PORTFOLIO
KAWALJEET KAUR
2023-11-01
INTRODUCTION This portfolio aims to investigate my transaction history that i made from Oct 1 to Oct 31, 2023.Out of 31 days, i had transactions just for 23 days and 8 days were without any transactions.These 8 days were probably those days when i did not come to university and stayed at home without any expense. This portfolio is important for me as I am an international student and i need to manage my expenses and take care of deposits as well to avoid any breaching of my work right entitlements.To do the investigation, i made a data set of my transactions in which every row represents the transaction on a particular day and include following columns. 1.Date - Represented as Day 01, Day 02 etc 2.Total - This column contains the information of my account money in the beginning of the day. 3.Deposit - It includes the money deposited to my account on any particular day. I get money from my part time jobs as salary which are deposited to my account on weekly basis. That, why most of rows are 0 in this column. 4.Amount - This represents the money value that i spent on a particular item on a particular day. 5.Purpose - It give information about the item on which i spent amount.It is divided into 3 items which are Travel, Food, and Rent. 6.Time - This column highlights the time of day when i used that money.I considered Morning, Afternoon and Evening. 7.Balance - This column includes the money that is left in my account at the end of the day.
Among the columns given above, Date, Purpose and Time are categorical variables while rest are numerical.All the transactions are in Nz$. The data set includes just three purposes for the transactions that are Food, Rent and Travel. I did not use shopping and other expenses as purpose, in my data set. The reason for this is that, being a married girl, my husband is responsible for all other expenses. Also, Time column does not include the ‘Night’ as i do not buy food, pay rent or travel at night time.
IMPORT DATA TO R
Transaction_data =read.csv("C:/Users/ASUS/Desktop/C.COMPUTATIONAL STATISTICS/portfolio_dataN.csv")
Transaction_dataTabular form of categorical columns
table(table(Transaction_data$Date))##
## 1 2 3
## 18 4 1table(Transaction_data$Time)##
## Afternoon Evening Morning
## 4 13 12table(Transaction_data$Purpose)##
## Food Rent Travel
## 16 5 8Test 1 Basics statistics
#Histogram for Amount
hist(Transaction_data$Amount, col = "blue", xlab = "Amount used", main = "Histogram of Amount")
This histogram shows that most of the transactions include upto $50
only. I never made transactions between $100-200 and even $50-100
transactions were rare during the considered time.MY transactions or
spendings were less than $250.
#Histogram for Balance
hist(Transaction_data$Balance, col = "pink", xlab = "Balance in the account ", main = "Histogram of Account Balance")
As per the histogram, most of the times, my account balance was between
$400-800. It was never less than 200 and more than 1000 during the
considered period.Frequency for $400-500 is notably high.
#Barplot for Purpose of spending
barplot(table(Transaction_data$Purpose), main = "Purpose of spending", col='lightgreen')
From barplot, I used money for Food more frequently than Rent and
Travel.
summary(Transaction_data)## Date Total Deposit Amount
## Length:29 Min. :267.5 Min. : 0.00 Min. : 2.30
## Class :character 1st Qu.:417.4 1st Qu.: 0.00 1st Qu.: 3.50
## Mode :character Median :491.3 Median : 0.00 Median : 7.61
## Mean :495.4 Mean : 42.64 Mean : 33.73
## 3rd Qu.:596.9 3rd Qu.: 0.00 3rd Qu.: 20.00
## Max. :933.9 Max. :386.45 Max. :224.50
## Purpose Time Balance
## Length:29 Length:29 Min. :267.5
## Class :character Class :character 1st Qu.:420.6
## Mode :character Mode :character Median :493.6
## Mean :504.3
## 3rd Qu.:596.9
## Max. :933.9DISCUSSION The Aim to run the summary is particularly linked with the Deposit. This column is linked with my weekly salary. Being an international student, I am entitled to work for 20 hours only which gives me nearly $350. My deposit should be less than or equal to $350, otherwise, the immigration can cancel my student visa as well. By checking my mean deposit i can easily check whether i am breaching my work conditions or not. Data has provided me evidence that i am following conditions on my work entitlement. Mean deposit per day = 42.64 deposit for a week = 42.64*7 = $298.48 < $350 In addition to it, the mean balance in my account was approximately 505 dollars. My average spending per day is nearly $34. It can be seen from the output that Total money in my bank account in the morning is nearly equal to balance remain at night on average although i was spending some money on daily basis.This due to the money deposited to my account which balanced out the spending.
