FINAL EXAM PROJECT
Mustafa Hassan Dahir Ali
Question ONE : Data Structures in R
* Instruction
- Import the SDHS data set into R.
# Setup
##Install and load the necessary packages to reproduce the report
library(haven) # Useful for importing SPSS, SAS, STATA etc. data files
library(foreign) # Useful for importing SPSS, SAS, STATA etc. data filesImport the STATA file into R
R CODE
library(haven)
SLDHS<-read_dta("~/SLDHS12.dta")- Display the structure of the data set using the str() function.’
R CODE
str(SLDHS$V106)## dbl+lbl [1:17686] 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,...
## @ label : chr "Highest educational level"
## @ format.stata: chr "%1.0f"
## @ labels : Named num [1:4] 0 1 2 3
## ..- attr(*, "names")= chr [1:4] "No Education" "Primary" "Secondary" "Higher"- Identify and list all the variables provided in the variable code list.
R CODE
# Loading the dplyr package
library(dplyr)
# Select only the variable code list
sLDHS_DATA1<-SLDHS %>%
select("V190", "V024", "V025", "V106", "V152", "V151", "V136", "V201", "V501", "V113", "V116")- Determine the data types of each variable.
R CODE
str(sLDHS_DATA1)## tibble [17,686 x 11] (S3: tbl_df/tbl/data.frame)
