Beetles and beavers Do beavers benefit beetles? Researchers laid out 23 circular plots, every 4 meters in diameter, in an area where beavers were cutting down cottonwood trees. In each plot, they counted the number of stumps from trees cut by beavers and the number of clusters of beetle larvae. Ecologists believe that the new sprouts from stumps are more tender than other cottonwood growth, so beetles prefer them. If so, more stumps should

produce more beetle larvae.31 Here is the computer output for regression of y = number of beetle larvae on x = number of stumps:

a. Calculate and interpret the residual for the plot that had 2 stumps and 30 beetle larvae.

b. Interpret the slope.

c. By how much does the actual number of larvae typically vary from the values predicted by the least-squares regression line with x = number of stumps?

d. What percent of the variability in a number of larvae is accounted for by the least-squares regression line with x = number of stumps?

The relationship of number of larvae and the number of stumps is given in the question.

The formula used:$Residual=y-\stackrel{⏜}{y}$

The least square regression line's general equation is:

$\stackrel{⏜}{y}=b_0+b_{1}x$

As a result, the estimate of the constant $b_0$is given in the computer output's row "Intercept" and column "Estimate" as:

$b_0=-1.286104$

In the computer output, the estimate of the slope $b_1$is reported in the row "Number of stumps" and the column "Estimate" as:

$b_1=11.893733$

As a result, when we plug the values into the general equation, we get:

$$\begin{gathered} \stackrel{⏜}{y}=b_0+b_{1}x \\ \Rightarrow \stackrel{⏜}{y}=-1.286104+11.893733x \end{gathered}$$

As a result, the number of beetle larvae with two stumps was as follows:

$$\begin{gathered} \stackrel{⏜}{y}=-1.286104+11.893733x \\ =-1.286104+11.893733\left(2\right) \\ =22.501362 \end{gathered}$$

Thus the residual will be calculated as:

$$\begin{gathered} Residual=y-\stackrel{⏜}{y} \\ =30-22.501362 \\ =7.498638 \end{gathered}$$

This means that when using the regression line to forecast the number of larvae when there are two stumps, we underestimated the number of larvae by $7.498638$beetle larvae.

The question specifies the link between the quantity of larvae and the number of stumps. The regression line is as follows:

$\stackrel{⏜}{y}=-1.286104+11.893733x$

The slope is the coefficient of $x$ in the least-squares regression equation, and it reflects the average rise or decrease of $y$ per unit of $x$ as we all know. Thus,

$b_1=11.893733$

As a result, the average number of beetle larvae per stump increases by $11.893733$ beetle larvae.

The question specifies the link between the quantity of larvae and the number of stumps. The regression line is as follows:

$\stackrel{⏜}{y}=-1.286104+11.893733x$

The standard error of the estimate s is reported as follows in the computer output after "Root Mean square error":

$s=6.419386$

The standard error of the estimations, as we all know, is the average error of forecasts, and thus the average difference between actual and predicted values. As a result, the anticipated number of beetle larvae using the equation of least square regression line differed from the actual number of beetle larvae by an average of $6.419386$ insect larvae.

The question specifies the link between the number of larvae and the number of stumps. The regression line is as follows:

$\stackrel{⏜}{y}=-1.286104+11.893733x$

The coefficient of determination is given as follows in the computer output after "RSquare":

$$\begin{gathered} r^2=0.839144 \\ =83.9144\% \end{gathered}$$

The coefficient of determination, as we know, is a measurement of how much variation in the answers $y$ variable is explained by the least square regression model with the explanatory variable. As a result, the least square regression line utilizing the number of stumps as an explanatory variable can explain $83.9144$ percent of the variation in the number of beetle larvae.