DEFINITION OF P-VALUE Before starting any particular test on my data set, i just want to give definition of P-value. The P-value is defined as the probability , if null is true, of obtaining the observation or an observation more extreme.”More Extreme” means larger than the observed sample.
Test 2 Student t-test I tried to do student t-test on “Amount” and “Deposit”. But deposit data does not follow Gaussian distribution.
y=Transaction_data$Deposit
y## [1] 0.00 0.00 0.00 0.00 92.70 358.89 0.00 0.00 0.00 0.00
## [11] 0.00 0.00 305.74 0.00 0.00 0.00 0.00 0.00 0.00 92.70
## [21] 386.45 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00hist(y)
From histogram, it is clear that Distribution of Deposit is not
Gaussian.So, i can not apply Student t-test here. To apply student
t-test, i chose Amount column and checked the amount used using two
cases. 1.Amount used when deposit was zero. 2.Amount used when deposit
was more than zero.
Null hypothesis(H0) The true difference in means between the “Amount” spent on days when there was a “Deposit” (money deposited) and the “Amount” spent on days when there was no “Deposit” is equal to zero.In simple words, there is no significant difference in the spending when there was a deposit in my bank account versus there was no deposit. Alternative Hypothesis(Ha) There is significant difference in the amount spent on days with deposit and days without a deposit.
data_a = Transaction_data$Amount[Transaction_data$Deposit>0]
data_b = Transaction_data$Amount[Transaction_data$Deposit==0]
Result_t.test = t.test(data_a, data_b)
Result_t.test##
## Welch Two Sample t-test
##
## data: data_a and data_b
## t = 1.3297, df = 4.2598, p-value = 0.2504
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## -73.27995 214.37012
## sample estimates:
## mean of x mean of y
## 92.10800 21.56292Discussion The P-value is 0.2504 which is non-significant.It suggests that there is no strong evidence to reject the null hypothesis at the significance level of 0.05. In other words, the transaction data does not provide enough evidence to conclude that the spending on days with a deposit is significantly different from spending on days without a deposit. Also the 95% confidence interval is -73.27995 214.37012, which includes 0, further supporting the idea that there is no significant difference between Amount[Deposit > 0] and Amount[Deposit == 0].
Test 3 Null Hypothesis (H0): There is no significant association between the time of day and the Purpose. In other words, the choice of Purpose of spending money is independent of the Time of the day. Alternative Hypothesis (Ha): Purpose and time of the day are not independent.Or we can say that there is strong association between Time of the day and the purpose for which i am spending money.
# Define a contingency table
Transaction_data = matrix(c(3, 3, 5, 3, 0, 1, 10, 2, 2), 3, 3)
rownames(Transaction_data) = c("Food", "Rent", "Travel")
colnames(Transaction_data) = c("Morning","Afternoon", "Evening")
Transaction_data## Morning Afternoon Evening
## Food 3 3 10
## Rent 3 0 2
## Travel 5 1 2#fisher test
fisher_result=fisher.test(Transaction_data, alternative="greater")
fisher_result##
## Fisher's Exact Test for Count Data
##
## data: Transaction_data
## p-value = 0.1835
## alternative hypothesis: greaterP-value is greater than 0.05 which is non-significant.So there are no strong evidence to reject the null or we are fail to reject the Null Hypothesis. In other words, Purpose for which I spend money is not associated with the time of the day. It means I cannot say that whenever i spend money in the morning , it is usually for the food or other particular purpose. Purpose of spending and Time of the day are Independent.
Test 4 Chi-square test and the Poisson distribution This test is for
Number of transactions that i made on daily basis during the considered
period. As per the data set, I had transactions for 23 days and 8 days
were without transactions in month of October which has 31 days and data
is collected for all these days. The number of transactions on a
particular day were less. So, the poisson distribution analysis is
appropriate for this data set.
