1. Student Details

Rashbir Singh Kohli (s3810585)

2. Problem Statement

For this assignment the student is required to choose only one measurement from the given data having 25 variables and 507 observations.
After selecting the variable of choice we have to determine that does it fits a normal distribution or not seperately for male and female.
For this assignment I have taken variable names ‘bic.gi’ (Bicep Girth) and created different dataframe for both the gender groups.
For the normal distribution fitting I have used an histogram and shown the density distribution. along with normal distribution.

3. Load Packages

For this assignment two libraries are used:

data.table - It is an sub package of package named data.frame. It helps in fast reading of large datasets, is smart and smarty manages all the data points, identify seperator and do not import string as factors by default. Makes over all running of code faster.
ggpubr - visualisation library based on ggplot.

library('data.table')
library("ggpubr")

4. Data

For reading the bdims.csv data I have used fread() function as it has a greater reading speed and can smartly manage the variables.
After reading the data I created seperate data frames for each gender group by filtering on the sex column of the data DataFrame, where 1 denotes male and 0 denotes female.

data <- fread('bdims.csv')
df <- transform(data[, c('bic.gi', 'sex')], sex=factor(sex))
df_male <- df[sex == 1, 'bic.gi']
df_female <- df[sex == 0, 'bic.gi']

5. Summary Statistics

For the purpose of calculating the summary statistics (i.e., mean, median, standard deviation, first and third quartile, interquartile range, minimum and maximum values) for the assignment, I used summary() function and stored the data seperately for both the genders into a DataFrame.

5.1 Creating Female information dataframe

infoDfFe <- data.frame(trimws(sub(".*:", "", summary(df_female)[1:length(summary(df_female))])), metric <- c('Min', '1Q', 'Median', 'Mean', '3Q', 'Max'), stringsAsFactors = FALSE)
names(infoDfFe) <- c('Value', 'metric')
infoDfFe <- transform(infoDfFe, Value = as.numeric(Value))

5.2 Creating Male information dataframe

infoDfMale <- data.frame(trimws(sub(".*:", "", summary(df_male)[1:length(summary(df_male))])), metric <- c('Min', '1Q', 'Median', 'Mean', '3Q', 'Max'), stringsAsFactors = FALSE)
names(infoDfMale) <- c('Value', 'metric')
infoDfMale <- transform(infoDfMale, Value = as.numeric(Value))

5.3 Adding standard deviation

Calculated the Standar Deviation using sd() function.
Stored the result into the DataFrame.

infoDfFe <- rbind(infoDfFe, list(sd(df_female$bic.gi), 'Std'))
infoDfMale <- rbind(infoDfMale, list(sd(df_male$bic.gi), 'Std'))

5.4 Adding IQR ( IQR = Q3 - Q1)

Calculated the Inter Quartile Range using IQR() function.
Stored the result into the DataFrame.

infoDfFe <- rbind(infoDfFe, list(IQR(df_female$bic.gi), 'IQR'))
infoDfMale <- rbind(infoDfMale, list(IQR(df_male$bic.gi), 'IQR'))

5.5 OUTPUT

The following section shows the Summary Statistics for the ‘bic.gi’ (Bicep Girth) of Male and Female

5.5.1 Female

For Female Summary Statistics are as follows:
1. Mean - 28.1
2. Median - 27.8
3. Standard Deviation - 2.709477
4. First Quartile - 26.4
5. Third Quartile - 29.8
6. Interquartile Range - 3.4
7. Minimum value - 22.4
8. Maximum value - 40.3

cat(' Mean value for female: ',infoDfFe$Value[infoDfFe$metric == 'Mean'], '\n', 'Median value for female: ',infoDfFe$Value[infoDfFe$metric == 'Median'], '\n', 'Standard Deviation value for female: ',infoDfFe$Value[infoDfFe$metric == 'Std'], '\n', 'First Quartile value for female: ',infoDfFe$Value[infoDfFe$metric == '1Q'], '\n', 'Third Quartile value for female: ',infoDfFe$Value[infoDfFe$metric == '3Q'], '\n', 'Interquartile Range value for female: ',infoDfFe$Value[infoDfFe$metric == 'IQR'], '\n', 'Minimum value for female: ',infoDfFe$Value[infoDfFe$metric == 'Min'], '\n', 'Maximum value for female: ',infoDfFe$Value[infoDfFe$metric == 'Max'])

