Question

Boyle’s law Refers to Exercise 34. We took the logarithm (base $10$) of the values for both volume and pressure. 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 pressure and volume.

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 pressure in the syringe when the volume is $17$cubic centimeters.

Short answer

a). The scatter plot of log (volume) and log (pressure) is linear, and the residual figure reflects no evident remaining patterns, according to the equation.

b). The equation of the least-squares regression line is
$\log \left(\text{pressure}\right)=1.11116-0.81344\log \left(\text{volume}\right)$.

c). The expected pressure is $1.28914$ atm.

Step-by-step solution

Part (a) Step 1: Given Information

Given data:

Part (a) Step 2: Explanation

According to the equation, the scatter plot of log (volume) and log (pressure) is linear, and the residual figure shows no discernible patterns.

Part (b) Step 1: Given Information

Given data:

Part (b) Step 2: Explanation

The general regression equation is derived from the data.

$\log \left(\text{pressure}\right)=\alpha +\beta \log \left(\text{volume}\right)$

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

$\alpha =1.11116$

To find the $\beta$(constant) in the row " log (volume) " and column "Coef" of the computer output.

$\beta =-0.81344$

Substituting the value of $\alpha$and $\beta$in the equation:

$\log \left(\text{pressure}\right)=\alpha +\beta \log \left(\text{volume}\right)$$\log \left(\text{pressure}\right)=1.11116-0.81344\log \left(\text{volume}\right)$

Part (c) Step 1: Given Information

Given data:

Part (c) Step 2: Explanation

The equation of the least-squares regression line is

$\log \left(pressure\right)=1.11116-0.81344\log \left(volume\right)$

Calculate by multiplying the volume amount by $17$:

$\log \left(\text{pressure}\right)=1.11116-0.81344\log \left(17\right)$

$\log \left(\text{pressure}\right)=0.1103$

Using the exponential with a value of $10$:

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

$={10}^{0.1103}$

$=1.28914$