Chapter 3.4, Problem 63E

Solution

(a)

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

The power regression model for the given data is $y=2.75x^5$ .

Given information :

The data:

$x$	$y$
$\begin{array}{l}4\\ 6.5\\ 8.5\\ 10\end{array}$	$\begin{array}{l}2816\\ 31,098\\ 122,019\\ 275,000\end{array}$

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.

  $\begin{array}{l}a\approx 2.75\\ b\approx 5\end{array}$

Therefore, the power regression model is $y=2.75x^5$ .

(b)

To calculate : The $y$ value associated with $x=7.1$ using power regression model.

The required value is $49,616$ .

Given information :

The data:

$x$	$y$
$\begin{array}{l}4\\ 6.5\\ 8.5\\ 10\end{array}$	$\begin{array}{l}2816\\ 31,098\\ 122,019\\ 275,000\end{array}$

The power regression model is $y=2.75x^5$

Calculation :

Consider the power regression model $y=2.75x^5$ .

Substitute $x=7.1$ in the power regression model.

  $\begin{array}{c}y=2.75{\left(7.1\right)}^5\\ \approx 49616\end{array}$

Therefore, the required value is $49,616$ .

(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:

$\ln x$	$\ln y$
$\begin{array}{l}1.39\\ 1.87\\ 2.14\\ 2.30\end{array}$	$\begin{array}{l}7.94\\ 10.37\\ 11.71\\ 12.52\end{array}$

Given information :

The data:

$x$	$y$
$\begin{array}{l}4\\ 6.5\\ 8.5\\ 10\end{array}$	$\begin{array}{l}2816\\ 31,098\\ 122,019\\ 275,000\end{array}$

The power regression model is $y=2.75x^5$

Calculation :

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

$\ln x$	$\ln y$
$\begin{array}{l}1.39\\ 1.87\\ 2.14\\ 2.30\end{array}$	$\begin{array}{l}7.94\\ 10.37\\ 11.71\\ 12.52\end{array}$

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 $\ln y=5\left(\ln x\right)+1.011$

Given information :

The data:

$x$	$y$
$\begin{array}{l}4\\ 6.5\\ 8.5\\ 10\end{array}$	$\begin{array}{l}2816\\ 31,098\\ 122,019\\ 275,000\end{array}$

Calculation :

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

$\ln x$	$\ln y$
$\begin{array}{l}1.39\\ 1.87\\ 2.14\\ 2.30\end{array}$	$\begin{array}{l}7.94\\ 10.37\\ 11.71\\ 12.52\end{array}$

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.

  $\begin{array}{l}a\approx 5\\ b\approx 1.011\end{array}$

Therefore, the linear regression model is $\ln y=5\left(\ln x\right)+1.011$ .

(e)

To prove : $y=e^b\cdot x^a$ is the power regression model $y=2.75x^5$ .

Given information :

The data:

$x$	$y$
$\begin{array}{l}4\\ 6.5\\ 8.5\\ 10\end{array}$	$\begin{array}{l}2816\\ 31,098\\ 122,019\\ 275,000\end{array}$

The linear regression model is $\ln y=5\left(\ln x\right)+1.011$ .

Proof :

Consider the linear regression model $\ln y=5\left(\ln x\right)+1.011$ .

  $\begin{array}{c}\ln y=5\left(\ln x\right)+1.011\\ \ln y=\ln x^5+1.011\\ y=e^{\ln x^5+1.011}\\ y=e^{\ln x^5}\times e^{1.011}\\ y=x^5\left(2.75\right)\\ y=2.75x^5\end{array}$

Therefore, it is proved that $y=e^b\cdot x^a$ is the power regression model $y=2.75x^5$ .