Descriptive Statistics

Bivariate Analysis

Introduction

This is our first lab when we are considering 2 dimensions and instead of calculating univariate statistics by groups (or factors) of other variable - we will measure their common relationships based on co-variance and correlation coefficients.

*Please be very careful when choosing the measure of correlation! In case of different measurement scales we have to recode one of the variables into weaker scale.

It would be nice to add some additional plots in the background. Feel free to add your own sections and use external packages.

Data

This time we are going to use a typical credit scoring data with predefined “default” variables and personal demographic and income data. Please take a look closer at headers and descriptions of each variable.

Scatterplots

First let’s visualize our quantitative relationships using scatter plots.

You can also normalize the skewed distribution of incomes using log:

## Warning: Using size for a discrete variable is not advised.

## [1] 0.4251568

## [1] 0.1577567

##    estimate      p.value statistic   n gp  Method
## 1 0.5842774 3.210675e-65   19.0072 700  1 pearson

##    estimate      p.value statistic   n gp  Method
## 1 0.3194263 4.805085e-18  8.899323 700  1 pearson

We can add an estimated linear regression line:

## `geom_smooth()` using formula = 'y ~ x'

Scatterplots by groups

We can finally see if there any differences between risk status:

## `geom_smooth()` using formula = 'y ~ x'

We can also see more closely if there any differences between those two distributions adding their estimated density plots:

## Warning: pakiet 'ggExtra' został zbudowany w wersji R 4.3.3

We can also put those plots together:

Scatterplots with density curves

We can also see more closely if there any differences between those two distributions adding their estimated density plots:

Correlation coefficients - Pearson’s linear correlation

Ok, let’s move to some calculations. In R, we can use the cor() function. It takes three arguments and the method: cor(x, y, method) For 2 quantitative data, with all assumptions met, we can calculate simple Pearson’s coefficient of linear correlation:

## [1] 0.574346

Ok, what about the percentage of the explained variability?

## 32.98734 %

So as we can see almost 32.99% of total log of incomes’ variability is explained by differences in age. The rest 67% is probably explained by other factors.

Partial and semipartial correlation

The partial and semi-partial (also known as part) correlations are used to express the specific portion of variance explained by eliminating the effect of other variables when assessing the correlation between two variables.

Partial correlation holds constant one variable when computing the relations to others. Suppose we want to know the correlation between X and Y holding Z constant for both X and Y. That would be the partial correlation between X and Y controlling for Z.

Semipartial correlation holds Z constant for either X or Y, but not both, so if we wanted to control X for Z, we could compute the semipartial correlation between X and Y holding Z constant for X.

Suppose we want to know the correlation between the log of income and age controlling for years of employment. How highly correlated are these after controlling for tenure?

**There can be more than one control variable.

## Partial Correlation (controlling for years of employment): 0.3194263

## Semipartial Correlation (controlling for years of employment on age): 0.2203711

How can we interpret the obtained partial correlation coefficient? What is the difference between that one and the semi-partial coefficient:

The partial correlation between logged income and age, controlling for years of employment, is 0.3194263 . This value indicates that the unique relationship between age and logged income, independent of the years of employment, is positive , suggesting that older individuals tend to have higher logged incomes, after accounting for their years of employment.

The semipartial correlation, on the other hand, shows how much of the variance in logged income can be explained uniquely by age, after controlling for the effect of years of employment only on the age variable. The semipartial correlation coefficient is 0.2203711. This measure highlights the direct influence of age on logged income without the mixed effects of employment duration on age.

Rank correlation

For 2 different scales - like for example this pair of variables: income vs. education levels - we cannot use Pearson’s coefficient. The only possibility is to rank also incomes… and lose some more detailed information about them.

First, let’s see boxplots of income by education levels.

Now, let’s see Kendal’s coefficient of rank correlation (robust for ties).

