Question

Ecologists look at data to learn about nature’s patterns. One pattern they have found relates the size of a carnivore (body mass in kilograms) to how many of those carnivores there are in an area. The right measure of “how many” is to count carnivores per $10,000$kilograms (kg) of their prey in the area. The table below gives data for $25$carnivore species

$\begin{array}{lcc}\text{Carnivore species}& \begin{array}{c}\text{Body mass}\\ \left(\mathbf{kg}\right)\end{array}& \begin{array}{c}\text{Abundance}\\ \text{(per 10,000 kg of prey)}\end{array}\\ \text{Least weasel}& 0.14& 1656.49\\ \text{Ermine}& 0.16& 406.66\\ \text{Small Indian mongoose}& 0.55& 514.84\\ \text{Pine marten}& 1.3& 31.84\\ \text{Kit fox}& 2.02& 15.96\\ \text{Channel Island fox}& 2.16& 145.94\\ \text{Arctic fox}& 3.19& 21.63\\ \text{Red fox}& 4.6& 32.21\\ \text{Bobcat}& 10.0& 9.75\\ \text{Canadian lynx}& 11.2& 4.79\\ \text{European badger}& 13.0& 7.35\\ \text{Coyote}& 13.0& 11.65\\ \text{Ethiopian wolf}& 14.5& 2.70\\ \text{Eurasian lynx}& 20.0& 0.46\\ \text{Wild dog}& 25.0& 1.61\\ \text{Dhole}& 25.0& 0.81\\ \text{Snow leopard}& 40.0& 1.89\\ \text{Wolf}& 46.0& 0.62\\ \text{Leopard}& 46.5& 6.17\\ \text{Cheetah}& 50.0& 2.29\\ \text{Puma}& 51.9& 0.94\\ \text{Spotted hyena}& 58.6& 0.68\\ \text{Tiger}& 182.0& 3.40\\ \text{Polar bear}& 310.0& 0.33\\ & & \end{array}$

Here is a scatterplot of the data.

(a) The following graphs show the results of two different transformations of the data. Would an exponential model or a power model provide a better description of the relationship between body mass and abundance? Justify your answer.

(b) Minitab output from a linear regression analysis on the transformed data of log(abundance) versus log(body mass) is shown below. Give the equation of the least-squares regression line. Be sure to define any variables you use.

$\begin{array}{lclrc}\text{Predictor}& \text{Coef}& \text{SE Coef}& \text{T}& \text{P}\\ \text{Constant}& 1.9503& 0.1342& 14.53& 0.000\\ \log (\text{body}& -1.04811& 0.09802& -10.69& 0.000\\ \text{mass)}& & & & \end{array}$

$\mathrm{S}=0.423352\mathrm{R}-\mathrm{Sq}=83.3\%\mathrm{R}-\mathrm{Sq}\left(\mathrm{adj}\right)=82.5\%$

(c) Use your model from part (b) to predict the abundance of black bears, which have a body mass of $92.5$kilograms. Show your work.

(d) A residual plot for the linear regression in part (b) is shown below. Explain what this graph tells you about how well the model fits the data.

Short answer

(a) The link between body mass and abundance is better described by a power model.

(b) The equation is $\log \hat{y}=1.9503-1.04811\log x$.

(c) The abundance of black bears which has a body mass of $92.5$kilograms is $0.775474$per $10000kg$of prey.

(d) The model appears to be an excellent match.

Step-by-step solution

Part(a) Step 1: Given Information

Part(a) Step 2: Explanation

The data on body mass and abundance is provided in the inquiry. As a result, the scatterplot pattern in the model that best describes the connection between body mass and abundance must be roughly linear. Because the top graph corresponds to an exponential model and the bottom graph corresponds to a power model, the power model will provide a better representation of the link between body mass and abundance.

Part(b) Step 1: Given Information

$\begin{array}{lclrc}\text{Predictor}& \text{Coef}& \text{SE Coef}& \text{T}& \text{P}\\ \text{Constant}& 1.9503& 0.1342& 14.53& 0.000\\ \log (\text{body}& -1.04811& 0.09802& -10.69& 0.000\\ \text{mass)}& & & & \end{array}$

$\mathrm{S}=0.423352\mathrm{R}-\mathrm{Sq}=83.3\%\mathrm{R}-\mathrm{Sq}\left(\mathrm{adj}\right)=82.5\%$

Part(b) Step 2: Explanation

Now, the question specifies that variable $x$represents body mass and variable $y$represents abundance. As a result, the general equation will be:

localid="1652804782326" $\log \hat{y}=a+b\log x$

The slope and constant of the regression line are given in the question:

$a=1.9503$

$b=-1.04811$

As a result, the regression line appears to be as follows:

localid="1652805060619" $$\begin{gathered} \log \hat{y}=a+b\log x \\ =1.9503-1.04811\log x \end{gathered}$$

Part(c) Step 1: Given Information

$\begin{array}{lclrc}\text{Predictor}& \text{Coef}& \text{SE Coef}& \text{T}& \text{P}\\ \text{Constant}& 1.9503& 0.1342& 14.53& 0.000\\ \log (\text{body}& -1.04811& 0.09802& -10.69& 0.000\\ \text{mass)}& & & & \end{array}$

$\mathrm{S}=0.423352\mathrm{R}-\mathrm{Sq}=83.3\%\mathrm{R}-\mathrm{Sq}\left(\mathrm{adj}\right)=82.5\%$

Part(c) Step 2: Explanation

The regression line is

$$\begin{gathered} \log \hat{y}=a+b\log x \\ =1.9503-1.04811\log x \end{gathered}$$

As a result, we must forecast the number of black bears with a body mass of $92.5$kg. As a result of examining part (b), we have:

$$\begin{gathered} \log \hat{y}=a+b\log x \\ =1.9503-1.04811\log x \\ =1.9503-1.04811\log 92.5 \\ =-0.110433 \\ ={10}^{-0.110433} \\ =0.775474 \end{gathered}$$

Part(d) Step 1: Given Information

$\begin{array}{lclrc}\text{Predictor}& \text{Coef}& \text{SE Coef}& \text{T}& \text{P}\\ \text{Constant}& 1.9503& 0.1342& 14.53& 0.000\\ \log (\text{body}& -1.04811& 0.09802& -10.69& 0.000\\ \text{mass)}& & & & \end{array}$

$\mathrm{S}=0.423352\mathrm{R}-\mathrm{Sq}=83.3\%\mathrm{R}-\mathrm{Sq}\left(\mathrm{adj}\right)=82.5\%$

Part(d) Step 2: Explanation

The regression line is

$$\begin{gathered} \log \hat{y}=a+b\log x \\ =1.9503-1.04811\log x \end{gathered}$$

The question also includes a residual plot of the linear regression in section (b). Because the residuals in the residual plot are all centered about zero, there is no clear pattern in the residual plot, and the vertical spread of the residuals appears to be roughly the same everywhere, the model appears to be a good fit.