CRYPTOGRAPHY ADVERTISING

## 1a). Defining the Question

###–>> Which individuals are more likely to click on adverts on cryptography?

b). Defining the Metric of Success

###–>> The project will be considered a success when we can identify which individuals will click on the advert.

c). Understanding the Context

###–>> A Kenyan entrepreneur has created an online cryptography course and would want to advertise it on her blog. She currently targets audiences originating from various countries. In the past, she ran ads to advertise a related course on the same blog and collected data in the process. She would now like to employ your services as a Data Science Consultant to help her identify which individuals are most likely to click on her ads.

d). Recording the Experimental Design

###–>> (i) Find and deal with outliers, anomalies, and missing data within the dataset. (ii) Perform uni variate and bivariate analysis. (iii) From your insights provide a conclusion and recommendation

e). Data Relevance

###–>> The data is valid and has been provided by the entrepreneur, it was collected from the previous adverts.

# 2. Reading the data

let’s import the dataset

adverts <- read.csv("advertising.csv")

# 3. Checking the Data

let’s preview the top 6 records of the dataset

head(adverts)

let’s check the last 6 records of the dataset

tail(adverts)

let’s see the shape of our dataset

dim(adverts)
## [1] 1000   10

###–>> The dataframe has 1000 observations and 10 variables

let’s see the data types of the variables

str(adverts)
## 'data.frame':    1000 obs. of  10 variables:
##  $ Daily.Time.Spent.on.Site: num  69 80.2 69.5 74.2 68.4 ...
##  $ Age                     : int  35 31 26 29 35 23 33 48 30 20 ...
##  $ Area.Income             : num  61834 68442 59786 54806 73890 ...
##  $ Daily.Internet.Usage    : num  256 194 236 246 226 ...
##  $ Ad.Topic.Line           : chr  "Cloned 5thgeneration orchestration" "Monitored national standardization" "Organic bottom-line service-desk" "Triple-buffered reciprocal time-frame" ...
##  $ City                    : chr  "Wrightburgh" "West Jodi" "Davidton" "West Terrifurt" ...
##  $ Male                    : int  0 1 0 1 0 1 0 1 1 1 ...
##  $ Country                 : chr  "Tunisia" "Nauru" "San Marino" "Italy" ...
##  $ Timestamp               : chr  "2016-03-27 00:53:11" "2016-04-04 01:39:02" "2016-03-13 20:35:42" "2016-01-10 02:31:19" ...
##  $ Clicked.on.Ad           : int  0 0 0 0 0 0 0 1 0 0 ...

###–>> R stores the dataframe and views the variables as lists so to see the the various data types of this list we use the str function.

let’s check for duplicates in the dataframe

duplicates <- adverts[duplicated(adverts), ]
duplicates

–>> The dataframe does not contain duplicate values.

let’s check for missing data in each column

colSums(is.na(adverts))
## Daily.Time.Spent.on.Site                      Age              Area.Income 
##                        0                        0                        0 
##     Daily.Internet.Usage            Ad.Topic.Line                     City 
##                        0                        0                        0 
##                     Male                  Country                Timestamp 
##                        0                        0                        0 
##            Clicked.on.Ad 
##                        0

###–>> The dataset’s columns does not have missing data.

let’s check for outliers in the dataset

selecting only numeric columns

num_cols <- adverts[,unlist(lapply(adverts, is.numeric))]
num_cols

###–>> 6 columns are numerical in nature

let’s check for outliers in the numerical columns using BOXPLOT

boxplot(num_cols, notch = TRUE)
## Warning in (function (z, notch = FALSE, width = NULL, varwidth = FALSE, : some
## notches went outside hinges ('box'): maybe set notch=FALSE

###–>> The Area.Income variable has outliers which will be imputed.

let’s see the values which are outliers in the Area.Income variable

boxplot.stats(adverts$Area.Income)$out
## [1] 17709.98 18819.34 15598.29 15879.10 14548.06 13996.50 14775.50 18368.57

let’s check for outliers using Z-SCORES

The z-score indicates the number of standard deviations a given value deviates from the mean.

z_scores <- as.data.frame(sapply(num_cols, function(num_cols) (abs(num_cols-mean(num_cols))/sd(num_cols))))
z_scores

###–>> We will drop values with a Z-Score of more than 3 or -3. They are the outliers

