Introduction

The study below will seek to understand the if there is any statistical significance between body parts. For instance, the diameter of the chest can depend on how tall the person is. However, there can be other factors which can affect this measurement. Therefore, by knowing all this factors we will have a better understanding of our chest diameter is dependant on height.

In this study, we will seek to understand if there is a statistical significant relationship between a respondent’s chest and height. We have gathered a dataset of body girth measurements and skeletal diameter measurements, as well as age, weight, height, and gender. We analyse these datasets and apply statistical tests and techniques to determine factors which the chest measurements depend on.

Problem Statement

The question that we have to ask is “Can we predict the chest diameter of a person with their height measurement?”. This investigation will seek to understand if there is any statistical significant relationship between a person’s chest diameter (che.di) and height (hgt).

\(Chest Diameter =\beta (Height) + \alpha\)

We will use Linear Regression to find out if any such relation exists and can be predicted as govern by the equation, \(y=\beta x + \alpha\) where \(\alpha\) is the intercept, \(\beta\) is the slope \(y\) is the dependent variable and \(x\) is independent variable.

Data

Data Preprocessing

• In this section, we clean the dataset in order to do statistical analysis.
• We check the structure of the data and subset the variable that are of interest in this investigation.
• Extract on relevant variables from the dataset (i.e. Chest Diameter and height)
• Detecting and identify the missing values and outliers in the dataset and eliminating them if they exists.

Data cont.

Import Data

body <- read.csv("bdims.csv")

Subsetting the Data

We subset the data to obtain only the relevant variable to our study which are:

• che.di - Respondent’s chest diameter in centimeters, measured at nipple level, mid-expiration.
• hgt - Respondent’s height in centimeters.

body_inves <- body[,c("che.di","hgt")]
head(body_inves)

  che.di   hgt
1   28.0 174.0
2   30.8 175.3
3   31.7 193.5
4   28.2 186.5
5   29.4 187.2
6   31.3 181.5

Descriptive Statistics and Visualisation

Summary Statistics

Summary Statistics for chest diameter of respondants

body_inves %>% summarise(Min = min(body_inves$che.di,na.rm = TRUE),
                  Q1 = quantile(body_inves$che.di,probs = .25,na.rm = TRUE),
                  Median = median(body_inves$che.di, na.rm = TRUE),
                  Q3 = quantile(body_inves$che.di,probs = .75,na.rm = TRUE),
                  Max = max(body_inves$che.di,na.rm = TRUE),
                  Mean = mean(body_inves$che.di, na.rm = TRUE) %>% round(3),
                  SD = sd(body_inves$che.di, na.rm = TRUE) %>% round(3),
                  n = n(),
                  Missing = sum(is.na(body_inves$che.di)))

   Min    Q1 Median    Q3  Max   Mean    SD   n Missing
1 22.2 25.65   27.8 29.95 35.6 27.974 2.742 507       0

Summary Statistics for height of respondants

body_inves %>% summarise(Min = min(body_inves$hgt,na.rm = TRUE),
                  Q1 = quantile(body_inves$hgt,probs = .25,na.rm = TRUE),
                  Median = median(body_inves$hgt, na.rm = TRUE),
                  Q3 = quantile(body_inves$hgt,probs = .75,na.rm = TRUE),
                  Max = max(body_inves$hgt,na.rm = TRUE),
                  Mean = mean(body_inves$hgt, na.rm = TRUE) %>% round(3),
                  SD = sd(body_inves$hgt, na.rm = TRUE) %>% round(3),
                  n = n(),
                  Missing = sum(is.na(body_inves$hgt)))

    Min    Q1 Median    Q3   Max    Mean    SD   n Missing
1 147.2 163.8  170.3 177.8 198.1 171.144 9.407 507       0

Descriptive Statistics and Visualisation cont.

Scatter plot

plot(che.di ~ hgt, data = body_inves, ylab = "Chest Diameter (cm)", xlab = "Height (cm)")

Hypothesis Testing

In this study, we will be using the F-test for Linear Regression.

The hypothesis for the overall Linear Regression Model.

• \(H_0\) : The data does not fit a linear regression model.
• \(H_A\) : The data does fit a linear regression model.

Assumptions:

• Independence (check research design)
• Linearity (check scatter plot)
• Normality of residuals (check after model is fitted)
• Homoscedasticity (check after model is fitted)

Hypothesis Testing cont.

Linear regression models are fitted using the lm() function.

lin_model <- lm(che.di ~ hgt, data = body_inves)
lin_model %>% summary()

Call:
lm(formula = che.di ~ hgt, data = body_inves)

Residuals:
    Min      1Q  Median      3Q     Max 
-6.3102 -1.4326 -0.0696  1.4168  6.8929 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept)  -3.2947     1.7319  -1.902   0.0577 .  
hgt           0.1827     0.0101  18.082   <2e-16 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 2.138 on 505 degrees of freedom
Multiple R-squared:  0.393, Adjusted R-squared:  0.3918 
F-statistic:   327 on 1 and 505 DF,  p-value: < 2.2e-16

Further confirming the p-value:

pf(q = 327, 1, 505, lower.tail = FALSE)

[1] 1.00343e-56

With the p-value we see that it is p < 0.001. Thus, as the test is statistically significant, we reject \(H_0\).

