Homework 1 - Blanka Pucer
2025-01-11
Part 1
mydata <- read.table("./HW_1.csv",
header = TRUE,
sep = ",",
dec = ".")
head(mydata)## Group Customer_Segment Sales_Before Sales_After
## 1 Control High Value 240.5484 300.0076
## 2 Treatment High Value 246.8621 381.3376
## 3 Control High Value 156.9781 179.3305
## 4 Control Medium Value 192.1267 229.2780
## 5 Control High Value 229.6856 270.1677
## 6 Treatment Low Value 135.5730 218.5600
## Customer_Satisfaction_Before Customer_Satisfaction_After Purchase_Made
## 1 74.68477 74.09366 No
## 2 100.00000 100.00000 Yes
## 3 98.78073 100.00000 No
## 4 49.33377 39.81184 Yes
## 5 83.97485 87.73859 Yes
## 6 58.07534 69.40492 Noset.seed(123)
mydata <- mydata[sample(nrow(mydata), 500), ]I have reduced the number of observations to 500 so that I will be able to perform the necessary tests.
This data set contains information on sales and customer satisfaction before and after an intervention and purchase data for control and treatment groups.
Unit of observation is a customer transaction or purchase. The explanation of variables:
- Group: either control or treatment
- Customer_Segment: categorizes customers based on their value into high value, medium value, and low value. The description of the data set does not specify what a customer being high value, for example, means, but the classification may be based on their income, the high-value customers having the highest income.
- Sales_Before: sales before intervention
- Sales_After: sales after intervention
- Customer_Satisfaction_Before: customer satisfaction scores before the intervention
- Customer-Satisfaction_After: customer satisfaction scores after the intervention
- Purchase_Made: indicates whether a purchase was made after the intervention
Data set source: https://www.kaggle.com/datasets/matinmahmoudi/sales-and-satisfaction
Let us start with some descriptive statistics.
library(psych)
summary(mydata$Sales_Before)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 36.34 166.36 203.74 202.30 238.15 380.45I will explain a few estimates of parameters:
- The average sales before the intervention were 202.30
- The median is 203.74, which means that half of the sales before the intervention were larger than 203.74, and half were smaller
- The minimum sales before the intervention were 36.34, and the maximum sales before the intervention were 380.45
Let us start with the first research question.
Research question 1: I am interested in whether there is a difference in sales before and after intervention.
My hypotheses are:
H0: the average sales before the intervention are the same as the average sales after the intervention; the difference is zero. μ sales_before = μ sales_after or μ difference = 0.
H1: the average sales before the intervention are not the same as the average sales after the intervention; the difference is not zero. μ sales_before ≠ μ sales_after or μ difference ≠ 0.
library(pastecs)
round(stat.desc(mydata[ , c(3,4)]), 2)## Sales_Before Sales_After
## nbr.val 500.00 500.00
## nbr.null 0.00 0.00
## nbr.na 0.00 0.00
## min 36.34 46.64
## max 380.45 587.44
## range 344.11 540.80
## sum 101148.36 139380.90
## median 203.74 271.86
## mean 202.30 278.76
## SE.mean 2.40 3.73
## CI.mean.0.95 4.71 7.33
## var 2878.68 6954.08
## std.dev 53.65 83.39
## coef.var 0.27 0.30mydata$Difference <- mydata$Sales_After - mydata$Sales_BeforeI have created a new variable, Difference, which is the difference between sales after the intervention and sales before the intervention.
library(ggplot2)##
## Attaching package: 'ggplot2'## The following objects are masked from 'package:psych':
##
## %+%, alphaggplot(mydata, aes (x = Difference)) +
geom_histogram(position = "identity", binwidth = 25, colour = "red") +
ylab("Frequency") +
xlab("Difference")
Observing the histogram, the differences do not seem to be normally distributed. However, to be sure of that, we have to perform the Shapiro-Wilk normality test.
The hypotheses are:
- H0: the differences are normally distributed.
