gender <dbl>, Location <dbl>,
## # Education <dbl>, Job <dbl>, Bank <dbl>, EmplStatus <dbl>, EducFixed <fct>,
## # BankFixed <fct>, EmplFixed <fct>, JobFixed <fct>, LocationFixed <dbl>,
## # PreferCash <dbl>, KeepCash <dbl>, FindSeller <dbl>, SmallCash <dbl>,
## # DigitalEasy <dbl>, ConvenientFlik <dbl>, FriendsFlik <dbl>,
## # Hypothetical <dbl>, Income <dbl>, GenderFactor <fct>, …Euclidian Distance
Let us calculate the Euclidian distances, and show the matrix of distances.
library(factoextra) ## Warning: package 'factoextra' was built under R version 4.4.2## Loading required package: ggplot2## Welcome! Want to learn more? See two factoextra-related books at https://goo.gl/ve3WBaDistances <- get_dist(mydata_clu_new,
method = "euclidian")
fviz_dist(Distances,
gradient = list(low = "slateblue4",
mid = "green",
high = "white"))
Hopkins Statistics
Let us calculate the Hopkins statistics. If it is higher than 0.5, this means that data is clusterable.
library(factoextra)
get_clust_tendency(mydata_clu_new,
n = nrow(mydata_clu_new) - 1,
graph = FALSE)## $hopkins_stat
## [1] 0.6418886
##
## $plot
## NULLHopkins statistics is above 0.5 - data is clusterable.
K Means Approach
The graphs below - the elbow method and the silhouette analysis - are used to determine the optimal number of clusters. This is the K-means approach to clustering.
library(factoextra)
library(NbClust)
fviz_nbclust(mydata_clu_new, kmeans, method = "wss") +
labs(subtitle = "Elbow method")
fviz_nbclust(mydata_clu_new, kmeans, method = "silhouette")+
labs(subtitle = "Silhouette analysis")
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, unionlibrary(factoextra)
WARD <- mydata_clu_new %>%
get_dist(method = "euclidean") %>%
hclust(method = "ward.D2")
WARD##
## Call:
## hclust(d = ., method = "ward.D2")
##
## Cluster method : ward.D2
## Distance : euclidean
## Number of objects: 152Hierarchical Approach
Let us check the hierarchical approach to clustering. It shows the option of creating five clusters, which is what we have opted for.
library(factoextra)
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.
library(NbClust)
NbClust(mydata_clu_new,
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:
## * 8 proposed 2 as the best number of clusters
## * 5 proposed 3 as the best number of clusters
## * 3 proposed 5 as the best number of clusters
## * 2 proposed 6 as the best number of clusters
## * 2 proposed 9 as the best number of clusters
## * 3 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
## 2 16.5370 59.5648 22.8451 -1.2199 190.7707 4.307472e+12 12926.431 648.4867
## 3 0.1354 45.4359 23.4358 -1.8840 367.1832 3.036401e+12 9223.823 562.7755
## 4 0.7123 42.5792 23.8440 -1.6591 474.6207 2.662356e+12 6927.557 486.2886
## 5 2.2046 42.7495 14.9309 0.1108 564.7277 2.299495e+12 4648.256 418.8142
## 6 2.5760 40.3828 9.9144 0.8422 663.8276 1.725228e+12 3722.664 380.1973
## 7 0.8891 37.3325 9.0992 0.7345 727.1115 1.548550e+12 3098.369 356.0210
## 8 0.4194 35.0637 11.8946 0.7330 782.5853 1.404137e+12 2947.080 334.9989
## 9 9.5096 34.4609 5.7515 1.5349 847.1365 1.162195e+12 2492.866 309.4387
## 10 0.0898 32.2757 11.9974 1.1503 877.7749 1.172881e+12 2235.064 297.4742
## Friedman Rubin Cindex DB Silhouette Duda Pseudot2 Beale Ratkowsky
## 2 3.2372 1.3971 0.4534 1.5833 0.2516 1.2250 -18.5509 -0.6976 0.3433
## 3 5.4346 1.6099 0.4247 1.7077 0.2047 1.5054 -24.1722 -1.2658 0.3359
## 4 6.8884 1.8631 0.4612 1.5760 0.2104 1.1178 -5.7980 -0.3969 0.3241
## 5 7.5885 2.1633 0.4244 1.3974 0.2328 1.2897 -11.6811 -0.8421 0.3275
## 6 9.3895 2.3830 0.3998 1.3793 0.2277 1.4435 -10.4458 -1.1283 0.3105
## 7 10.5507 2.5448 0.4490 1.4012 0.2230 1.0142 -0.5887 -0.0525 0.2941
## 8 11.6282 2.7045 0.4165 1.4359 0.2077 1.6961 -10.2607 -1.4862 0.2804
## 9 12.8642 2.9279 0.4229 1.4137 0.2152 1.1606 -2.4907 -0.4969 0.2704
## 10 13.4092 3.0456 0.4162 1.3749 0.2092 1.8949 -12.7511 -1.6655 0.2591
## Ball