Question

If $n=10,\sum x=50,\sum y=-30,\sum x^2=290,\sum y^2=300,\sum xy=-115$, then find the coefficient of correlation.

Answer 1

Hint: First find the standard deviation of $x$ by the formula ${\sigma }_x=\sqrt{{\displaystyle \frac{\sum x^2}{n}}-{\left({\displaystyle \frac{\sum x}{n}}\right)}^2}$. Similarly, find the standard deviation of $y$. After that find the covariance by the formula $Cov\left(x,y\right)={\displaystyle \frac{\sum xy}{n}}-{\displaystyle \frac{\sum x}{n}}\cdot {\displaystyle \frac{\sum y}{n}}$. Then use the values to calculate the coefficient of correlation by the formula ${\displaystyle \frac{Cov\left(x,y\right)}{{\sigma }_x{\sigma }_y}}$.

Complete Step by Step Solution:
The correlation coefficient is a statistical measure of the strength in the relationship between the relative movements of two variables. The values range between -1 and 1. If the calculated number is greater than 1 or less than -1 means that there is an error in the correlation measurement. A correlation of -1 shows a perfect negative correlation, while a correlation of 1 shows a perfect positive correlation. A correlation of 0 shows no linear relationship between the movement of the two variables.
To calculate the product-moment correlation, one must first determine the covariance of the two variables in question. Next, one must calculate each variable's standard deviation. The correlation coefficient is determined by dividing the covariance by the product of the two variables' standard deviations.
Now, calculate the standard deviation of $x$ by the formula,
${\sigma }_x=\sqrt{{\displaystyle \frac{\sum x^2}{n}}-{\left({\displaystyle \frac{\sum x}{n}}\right)}^2}$
Substitute the values,
$\Rightarrow {\sigma }_x=\sqrt{{\displaystyle \frac{290}{10}}-{\left({\displaystyle \frac{50}{10}}\right)}^2}$
Cancel out the common factors,
$\Rightarrow {\sigma }_x=\sqrt{29-{\left(5\right)}^2}$
Square the term and subtract from 29,
$\Rightarrow {\sigma }_x=\sqrt{4}$
So, the standard deviation of $x$ is,
$\therefore {\sigma }_x=2$
Now, calculate the standard deviation of $y$ by the formula,
${\sigma }_y=\sqrt{{\displaystyle \frac{\sum y^2}{n}}-{\left({\displaystyle \frac{\sum y}{n}}\right)}^2}$
Substitute the values,
$\Rightarrow {\sigma }_y=\sqrt{{\displaystyle \frac{300}{10}}-{\left({\displaystyle \frac{-30}{10}}\right)}^2}$
Cancel out the common factors,
$\Rightarrow {\sigma }_y=\sqrt{30-{\left(-3\right)}^2}$
Square the term and subtract from 30,
$\Rightarrow {\sigma }_y=\sqrt{21}$
So, the standard deviation of $y$ is,
$\therefore {\sigma }_y=4.58$
Now calculate the covariance by the formula,
$Cov\left(x,y\right)={\displaystyle \frac{\sum xy}{n}}-{\displaystyle \frac{\sum x}{n}}\cdot {\displaystyle \frac{\sum y}{n}}$
Substitute the values,
$\Rightarrow Cov\left(x,y\right)={\displaystyle \frac{-115}{10}}-{\displaystyle \frac{50}{10}}\cdot {\displaystyle \frac{-30}{10}}$
Cancel out the common factors,
$\Rightarrow Cov\left(x,y\right)=-11.5+5\times 3$
Simplify the term,
$\Rightarrow Cov\left(x,y\right)=3.5$
Now, calculate the coefficient of correlation by the formula,
${\displaystyle \frac{Cov\left(x,y\right)}{{\sigma }_x{\sigma }_y}}$
Substitute the values,
$\Rightarrow$Coefficient of correlation $={\displaystyle \frac{3.5}{2\times 4.58}}$
Multiply the terms in denominator,
$\Rightarrow$Coefficient of correlation $={\displaystyle \frac{3.5}{9.16}}$
Convert the fraction into a decimal,
$\therefore$Coefficient of correlation $=0.38$

Hence, the coefficient of correlation is $0.38$.

Note: Correlation coefficients are used to measure the strength of the relationship between two variables. Pearson correlation is the one most commonly used in statistics. This measures the strength and direction of a linear relationship between two variables that we use in this question.