## 0.1577567

Point-biserial correlation

Let’s try to verify if there is a significant relationship between incomes and risk status. First, let’s take a look at the boxplot:

## Warning: The following aesthetics were dropped during statistical transformation: fill.
## ℹ This can happen when ggplot fails to infer the correct grouping structure in
##   the data.
## ℹ Did you forget to specify a `group` aesthetic or to convert a numerical
##   variable into a factor?

If you would like to compare 1 quantitative variable (income) and 1 dychotomous variable (default status - binary), then you can use point-biserial coefficient:

## 
##  Pearson's product-moment correlation
## 
## data:  bank$income and as.numeric(bank$default)
## t = -1.8797, df = 698, p-value = 0.06056
## alternative hypothesis: true correlation is not equal to 0
## 95 percent confidence interval:
##  -0.144313546  0.003149751
## sample estimates:
##         cor 
## -0.07096966

The point-biserial correlation coefficient is ~ -0.071. It is negative, so it means that when income increases, likelihood of defaulting on a loan decreases. It makes sense, because people with higher incomes are usually less likely to default. But also, p-value can suggest that this relationship is not so strong as I would thought it is, because it is above level of 0.05.

Nonlinear correlation - eta coefficient

If you would like to check if there are any nonlinearities between 2 variables, the only possibility (beside transformations and linear analysis) is to calculate “eta” coefficient and compare it with the Pearson’s linear coefficient.

## [1] 0.07096966

Eta coefficient is ~ 0.071, so it is the same as previously calculated point-biserial coefficient. It means the relationship between our 2 variables is entirely linear.

Correlation matrix

We can also prepare the correlation matrix for all quantitative variables stored in our data frame.

We can use ggcorr() function:

As you can see - the default correlation matrix is not the best idea for all measurement scales (including binary variable “default”).

That’s why now we can perform our bivariate analysis with ggpair with grouping.

Correlation matrix with scatterplots

Here is what we are about to calculate: - The correlation matrix between age, log_income, employ, address, debtinc, creddebt, and othdebt variable grouped by whether the person has a default status or not. - Plot the distribution of each variable by group - Display the scatter plot with the trend by group

# first correlation matrix still using ggcorr

bank$log_income <- log(bank$income)

data_for_matrix <- c("age", "log_income", "employ", "address", "debtinc", "creddebt", "othdebt")

default_groups <- split(bank, bank$default)

corr_matrix_plots <- lapply(names(default_groups), function(group) {
  data_numeric <- default_groups[[group]][, data_for_matrix, drop = FALSE]
  data_numeric <- na.omit(data_numeric)
  plot <- ggcorr(data_numeric, label = TRUE,  hjust = 1) +
    ggtitle(paste("Correlation matrix: ", group))
  return(plot)
})

# displaying matrices
grid.arrange(grobs = corr_matrix_plots, ncol = 1)

# and now using ggpairs :)

data <- bank[, data_for_matrix]

pairs_plot <- ggpairs(data, 
                      aes(color = as.factor(bank$default), alpha = 0.5),
                      upper = list(continuous = wrap("cor", size = 4)),
                      lower = list(continuous = wrap("points", alpha = 0.5)),
                      diag = list(continuous = wrap("barDiag", fill = "white")))

print(pairs_plot)

## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.

This matrix shows: - scatter plots in the lower triangle - correlation coefficients in the upper triangle - density plots on the diagonal

Qualitative data

In case of two variables measured on nominal or ordinal&nominal scale - we are forced to organize so called “contingency” table with frequencies and calculate some kind of the correlation coefficient based on them. This is so called “contingency analysis”.

Let’s consider one example based on our data: verify, if there is any significant correlation between education level and credit risk.

contingency_table <- table(bank$ed, bank$default)

chi_square_result <- chisq.test(contingency_table)

## Warning in chisq.test(contingency_table): Aproksymacja chi-kwadrat może być
## niepoprawna

print(chi_square_result)

## 
##  Pearson's Chi-squared test
## 
## data:  contingency_table
## X-squared = 11.492, df = 4, p-value = 0.02155

From the result we can assume that these two variables are not independent from each other, but probably more tests are needed to tell exactly how dependent they are.

Exercise 1. Contingency analysis.