Ptbiserial Frey McClain Dunn Hubert SDindex Dindex SDbw
## 2 324.2434 0.4586 0.8104 0.7394 0.1312 0.0023 1.6254 1.9756 1.2828
## 3 187.5918 0.4468 0.1855 1.4472 0.0831 0.0026 1.6291 1.8218 1.0393
## 4 121.5722 0.4777 0.0745 1.9274 0.0977 0.0028 1.4802 1.6897 0.6214
## 5 83.7628 0.5135 0.2775 2.2588 0.1090 0.0032 1.3569 1.5742 0.5371
## 6 63.3662 0.5105 0.7706 2.6311 0.1078 0.0033 1.3411 1.5061 0.4869
## 7 50.8601 0.4811 1.1917 3.1800 0.1112 0.0034 1.5588 1.4607 0.4626
## 8 41.8749 0.4367 0.0850 4.0759 0.1051 0.0036 1.6493 1.4036 0.4293
## 9 34.3821 0.4416 0.4428 4.3256 0.1112 0.0038 1.5593 1.3558 0.3930
## 10 29.7474 0.4290 -0.1904 4.7313 0.1480 0.0039 1.5085 1.3214 0.3768
##
## $All.CriticalValues
## CritValue_Duda CritValue_PseudoT2 Fvalue_Beale
## 2 0.7006 43.1654 1
## 3 0.6533 38.2096 1
## 4 0.6459 30.1530 1
## 5 0.6222 31.5710 1
## 6 0.5356 29.4770 1
## 7 0.6188 25.8772 1
## 8 0.4889 26.1338 1
## 9 0.4643 20.7641 1
## 10 0.4174 37.6933 1
##
## $Best.nc
## KL CH Hartigan CCC Scott Marriot TrCovW
## Number_clusters 2.000 2.0000 5.0000 9.0000 3.0000 3 3.000
## Value_Index 16.537 59.5648 8.9131 1.5349 176.4125 897026528841 3702.608
## TraceW Friedman Rubin Cindex DB Silhouette Duda
## Number_clusters 5.0000 3.0000 9.0000 6.0000 10.0000 2.0000 2.000
## Value_Index 28.8575 2.1974 -0.1056 0.3998 1.3749 0.2516 1.225
## PseudoT2 Beale Ratkowsky Ball PtBiserial Frey McClain
## Number_clusters 2.0000 2.0000 2.0000 3.0000 5.0000 1 2.0000
## Value_Index -18.5509 -0.6976 0.3433 136.6515 0.5135 NA 0.7394
## Dunn Hubert SDindex Dindex SDbw
## Number_clusters 10.000 0 6.0000 0 10.0000
## Value_Index 0.148 0 1.3411 0 0.3768
##
## $Best.partition
## [1] 1 2 1 2 1 1 2 2 2 1 2 1 1 2 2 1 1 2 2 1 2 1 1 1 1 1 1 1 1 1 2 1 2 2 2 2 1
## [38] 1 1 2 2 2 1 2 2 2 1 1 1 2 1 1 2 1 1 2 2 2 2 2 1 2 2 1 1 1 1 1 2 2 1 1 2 1
## [75] 2 1 2 2 1 2 2 1 2 2 1 2 2 2 2 2 1 2 1 1 1 1 2 2 2 1 2 2 1 2 1 1 1 1 1 1 2
## [112] 1 2 1 2 1 1 1 1 1 1 1 2 1 1 2 2 1 2 2 2 1 1 2 1 1 2 2 1 2 2 1 2 1 2 2 2 2
## [149] 1 2 1 1Making Clusters
In the first step, we will make five clusters with 30, 17, 24, 41, and 40 units, respectively.
Clustering <- kmeans(mydata_clu_new,
centers = 5,
nstart = 25)
Clustering## K-means clustering with 5 clusters of sizes 30, 17, 24, 41, 40
##
## Cluster means:
## Q2a_1 Q3a_1 Q4a_1 Q5a_1 Q6a_1 Q7a_1
## 1 0.30294872 -1.08988634 -1.2737371 -1.1121377 0.2349868 -0.4492697
## 2 -2.04299453 -0.15283293 0.1288002 -0.2342229 -0.1969184 -0.1982072
## 3 0.46619735 -0.27363453 -0.4237942 -0.3527855 -1.7529088 -0.1005993
## 4 0.32831346 0.01992193 0.4436057 0.4646443 0.5814218 -0.6697631
## 5 0.02482143 1.02612950 0.7001434 0.6690589 0.3632381 1.1680571
##
## Clustering vector:
## [1] 5 1 4 1 5 4 3 1 3 4 1 4 4 3 1 5 4 1 4 4 1 4 5 5 4 5 5 5 5 4 1 5 3 1 2 3 5
## [38] 4 4 3 2 1 4 1 1 2 5 4 4 3 2 5 3 5 5 1 3 2 2 4 5 2 3 5 5 5 4 5 4 3 5 5 3 5
## [75] 3 4 1 1 2 4 1 4 1 2 4 4 1 2 1 1 2 3 4 5 5 3 1 4 1 4 4 1 5 4 5 5 5 5 5 4 3
## [112] 5 1 5 4 4 4 4 4 2 5 4 1 4 2 4 3 5 3 3 3 4 5 3 2 4 1 3 2 1 2 5 1 5 4 1 3 3
## [149] 2 1 5 5
##
## Within cluster sum of squares by cluster:
## [1] 101.21820 62.11656 76.96477 92.61780 85.89687
## (between_SS / total_SS = 53.8 %)
##
## Available components:
##
## [1] "cluster" "centers" "totss" "withinss" "tot.withinss"
## [6] "betweenss" "size" "iter" "ifault"library(factoextra)
fviz_cluster(Clustering,
palette = "Set1",
repel = FALSE,
ggtheme = theme_bw(),
data = mydata_clu_new)
Some units seem to be far away from the center, which means that they are outliers, so we will remove them. As we remove the units, we standardize the data again.
mydata <- mydata %>%
filter(!ID %in% c(133, 108, 7))
mydata$ID <- seq(1, nrow(mydata))
mydata_clu_new <- as.data.frame(scale(mydata[c("Q2a_1", "Q3a_1", "Q4a_1", "Q5a_1", "Q6a_1", "Q7a_1")])) As we have removed three units, the sizes of our five clusters changed. They now have 29, 41, 25, 37, and 17 units, respectively.
