Chapter 12.2, Problem 37E

(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 37E

There is between scatter plot of log (length) versus log (period) is linear and the residual plot represents no obvious patterns

Explanation of Solution

Given:

  

  

  

Calculation:

  $\begin{array}{l}\log \left(y\right)hat=0.73675+0.51701\log \left(x\right)\\ \end{array}$

From the equation, it is noticed that this is a power function the reason is that there is a log of x and y. There is between scatter plot of log (length) versus log (period) is linear and the residual plot represents no obvious patterns

(b)

To determine

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

Answer to Problem 37E

  $\log \left(period\right)=-0.73675+0.51701\log (length)$

Explanation of Solution

Given:

  

  

  

Calculation:

On the basis of given information the general regression equation is

  $\log \left(period\right)=\alpha +\beta \log (length)$

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

  $\alpha$ =-0.73675

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

  $\beta$ =0.51701

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

  $\begin{array}{c}\log \left(period\right)=\alpha +\beta \log (length)\\ \log \left(period\right)=-0.73675+0.51701\log (length)\end{array}$

(c)

To determine

To Explain: the prediction of a pendulum with length of 80 cm using the part (a)

Answer to Problem 37E

1.76685 seconds

Explanation of Solution

Given:

  

  

  

Calculation:

From the part (b)

  $\log \left(period\right)=-0.73675+0.51701\log (length)$

Putting the value length by 80 and calculate

  $\begin{array}{c}\log \left(period\right)=-0.73675+0.51701\log (80)\\ \log \left(period\right)=0.2472\end{array}$

Taking the exponential with 10

  $\begin{array}{c}period={10}^{\log \left(period\right)}\\ ={10}^{0.2472}\\ =1.76685\end{array}$

Therefore the expected period is 1.76685 seconds.