91Ó°ÊÓ

Given data:

$\log \left(y\right)\text{hat}=0.73675+0.51701\log \left(x\right)$

The fact that there is a log of $x$and $y$in 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.

Given data:

The generic regression equation is built using the data.

$\log \left(\text{period}\right)=\mathrm{Î\pm }+\mathrm{β}\log \left(\text{length}\right)$

To find the $\mathrm{Î\pm }$(constant) in the row "constant" and column "Coef" of the computer output.

$\mathrm{Î\pm }=-0.73675$

To find the $\mathrm{β}$(constant) in the row " log (length) " and column "Coef" of the computer output.

$\mathrm{β}=0.51701$

Substituting the value of $\mathrm{Î\pm }$and $\mathrm{β}$in the equation

$\log \left(\text{period}\right)=\mathrm{Î\pm }+\mathrm{β}\log \left(\text{length}\right)$
localid="1654260405887" $\log \left(\text{period}\right)=-0.73675+0.51701\log \left(\text{length}\right)$

Given data:

The equation of the least-squares regression line is

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

Calculate by multiplying the length by $80$.

$\log \left(\text{period}\right)=-0.73675+0.51701\log \left(80\right)$

$\log \left(\text{period}\right)=0.2472$

Using the exponential with a value of $10$

$\text{period}={10}^{\log \left(\text{period}\right)}$

$={10}^{0.2472}$

$=1.76685$