Chapter 12.2, Problem 48E

(a)

To determine

To Explain: the better relationship between abundance and body mass.

Answer to Problem 48E

Power model

Explanation of Solution

Given:

  

  

  

It is observed that the given scatter plot between log(abundance) and log(body mass). is not having strong curvature and therefore it would be appropriate to utilize a linear model between the two variables of the scatter plot. Therefore it is appropriate to utilize a linear model between log(abundance) and log(body mass).

General linear model to expect log(abundance) and log(body mass).

  $\log \left(\text{abundance}\right)=a+b\left(\text{body mass}\right)$

Taking the exponential

  $\begin{array}{c}\text{abundance}=e^{\log \left(\text{abundance}\right)}=e^{a+b\left(\text{body mass}\right)}\\ =e^a{e}^{b\left(\text{body mass}\right)}\end{array}$

Therefore the model associate with the model $\text{abundance}=e^a{e}^{b\left(\text{body mass}\right)}$ .It is power model between abundance and body mass. Although the linear model between abundance and $\log \left(\text{body mass}\right)$ is appropriate, the power model between abundance and body mass is also appropriate.

(b)

To determine

To find: the equation of the least-squares regression line on the basis of given computer output.

Answer to Problem 48E

  $\log y=1.9503-1.04811\log x$

Explanation of Solution

Given:

  

Calculation:

General equation for the least square regression line

  $\widehat{y}=b_0+b_{1}x$

The calculated of the constant $b_0$ is mention in the row “constant” and in the column “Coef” of the output of the computer

  $b_0$ =1.9503

The calculated of the slope $b_1$ is mention in the row “Distance” and in the column “Coef” of the output of the computer

  $b_1$ =-1.04811

Putting the value of $b_0$ and $b_1$

  $\begin{array}{l}\widehat{y}=b_0+b_{1}x\\ \widehat{y}=1.9503-1.04811x\end{array}$

Taking the logarithm

  $\log y=1.9503-1.04811\log x$

Where $x$ is representing the body mass and y is representing the abundance.

(c)

To determine

To Explain: the prediction of the abundance of black bears on the basis of part (b).

Answer to Problem 48E

0.775532 per 10000kg prey

Explanation of Solution

Given:

  

Calculation:

From the part (b)

  $\log y=1.9503-1.04811\log x$

Where $x$ is representing the body mass and y is representing the abundance.

Putting the value of $x$ into the equation

  $\begin{array}{l}\log y=1.9503-1.04811\log x\\ \log y=1.9503-1.04811\log \left(92.5\right)\\ \log y=-0.1104\end{array}$

Taking the exponential

  $\begin{array}{c}\widehat{y}={10}^{\log y}={10}^{-0.1104}\\ =0.775532\end{array}$

Therefore the expected abundance is 0.775532 per 10000kg prey

(d)

To determine

To Calculate: here is given linear regression in part (b), is it expected that prediction in part (c) to be too large, too small or about right, justify the answer.

Answer to Problem 48E

About right

Explanation of Solution

Given:

  

From the part (c) it is prediction for the body mass of 92.5 kilogram.

  $\log \left(92.5\right)=1.966142$

In the residual figure, it is observed that the dots between 1.5 and 2.0 are both below and above the horizontal line at 0. Moreover it is noticed that the horizontal line 0 lies in the mid between these dots and therefore it is expected that prediction at about 2 to be about right.