Chapter 16.4, Problem 23E

(a)

To determine

Draw the scatter diagram.

Explanation of Solution

Figure 1 shows the scatter diagram of Y and X.

In the above figure, the position of the scatters shows a positive relationship between variables Y and X.

Economics Concept Introduction

Scatter diagram: The scatter diagram shows the value of two variables plotted along two axes. Each scatter represents the relationship between the variables.

(b)

To determine

Calculate the linear relationship.

Explanation of Solution

The least squares regression line is shown below:

$\stackrel{\wedge }{\text{y}}={\text{b}}_0+{\text{b}}_1\text{x}$

Where b₀ is the y-intercept and b₁ is the slope of the regression line.

The least squares regression line is calculated as follows:

Table 1

	Xi	Yi	Xi2	Yi2	XiY1
1	3	25	9	625	75
2	5	110	25	12100	550
3	2	9	4	81	18
4	6	250	36	62500	1500
5	1	3	1	9	3
6	4	71	16	5041	284
Total	21	468	91	80,356	2,430

These table values are obtained using a spreadsheet.

$\begin{array}{c}{\displaystyle \sum _{\text{i=1}}^{\text{n}}{\text{x}}_{\text{i}}}=21,{\displaystyle \sum _{\text{i=1}}^{\text{n}}{\text{y}}_{\text{i}}}=486\\ {\displaystyle \sum _{\text{i=1}}^{\text{n}}{\text{x}}_{\text{i}}^{\text{2}}}=91,{\displaystyle \sum _{\text{i=1}}^{\text{n}}{\text{y}}_{\text{i}}^{\text{2}}}=80,356\\ {\displaystyle \sum _{\text{i=1}}^{\text{n}}{\text{x}}_{\text{i}}{\text{y}}_{\text{i}}}=2,430\\ {\text{s}}_{\text{xy}}=\frac{\text{1}}{\text{n}-\text{1}}\left[{\displaystyle \sum _{\text{i=1}}^{\text{n}}{\text{x}}_{\text{i}}{\text{y}}_{\text{i}}}-\frac{{\displaystyle \sum _{\text{i=1}}^{\text{n}}{\text{x}}_{\text{i}}}{\displaystyle \sum _{\text{i=1}}^{\text{n}}{\text{y}}_{\text{i}}}}{\text{n}}\right]\\ =\frac{1}{6-1}\left[2,430-\frac{(21)(468)}{6}\right]=158.4\\ {\text{s}}_{\text{x}}^{\text{2}}=\frac{\text{1}}{\text{n}-\text{1}}\left[{\displaystyle \sum _{\text{i=1}}^{\text{n}}{\text{x}}_{\text{i}}^{\text{2}}}-\frac{{\left({\displaystyle \sum _{\text{i=1}}^{\text{n}}{\text{x}}_{\text{i}}}\right)}^{\text{2}}}{\text{n}}\right]\\ =\frac{1}{6-1}\left[91-\frac{{(21)}^2}{6}\right]=3.50\\ {\text{s}}_{\text{y}}^{\text{2}}\text{=}\frac{\text{1}}{\text{n}-\text{1}}\left[{\displaystyle \sum _{\text{i=1}}^{\text{n}}{\text{y}}_{\text{i}}^{\text{2}}}-\frac{{\left({\displaystyle \sum _{\text{i=1}}^{\text{n}}{\text{y}}_{\text{i}}}\right)}^{\text{2}}}{\text{n}}\right]\\ =\frac{1}{6-1}\left[80,356-\frac{{(468)}^2}{6}\right]=8,770\\ {\text{b}}_{\text{1}}\text{=}\frac{{\text{s}}_{\text{xy}}}{{\text{s}}_{\text{x}}^{\text{2}}} =\frac{158.4}{3.5}=45.26\\ \text{SSE}=\text{(n}-\text{1)}\left({\text{s}}_{\text{y}}^{\text{2}}-\frac{{\text{s}}_{\text{xy}}^{\text{2}}}{{\text{s}}_{\text{x}}^{\text{2}}}\right)\\ =(6-1)\left(8,770-\frac{{(158.4)}^2}{3.50}\right)=8,006\\ {\text{s}}_{\text{ε}}\text{=}\sqrt{\frac{\text{SSE}}{\text{n}-\text{2}}}=\sqrt{\frac{8,002}{6-2}}=844.74\\ {\text{H}}_{\text{0}}:{\text{β}}_{\text{1}}={\text{0, H}}_{\text{1}}:{\text{β}}_{\text{1}}\text{¹0}\\ {\text{s}}_{{\text{b}}_{\text{1}}}\text{=}\frac{{\text{s}}_{\text{ε}}}{\sqrt{\text{(n}-{\text{1)s}}_{\text{x}}^{\text{2}}}}=\frac{44.74}{\sqrt{(6-1)(3.50)}}=10.69\\ \text{t}=\frac{{\text{b}}_{\text{1}}-{\text{β}}_{\text{1}}}{{\text{s}}_{{\text{b}}_{\text{1}}}}=\frac{45.26-0}{10.69}=4.23\end{array}$

This calculation is enough evidence to conclude that variables X and Y are in a linear relationship.