Chapter 12.2, Problem 38E

(a)

To determine

To find: that is it reasonable to use a power model to explain the relation between the period of pendulum and the length.

Answer to Problem 38E

From the equation, it is noticed that the scatter plot of log (volume) and log (pressure) is linear and the residual figure represents no obvious leftover patterns.

Explanation of Solution

Given:

  

  

  

From the equation, it is noticed that the scatter plot of log (volume) and log (pressure) is linear and the residual figure represents no obvious leftover patterns.

(b)

To determine

To find: the equation of the least squares of regression line.

Answer to Problem 38E

  $\log \left(pressure\right)=1.11116-0.81344\log (volume)$

Explanation of Solution

Given:

  

  

  

Calculation:

On the basis of given information the general regression equation is

  $\log \left(pressure\right)=\alpha +\beta \log (volume)$

Calculate the $\alpha$ (constant) is mention in the row “constant” and in the column “Coef” of the mention output from the computer

  $\alpha$ =1.11116

Calculate the $\beta$ (constant) is mention in the row “ $\log (volume)$ ” and in the column “Coef” of the mention output from the computer

  $\beta$ =-0.81344

Putting the value of $\alpha$ and $\beta$ in the equation

  $\begin{array}{c}\log \left(pressure\right)=\alpha +\beta \log (volume)\\ \log \left(pressure\right)=1.11116-0.81344\log (volume)\end{array}$

(c)

To determine

To Explain: the prediction of the pressure in the syringe if the volume is 17 cubic cm with using the part (a)

Answer to Problem 38E

1.28914 atm

Explanation of Solution

Given:

  

  

  

Calculation:

From the part (b)

  $\log \left(pressure\right)=1.11116-0.81344\log (volume)$

Putting the value of volume by 17 and calculate

  $\begin{array}{c}\log \left(pressure\right)=1.11116-0.81344\log (17)\\ \log \left(pressure\right)=0.1103\end{array}$

Taking the exponential with 10

  $\begin{array}{c}pressure={10}^{\log \left(pressure\right)}\\ ={10}^{0.1103}\\ =1.28914\end{array}$

Therefore the expected pressure is 1.28914 atm.