Chapter 12.2, Problem 35E

(a)

To determine

To give the equation of the least squares regression line.

Answer to Problem 35E

Transformation $1$ : $\text{<mover accent="true"><mo> P</mo><mo>^</mo></mover> eriod}=-0.08594+0.209999\sqrt{\text{length}}$

Transformation $2$ : ${\text{<mover accent="true"><mo> P</mo><mo>^</mo></mover> eriod}}^2\text{=}-0.15465\text{+}0.042836\times \text{length}$

Explanation of Solution

In the question there are two transformations given and the computer output for each is also given. Thus, we have,

For transformation $1$ :

Thus, the general equation of the regression equation is as:

  $\text{<mover accent="true"><mo> P</mo><mo>^</mo></mover> eriod=a+b}\sqrt{\text{length}}$

Thus, the value of the slope and the constant is given in the computer output as:

  $\begin{array}{l}a=-0.08594\\ b=0.209999\end{array}$

Thus, the regression equation is as follows:

  $\begin{array}{l}\text{<mover accent="true"><mo> P</mo><mo>^</mo></mover> eriod=a+b}\sqrt{\text{length}}\\ \Rightarrow \text{<mover accent="true"><mo> P</mo><mo>^</mo></mover> eriod}=-0.08594+0.209999\sqrt{\text{length}}\end{array}$

For transformation $2$ :

Thus, the general equation of the regression equation is as:

  ${\text{<mover accent="true"><mo> P</mo><mo>^</mo></mover> eriod}}^2\text{=a+b}\times \text{length}$

Thus, the value of the slope and the constant is given in the computer output as:

  $\begin{array}{l}a=-0.15465\\ b=0.042836\end{array}$

Thus, the regression equation is as follows:

  $\begin{array}{l}{\text{<mover accent="true"><mo> P</mo><mo>^</mo></mover> eriod}}^2\text{=a+b}\times \text{length}\\ \Rightarrow {\text{<mover accent="true"><mo> P</mo><mo>^</mo></mover> eriod}}^2\text{=}-0.15465\text{+}0.042836\times \text{length}\end{array}$

(b)

To determine

To use the model from part (a) to predict the period of a pendulum with the length $80$ centimeters.

Answer to Problem 35E

Transformation $1$ : $1.7924$

Transformation $2$ : $1.8089$

Explanation of Solution

Thus, we need to find out the period of a pendulum with the length $80$ centimeters using the part (a) as:

For transformation $1$ :

Thus, the regression equation is as follows:

  $\begin{array}{l}\text{<mover accent="true"><mo> P</mo><mo>^</mo></mover> eriod=a+b}\sqrt{\text{length}}\\ \Rightarrow \text{<mover accent="true"><mo> P</mo><mo>^</mo></mover> eriod}=-0.08594+0.209999\sqrt{\text{length}}\end{array}$

Then by evaluating we have,

  $\begin{array}{l}\text{<mover accent="true"><mo> P</mo><mo>^</mo></mover> eriod=a+b}\sqrt{\text{length}}\\ \Rightarrow \text{<mover accent="true"><mo> P</mo><mo>^</mo></mover> eriod}=-0.08594+0.209999\sqrt{\text{length}}\\ =-0.08594+0.209999\sqrt{\text{80}}\\ =1.7924\end{array}$

For transformation $2$ :

Thus, the regression equation is as follows:

  $\begin{array}{l}{\text{<mover accent="true"><mo> P</mo><mo>^</mo></mover> eriod}}^2\text{=a+b}\times \text{length}\\ \Rightarrow {\text{<mover accent="true"><mo> P</mo><mo>^</mo></mover> eriod}}^2\text{=}-0.15465\text{+}0.042836\times \text{length}\end{array}$

Then by evaluating we have,

  $\begin{array}{l}{\text{<mover accent="true"><mo> P</mo><mo>^</mo></mover> eriod}}^2\text{=a+b}\times \text{length}\\ \Rightarrow {\text{<mover accent="true"><mo> P</mo><mo>^</mo></mover> eriod}}^2\text{=}-0.15465\text{+}0.042836\times \text{length}\\ \Rightarrow {\text{<mover accent="true"><mo> P</mo><mo>^</mo></mover> eriod}}^2\text{=}-0.15465\text{+}0.042836\times \text{80}\\ \Rightarrow {\text{<mover accent="true"><mo> P</mo><mo>^</mo></mover> eriod}}^2\text{=}3.27223\\ \Rightarrow \text{<mover accent="true"><mo> P</mo><mo>^</mo></mover> eriod}=\sqrt{3.27223}\\ \Rightarrow \text{<mover accent="true"><mo> P</mo><mo>^</mo></mover> eriod}=1.8089\end{array}$

(c)

To determine

To interpret the value of s in this context.

Explanation of Solution

As we need to interpret the value of s in this context. Thus, we have,

For transformation $1$ :

Thus, the regression equation is as follows:

  $\begin{array}{l}\text{<mover accent="true"><mo> P</mo><mo>^</mo></mover> eriod=a+b}\sqrt{\text{length}}\\ \Rightarrow \text{<mover accent="true"><mo> P</mo><mo>^</mo></mover> eriod}=-0.08594+0.209999\sqrt{\text{length}}\end{array}$

And the value of s is given as,

  $s=0.0464223$

This means that the expected error from the prediction of the period is $0.0464223$ .

For transformation $2$ :

Thus, the regression equation is as follows:

  $\begin{array}{l}{\text{<mover accent="true"><mo> P</mo><mo>^</mo></mover> eriod}}^2\text{=a+b}\times \text{length}\\ \Rightarrow {\text{<mover accent="true"><mo> P</mo><mo>^</mo></mover> eriod}}^2\text{=}-0.15465\text{+}0.042836\times \text{length}\end{array}$

And the value of s is given as,

  $s=0.105469$

This means that the expected error from the prediction of the square of the period is $0.105469$ .