Computational Statistics in Business Analytics - Introduction to Computation

Author: Daniele Melotti
Date: Nov 2023

See the related GitHub repository for code, data and list of tasks.

1. Retrieve Data Points

We import a dataset containing 1 variable, which is the age of a set of customers, and move the ages into a vector called ages.

dataset <- read.table(file = "../data/customers.txt", header = TRUE, colClasses = "numeric")
ages <- dataset$age

Then, we want to determine how many customers (or ages of customers) are present within our data.

length(ages)

## [1] 399

Now we know that there are 399 customers in total. Next, we check what is the age of certain elements of the vector; for instance, the 1st, 5th and 100th element of ages.

ages[c(1, 5, 100)]

## [1] 49 45 34

We see that the the 1st, 5th and 100th are 49, 45, and 34 years old respectively. Now, we are interested in finding the age of the youngest and oldest customer.

min(ages)

## [1] 18

max(ages)

## [1] 85

We can see that the youngest and oldest customers present in the dataset are aged 18 and 85 respectively.

2. Sorting and High-Level Analysis

We sort the ages into ascending order. We can take a peek into the five lowest ages:

sorted_ages <- sort(ages)
sorted_ages[1:5]

## [1] 18 19 19 19 19

As we can see, there is only one customer aged 18, while there are at least four customers aged 19. If we were interested in seeing uniquely the 5 smallest ages without seeing how many customers have said ages, we could do it in the following way:

head(unique(sorted_ages), n = 5)

## [1] 18 19 20 21 22

Hence, we know that the 5 smallest ages are 18, 19, 20, 21 and 22. We repeat the process for the 5 highest ages:

tail(unique(sorted_ages), n = 5)

## [1] 80 81 82 83 85

Next, we compute a statistic of central tendency, the mean:

mean(ages)

## [1] 46.80702

The average age of the customers is 46.807, close to 47. As for the standard deviation:

sd(ages)

## [1] 16.3698

As we can see, the standard deviation is 16.3698, which is roughly a third of the mean.

3. Analysis of Differences

We compute the difference of each age value and the mean:

age_diff <- ages - mean(ages)

The values contained in ages_diff represents deviation from the mean. Now, we compute the average of the differences above.

mean(age_diff)

## [1] -1.623275e-15

The result is a zero. Why is that? Usually, the mean of differences is zero because in theory for every positive deviation, there is always a corresponding negative deviation of equal magnitude. This is the reason why we usually compute standard deviation, which is based on squared differences from the mean rather than raw differences; squaring the differences does two things: 1. It gets rid of negative signs. 2. It gives more weight to larger differences.

4. Visualization

In order to get a better view of how the values of ages are distributed, we can plot a histogram.

hist(ages, main = "Customer Ages Histogram", xlab = "Age", ylab = "Frequency", breaks = 35, col = "cornflowerblue")

The histogram allows us to see that there is a relatively large number of customers aged between 48 and 50. Generally, customers aged between 44 and 50 are more largely present in the dataset compared to other ages. We can also draw a density plot of ages:

plot(density(ages), main = "Customer Ages Histogram", xlab = "Age", ylab = "Density",
     col = "cornflowerblue", lwd = 2)

We can see that the shape of the density plot resembles the histogram. Finally, we show a boxplot with a stripchart:

boxplot(ages, main = "Boxplot + Stripchart", horizontal = TRUE,
        xlab = "Age", ylab = "", col = "cornflowerblue")
stripchart(ages, method = "stack", add = TRUE, col = "darkblue", bg = "lightblue", at = 0.6)

The plot above clearly helps us to visualize plenty of information, such as mean, outliers, data distribution, and more.

Summary

This project is helpful in getting a gentle introduction to data retrieval techniques, measures of central tendency such as the mean and standard deviation, as well as on why we tend to use standard deviation rather than raw deviations.

We analyzed a dataset of customer ages, which includes 399 data entries, where the youngest customer is 18 and the older customer is 85 years old. We found that the mean age is 46.807 and the standard deviation is 16.3698. When computing the average value of raw differences, we noticed that said value was zero, making us point out that it is better to use standard deviation as a measure of dispersion. Finally, we displayed a histogram, a density plot, and a boxplot with stripchart, which are all informative and allowed us to get a grasp of the shape of the data distribution and other important statistics.