mydata_clu_new[is.na(mydata_clu_new)] <- 0
Clustering <- kmeans(mydata_clu_new,
centers = 5,
nstart = 25)
Clustering## K-means clustering with 5 clusters of sizes 29, 41, 25, 37, 17
##
## Cluster means:
## Q2a_1 Q3a_1 Q4a_1 Q5a_1 Q6a_1 Q7a_1
## 1 0.32527292 -1.099435794 -1.2365294 -1.1969146 0.2499536 -0.4063642
## 2 0.34475122 -0.004020726 0.4124110 0.4922125 0.5570996 -0.7193229
## 3 0.46902278 -0.119537224 -0.4047331 -0.2429699 -1.7180355 0.0505640
## 4 -0.02546811 1.014802901 0.7180670 0.6648481 0.4418536 1.1696604
## 5 -2.02064493 -0.147693467 0.1470795 -0.2350188 -0.2051434 -0.1920432
##
## Clustering vector:
## [1] 4 1 2 1 4 2 1 3 2 1 2 2 3 1 4 2 1 2 4 1 2 4 4 2 4 4 4 4 2 1 4 3 1 5 3 4 2
## [38] 2 3 5 1 2 1 1 5 4 2 2 3 5 4 3 3 4 1 3 5 5 2 4 5 3 4 4 4 2 4 2 3 4 4 3 4 3
## [75] 2 1 1 5 2 1 2 1 5 2 2 1 5 1 1 5 3 2 4 4 3 1 2 1 2 2 1 4 2 4 4 4 4 2 3 4 1
## [112] 4 2 2 2 2 2 5 4 2 1 2 5 2 3 4 3 3 3 2 3 5 2 1 3 5 2 5 4 1 3 2 1 3 3 5 1 4
## [149] 4
##
## Within cluster sum of squares by cluster:
## [1] 94.10655 99.94698 77.03574 78.50395 62.95983
## (between_SS / total_SS = 53.5 %)
##
## Available components:
##
## [1] "cluster" "centers" "totss" "withinss" "tot.withinss"
## [6] "betweenss" "size" "iter" "ifault"library(factoextra)
fviz_cluster(Clustering,
palette = "Set1",
repel = FALSE,
ggtheme = theme_bw(),
data = mydata_clu_new)
Averages <- Clustering$centers
Averages ## Q2a_1 Q3a_1 Q4a_1 Q5a_1 Q6a_1 Q7a_1
## 1 0.32527292 -1.099435794 -1.2365294 -1.1969146 0.2499536 -0.4063642
## 2 0.34475122 -0.004020726 0.4124110 0.4922125 0.5570996 -0.7193229
## 3 0.46902278 -0.119537224 -0.4047331 -0.2429699 -1.7180355 0.0505640
## 4 -0.02546811 1.014802901 0.7180670 0.6648481 0.4418536 1.1696604
## 5 -2.02064493 -0.147693467 0.1470795 -0.2350188 -0.2051434 -0.1920432Describing the Groups
The graph below shows how different clusters perceive cash in terms of safety, speed, ease of use, convenience, privacy, and tracking expenses.
Group 1: cash is perceived as above average in safety and privacy. However, it is below average in speed, ease of use, convenience, and tracking expenses.
Group 2: cash is perceived as above average in safety, ease of use, convenience, and privacy, and about average in speed. However, it is below average in tracking expenses.
Group 3: cash is perceived as above average in safety and about average in tracking expenses. However, it is below average in speed, ease of use, convenience, and privacy.
Group 4: cash is about average in safety, and above average in speed, ease of use, convenience, privacy, and tracking expenses.
Group 5: cash is below average in safety, privacy, and tracking expenses. It is about average in speed and above average in ease of use.
Figure <- as.data.frame(Averages)
Figure$ID <- 1:nrow(Figure)
library(tidyr)
Figure <- pivot_longer(Figure, cols = c("Q2a_1", "Q3a_1", "Q4a_1", "Q5a_1", "Q6a_1", "Q7a_1"))
Figure$Group <- factor(Figure$ID,
levels = c(1, 2, 3, 4, 5),
labels = c("1", "2", "3", "4", "5"))
Figure$NameF <- factor(Figure$name,
levels = c("Q2a_1", "Q3a_1", "Q4a_1", "Q5a_1", "Q6a_1", "Q7a_1"),
labels = c("Cash_Safety", "Cash_Speed", "Cash_Ease of Use", "Cash_Convenience", "Cash_Privacy", "Cash_Tracking Expenses"))
library(ggplot2)
ggplot(Figure, aes(x = NameF, y = value)) +
geom_hline(yintercept = 0) +
theme_bw() +
geom_point(aes(shape = Group, col = Group), size = 5) +
geom_line(aes(group = ID), linewidth = 1) +
ylab("Averages") +
xlab("Cluster variables")+
ylim(-2.5, 2.5) +
theme(axis.text.x = element_text(angle = 45, vjust = 0.50, size = 10))
mydata$Group <- Clustering$clusterChecking the Fit
Below are the ANOVA tests, which test the differences between group means for the cluster variables. As p < 0.05, the H0 that the means are the same for all groups can be rejected. This means that the groups are statistically different in the mean values of cluster variables.
fit <- aov(cbind(Q2a_1, Q3a_1, Q4a_1, Q5a_1, Q6a_1, Q7a_1) ~ as.factor(Group),
data = mydata)
summary(fit)## Response Q2a_1 :
## Df Sum Sq Mean Sq F value Pr(>F)
## as.factor(Group) 4 158.22 39.555 45.813 < 2.2e-16 ***
## Residuals 144 124.33 0.863
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Response Q3a_1 :
## Df Sum Sq Mean Sq F value Pr(>F)
## as.factor(Group) 4 186.27 46.568 35.889 < 2.2e-16 ***
## Residuals 144 186.84 1.298
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Response Q4a_1 :
## Df Sum Sq Mean Sq F value Pr(>F)
## as.factor(Group) 4 166.72 41.681 36.842 < 2.2e-16 ***
## Residuals 144 162.91 1.131
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Response Q5a_1 :
## Df Sum Sq Mean Sq F value Pr(>F)
## as.factor(Group) 4 221.47 55.367 32.526 < 2.2e-16 ***
## Residuals 144 245.12 1.702
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Response Q6a_1 :
## Df Sum Sq Mean Sq F value Pr(>F)
## as.factor(Group) 4 102.047 25.5117 66.99 < 2.2e-16 ***
## Residuals 144 54.839 0.3808
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Response Q7a_1 :
## Df Sum Sq Mean Sq F value Pr(>F)
## as.factor(Group) 4 296.74 74.185 39.376 < 2.2e-16 ***
## Residuals 144 271.30 1.884
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1Validation - Demographics
Now we will use demographic variables to check what we can describe our clusters with.
Age
aggregate(mydata$Age,
by = list(mydata$Group),
FUN = mean)## Group.1 x
## 1 1 36.03448
## 2 2 35.48780
## 3 3 38.28000
## 4 4 45.78378
## 5 5 40.41176The table above shows the average age for each of the clusters. The average age in group 1 is 36.03, and 40.41 in group 5.
Let us test the normal distribution of the variable.
H0: normality is met.
H1: normality is not met.
H0 cannot be rejected for all clusters. Thus, we turn to the non-parametric test.
library(dplyr)
library(rstatix)## Warning: package 'rstatix' was built under R version 4.4.2##
## Attaching package: 'rstatix'## The following object is masked from 'package:stats':
##
## filtermydata %>%
group_by(as.factor(mydata$Group)) %>%
shapiro_test(Age)## # A tibble: 5 × 4
## `as.factor(mydata$Group)` variable statistic p
## <fct> <chr> <dbl> <dbl>
## 1 1 Age 0.890 0.00569
## 2 2 Age 0.913 0.00399
## 3 3 Age 0.871 0.00457
## 4 4 Age 0.879 0.000818
## 5 5 Age 0.929 0.207The hypotheses for the non-parametric test are the following:
H0: all distribution locations of the variable are the same.
