Chapter 16.6, Problem 113E

a:

To determine

Regression equation.

Explanation of Solution

The general regression equation can be written as follows:

$\widehat{\text{y}}={\text{b}}_{\text{0}}+{\text{b}}_{\text{1}}\text{x}$

The term $\widehat{\text{y}}$ is the dependent variable (estimated value of the dependent value), the term b₀ is the indicates intercept, the term x indicates the independent variable, and the term b₁ indicates the coefficient of the independent variable.

Table 1 shows the processing of data as follows:

Table 1

Sl	${\text{x}}_{\text{i}}$	${\text{y}}_{\text{i}}$	${\text{x}}_{\text{i}}^2$	${\text{y}}_{\text{i}}^2$	${\text{x}}_{\text{i}}\times {\text{y}}_{\text{i}}$
1	-5	15	25	225	-75
2	-2	9	4	81	-18
3	0	7	0	49	0
4	3	6	9	36	18
5	4	4	16	16	16
6	7	1	49	1	7
Total	7	42	103	408	-52

Standard deviation $\left({\text{s}}_{\text{xy}}\right)$ can be calculated as follows:

$\begin{array}{c}{\text{s}}_{\text{xy}}=\frac{\text{1}}{\text{n}-\text{1}}\left[{\displaystyle \sum _{\text{i}=\text{1}}^{\text{n}}{\text{x}}_{\text{i}}{\text{y}}_{\text{i}}}-\frac{{\displaystyle \sum _{\text{i}=\text{1}}^{\text{n}}{\text{x}}_{\text{i}}}{\displaystyle \sum _{\text{i}=\text{1}}^{\text{n}}{\text{y}}_{\text{i}}}}{\text{n}}\right]\\ =\frac{1}{6-1}\left[-52-\frac{(7)(42}{6}\right]\\ =-20.20\end{array}$

Standard deviation is -202.2.

The variance of x $\left({\text{s}}_{\text{x}}^{\text{2}}\right)$ can be calculated as follows:

$\begin{array}{c}{\text{s}}_{\text{x}}^{\text{2}}=\frac{\text{1}}{\text{n}-\text{1}}\left[{\displaystyle \sum _{\text{i}=\text{1}}^{\text{n}}{\text{x}}_{\text{i}}^{\text{2}}}-\frac{{\left({\displaystyle \sum _{\text{i}=\text{1}}^{\text{n}}{\text{x}}_{\text{i}}}\right)}^{\text{2}}}{\text{n}}\right]\\ =\frac{1}{6-1}\left[103-\frac{{(7)}^2}{6}\right]\\ =18.97\end{array}$

The variance of x is 18.97.

The value of coefficient ( b₁) can be calculated as follows:

$\begin{array}{c}{\text{b}}_{\text{1}}=\frac{{\text{s}}_{\text{xy}}}{{\text{s}}_{\text{x}}^{\text{2}}}\\ =\frac{-20.2}{18.97}\\ =-1.065\end{array}$

The value of b₁ is -1.065.

The average value $\left(\overline{\text{x}}\right)$ of the independent variable x can be calculated as follows:

$\begin{array}{c}\overline{\text{x}}=\frac{{\displaystyle \sum {\text{x}}_{\text{i}}}}{\text{n}}\\ =\frac{7}{6}\\ =1.167\end{array}$

The average value of x is 1.167.

The average value $\left(\overline{\text{y}}\right)$ of the independent variable x can be calculated as follows:

$\begin{array}{c}\overline{\text{y}}=\frac{{\displaystyle \sum {\text{y}}_{\text{i}}}}{\text{n}}\\ =\frac{42}{6}\\ =7\end{array}$

The average value of x is 7.

Intercept b₀ can be calculated as follows:

$\begin{array}{c}{\text{b}}_{\text{0}}=\overline{\text{y}}-{\text{b}}_{\text{1}}\overline{\text{x}}\\ \text{=}7–\left(–1.065\right)\left(1.167\right)\\ =8.253\end{array}$

The intercept value is 8.253.

Thus, the regression equation is given below:

$\widehat{\text{y}}=\text{8}\text{.253}-\text{1}\text{.065x}$

b:

To determine

The predicted value of y.