## $ V190: dbl+lbl [1:17686] 5, 5, 5, 5, 5, 3, 3, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3...
## ..@ label : chr "Wealth index combined"
## ..@ format.stata: chr "%1.0f"
## ..@ labels : Named num [1:5] 1 2 3 4 5
## .. ..- attr(*, "names")= chr [1:5] "Lowest" "Second" "Middle" "Fourth" ...
## $ V024: dbl+lbl [1:17686] 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, ...
## ..@ label : chr "Region"
## ..@ format.stata: chr "%2.0f"
## ..@ labels : Named num [1:6] 11 12 13 14 15 16
## .. ..- attr(*, "names")= chr [1:6] "Awdal" " Marodijeh" "Sahil" "Togdheer" ...
## $ V025: dbl+lbl [1:17686] 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2...
## ..@ label : chr "Type of place of residence"
## ..@ format.stata: chr "%1.0f"
## ..@ labels : Named num [1:6] 1 2 3 4 5 6
## .. ..- attr(*, "names")= chr [1:6] "Rural" "Urban" "Nomadic" "Rural IDP" ...
## $ V106: dbl+lbl [1:17686] 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0...
## ..@ label : chr "Highest educational level"
## ..@ format.stata: chr "%1.0f"
## ..@ labels : Named num [1:4] 0 1 2 3
## .. ..- attr(*, "names")= chr [1:4] "No Education" "Primary" "Secondary" "Higher"
## $ V152: num [1:17686] 23 23 23 23 23 61 61 23 23 23 ...
## ..- attr(*, "label")= chr "Age of household head"
## ..- attr(*, "format.stata")= chr "%2.0f"
## $ V151: num [1:17686] 1 1 1 1 1 1 1 1 1 1 ...
## ..- attr(*, "label")= chr "Sex of household head"
## ..- attr(*, "format.stata")= chr "%1.0f"
## $ V136: num [1:17686] 6 6 6 6 6 4 4 6 6 6 ...
## ..- attr(*, "label")= chr "Number of household members (listed)"
## ..- attr(*, "format.stata")= chr "%1.0f"
## $ V201: num [1:17686] 5 5 5 5 5 2 2 4 4 4 ...
## ..- attr(*, "label")= chr "Total children ever born"
## ..- attr(*, "format.stata")= chr "%2.0f"
## $ V501: dbl+lbl [1:17686] 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1...
## ..@ label : chr "Current marital status"
## ..@ format.stata: chr "%1.0f"
## ..@ labels : Named num [1:4] 0 1 2 3
## .. ..- attr(*, "names")= chr [1:4] "Never Married" "Married" "Divorced" "Widowed"
## $ V113: num [1:17686] 11 11 11 11 11 13 13 13 13 13 ...
## ..- attr(*, "label")= chr "Source of drinking water"
## ..- attr(*, "format.stata")= chr "%2.0f"
## $ V116: num [1:17686] 23 23 23 23 23 61 61 23 23 23 ...
## ..- attr(*, "label")= chr "Type of toilet facility"
## ..- attr(*, "format.stata")= chr "%2.0f"- briefly explain the difference between numeric and factor variables. #### Numeric variables Numeric variables represent quantities, and can be used in arithmetic operations (e.g., age, height, temperature). #### Factor variables factor variables represent categories or groups, and are typically represented by labels or levels (e.g., gender, color, species). While a factor might use numbers to represent categories, those numbers don’t have inherent mathematical meaning; they’re just labels.
Question TWO : Descriptive Statistics in R
* Instructions:
- Calculate the mean, median and standard deviation for the variable “V136” (number of household members).
mean(sLDHS_DATA1$V136,na.rm = TRUE )## [1] 5.107781median(sLDHS_DATA1$V136, na.rm = TRUE)## [1] 5sd(sLDHS_DATA1$V136, na.rm = TRUE)## [1] 2.475511- Create a function table for the variable “V106” (education level) using the table() function.
R code
table(sLDHS_DATA1$V106)##
## 0 1 2 3
## 15287 1991 311 97- Calculate the proportion of households in each wealth quintile (“V190”).
# Create a Proportion of household for V106
prop.table(table(sLDHS_DATA1$V106))##
## 0 1 2 3
## 0.864355988 0.112574918 0.017584530 0.005484564- Explain how you would use R to calculate the correlation coefficient between age of household head (“V151”) and the number of living children (“V201”)
sum(is.na(sLDHS_DATA1$V152))## [1] 862sum(is.na(sLDHS_DATA1$V201))## [1] 13data_clean <- na.omit(sLDHS_DATA1[, c("V152", "V201")])
correlation <- cor(data_clean$V152, data_clean$V201, method = "pearson")
print(correlation)## [1] 0.04612714# There is weak direct correlation between V152 and V201 it means if V152 increases the V201 increase very little value.Question THREE : Data Cleaning in R
* Instructions:
- Create a new variable called “poverty_status” based on the “V190” variable (wealth quintile) and categorize household into two groups: “Poor”: Poorer, and Middle quintiles ”Non-Poor”: Richer and Richest quintiles
# Load the dplyr package
library(dplyr)
sLDHS_DATA1$poverty_status <- ifelse(sLDHS_DATA1$V190 %in% c(1, 2,3), "Poor",
ifelse(sLDHS_DATA1$V190 %in% c(4, 5), "Non-Poor", NA))
table(sLDHS_DATA1$poverty_status)##
## Non-Poor Poor
## 6293 11393- Check for missing values in all variables using the summary() function.
R CODE
summary(sLDHS_DATA1)## V190 V024 V025 V106
## Min. :1.000 Min. :11.00 Min. :1.000 Min. :0.0000
## 1st Qu.:1.000 1st Qu.:13.00 1st Qu.:1.000 1st Qu.:0.0000
## Median :2.000 Median :14.00 Median :1.000 Median :0.0000
## Mean :2.662 Mean :13.95 Mean :1.498 Mean :0.1642
## 3rd Qu.:4.000 3rd Qu.:15.00 3rd Qu.:2.000 3rd Qu.:0.0000
## Max. :5.000 Max. :16.00 Max. :2.000 Max. :3.0000
##
## V152 V151 V136 V201
## Min. :11.0 Min. :1.000 Min. :0.000 Min. : 0.000
## 1st Qu.:13.0 1st Qu.:1.000 1st Qu.:3.000 1st Qu.: 4.000
## Median :22.0 Median :1.000 Median :5.000 Median : 6.000
## Mean :26.2 Mean :1.376 Mean :5.108 Mean : 6.274
## 3rd Qu.:23.0 3rd Qu.:2.000 3rd Qu.:7.000 3rd Qu.: 8.000
## Max. :96.0 Max. :2.000 Max. :9.000 Max. :19.000
## NA's :862 NA's :858 NA's :2303 NA's :13
## V501 V113 V116 poverty_status
## Min. :1.000 Min. :11.00 Min. :11.0 Length:17686
## 1st Qu.:1.000 1st Qu.:12.00 1st Qu.:13.0 Class :character
## Median :1.000 Median :31.00 Median :22.0 Mode :character
## Mean :1.138 Mean :33.44 Mean :26.2
## 3rd Qu.:1.000 3rd Qu.:61.00 3rd Qu.:23.0
## Max. :3.000 Max. :96.00 Max. :96.0
## NA's :862 NA's :862- Recode the variables “V113” (source of drinking water) and “V116” (toilet facility) into “Improved” and “Unimproved” categories based on the definitions provided in the DHS data.
