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The swinging pendulum Refer to Exercise 33. We took the logarithm (base 10) of the values for both length and period. Here is some computer output from a linear regression analysis of the transformed data.


a. Based on the output, explain why it would be reasonable to use a power model to describe the relationship between the length and period of a pendulum.

b. Give the equation of the least-squares regression line. Be sure to define any variables you use.

c. Use the model from part (b) to predict the period of a pendulum with a length of 80cm.

Short Answer

Expert verified

a). There is a linear relationship between the scatter plots of log (length) vs log (period), and the residual plot shows no identifiable patterns.

b). The equation of the least-squares regression line islog(period)=-0.73675+0.51701log(length).

c). The expected period is 1.76685seconds.

Step by step solution

01

Part (a) Step 1: Given Information

Given data:

02

Part (a) Step 2: Explanation

log(y)hat=0.73675+0.51701log(x)

The fact that there is a log of xand yin the equation indicates that it is a power function. The scatter plots of log (length) vs log (time) demonstrate a linear connection, while the residual plot exhibits no identifiable patterns.

03

Part (b) Step 1: Given Information

Given data:

04

Part (b) Step 2: Explanation

The generic regression equation is built using the data.

log(period)=α+βlog(length)

To find the α(constant) in the row "constant" and column "Coef" of the computer output.

α=-0.73675

To find the β(constant) in the row " log (length) " and column "Coef" of the computer output.

β=0.51701

Substituting the value of αand βin the equation

log(period)=α+βlog(length)
localid="1654260405887" log(period)=-0.73675+0.51701log(length)

05

Part (c) Step 1: Given Information

Given data:

06

Part (c) Step 2: Explanation

The equation of the least-squares regression line is

log(period)=-0.73675+0.51701log(length)

Calculate by multiplying the length by 80.

log(period)=-0.73675+0.51701log(80)

log(period)=0.2472

Using the exponential with a value of 10

period=10log(period)

=100.2472

=1.76685

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Most popular questions from this chapter

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