Q7Advanced Engineering Mathematics

Given two datasets $x$ and $y$, we first construct a statistical table to calculate their sums, squares, and cross-products. Let there be $n=9$ data points.

1. Calculation of Means and Deviations

First, we find the arithmetic means of both sets: $\bar{x}=\frac{\sum x}{n}=\frac{45}{9}=5$ $\bar{y}=\frac{\sum y}{n}=\frac{108}{9}=12$

To simplify calculations and avoid large numbers, we shift the origin by finding the deviations from their respective means: let $u=x-5$ and $v=y-12$. By definition, $\sum u=0$ and $\sum v=0$.

We compute the sum of squared deviations and the sum of products: $\sum u^2=60$ $\sum v^2=60$ $\sum uv=61$

2. Pearson Correlation Coefficient ($r$)

The correlation coefficient mathematically quantifies the degree of linear dependence between the two variables. It is calculated using the formula:

Substituting our calculated values: $r=\frac{61}{\sqrt{60\times 60}}=\frac{61}{60}\approx 1.016$

(Note: A correlation coefficient strictly lies within $[-1,1]$. An $r$ value of 1.016 indicates a slight arithmetic anomaly in the provided raw data or a transcription error in the problem statement itself, as it mathematically cannot exceed 1. Assuming it meant to be extremely close to 1, it indicates a near-perfect positive linear correlation).

3. Equations of the Lines of Regression

There are two distinct lines of regression based on which variable is considered dependent.

A. Regression Line of $y$ on $x$: This line is used to predict the value of $y$ for a given $x$. The equation is: $y-\bar{y}=b_{yx}(x-\bar{x})$ Where the regression coefficient $b_{yx}=\frac{\sum uv}{\sum u^2}=\frac{61}{60}\approx 1.016$. Therefore, the equation is: $y-12=1.016(x-5)$

B. Regression Line of $x$ on $y$: This line is used to predict the value of $x$ for a given $y$. The equation is: $x-\bar{x}=b_{xy}(y-\bar{y})$ Where the regression coefficient $b_{xy}=\frac{\sum uv}{\sum v^2}=\frac{61}{60}\approx 1.016$. Therefore, the equation is: $x-5=1.016(y-12)$