Chapter 12.2, Problem 41E

(a)

To determine

To explain would an exponential model or a power model provide a better description of the relationship between body mass and abundance.

Answer to Problem 41E

The power model.

Explanation of Solution

In the question the data about the body mass and the abundance in given. So, the model that gives the better description of the relationship between the body mass and abundance needs to have a roughly linear pattern in the scatterplot. Thus, the power model will provide a better description of the relationship between body mass and abundance since the top graph corresponds with an exponential model and the bottom graph corresponds with a power model.

(b)

To determine

To give the equation of the least squares regression line.

Answer to Problem 41E

  $\log \widehat{y}=1.9503-1.04811\log x$ .

Explanation of Solution

Now, it is given in the question that variable x be the body mass and the variable y be abundance. Thus, the general equation will be as:

  $\log \widehat{y}=a+b\log x$

And the slope and the constant of the regression line is give in the question as:

  $\begin{array}{l}a=1.9503\\ b=-1.04811\end{array}$

Thus, the regression line is as:

  $\begin{array}{l}\log \widehat{y}=a+b\log x\\ \Rightarrow \log \widehat{y}=1.9503-1.04811\log x\end{array}$

(c)

To determine

To use your model in part (b) to predict the abundance of black bears which has a body mass of $92.5$ kilograms.

Answer to Problem 41E

It is $0.775474$ per $10000$ kg of prey.

Explanation of Solution

Now, the regression line is as:

  $\begin{array}{l}\log \widehat{y}=a+b\log x\\ \Rightarrow \log \widehat{y}=1.9503-1.04811\log x\end{array}$

Thus, we need to predict the abundance of black bears which has a body mass of $92.5$ kilograms thus, evaluating the equation in part (b), we have,

  $\begin{array}{l}\log \widehat{y}=a+b\log x\\ \Rightarrow \log \widehat{y}=1.9503-1.04811\log x\\ \Rightarrow \log \widehat{y}=1.9503-1.04811\log 92.5\\ \Rightarrow \log \widehat{y}=-0.110433\\ \Rightarrow \widehat{y}={10}^{-0.110433}\\ \Rightarrow \widehat{y}=0.775474\end{array}$

(d)

To determine

To explain what is the graph tells you about how well the model fits the data.

Answer to Problem 41E

The model seems to be a good fit.

Explanation of Solution

Now, the regression line is as:

  $\begin{array}{l}\log \widehat{y}=a+b\log x\\ \Rightarrow \log \widehat{y}=1.9503-1.04811\log x\end{array}$

And the residual plot of the linear regression in part (b) is shown in the question. Thus, the model seems to be a good fit, because the residuals in the residual plot are all centered about zero, there is no obvious pattern in the residual plot and the vertical spread of the residuals seems to be roughly the same everywhere.