Chapter 12.2, Problem 33E

(a)

To determine

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

Answer to Problem 33E

  $\widehat{y}=-0.08594+0.209999\sqrt{x}$

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

  $b_0$ =-0.08594

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

  $b_1$ =0.209999

Putting the value of $b_0$ and $b_1$

  $\begin{array}{l}\widehat{y}=b_0+b_{1}x\\ \widehat{y}=-0.08594+0.209999x\end{array}$

Where x shows the square root of the length and y shows the period.

  $\widehat{y}=-0.08594+0.209999\sqrt{x}$

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 33E

1.7923 seconds

Explanation of Solution

Given:

  

  

Calculation:

From the part (a)

  $\widehat{y}=-0.08594+0.209999\sqrt{x}$

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

Putting the value of pendulum length in the equation

  $\begin{array}{l}\widehat{y}=-0.08594+0.209999\sqrt{80}\\ \widehat{y}=1.7923\end{array}$

Therefore the expected period is 1.7923 seconds.