Chapter 3.2, Problem 62E

(a)

To determine

To calculate and interpret the residual for the plot that had $2$ stumps and $30$ beetle larvae.

Answer to Problem 62E

Residual is $7.498638$ .

Explanation of Solution

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

The general equation of the least square regression line is:

  $\widehat{y}=b_0+b_{1}x$

Thus, the estimate of the constant $b_0$ is given in the row “Intercept” and in the column “Estimate” of the computer output as:

  $b_0=-1.286104$

And the estimate of the slope $b_1$ is given in the row “Number of stumps” and in the column “Estimate” of the computer output as:

  $b_1=11.893733$

Thus, putting the values in the general equation we will get,

  $\begin{array}{l}\widehat{y}=b_0+b_{1}x\\ \Rightarrow \widehat{y}=-1.286104+11.893733x\end{array}$

Thus, the number of beetle larvae that had $2$ stumps are as:

  $\begin{array}{l}\widehat{y}=-1.286104+11.893733x\\ =-1.286104+11.893733(2)\\ =22.501362\end{array}$

Thus the residual will be calculated as:

  $\begin{array}{l}\text{Residual}=y-\widehat{y}\\ =30-22.501362\\ =7.498638\end{array}$

This then implies that we underestimated the number of larvae when there are $2$ stumps by $7.498638$ beetle larvae when making a prediction using the regression line.

(b)

To determine

To interpret the slope.

Explanation of Solution

The relationship of number of larvae and the number of stumps is given in the question. And the regression line is as:

  $\widehat{y}=-1.286104+11.893733x$

As we know that the slope is coefficient of x in the least squares regression equation and represents the average increase or decrease of y per unit of x . Thus,

  $b_1=11.893733$

Thus this implies that on average, the number of beetle larvae increases by $11.893733$ beetle larvae per stumps.

(c)

To determine

To find out by how much do the actual number of larvae typically vary from the values predicted by the least square regression line with x as number of stumps.

Answer to Problem 62E

The predicted number of beetle larvae using the equation of least square regression line deviated on average by $6.419386$ beetle larvae from the actual number of beetle larvae.

Explanation of Solution

The relationship of number of larvae and the number of stumps is given in the question. And the regression line is as:

  $\widehat{y}=-1.286104+11.893733x$

As, it is given that the standard error of the estimate s is given in the computer output after “Root Mean square error” as:

  $s=6.419386$

As we know that the standard error of the estimate s represents the average error of predictions thus the average deviation between actual and the predicted values. Thus the predicted number of beetle larvae using the equation of least square regression line deviated on average by $6.419386$ beetle larvae from the actual number of beetle larvae.

(c)

To determine

To find out what percent of the variability in number of larvae is accounted for by the least squares regression line with x as number of stumps.

Answer to Problem 62E

  $83.9144\%$ of the variation in the number of beetle larvae is explained by the least square regression line using the number of stumps as explanatory variable.

Explanation of Solution

The relationship of number of larvae and the number of stumps is given in the question. And the regression line is as:

  $\widehat{y}=-1.286104+11.893733x$

As it is given that the coefficient of determination is given in the computer output after “RSquare” as:

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

As we know that the coefficient of determination measures the proportion of variation in the responses y variable that is explained by the least square regression model using the explanatory x variable. Thus, we can say that $83.9144\%$ of the variation in the number of beetle larvae is explained by the least square regression line using the number of stumps as explanatory variable.