H1: at least one distribution location of the variable is different.
kruskal.test(Age ~ as.factor(Group),
data = mydata)##
## Kruskal-Wallis rank sum test
##
## data: Age by as.factor(Group)
## Kruskal-Wallis chi-squared = 11.386, df = 4, p-value = 0.02256As p < 0.05, H0 can be rejected. This means that the variable is statistically significant and can be used to describe clusters.
Location (either urban or rural)
Let us perform the Pearson Chi-Squared test. The hypotheses are the following:
H0: there is no association between location and classification of people into clusters.
H1: there is association between location and classification of people into clusters.
chi_square <- chisq.test(mydata$LocationFactor, as.factor(mydata$Group))## Warning in chisq.test(mydata$LocationFactor, as.factor(mydata$Group)):
## Chi-squared approximation may be incorrectchi_square##
## Pearson's Chi-squared test
##
## data: mydata$LocationFactor and as.factor(mydata$Group)
## X-squared = 5.2367, df = 4, p-value = 0.2639As the Pearson Chi-Squared test approximation may be incorrect, we have performed the Fisher’s test.
fisher_result <- fisher.test(table(mydata$LocationFactor, mydata$Group), simulate.p.value = TRUE, B = 10000)
fisher_result##
## Fisher's Exact Test for Count Data with simulated p-value (based on
## 10000 replicates)
##
## data: table(mydata$LocationFactor, mydata$Group)
## p-value = 0.2848
## alternative hypothesis: two.sidedLocation is statistically not significant.
table_data_edu <- table(mydata$LocationFactor, mydata$Group)
print(table_data_edu)##
## 1 2 3 4 5
## Urban 23 35 22 25 13
## Rural 6 6 3 12 4The table above shows the number of people who live in urban or rural areas. In cluster 1, for example, 23 people live in urban areas, and 6 in rural areas.
Gender (either male or female)
Let us perform the Pearson Chi-Squared test. The hypotheses are the following:
H0: there is no association between gender and classification of people into clusters.
H1: there is association between gender and classification of people into clusters.
chi_square <- chisq.test(mydata$GenderFactor, as.factor(mydata$Group))
chi_square##
## Pearson's Chi-squared test
##
## data: mydata$GenderFactor and as.factor(mydata$Group)
## X-squared = 3.0375, df = 4, p-value = 0.5516As p > 0.05, H0 cannot be rejected. Gender is statistically not significant.
table_data_edu <- table(mydata$GenderFactor, mydata$Group)
print(table_data_edu)##
## 1 2 3 4 5
## Male 10 21 10 13 8
## Female 19 20 15 24 9The table above shows the number of male and female in different clusters. In cluster 1, for example, 10 people are male, and 19 people are female.
Education (either up to high school or undergraduate degree and more)
Let us perform the Pearson Chi-Squared test. The hypotheses are the following:
H0: there is no association between education and classification of people into clusters.
H1: there is association between education and classification of people into clusters.
chi_square <- chisq.test(mydata$EducFixed, as.factor(mydata$Group), correct=TRUE)
chi_square##
## Pearson's Chi-squared test
##
## data: mydata$EducFixed and as.factor(mydata$Group)
## X-squared = 4.5211, df = 4, p-value = 0.3401As p > 0.05, H0 cannot be rejected. Education is statistically not significant.
table_data_edu <- table(mydata$EducFixed, mydata$Group)
print(table_data_edu)##
## 1 2 3 4 5
## Up to high school 7 16 6 16 7
## Undergrad and more 22 25 19 21 10The table above shows the number of people whose highest level of education obtained is high school, and those who have obtained at least a bachelor’s degree. In cluster 1, for example, finished high school is the highest level of education obtained for 7 people, and 22 have obtained at least a bachelor’s degree.
Employment status (either employed or self-employed)
Let us perform the Pearson Chi-Squared test. The hypotheses are the following:
H0: there is no association between employment status and classification of people into clusters.
H1: there is association between employment status and classification of people into clusters.
chi_square <- chisq.test(mydata$EmplStatusFactor, as.factor(mydata$Group), correct=TRUE)## Warning in chisq.test(mydata$EmplStatusFactor, as.factor(mydata$Group), :
## Chi-squared approximation may be incorrectchi_square##
## Pearson's Chi-squared test
##
## data: mydata$EmplStatusFactor and as.factor(mydata$Group)
## X-squared = 16.828, df = 4, p-value = 0.002087As p < 0.05, H0 can be rejected.
table_data_emp <- table(mydata$EmplStatusFactor, mydata$Group)
print(table_data_emp)##
## 1 2 3 4 5
## Employed 25 37 19 20 14
## Self-employed 4 4 6 17 3The table above shows the number of people who are employed or self-employed in each of the clusters. In cluster 1, for example, 25 people are employed, and 4 self-employed.
As the Pearson Chi-Squared test approximation may be incorrect, we have performed the Fisher’s test.
fisher_test_emp <- fisher.test(table_data_emp, simulate.p.value = TRUE, B = 10000)
print(fisher_test_emp)##
## Fisher's Exact Test for Count Data with simulated p-value (based on
## 10000 replicates)
##
## data: table_data_emp
## p-value = 0.0033
## alternative hypothesis: two.sidedEmployment status is statistically significant and can be used to describe clusters.
Bank (either NLB or other)
Let us perform the Pearson Chi-Squared test. The hypotheses are the following:
H0: there is no association between bank and classification of people into clusters.
H1: there is association between bank and classification of people into clusters.
chi_square <- chisq.test(mydata$BankFixed, as.factor(mydata$Group), correct=TRUE)
chi_square##
## Pearson's Chi-squared test
##
## data: mydata$BankFixed and as.factor(mydata$Group)
## X-squared = 6.6039, df = 4, p-value = 0.1584As p > 0.05, H0 cannot be rejected. Bank is not statistically significant.
table_data_job <- table(mydata$BankFixed, mydata$Group)
print(table_data_job)##
## 1 2 3 4 5
## NLB 11 15 5 14 10
## Other banks 18 26 20 23 7The table above shows the number of people who use NLB and those who use other banks. In cluster 1, for example, 11 people use NLB, and 18 use other banks.
