Chapter 16.4, Problem 22E

(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	1	1	1	1	1
2	3	8	9	64	24
3	4	15	16	225	60
4	6	33	36	1089	198
5	9	75	81	5625	675
6	8	70	64	4900	560
7	10	95	100	9025	950
Total	41	297	307	20,929	2,468

These table values are obtained using a spreadsheet.

$\begin{array}{c}{\displaystyle \sum _{\text{i=1}}^{\text{n}}{\text{x}}_{\text{i}}}=41,{\displaystyle \sum _{\text{i=1}}^{\text{n}}{\text{y}}_{\text{i}}}=297\\ {\displaystyle \sum _{\text{i=1}}^{\text{n}}{\text{x}}_{\text{i}}^{\text{2}}}=307,{\displaystyle \sum _{\text{i=1}}^{\text{n}}{\text{y}}_{\text{i}}^{\text{2}}}=20,929\\ {\displaystyle \sum _{\text{i=1}}^{\text{n}}{\text{x}}_{\text{i}}{\text{y}}_{\text{i}}}=2,468\\ {\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}{7-1}\left[2,468-\frac{(41)(297}{7}\right]=121.4\\ {\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}{7-1}\left[307-\frac{{(41)}^2}{7}\right]=11.14\\ {\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}{7-1}\left[20,929-\frac{{(297)}^2}{7}\right]=1,388.0\\ {\text{b}}_{\text{1}}\text{=}\frac{{\text{s}}_{\text{xy}}}{{\text{s}}_{\text{x}}^{\text{2}}} =\frac{121.4}{11.14}=10.90\\ \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)=(7-1)\left(1,388.0-\frac{{(121.4)}^2}{11.14}\right)=390.1\\ {\text{s}}_{\text{ε}}\text{=}\sqrt{\frac{\text{SSE}}{\text{n}-\text{2}}}=\sqrt{\frac{390.1}{7-2}}=8.83\\ {\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{8.83}{\sqrt{(7-1)(11.14)}}=1.08\\ \text{t}=\frac{{\text{b}}_{\text{1}}-{\text{β}}_{\text{1}}}{{\text{s}}_{{\text{b}}_{\text{1}}}}=\frac{10.90-0}{1.08}=10.09\end{array}$

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