Boyle’s law Refers to Exercise 34. Here is Minitab output from separate regression analyses of the two sets of transformed pressure data:

Do each of the following for both transformations.

(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 pressure in the syringe when the volume is $17$ cubic centimeters. Show your work.

(c) Interpret the value of s in context.

For transformation 1 :

Thus, the general equation of the regression equation is as:

$\hat{\mathrm{P}}\text{ressure}=\mathrm{a}+\mathrm{b}\times \frac{1}{\text{Volume}}$

Thus, the value of the slope and the constant is given in the computer output as:

$a=0.36774$

$b=15.8994$

Thus, the regression equation is as follows:

localid="1650605639483" $\hat{\mathrm{P}}\mathrm{ressure}=a+b\times \frac{1}{\text{Volume}}$

$\Rightarrow \hat{\mathrm{P}}\text{ressure}=0.36774+15.8994\times \frac{1}{\text{Volume}}$

For transformation $2$:

Thus, the general equation of the regression equation is as:

localid="1650605182361" $\frac{1}{\hat{P}\text{ressure}}=a+b\times \text{Volume}$

Thus, the value of the slope and the constant is given in the computer output as:

$a=0.100170$

$b=0.0398119$

Thus, the regression equation is as follows:

localid="1650605211910" $\frac{1}{\hat{P}\text{ressure}}=\mathrm{a}+\mathrm{b}\times \text{Volume}$

localid="1650605239139" $\Rightarrow \frac{1}{\hat{P}ressure}=-0.15465+0.042836\times \text{Volume}$

We need to find out the period of a pendulum with the length $80$centimeters using the part (a) as:

For transformation $1:$

Thus, the regression equation is as follows:

localid="1650605657560" $\hat{\mathrm{P}}\mathrm{ressure}=\mathrm{a}+\mathrm{b}\times \frac{1}{\text{Volume}}$

localid="1650605667747" $\Rightarrow \hat{\mathrm{P}}\mathrm{ressure}=0.36774+15.8994\times \frac{1}{\text{Volume}}$

Then by evaluating we have,

localid="1650605678029" $\hat{\mathrm{P}}\mathrm{ressure}=\mathrm{a}+\mathrm{b}\times \frac{1}{\text{Volume}}$

$\Rightarrow \hat{\mathrm{P}}\text{ressure}=0.36774+15.8994\times \frac{1}{\text{Volume}}$

$=0.36774+15.8994\times \frac{1}{7}$

$=1.303$

For transformation 2 :

Thus, the regression equation is as follows:

$\frac{1}{\hat{P}ressure}=\mathrm{a}+\mathrm{b}\times \text{Volume}$$\Rightarrow \frac{1}{\hat{P}ressure}=-0.15465+0.042836\times \text{Volume}$

Then by evaluating we have,

$\frac{1}{\hat{P}ressure}=a+b\times$Volume

$\Rightarrow \frac{1}{\hat{P}ressure}=-0.15465+0.042836\times$Volume

$\Rightarrow \frac{1}{\hat{P}ressure}=-0.15465+0.042836\times 17$

$\Rightarrow \frac{1}{\hat{P}ressure}=0.7769723$

$\Rightarrow \hat{P}ressure=\frac{1}{0.7769723}$

$\Rightarrow \hat{\mathrm{P}}\mathrm{ressure}=1.287$

We need to interpret the value of $s$in this context. Thus, we have,

For transformation 1:

Thus, the regression equation is as follows:

localid="1650605513602" $\hat{P}ressure=\mathrm{a}+\mathrm{b}\times \frac{1}{\text{Volume}}$

$\Rightarrow \hat{\mathrm{P}}\text{ressure}=0.36774+15.8994\times \frac{1}{\text{Volume}}$

And the value of $s$is given as,

$s=0.044205$

This means that the expected error from the prediction of the pressure is $0.044205$.

For transformation 2 :

Thus, the regression equation is as follows:

$\frac{1}{\hat{\mathrm{P}}\mathrm{ressure}}=a+b\times Volume$

$\Rightarrow \frac{1}{\hat{\mathrm{P}}\mathrm{ressure}}=-0.15465+0.042836\times \text{Volume}$

And the value of $s$is given as,

$s=0.00398119$

This means that the expected error from the prediction of the reciprocal of the pressure is $0.00398119$.