Chapter 4.4, Problem 66E

(a)

To determine

Determine the 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:

${\text{x}}_{\text{i}}$	${\text{y}}_{\text{i}}$	${\text{x}}_{\text{i}}{}^2$	${\text{y}}_{\text{i}}{}^2$	${\text{x}}_{\text{i}}{\text{y}}_{\text{i}}$
40	77	1600	5929	3080
42	63	1764	3969	2646
37	79	1369	6241	2923
47	86	2209	7396	4041
25	51	625	2601	1276
44	78	1936	6084	3432
41	83	1681	6889	3403
48	90	2304	8100	4320
35	65	1225	4225	2275
28	47	748	2209	1316
${\displaystyle \sum _{\text{i}=1}^{\text{n}}{\text{x}}_{\text{i}}}=387$	${\displaystyle \sum _{\text{i}=1}^{\text{n}}{\text{y}}_{\text{i}}}=719$	${\displaystyle \sum _{\text{i}=1}^{\text{n}}{\text{x}}_{\text{i}}{}^2}=15,497$	${\displaystyle \sum _{\text{i}=1}^{\text{n}}{\text{y}}_{\text{i}}{}^2}=53,643$	${\displaystyle \sum _{\text{i}=1}^{\text{n}}{\text{x}}_{\text{i}}{\text{y}}_{\text{i}}}=28,712$

Covariance can be calculated as follows:

$\begin{array}{c}{s}_{xy}=\frac{1}{n-1}\left[{\displaystyle \sum _{i=1}^n{x}_i{y}_i}-\frac{{\displaystyle \sum _{i=1}^n{x}_i}{\displaystyle \sum _{i=1}^n{y}_i}}{n}\right]\\ =\frac{1}{10-1}\left[28,712-\frac{(387)(719)}{10}\right]\\ =98.52\end{array}$

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

$\begin{array}{c}{s}_x^2=\frac{1}{n-1}\left[{\displaystyle \sum _{i=1}^n{x}_i^2}-\frac{{\left({\displaystyle \sum _{i=1}^n{x}_i}\right)}^2}{n}\right]\\ =\frac{1}{10-1}\left[15,497-\frac{{(387)}^2}{10}\right]\\ =57.79\end{array}$

$\begin{array}{c}{s}_y^2=\frac{1}{n-1}\left[{\displaystyle \sum _{i=1}^n{y}_i^2}-\frac{{\left({\displaystyle \sum _{i=1}^n{y}_i}\right)}^2}{n}\right]\\ =\frac{1}{10-1}\left[53,643-\frac{{(719)}^2}{10}\right]\\ =216.32\end{array}$

Economics Concept Introduction

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

(b)

To determine

Determine the coefficient of correlation.

Explanation of Solution

The coefficient of correlation can be calculated by dividing the covariance with the product of the two sample deviations as follows:

$\begin{array}{c}\text{Coefficient of correlation}=\frac{{\text{Covariance}}_{\text{xy}}}{{\text{Standard deviation}}_{\text{x}}{\text{Standard deviation}}_{\text{y}}}\\ =\frac{{\text{S}}_{\text{xy}}}{{\text{S}}_{\text{x}}{\text{S}}_{\text{y}}}\\ =\frac{98.52}{\sqrt{\left(57.79\right)\left(216.32\right)}}\\ =0.8811\end{array}$

Thus, the coefficient of correlation is calculated to be 0.8811.

(c)

To determine

Determine the coefficient of determination.

Explanation of Solution

The coefficient of determination can be calculated by squaring the coefficient of correlation as follows:

$\begin{array}{c}\text{Coefficient of determination}={\left(\text{Coefficient of correlation}\right)}^2\\ ={\left(0.8811\right)}^2\\ =0.7763\end{array}$

Thus, the coefficient of determination is calculated to be 0.7763.

(d)

To determine

Determine the least squares line.

Explanation of Solution

The covariance is calculated to be 98.52 and the value of ${\text{S}}^2{}_{\text{x}}$ is 57.79. Thus, the value of b₁ can be calculated as follows:

$\begin{array}{c}{\text{b}}_1=\frac{{\text{S}}_{\text{xy}}}{{\text{S}}^2{}_{\text{x}}}\\ =\frac{98.52}{57.79}\\ =1.705\end{array}$

The values of $\overline{\text{X}}$ and $\overline{\text{Y}}$ can be calculated as follows:

$\begin{array}{c}\overline{\text{X}}=\frac{{\displaystyle \sum {\text{X}}_{\text{i}}}}{\text{n}}\\ =\frac{387}{10}\\ =38.7\end{array}$

Similarly,

$\begin{array}{c}\overline{\text{Y}}=\frac{{\displaystyle \sum {\text{Y}}_{\text{i}}}}{\text{n}}\\ =\frac{719}{10}\\ =71.9\end{array}$

The value of ${\text{b}}_0$ can be calculated as follows:

$\begin{array}{c}{\text{b}}_0=\overline{\text{Y}}-{\text{b}}_1\overline{\text{X}}\\ =71.9-\left(1.705\times 38.7\right)\\ =5.917\end{array}$

The least square line will thus be $\widehat{\text{y}}=5.917+1.705\text{x}$.

(e)

To determine

Determine the indication of the least square line.

Explanation of Solution

The results show that there is a very strong positive linear relationship that exists between the two variables taken to analysis. This means that there is a strong positive linear relationship between the time taken to study and the marks obtained. Each additional hour for the study leads to an increase in the marks, and it is on an average by 1.705.