Hypothesis Testing cont. - Interpreting the Intercept

The intercepting point is at \(\alpha = -3.2947\). The intercept is the average value for Chest Diameter when height=0. We confrim the statistical significance of the intercept.

• \(H_0\) : \(\alpha = 0\)
• \(H_A\) : \(\alpha \neq 0\)

Assumptions:

• Same as overall regression model.

lin_model %>% confint()

                 2.5 %    97.5 %
(Intercept) -6.6972252 0.1079121
hgt          0.1628512 0.2025541

Since the 95% Confidence Interval does not capture \(H_0\) and p-value < 0.001, we reject \(H_0\).

Hypothesis Testing cont. - Interpreting the Slope

The slope of the regression line was reported as \(\beta=0.1827\). The slope represents the average increase in Chest Diameter following a one unit increase in height. The hypothesis test of the slope, \(\beta\), was as follows:

• \(H_0\) : \(\beta = 0\)
• \(H_A\) : \(\beta \neq 0\)

Assumptions:

• Same as overall regression model.

lin_model %>% summary() %>% coef()

##               Estimate Std. Error   t value     Pr(>|t|)
## (Intercept) -3.2946566  1.7318756 -1.902363 5.769227e-02
## hgt          0.1827027  0.0101042 18.081859 1.017694e-56

2*pt(q = 18.08, df = 507-2, lower.tail = FALSE)

## [1] 1.038739e-56

Hypothesis Testing cont. - Interpreting the Slope

Calculating two-tailed p-value for the slope, we confirm that p < 0.001. We reject the \(H_0\). The 95% CI for \(\beta\) to be [0.163, 0.203]. This 95% CI does not capture \(H_0\), therefore it was rejected. Hence, there was a statistically significant positive relationship between the chest diameter and the height of the respondants.

Hypothesis Testing Cont. - Model Fitting

We now plot line of best fit on the linear regression model

plot(che.di ~ hgt, data = body_inves, ylab = "chest diameter", xlab = "height")
abline(lin_model, col=2, lw=2)

Hypothesis Testing Cont. - Testing the Assumptions

par(mfrow = c(2,2))
plot(lin_model)

Hypothesis Testing Cont. - Testing the Assumptions

• Residual Vs Fitted: Trend line is closed to flat which means that there are no signs of non-linearity. However, the variable line along the y-axis is not constant across the range of values of the x-axis, which indicates the signs of heteroscedasity.
• Normal Q-Q: Majority of the residuals fall on the line, showing that the residuals follow a normal distribution.
• Scale-Location: The red line line does not appear to be flat. Hence, indicating that there is homoscedasticity.
• Residual Vs Leveraged: Majority of the residual falls within the Cook’s Distance. However, there a very small amount of residuals that fall outside the Cook Distance. Since, the dataset is large enough, those values can be ignored.

Hypothesis Testing Cont. - Correlation Test

A Pearson’s correlation was calculated to measure the strength of the linear relationship between Chest Diameter and Height of the Respondant.

library(psychometric)

r=cor(body_inves$che.di, body_inves$hgt)
r

## [1] 0.6268931

CIr(r = r, n = 507, level = .95)

## [1] 0.5709813 0.6770164

Therefore, r=0.627. 95% CI [0.571, 0.677]. This confidence interval does not capture \(H_0\). Therefore, \(H_0\) was rejected. There was a statistically significant positive correlation between Chest Diameter and Height of the Respondant.

Discussion

Results:

• Linearity was assumed,
• Residuals was normally distributed.
• Existence of Homoscedasticity.
• No influential cases.

Decisions:

• Overall model: Reject \(H_0\).
• Intercept: Reject \(H_0\).
• Slope: Reject \(H_0\).

Hence, we concluded that there was a statistically significant positive linear relationship between a Chest Diameter and Height of the Respondant.

Discussion cont.

Interpretations are as follows:

• A linear regression model was fitted to predict the dependent variable, Chest Diameter, using Heights of the Respondants.
• Prior to fitting the regression, a scatter plot assessing the bivariate relationship between Chest Diameter and Height of the Respondant. The scatter plot demonstrated evidence of a positive linear relationship.
• The overall regression model was statistically significant, F(1,505) = 327], p < 0.001, and explained 39.3% of the variability in Chest Diameter, \(R^2\)=0.393.
• The estimated regression equation was Chest Diameter = 0.1827*Height - 3.2947 following the equation \(y=\beta x+\alpha\) where \(\alpha\) is the intercept and \(\beta\) is the slope.
• Intercept \(\alpha= -3.2947\) at 95% CI [-6.697 0.108] was statistically significant.
• The positive slope for height was statistically significant, \(\beta=0.1827\), t(505) = 18.08, p < 0.001, 95% CI [0.163, 0.203].
• Finally, the inspection of the residuals supported normality but not homoscedasticity.

References

• https://drive.google.com/drive/u/1/folders/0Bwtqn_QygJ8_SGxMR1RmWTVWTGM
• https://astral-theory-157510.appspot.com/secured/MATH1324_Module_09.html#linear_regression