Homework assignment 2 at the course Applied Data Analysis in Business with R
Wong, Hing Yee Yashelyn
2024-04-26
Author: Wong, Hing Yee Yashelyn
1 Research question
1.1 Are there differences in the number of purchases on web for married and single customers?
1.2 Variables to be collected: Married customers, single customers, number of web purchase
mydata <- read.table("./FoodAppBusiness.csv", header=TRUE, sep=",", dec=".")
head(mydata[c(-10:-12)], )## MonthlyIncome ActiveSinceDays Age Married Single NoOfDaysSinceLastPurchase
## 1 58138 2822 63 0 1 58
## 2 46344 2272 66 0 1 38
## 3 71613 2471 55 1 0 26
## 4 26646 2298 36 1 0 26
## 5 58293 2320 39 1 0 94
## 6 62513 2452 53 1 0 16
## NoOfWebPurchase NoOfStorePurchase NoOfWebVisitsMonth
## 1 8 4 7
## 2 1 2 5
## 3 8 10 4
## 4 2 4 6
## 5 5 6 5
## 6 6 10 62 Data
2.1 Definition of variables:
MonthlyIncome - the observed customers’ income per month
ActiveSinceDays - the observed customers’ active number of days since using the app
Age - the age of the observed customers
Married - the customer’s marital status being married, where 0 stands for No and 1 stands for Yes
Single - the customer’s marital status being single, where 0 stands for No and 1 stands for Yes
NoOfDaysSinceLastPurchased - the number of days since the last purchase from the web or store
NoOfWebPurchase - number of times the customer purchase from the web
NoOfStorePurchase - number of times the customer purchase from the store
NoOfWebVisitsMonth - number of times the customer visited the web
2.2 Unit of observation and sample size: 2205
2.3 Source of the data set: Kaggle - https://www.kaggle.com/datasets/ybifoundation/food-app-business
mydata$MarriedF <- factor(mydata$Married, levels = c(0,1), labels = c("No", "Yes"))
mydata$SingleF <- factor(mydata$Single, levels = c(0,1), labels = c("No", "Yes"))summary(mydata[c(-2,-4:-6,-9:-12)], )## MonthlyIncome Age NoOfWebPurchase NoOfStorePurchase MarriedF
## Min. : 1730 Min. :24.0 Min. : 0.000 Min. : 0.000 No : 783
## 1st Qu.: 35196 1st Qu.:43.0 1st Qu.: 2.000 1st Qu.: 3.000 Yes:1422
## Median : 51287 Median :50.0 Median : 4.000 Median : 5.000
## Mean : 51622 Mean :51.1 Mean : 4.101 Mean : 5.824
## 3rd Qu.: 68281 3rd Qu.:61.0 3rd Qu.: 6.000 3rd Qu.: 8.000
## Max. :113734 Max. :80.0 Max. :27.000 Max. :13.000
## SingleF
## No :1422
## Yes: 783
##
##
##
## 2.4 Basic descriptive statistics - estimate a few parameters (e.g., functions summary, describe, etc.) and explanation
Min. of MonthlyIncome: the least monthly income of the observed customers is 1730
Median of Age: 50% or less of the observed customers are 50 years old
Max. of NoOfWebPurchase; the most number of web purchase is 27 times
Mean of NoOfStorePurchase: the average number of store purchase is 5.824 times
MarriedF and SingleF: 1422 of the observed customers are married and 783 of the observed customers are single
3 Analysis
3.1 Independent t-test will be used because the hypothes is about the difference between two population arithmetic means (married and single customers) and sample size is large.
3.2 Assumptions: the variables are numerical, there is normal distribution and equal variances in both groups single and married customers, the data is from two independent population.
3.3 Statistical test based on the results of the assumption evaluation and its interpretation:
library(psych)
describeBy(mydata$NoOfWebPurchase, mydata$MarriedF)##
## Descriptive statistics by group
## group: No
## vars n mean sd median trimmed mad min max range skew kurtosis se
## X1 1 783 4.08 2.82 4 3.78 2.97 0 27 27 1.6 7.18 0.1
## ------------------------------------------------------------
## group: Yes
## vars n mean sd median trimmed mad min max range skew kurtosis se
## X1 1 1422 4.11 2.69 4 3.86 2.97 0 25 25 0.94 1.98 0.07The results for Single and Married customers regarding Number of Web Purchase are similar.