Removing the outliers

no_outliers <- z_scores[!rowSums(z_scores>3), ]
head(no_outliers)

let’s check the number of observations after removing outliers

dim(num_cols)
## [1] 1000    6
dim(no_outliers)
## [1] 998   6

###–>> We removed 2 observations.

let’s check for outliers in the new dataframe after removing them

boxplot(no_outliers)

###–>> There are still outliers so we will use interquantile range method to remove outliers

checking and removing outliers using IQR

The Area.Income column had outliers so we focus on it

# Apply the Interquartile Range, IQR(), function on the area income column
income.IQR <- 65471-47032
income.IQR <-IQR(adverts$`Area.Income`)
income.IQR
## [1] 18438.83

let’s save the dataframe without outliers into a new dataframe by assigning it to a variable

adverts_2 <- subset(adverts, adverts$`Area.Income`> (47032 - 1.5*income.IQR) & adverts$`Area.Income`<(65471 + 1.5*income.IQR))

let’s see the shape of the new dataframe

dim(adverts_2)
## [1] 991  10

###–>> We have lost 9 observations that included the outliers. We proceed with analysis.

### 4. {UNIVARIATE ANALYSIS}

let’s get the mean of the numerical columns

summary(num_cols)
##  Daily.Time.Spent.on.Site      Age         Area.Income    Daily.Internet.Usage
##  Min.   :32.60            Min.   :19.00   Min.   :13996   Min.   :104.8       
##  1st Qu.:51.36            1st Qu.:29.00   1st Qu.:47032   1st Qu.:138.8       
##  Median :68.22            Median :35.00   Median :57012   Median :183.1       
##  Mean   :65.00            Mean   :36.01   Mean   :55000   Mean   :180.0       
##  3rd Qu.:78.55            3rd Qu.:42.00   3rd Qu.:65471   3rd Qu.:218.8       
##  Max.   :91.43            Max.   :61.00   Max.   :79485   Max.   :270.0       
##       Male       Clicked.on.Ad
##  Min.   :0.000   Min.   :0.0  
##  1st Qu.:0.000   1st Qu.:0.0  
##  Median :0.000   Median :0.5  
##  Mean   :0.481   Mean   :0.5  
##  3rd Qu.:1.000   3rd Qu.:1.0  
##  Max.   :1.000   Max.   :1.0

###–>> The summary shows: ###–>> 1. The minimum value for each numerical variable. ###–>> 2. The first quantile for each numerical variable ###–>> 3. The median value for all numeric variables across the dataframe. ###–>> 4. The mean value for all numeric variables. ###–>> 5. The third quantile. ###–>> 6. The maximum value for all numerical columns.

let’s get the variance for the numeric variables

variance <- var(num_cols)
variance
##                          Daily.Time.Spent.on.Site           Age   Area.Income
## Daily.Time.Spent.on.Site              251.3370949 -4.617415e+01  6.613081e+04
## Age                                   -46.1741459  7.718611e+01 -2.152093e+04
## Area.Income                         66130.8109082 -2.152093e+04  1.799524e+08
## Daily.Internet.Usage                  360.9918827 -1.416348e+02  1.987625e+05
## Male                                   -0.1501864 -9.242142e-02  8.867509e+00
## Clicked.on.Ad                          -5.9331431  2.164665e+00 -3.195989e+03
##                          Daily.Internet.Usage        Male Clicked.on.Ad
## Daily.Time.Spent.on.Site         3.609919e+02 -0.15018639 -5.933143e+00
## Age                             -1.416348e+02 -0.09242142  2.164665e+00
## Area.Income                      1.987625e+05  8.86750903 -3.195989e+03
## Daily.Internet.Usage             1.927415e+03  0.61476667 -1.727409e+01
## Male                             6.147667e-01  0.24988889 -9.509510e-03
## Clicked.on.Ad                   -1.727409e+01 -0.00950951  2.502503e-01

###–>> variance is a measure of how far the set of data points per column is spread out from their mean eg. those of the area income seem to be far spread out from their mean when compared to that of the age column.