R CODE
## Recode V116 into new variable 'toilet_facility' using base R
## Recode V116 into "improved" and "unimproved"
sLDHS_DATA1$toilet_facility <- NA
sLDHS_DATA1$toilet_facility[sLDHS_DATA1$V116 %in% c(11, 12, 13, 21, 22, 31)] <- "improved"
sLDHS_DATA1$toilet_facility[sLDHS_DATA1$V116 %in% c(14, 15, 23, 41, 51, 61, 96)] <- "unimproved"
table(sLDHS_DATA1$toilet_facility)##
## improved unimproved
## 9520 7304R CODE
#Recode V113 into a new variable 'Source_of_drinking_water' using base R
#Recode V113 into "improved" and "unimproved"
sLDHS_DATA1$Source_of_drinking_water <- NA
# Initialize the 'Source_of_drinking_water' variable as NA
sLDHS_DATA1$Source_of_drinking_water[sLDHS_DATA1$V113 %in% c(11, 12, 13, 14, 21, 51, 61, 71, 72)] <- "improved"
sLDHS_DATA1$Source_of_drinking_water[sLDHS_DATA1$V113 %in% c(32, 42, 81, 96)] <- "unimproved"
table(sLDHS_DATA1$Source_of_drinking_water)##
## improved unimproved
## 12355 2311- Explain how you would handle missing values in your dataset.
R CODE
# To handle missing values in dataset you may follow the below steps:
any(is.na(sLDHS_DATA1))## [1] TRUEsum(is.na(sLDHS_DATA1))## [1] 9642Question FOUR : Data Visualization in R
* Instructions:
- create a histogram to show the distribution of the variable “V136” (number of household members).
R CODE
library(ggplot2)
hist(sLDHS_DATA1$V136, breaks = 5 )
- Create a bar chart to visualize the proportion of households in each poverty status category (“poverty_status”).
R CODE
barplot(table(sLDHS_DATA1$poverty_status))
2. Create a boxplot to compare the number of living children (“V201”)
between poor and non-poor households (“poverty_status”)
R CODE
boxplot(sLDHS_DATA1$V201,main="SLDHS$poverty_status",col="blue")
- Briefly explain the importance of choosing appropriate visualization techniques for different types of data.
Choosing the right visualization in R (or any data visualization tool) is crucial for effective communication. Different visualization techniques highlight different aspects of the data, and using the wrong one can mislead the audience or obscure important information.
Histograms:
Best for showing the distribution of a single numeric variable. They display the frequency of data points falling within specific ranges (bins). Useful for identifying patterns like skewness, modality (number of peaks), and outliers.
Bar charts:
Best for comparing the frequencies or values of different categorical (factor) variables or groups. Each bar represents a category, and its height shows its corresponding value. Avoid using bar charts for continuous numeric data.
Boxplots:
Excellent for comparing the distribution of a numeric variable across different categorical (factor) groups. They show the median, quartiles, and potential outliers for each group, providing a concise summary of the central tendency, spread, and potential extreme values