library(ggplot2)##
## Attaching package: 'ggplot2'## The following objects are masked from 'package:psych':
##
## %+%, alphaMarried_no <- ggplot(mydata[mydata$MarriedF == "No", ], aes(x = NoOfWebPurchase)) + theme_linedraw() + geom_histogram() + ggtitle("Single")
Married_yes <- ggplot(mydata[mydata$MarriedF == "Yes", ], aes(x = NoOfWebPurchase)) + theme_linedraw() + geom_histogram() + ggtitle("Married")
install.packages("ggpubr", repos = "https://cloud.r-project.org/", dependencies = TRUE)## Installing package into 'C:/Users/user/AppData/Local/R/win-library/4.3'
## (as 'lib' is unspecified)## package 'ggpubr' successfully unpacked and MD5 sums checked
##
## The downloaded binary packages are in
## C:\Users\user\AppData\Local\Temp\RtmpiQrUzf\downloaded_packageslibrary(ggpubr)
ggarrange(Married_no, Married_yes, ncol = 2, nrow = 1)## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
The graphs for Single and Married are right-skewed, therefore Wilcoxon rank sum test will be used.
library(rstatix)##
## Attaching package: 'rstatix'## The following object is masked from 'package:stats':
##
## filtermydata %>%
group_by(MarriedF) %>%
shapiro_test(NoOfWebPurchase)## # A tibble: 2 × 4
## MarriedF variable statistic p
## <fct> <chr> <dbl> <dbl>
## 1 No NoOfWebPurchase 0.885 1.13e-23
## 2 Yes NoOfWebPurchase 0.920 9.69e-27H0: NoOfWebPurchase is normally distributed for Singled.
H1: NoOfWebPurchase is not normally distributed for Singled.
We reject H0 at p<0.001.
H0: NoOfWebPurchase is normally distributed for Married.
H1: NoOfWebPurchase is not normally distributed for Married.
We reject H0 at p<0.001.
wilcox.test(mydata$NoOfWebPurchase ~ mydata$MarriedF,
paired = FALSE,
correct = FALSE,
exact = FALSE,
alternative = "two.sided")##
## Wilcoxon rank sum test
##
## data: mydata$NoOfWebPurchase by mydata$MarriedF
## W = 549996, p-value = 0.636
## alternative hypothesis: true location shift is not equal to 0wilcox.test(mydata$NoOfWebPurchase ~ mydata$SingleF,
paired = FALSE,
correct = FALSE,
exact = FALSE,
alternative = "two.sided")##
## Wilcoxon rank sum test
##
## data: mydata$NoOfWebPurchase by mydata$SingleF
## W = 563431, p-value = 0.636
## alternative hypothesis: true location shift is not equal to 0H0: locations of distribution of web purchase are the same for Single and Married customers.
H1: locations of distribution of web purchases are not the same for Single and Married customers.
We cannot reject H0.
3.4 Calculation of the effect size and its interpretation:
install.packages('effectsize', repos = "https://cloud.r-project.org/", dependencies = TRUE)## Installing package into 'C:/Users/user/AppData/Local/R/win-library/4.3'
## (as 'lib' is unspecified)## package 'effectsize' successfully unpacked and MD5 sums checked
##
## The downloaded binary packages are in
## C:\Users\user\AppData\Local\Temp\RtmpiQrUzf\downloaded_packageslibrary(effectsize)##
## Attaching package: 'effectsize'## The following objects are masked from 'package:rstatix':
##
## cohens_d, eta_squared## The following object is masked from 'package:psych':
##
## phieffectsize(wilcox.test(mydata$NoOfWebPurchase ~ mydata$MarriedF,
paired = FALSE,
correct = FALSE,
exact = FALSE,
alternative = "two.sided"))## r (rank biserial) | 95% CI
## ---------------------------------
## -0.01 | [-0.06, 0.04]interpret_rank_biserial(0.01)## [1] "tiny"
## (Rules: funder2019)4 Conclusion
Based on the sample data, it cannot prove that there are differences in the number of web purchases between Married and Single customers (p>0.05). Single and Married customers have similar number of purchases on web. The difference in distribution is tiny (r = 0.01). ___________________________________________
*Source to deal with error “trying to use CRAN without mirror”: https://stackoverflow.com/questions/62357732/installation-issues-of-ggpubr