Do you believe in the Afterlife? https://nationalpost.com/news/canada/millennials-do-you-believe-in-life-after-life A survey was conducted and a random sample of 1091 questionnaires is given in the form of the following contingency table:

##         Believe
## Gender   Yes  No
##   Female 435 375
##   Male   147 134

Our task is to check if there is a significant relationship between the belief in the afterlife and gender. We can perform this procedure with the simple chi-square statistics and chosen qualitative correlation coefficient (two-way 2x2 table).

## 
##  Pearson's Chi-squared test with Yates' continuity correction
## 
## data:  dane
## X-squared = 0.11103, df = 1, p-value = 0.739

##         Believe
## Gender         Yes        No
##   Female 0.3987168 0.3437214
##   Male   0.1347388 0.1228231

As you can see we can calculate our chi-square statistic really quickly for two-way tables or larger. Now we can standardize this contingency measure to see if the relationship is significant.

## [1] 0.01218871

Exercise 2. Contingency analysis for the ‘Titanic’ data.

Let’s consider the titanic dataset which contains a complete list of passengers and crew members on the RMS Titanic. It includes a variable indicating whether a person did survive the sinking of the RMS Titanic on April 15, 1912. A data frame contains 2456 observations on 14 variables.

The website http://www.encyclopedia-titanica.org/ offers detailed information about passengers and crew members on the RMS Titanic. According to the website 1317 passengers and 890 crew member were aboard.

8 musicians and 9 employees of the shipyard company are listed as passengers, but travelled with a free ticket, which is why they have NA values in fare. In addition to that, fare is truely missing for a few regular passengers.

head(titanic)

##                          Status Disembarked.at Home.Country Age Year.of.Birth
## DE GRASSE, Mr J.                     Cherbourg               NA            NA
## EVANS, Miss                          Cherbourg               NA            NA
## MULLEN,                              Cherbourg               NA            NA
## WOTTON, Mr Henry Swaffin             Cherbourg               54          1858
## BRAND, Mr                            Cherbourg               NA            NA
## FLETCHER, Miss N.                    Cherbourg               NA            NA
##                          Crew.or.Passenger. Gender Class...Department
## DE GRASSE, Mr J.                  Passenger   Male          2nd Class
## EVANS, Miss                       Passenger Female          2nd Class
## MULLEN,                           Passenger Female          2nd Class
## WOTTON, Mr Henry Swaffin          Passenger   Male          1st Class
## BRAND, Mr                         Passenger   Male          1st Class
## FLETCHER, Miss N.                 Passenger Female          1st Class
##                             Embarked     Job               Job.details
## DE GRASSE, Mr J.         Southampton                                  
## EVANS, Miss              Southampton                                  
## MULLEN,                  Southampton                                  
## WOTTON, Mr Henry Swaffin Southampton Butcher Butcher's Shop Proprietor
## BRAND, Mr                Southampton                                  
## FLETCHER, Miss N.        Southampton                                  
##                          Ticket.Number Fare.Price Fare_GBP Fare_today
## DE GRASSE, Mr J.                   761         P1      1.0     82.110
## EVANS, Miss                         88         P1      1.0     82.110
## MULLEN,                            404         P1      1.0     82.110
## WOTTON, Mr Henry Swaffin            86     P1 10s      1.5    123.165
## BRAND, Mr                            8     P1 10s      1.5    123.165
## FLETCHER, Miss N.                  405     P1 10s      1.5    123.165
##                                                                                        Profile.on.Encyclopedia.Titanica
## DE GRASSE, Mr J.                                http://www.encyclopedia-titanica.org/titanic-biography/j-de-grasse.html
## EVANS, Miss                                           http://www.encyclopedia-titanica.org/titanic-biography/evans.html
## MULLEN,                                              http://www.encyclopedia-titanica.org/titanic-biography/mullen.html
## WOTTON, Mr Henry Swaffin http://www.encyclopedia-titanica.org/titanic-cross-channel-passenger/henry-swaffin-wotton.html
## BRAND, Mr                                             http://www.encyclopedia-titanica.org/titanic-biography/brand.html
## FLETCHER, Miss N.                                http://www.encyclopedia-titanica.org/titanic-biography/n-fletcher.html

summary(titanic)

