Chapter 12.2, Problem 35E

(a)

To determine

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

Answer to Problem 35E

  ${\widehat{y}}^2=-0.15465+0.042836x$

Explanation of Solution

Given:

  

  

Calculation:

General equation for the least square regression line

  $\widehat{y}=b_0+b_{1}x$

The calculated of the constant $b_0$ is mention in the row “constant” and in the column “Coef” of the output of the computer

  $b_0$ =-0.15465

The calculated of the slope $b_1$ is mention in the row “length” and in the column “Coef” of the output of the computer

  $b_1$ =0.042836

Putting the value of $b_0$ and $b_1$

  $\begin{array}{l}\widehat{y}=b_0+b_{1}x\\ \widehat{y}=-0.15465+0.042836x\end{array}$

Where x shows the length and y shows the period squared

  ${\widehat{y}}^2=-0.15465+0.042836x$

Where $x$ is representing the length and y is representing the period

(b)

To determine

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

Answer to Problem 35E

1.8089 seconds

Explanation of Solution

Given:

  

  

Calculation:

From the part (a)

  ${\widehat{y}}^2=-0.15465+0.042836x$

Where $x$ is representing the length and y is representing the period

Putting the value of pendulum length in the equation

  $\begin{array}{c}{\widehat{y}}^2=-0.15465+0.042836x\\ =-0.15465+0.042836\left(80\right)\\ =3.27223\\ \widehat{y}=\sqrt{3.27223}\\ =1.8089\end{array}$

Therefore the expected period is 1.8089 seconds.