- H1: the differences are not normally distributed.
shapiro.test(mydata$Difference)##
## Shapiro-Wilk normality test
##
## data: mydata$Difference
## W = 0.92626, p-value = 5.866e-15H0 can be rejected at p < 0.001. With the help of the statistical test, I have formally shown that the differences are not normally distributed.
t.test(mydata$Sales_Before, mydata$Sales_After,
paired = TRUE,
alternative = "two.sided")##
## Paired t-test
##
## data: mydata$Sales_Before and mydata$Sales_After
## t = -39.659, df = 499, p-value < 2.2e-16
## alternative hypothesis: true mean difference is not equal to 0
## 95 percent confidence interval:
## -80.25323 -72.67692
## sample estimates:
## mean difference
## -76.46508I have performed a paired samples t-test, where the assumptions are the following:
- The variable is numeric; and
- The differences on the population are normally distributed.
Since the p-value < 0.001, we can reject the H0, which says that the average sales before the intervention are the same as the average sales after the intervention; the difference is zero. The mean difference is about -76.5.
Let us perform the non-parametric alternative. The non-parametric alternative to the paired t-test is the Wilcoxon signed rank test.
My hypotheses are the following:
- H0: the distribution location of sales is the same.
- H1: the distribution location of sales not the same.
wilcox.test(mydata$Sales_Before, mydata$Sales_After,
paired = TRUE,
correct = FALSE,
exact = FALSE,
alternative = "two.sided")##
## Wilcoxon signed rank test
##
## data: mydata$Sales_Before and mydata$Sales_After
## V = 0, p-value < 2.2e-16
## alternative hypothesis: true location shift is not equal to 0Again, the p-value < 0.001, and the H0 can be rejected.
library(effectsize)##
## Attaching package: 'effectsize'## The following object is masked from 'package:psych':
##
## phicohens_d(mydata$Difference)## Cohen's d | 95% CI
## ------------------------
## 1.77 | [1.63, 1.91]interpret_cohens_d(1.77, rules = "sawilowsky2009")## [1] "very large"
## (Rules: sawilowsky2009)The difference in sales before the intervention and after the intervention is very large.
Part 2
head(mydata)## Group Customer_Segment Sales_Before Sales_After
## 2463 Control Medium Value 141.1834 178.6317
## 2511 Treatment Medium Value 182.3470 292.2028
## 8718 Treatment High Value 380.4529 587.4384
## 2986 Treatment Medium Value 248.8122 384.3708
## 1842 Treatment Low Value 177.7405 273.7634
## 9334 Control High Value 246.6162 298.9664
## Customer_Satisfaction_Before Customer_Satisfaction_After Purchase_Made
## 2463 57.47312 62.91412 No
## 2511 53.03289 35.49490 No
## 8718 82.97648 76.14568 No
## 2986 57.98888 38.54076 Yes
## 1842 68.14757 75.70201 Yes
## 9334 93.00714 86.44396 No
## Difference
## 2463 37.44828
## 2511 109.85582
## 8718 206.98546
## 2986 135.55857
## 1842 96.02289
## 9334 52.35014library(car)## Loading required package: carData##
## Attaching package: 'car'## The following object is masked from 'package:psych':
##
## logitscatterplotMatrix(mydata[, c(5,6)], smooth = FALSE)
Research question 2: Let us say that I would like to test whether there is some correlation between customer satisfaction before the intervention and customer satisfaction after the intervention.
My hypotheses are the following:
- H0: ρ satisfaction_before, satisfaction_after = 0
- H1: ρ satisfaction_before, satisfaction_after =/= 0.
I will test these hypotheses with the help of Pearson correlation coefficient.
library(Hmisc)##
## Attaching package: 'Hmisc'## The following object is masked from 'package:psych':
##
## describe## The following objects are masked from 'package:base':
##
## format.pval, unitscor.test(mydata$Customer_Satisfaction_Before, mydata$Customer_Satisfaction_After,
method = "pearson",
use = "complete.obs")##
## Pearson's product-moment correlation
##
## data: mydata$Customer_Satisfaction_Before and mydata$Customer_Satisfaction_After
## t = 35.651, df = 498, p-value < 2.2e-16
## alternative hypothesis: true correlation is not equal to 0
## 95 percent confidence interval:
## 0.8209649 0.8706121
## sample estimates:
## cor
## 0.8476336H0 is rejected at p < 0.001.
cor(mydata$Customer_Satisfaction_Before, mydata$Customer_Satisfaction_After,
method = "pearson",
use = "complete.obs")## [1] 0.8476336The linear relationship between customer satisfaction before the intervention and customer satisfaction after the intervention is positive and strong.
