Chapter 3.4, Problem 64E

Solution

(a)

To calculate : The power regression model for the given data.

The power regression model for the given data is $y=8.0955x^{-0.113}$ .

Given information :

The data:

$x$	$y$
$\begin{array}{l}2\\ 3\\ 4.8\\ 7.7\end{array}$	$\begin{array}{l}7.48\\ 7.14\\ 6.81\\ 6.41\end{array}$

Calculation :

The power regression model is given by:

  $y=a\times x^b$

Use calculator to find the value of $a\text{ and }b$ for the given data.

  $\begin{array}{l}a\approx 8.0955\\ b\approx -0.113\end{array}$

Therefore, the power regression model is $y=8.0955x^{-0.113}$ .

(b)

To calculate : The $y$ value associated with $x=9.2$ using power regression model.

The required value is $6.3$ .

Given information :

The data:

$x$	$y$
$\begin{array}{l}2\\ 3\\ 4.8\\ 7.7\end{array}$	$\begin{array}{l}7.48\\ 7.14\\ 6.81\\ 6.41\end{array}$

The power regression model is $y=8.0955x^{-0.113}$

Calculation :

Consider the power regression model $y=8.0955x^{-0.113}$ .

Substitute $x=9.2$ in the power regression model.

  $\begin{array}{c}y=8.0955{\left(9.2\right)}^{-0.113}\\ \approx 6.3\end{array}$

Therefore, the required value is $6.3$ .

(c)

To express : The data in terms of their natural logarithms and draw a scatter plot of $\left(\ln x,\ln y\right)$ .

The required data is:

$\ln x$	$\ln y$
$\begin{array}{l}0.69\\ 1.10\\ 1.57\\ 2.04\end{array}$	$\begin{array}{l}2.01\\ 1.97\\ 1.92\\ 1.86\end{array}$

Given information :

The data:

$x$	$y$
$\begin{array}{l}2\\ 3\\ 4.8\\ 7.7\end{array}$	$\begin{array}{l}7.48\\ 7.14\\ 6.81\\ 6.41\end{array}$

Calculation :

Consider the given table and express the given data in terms of their natural logarithms.

$\ln x$	$\ln y$
$\begin{array}{l}0.69\\ 1.10\\ 1.57\\ 2.04\end{array}$	$\begin{array}{l}2.01\\ 1.97\\ 1.92\\ 1.86\end{array}$

Plot $\ln x$ on $x-$ axis and $\ln y$ on $y-$ axis and make scatter diagram.

  

(d)

To calculate : The linear regression model $\left(\ln y\right)=a\left(\ln x\right)+b$ for $\left(\ln x,\ln y\right)$ .

The linear regression model is $\ln y=-0.1131\left(\ln x\right)+2.091$ .

Given information :

The data:

$x$	$y$
$\begin{array}{l}2\\ 3\\ 4.8\\ 7.7\end{array}$	$\begin{array}{l}7.48\\ 7.14\\ 6.81\\ 6.41\end{array}$

Calculation :

Consider the table expressed in terms of natural logarithms of given data.

$\ln x$	$\ln y$
$\begin{array}{l}0.69\\ 1.10\\ 1.57\\ 2.04\end{array}$	$\begin{array}{l}2.01\\ 1.97\\ 1.92\\ 1.86\end{array}$

The linear regression model is given by:

  $\left(\ln y\right)=a\left(\ln x\right)+b$

Use calculator to find the value of $a\text{ and }b$ for the given data.

  $\begin{array}{l}a\approx -0.1131\\ b\approx 2.091\end{array}$

Therefore, the linear regression model is $\ln y=-0.1131\left(\ln x\right)+2.091$ .

(e)

To prove : $y=e^b\cdot x^a$ is the power regression model $y=8.0955x^{-0.113}$ .

Given information :

The data:

$x$	$y$
$\begin{array}{l}2\\ 3\\ 4.8\\ 7.7\end{array}$	$\begin{array}{l}7.48\\ 7.14\\ 6.81\\ 6.41\end{array}$

The linear regression model is $\ln y=-0.1131\left(\ln x\right)+2.091$ .

Proof :

Consider the linear regression model $\ln y=-0.1131\left(\ln x\right)+2.091$ .

  $\begin{array}{c}\ln y=-0.1131\left(\ln x\right)+2.091\\ \ln y=\ln x^{-0.1131}+2.091\\ y=e^{\ln x^{-0.1131}+2.091}\\ y=e^{\ln x^{-0.1131}}\times e^{2.091}\\ y=x^{-0.1131}\left(8.0955\right)\\ y=8.0955x^{-0.113}\end{array}$

Therefore, it is proved that $y=e^b\cdot x^a$ is the power regression model $y=8.0955x^{-0.113}$ .