let’s get the standard deviation of the numeric variables

let’s create a function to get the standard deviations

sd.function <- function(column) {
  standard.deviations <- sd(column)
  print(standard.deviations)
}

standard deviation for daily time spent on site

sd.function(adverts_2$Daily.Time.Spent.on.Site)
## [1] 15.9005

standard deviation for Age

sd.function(adverts_2$Age)
## [1] 8.804716

standard deviation for Area.Income

sd.function(adverts_2$Area.Income)
## [1] 12961.5

standard deviation for Daily.Internet.Usage

sd.function(adverts_2$Daily.Internet.Usage)
## [1] 44.05386

###–>> Where a low standard deviation indicates that values are closer to the mean a high one indicates the standard deviation is far from the mean e.g the age column standard deviation of 8.8 displays that its values are closer to their mean than that of the Area income column whose value is 12961

let’s get the skewness of the numerical column

library(moments)
skewness(num_cols)
## Daily.Time.Spent.on.Site                      Age              Area.Income 
##              -0.37120261               0.47842268              -0.64939670 
##     Daily.Internet.Usage                     Male            Clicked.on.Ad 
##              -0.03348703               0.07605493               0.00000000

###–>> The skewness of the Age variable being positive indicates that its distribution has a longer right tail than left tail while the rest of the columns’ left tails.

### 5. {BIVARIATE ANALYSIS}

let’s get the covariance of the numeric variables

cov(num_cols)
##                          Daily.Time.Spent.on.Site           Age   Area.Income
## Daily.Time.Spent.on.Site              251.3370949 -4.617415e+01  6.613081e+04
## Age                                   -46.1741459  7.718611e+01 -2.152093e+04
## Area.Income                         66130.8109082 -2.152093e+04  1.799524e+08
## Daily.Internet.Usage                  360.9918827 -1.416348e+02  1.987625e+05
## Male                                   -0.1501864 -9.242142e-02  8.867509e+00
## Clicked.on.Ad                          -5.9331431  2.164665e+00 -3.195989e+03
##                          Daily.Internet.Usage        Male Clicked.on.Ad
## Daily.Time.Spent.on.Site         3.609919e+02 -0.15018639 -5.933143e+00
## Age                             -1.416348e+02 -0.09242142  2.164665e+00
## Area.Income                      1.987625e+05  8.86750903 -3.195989e+03
## Daily.Internet.Usage             1.927415e+03  0.61476667 -1.727409e+01
## Male                             6.147667e-01  0.24988889 -9.509510e-03
## Clicked.on.Ad                   -1.727409e+01 -0.00950951  2.502503e-01

###–>> The age variable is the only column with a positive covariance with the ad click variable, the rest have negative covariances.

let’s get the correlation coefficient

cor(num_cols)
##                          Daily.Time.Spent.on.Site         Age  Area.Income
## Daily.Time.Spent.on.Site               1.00000000 -0.33151334  0.310954413
## Age                                   -0.33151334  1.00000000 -0.182604955
## Area.Income                            0.31095441 -0.18260496  1.000000000
## Daily.Internet.Usage                   0.51865848 -0.36720856  0.337495533
## Male                                  -0.01895085 -0.02104406  0.001322359
## Clicked.on.Ad                         -0.74811656  0.49253127 -0.476254628
##                          Daily.Internet.Usage         Male Clicked.on.Ad
## Daily.Time.Spent.on.Site           0.51865848 -0.018950855   -0.74811656
## Age                               -0.36720856 -0.021044064    0.49253127
## Area.Income                        0.33749553  0.001322359   -0.47625463
## Daily.Internet.Usage               1.00000000  0.028012326   -0.78653918
## Male                               0.02801233  1.000000000   -0.03802747
## Clicked.on.Ad                     -0.78653918 -0.038027466    1.00000000

###–>> The variables have a negative correlation with the target variable apart from the age variable which has a positive correlation. Let’s see that in the correlogram below

let’s see the corrplot of the numeric variables

library(corrplot)
## corrplot 0.92 loaded
corr_ <- cor(num_cols)

corrplot(corr_, method = 'color')

### {RECOMMENDATIONS}

a. The entrepreneur should focus on the older population as the correlation between age and advert clicks is slightly positive indicating that as age increases the more likely the clicks are made.

b. The entreoreneur should focus on regions with bigger area coverage as those with a smaller area since the correlation between area income and advert clicks is negatively weak one indicating that as area income decreases the more likely the clicks are made and vice versa.

c. She should focus on the regions with low daily internet usage because the correlation between the daily internet usage and clicks on ads is negative indicating that as internet use decreases the more likely the clicks will be made.