91影视

The given data is

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.

The given data is

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.

The given data is

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.

The given data is

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.