91影视

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.

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

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

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

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

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

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:

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

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

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

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.