Question

Counting carnivores Ecologists look at data to learn about nature’s patterns. One pattern they have identified relates the size of a carnivore (body mass in kilograms) to how many of those carnivores exist in an area. A good measure of “how many” is to count carnivores per 10,000 kg of their prey in the area. The scatterplot shows this relationship between body mass and abundance for 25 carnivore species.

The following graphs show the results of two different transformations of the data. The first graph plots the logarithm (base 10) of abundance against body mass. The second graph plots the logarithm (base 10) of abundance against the logarithm (base 10) of body mass.

a. Based on the scatterplots, would an exponential model or a power model provide a better description of the relationship between abundance and body mass? Justify your answer.

b. Here is a computer output from a linear regression analysis of log(abundance) and log(body mass). Give the equation of the least-squares regression line. Be sure to define any variables you use.

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

d. Here is a residual plot for the linear regression in part (b). Do you expect your prediction in part (c) to be too large, too small, or about right? Justify your answer.

Short answer

(a) Power model

(b)$\log y=1.9503-1.04811x$

(c) The predicted abundance is $0.775532$per $10000kg$per prey

(d) The forecast of $2$is nearly right.

Step-by-step solution

Part (a) Step 1: Given information

The given data is

Part (a) Step 2: Explanation

The scatter plot between log(abundance) and log(abundance) is seen (body mass). Because the scatter plot's two variables do not have a lot of curvatures, a linear model between them would be appropriate. As a result, a linear model between log(abundance) and log(abundance) is adequate (body mass).

Expect log(abundance) and log(abundance) using a general linear model (body mass).

$\log \left(abundance\right)=a+b(bodymass)$

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

$=e^a{e}^{b\left(\text{body mass}\right)}$

As a result, the model is associated with abundance$=e^a{e}^b$(body mass). It's a powerful model that combines abundance with body bulk.

Although the linear model of abundance and log (body mass) is useful, the power model of abundance and body mass is equally useful.

Part (b) Step 1: Given information

The given data is

Part (b) Step 2: Explanation

The equation for square regression line is

$\hat{y}=b_0+b_{1}x$

In the row "constant" and the column "Coef" of the computer output, the calculated constant $b_0$is mentioned.

$b_0=1.9503$

In the row "Distance" and the column "Coef" of the computer output, the calculated slope $b_1$is mentioned.

$b_1=-1.04811$

On substituting the values

$\hat{y}=1.9503-1.04811x$

Calculating the log value

$\log y=1.9503-1.04811\log x$

Where $x$ stands for body mass and$y$ stands for abundance.

Part (c) Step 1: Given information

The given data is

Part (c) Explanation

From part (b)

$\log y=1.9503-1.04811\log x$

Where $x$stands for body mass and $y$stands for abundance.

Substituting the value of x

$$\begin{gathered} \log y=1.9503-1.04811\log (92.5) \\ \log y=-0.1104 \end{gathered}$$

Then take the exponential

$$\begin{gathered} \hat{y}={10}^{\log y} \\ \hat{y}={10}^{-0.1104} \\ \hat{y}=0.775532 \end{gathered}$$

As a result, $0.775532$is the predicted abundance per$10000kg$per prey.

Part (d) Step 1: Given information

The given data is

Part (d) Step 2: Explanation

It is estimated that the body mass will be $92.5$kg based on portion (c).

$\log (92.5)=1.966142$

The dots between $1.5$and $2.0$are both below and above the horizontal line at $0$, as shown in the residual figure. Furthermore, the horizontal line 0 is in the middle of these dots, implying that the forecast of $2$is nearly right.