Chapter 12, Problem T12.12SPT

(a)

To determine

To explain why a linear model may not be appropriate in this case.

Explanation of Solution

Foresters are interested in predicting the amount of usable lumber they can harvest from various tree species. They collected data on the diameter at breast height in inches and the yield in broad feet of a random sample of Ponderosa pine trees. The computer output and a residual plot from a least square regression is given in the question. Thus, a linear model will not be appropriate because the pattern in the residual plot contains strong curvature and this then indicates that the variables have a curved relationship.

(b)

To determine

To use both the models to predict the amount of usable lumber from a Ponderosa Pine with diameter $30$ inches.

Answer to Problem T12.12SPT

Option $1$ : $117.0899$ board feet.

Option $2$ : $102.967$ board feet.

Explanation of Solution

Foresters are interested in predicting the amount of usable lumber they can harvest from various tree species. They collected data on the diameter at breast height in inches and the yield in broad feet of a random sample of Ponderosa pine trees. The computer output and a residual plot from a least square regression is given in the question. The foresters are considering two possible transformations of the original data by cubing the diameter values or taking the natural logarithm of the yields measurements. Thus, for option $1$ general equation of the least square regression line is:

  $\widehat{y}=b_0+b_{1}x$

The estimate of the constant $b_0$ is given in the row “Constant” and in the column “Coef” of the computer output as:

  $b_0=2.078$

The estimate of the constant $b_1$ is given in the row “DBH $3$ ” and in the column “Coef” of the computer output as:

  $b_1=0.0042597$

Replace $b_0$ by $2.078$ and $b_1$ by $b_1=0.0042597$ , in the general equation, we have,

  $\begin{array}{l}\widehat{y}=b_0+b_{1}x\\ \Rightarrow \widehat{y}=2.078+0.0042597x\end{array}$

Thus the cubic equation is:

  $\widehat{y}=2.078+0.0042597x^3$

Now, replace $x\text{ by 30}$ and evaluate,

  $\begin{array}{l}\widehat{y}=2.078+0.0042597x^3\\ $ =2.078+0.0042597{(30)}^3$=117.0899\end{array}$

Thus the predicted yield is $117.0899$ board feet.

Thus, for option $2$ : general equation of the least square regression line is:

  $\widehat{y}=b_0+b_{1}x$

The estimate of the constant $b_0$ is given in the row “Constant” and in the column “Coef” of the computer output as:

  $b_0=1.2319$

The estimate of the constant $b_1$ is given in the row “DBH” and in the column “Coef” of the computer output as:

  $b_1=0.113417$

Replace $b_0=1.2319$ and $b_1$ by $b_1=0.113417$ , in the general equation, we have,

  $\begin{array}{l}\widehat{y}=b_0+b_{1}x\\ \Rightarrow \widehat{y}=1.2319+0.113417x\end{array}$

Let us define the logarithm in the equation as:

  $\ln \widehat{y}=1.2319+0.113417x$

And then replace $x\text{ by 30}$ as:

  $\begin{array}{l}\ln \widehat{y}=1.2319+0.113417x\\ =1.2319+0.113417(30)\\ =4.63441\end{array}$

Take the exponential of each side:

  $\begin{array}{l}\widehat{y}=e^{\ln \widehat{y}}\\ =e^{4.63441}\\ =102.967\end{array}$

Thus the predicted yield is $102.967$ board feet.

(c)

To determine

To explain which of the predictions in part (b) seems more relatable.

Answer to Problem T12.12SPT

Prediction using option $1$ is better.

Explanation of Solution

Foresters are interested in predicting the amount of usable lumber they can harvest from various tree species. They collected data on the diameter at breast height in inches and the yield in broad feet of a random sample of Ponderosa pine trees. The computer output and a residual plot from a least square regression is given in the question. The foresters are considering two possible transformations of the original data by cubing the diameter values or taking the natural logarithm of the yields measurements. Thus, there is no strong curvature in the residual plot of option $1$ , while there is strong curvature in the residual plot of option. This then indicate that the model in option $1$ is appropriate to make predictions and the model in option $2$ is not appropriate to make predictions. We then expect prediction using option $1$ is better.