The swinging pendulum Refer to Exercise 33. Here is Minitab output from separate regression analyses of the two sets of transformed pendulum 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 period of a pendulum with length of $80$ centimeters. Show your work.

(c) Interpret the value of s in context

In the question there are two transformations given and the computer output for each is also given. Thus, we have,

For transformation $1:$

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

localid="1650609318117" $\hat{P}\text{eriod}=a+b\sqrt{\text{length}}$

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

$a=-0.08594$

$b=0.209999$

Thus, the regression equation is as follows:

localid="1650609326733" $\hat{P}eriod=a+b\sqrt{\text{length}}$

localid="1650609346683" $\hat{P}eriod=-0.08594+0.209999\sqrt{\text{length}}$

For transformation 2 :

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

$\hat{P}{\text{eriod}}^2=a+b\times \text{length}$

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

$a=-0.15465$

$b=0.042836$

Thus, the regression equation is as follows:

localid="1650609373265" $\hat{P}{\text{eriod}}^2=a+b\times \text{length}$

localid="1650609386019" $\Rightarrow \hat{P}{\text{eriod}}^2=-0.15465+0.042836\times \text{length}$

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="1650609408367" $\hat{P}\text{eriod}=\mathrm{a}+\mathrm{b}\sqrt{\text{length}}$

$\Rightarrow \hat{\mathrm{P}}\text{eriod}=-0.08594+0.209999\sqrt{\text{length}}$

Then by evaluating we have,

localid="1650609425383" $\hat{P}\mathrm{eriod}=\mathrm{a}+\mathrm{b}\sqrt{\text{length}}$

localid="1650609440792" $\Rightarrow \hat{P}\mathrm{eriod}=-0.08594+0.209999\sqrt{\text{length}}$

$=-0.08594+0.209999\sqrt{80}$

$=1.7924$

For transformation 2 :

Thus, the regression equation is as follows:

$\mathrm{P}\text{eriod}{}^2=\mathrm{a}+\mathrm{b}\times \text{length}$

$\Rightarrow \mathrm{P}{\text{eriod}}^2=-0.15465+0.042836\times \text{length}$

Then by evaluating we have,

localid="1650609465411" $\mathrm{P}\text{eriod}=\mathrm{a}+\mathrm{b}\times \text{length}$

$\Rightarrow \mathrm{P}{\text{eriod}}^2=-0.15465+0.042836\times 80$

$\Rightarrow \mathrm{P}{\text{eriod}}^2=3.27223$

$\Rightarrow \mathrm{P}\text{eriod}=\sqrt{3.27223}$

$\Rightarrow \mathrm{P}\text{eriod}=1.8089$

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

For transformation 1:

Thus, the regression equation is as follows:

$\hat{P}\mathrm{eriod}=\mathrm{a}+\mathrm{b}\sqrt{\text{length}}$

$\Rightarrow \hat{\mathrm{P}}\text{eriod}=-0.08594+0.209999\sqrt{\text{length}}$

And the value of $s$is given as,

$s=0.0464223$

This means that the expected error from the prediction of the period is $0.0464223$.

For transformation $2$:

Thus, the regression equation is as follows:

$P{eriod}^2=a+b\times \text{length}$

localid="1650609508643" $\Rightarrow \mathrm{P}{\text{eriod}}^2=-0.15465+0.042836\times \text{length}$

And the value of $s$is given as,

$s=0.105469$

This means that the expected error from the prediction of the square of the period is $0.105469$.