##  Mean value for female:  28.1 
##  Median value for female:  27.8 
##  Standard Deviation value for female:  2.709477 
##  First Quartile value for female:  26.4 
##  Third Quartile value for female:  29.8 
##  Interquartile Range value for female:  3.4 
##  Minimum value for female:  22.4 
##  Maximum value for female:  40.3

5.5.2 Male

For Male Summary Statistics are as follows:
1. Mean - 34.4
2. Median - 34.4
3. Standard Deviation - 2.982037
4. First Quartile - 32.5
5. Third Quartile - 36.4
6. Interquartile Range - 3.9
7. Minimum value - 25.6
8. Maximum value - 42.4

cat(' Mean value for male: ',infoDfMale$Value[infoDfMale$metric == 'Mean'], '\n', 'Median value for male: ',infoDfMale$Value[infoDfMale$metric == 'Median'], '\n', 'Standard Deviation value for male: ',infoDfMale$Value[infoDfMale$metric == 'Std'], '\n', 'First Quartile value for male: ',infoDfMale$Value[infoDfMale$metric == '1Q'], '\n', 'Third Quartile value for male: ',infoDfMale$Value[infoDfMale$metric == '3Q'], '\n', 'Interquartile Range value for male: ',infoDfMale$Value[infoDfMale$metric == 'IQR'], '\n', 'Minimum value for male: ',infoDfMale$Value[infoDfMale$metric == 'Min'], '\n', 'Maximum value for male: ',infoDfMale$Value[infoDfMale$metric == 'Max'])

##  Mean value for male:  34.4 
##  Median value for male:  34.4 
##  Standard Deviation value for male:  2.982037 
##  First Quartile value for male:  32.5 
##  Third Quartile value for male:  36.4 
##  Interquartile Range value for male:  3.9 
##  Minimum value for male:  25.6 
##  Maximum value for male:  42.4

6. Distribution Fitting

Histogram is used to aggregate the data into bins of different sizes.
Histogram is a one dimentional plot that shows the aggregated frequency.
The green dotted line shows the actual normal distribution curve.
The blue curev shows the density curve of the sample data set having ‘bic.gi’ (Bicep Girth) as the variable of interest.
The X-Axis shows the variable of interest i.e ‘bic.gi’ (Bicep Girth).
The Y-Axis shows the frequency of the each value on X-Axis.

6.1 Distribution Fitting for Female

hist(df_female$bic.gi, breaks = 20, probability = TRUE, xlab = 'Bicep girth', ylab = 'Frequency', main = 'Histogram for Female Bicep Girth')
lines(density(df_female$bic.gi), col = 'Blue', lwd=2)
curve(dnorm(x, mean=mean(df_female$bic.gi), sd=sd(df_female$bic.gi)), yaxt="n", lty="dotted", col="darkgreen", lwd=4, add=TRUE)
op <- par(cex = 0.7)
legend("topright", legend = c("Density Curve for Female Sample", "Normal Curve"), bty = "n", text.col = "black", horiz = F, pch=c(15,16), col = c('Blue', "darkgreen"))

6.2 Distribution Fitting for Male

hist(df_male$bic.gi, breaks = 20, probability = TRUE, xlab = 'Bicep girth', ylab = 'Frequency', main = 'Histogram for Male Bicep Girth')
lines(density(df_male$bic.gi), col = 'Blue', lwd=2)
curve(dnorm(x, mean=mean(df_male$bic.gi), sd=sd(df_male$bic.gi)), add=TRUE, yaxt="n", lty="dotted", col="darkgreen", lwd=4)
op <- par(cex = 0.7)
legend("topright", legend = c("Density Curve for Male Sample", "Normal Curve"), bty = "n", text.col = "black", horiz = F, pch=c(15,16), col = c('Blue', "darkgreen"), )