Explanation of Solution

The estimated value of y $\left(\widehat{\text{y}}\right)$ can be calculated using the below equation:

$\widehat{\text{y}}=\text{8}\text{.253}-\text{1}\text{.065x}$

Table 2 shows the predicted value of x that is obtained using the regression equation with different levels of x values as follows:

Table 2

Sl	${\text{x}}_{\text{i}}$	$\widehat{\text{y}}=\text{8}\text{.253}-\text{1}\text{.065x}$
1	-5	13.58
2	-2	10.38
3	0	8.253
4	3	5.058
5	4	3.993
6	7	0.758

c:

To determine

Calculate the residual.

Explanation of Solution

The value of residual $\left({\text{e}}_{\text{i}}\right)$ can be calculated using the below equation:

${\text{e}}_{\text{i}}={\text{y}}_{\text{i}}-{\widehat{\text{y}}}_{\text{i}}$        (1)

Table 3 shows the residual value that is obtained using Equation (1) as follows:

Table 3

Sl	${\text{x}}_{\text{i}}$	$\widehat{\text{y}}=\text{8}\text{.253}-\text{1}\text{.065x}$	${\text{e}}_{\text{i}}={\text{y}}_{\text{i}}-{\widehat{\text{y}}}_{\text{i}}$
1	-5	13.58	1.42
2	-2	10.38	-1.38
3	0	8.253	-1.253
4	3	5.058	0.942
5	4	3.993	0.007
6	7	0.758	0.202

d:

To determine

Calculate the standardized residual.

Explanation of Solution

The variance of y $\left({\text{s}}_{\text{y}}^{\text{2}}\right)$ can be calculated as follows:

$\begin{array}{c}{\text{s}}_{\text{y}}^{\text{2}}=\frac{\text{1}}{\text{n}-\text{1}}\left[{\displaystyle \sum _{\text{i}=\text{1}}^{\text{n}}{\text{y}}_{\text{i}}^{\text{2}}}-\frac{{\left({\displaystyle \sum _{\text{i}=\text{1}}^{\text{n}}{\text{y}}_{\text{i}}}\right)}^{\text{2}}}{\text{n}}\right]\\ =\frac{1}{6-1}\left[408-\frac{{(42)}^2}{6}\right]\\ =22.80\end{array}$

The variance of y is 22.8.

Error sum of square (SSE) can be calculated as follows:

$\begin{array}{c}\text{SSE}=\text{(n}-\text{1)}\left({\text{s}}_{\text{y}}^{\text{2}}-\frac{{\text{s}}_{\text{xy}}^{\text{2}}}{{\text{s}}_{\text{x}}^{\text{2}}}\right)\\ =(6-1)\left(22.80-\frac{{(20.20)}^2}{18.97}\right)\\ =6.451\end{array}$

Error sum of square is 6.451.

The value of ${\text{s}}_{\text{ε}}$ can be calculated as follows:

$\begin{array}{c}{\text{s}}_{\text{ε}}=\sqrt{\frac{\text{SSE}}{\text{n}-\text{2}}}\\ =\sqrt{\frac{6.451}{6-2}}\\ =1.270\end{array}$

The value of ${\text{s}}_{\text{ε}}$ is 1.27.

The value of standardized residual (se) can be calculated using the below equation:

$\text{se}=\left(\frac{{\text{e}}_{\text{i}}}{{\text{s}}_{\text{ε}}}\right)$        (2)

Table 5 shows the standardized residual value that is obtained using Equation (2) as follows:

Table 5

Sl	${\text{x}}_{\text{i}}$	$\widehat{\text{y}}=\text{8}\text{.253}-\text{1}\text{.065x}$	${\text{e}}_{\text{i}}={\text{y}}_{\text{i}}-{\widehat{\text{y}}}_{\text{i}}$	se
1	-5	13.58	1.42	1.18
2	-2	10.38	-1.38	-1.087
3	0	8.253	-1.253	-0.987
4	3	5.058	0.942	0.742
5	4	3.993	0.007	0.0055
6	7	0.758	0.202	0.159

e:

To determine

Identification of outliers.

Explanation of Solution

From the above calculations, it is known that there are no outliers.