Braking distance, again How is the braking distance for a car related to the amount of tread left on the tires? Here are the braking distances (measured in car lengths) for a car making a panic stop in standing water, along with the tread depth of the tires (in $1/32$inch):

$\begin{array}{|ll|}\hline \text{Tread depth (1/32 inch)}& \text{Breaking distance (car lengths)}\\ 11& 9.7\\ 10& 9.8\\ 9& 10.1\\ 8& 10.4\\ 7& 10.8\\ 6& 11.2\\ 5& 11.8\\ 4& 12.4\\ 3& 13.6\\ 1& 15.2\\ \hline\end{array}$

a. Transform both variables using logarithms. Then calculate and state the least-squares regression line using the transformed variables.

b. Use the model from part (a) to calculate and interpret the residual for the trial when the tread depth was $3/32$ inch.

Given data:

$\begin{array}{|ll|}\hline \text{Tread depth (1/32 inch)}& \text{Breaking distance (car lengths)}\\ 11& 9.7\\ 10& 9.8\\ 9& 10.1\\ 8& 10.4\\ 7& 10.8\\ 6& 11.2\\ 5& 11.8\\ 4& 12.4\\ 3& 13.6\\ 1& 15.2\\ \hline\end{array}$

Log of given data:

$\begin{array}{|llll|}\hline \text{Tread depth (1/32 inch)}& \text{Breaking distance (car lengths)}& \log (\text{Tread depth)}& \log (\text{length)}\\ 11& 9.7& 1.041392685& 0.98677173\\ 10& 9.8& 1& 0.99122608\\ 9& 10.1& 0.954242509& 1.00432137\\ 8& 10.4& 0.903089987& 1.01703334\\ 6& 10.8& 0.84509804& 1.03342376\\ 6& 11.2& 0.77815125& 1.04921802\\ 4& 11.8& 0.698970004& 1.07188201\\ 3& 12.4& 0.602059991& 1.09342169\\ 1& 13.6& 0.477121255& 1.13353891\\ 15.2& 0.301029996& 1.18184359& \\ 1& 18.3& 0& 1.26245109\\ \hline\end{array}$

Using the calculator $Ti83/84$

Step 1: Select STAT;

Step 2: Select 1: EDIT

Step 3: in list L1, type the data for logarithmic Tread depth, and in list L2, type the data for logarithmic Braking Distance.

Step 4: hit STAT once more, select CALC, and then $LinReg(a+bx)$.

The final result

$y=a+bx$

$a=-0.2690$

$b=1.2609$

Substituting the value in $a$and$b$:

$y=-0.2690+1.2609x$

The logarithm of tread depth is $x$, whereas the logarithm of braking distance is $y$.
role="math" localid="1654266116493" $\log y=-0.2690+1.2609\log x$

Given data:

$\begin{array}{|ll|}\hline \text{Tread depth (1/32 inch)}& \text{Breaking distance (car lengths)}\\ 11& 9.7\\ 10& 9.8\\ 9& 10.1\\ 8& 10.4\\ 7& 10.8\\ 6& 11.2\\ 5& 11.8\\ 4& 12.4\\ 3& 13.6\\ 1& 15.2\\ \hline\end{array}$

Substituting the value $x$by $3$:

$\log y=-0.2690+1.2609\log x$

$\log y=-0.2690+1.2609\log 3$

$\log y=0.3326$

Using the exponential function with a base of ten:

$y={10}^{\log y}$

$={10}^{0.33263}$

$=2.1508$