Chapter 4.4, Problem 85E

(a)

To determine

Determine the covariance and coefficient of correlation.

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:

${\text{x}}_{\text{i}}$	${\text{y}}_{\text{i}}$	${\text{x}}_{\text{i}}{}^2$	${\text{y}}_{\text{i}}{}^2$	${\text{x}}_{\text{i}}{\text{y}}_{\text{i}}$
20	14	400	196	280
40	16	1600	256	640
60	18	3600	324	1080
50	17	2500	289	850
50	18	2500	324	900
55	18	3025	324	990
60	18	3600	324	1080
70	20	4900	400	1400
${\displaystyle \sum _{\text{i}=1}^{\text{n}}{\text{x}}_{\text{i}}}=405$	${\displaystyle \sum _{\text{i}=1}^{\text{n}}{\text{y}}_{\text{i}}}=139$	${\displaystyle \sum _{\text{i}=1}^{\text{n}}{\text{x}}_{\text{i}}{}^2}=22,125$	${\displaystyle \sum _{\text{i}=1}^{\text{n}}{\text{y}}_{\text{i}}{}^2}=2,437$	${\displaystyle \sum _{\text{i}=1}^{\text{n}}{\text{x}}_{\text{i}}{\text{y}}_{\text{i}}}=7,220$

Covariance can be calculated as follows:

$\begin{array}{c}{s}_{xy}=\frac{1}{n-1}\left[{\displaystyle \sum _{i=1}^n{x}_i{y}_i}-\frac{{\displaystyle \sum _{i=1}^n{x}_i}{\displaystyle \sum _{i=1}^n{y}_i}}{n}\right]\\ =\frac{1}{8-1}\left[7,220-\frac{(405)(139)}{8}\right]\\ =26.16\end{array}$

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

$\begin{array}{c}{s}_x^2=\frac{1}{n-1}\left[{\displaystyle \sum _{i=1}^n{x}_i^2}-\frac{{\left({\displaystyle \sum _{i=1}^n{x}_i}\right)}^2}{n}\right]\\ =\frac{1}{8-1}\left[22,125-\frac{{(405)}^2}{8}\right]\\ =231.7\end{array}$

$\begin{array}{c}{s}_x=\sqrt{s_x^2}\\ =\sqrt{231.7}\\ =15.22\end{array}$

Similarly,

$\begin{array}{c}{s}_y^2=\frac{1}{n-1}\left[{\displaystyle \sum _{i=1}^n{y}_i^2}-\frac{{\left({\displaystyle \sum _{i=1}^n{y}_i}\right)}^2}{n}\right]\\ =\frac{1}{8-1}\left[2,437-\frac{{(139)}^2}{8}\right]\\ =3.13\end{array}$

$\begin{array}{c}{s}_y=\sqrt{s_y^2}\\ =\sqrt{3.13}\\ =1.77\end{array}$

$\begin{array}{c}\text{Coefficient of correlation}=\frac{{\text{Covariance}}_{\text{xy}}}{{\text{Standard deviation}}_{\text{x}}{\text{Standard deviation}}_{\text{y}}}\\ =\frac{{\text{S}}_{\text{xy}}}{{\text{S}}_{\text{x}}{\text{S}}_{\text{y}}}\\ =\frac{26.16}{\left(15.22\right)\left(1.77\right)}\\ =\frac{26.16}{26.9394}\\ =0.97106\end{array}$

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

$\begin{array}{c}\text{Coefficient of determination}={\left(\text{Coefficient of correlation}\right)}^2\\ ={\left(0.9711\right)}^2\\ =0.9430\end{array}$

Thus, covariance is calculated to be 26.16 and the coefficient of correlation is calculated to be 0.9711. The coefficient of determination is also calculated, and it is 0.9430. These values indicate that 94.30 percent of variation in variable ‘y’ is explained by the variations in variable ‘x’.

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 least squares line and estimate the retailer wants.

Explanation of Solution

Covariance is calculated to be 26.16, and the value of ${\text{S}}^2{}_{\text{x}}$ is 231.7. Thus, the value of b1 can be calculated as follows:

$\begin{array}{c}{\text{b}}_1=\frac{{\text{S}}_{\text{xy}}}{{\text{S}}^2{}_{\text{x}}}\\ =\frac{26.16}{231.7}\\ =0.113\end{array}$

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

$\begin{array}{c}\overline{\text{X}}=\frac{{\displaystyle \sum {\text{X}}_{\text{i}}}}{\text{n}}\\ =\frac{405}{8}\\ =50.63\end{array}$

Similarly,

$\begin{array}{c}\overline{\text{Y}}=\frac{{\displaystyle \sum {\text{Y}}_{\text{i}}}}{\text{n}}\\ =\frac{139}{8}\\ =17.38\end{array}$

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

$\begin{array}{c}{\text{b}}_0=\overline{\text{Y}}-{\text{b}}_1\overline{\text{X}}\\ =17.38-\left(0.113\times 50.63\right)\\ =11.66\end{array}$

The least square line will thus be $\widehat{\text{y}}=11.66+0.113\text{x}$. This means that the estimated variable cost is 0.113 and the estimated fixed cost is 11.66.