Chapter 4.4, Problem 67E

To determine

Determine covariance.

Explanation of Solution

Covariance is a measure that is used to measure the relationship between the movements of two variables. Thus, covariance measures how changes in one variable are associated with the changes in a second variable. Therefore, covariance measures the degree to which two variables are linearly associated. In this case, covariance can be measured after calculating the square of x and y and the product of xy as follows:

${\text{x}}_{\text{i}}$	${\text{y}}_{\text{i}}$	${\text{x}}_{\text{i}}{}^2$	${\text{y}}_{\text{i}}{}^2$	${\text{x}}_{\text{i}}{\text{y}}_{\text{i}}$
599	9.6	358,801	92.16	5750.4
689	8.8	474,721	77.44	6063.2
584	7.4	341,056	54.76	4321.6
631	10.0	398,161	100.00	6310.0
594	7.8	352,836	60.84	4632.2
643	9.2	413,449	84.64	5915.6
656	9.6	430,336	92.16	6297.6
594	8.4	352,836	70.56	4989.6
710	11.2	504,100	125.44	7952.0
611	7.6	373,321	57.76	4643.6
593	8.8	351,649	77.44	5218.4
683	8.0	466,489	64.00	5464.0
${\displaystyle \sum _{\text{i}=1}^{\text{n}}{\text{x}}_{\text{i}}}=7587$	${\displaystyle \sum _{\text{i}=1}^{\text{n}}{\text{y}}_{\text{i}}}=106.4$	${\displaystyle \sum _{\text{i}=1}^{\text{n}}{\text{x}}_{\text{i}}{}^2}=4817755$	${\displaystyle \sum _{\text{i}=1}^{\text{n}}{\text{y}}_{\text{i}}{}^2}=957.2$	${\displaystyle \sum _{\text{i}=1}^{\text{n}}{\text{x}}_{\text{i}}{\text{y}}_{\text{i}}}=67559.2$

Covariance can be calculated as follows:

$\begin{array}{c}{s}_{xy}=\frac{1}{n-1}\left[{\displaystyle \sum _{i=1}^n{x}_i{y}_i}-\frac{{\displaystyle \sum _{i=1}^n{x}_i}{\displaystyle \sum _{i=1}^n{y}_i}}{n}\right]\\ =\frac{1}{12-1}\left[67,559.2-\frac{(7,587)(106.4)}{12}\right]\\ =26.16\end{array}$

The values of two standard deviations can be calculated as follows:

$\begin{array}{c}{s}_x^2=\frac{1}{n-1}\left[{\displaystyle \sum _{i=1}^n{x}_i^2}-\frac{{\left({\displaystyle \sum _{i=1}^n{x}_i}\right)}^2}{n}\right]\\ =\frac{1}{12-1}\left[4,817,755-\frac{{(7,587)}^2}{12}\right]\\ =1,897.7\end{array}$

$\begin{array}{c}{\text{S}}_{\text{X}}=\sqrt{{\text{S}}^2{}_{\text{X}}}\\ =\sqrt{1897.7}\\ =43.56\end{array}$

Similarly,

$\begin{array}{c}{s}_y^2=\frac{1}{n-1}\left[{\displaystyle \sum _{i=1}^n{y}_i^2}-\frac{{\left({\displaystyle \sum _{i=1}^n{y}_i}\right)}^2}{n}\right]\\ =\frac{1}{12-1}\left[957.2-\frac{{(106.4)}^2}{12}\right]\\ =1.25\end{array}$

$\begin{array}{c}{\text{S}}_{\text{Y}}=\sqrt{{\text{S}}^2{}_{\text{Y}}}\\ =\sqrt{1.25}\\ =1.12\end{array}$

The value of ‘r’ can be calculated as follows:

$\begin{array}{c}\text{Coefficient of correlation}=\frac{{\text{Covariance}}_{\text{xy}}}{{\text{Standard deviation}}_{\text{x}}{\text{Standard deviation}}_{\text{y}}}\\ \text{r}=\frac{{\text{S}}_{\text{xy}}}{{\text{S}}_{\text{x}}{\text{S}}_{\text{y}}}\\ =\frac{26.16}{\left(43.56\right)\left(1.12\right)}\\ =0.5362\end{array}$

The value of coefficient of correlation can be calculated as follows:

$\begin{array}{c}\text{Coefficient of determination}={\left(\text{Coefficient of correlation}\right)}^2\\ ={\left(0.5362\right)}^2\\ =0.2875\end{array}$

The calculation indicates that the covariance of the data set is 26.16 and the coefficient of correlation is 0.5362. The coefficient of determination is 0.2875, which means that 28.75 percent of the variation in MBA GPA is explained by the variation in the GMAT scores.

Economics Concept Introduction

Covariance: Covariance is a measure that is used to measure the relationship between the movements of two variables.