##     Status          Disembarked.at     Home.Country            Age       
##  Length:2456        Length:2456        Length:2456        Min.   : 0.17  
##  Class :character   Class :character   Class :character   1st Qu.:23.00  
##  Mode  :character   Mode  :character   Mode  :character   Median :29.00  
##                                                           Mean   :30.59  
##                                                           3rd Qu.:38.00  
##                                                           Max.   :74.00  
##                                                           NA's   :32     
##  Year.of.Birth  Crew.or.Passenger.    Gender          Class...Department
##  Min.   :1837   Length:2456        Length:2456        Length:2456       
##  1st Qu.:1874   Class :character   Class :character   Class :character  
##  Median :1882   Mode  :character   Mode  :character   Mode  :character  
##  Mean   :1881                                                           
##  3rd Qu.:1889                                                           
##  Max.   :1973                                                           
##  NA's   :32                                                             
##    Embarked             Job            Job.details        Ticket.Number     
##  Length:2456        Length:2456        Length:2456        Length:2456       
##  Class :character   Class :character   Class :character   Class :character  
##  Mode  :character   Mode  :character   Mode  :character   Mode  :character  
##                                                                             
##                                                                             
##                                                                             
##                                                                             
##   Fare.Price           Fare_GBP        Fare_today      
##  Length:2456        Min.   :  1.00   Min.   :   82.11  
##  Class :character   1st Qu.:  7.90   1st Qu.:  648.33  
##  Mode  :character   Median : 14.45   Median : 1186.83  
##                     Mean   : 33.16   Mean   : 2722.71  
##                     3rd Qu.: 31.07   3rd Qu.: 2551.06  
##                     Max.   :512.33   Max.   :42067.35  
##                     NA's   :1136     NA's   :1136      
##  Profile.on.Encyclopedia.Titanica
##  Length:2456                     
##  Class :character                
##  Mode  :character                
##                                  
##                                  
##                                  
## 

# Replace NA in fare with the median fare
titanic$Fare[is.na(titanic$Fare)] <- median(titanic$Fare, na.rm = TRUE)
titanic$Age[is.na(titanic$Age)] <- median(titanic$Age, na.rm = TRUE)

ggplot(titanic, aes(x = Age, fill = factor(Status))) + 
    geom_histogram(binwidth = 5, position = "fill", color = "black") + 
    labs(y = "Proportion", x = "Age", fill = "Status", title = "Survival by Age") +
    theme_minimal()

# contingency table of survival by country
survival_by_country <- table(titanic$Status, titanic$Home.Country)
chisq.test(survival_by_country)

## Warning in chisq.test(survival_by_country): Aproksymacja chi-kwadrat może być
## niepoprawna

## 
##  Pearson's Chi-squared test
## 
## data:  survival_by_country
## X-squared = 1696.2, df = 84, p-value < 2.2e-16

# contingency table of survival by gender
survival_by_gender <- table(titanic$Status, titanic$Gender)
chisq.test(survival_by_gender)

## 
##  Pearson's Chi-squared test
## 
## data:  survival_by_gender
## X-squared = 552.06, df = 2, p-value < 2.2e-16

# contingency table of survival by age
survival_by_age <- table(titanic$Status, titanic$Age)
chisq.test(survival_by_age)

## Warning in chisq.test(survival_by_age): Aproksymacja chi-kwadrat może być
## niepoprawna

## 
##  Pearson's Chi-squared test
## 
## data:  survival_by_age
## X-squared = 308.5, df = 154, p-value = 2.135e-12

From the plot we can see, that there is some small relation between age and surviving - generally there were more younger than older people who survived, but also there is some increase when looking at old people - that were about 60 years old. Looking at chi-squared test results, for every tested variable p-value is extremely low. It means, that there is significant dependence between survival status and those variables.