91Ó°ÊÓ

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

The formula used:$Residual=y−\stackrel{Áåœ}{y}$

The least square regression line's general equation is:

$\stackrel{Áåœ}{y}=b_0+b_{1}x$

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:

$$\begin{gathered} \stackrel{Áåœ}{y}=b_0+b_{1}x \\ ⇒\stackrel{Áåœ}{y}=−1.286104+11.893733x \end{gathered}$$

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

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

Thus the residual will be calculated as:

$$\begin{gathered} Residual=y−\stackrel{Áåœ}{y} \\ =30−22.501362 \\ =7.498638 \end{gathered}$$

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:

$\stackrel{Áåœ}{y}=−1.286104+11.893733x$

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:

$\stackrel{Áåœ}{y}=−1.286104+11.893733x$

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:

$\stackrel{Áåœ}{y}=−1.286104+11.893733x$

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.