Question

The swinging pendulum Refer to Exercise 33. Here is a graph of the period versus length, along with output from a linear regression analysis using these variables.

a. Give the equation of the least-squares regression line. Define any variables you use. b. Use the model from part (a) to predict the period of a pendulum with length $80$centimeters.

Short answer

a). The equation of the least-squares regression line is${\hat{y}}^2=-0.15465+0.042836x$.

b). The expected period is $1.8089$seconds.

Step-by-step solution

Part (a) Step 1: Given Information

Given data:

Part (a) Step 2: Explanation

Least square regression line's general equation

$\hat{y}=b_0+b_{1}x$

In the row "constant" and the column "Coef" of the computer output, the calculated constant $b_0$is indicated.

$b_0=-0.15465$

In the row "length" and the column "Coef" of the computer's output, the calculated slope $b_1$ is mentioned.

$b_1=0.042836$

Part (a) Step 3: Explanation

Substituting the value of $b_0$and $b_1$:

$\hat{y}=b_0+b_{1}x$

$\hat{y}=-0.15465+0.042836x$

Where $x$denotes the length and $y$denotes the squared period.

localid="1654254443631" ${\hat{y}}^2=-0.15465+0.042836x$

Where $x$denotes length and $y$denotes time.

Part (b) Step 1: Given Information

Given data:

Part (b) Step 2: Explanation

The least-squares regression line's equation is:

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

Where $x$denotes length and $y$denotes time.

Using the pendulum length as a variable in the equation

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

$=-0.15465+0.042836\left(80\right)$

$=3.27223$

$\hat{y}=\sqrt{3.27223}$

$=1.8089$