Chapter 12.2, Problem 36E

(a)

To determine

To give the equation of the least squares regression line.

Answer to Problem 36E

Transformation $1$ : $\text{<mover accent="true"><mo> P</mo><mo>^</mo></mover> ressure}=0.36774+15.8994\times \frac{1}{\text{Volume}}$

Transformation $2$ : $\frac{1}{\text{<mover accent="true"><mo> P</mo><mo>^</mo></mover> ressure}}\text{=}-0.15465\text{+}0.042836\times \text{Volume}$

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:

  $\text{<mover accent="true"><mo> P</mo><mo>^</mo></mover> ressure=a+b}\times \frac{1}{\text{Volume}}$

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

  $\begin{array}{l}a=0.36774\\ b=15.8994\end{array}$

Thus, the regression equation is as follows:

  $\begin{array}{l}\text{<mover accent="true"><mo> P</mo><mo>^</mo></mover> ressure=a+b}\times \frac{1}{\text{Volume}}\\ \Rightarrow \text{<mover accent="true"><mo> P</mo><mo>^</mo></mover> ressure}=0.36774+15.8994\times \frac{1}{\text{Volume}}\end{array}$

For transformation $2$ :

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

  $\frac{1}{\text{<mover accent="true"><mo> P</mo><mo>^</mo></mover> ressure}}\text{=a+b}\times \text{Volume}$

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

  $\begin{array}{l}a=0.100170\\ b=0.0398119\end{array}$

Thus, the regression equation is as follows:

  $\begin{array}{l}\frac{1}{\text{<mover accent="true"><mo> P</mo><mo>^</mo></mover> ressure}}\text{=a+b}\times \text{Volume}\\ \Rightarrow \frac{1}{\text{<mover accent="true"><mo> P</mo><mo>^</mo></mover> ressure}}\text{=}-0.15465\text{+}0.042836\times \text{Volume}\end{array}$

(b)

To determine

To use the model from part (a) to predict the pressure in the syringe when the volume is $17$ cubic centimeter.

Answer to Problem 36E

Transformation

  $1$ : $1.303$

Transformation $2$ : $1.287$

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:

  $\begin{array}{l}\text{<mover accent="true"><mo> P</mo><mo>^</mo></mover> ressure=a+b}\times \frac{1}{\text{Volume}}\\ \Rightarrow \text{<mover accent="true"><mo> P</mo><mo>^</mo></mover> ressure}=0.36774+15.8994\times \frac{1}{\text{Volume}}\end{array}$

Then by evaluating we have,

  $\begin{array}{l}\text{<mover accent="true"><mo> P</mo><mo>^</mo></mover> ressure=a+b}\times \frac{1}{\text{Volume}}\\ \Rightarrow \text{<mover accent="true"><mo> P</mo><mo>^</mo></mover> ressure}=0.36774+15.8994\times \frac{1}{\text{Volume}}\\ =0.36774+15.8994\times \frac{1}{\text{17}}\\ =1.303\end{array}$

For transformation $2$ :

Thus, the regression equation is as follows:

  $\begin{array}{l}\frac{1}{\text{<mover accent="true"><mo> P</mo><mo>^</mo></mover> ressure}}\text{=a+b}\times \text{Volume}\\ \Rightarrow \frac{1}{\text{<mover accent="true"><mo> P</mo><mo>^</mo></mover> ressure}}\text{=}-0.15465\text{+}0.042836\times \text{Volume}\end{array}$

Then by evaluating we have,

  $\begin{array}{l}\frac{1}{\text{<mover accent="true"><mo> P</mo><mo>^</mo></mover> ressure}}\text{=a+b}\times \text{Volume}\\ \Rightarrow \frac{1}{\text{<mover accent="true"><mo> P</mo><mo>^</mo></mover> ressure}}\text{=}-0.15465\text{+}0.042836\times \text{Volume}\\ \Rightarrow \frac{1}{\text{<mover accent="true"><mo> P</mo><mo>^</mo></mover> ressure}}\text{=}-0.15465\text{+}0.042836\times \text{17}\\ \Rightarrow \frac{1}{\text{<mover accent="true"><mo> P</mo><mo>^</mo></mover> ressure}}\text{=}0.7769723\\ \Rightarrow \text{<mover accent="true"><mo> P</mo><mo>^</mo></mover> ressure}=\frac{1}{0.7769723}\\ \Rightarrow \text{<mover accent="true"><mo> P</mo><mo>^</mo></mover> ressure}=1.287\end{array}$

(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:

  $\begin{array}{l}\text{<mover accent="true"><mo> P</mo><mo>^</mo></mover> ressure=a+b}\times \frac{1}{\text{Volume}}\\ \Rightarrow \text{<mover accent="true"><mo> P</mo><mo>^</mo></mover> ressure}=0.36774+15.8994\times \frac{1}{\text{Volume}}\end{array}$

And the value of s is given as,

  $s=0.044205$

This means that the expected error from the prediction of the pressure is $0.044205$ .

For transformation $2$ :

Thus, the regression equation is as follows:

  $\begin{array}{l}\frac{1}{\text{<mover accent="true"><mo> P</mo><mo>^</mo></mover> ressure}}\text{=a+b}\times \text{Volume}\\ \Rightarrow \frac{1}{\text{<mover accent="true"><mo> P</mo><mo>^</mo></mover> ressure}}\text{=}-0.15465\text{+}0.042836\times \text{Volume}\end{array}$

And the value of s is given as,

  $s=0.00398119$

This means that the expected error from the prediction of the reciprocal of the pressure is $0.00398119$ .