Null Hypothesis(H0:) The number of transactions follow poisson distribution. Alternative Hypothesis The number of transactions are not poisson. The first step is to tabulate the data.
The table below shows that Out of 31 days, there was just 1 transaction for 18 days, 2 transactions for 4 days and 3 transactions for 1 days and 8 days with 0 transactions.
o = c(8, 18, 4, 1)
Total_transactions = 8*0 + 18*1 + 4*2 + 1*3
Total_transactions## [1] 29transactions_per_day = 29/31
transactions_per_day## [1] 0.9354839probs = dpois(0:3, 0.95)
probs## [1] 0.38674102 0.36740397 0.17451689 0.05526368# verification
sum(probs)## [1] 0.9839256probabilities = c(probs, 1-sum(probs))
sum(probabilities)## [1] 1sum(probabilities)*29## [1] 29On daily basis i have 0.93 transactions. The probabilities of 0, 1, 2 and 3 transactions on any day are 0.38674102, 0.36740397, 0.17451689, and 0.05526368 respectively. The probability of 3 transactions is quite low. Now we need to find expected value(e)
#expected value of number of transactions
e = probs*29
e## [1] 11.215490 10.654715 5.060990 1.602647round(e)## [1] 11 11 5 2# observed value
o## [1] 8 18 4 1#Badness of fit
B = sum((o-e)^2/e)
B## [1] 6.434712# P-value
P_value=pchisq(B, df=2, lower.tail= FALSE)
P_value## [1] 0.04006084From chi-square test, P-value is significant(0.04006084 < 0.05). It means we have strong evidence to reject the Null hypothesis which states the distribution of number of transactions as poisson. In other words, Number of transactions do not follow poisson distribution or are not consistent with poisson distribution.
For next test i.e. Linear regression, i make a dataset including number of transactions.
Transaction_data1 =read.csv("C:/Users/ASUS/Desktop/C.COMPUTATIONAL STATISTICS//portfolio data 2N.csv")
Transaction_data1Test 5 Linear regression model I want to check whether my ‘number of transactions’ depend upon ‘deposit’ and ‘amount’ or not. Here number of transactions(dependent) are response and other variables are predictor(independent).Here i considered ‘deposit’ as predictor of ‘number of transaction’ because it is common human nature to make more purchases after getting salary or any other deposit. I just want to check this behavior is found in me as well or not. To do so, i will first plot the the graph of predictors and response but it is difficult to draw a scatter plot with more than one independent variables.
#Linear regression model
model <- lm(Number_of_transactions ~ deposit + amount, data =Transaction_data1 )
# Summary of model
summary(model)##
## Call:
## lm(formula = Number_of_transactions ~ deposit + amount, data = Transaction_data1)
##
## Residuals:
## Min 1Q Median 3Q Max
## -0.6703 -0.1668 -0.1476 0.1147 0.8048
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 1.1312366 0.0951612 11.888 1.61e-10 ***
## deposit -0.0016302 0.0007776 -2.096 0.048979 *
## amount 0.0051094 0.0011139 4.587 0.000179 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.3959 on 20 degrees of freedom
## Multiple R-squared: 0.5127, Adjusted R-squared: 0.464
## F-statistic: 10.52 on 2 and 20 DF, p-value: 0.0007546hist(residuals(model), col='blue')
Discussion The model summary make it clear that both ‘deposit’ and
‘amount’ are statistically significant. The deposit variable, with an
estimated coefficient of approximately -0.0016, suggests that as deposit
amounts increase, there is a negative association with the number of
transactions, implying that higher deposits are associated with slightly
fewer transactions which is opposite to my belief. Conversely, the
positive coefficient for the amount variable (approximately 0.0051)
indicates that an increase in the transaction amount is associated with
a higher number of transactions. The model as a whole is statistically
significant, with an F-statistic of 10.52 and a low p-value, suggesting
that the combined effect of deposit and amount is a meaningful predictor
of transaction numbers. The R-squared value of 0.5127 indicates that
51.27% of the variance in the number of transactions can be explained by
these two variables, indicating a moderate fit of the model to the
data.I need more data to get good fit of model and a proper
histogram.