Job (either white-collar or blue-collar)
Let us perform the Pearson Chi-Squared test. The hypotheses are the following:
H0: there is no association between job and classification of people into clusters.
H1: there is association between job and classification of people into clusters.
chi_square <- chisq.test(mydata$JobFixed, as.factor(mydata$Group), correct=TRUE)
chi_square##
## Pearson's Chi-squared test
##
## data: mydata$JobFixed and as.factor(mydata$Group)
## X-squared = 12.681, df = 4, p-value = 0.01295As p < 0.05, H0 can be rejected. Job is statistically significant and can be used to describe clusters.
table_data_job <- table(mydata$JobFixed, mydata$Group)
print(table_data_job)##
## 1 2 3 4 5
## White Collar 17 22 20 32 11
## Blue Collar 12 19 5 5 6The table above shows the number of white-collar and blue-collar workers in each of the clusters. In cluster 1, for example, 17 people have white-collar jobs, and 12 have blue-collar jobs.
Location (either urban or rural)
Let us perform the Pearson Chi-Squared test. The hypotheses are the following:
H0: there is no association between location and classification of people into clusters.
H1: there is association between location and classification of people into clusters.
chi_square <- chisq.test(mydata$LocationFactor, as.factor(mydata$Group), correct=TRUE)## Warning in chisq.test(mydata$LocationFactor, as.factor(mydata$Group), correct =
## TRUE): Chi-squared approximation may be incorrectchi_square##
## Pearson's Chi-squared test
##
## data: mydata$LocationFactor and as.factor(mydata$Group)
## X-squared = 5.2367, df = 4, p-value = 0.2639As p > 0.05, H0 cannot be rejected. Location is not statistically significant.
table_data_loc <- table(mydata$LocationFactor, mydata$Group)
print(table_data_loc)##
## 1 2 3 4 5
## Urban 23 35 22 25 13
## Rural 6 6 3 12 4The table above shows the number of people who live in urban and rural areas in each of the clusters. In cluster 1, for example, 23 people live in urban areas, and 6 in rural.
As the Pearson Chi-Squared test approximation may be incorrect, we have performed the Fisher’s test.
fisher_test_loc <- fisher.test(table_data_loc, simulate.p.value = TRUE, B = 10000)
print(fisher_test_loc)##
## Fisher's Exact Test for Count Data with simulated p-value (based on
## 10000 replicates)
##
## data: table_data_loc
## p-value = 0.2793
## alternative hypothesis: two.sidedLocation is not statistically significant.
Statistically significant are the following variables:
Age
Job (white- or blue-collar)
Employment status (employed or self-employed)
We have shown that other variables describing demographics are not statistically significant, but we will still use them to describe the clusters.
Validation - Likert Scale
Raje imam pri sebi gotovino, ker ni vedno mogoče plačati z drugimi plačilnimi sredstvi. (angl. I prefer to carry cash becazse it is not always possible to pay with other means of payment.)
aggregate(mydata$PreferCash,
by = list(mydata$Group),
FUN = mean)## Group.1 x
## 1 1 1.862069
## 2 2 3.731707
## 3 3 3.560000
## 4 4 4.351351
## 5 5 3.117647The table above shows the average answer to the statement “Raje imam pri sebi gotovino, ker ni vedno mogoče plačati z drugimi plačilnimi sredstvi”, where 1 means strongly disagree and 7 strongly agree.
Let us perform the Shapiro-Wilk normality test to check the normal distribution of the variable. Our hypotheses are the following:
H0: normality is met.
H1: normality is not met.
library(dplyr)
library(rstatix)
mydata %>%
group_by(as.factor(mydata$Group)) %>%
shapiro_test(PreferCash)## # A tibble: 5 × 4
## `as.factor(mydata$Group)` variable statistic p
## <fct> <chr> <dbl> <dbl>
## 1 1 PreferCash 0.684 0.00000124
## 2 2 PreferCash 0.890 0.000864
## 3 3 PreferCash 0.890 0.0111
## 4 4 PreferCash 0.865 0.000359
## 5 5 PreferCash 0.877 0.0281As p < 0.05, H0 can be rejected. Thus, we have to perform the non-parameteric alternative. Our hypotheses are:
H0: all distribution locations of the variable are the same.
H1: at least one distribution location of the variable is different.
kruskal.test(PreferCash ~ as.factor(Group),
data = mydata)##
## Kruskal-Wallis rank sum test
##
## data: PreferCash by as.factor(Group)
## Kruskal-Wallis chi-squared = 28.631, df = 4, p-value = 9.29e-06As p < 0.05, H0 can be rejected. There are differences between clusters.
Below is the visual representation of the attitude towards the statement “Raje imam pri sebi gotovino, ker ni vedno mogoče plačati z drugimi plačilnimi sredstvi.”
library(ggplot2)
library(dplyr)
likert_labels <- c(
"1" = "Strongly Disagree",
"2" = "Disagree",
"3" = "Somewhat Disagree",
"4" = "Neutral",
"5" = "Somewhat Agree",
"6" = "Agree",
"7" = "Strongly Agree"
)
table_clusters <- table(mydata$Group, mydata$PreferCash)
prop_table_clusters <- prop.table(table_clusters, margin = 1)
prop_df <- as.data.frame(as.table(prop_table_clusters))
prop_df$Var2 <- factor(prop_df$Var2, levels = names(likert_labels), labels = likert_labels)
ggplot(prop_df, aes(x = Var1, y = Freq * 100, fill = Var2)) +
geom_bar(stat = "identity", position = "stack") +
labs(
x = "Group",
y = "Percentage (%)",
fill = "Response",
title = "Percentage Distribution of Those Who Keep Cash"
) +
theme_minimal()
Raje plačujem z gotovino kot z digitalnimi plačilnimi sredstvi. (angl. I prefer to pay with cash rather than digital means of payment.)
aggregate(mydata$KeepCash,
by = list(mydata$Group),
FUN = mean)## Group.1 x
## 1 1 3.793103
## 2 2 4.731707
## 3 3 4.040000
## 4 4 4.351351
## 5 5 4.000000The table above shows the average answer to the statement “Raje plačujem z gotovino kot z digitalnimi plačilnimi sredstvi”, where 1 means strongly disagree and 7 strongly agree.
Let us perform the Shapiro-Wilk normality test to check the normal distribution of the variable. Our hypotheses are the following:
H0: normality is met.