7. Interpretation

7.1 Interpretation based on section 5 and 6

From section 5.5.1 that shows summary statistics for bicep girth of females it can be interpreted that:
1. Mean is greater than Median it shows that data is Right Skewed.
2. Standard Deviation is 2.709477, which shows that data is less spread.
3. First Quartile is 26.4 that shows that 25% data is less than 26.4.
4. Third Quartile is 29.8 that shows that 75% data is less than 29.8.
5. Interquartile Range value is 3.4, it shows that the Spread is less
From section 5.5.2 that shows summary statistics for bicep girth of males it can be interpreted that:
1. Mean is equal to the Median it shows that data is Symmetric.
2. Standard Deviation is 2.982037, which shows that data is less spread.
3. First Quartile is 32.5 that shows that 25% data is less than 32.5.
4. Third Quartile is 36.4 that shows that 75% data is less than 36.4.
5. Interquartile Range value is 3.9, it shows that the Spread is less
From section 6.1 that shows the histogram distribution along with the density curve and normal curve it can be interpreted that:
1. The histogram is Right Skewed.
2. The histogram have Less spread that shows lower standard deviation.
3. Single Higher peak showes where the Mean is and is close to that of bell curve shape.
4. It can also be interpretated that majority of data lies between 25 to 30, that resembles the interpretation of section 5.5.1.
5. According to 68-95-99.7 rule for normal distribution:
  - For this histogram 69.62% values lies between 1 standard deviation. distance of mean.
  - For this histogram 95.38% values lies between 2 standard deviation. distance of mean.
  - For this histogram 99.23% values lies between 3 standard deviation. distance of mean.
6. It can be said that bicep girth for females Closely Follow the Rules of Normal Distribution.

print('68-95-99.7 rule')

## [1] "68-95-99.7 rule"

(length(df_female$bic.gi[df_female$bic.gi > (28.1 - 2.709477) & df_female$bic.gi < (28.1 + 2.709477)]) / length(df_female$bic.gi)) * 100

## [1] 69.61538

(length(df_female$bic.gi[df_female$bic.gi > (28.1 - 2*(2.709477)) & df_female$bic.gi < (28.1 + 2*(2.709477))]) / length(df_female$bic.gi)) * 100

## [1] 95.38462

(length(df_female$bic.gi[df_female$bic.gi > (28.1 - 3*(2.709477)) & df_female$bic.gi < (28.1 + 3*(2.709477))]) / length(df_female$bic.gi)) * 100

## [1] 99.23077

From section 6.2 that shows the histogram distribution along with the density curve and normal curve it can be interpreted that:
1. The histogram is Symmetric.
2. The histogram have Less spread that shows lower standard deviation.
3. Single Higher peak showes where the Mean is and is close to that of **bell curve shape*.
4. It can also be interpretated that majority of data lies between 31 to 38, that resembles the interpretation of section 5.5.2.
5. According to 68-95-99.7 rule for normal distribution:
  - For this histogram roughly 68.02% values lies between 1 standard deviation. distance of mean.
  - For this histogram roughly 96.36% values lies between 2 standard deviation. distance of mean.
  - For this histogram roughly 100% values lies between 3 standard deviation. distance of mean.
6. It can be said that bicep girth for males Follow the Rules of Normal Distribution.

print('68-95-99.7 rule')

## [1] "68-95-99.7 rule"

(length(df_male$bic.gi[df_male$bic.gi > (34.4 - 2.982037) & df_male$bic.gi < (34.4 + 2.982037)]) / length(df_male$bic.gi)) * 100

## [1] 68.01619

(length(df_male$bic.gi[df_male$bic.gi > (34.4 - 2*(2.982037)) & df_male$bic.gi < (34.4 + 2*(2.982037))]) / length(df_male$bic.gi)) * 100

## [1] 96.35628

(length(df_male$bic.gi[df_male$bic.gi > (34.4 - 3*(2.982037)) & df_male$bic.gi < (34.4 + 3*(2.982037))]) / length(df_male$bic.gi)) * 100

## [1] 100

7.2 Interpretation based on Q-Q Plot

Another way of testing the normality of data is using QQ plot.
QQ plot is a scatter plot with diagnol line.
If the data is normal then all the scatter points lies on the diagnol line.

7.2.1 Q-Q Plot for female

Some of the point are away from the diagnol line in the top-right corner, that shows the right skewness in section 6.1.
The majority of the scatter is close to the diagnol line so it is safe to say that bicep girth for females Closely Follow the Rules of Normal Distribution.

ggqqplot(df_female$bic.gi, main = '                                  QQ Plot for Female Bicep Girth')

7.2.2 Q-Q Plot for male

The majority of the scatter is close to the diagnol line so it is safe to say that bicep girth for males Follow the Rules of Normal Distribution.

ggqqplot(df_male$bic.gi, main = '                                 QQ Plot for Male Bicep Girth')

MATH1324 Assignment 1

Modeling Body Measurements