Part 3
head(mydata)## Group Customer_Segment Sales_Before Sales_After
## 2463 Control Medium Value 141.1834 178.6317
## 2511 Treatment Medium Value 182.3470 292.2028
## 8718 Treatment High Value 380.4529 587.4384
## 2986 Treatment Medium Value 248.8122 384.3708
## 1842 Treatment Low Value 177.7405 273.7634
## 9334 Control High Value 246.6162 298.9664
## Customer_Satisfaction_Before Customer_Satisfaction_After Purchase_Made
## 2463 57.47312 62.91412 No
## 2511 53.03289 35.49490 No
## 8718 82.97648 76.14568 No
## 2986 57.98888 38.54076 Yes
## 1842 68.14757 75.70201 Yes
## 9334 93.00714 86.44396 No
## Difference
## 2463 37.44828
## 2511 109.85582
## 8718 206.98546
## 2986 135.55857
## 1842 96.02289
## 9334 52.35014Research question 3: Let us say that I would like to test whether there is some association between the customer segment (whether they are low, medium, or high value) and whether they have made a purchase.
My hypotheses are:
- H0: Customer segment and making a purchase are not associated.
- H1: Customer segment and making a purchase are associated.
results <- chisq.test(mydata$Customer_Segment, mydata$Purchase_Made)
results##
## Pearson's Chi-squared test
##
## data: mydata$Customer_Segment and mydata$Purchase_Made
## X-squared = 0.66079, df = 2, p-value = 0.7186As the p-value > 0.05 (p-value = 0.7186), we cannot reject the H0.
addmargins(results$observed)## mydata$Purchase_Made
## mydata$Customer_Segment No Yes Sum
## High Value 83 70 153
## Low Value 92 77 169
## Medium Value 90 88 178
## Sum 265 235 500This table shows observed (empirical) frequencies. Out of 153 high-value customers, 83 have not made a purchase and 70 have. Out of 169 low-value customers, 92 have not made a purchase and 77 have. Out of 178 medium-value customers, 90 have not made a purchase and 88 have.
addmargins(round(results$expected, 2))## mydata$Purchase_Made
## mydata$Customer_Segment No Yes Sum
## High Value 81.09 71.91 153
## Low Value 89.57 79.43 169
## Medium Value 94.34 83.66 178
## Sum 265.00 235.00 500This table shows expected (theoretical) frequencies. If there were no association between customer segment and making a purchase, we would expect that 81.09 high-value customers would not make a purchase and 71.91 would make a purchase. If there were no association between customer segment and making a purchase, we would expect that 89.57 low-value customers would not make a purchase and 79.43 would make a purchase. If there were no association between customer segment and making a purchase, we would expect that 94.34 medium-value customers would not make a purchase and 83.66 would make a purchase.
round(results$residuals, 2)## mydata$Purchase_Made
## mydata$Customer_Segment No Yes
## High Value 0.21 -0.23
## Low Value 0.26 -0.27
## Medium Value -0.45 0.47Let us calculate the standardized residuals. As the p-value of the Pearson’s Chi-squared test is higher than 0.05, it can be expected that the standardized residuals will not be significant either. None of the combinations is 1.96 or higher in absolute terms, they are all insignificant.
library(effectsize)
effectsize::cramers_v(mydata$Customer_Segment, mydata$Purchase_Made)## Cramer's V (adj.) | 95% CI
## --------------------------------
## 0.00 | [0.00, 1.00]
##
## - One-sided CIs: upper bound fixed at [1.00].interpret_cramers_v(0.00)## [1] "tiny"
## (Rules: funder2019)The effect size is tiny. This means that there is no real difference - or the difference is tiny - between the expected (theoretical) and observed (empirical) frequencies.