H1: normality is not met.
library(dplyr)
library(rstatix)
mydata %>%
group_by(as.factor(mydata$Group)) %>%
shapiro_test(KeepCash)## # A tibble: 5 × 4
## `as.factor(mydata$Group)` variable statistic p
## <fct> <chr> <dbl> <dbl>
## 1 1 KeepCash 0.892 0.00638
## 2 2 KeepCash 0.877 0.000368
## 3 3 KeepCash 0.855 0.00216
## 4 4 KeepCash 0.897 0.00250
## 5 5 KeepCash 0.843 0.00855As p < 0.05, H0 can be rejected. Thus, we have to perform the non-parameteric alternative. Our hypotheses are:
H0: all distribution locations of the variable are the same.
H1: at least one distribution location of the variable is different.
kruskal.test(KeepCash ~ as.factor(Group),
data = mydata)##
## Kruskal-Wallis rank sum test
##
## data: KeepCash by as.factor(Group)
## Kruskal-Wallis chi-squared = 5.7094, df = 4, p-value = 0.2219As p > 0.05, H0 cannot be rejected. We cannot say that there are statistically significant differences between the groups.
Below is the visual representation of the attitude towards the statement “Raje plačujem z gotovino kot z digitalnimi plačilnimi sredstvi.”
library(ggplot2)
library(dplyr)
likert_labels <- c(
"1" = "Strongly Disagree",
"2" = "Disagree",
"3" = "Somewhat Disagree",
"4" = "Neutral",
"5" = "Somewhat Agree",
"6" = "Agree",
"7" = "Strongly Agree"
)
table_clusters <- table(mydata$Group, mydata$KeepCash)
prop_table_clusters <- prop.table(table_clusters, margin = 1)
prop_df <- as.data.frame(as.table(prop_table_clusters))
prop_df$Var2 <- factor(prop_df$Var2, levels = names(likert_labels), labels = likert_labels)
ggplot(prop_df, aes(x = Var1, y = Freq * 100, fill = Var2)) +
geom_bar(stat = "identity", position = "stack") +
labs(
x = "Group",
y = "Percentage (%)",
fill = "Response",
title = "Percentage Distribution of Those Who Prefer Cash"
) +
theme_minimal()
Zlahka najdem ponudnike, ki sprejemajo digitalna placila za vsakodnevne nakupe. (angl. I can easily find providers that accept digital payments for everyday purchases.)
aggregate(mydata$FindSeller,
by = list(mydata$Group),
FUN = mean)## Group.1 x
## 1 1 6.172414
## 2 2 6.000000
## 3 3 5.240000
## 4 4 5.054054
## 5 5 5.470588The table above shows the average answer to the statement “Zlahka najdem ponudnike, ki sprejemajo digitalna plačila za vsakodnevne nakupe”, where 1 means strongly disagree and 7 strongly agree.
Let us perform the Shapiro-Wilk normality test to check the normal distribution of the variable. Our hypotheses are the following:
H0: normality is met.
H1: normality is not met.
library(dplyr)
library(rstatix)
mydata %>%
group_by(as.factor(mydata$Group)) %>%
shapiro_test(FindSeller)## # A tibble: 5 × 4
## `as.factor(mydata$Group)` variable statistic p
## <fct> <chr> <dbl> <dbl>
## 1 1 FindSeller 0.701 0.00000216
## 2 2 FindSeller 0.850 0.0000745
## 3 3 FindSeller 0.887 0.00989
## 4 4 FindSeller 0.887 0.00129
## 5 5 FindSeller 0.888 0.0434As p < 0.05, H0 can be rejected. Thus, we have to perform the non-parameteric alternative. Our hypotheses are:
H0: all distribution locations of the variable are the same.
H1: at least one distribution location of the variable is different.
kruskal.test(FindSeller ~ as.factor(Group),
data = mydata)##
## Kruskal-Wallis rank sum test
##
## data: FindSeller by as.factor(Group)
## Kruskal-Wallis chi-squared = 15.429, df = 4, p-value = 0.00389As p < 0.05, H0 can be rejected. There are differences between clusters.
Below is the visual representation of the attitude towards the statement “Zlahka najdem ponudnike, ki sprejemajo digitalna plačila za vsakodnevne nakupe.”
library(ggplot2)
library(dplyr)
likert_labels <- c(
"1" = "Strongly Disagree",
"2" = "Disagree",
"3" = "Somewhat Disagree",
"4" = "Neutral",
"5" = "Somewhat Agree",
"6" = "Agree",
"7" = "Strongly Agree"
)
table_clusters <- table(mydata$Group, mydata$FindSeller)
prop_table_clusters <- prop.table(table_clusters, margin = 1)
prop_df <- as.data.frame(as.table(prop_table_clusters))
prop_df$Var2 <- factor(prop_df$Var2, levels = names(likert_labels), labels = likert_labels)
ggplot(prop_df, aes(x = Var1, y = Freq * 100, fill = Var2)) +
geom_bar(stat = "identity", position = "stack") +
labs(
x = "Group",
y = "Percentage (%)",
fill = "Response",
title = "Percentage Distribution of Those Who Find Vendors Accepting Digital Payments"
) +
theme_minimal()
Pri plačilih manjše vrednosti (do 10€) se mi zdi gotovina hitrejši način plačila od digitalnih plačil. (angl. For smaller payments (up to €10), I find cash a faster way to pay than digital payments.)
aggregate(mydata$SmallCash,
by = list(mydata$Group),
FUN = mean)## Group.1 x
## 1 1 1.862069
## 2 2 2.902439
## 3 3 3.200000
## 4 4 4.351351
## 5 5 3.176471The table above shows the average answer to the statement “Pri plačilih manjše vrednosti (do 10€) se mi zdi gotovina hitrejši način plačila od digitalnih plačil”, where 1 means strongly disagree and 7 strongly agree.
Let us perform the Shapiro-Wilk normality test to check the normal distribution of the variable. Our hypotheses are the following:
H0: normality is met.
H1: normality is not met.
library(dplyr)
library(rstatix)
mydata %>%
group_by(as.factor(mydata$Group)) %>%
shapiro_test(SmallCash)## # A tibble: 5 × 4
## `as.factor(mydata$Group)` variable statistic p
## <fct> <chr> <dbl> <dbl>
## 1 1 SmallCash 0.810 0.000126
## 2 2 SmallCash 0.879 0.000428
## 3 3 SmallCash 0.894 0.0138
## 4 4 SmallCash 0.854 0.000197
## 5 5 SmallCash 0.844 0.00885As p < 0.05, H0 can be rejected. Thus, we have to perform the non-parameteric alternative. Our hypotheses are:
H0: all distribution locations of the variable are the same.
H1: at least one distribution location of the variable is different.
kruskal.test(SmallCash ~ as.factor(Group),
data = mydata)##
## Kruskal-Wallis rank sum test
##
## data: SmallCash by as.factor(Group)
## Kruskal-Wallis chi-squared = 22.6, df = 4, p-value = 0.0001522As p < 0.05, H0 can be rejected. There are differences between clusters.
Below is the visual representation of the attitude towards the statement “Pri plačilih manjše vrednosti (do 10€) se mi zdi gotovina hitrejši način plačila od digitalnih plačil.”
library(ggplot2)
library(dplyr)
likert_labels <- c(
"1" = "Strongly Disagree",
"2" = "Disagree",
"3" = "Somewhat Disagree",
"4" = "Neutral",
"5" = "Somewhat Agree",
"6" = "Agree",
"7" = "Strongly Agree"
)
table_clusters <- table(mydata$Group, mydata$SmallCash)
prop_table_clusters <- prop.table(table_clusters, margin = 1)
prop_df <- as.data.frame(as.table(prop_table_clusters))
prop_df$Var2 <- factor(prop_df$Var2, levels = names(likert_labels), labels = likert_labels)
ggplot(prop_df, aes(x = Var1, y = Freq * 100, fill = Var2)) +
geom_bar(stat = "identity", position = "stack") +
labs(
x = "Group",
y = "Percentage (%)",
fill = "Response",
title = "Percentage Distribution of Those Who Prefer Cash For Payments up to 10€"
) +
theme_minimal()
Uporaba digitalnih plačil za transakcije z majhno vrednostjo (do 10€) je priročna in enostavna, z minimalnimi dodatnimi koraki (npr. vnos PIN-a, skeniranje QR kode) (angl. Using digital payments for small value transactions (up to €10) is convenient and easy, with minimal additional steps (e.g., PIN entry, QR code scanning.)
aggregate(mydata$DigitalEasy,
by = list(mydata$Group),
FUN = mean)## Group.1 x
## 1 1 6.068966
## 2 2 5.878049
## 3 3 5.200000
## 4 4 4.432432
## 5 5 5.705882The table above shows the average answer to the statement “Uporaba digitalnih placil za transkacije z majhno vrednostjo (do 10) je prirocna in enostavna, z minimalnimi dodatnimi koraki (npr. vnos PIN-a, skeniranje QR kode)”, where 1 means strongly disagree and 7 strongly agree.
Let us perform the Shapiro-Wilk normality test to check the normal distribution of the variable. Our hypotheses are the following:
H0: normality is met.
H1: normality is not met.
library(dplyr)
library(rstatix)
mydata %>%
group_by(as.factor(mydata$Group)) %>%
shapiro_test(DigitalEasy)## # A tibble: 5 × 4
## `as.factor(mydata$Group)` variable statistic p
## <fct> <chr> <dbl> <dbl>
## 1 1 DigitalEasy 0.668 0.000000744
## 2 2 DigitalEasy 0.812 0.0000100
## 3 3 DigitalEasy 0.878 0.00640
## 4 4 DigitalEasy 0.901 0.00305
## 5 5 DigitalEasy 0.877 0.0288As p < 0.05, H0 can be rejected. Thus, we have to perform the non-parameteric alternative. Our hypotheses are:
H0: all distribution locations of the variable are the same.
H1: at least one distribution location of the variable is different.
kruskal.test(DigitalEasy ~ as.factor(Group),
data = mydata)##
## Kruskal-Wallis rank sum test
##
## data: DigitalEasy by as.factor(Group)
## Kruskal-Wallis chi-squared = 20.426, df = 4, p-value = 0.0004114As p < 0.05, H0 can be rejected. There are differences between clusters.
Below is the visual representation of the attitude towards the statement “Uporaba digitalnih plačil za transkacije z majhno vrednostjo (do 10€) je priročna in enostavna, z minimalnimi dodatnimi koraki (npr. vnos PIN-a, skeniranje QR kode)”
library(ggplot2)
library(dplyr)
likert_labels <- c(
"1" = "Strongly Disagree",
"2" = "Disagree",
"3" = "Somewhat Disagree",
"4" = "Neutral",
"5" = "Somewhat Agree",
"6" = "Agree",
"7" = "Strongly Agree"
)
table_clusters <- table(mydata$Group, mydata$DigitalEasy)
prop_table_clusters <- prop.table(table_clusters, margin = 1)
prop_df <- as.data.frame(as.table(prop_table_clusters))
prop_df$Var2 <- factor(prop_df$Var2, levels = names(likert_labels), labels = likert_labels)
ggplot(prop_df, aes(x = Var1, y = Freq * 100, fill = Var2)) +
geom_bar(stat = "identity", position = "stack") +
labs(
x = "Group",
y = "Percentage (%)",
fill = "Response",
title = "Percentage Distribution of Those Who Prefer Digital For Payments up to 10€"
) +
theme_minimal()
Rad/a uporabljam mobilne plačilne platforme (kot so Apple Pay, Revolut, PayPal, Flik), ker so priročne.(angl. I like to use mobile payment platforms (such as Apple Pay, Revolut, PayPal, Flik) because they are convenient.)
aggregate(mydata$ConvenientFlik,
by = list(mydata$Group),
FUN = mean)## Group.1 x
## 1 1 6.413793
## 2 2 5.292683
## 3 3 5.640000
## 4 4 4.081081
## 5 5 5.588235The table above shows the average answer to the statement “Rad/a uporabljam mobilne plačilne platforme (kot so Apple Pay, Revolut, PayPal, Flik), ker so priročne”, where 1 means strongly disagree and 7 strongly agree.
Let us perform the Shapiro-Wilk normality test to check the normal distribution of the variable. Our hypotheses are the following:
H0: normality is met.
H1: normality is not met.
library(dplyr)
library(rstatix)
mydata %>%
group_by(as.factor(mydata$Group)) %>%
shapiro_test(ConvenientFlik)## # A tibble: 5 × 4
## `as.factor(mydata$Group)` variable statistic p
## <fct> <chr> <dbl> <dbl>
## 1 1 ConvenientFlik 0.528 0.0000000154
## 2 2 ConvenientFlik 0.781 0.00000224
## 3 3 ConvenientFlik 0.777 0.0000966
## 4 4 ConvenientFlik 0.872 0.000532
## 5 5 ConvenientFlik 0.817 0.00351As p < 0.05, H0 can be rejected. Thus, we have to perform the non-parameteric alternative. Our hypotheses are:
H0: all distribution locations of the variable are the same.
H1: at least one distribution location of the variable is different.
kruskal.test(ConvenientFlik ~ as.factor(Group),
data = mydata)##
## Kruskal-Wallis rank sum test
##
## data: ConvenientFlik by as.factor(Group)
## Kruskal-Wallis chi-squared = 24.896, df = 4, p-value = 5.28e-05As p < 0.05, H0 can be rejected. There are differences between clusters.
Below is the visual representation of the attitude towards the statement “Rad/a uporabljam mobilne plačilne platforme (kot so Apple Pay, Revolut, PayPal, Flik), ker so priročne.”
library(ggplot2)
library(dplyr)
likert_labels <- c(
"1" = "Strongly Disagree",
"2" = "Disagree",
"3" = "Somewhat Disagree",
"4" = "Neutral",
"5" = "Somewhat Agree",
"6" = "Agree",
"7" = "Strongly Agree"
)
table_clusters <- table(mydata$Group, mydata$ConvenientFlik)
prop_table_clusters <- prop.table(table_clusters, margin = 1)
prop_df <- as.data.frame(as.table(prop_table_clusters))
prop_df$Var2 <- factor(prop_df$Var2, levels = names(likert_labels), labels = likert_labels)
# Plot with updated labels
ggplot(prop_df, aes(x = Var1, y = Freq * 100, fill = Var2)) +
geom_bar(stat = "identity", position = "stack") +
labs(
x = "Group",
y = "Percentage (%)",
fill = "Response",
title = "Percentage Distribution of Those Who Use Mobile Payment Platforms due to Convenience"
) +
theme_minimal()
Verjetno bom uporabljal/a mobilne plačilne platforme (kot so Apple Pay, Revolut, PayPal, Flik), če mi jih priporočijo prijatelji, družina ali vrstniki.(angl. I am likely to use mobile payment platforms (such as Apple Pay, Revolut, PayPal, Flik) if they are recommended to be by friends, family, or peers.)
aggregate(mydata$FriendsFlik,
by = list(mydata$Group),
FUN = mean)## Group.1 x
## 1 1 5.413793
## 2 2 5.000000
## 3 3 5.040000
## 4 4 4.378378
## 5 5 5.235294The table above shows the average answer to the statement “Verjetno bom uporabljal/a mobilne plačilne platforme (kot so Apple Pay, Revolut, PayPal, Flik), če mi jih priporočijo prijatelji, družina ali vrstniki”, where 1 means strongly disagree and 7 strongly agree.
Let us perform the Shapiro-Wilk normality test to check the normal distribution of the variable. Our hypotheses are the following:
H0: normality is met.
H1: normality is not met.
library(dplyr)
library(rstatix)
mydata %>%
group_by(as.factor(mydata$Group)) %>%
shapiro_test(FriendsFlik)## # A tibble: 5 × 4
## `as.factor(mydata$Group)` variable statistic p
## <fct> <chr> <dbl> <dbl>
## 1 1 FriendsFlik 0.808 0.000117
## 2 2 FriendsFlik 0.846 0.0000602
## 3 3 FriendsFlik 0.875 0.00558
## 4 4 FriendsFlik 0.898 0.00255
## 5 5 FriendsFlik 0.838 0.00697As p < 0.05, H0 can be rejected. Thus, we have to perform the non-parameteric alternative. Our hypotheses are:
H0: all distribution locations of the variable are the same.
H1: at least one distribution location of the variable is different.
kruskal.test(FriendsFlik ~ as.factor(Group),
data = mydata)##
## Kruskal-Wallis rank sum test
##
## data: FriendsFlik by as.factor(Group)
## Kruskal-Wallis chi-squared = 7.617, df = 4, p-value = 0.1067As p > 0.05, H0 can be rejected. There are differences between clusters.
Below is the visual representation of the attitude towards the statement “Verjetno bom uporabljal/a mobilne plačilne platforme (kot so Apple Pay, Revolut, PayPal, Flik), če mi jih priporočijo prijatelji, družina ali vrstniki.”
library(ggplot2)
library(dplyr)
likert_labels <- c(
"1" = "Strongly Disagree",
"2" = "Disagree",
"3" = "Somewhat Disagree",
"4" = "Neutral",
"5" = "Somewhat Agree",
"6" = "Agree",
"7" = "Strongly Agree"
)
table_clusters <- table(mydata$Group, mydata$FriendsFlik)
prop_table_clusters <- prop.table(table_clusters, margin = 1)
prop_df <- as.data.frame(as.table(prop_table_clusters))
prop_df$Var2 <- factor(prop_df$Var2, levels = names(likert_labels), labels = likert_labels)
ggplot(prop_df, aes(x = Var1, y = Freq * 100, fill = Var2)) +
geom_bar(stat = "identity", position = "stack") +
labs(
x = "Group",
y = "Percentage (%)",
fill = "Response",
title = "Percentage Distribution of Those Who Use Mobile Payment Platforms due to Recommendations"
) +
theme_minimal()
Statistically significant are the following variables:
“Raje imam pri sebi gotovino, ker ni vedno mogoče plačati z drugimi plačilnimi sredstvi.”
“Zlahka najdem ponudnike, ki sprejemajo digitalna plačila za vsakodnevne nakupe.”
“Pri plačilih za manjše vrednosti (do 10€) se mi zdi gotovina hitrejši način plačila od digitalnih plačil.”
“Uporaba digitalnih plačil za transakcije z majhno vrednostjo (do10€) je priročna in enostavna, z minimalnimi dodatnimi koraki.”
“Rad/a uporabljam mobilne plačilne platforme (kot so ApplePay, Revolut, PayPal, Flik), ker so priročne.”
We have shown that other two variables - “Raje plačujem z gotovino kot z digitalnimi plačilnimi sredstvi.” and “Verjetno bom uporabljal/a mobilne plačilne platforme (kot so ApplePay, Revolut, PayPal, Flik), če mi jih priporočijo prijatelji, družina ali vrstniki.” are not statistically significant, but we will still use them to describe the clusters.